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FOR COLLEGIATE STUDENTS OF
AGRICULTURE AND GENERAL SCIENCE
REVISED EDITION
A SERIES OF MATHEMATICAL TEXTS
EDITED BY
EARLE RAYMOND HEDRICK
THE CALCULUS
By ELLERY WILLIAMS DAVIS and WILLIAM CHAKLES BRENKE.
ANALYTIC GEOMETRY AND ALGEBRA
By ALEXANDER ZIWET and Louis ALLEN HOPKINS.
ELEMENTS OF ANALYTIC GEOMETRY
By ALEXANDER ZIWET and Louis ALLEN HOPKINS.
PLANE AND SPHERICAL TRIGONOMETRY WITH
COMPLETE TABLES
By ALFRED MONROE KENYON and Louis INGOLD.
PLANE AND SPHERICAL TRIGONOMETRY WITH
BRIEF TABLES
By ALFRED MONROE KENYON and Louis INGOLD.
ELEMENTARY MATHEMATICAL ANALYSIS
By JOHN WESLEY YOUNG and FRANK MILLETS MORGAN.
COLLEGE ALGEBRA
By ERNEST BROWN SKINNER.
MATHEMATICS FOR COLLEGIATE STUDENTS OF
AGRICULTURE AND GENERAL SCIENCE
By ALFRED MONROE KENYON and WILLIAM VERNON LOVITT.
PLANE TRIGONOMETRY
By ALFRED MONROE KENYON and Louis INGOLD.
THE MACMILLAN TABLES
Prepared under the direction of EARLE RAYMOND HEDRICK.
PLANE GEOMETRY
By WALTER BURTON FORD and CHARLES AMMERMAN.
PLANE AND SOLID GEOMETRY
By WALTER BURTON FORD and CHARLES AMMERMAN.
SOLID GEOMETRY
By WALTER BURTON FORD and CHARLES AMMERMAN.
CONSTRUCTIVE GEOMETRY
Prepared under the direction of EARLE RAYMOND HEDRICK.
JUNIOR HIGH SCHOOL MATHEMATICS
By WILLIAM LEDLEY VOSBURGH and FREDERICK WILLIAM
GENTLEMAN.
MATHEMATICS
FOB COLLEGIATE STUDENTS OF
REVISED EDITION
BY
ALFRED MONROE KENYON
PROFESSOR OF MATHEMATICS IN PURDUE UNIVERSITY
AND
WILLIAM VERNON LOVITT
ASSOCIATE PROFESSOR OF MATHEMATICS IN COLORADO COLLEGE
Nefa gorfc
THE MACMILLAN COMPANY
LONDON: MACMILLAN & CO., LTD.
1918
COPYBIQHT, 1917,
BY THE MACMILLAN COMPANY.
Set up and electrotyped. Published December, 1917.
Nortoooli
Printed by Berwick & Smith Co., Norwood, Mass., U.S.A.
ft 4- 2
PREFACE
This book is designed as a text in freshman mathematics for
students specializing in agriculture, biology, chemistry, and
physics, in colleges and in technical schools.
The selection of topics has been determined by the definite
needs of these students. An attempt has been made to treat
these topics and to select material for illustration so as to put
in evidence their close and practical relations with everyday
life, both in and out of college. It is certain that the interest
of the student can be aroused and sustained in this way. We
believe also that he can be trained to understand and to solve
those mathematical problems which will confront him in the
subsequent years of his college work and in after-life, without
losing anything in orderly arrangement or in clear and accurate
logical thinking.
Reference to the table of contents will indicate the scope and
proportions of the material presented and something of the
means employed in relating the material to the vital interests
of the student and of correlating it to his experience and his
intellectual attainments. Many of the chapter subjects and
paragraph headings are traditional. Nothing has been intro-
duced merely for novelty. Since this course is to constitute the
entire mathematical equipment of some students, some chapters
have been inserted which have seldom been available to fresh-
men; for example, the chapters on annuities, averages, and
correlation, and the exposition of Mendel's law in the chapter
on the binomial expansion.
Particular attention has been given to the illustrative examples
and figures, and to the grading of the problems in the lists.
The exercises constitute about one fifth of the text and contain
yi PREFACE
a wealth of material. They include much data taken from
agricultural and other experiments, carefully selected to stimu-
late thinking and to show the application of general principles
to problems which actually arise in real life, and in the solution
of which ordinary men and women are vitally interested.
The book is intended for a course of three hours a week for
one year, but it can be shortened to a half-year course. The
chapters on statics, small errors, land surveying, annuities,
compound interest law, and as many as is desired at the end,
can be omitted without breaking the continuity of the course.
The first two chapters are more than a mere review. This
matter is so presented as to give the student a new point of
view. The treatment will show the significance and importance
of certain fundamental relations among the concepts and
processes of arithmetic and algebra which the student may
have used somewhat mechanically .in secondary school work.
Well prepared students can read these chapters rather rapidly,
however.
The four place mathematical tables printed at the end of the
text have been selected and arranged for practical use as the
result of long experience and actual use in computing, and are
adapted to the requirements of the examples and exercises in
the book.
The first edition of this book contained problems, formulas
and other matter taken from a large number of sources. Those
passages that were directly from other books have now been
entirely rewritten ; but the book remains indebted to a num-
ber of others, notably SKINNER, Mathematical Theory of Invest-
ment, and DAVENPORT, Principles of Breeding. Other references
occur throughout the text.
A. M. KENYON,
W. V. LOVITT.
CONTENTS
CHAPTER PAGES ARTICLE
I. Introduction 1-10 1
II. Review of Equations 11-35 12
III. Graphic Representation 36-71 34
IV. Logarithms 72-90 63
V. Trigo'nometry 91-138 74
VI. Land Surveying 139-152 108
VII. Statics 153-176 122
VIII. Small Errors 177-189 139
IX. Conic Sections 190-217 149
X. Variation . .> 218-225 167
XI. Empirical Equations 226-242 173
XII. The Progressions 243-253 178
XIII. Annuities 254-261 186
XIV. Averages 262-268 194
XV. Permutations and Combinations 269-274 201
XVI. The Binomial Expansion — Laws of
Heredity 275-285 207
XVII. The Compound Interest Law 286-290 217
XVIII. Probability 291-303 219
XIX. Correlation 304-313 232
TABLES 314-333
INDEX;. . 335-337
vu
MATHEMATICS
CHAPTER I
INTRODUCTION
1. Uses of Mathematics. The applications of mathematics
are chiefly to determine the magnitude of some quantity such
as length, angle, area, volume, mass, weight, value, speed, etc.,
from its relations to other quantities whose magnitudes are
known, or to determine what magnitude of some such quantity
will be required in order to have certain prescribed relations
to other known quantities.
2. Measurement. To measure a quantity is to find its ratio
to a conveniently chosen unit of the same kind. This number
is called the numerical measure of the quantity measured.
• The expression of every measured quantity consists of two
components: a number (the numerical measure), and a name
(that of the unit employed). For example, we write: 10 inches,
27 acres, 231 cubic inches, 16 ounces, 22 feet per second.
3. Arithmetic and Algebra. In arithmetic we study the
rules of reckoning with positive rational numbers. In algebra
negative, irrational, and imaginary numbers are introduced,
letters are used to represent classes of numbers, and the rules
of reckoning are extended and generalized. Algebra differs from
arithmetic also in making use of equations for the solution of
problems requiring the discovery of numbers which shall satisfy
certain prescribed conditions.
2 1
2 MATHEMATICS [I, §4
4. Positive Numbers. The natural numbers 1, 2, 3, 4, etc.,
are the foundation on which the whole structure of mathe-
matics is built. They are also called whole numbers, or
positive integers. Together with the fractions, of which 1/2,
5/3, .9, 2.31, are examples, they form the class of positive
rational numbers.
Every positive rational number can be expressed as a fraction
whose numerator and denominator are whole numbers.
Two quantities of the same kind are said to be commensur-
able when there is a unit in terms of which each has for numer-
ical measure a whole number. Consequently, their ratio is
a rational number. If two quantities are not commensurable,
they are said to be incommensurable.
The ratio of two quantities which are incommensurable,
such as the side and the diagonal of a square, or the diameter
and the circumference of a circle, is an irrational number.
No irrational number can be expressed as a fraction whose
numerator and denominator are whole numbers. However, it
is always possible to find two rational numbers, one less and the
other greater than a given irrational number, whose difference
is as small as we please. For example,
3.162277 < VlO < 3.162278
and the difference between the first and the last of these num-
bers is only .000001. Two such rational numbers whose dif-
ference is still less can easily be found. In all practical appli-
cations, one of these rational numbers is used as an approximation
for the irrational number. Thus, we may find the length of the
circumference of a circle approximately by multiplying its di-
ameter by 3f. If a closer approximation is needed, the value
3.1416 is often used.
The (positive) rational and the (positive) irrational numbers
make up the class of (positive) real numbers.
I, §7] INTRODUCTION 3
5. Negative Numbers. Zero. To every positive real
number r, there corresponds a negative real number — r, called
negative r. The negatives of the natural numbers are called
negative integers. The real number zero separates the negative
numbers from the positive numbers. It is neither positive nor
negative and corresponds to itself.
The negatives of negative numbers are the corresponding
positive numbers; thus, — (— 2) =2.
6. The Four Fundamental Operations. The direct opera-
tions of addition and multiplication of real numbers are so
defined that they are always possible, and so that the result
in each case is a unique real number. These operations are
subject to the rules of signs and to the following fundamental
laws of algebra.
I. The commutative law:
a + 6 = 6 + a, ab = ba.
II. The associative law:
(a + &) + c = a + (b + c), (a6)c = a(6c).
III. The distributive law:
a(b + c) = ab + ac-
The indirect operations of subtraction and division of real
numbers are always possible, division by zero excepted,* and
the result is a unique real number.
7. Involution and Evolution. Involution, or raising to pow-
ers, is always possible, and the result is unique when the base
is any real number provided the exponent is a positive integer.
* Division by zero is excluded because, in general, it is impossible, and when possible
it is trivial. Thus there is no real number which will satisfy the equation 0 -x = o 4=
0, and every real number satisfies the equation 0 -x = 0.
4 MATHEMATICS [I, §7
Evolution, or extraction of roots,* is not always possible.
Even when possible, it is not always unique. In particular,
the square of every real number is a positive real number.
Hence no negative number can have a real square root. On
the other hand, every positive real number, a, has two real
square roots: a positive one, which is denoted by the symbol
•^a; and a negative one, which is denoted by — Va. In fact,
every positive real number has exactly two real nth roots of
every even index n, denoted by Va and — "N/o, respectively.
Every real number, r, has a unique real nth root of every odd
index n, denoted by Vr; it is positive when r is positive, and
negative when r is negative.
8. Rational Exponents. Involution is extended to frac-
tional exponents as follows. If a is a positive real number,
and if m and n are natural numbers, we define amln by the
equation
. «i —
amln — -\am.
For example,
82/3 = ^ = 4>
ni—
In particular, a1'" = Va denotes the unique real positive nth
root of a.
If r is a positive rational number, (— a)r is defined only when
r, expressed as a fraction m/n, in its lowest terms, has an odd
denominator and in this case,
(- a)r = (- l)mar.
For example, (- 32) •« = (- 32)3/5 = (- 1)3(32)3/5 = - 8, and
(- 32) •« = (- 32)4/5 = (- 1)«(32)4/5 = 16.
In particular, (— a)1/n = — a1/n = — Va, if n is odd.
* The index of a root is always a positive integer.
I, §9] INTRODUCTION 5
9. Negative and Zero Exponents. By definition, we write
a~b = —b and a° = 1,
Q
provided a =|= 0. Thus a~b is defined for the same real values
of a and 6 as is a6 and the two are reciprocals.* For example,
1 1 /I \~5/2 1
g-2/3 _ _!_ = L *
82/s 4>
A consequence of this definition is the rule: A factor may be
moved from the numerator to the denominator of a fraction, or vice
versa, on changing the sign of its exponent. For example,
a26c~3 _ 2-1d~1e _
2de~l " a-^-'c3 ~ 2c3d '
= (a? - z2)-i/2,
Va2 — x
or2 + 2x-*y~l + 7/-2 = - + — + - .
xz xy y2
EXERCISES
1. Verify the fundamental laws of algebra by making use of the
three numbers f, — 5-J-, |.
2. How many real square roots has 24.5? — 4.5?
3. How many real cube roots has 6f? — 12£?
4. Find the numerical value of each of the following expressions,
exactly when rational, correct to three decimal places when irrational.
(a) 9s/2. (e) (32)s/5. (i) (-0.027) »/8.
(*>) Gfr)2/s. (/)(-32)«». (j) (H)"4-
(c) v<3. (g) (-2)"'. (*) (t)1".
(d) (- 2)«». (A) (- 0.375)^. (1) (- «)w.
* Two numbers are reciprocals when their product is + 1. Every real number has a
reciprocal except 0, which has none.
6 MATHEMATICS [I, § 9
5. Write each of the following expressions without radical signs.
(a) -^32. (b) Vl28. _c)
6)2. (i) ^4?-84. 0') -v-^
6. Write each of the following expressions without negative expo-
nents and simplify when possible.
(27 x-V2"12)~1/3- (c) (a^ + fc-2)-1. (/) (a
ar1 + y-1 a"1 - fe-1 a;-1?/0 + x°y
,.. 30a-W -S-^tfb0 ffcx x-ly-*z~3 + x*y*z
32a°6-1+3a6 ' ar3?/-^-1 + XT/V '
10. Laws of Exponents. The following five laws are useful
for the reduction of exponential and radical expressions to sim-
pler forms. They are valid, (1) when the bases are any real
numbers whatever, provided the exponents are integers or zero,
and (2) when the exponents are any real numbers whatever,
provided the bases are positive.
I. ab - ac = ab+c.
EXAMPLES. 32 • tr* = 3~2. ( - |)B( - f )-3 = ( - f)2.
(2)-l/3(|)l/2 = (!)l/6- g-l/3 . g2 = 85/3.
11. acbc = (ab)c.
EXAMPLES. 2353 = 103. (- 3)-2(- 5)~2 = (15)~2.
(17)1/3(^)1/3 = 51/8,
III. (ab)c = abc.
EXAMPLES. (28)2 = 26. [(- |)-2]-3 = (- f)6.
[(f)1/3l6 = (I)2-
I, §11] INTRODUCTION
ab
IV. r = ab~c.
ac
Q2 (_ 5N1/2
EXAMPLES. — = 3'. (_ |j-.,. = (- f>"°-
V. £.
4-2/5 (2\-2
T?YAU*T>ri?a — — f4\-2/5 v"x _ f2\-2
EXAMPLES. 3_2/5 - ($) . 3^_2 - UJ .
These laws are readily proved when the exponents are positive
integers. Thus, to prove law II, when the exponent is a positive
integer n, we write
(1)(2)(3) (n)(n (2) (3) (n)
an-bn = a- a- a • • • a-b-b-b • • • b
(1) (2) (3) (n)
= ab-ab-ab • • • ab = (ab)n.
Similarly, each of the other laws can be proved when the
exponents are positive integers. When the exponents are
negative, we make use of the definition of § 9. If they are
positive fractions we make use of the following lemma: // a
and b are real numbers of like sign, and if an = bn, where n is a
positive integer, then a = b.
11. Binomial Theorem. By multiplying out, we find the
following equalities:
(x + 7/)2 = x1 + 2#y + ?/2,
(x + y)3 = x3 + 3z2?/ + 3z?/2 + y3,
(x + 2/)4 = a;4 + 4:X3y + 6.r27/2 + 4xy3 + y4,
(x + yY = x5 + 5x4y + lOo;3?/2 + Wx2y3 + 5xy4 + y5.
By observing the coefficients and the exponents of x and of y
in the various terms, we observe the law by which these results
can be written down without the work of multiplying them out.
In the expansion of (x + y)n for n — 2, 3, 4, 5, we note the
following facts:
(1) The number of terms is n + 1-
8 MATHEMATICS [I, §11
(2) The exponent of x in the first term is n and it decreases
by 1 in each succeeding term; the exponent of y in the second
term is 1 and it increases by 1 in each succeeding term.
(3) The first coefficient is 1; the second is n; the coefficient
of any term after the second may be found from the preceding
term by multiplying the coefficient by the exponent of x and dividing
by a number 1 greater than the exponent of y.
These three statements constitute the binomial theorem,
which will be proved in § 208, Chapter XVI, for all values of
x and y no matter how large the positive integer n may be.
The coefficients which appear in these expansions are called
binomial coefficients. For example, the numbers
1, 5, 10, 10, 5, 1
are the binomial coefficients for the fifth power. The binomial
coefficients for the second, third, fourth, and fifth powers should
be memorized.
EXERCISES
Use the laws of exponents to combine and simplify the following
expressions.
1. g-^-S^-S-^-S2 -j- 83/4-81/12. 2. 32/B-42/B-52/B 4- 152/B-82/5.
3. (32-31/2-56/2)2 -i- (73 - 102). 4. (11 -32 + 74)1/2.
C3/4 1 03/2
K (K4 _ 92.96^1/2 A ° _ 7 •*"*
D< 88'12' 33/2 '
402" V48 V54 Vl2
O. iro/o • a. ;r~ . 1U. A - . 11. — — .
52/3 V3 M36 >/6
12.
I, § 11] INTRODUCTION 9
Perform the indicated operations and simplify each of the following
expressions when possible.
\
16. . 7 •ax- 18
\Wy) UW '
19. f"-l™-*Yl. 20 f^blY3' 21
\ *V2 / ' \8aV/
22 / x + 2 \-i. /r«±&V
\x*+x-2j • \a-b) '
24. 63a4x5 -^ 9a3x2 H- 3a2x. 25. (o° + 6) (a + 6°).
'
3a262
28
30 2a2 + 7ax + 3x2 ^ 3a2 + 7ax + 2x2
2a + x a + 3x
31
a2 - a - 20 a2 - 2a - 15 a2 - a a2
33.
x2 - 5x + 4 x2 - lOx +21 x2 - 9x + 20
4. a + 3 va2 +a-2 . a2 +3a+2
6
Multiply :
34. a5/6 _
35. a1/2 + 2&1/2 - 3c1/2 by a1/2 - 261/2 + 3c1/2.
36. x4/3 + 2x + 3x2/3 + 2x1/3 + x° by x2/3 - 2X1/3 + x°.
37. Vx3 - x2y - XT/2 + y3 by Vx3 + 3x2y + 3xy2 + y3.
38. o1/4 - 61/4 by a3/4 + &3/4. 39. x3/5 - ?/2/8 by x2/5 + y3'*.
40. (a1/2^/2 + c1/2)2 by (a1/2*)1/2 - c1/2)2.
41. (a-1/2 - 3)2 by (3O1/2 + I)2.
Divide :
42. x8/2 + x2 - 2X1/2 + 1 by x + x1/2 - 1.
43. x3 + 27x - 9x1/2 - 10 by x - 3x1/2 + 5.
44. x-y- 6x2/3 + 12X1/3 - 8 by x1/3 - y1'3 - 2.
45. a6/2 - a26 + a3 2c - oc + a1/2& - 1 by o1/2 - 1.
46. a2 + Sa1/4 + 7 by a1/2 + 2o1/4 + 1.
10 MATHEMATICS [I, § 11
Reduce each of the following to its simplest form :
47. Vl2.25zV. 48. v/15.625a669.
49. V343a10625. 50. V3a26 - 2a2c.
51. V(a3 + 53) (02 _|_ 52 _ atyf 52. Vx4?/2 - 2xy + x2?/4.
53. VVl024. 54.
55. V27av/27a64.
57. •\/1.35a2V6.25a;!.
59. V|05 -2V605+V845. 60. *V 192 - 2v/375 + #648
61. V72-V8-V50. 62. ^81 - 2\yl92 + ^375.
63. (VI53 - VTI7 + A/52 - V68)(V/5T + V39).
64. (Vl2 + V3 + Vs})(vls
65. (2 + V3 + v/4)(2 - V3
66. (3V20 -4V5 + 5V2 -3V8)(V5 + VCL5).
67. V19 + 3V2- V19-3V2. 68. Vl6 + Vl3-
Rationalize the denominator of each of the following fractions :
69. -Iz^L 70. 1^. 71. -
3-2V2 2+V5 2V5+3V2
72 x/3+v/2 + \/2 -V3, 73 2+V5x5 -V2.
\/3-V2 V2+V3 2-Vg 5+V2
74 V2 - \73 75 Vl89+3V20_
V3^_|_V48-V'50-v/75 V84-V80
Expand each of the following expressions :
76. (2p
79. (1 -
82.
85. (1 + x2)2.
88. (fc2+3)2.
91. (Va+Va;
94. (2x - 37/)3
97. (1 + x)3.
100.
77.
(5c - 9d)2.
78.
(4m - 3n)2.
80.
(1 - *)2.
81.
(fa + f 6)2.
83.
(l-i)2-
84.
(^ - tl/)2.
86.
(1 - X2)2.
87.
(1 +Vx)2.
89.
(2t2 + 5)2.
90.
(a2 + a6)2.
92.
(a-l/2 + xl/t)».
93.
(ftl/3 _ 2/l/2)2_
95.
(a + ^b)3.
96.
(v^+Vm)
98.
(1 - x)3.
99.
(1 + x2)3.
101.
(x+y- a)2.
102.
(a2 + ab + W
CHAPTER II
REVIEW OF EQUATIONS*
12. Use of Equations. As indicated before, the chief ad-
vantage of algebra over arithmetic in solving problems lies in
the method of attack. The algebraic method is to translate
the problem into an equation and then to solve the equation by
general methods.
13. Definition of an Equation. An equation is a statement
of the equality of two expressions. Each of the expressions
may contain letters and figures called knowns, representing
numbers supposed to be given or known; letters called unknowns,
representing numbers to be found; and symbols of operation
and combination, such as +, — , etc.
As examples of equations in one unknown, we may write
(1) x + 13 = 2x - 7,
(2) x(S - x) = 2(x + l)(z2 - x + 1),
(3) x(x + 2) = (x - l)(x - 2) + 5x - 2,
2x + l 2x - 1 _
x- 1 x + l
(5) 7 Vx — 6 + 6 >/3x + 4 = 4x + 3.
As examples of equations in two unknowns, we may write
(6) x2 - if + 2y = 1,
(7) (2x — ?/)2 — 5x2 = 5?/2 — (x + 2y)z.
Similarly, we may have equations in more than two unknowns.
*This chapter is intended for review work. Parts of it may bo omitted at the dis-
cretion of the instructor, if it appears that the students do not need to review some of
the topics.
11
12 MATHEMATICS [II, §13
The expression on the left of the equality sign is called the
left member, or the left side, of the equation. The other is called
the right member, or right side.
14. Substitution. It is often necessary to substitute for the
unknowns in an expression such as one of the members of the
above equations, certain definite numbers, called values of the
unknowns. The result of such substitution is, in general, to
reduce the expression to a single number.
Thus, if we put 10 for x in equation (1), the left side reduces to 23
and the right side to 13. If we put 20 for x, each member reduces to
the same number, 33.
Again, if we put 1 for x and 1 for y in equation (6), the left side
reduces to 2 and the right side to 1; but if we put 2 for x and — 1 for
y, each member reduces to 1.
15. Solution of an Equation. Any set of values of the un-
knowns which reduces each of the two members of an equation
to the same number is said to satisfy the equation, and to be a
solution of the equation. A solution of an equation in one
unknown is also called a root of the equation.
The final test to determine whether a set of values of the
unknowns in an equation is a solution or not, is to substitute
these values for the unknowns and see whether the equation
is satisfied or not.
For example, x = 20 is a solution of equation (1), § 13. The value
x = 10 does not satisfy it. Again, z = |, x = 1, x = — 2, are three
solutions of (2). Every real number is a solution of (3). The value
x = 2 is a solution of (4). The value x = 15 is a solution of (5). The
values x = 2, y = — 1 constitute a solution of (6). Every pair of real
numbers constitutes a solution of (7).
16. Identities. An equation which is satisfied by all values
of the unknowns (excepting those values if there are any for
which either member is not defined) is called an identity. An
II, § 16] REVIEW OF EQUATIONS 13
equation which is not an identity is called a conditional equation,
or when no ambiguity is likely to arise, simply an equation.
EXAMPLES. Of the equations in § 13, (3) and (7) are identities, the
others are conditional equations. Also,
is an identity; it is satisfied by all values of x, except x = 1 for which
neither side is defined.
The distinction in point of view between identities and con-
ditional equations is fundamental. To show that an equation
is not an identity, we need only find a single set of values of
the unknown quantities for which both sides are defined, and
for which the equation is not true.
EXERCISES
1. Which of the numbers — 3.5, — 2, — 1, 0, \, 2, satisfy the
equation
ix+«±2_ 10 ?
3 ~2x~+~l{
2. Which of the numbers TV V7, 2 + V3, Vl4, 2 — V3, are solutions
of the equation x2 + 1 = 4x?
3. Which of the following pairs of numbers (0, 0), (1, 3), (4, 2),
(0, 2), (1, - 1), (3, - 1), (4, 0), (3, 3), satisfy the equation
Is this equation an identity?
4. Which of the following pairs of numbers (0, 1), (1, 1), (— 1, 0),
(2, 3), (— 2, 1), (1, — 1), (3, — 2), are solutions of the equation
* + 2y = 1+V(1 __ V_ \ ?
x + y x\ x + y) '
Is this equation an identity?
14 MATHEMATICS [II, §16
5. Which of the following equations are identities?
(a) x(x2 — y2) = (x + y)(x2 — xy).
(6) x(x2 + y2) = (x - y)(x2 + xy).
(c) x(x + 7) - (x + 3)(x + 4) + 12 = 0.
(d) x(7 - x) + (3 - x)(4 - x) = 12.
(c) 4x2 + 7x + 2y = 0. (/) 4x2 + 7x - 2y = 0.
(g) x* = (x2 + l)(x + l)(x - 1) + 1.
(h) tf = (1 + x2)(l + x)(l -x) + 1.
(i) (ax - b)2 + (6x + a)2 = (a2 + 62)(1 + x2).
0") (ax - 6)2 + (ax + &)2 = (a2 + 62)(1 + x2).
(fc) (x - t/)3 + (y - zY + (z- X? = 3(x - y)(y - 2) (a - x).
(0 (x + 2/ + z)3 - (x3 + y3 + 0s) = 3(x + y)(y + 2) (2 + x).
fm) V* + *^ . + W = 1
^ ' (x - y)(x - z) ^ (y - z)(y - x) ^ (z - x)(z - y)
17. Equivalent Equations. Two equations are said to be
equivalent when every solution of the first is a solution of the
second and conversely, every solution of the second is a solu-
tion of the first.
For example, the equations
5-5 = o
3 7
and
7x = 15
are equivalent; each has the unique solution x = 2f.
On the other hand
2x - 3 = x - 1
and
(2x - 3)2 = (x - l)z
are not equivalent; the latter has the solution 1^, which does not satisfy
the first.
18. Transformations of Equations. The following changes
in an equation lead always to an equivalent equation:
1. Transposition of terms with change of sign.
2. Multiplication, or division, of all the terms by the same
constant (not zero).
II, §18] REVIEW OF EQUATIONS 15
If all the terms of an equation be transposed to the left side
(so that the right member is zero), if the left member be factored,
and if each of the factors be equated to zero, then the solutions
of the separate equations so formed are all solutions of the
original equation, and it has no others.
EXAMPLE. The equations
r3 -4- "vr
'-~--=x*-x + l and (x - l)(x - 2)(x - 3) = 0
are equivalent, and the solutions of the latter are seen by inspection to
be z = 1, x = 2, x = 3.
The following changes in an equation lead to a new equation
which is satisfied by every solution of the given equation, but
which generally has other solutions also.
3. Multiplying through by an expression containing un-
knowns (defined for all values of these unknowns).
4. Squaring both members, or raising both members to the
same positive integral power.
Since the new equation is not, in general, equivalent to the
given equation, it is necessary to test all results by substituting
them in the given equation in its original form.
EXAMPLES. Every solution of the equation
x2 1 _ 5x
6~H ~ 6"
is a solution of the equation
x3 + 6z = 5x*,
which is formed by multiplying the first through by 6x; but they are
not equivalent, since x = 0 satisfies the second but does not satisi'y
the first.
Every solution of the equation
3x - 1 =x - 1
x + 1 ~ x - 2
16 MATHEMATICS [II, §18
is a solution of the equation
3x2 - 7x + 2 = x2 - 1,
which results from clearing the former of fractions. These two equa-
tions are in fact equivalent. Each is satisfied by x — |, and by x — 3,
and by these only.
Every solution of the equation
x - 4 = Vz + 2
is a solution of the equation
x2 - 9z + 14 = 0,
which results from squaring and transposition in the former; but they
are not equivalent; the latter equation has the two solutions x = 2,
x = 7, while the former has only one, x = 7.
19. Simultaneous Equations. When a common solution of
two or more equations is sought, the equations are said to be
simultaneous. For example, each of the equations
(8) 3z - 2y = 4
and
(9) 2x - y = 3
has an infinite number of solutions: (0, — 2), (2, 1), (4, 4),
(G, 7), etc., satisfy (8), and (1, - 1), (2, 1), (3, 3), (4, 5), etc.,
satisfy (9). But (2, 1) is the only common solution.
By a solution of a set of equations is meant a common solu-
tion of all the equations of the set, regarded as simultaneous
equations. Thus, the set of equations (8) and (9) has a unique
solution, namely, x = 2, y = 1.
Two sets of simultaneous equations are equivalent when
each set is satisfied by all of the solutions of the other set.
If each of two or more equations from a set of simultaneous
equations be multiplied through by any constant, or by any
expression containing unknowns,* and if the resulting equations
* Defined for all values of the unknowns.
II, §20] REVIEW OF EQUATIONS 17
be added or multiplied together, the new equation will be
satisfied by all the (common) solutions of the given set.
EXAMPLE. If in the set of simultaneous equations,
2x2 + 2y2 - 3x + y = 9,
3x2 + 3yi + x _ y = 14>
we multiply the first by — 3, the second by 2, and add, the resulting
equation
llx - 5y = 1
is satisfied by every solution of the given set. One such solution is
x = 1, y = 2.
20. Elimination. By a proper choice of multipliers we can
use the above principle to secure a new equation lacking a
certain term, or certain terms, which occur in the given set of
equations. The missing terms are said to have been eliminated
and this process is called elimination by addition.
EXAMPLES. We can eliminate the term in x2 from the equations,
- 2
5
5x2 - 9x = 2,
2x2 - x = 6,
by multiplying by — 2 and + 5, respectively, and adding. The result
is 13x = 26. We conclude that if the given equations have a common
solution, it is x = 2, and we verify that this is a solution of each.
If we eliminate x2 from the equations,
- 2j|5x2 + 9x = 2,
5||2z2 + 5x = 5,
we obtain
7x = 21.
Since x = 3 is not a solution of the given equations, they have none.
When y is eliminated from the equations,
2||3x2 - 4x - 15y + 1 =0,
- 3 || 2x2 - 3x - 10y + 1 =0,
the result is
x - 1 = 0
and on substituting x = 1 in either of the given equations, we find
y = 0. Therefore (1, 0) is the unique common solution.
3
18 MATHEMATICS [II, §20
When it is possible to solve one of a set of simultaneous equa-
tions for one of the unknowns, we can eliminate this unknown
by substituting the value thus found in the other equations of
the set. This is called elimination by substitution.
For example, to eliminate t from the set of equations,
x = a(l+ <2),
y = o(l + 0,
solve the second for t and substitute this value in the first. The result is
x = ^ - 2y + 2a,
which is equivalent to the equation
y2 — ax — lay + 2a2 = 0.
If we can solve each of two simultaneous equations for the
same unknown, this unknown will be eliminated by equating
these two values to each other. This is called elimination by
comparison.
Thus, if we solve each of the equations
z2 - xy - 4x + 2y + 1 =0,
2x* - 2xy + 3x - 2y + 3 =0,
for y, and equate these values, the result is
x* - 4x + 1 = 2x* + 3x + 3
x -2 2x + 2
which is equivalent to the equation
(5z + 8)(x - 1) = 0.
21. Linear Equations. An equation of the first degree in
the unknown quantities is called a linear equation. A set of
linear simultaneous equations can be solved, if they have a
solution, by successively eliminating the unknowns until a single
equation in one unknown is obtained.
II, §21] REVIEW OF EQUATIONS 19
EXAMPLES.
3 2x + y = 4,
1 x - 3y = 9.
Eliminating y by addition, we obtain
7x + 0-y = 21.
Eliminating x, we get
Q-x + 7y = - 14.
We conclude that if the given equations have a solution it is x = 3,
y = — 2, and we verify that this is a solution.
To eliminate x by substitution from the equations
7x - Qy = 15,
5x - Sy = 17,
solve the first for x and substitute this value in the second. The
result is
/ n,,. i 1 c \
- 8y = 17,
which is equivalent to y = — 4. Substituting — 4 f or y in either of the
given equations, we fird x = — 3. Finally, we verify that x = — 3,
y = — 4, is a solution of the given set.
To eliminate x by comparison from the equations
3x - 7y = 19,
2x - 5y = 13,
solve each equation for x, and equate the results. This gives
7y + 19 5y + 13
3 2
which is equivalent to y = —I. Substituting this value for y in either
of the given equations leads to x = 4.
20 MATHEMATICS [II, §21
EXERCISES
1. Solve the following equations and determine whether or not the
two equations in each pair are equivalent.
x + 5 x + 1 _ x - 3 _ 1 3x - 7
(a) ~2~ T~ ~2x~~ ~ 3 ~ ~^x~~
n) y~7 + 2 - y + 8 2(y ~ 7) 4. ^^ - y + 3
() 5 10 ' jf + 3i,_28 + y-4~y + 7'
3£ - 5 _ 9< - 7 = 2_ 5t + 4 _ lit - 2
4 12 ~ 3t' 2t 6t
„ 6x - 1 8x + 3 4x - 3 3.7
(d)~l- -fflr -5—' 4i + x"15^
, 5x + 1 , a; 15x2 — 5x — 8
w'2x~3T 2x-3' 3x2 + 6x + 4
(?) x - 1 = V3x - 5, x2 - 5x + 6 = 0.
= 5.
(0 x = 2, x(x - 1) = 2(x - 1).
0') 2x = 1, Sx3 - 12x2 + 6x = 1.
2. Solve the following simultaneous equations and determine whether
or not the two sets in each pair are equivalent.
f 3x + 2y = 32, f 7x - y = 1,
(a) \ 20x - 3y = 1. t 9x + 4y = 70.
Ans. (2, 13). The two sets are equivalent.
3x + 7y = 2, f 2x + 3y = 0,
7x + 8y = - 2. t 4x + y = - 4.
2t = - 3, f s + 5t = 3,
f
I
1 i 5s - 3t = - 6. (. s + t = 0.
f f * + y = i, f 5x + 4y = 22,
1 1* - fjr - H. I 3x + t/ = 9.
Ans. These two sets are not equivalent.
3x - 2y = 1, t x - y = 1,
(e) { 3x + 4z = 5, | x + z = 1,
3y + 5z = 4. I y + z = 0.
Ans. (—1, — 2, 2). Ans. An infinite number of solutions.
II, § 21] REVIEW OF EQUATIONS 21
3. Eliminate the x2 term from the equations
x2 - 2y* + 13x +2y = l, 3z2 + 4?/2 - x + Qy = 3.
4. Eliminate y from the equations
x2 + 3xy - x + 1 = 0, 2x + y + 1 = 0.
5. Eliminate t from the equations
,
l+t*' 1+1? .
6. Eliminate t from the equations
J2x = <4 + <2 + 1, ty = P - 1.
7. Eliminate ra from the equations
w = — — mx, x = my.
m
8. Clear the following equations of fractions and radicals and
determine in each whether the resulting equation is equivalent to the
given one :
2-3x 3x-l x -
(d)
4x9x x2 — 9 a; + 3
(e) x + Vx + 6 = 0. Cf) V'6 - 5x =
(i)
9. How must 1% ammonia and 28% ammonia be mixed to get 12
pints of 10% ammonia?
Ans. 8 and 4 pints.
10. Two given mixtures contain respectively p% and q% of a cer-
tain ingredient. Show that if x units of the first be combined with y
units of the second so that the resulting mixture contains r% of this
ingredient, then x:y = r — p:q — r.
22 MATHEMATICS [II, §21
10. Assume that gravel has 45% voids and sand 33%, and that 4
bags of cement make 3.8 cu. ft., how much cement, sand, and gravel
are necessary to make 1 cu. yd. of concrete?
(a) in a 1:2:4 mixture. (fe) in a 1 : 3 : 6 mixture.
(c) in a 1 : 2 : 3 mixture. (d) in a 1 : 3 : 5 mixture.
11. How many pounds of skimmilk must be extracted from 12000
Ibs. of 4% milk to raise the test to 4.5%?
Am. 1333| Ibs.
12. How many pounds each of 40% cream and skimmilk are required
to make 125*pounds of 18% cream?
Ans. 56.25 Ibs. cream, 68.75 Ibs. skimmilk.
13. How many pounds each of 25% cream and 3.5% milk are
required to make 130 pounds of 22.5% cream?
Ans. 114.8 Ibs. of 25%, 15.2 Ibs. of 3.5%.
14. How much 25% cream must be added to 1000 pounds of 50%
cream to reduce it to 40% cream?
Ans. 666f Ibs.
15. How many pounds each of 50% and 25% cream must be mixed
together to produce 1000 pounds of 40% cream?
Ans. 600 Ibs. of 50%, 400 Ibs. of 25%.
22. Polynomials. Expressions of the form
1 - x, xz - 3% + 2, x+ A/3.T3 + 3.4 + lxz,
x* - 2x* + x - 5
, 3x«,
are examples of polynomials in x;
y - 5 + 4y2, z5 + ^ + A/2z2 - \ , ^ + a3 - 1 + 2a,
5 / o
are polynomials in y, z, and «, respectively
A polynomial, in x for example, is a sum of terms each con-
taining a positive integral power of x multiplied by a coefficient
independent of x, and usually also an absolute term.
If any number (value of x) be substituted for x, the poly-
nomial reduces to a number called a value of the polynomial.
II, §23] REVIEW OF EQUATIONS 23
To each value of x, which is called the variable, there corre-
sponds a unique value of the polynomial. For example, the
values of x2 — 3x + 2 which correspond to x = 0, x = 1,
x = \, are 2, 0, f .
The degree of any term in a polynomial is the exponent of
the variable in that term. The degree of a polynomial is the
degree of the term of highest degree in it. Polynomials are
usually arranged according to the degrees of the terms and it is
sometimes convenient to supply with zero coefficients missing
terms of degree lower than the degree of the polynomial; thus
3x2 + 0-x + 2, z5 + 0-z4 + 0-z3 + V2> + fz - f .
A sharp distinction is to be made between the coefficients
and the exponents in a polynomial. The coefficients are very
general: they may be any real numbers whatever, natural
numbers, rational or irrational numbers, positive, negative, or
zero. On the other hand the exponents are very special: they
must be positive integers. Thus while the expressions
z2 + 1, ?/ + y/2, z2 - TTZ + V2,
are polynomials, the expressions
3.1/2 + 1? yt _j_ 2/y, z2 - 32 + Vz,
are not.
23. Polynomial of the nth Degree. A polynomial of de-
gree n in x (n being any given natural number 1, 2, 3, • • •) can
be reduced by merely rearranging its terms and adding the
coefficients of like powers of x to the form
in which the o's (coefficients) are any real numbers (oo =J= 0
but any or all the others may be zero), and the exponent n is a
positive whole number.
24 MATHEMATICS [II, §24
24. Linear Equations. An equation of the first degree, or a
linear equation, in one unknown, x for example, is the result
of equating to zero a polynomial of the first degree in x,
(10) ax + b = 0 (a + 0).
This equation has one and only one solution. The method of
finding the solution is already known to the student.
25. Quadratic Equations. An equation of the second de-
gree, or a quadratic equation, in x for example, is the result of
equating to zero a polynomial of the second degree in x,
(11) ax2 + bx+c = Q (a 4= 0).
Any equation which can be reduced to this form by merely
transposing and combining like terms is also called a quadratic.
Thus,
(x - l)(x - 2) = Q(x - 3)
is a quadratic.
SOLUTION BY FACTORING. If the polynomial axz + bx + c
can be factored into two linear factors in x (i. e., polynomials
of the first degree in x) the roots of the quadratic equation
ax2 + bx + c =0 can be found by inspection.
EXAMPLE 1. Solve 6x2 + x = 15. Transpose all terms to the
left side and factor. In order to do this we seek a pair of numbers
whose product is 6 and another pair whose product is — 15 and such
that the cross product is 1. The work may be put down as follows:
6z2 + x - 15 = 0
-5
This gives the cross product 13, but a few trials of other factors and
other arrangements quickly leads to the combination
-3
II, §25] REVIEW OF EQUATIONS 25
which gives the cross product 1 as desired. Hence the factors are
3x + 5 and 2x — 3, and we have to solve the equation
(3x + 5)(2x - 3) = 0.
On equating the first factor to zero (mentally) and solving we get
Xi = — 5/3 and similarly from the second factor xz = +3/2, and
these are the two solutions of the given quadratic equation.
EXAMPLE 2.
Q o 2_ "" "^
3x2 — i x = • — = — - .
Transposing and combining terms this reduces to
¥ & + & x = 0
which factors by inspection into
Z(-27Q X + *V) = 0
whence xi = 0 and xz = — ?V-
If there are fractional coefficients in a quadratic it is usually best to
reduce it to an equivalent equation free from fractions by multiplying
every term by the least common multiple of all the denominators.
Thus in Example 2, we could multiply every term by 21 and obtain,
63x2 - 14x = 3x2 - 15x.
EXERCISES
Solve the following quadratic equations.
1. 2z2 - 5x = 3. 2. 10x2 + x = 2.
3. 6x2 + 5x = 6. 4. 15x2 - x = 6.
5. 6x2 - 5 = 7x. 6. 28x2 - 15 = x.
7. 135x2 + 3x = 28. 8. 78x2 - x = 2.
9. 3?/2 + y = 10. 10. 147/2 + y = 168.
11. Gy2 + lly = 35. 12. 15?/2 + 4 = 16y.
13. 6a2 + a = 5. 14. 2a + 3 = 8a2.
15. 9a(2a + 1) = 14. 16. 10(2a2 - 3) + a = 0.
17. 3(2s2 - 7) = 5s. 18. 15(2i2 - 1) + 7t = 0.
19. p(12p - 7) = 10. 20. 5(3r2 - 8) + r = 0.
26 MATHEMATICS [II, §25
21. A z2 + 2x + 1T47 = 0. 22. v(y ~ 1} = 3(y + 1) - 2/2.
23. 2(2 - 1) = ;ft(6z - 1). 24. P + 3« - 1 = -
25. (1 - e2)z2 - 2px + p2 = 0.
Clear the following equations of fractions, solve the resulting equa-
tions and test their solutions in the given equations.
07
2x - 7 x - 3 ' x2 - 1 2(x + 1) 4 '
3Oi» 1 O O A
M,X i_ zx — K OQ z _ d
z-5z-3 z-1 x-2 x - 3 x-4'
26. Solution of a Quadratic by Completing the Suqare.
If the polynomial on the left of the quadratic equation
ax2 + bx + c — 0
cannot readily be factored by inspection, the equation can
be solved by transposing the absolute term c, completing the
square of the terms in x and extracting the square roots of
both sides.
To complete the square of ax2 + bx is to find a number d
such that ax2 + bx + d is the square of a linear factor in x and
it can always be done as follows: 1) extract the square root of
the first term; 2) double this; 3) divide this into the second term;
4) square the quotient.
EXAMPLE. Solve 6x2 — 4x — 1 =0.
Transpose — 1, and find the number to complete the square of
6x2 - 4x by the above four steps: 1) xV6, 2) 2x>/6, 3) 2/V6, 4) 2/3;
add this to both sides:
6x2 - 4x + f = f .
Extracting the square roots, we have
-v-v/fi — A/1 = 4- A/£
•*• 'u »3 — 3I "?»
whence solving for x, we find
r, — i -L. IA/TO r« = l i •\/T7)
•^l — 3 i^ 6 '-*-", ^2 3 ^^ § i(XV/.
II, §27] REVIEW OF EQUATIONS 27
The computations are more easily made, if we multiply the given
equation through by a number which will make the coefficient of x2 a
perfect square. In the above example we should have to solve the
equivalent equation,
36z2 - 24x + ( ) = 6.
The number required to complete the square is 4,
36z2 - 24z + 4 = 10.
Whence
6z - 2 = ± VlO
and
Si -i- |ViO.
EXERCISES
Solve these equations by completing the square.
1. 4z2 + 3z = 9. 2. 25z2 - 14z + 1 = 0.
3. 50z2 + 12x = x2 - i 4. (x2 + 1) V3 = 4z.
5. 12z2 + 5x = 1. 6. 6^2 + 1 = 6y.
7. 322 = 13(2 - 1). 8. x + 2 = llar(l - x).
9. 12<2 - 4(o + b)t + a6 = 0. 10. 2(?/2 + c2) = 5ey.
27. Solution of a Quadratic by a Formula. By the process
of completing the square, a formula for the roots of the general
quadratic equation can be found as follows. Given the equation
(12) ax* + bx + c = 0,
multiply through by 4a, transpose 4ac, and complete the square,
4a2z2 + 4a6x + 62 = - 4ac + 62,
extracting the square roots, we have
2ax + b = ± V62 - 4oc,
whence we find
- & =fc Vfe2 - 4ac
~^a^ -•
which gives the two roots
- 6 + &2 - 4ac - 6 -
28 MATHEMATICS [II, §27
This result may be used as a formula for the solution of any
quadratic equation by substituting for a, 6, c, of this formula
their values from the given equation.
EXAMPLE. Solve 3x2 + 4x — 15 = 0.
Here a = 3, 6 = 4, c = — 15, and by the formula
_ - 4 ± V16 + 180
6
whence
_ -2+7_5
3/1 "~ « — * «-. cLIlCi 3/9 "
33 a
EXERCISES
Solve the following equations by the formula.
1. 2x2 + 3x = 4. 2. x2 = 220 + 9x.
3. 5x2 + 3x = 3. 4. 5x2 + 5x + 1 = 0.
5. 15?/2 = 86y + 64. 6. 5z2 = 80 + 21.
7. a2 + a = 3. 8. p2 + 3p = 40.
9. I2 + 3a2 = 4o< - 1. 10. 5m2 + 21m + 4=0.
Solve the following equations by any method and test all results in
the given equation.
11. (2x - 3)2 = 8x. 12. x2 - 2 A
10 "x , X ~r •" o 14*C'l
!3x + 2 = 0.
J' x + 2 ' 2x x
I K X 1 , 1 * f)
x - 1*
x(x - 2) 2x - 2 ' 2x
ifi 4 ! 3 2
'x — 1 4 — x x — 2 3— x'
17. 3x2 + (9o - l)x - 3a = 0. 18. x2 - 2ax + a2 - fc2 = 0.
19. c2x2 + c(a - 6)x - afc = 0. 20. x2 - 4ax + 4a2 - 62 = 0.
21. x2 - 6acx + a2(9c2 - 462) = 0.
22. (a2 - 62)x2 - 2 (a2 + 62)x + a2 - ft2 = 0.
II, §28] REVIEW OF EQUATIONS 29
Solve for y in terms of x.
23. x2 + 12xy + 9y2 + 3 = 0. 24. x2 - 4xy - 4y* + x = 0.
25. llx2 + 3Qxy + 25y* = 3. 26. 8x2 - 12x?/ + 4y2 = x + 1.
27. Gx2 - XT/ - 2y2 = 0. 28. 21x2 = xy + lOy2.
29. 30x2 + 150* = 43xy. 30. 12x2 + 41xy + 35y2 = 0.
31. 2x2 + 3xy - 2y2 + x + 7y - 3 = 0.
32. 3x2 + lOxy + 8?/2 + 4x + 2y - 15 = 0.
33. 10x2 + 7xy + r/2 - x - 2?/ - 3 = 0.
34. 12x2 = 4xy + 2ly2 + 2x + 29y + 10.
35. A farmer mows around a meadow 18 X 80 rods. If the swath
averages 5 ft. 6 in., how many circuits will cut half the grass?
Ans. 12.
38. What are eggs worth when 2 more for a quarter lowers the price
5 cents a dozen?
37. If the radius of a circle be divided in extreme and mean ratio
the greater part is the side of the regular inscribed decagon. What
is the perimeter of the regular decagon inscribed in a circle 2 feet in
diameter? Ans. 6.180
38. When a heavy body is thrown upward with an initial velocity
v ft. per second, its distance from the earth's surface at the end of t
seconds is given by the equation d — vt — 16<2. If a projectile is shot
upward with a muzzle velocity of 1000 ft. per second, when will it be
15,600 ft. high? Ans. 30 and 32^ sec.
28. Equations in Quadratic Form. The terms of an equa-
tion which is not a quadratic in the unknown can sometimes
be grouped so as to make it a quadratic in an expression con-
taining the unknown. Thus, x4 — 13z2 + 36 = 0 is not a
quadratic in x but it is a quadratic in x2] again if the terms of
x* — 6x3 + 7x2 + Qx = 8 be grouped in the form
- 2(x2 - 3z) = 8
it is seen to be a quadratic in (x2 — 3x).
30 MATHEMATICS [II, §28
EXAMPLE 1. Solve 6x - 7Vx = 20.
Transpose 20 and this can be solved by the formula as a quadratic
in Vx; whence,
r 7 ± A/49 + 480
Vx= "IT"
and, since the positive square root cannot be negative,
>lx = 2.5 and a; = 6.25.
We verify that this satisfies the given equation.
EXERCISES
1.x4- 13x2 + 36 = 0.
2. x + Vx + 6 = 14. Ans. 10.
3. 2x2 + 3 Vx2 - 2x + 6 = 4x + 15. Ans. - 1 and 3.
= 2. Ans. 0 and 1.
x + Vl -x2
5. sVx + 18 = 5 3Vx~2. Ans. 8 and - 729/125.
6. x4 - 6X3 + 7x2 + 6x = 8. Ans. - 1, 1, 2, 4.
29. Imaginary Roots. . There are quadratics which are not
satisfied by any real number. For example, z2 = — 4, x2 + 2x
+ 2 = 0. This is because the square of every real number
(except 0) is positive. If we attempt to solve the equation
x2 + 2x + 2 = 0 either by completing the square or by the
formula we are led to the indicated square root of a negative
number, and this is not a real number; thus
These, and other considerations have led to the invention
of numbers whose squares are negative real numbers; they are
called imaginary numbers. The imaginary unit is usually
denoted by i. Hy definition, we have
II, §29] REVIEW OF EQUATIONS 31
The number r-i, where r is any real number is called a pure
imaginary number; e. g., 2i, 5i, — 3i, -- %i, i V3, etc. The
squares of pure imaginary numbers are negative real numbers;
e. g., (2i)z = 22i2 = - 4; (- 3i)2 = (- 3)2i2 = - 9; (i VJF)2
= - 3.
Conversely, the square roots of negative real numbers are
imaginary numbers; the square roots of — 4 are 2i and — 2i;
i. e., V^l =_i VI" = 2i, - - V- 4 = - i VI" = - 2i; V^3
= i V3, — -V — 3 = • i V3 ; in general, V— p = i ^p, where
p is a real positive number.
Expressions of the form 2 + 5i, 1 — i, 3 — 2i, — I + i,
etc., indicating the sum of a real and an imaginary number are
called complex numbers. They may be added, subtracted,
multiplied, and divided by the laws of algebra as though i were
a real number and the results simplified by putting — 1 for i-,
- i for i3, + 1 for i4, etc.
We can now say that every quadratic equation can be solved.
The solutions of the equation
x2 - 2x + 2 = 0
may be found by the formula,
2 ± A/4 - 8
,-- --^
whence
x\ = 1 + i and xz = 1 — i;
and we verify both these answers as follows:
(1 + i')2 - 2(1 + i) + 2 = 0, (1 - *)* - 2(1 -0+2 =
EXERCISES
i. x- - 4x + 5 = 0. 2. xz + <ox + 13 = 0.
-4ns. 2 ± i. Ans. — 3 ± 2i.
3. 36x2 - 36x + 13 = 0. 4. 2x2 + 2x + 1 = 0.
Ans. 1/2 ± t'/3. -4ns. — 1/2 ± t/2.
32 MATHEMATICS [II, §29
5. x2 + 4 = 0. 6. x2 + x + 1 = 0.
Ans. ± 2i. Ans. - 1/2 ±i V3/2.
7. z2 - 2z + 3 = 0. 8. x2 - |x + 1 = 0.
9. x2 - 2x VJj + 7 = 0. 10. 2z2 - 2x + 5 = 0.
11. x2 + 3x + 2.5 = 0. 12. 49z2 - 56x + 19 = 0.
30. The Sum and Product of the Roots. The two roots
of the quadratic equation
ax2 + bx + c = 0
are by (14), §27,
- 6 + V&2 - 4oc - 6 - Vb2 - 4oc
zi = and x2 = — - .
2a 2a
The sum of these roots is — b/a, and their product is + c/a, as
may be seen by adding and multiplying them together.
We can thus find the sum and the product of the roots of a
given quadratic equation without solving it. Thus in the
equation,
36z2 - 36z + 13 = 0
the sum of the roots is 1, and their product is 13/36.
Again in the equation,
myz — 4ay + 4a6 = 0
the sum of the roots is 4a/m, and their product is 4ab/m.
31. Equation having Given Roots. We have seen that if
the left member of the quadratic equation
ax2 + bx + c = 0
can be separated into linear factors, its roots can be found by
inspection. Therefore if we wish to make up a quadratic
equation whose roots shall be two given numbers, r and s for
example, we hare only to write
a(x — r)(x — s) =0
II, §33] REVIEW OF EQUATIONS. 33
and multiply out. The factor a is arbitrary and may be chosen
so as to clear the equation of fractions if desired; thus, to make
an equation whose roots shall be f and — f , we write,
a(z-f)(z+f) =0,
and if we take a = 10, the resulting equation is
10z2 - llz - 6 = 0.
32. Number of Roots. Conversely, it is readily shown that
if r and s are roots of the quadratic equation,
ox2 + bx + c = 0
then the left member can be factored in the form,
(15) a(x - r)(x - s) = 0
and this shows that no quadratic can have more than two
roots. Some quadratics have only one root; for example
4z2 + 9 = 12x is satisfied only by x = 3/2.
If 62 — 4ct£ = 0, then the polynomial ax2 + bx + c is a
perfect square and the equation ox2 + bx + c = 0 has only
one root, and conversely.
For, if 62 — 4oc = 0, then c = 62/4a and this is precisely
the number necessary to complete the square of ox2 + bx.
Also if 62 — 4oc = 0, the formula (13), § 27, gives not two but
one root.
33. Kind of Roots. If a, b, and c, are real numbers and if
62 — 4ac > 0, then the quadratic equation ax* + bx + c = 0
has two real roots; but if 62 — 4oc < 0, the equation has two
imaginary roots.
This is seen at once on noting the formula (13), § 27, which
gives the roots.
4
34 MATHEMATICS [II, §33
EXAMPLE 1. 4z2 - 12x + 9 = 0.
Here b2 — 4ac = 144 — 144 = 0, the left member is a perfect square
and the equation has only one root.
EXAMPLE 2. 3x2 - 5x + 2 = 0.
Compute b2 — 4ac = 25 — 24 = + 1, which shows that the equa-
tion has two real roots.
EXAMPLE 3. x2 + x + I = 0.
Here b2 — 4oc = — 3, which shows that the equation has imaginary
roots.
If a, b, and c, are rational numbers, then the roots of the equation
ax2 + bx + c = 0 are rational if b2 — 4ac is a perfect square, i. e., the
square of a rational number: in particular if a, b, and c, are integers
and if b2 — 4ac is a perfect square the left member of the equation can
be factored by inspection.
EXAMPLE 4. 2x2 - x - 6 = 0.
Here b2 — 4ac = 1 + 48 = 49; the left member factors into
(2x + 3)(x - 2) = 0
whence the roots are — 3/2 and 2.
EXERCISES
1. Form the equations whose roots are
(a) 1, 3, - 5. (6) - 2, 3, - 4, 6.
(c) 1/3, - 7/2, 3/5. (d) ± 1, ± 4.
(e) ± V2, ± V5. (/) 0, - 2, ± V^"2.
(?) 3, 5 ± >/5. (ft) 4 ± A/3, - 1 ± VS.
(i) - a, - 0, - 7. 0') ka> k-P, ky.
(k) <x+k,0^k y+k. (/) I/a, 1//S, 1/7.
(TO) a, ft 7. (») a2, /S2, 72-
(0) a - 0, 0 - 7, 7 - «• (P) "A 07, 7«-
2. Determine the nature of the roots of the following equations.
(a) 3x2 - 4.r - 1 =0. (6) 5x2 + 6z + 1 = 0.
(c) 2x2 + x - 6 = 0. (d) x* - 2x - 1 = 0.
(e) 5x2 - 6x + 5 = 0. (/ ) x2 - 6 V3x - 5 = 0.
(g) x2 + x + 1 = 0. (h) 13x2 - 6 >/3x + 7 = C.
(1) 3x2 + 2x + 1 = 0. 0') 2*2 - 16x + 9 = 0.
(*) 5x2 - 12x - 8 = 0. (0 6x2 + 4x - 5 = 0.
II, §33] REVIEW OF EQUATIONS 35
(m) 5x + 7 = (3x + 2)(x - 1). (n) 5(x2 + x + 1) = 1 - 16x.
(o) 2x(x - 3) = 7(3x + 2). (p) 7(x2 + 5x + 3) = x(l - x).
(9) 3x(x + 1) = (3 - x)(3 + x). (r) 3x2 = 13(x - 1).
t(s) 0, + 2)(V - 2) = 2y - 7. (0 3fo + l)fo - 1) = 4y.
(M) 60(3 - %) = 19(0 - I)2- (») 2/2 - 2yV3 + 7 = 0.
3. Without solving find the sum and the product of the roots of each
of the equations in Ex. 2.
4. Determine the nature of the roots of the following equations in
which a, b, c are known real numbers.
(a) (x - a)2 = 62 + c2. (6) (x + a)2 = 862.
(c) a (ax2 + 26x - a) = b(bx2 - lax - b).
(d) 7/2 = 2a(y -b) + 2b(y - a).
(e) (a + 6 - c)?/2 - 2cy = (a + b + c).
(/) (a + b - c)x2 + 4(o + b)x + (a + b + c) = 0.
(0) (b + c - 2a)x2 + (c + a - 26)x + (a + 6 - 2c) =0.
5. Determine values of a for which each of the following quadratic
equations will have equal roots.
(a) x(x + 4) + 2a(2x - 1) = 0. (6) (x - I)2 = 2a(3x - 7) - 20.
(c) (x - a)2 = a2 - 8a + 15. (d) x2 - 15 = 2a(x - 4).
(e) 3(x2 + Sax - a) = x. (/) 9x2 + 6(0 - 4)x + a2 = 0.
(g) 3ax(x - 1) = ax - 2. (h) (a + l)x2-(a + 2)x +fa = 0.
(t) (4a2 + 3)x2 + 8a(3 - 2a)x + 4(4a2 - 12a - 3) = 0.
6. Find a value of k such that the sum of the roots of the equation
x2 — 3(k + l)x + Qk = 0 shall be one half their product.
7. Construct equations whose roots shall be greater by 2 (also less
by 2) than the roots of the equations in Ex. 2.
8. Construct equations whose roots shall be twice (also half) the
roots of the equations in Ex. 2.
CHAPTER III
GRAPHIC REPRESENTATION
34. Graphic Methods. The methods studied in plane geome-
try for constructing various figures when certain of their dimen-
sions and angles are known are used extensively in making
designs for machines, plans for buildings and various other
structures, and also for solving problems that require the deter-
mination of unknown dimensions, angles, areas, etc.
These methods often give the desired results with sufficient
accuracy for practical purposes, and they are more direct and
rapid than numerical computation. Of even greater importance
however is their use in checking the results of calculations, since
there are always possibilities of error even when great care is
exercised. It should be emphasized that every practical calcu-
lation (i.e. one which is to be used in construction, or other ac-
tual work where time, material, and money will be wasted if
the calculation is incorrect) should always be checked by some
independent means.
Two rectilinear figures are similar if their corresponding angles
are equal and their corresponding dimensions are proportional.
35. Drawing to Scale. When two plane figures are similar,
each is said to be a scale drawing of the other. The smaller
is said to be reduced or drawn to a smaller scale. For example,
if a drawing be made of a floor plan of a house so that the angles
in the drawing are equal to those in the house itself, and the
dimensions of the drawing are -^ of those of the house, it is said
to be drawn to a scale of ^ inch to one foot. From such a draw-
ing the builder can read off on a scale divided into quarter inches
the dimensions of the parts he is about to construct.
36
Ill, § 36] GRAPHIC REPRESENTATION 37
This method of drawing figures to scale, reading off their un-
known angles on a protractor, and their unknown dimensions
on a conveniently divided scale, furnishes a graphic solution of
many problems and it has many practical applications.
EXAMPLE. The distance AB = 98 yards,
Fig. 1, and the angles PAB = 51°, PBA = 63°,
having been measured from one side of a river,
the triangle can be drawn to scale and the
width PR of the river can be read off on the
scale, about 75 yards.
FIG. 1
EXERCISES
1. Find the length of the projection of the altitude of an equilateral
triangle upon one of its sides. Ans. .75s
2. Draw two diagonals through the centre of a regular hexagon.
Find the length of the projection of one of them upon the other.
Ans. s.
3. Draw two diagonals through the same vertex of a regular penta-
gon. Find the length of a projection of one of them upon the other.
Ans. 1.3s
4. The pitch of a roof is the ratio of the height of the ridge above
the plates to the distance between the plates. Find the length of the
rafters and their inclination for a f pitch roof on a building 28 ft. wide.
Ans. 21.8, 50°.
5. Find the length of the corner rafters, and also of the middle rafter
on each side of a square roof on a house 34 X 34 feet, the apex of the
roof being 12 feet above the top floor. Find also their inclinations.
Ans. 26.9, 20.8, 26°.5, 35°.
6. The roof of a building 36 ft. wide is inclined at an angle of 54°
to the horizon. Find the length of the rafters, allowing 2 ft. overhang.
Ans. 32.6 ft.
7. To determine the horizontal distance between two points P and Q
on the same level but separated by a hill, a point R is selected from which
P and Q are visible. Then PR = 200 ft., QR = 223 ft., and angle PRQ =
62° are measured. Draw the figure and scale off PQ. Ans. 210.
38 MATHEMATICS [III, § 36
8. The steps of a stairway have a tread of 10 in., and a rise of 7 in.
Find the inclination of the stringers to the floor. Ans. 35°.
9. Plot four points on a sheet of paper. Mark them A, B, C, D.
Construct a point P one-half the way from A to B, a point Q one-
third the way from P to C, and a point R one-fourth the way from
Q to D. Mark the four given points in some other order and repeat
the construction. What conclusion do you draw ?
36. Rectangular Coordinates of a Point in a Plane. The
position of any point in the plane is uniquely determined as
X' O
Y'
FIG. 2
soon as we know its distance and sense from each of the two per-
pendicular lines X' X and Y'Y. These lines are taken first, and
are drawn in any convenient position.
The distance from X' X (RP = b in the figure) is called the
ordinate of the point P. The distance from Y'Y (SP = a in
the figure) is the abscissa of P.
Abscissas measured to the right of Y'Y are positive, those to
the left of Y'Y are negative. Ordinates measured above X' X
are positive, those below negative.
The abscissa and ordinate taken together are called the
coordinates of the point, and are denoted by the symbol (a, 6).
In this symbol it is agreed that the number written first shall
stand for the abscissa.
The lines X' X and Y'Y are called the axes of coordinates,
X' X being the axis of abscissas or the axis of X, and Y'Y the
HI, §37]
GRAPHIC REPRESENTATION
39
axis of ordinates or the axis of Y; and the point 0 is called the
origin of coordinates.
The axes of coordinates divide the plane into four parts
called quadrants. Figure 3 indicates the proper signs of the
coordinates in the different quadrants.
III
IV
FIG. 3
FlG. 4
37. Plotting Points. To plot a point is to locate it with
reference to a set of coordinate axes. The most convenient
way to do this is to first count off from 0 along X'X a number
of divisions equal to the abscissa, to the right or left according
as the abscissa is positive or negative. Then from the point
so determined count off a number of divisions equal to the ordi-
nate, upward or downward according as the ordinate is positive
or negative. The work of plotting is much simplified by the
use of coordinate paper, or squared paper, which is made by ruling
off the plane into equal squares, the sides being parallel to the
axes. Thus, to plot the point (4, — 3), count off four divisions
from 0 on the axis of X to the right, and then three divisions
downward from the point so determined on a line parallel to
the axis of Y, as in Fig. 4.
If we let both x and y take on every possible pair of real
values, we have a point of the plane corresponding to each pair
of values of (x, ?/). Conversely, to every point of the plane
corresponds a pair of values of (x, y).
40
MATHEMATICS
[HI, §37
EXERCISES
1. Plot the following points (3, 3), (4, 5), (- 2, 3), (- 4, - 2),
(7, - 2), (0, 4), (0, - 4), (3, 0), (- 3, 0), (0, 0).
2. What is the y-coordinate of any point on the x-axis?
3. What is the z-coordinate of any point on the y-axis?
4. Show that the line joining (5, 4) and (—5, — 4) is bisected by
the origin.
5. Find the distance from the origin to each of the points in Ex. 1.
6. Find the lengths of the sides of the triangle whose vertices are
(1, 1), (5, 2), (3, 4). Ans'. Vl7; Vl3; 2V2.
7. What is the abscissa of any point upon a straight line parallel
to the y-axis and four units to its right?
8. What is the ordinate of any point upon a straight line parallel
to the re-axis and three units above it?
9. (a) What relation exists between the coordinates of any point
of a line bisecting the angle between the positive directions of the
two axes? (6) Between the positive direction of the y-axis and the
negative direction of the x-axis?
10. What relation would exist between the coordinates of any
point of the line in Ex. 9 (a), if it were raised four units parallel to
itself? If it were lowered five units?
38. Statistical Data. The following table shows the rainfall
in inches, as observed at the Agricultural Experiment Station at
LaFayette, Indiana, by months for 1916, 1917, and the average
for the past 30 years.
Jan.
Feb.
Mar.
i~08
Apr.
May
Jun.
Jul.
Aug.
Sep.
Oct.
2~25
Nov.
Dec.
1916 . .
740
1 16
1 57
582
527
3 56
1 81
222
225
4.79
1917 . .
1 54
1 25
409
432
475
541
1 47
409
1 03
522
0 13
0.68
Average .
3.11
2.88
3.78
3.38
4.05
3.75
3.54
3.32
3.03
2.46
3.23
2.71
While it is possible by a study of this table to compare the
rainfall month by month in the same year, or for the same month
in the two years, or any month with the normal for that month,
Ill, § 38]
GRAPHIC REPRESENTATION
41
these comparisons are more easily made and the facts are pre-
sented much more emphatically by the diagram shown in Fig. 5.
This is made from the data of the table as follows. The 24
vertical lines represent the months of the two-year period. The
altitudes of the horizontal lines represent inches of rainfall. The
height (ordinate) of the point marked on any vertical line shows
the rainfall for that month. The points are connected by lines
to aid the eye in following the march of the rainfall. The full
line represents the rainfall for 1916 and 1917, the dotted line
the normal rainfall as shown by the experience of 30 years.
J F M A M J JASONDJFMAMJJA S O N D
MONTHS 1910 .MONTHS 1817
FIG. 5
Rainfall is a discontinuous phenomenon. Moisture is not pre-
cipitated continuously, but intermittently. However, if we
make a similar diagram showing the temperature at each hour
of the day we might have inserted many other points. We
42
MATHEMATICS
[HI, § 38
think of the change in temperature as a continuous phenome-
non ; e.g. when the temperature rises from 42° at 8 A.M. to 51°
at 9 A.M., we think of it as having passed through every inter-
vening degree in that hour. Thus we can think of the points
which represent the temperature on the diagram from instant
to instant as lying thick on a continuous curved line. This
curve is called the temperature curve.
In making a graph of a discontinuous function like rainfall,
we connect the points with straight lines as in Fig. 5, but in case
of a continuous function like temperature, a smooth curve which
passes through all the plotted points is the best graphic repre-
sentation of the function.
EXERCISES
1. Make a temperature graph from the following data,
Hour, A. M.
Temperature
12
45
1
45
2
45
3
45
4
43
5
42
6
41
7
40
8
42
9
51
10
57
11
59
12
62
Hour, p. M.
Temperature
1
66
2
70
3
74
4
76
5
76
6
75
7
74
8
73
9
72
10
70
11
69
12
68
2. Determine from Fig. 5 which were the dry months in 1916. In
1917. To what extent do they agree with each other and with the
normal ?
3. Do as directed in Ex. 2 for the wet months.
4. What straight line in Fig. 5 would represent the average monthly
rainfall for 1916? For 1917? For the past 30 years?
5. To what extent does the dotted line in Fig. 5 enable you to pre-
dict the probable rainfall in any given month subsequent to 1917?
6. Procure the census data and plot the population graph of the
United States by decades for a century.
7. Plot a graph of the attendance of students at your college or Uni-
versity for as many years back as you can secure the data.
8. The following data give the Chicago price per bu. of No. 2 corn
by months from Jan., 1903, to May, 1908. Plot the data using years
as abscissas and price as ordinates.
Ill, §38]
GRAPHIC REPRESENTATION
43
Jan.
Feb.
Men.
Apr.
May.
June.
July.
Aug.
Sept.
Oct.
Nov.
Dec.
1903
1904
1905
1906
1907
1908
43
42
41
42
39
59
42
46
42
41
43
56
41
49
45
39
43
58
41
46
46
43
44
65
44
47
48
47
49
70
47
53
51
50
51
49
47
53
49
52
50
51
53
48
54
45
51
51
47
60
43
50
50
44
55
41
50
45
44
55
41
43
42
40
57
9. Find from the graph that month in each year in which the highest
price occurred. The lowest price. Find the difference for each year
between the highest and lowest price for that year. Does there appear
to be any relation between these prices and the period of harvest?
10. The following data gives the Chicago price of No. 2 oats by
months from Jan., 1903 to May, 1908. Plot the data using years as
abscissas and price as ordinates.
Jan.
Feb.
Mch.
Apr.
May.
June.
July.
Aug.
Sept.
Oct.
Nov.
Dec.
1903
1904
1905
1906
1907
1908
31
36
29
31
33|
48
33
39
29
29
37
48
31
38
29
28
39
52
32
36
28
30
41
51
33
39
28
32
44
53
35
39
30
33
41
33
38
27
30
41
33
31
25
29
44
35
29
25
30
51
34
28
27
32
45
33
29
29
33
44
34
28
29
33
46
11. Handle the data in Ex. 10 as directed in Ex. 9.
12. A restaurant keeper finds that if he has G guests a day his total
daily expenditure is E dollars and his total daily receipts are R dollars.
The following numbers are averages obtained from the books:
G..
210
270
320
360
E
16.70
19.40
21.60
23.40
R
15.80
21.20
26.40
29.80
Plot two curves on the same set of axes in each case using G as abscissas.
For one curve use E as ordinates, for the other use R as ordinates.
Below what value of G does the business cease to be profitable?
Connect the points (G, E) by a smooth curve. Continue this curve
until it cuts the line (7 = 0. What is the meaning of the ordinate E
for G = 0? Through what point ought the curve connecting the
points (G, R) to pass? Ans. (0, 0).
44
MATHEMATICS
[III, §38
13. The population of the United States by decades was as follows.
Plot, and estimate the population for 1920.
Year.
Population.
Year.
Population.
Year.
Population.
1790. .
1800....
1810....
1820
3,929,214
5,308,433
7,229,881
9,663,822
1840... .
1850. . . .
I860....
1870
17,069,453
23,191,876
31,443,321
38,558,371
1890..
1900. . . .
1910.. . .
62,669,756
76,295,200
91,972,266
1830
12,806,020
1880
50,155,783
14. The football accidents for the years given are as follows:
Year.
Deaths.
Injuries.
Year.
Deaths.
Injuries.
1901....
1902....
1903....
1904....
1905. . . .
1906
7
15
14
14
24
14
74
106
63
276
200
160
1907...
1908....
1909....
1910....
1911....
15
11
30
22
11
166
304
216
499
178
Plot two curves, using the years as abscissas and the deaths and injuries
respectively as ordinates.
15. The monthly wages in dollars of a man for each of his first 13
years of work was as follows: 28, 30, 37.50, 45, 60, 65, 90, 95, 95, 137,
162, 190, 210. Plot the curve showing the change. Estimate his
salary for the fourteenth and fifteenth years. Can you be certain of
his salaries for these years?
16. Of 100,000 persons born alive at the same time the following
table shows the number dying in the respective age intervals :
Months.
Deaths.
Months.
Deaths.
0-1
4,377
6- 7
579
1-2
1,131
7- 8
533
2-3
943
8- 9
492
3-4
801
9-10
456
4-5
705
10-11
421
5-6
635
11-12
389
Ill, §38]
GRAPHIC REPRESENTATION
45
Years.
Deaths.
Years.
Deaths.
0- 1
11,462
19- 20
344
1- 2
2,446
29- 30
479
2- 3
1,062
39- 40
644
3- 4
666
49- 50
873
4- 5
477
59- 60
1,404
5- 6
390
69- 70
, 1,974
6- 7
327
79- 80
1,854
7- 8
274
89- 90
571
8- 9
234
99-100
25
9-10
204
106-107
1
Plot the above data. Make two graphs. In each graph use deaths
as ordinates; in one use months as abscissas, in the other use years.
When is the ordinate smallest? largest? Does a small ordinate for the
years 99-100 and 106-107 indicate a low death rate? Explain. Note
the continuous decrease in the ordinate of the first curve.
17. Using the data below and on p. 46, plot a curve using years
as abscissas and price of corn as ordinates. Do you notice any reg-
ularity in the number of years elapsing between successive high prices?
successive low prices? Draw like graphs for the other crops listed?
18. Plot the prices for the yrs. 74,. 81, 87, 90, 94, 01, 08, 11, 1916.
What do you observe from this curve as to the tendency in the high
price of corn? Do you observe any tendency in the lowest prices of
corn that is in the prices for the yrs. 72, 78, 84, 89, 96, 02, 06, 1910?
AVERAGE FARM PRICE DECEMBER FIRST
Data from the year book of the Department of Agriculture 1916
Year.
Corn.
Wheat.
Oats.
Barley.
Rye.
Potatoes.
Hay, $ per
Ton.
1870.
49.4
94.4
39.0
79.1
73.2
65.0
12.47
1871.
43.4
114.5
36.2
75.8
71.1
53.9
14.30
1872.
35.3
111.4
29.9
68.6
67.6
53.5
12.94
1873.
44.2
106.9
34.6
86.7
70.3
65.2
12.53
1874.
58.4
86.3
47.1
86.0
77.4
61.5
11.94
1875.
36.7
89.5
32.0
74.1
67.1
34.4
10.78
1876.
34.0
97.0
32.4
63.0
61.4
61.9
8.97
1877.
34.8
105.7
28.4
62.5
57.6
43.7
8.37.
1878.
31.7
77.6
24.6
57.9
52.5
58.7
7.20
1879.
37.5
110.8
33.1
58.9
65.6
43.6
9.32
Continued on p. 46.
46
MATHEMATICS
[HI, §38
AVERAGE FARM PRICE, DECEMBER FIRST
Continued.
Year.
Corn.
Wheat.
Oats.
Barley.
Rye.
Potatoes.
Hay, S per
Ton.
1880.
39.6
95.1
36.0
66.6
75.6
48.3
11.65
1881.
63.6
119.2
46.4
82.3
93.3
91.0
11.82
1882.
48.5
88.4
37.5
62.9
61.5
55.7
9.73
1883.
42.4
91.1
32.7
58.7
58.1
42.2
8.19
1884.
35.7
64.5
27.7
48.7
51.9
39.6
8.17
1885.
32.8
77.1
28.5
56.3
57.9
44.7
8.71
1886.
36.6
68.7
29.8
53.6
53.8
46.7
8.46
1887.
44.4
68.1
30.4
51 9
54.5
68.2
9.97
1S88.
34.1
92.6
27.8
59.0
58.8
40.2
8.76
1889.
28.3
69.8
22.9
41.6
42.3
35.4
7.04
1890.
50.6
83.8
42.4
62.7
62.9
75.8
7.87
1891.
40.6
83.9
31.5
52.4
77.4
35.8
8.12
1892.
394
62.4
31.7
47.5
54.2
66.1
8.20
1893.
36.5
53.8
29.4
41.1
51.3
59.4
8.68
1894.
45.7
49.1
32.4
44.2
50.1
53.6
8.54
1895.
25.3
50.9
19.9
33.7
44.0
26.6
8.35
1896.
21.5
72.6
18.7
32.3
40.9
28.6
6.55
1897.
26.3
80.8
21.2
37.7
44.7
54.7
6.62
1898.
28.7
58.2
25.5
41.3
46.3
41.4
6.00
1899.
30.3
58.4
24.9
40.3
51.0
39.0
7.27
1900.
35.7
61.9
25.8
40.9
51.2
43.1
8.89
1901.
60.5
62.4
39.9
45.2
55.7
76.7
10.01
1902.
40.3
63.0
30.7
45.9
50.8
47.1
9.06
1903.
42.5
69.5
34.1
45.6
54.5
61.4
9.07
1904.
44.1
92.4
31.3
42.0
68.8
45.3
8.72
1905.
41.2
74.8
29.1
40.5
61.1
61.7
8.52
1906.
39.9
66.7
31.7
41.5
58.9
51.1
10.37
1907.
51.6
87.4
44.3
66.6
73.1
61.8
11.68
1908.
60.6
92.8
47.2
55.4
73.6
70.6
8.98
1909.
57.9
98.6
40.2
54.0
71.8
54.1
10.50
1910.
48.0
88.3
34.4
57.8
71.5
55.7
12.14
1911.
61.8
87.4
45.0
86.9
83.2
79.9
14.29
1912.
48.7
76.0
31.9
50.5
66.3
50.5
11.79
1913.
69.1
79.9
39.2
53.7
63.4
68.7
12.43
1914.
64.4
98.6
43.8
54.3
86.5
48.9
11.12
1915.
57.5
92
36.1
51.7
839
61.6
10.70
1916.
88.9
160.3
52.4
88.2
122.1
146.1
10.59
Ill, §40]
GRAPHIC REPRESENTATION
47
39. Other Graphic Methods. The statistical data given in
the preceding articles has been studied by means of curves or
graphs drawn on rectangular cross-section paper. There arc
other important methods of representing statistical data. Of
these methods we will give names to three:
(1) Bar diagrams or columnar charts.
(2) Dot diagrams.
(3) Circular diagrams.
These methods are best explained by means of examples.
40. Bar Diagrams. Below is given a bar diagram or chart
comparing the average size of farms for the years 1900 and
1910 for the states indicated.
Sizes of Farmsdn Hundreds of Ac-res
3 6 9 12 15
W (joining
California
Arizona
Nebraska
Missouri
Michigan
Georgia
Alabama
New York
Delaware
Legend:
mm&lO
c=l 1900
.
FIG. 6
EXERCISES
Make a bar diagram from the following data:
1. The number of cattle in millions on farms for 1900 and 1910 in
the following states were as follows.
Date.
Texas.
Iowa.
Nebraska.
New York.
Oklahoma.
Indiana.
1910...
7.0
4.5
2.9
2.4
2.0
1.3
1900. ..
10.0
5.5
3.2
2.6
3.3
1.7
48 MATHEMATICS [III, §40
2. The sheep on farms in millions in 1910 and 1900 were as follows.
Date.
Texas.
Iowa.
Nebraska.
New York.
Oklahoma.
Indiana.
1910...
1.6
1.1
0.3
0.9
0.1
1.3
1900. . .
1.8
1.0
0.5
1.7
0.15
1.7
3. Make a bar diagram comparing the number of hours work re-
quired by hand and machine labor in producing selected units (U. S.
labor bulletin 54).
Description of Unit.
Number of Hours Worked.
Hand.
Machiue.
Corn 50 bu. husked. Stalk left
48.44
223.78
284.00
160.63
247.54
125.00
137.50
115.28
171.05
18.91
78.70
92.63
7.43
86.36
12.50
28.33
80.67
94.30
Seed 1000 Ibs. cotton
Harvesting and baling 8 tons timothy. . . .
Wheat 50 bu
Potatoes 500 bu
Butter 500 Ibs. in tubs
5000 cotton flour sacks
Quarry 100 tons limestone
Mine 50 tons bituminous coal
4. Make a bar diagram comparing the value of farm property for
the two years 1900 and 1910.
Year
1910.
1900.
Land
28,475,674,169
13,058,007,995
Buildings
6,325,451,528
3,556,639,496
Implements and machinery
1,265,149,783
749,775,970
Domestic animals, poultry, and bees
4,925,173,610
3,075,477,703
5. Make a bar diagram of the population of the following states
for the years 1900 and 1910.
Date.
Colorado.
Nevada.
Idaho.
Washington.
Oregon.
California.
1910
1900
799,024
539,700
81,875
42,335
325,594
161,772
1.141,990
518,103
672,765
415,536
2,377,549
1,485,053
HI, §41]
GRAPHIC REPRESENTATION
41. Double Bar Diagrams. In certain diagrams it is ad-
vantageous to have the bars extend in both directions from the
base line as in the following figure which gives the distribution
by age and sex of the total population for 1910.
Males
Females
'12 10 8 0 4 2 0 2 4 6 8 10 12
Hundreds of Thousands
FlG. 7
EXERCISES
Make corresponding figures for the distribution by age periods and
sex for 1910, in per cents, of
1. Native Whites of Native Parentage.
2. Negroes.
Age.
Male.
Female.
Male.
Female.
Under 5
6.7
6.0
5.5
5.2
4.7
4.1
3.5
3.2
2.6
2.2
2.1
1.6
1.3
1.0
0.6
6.5
5.8
5.3
5.1
4.7
4.0
3.4
3.0
2.4
2.0
1.8
1.4
1.2
0.9
0.6
6.4
6.3
5.9
5.2
4.9
4.3
3.4
3.3
2.3
2.0
1.8
1.2
1.0
0.7
0.4
6.5
6.4
5.9
5.6
5.6
4.7
3.4
3.2
2.3
1.9
1.5
1.0
0.9
0.6
0.4
5-9
10-14
15-19
20-24
25-29
30-34
35-39
40-44
45-49
50-54
55-59
60-64
65-69
70-74 . ...
50
MATHEMATICS
[HI, §41
3. Make a diagram displaying the following data on the average
yields and values per acre of Iowa farm crops for 1909.
Crop.
Yield.
Value.
Corn
37.1 bu.
$18.60
Oats
27.5
10.54
Wheat
15.3
14.62
Barley
19.2
9.31
Rve
13.6
8.50
Flaxseed
9.1
11.74
Timothy seed
4.2
5.79
Hay and forage
32.0 cwt.
11.76
Potatoes . .
86.8 bu.
39.10
4. Make a diagram showing the weight in pounds and value of the
dairy products shipped from Humboldt County, California, in 1913.
Article.
Weight in Lbs.
Value.
Butter .
5,793,620
$1,796,190
Cheese
304,570
54,820
Condensed milk
1,302,560
112,720
Dry milk
1,692,100
157,430
Fresh cream and buttermilk
277,800
6,920
Casein ....
1,484,910
89,100
42. Dot Diagrams. The following diagram taken from the
U. S. census reports gives the number of all sheep on farms
April 15, 1910.
JV. Dak. ^
• Q '
S. Dak.
Arcb.
• 0
LEGEND
• 200,000
9 150,000 to 200,000
3 100,000 to 150,000
Q 50,000 to 100,000
O less than 50,000
FIG. 8
HI, §43]
GRAPHIC REPRESENTATION
51
EXERCISES
1. Make a corresponding chart showing all sheep on farms April 15,
1910 for
Wyoming
5,397,000
Utah
1,827,000
Montana
5,380,000
Colorado
1,400,000
Idaho
3,010,000
Nevada
1,150,000
2. Make a dot diagram showing all fowls on farms in the states
given on April 15, 1910. [Here it is convenient to let • stand for
1,000,000.]
North Dakota
3,268,000
Iowa
23,482,000
South Dakota
5,251,000
Minnesota
10,697,000
Nebraska
9,351,000
Montana
966,000
43. Circular Diagrams. The following diagram shows the
relative percentage of improved and unimproved land area in
farms for the total land area of the U. S. 1850-1880-1910.
(U. S. census report 1910.)
1850
1910
FIG. 9
FIG. 10
FIG. 11
The circles indicate by the size of their sectors the relative
ratio of lands improved and unimproved in farms to the
total land area of the U. S. Note the rapid decrease in the
area not in farms, also the increase in the proportion of improved
to the unimproved. In 1910 less than 50% of the total area is
in farms.
52
MATHEMATICS
[HI, §43
EXERCISES
1. Make a circular diagram showing in percents the relative im-
portance of the several countries in the production and consumption of
cotton.
United States 60.9
India 17.1
Egypt 6.6
China . . .5.4
Russia 4.5
Brazil 1.9
All others.. ..3.6
2. Make circular diagrams showing per cent, distribution of foreign
born population by principal countries of birth for the years indicated.
1850.
1870.
1890.
1910.
Germany
26.0
30.4
30.1
18.5
Ireland
42.8
33.3
20.2
10.0
Canada and New Foundland
6.6
8.9
10.6
9.0
Great Britain
16.9
13.8
13.5
9.0
Norway, Sweden and Denmark
0.8
4.3
10.1
9.3
Austria-Hungary
1.3
3.3
12.4
Russia and Finland
0.1
0.1
2.0
12.8
Italy
0.2
0.3
2.0
9.9
All others
6.6
7.6
8.2
9.1
3. Make circular diagrams for the years 1900 and 1910 showing per
cent, of total value of farm property represented by the items men-
tioned.
1910.
1900.
Land
69.5
63.9
Buildings
15.4
17.4
Implements and machinery T .
3.1
3.7
Domestic animals, poultry, and bees
12.0
15.0
(Compare with Ex. 4, p. 48.)
44. Different Shadings or Colors are sometimes used in
maps to represent different statistical facts. The annexed chart
gives the average value of farm land per acre in Delaware.
The average for the state is $33.63.
Ill, §45]
GRAPHIC REPRESENTATION
53
Legend.
H $10 to $25 per acre
'3 $50 to $75 per acre
[^ HJ $75 to $100 per acre
FIG. 12
EXERCISES
1. Draw a map of Connecticut showing the counties and mark to
show the average value of farm land per acre. Average value for state
is $33.03. Average value by counties is: Fairfield $75 to $100 per
acre. New Haven and Hartford $25 to $50 per acre. Litchfield,
Tolland, Windharn, Middlesex, and New London $10 to $25 per acre.
2. Draw a map, give legend, and mark to show per cent, of im-
proved land in farms operated by tenants by states in 1910. Utah,
less than 10 per cent. Wyoming, 10 to 20 per cent. Colorado and
Missouri, 20 to 30 per cent. Kansas, Nebraska, and Iowa, 30 to 40
per cent. Illinois, 40 to 50 per cent.
45 . Distance between two Points. Let PI and P2 be the end
points of a given segment in the plane. PI and P2 are given
points, i.e., their coordinates (Xi, YI) and (^2, ^2) are given
or known numbers.
We wish to find the length
of the segment PiP2 in terms
of xi, i/i, xz, yz ; or, in other
words to find the distance
between two given points.
Through PI draw a line
parallel to the a>axis, and
M,
FIG. 13
through PZ a line parallel to the i/-axis intersecting the first
54 MATHEMATICS [III, §46
in S. Then whatever the relative positions of PI and P% in the
plane, the measure of PiS is x* — x\, and the measure of SP2 is
7/2 - y\ ; also
pj% = !\s* + isp?.
Therefore if we let d represent the required distance PiP2,
(1) d = (x2 - xtf + (yz - 7/02.
It is clearly immaterial which of the two points is called PI
and which P2, so the formula may also be written in the equiva-
lent form
(1) d = V(Xl - x2Y + (y, - ytf,
and may be expressed in words thus : The distance between two
points given by their rectangular coordinates is equal to the square
root of the sum of the square of the difference of the abscissas and
. the square of the difference of the ordinates.
EXAMPLE. The distance from the point (2, — 7) to the point (7, 5) is
d = 52 + 122 = 13.
46. Ratio of Division. Let PI and P2 be two fixed points
on a line and P any third point. Then the point P is said to
divide the segment PiP2 in the ratio
This ratio X is called the ratio of division or the division ratio.
PI P PI
FIG. 14
If we choose an origin 0 on the given line then the abscissas
Xi of PI and x^ of P2 are known. Let us denote the abscissa of
P by x. Then we have
PiP = x - xi, PiP2 = x2 - Xi ;
HI, §47]
GRAPHIC REPRESENTATION
55
hence the abscissa x of P must satisfy the condition
(3)
X —
X2 -
whence solving for x,
(4) x = Xi + \(x2 - Xi).
If the segments PiP and PiP2 have the same sense, the
division ratio is positive and P and P2 lie on the same side of P\.
If the segments PiP and PiP2 are oppositely directed, then the
division ratio is negative and P and P2 are on opposite sides
of PI. Thus, if the abscissas of PI and P* are 2 and 14, the
abscissas of the points that divide PiP2 in the ratios 3, ^, f ,
- 1, - 1, - 2 are 6, 8, 10, - 4, - 10, - 22.
47. Point of Division. To find the coordinates of the point
which divides the line joining two given points in a given ratio \.
Let P\(XI, yi) and P2(o;2, y2) be the two given points, X the
given ratio, and P(x, y) the required point.
Draw PiQi, PQ, P2Q2 parallel to the y-axis, and PiRi, PR,
P2R2 parallel to the z-axis. Then Q and R divide Q\Q2 and
R\Ri, respectively, in the ratio X. Now as OQ\ = Xi, OQ2 = x2,
OQ = x, it follows from (4) § 46 that
(5) x = Xi + \(x2 - Xi).
In the same way we find
(6) y = yi + K(y2 - yi).
Thus, the coordinates x, y of P are expressed in terms of the
coordinates of PI and P2 and the division ratio X.
56 MATHEMATICS [III, § 48
48. Middle Point. If P be the middle point of PiP2, X = |,
and
* = |0i + a*), y = K*/i + 2/s).
That is, the abscissa of the mid-point of a segment is one half
the sum of the abscissas of its end points, and the ordinate is one
half the sum of the ordinates.
EXERCISES
1. Find the lengths of the sides of the following triangles :
(a) (4, 8), (- 4, - 8), (1, 4). (6) (4, 5), (4, - 5), (- 4, 5).
(c) (2, 1), (- 1, 2), (- 3, 0). (d) (- 2, 1), (- 3, - 4), (2, 0).
(e) (2, 3), (1, - 2), (3, 8). (/) (5, 2), (- 3, — 2), (7, 3).
What inference can be drawn from the answers to (e) and (/) ?
2. Find the lengths of the sides and of the diagonals of the quadri-
lateral (2, 1), (5, 4), (4, 7), (1, 4).
3. A (0,2), B (3, 0), and C (4, 8) are the vertices of a triangle.
Show that the distance from A to the mid-point of BC is one-half the
length of BC.
4. Show that two medians of the triangle (1, 2), (5, 5), (— 2, 6) are
equal. What inference can you draw?
5. The ends of one diagonal of a parallelogram are (4, — 2) and
(—4, — 4). One end of the other diagonal is (1, 2). Find the other
end.
6. The end points of a segment PQ are (1,— 3) and (5, 0). Find
the length of the segment, and the lengths of its projections on the
x and y axes.
7. Show that (0, 10), (1, 1), (5, 6) are the vertices of an isosceles
right triangle.
8. Find the coordinates of the point
(a) Two-thirds of the way from (— 1, 7) to (8, 1) ;
(6) Two-thirds of the way from (8, 1) to (- 1, 7) ;
(c) Four-sevenths of the way from (1, — 7) to (8, 0) ;
(d) Three-sevenths of the way from (8, 0) to (1, — 7).
9. The segment from (4, 5) to (2, 3) is produced half its length.
Find the end point.
Ill, §50] GRAPHIC REPRESENTATION 57
49. Locus of a Point in a Fixed Plane. If a point is forced
to move so as to be always equidistant from two fixed points,
we know that it must lie on the perpendicular bisector of the
segment joining these points. If a point must be at a constant
distance from a fixed point, it will lie on a circle. If a point
must be always equidistant from a fixed point and a fixed line,
it will lie on a certain curve, called a parabola, which we have not
yet studied.
If x and y are the coordinates of a point P, the values of x
and y change as P moves in the plane. For this reason they are
called variables. If P is subject to a condition which forces it to
lie on a certain curve, then x and y must satisfy a certain condi-
tion which can be expressed as an equation in x and y.
For example, if P is always equidistant from (1, 2) and (2, 1),
then, for all positions of P, x — y = 0. If P is always equi-
distant from (0, 2) and the x-axis then x2 — 4y + 4 = 0. If P
is always 3 units from the origin, then x2 + yz = 9.
Whenever a plane curve and an equation in x and y are so
related that every point on the curve has coordinates which
satisfy the equation, and conversely, every real solution of the
equation furnishes coordinates of a point on the curve, then the
equation is called the equation of the curve, and the curve is
called the locus of the equation. This dual relation between
equation and curve is the subject of study in Analytic Geometry.
50. Equation of a Locus. To find the equation of the locus
of a point which moves in a plane according to some stated law,
we proceed as follows: First, draw a pair of coordinate axes;
and locate and denote by appropriate numbers or letters all
fixed distances, including the coordinates of fixed points. Second,
mark a point P with coordinates x and y, to represent the mov-
ing point ; express the conditions of the problem in terms of x, y,
and the given constants ; and simplify the resulting equation.
58 MATHEMATICS [III, § 50
Third, show that every real solution of the equation so obtained
gives a point which satisfies the conditions governing the motion
of P.
EXAMPLE. Find the equation of the locus of a point which is always
equidistant from a fixed line and a fixed point.
First. We are free to choose the axes where we please. It is conven-
ient to take the fixed line for the z-axis, and to take the y-axis through
the fixed point. Then the coordinates of the fixed point may be called
(0, a).
Second. The distance from P(x, y) to the fixed line is y, and its dis-
tance to the fixed point (0, a) is ^x2 + (y — a)2. Hence the condition
expressed in the problem gives y = Vz2 -|- (y — a)2. This simplifies to
x2 + 2ay = a2.
Third. It is easy to show, by reversing the above prqcess, that if
x = h, y = k, is any solution of this equation, then the point Q (h, k) is
equidistant from the re-axis and the point (0, a).
Therefore x2 + 2ay = a2 is the required equation.
EXERCISES
1. Find the equation of the locus of a point which moves so that :
(a) itx is equidistant from the coordinate axes ;
(6) it is four times as far from the z-axis as from the y-axis ;
(c) the sum of its distances from the axes is 6 ;
(d) the square of its distance from the rr-axis is four times its distance
from the y-axis.
2. Find the equation of the locus of a point that is always equi-
distant from (4, - 2) and (7, 3). Ans. 3x + 5y = 19.
3. Find the equation of the perpendicular bisector of the segment
joining the two points (a, 6) and (c, d).
Ans. (a - c)x + (b - d)y = \(a? + b2 - c2 - d2}.
4. Find the equation of the locus of a point whose distance from the
point (— 3, 4) is always equal to 5. .4ns. x2 + y2 + 6x — 8y = 0.
5. Find the equation of the circle whose center is (a, b) and whose
radius is c.
Ill, § 51]
GRAPHIC REPRESENTATION
59
51. Locus of an Equation. In general a single equation in
x and y has an infinite number of real solutions. Each of these
solutions furnishes the coordinates of a point on the locus.
To find solutions and plot points on the curve, solve the equa-
tion, if possible, for y in terms of x, or vice versa. Determine and
tabulate a convenient number of solutions by assigning values
to x and computing the corresponding values of y. Using these
for coordinates, plot the points which they represent and draw
a smooth curve through the plotted points.
EXAMPLE 1. Construct the locus of the equation
x2 = 4(x + y).
Solving the given equation for y we have
Assigning to x the values 0, 1, 2, 3, etc., — 1, — 2, — 3, etc., and com-
puting the corresponding values of y, we have the following solutions.
X. . .
0
1
2
3
4
5
6 .
7
- 1
- 2
- 3
y. . •
0
- .75
- 1
- .75
0
1.25
3
5.25
1.25
3
5.25
We choose the axes, as in Fig. 16, so that all these points will go on
the sheet.
FIG. 16
On plotting the points and drawing a smooth curve through them,
we have a sketch of the locus as shown.
60
MATHEMATICS
[III, § 52
EXAMPLE 2. Plot the curve whose equation is
x2 + y2 = Qx + 2y.
Solving the given equation for y, we have
y = 1 ± Vl + 6z - x2,
and we tabulate solutions as follows.
X. ...
0
1
2
3
4
5
6
7
- 1
- 2
y. . . .
0
- 1.45
2
- 2.16
- 2
- 1.45
0
imag.
imag.
imag.
2
3.45
4
4.16
4
3.45
2
7
FIG. 16a
We note that each value of x gives two values of y, i.e. there are two
points on the curve having the same
abscissa. We find also that values of
x ^ 7 do not give real values of y and
that the same is true for values of
x ^ — 1.
When these points have been plotted
and a curve drawn through them we
have the locus as shown in Fig. 16a.
52. Study of the Equation.
Important facts about the shape
and extent of the locus can be
learned by a study of its equation. In the first example above,
the equation is of the first degree in y. From this we infer that
every value of x, without exception, gives exactly one value of y.
Therefore every vertical line cuts the curve in one and only
one point. As x increases beyond 2, y always increases, and
the curve goes off beyond all limit in the first quadrant. The
same is true in the second quadrant. On the other hand, the
equation is of the second degree in x. When solved, it gives
X = 2 ± 2\/l +y;
hence every value of y greater than — 1 gives two real values
of x but every value of y less than — 1 gives an imaginary value
of x. Hence every horizontal line above y = — 1 cuts the curve
in two points, but there are no points on the curve below y = — 1.
Ill, § 53] GRAPHIC REPRESENTATION 61
The equation of the second example, when solved for y as
above, shows that values of x which make 1 + 6x — x2 < 0
give imaginary values for y. Hence there are no points on the
curve to the left of the line x = 3 — V 10 = — 0.16, nor to the
right of the line x = 3 + V 10 = 6.16, but every vertical line
between these limits cuts the curve in two points.
If we solve the same equation for x, we find
x = 3 ± Q +2y -y2-,
hence there are no points below_the line y = 1 — VlO = — 2.16
nor above the line y = 1 + VlO = 4.16, but every horizontal
line between these lines cuts the curve in two points.
If the equation is a polynomial in x and y equated to zero,
a glance will show whether or not it passes through the origin.
The intercepts * can be found by the rule : To find the x-inter-
cepts let y =0 and solve for x. Similarly find the 7/-intercepts.
53. Symmetry. Two points A and B are said to be sym-
metric with respect to a point P when the line AB is bisected by P.
Two points A and B are said to be symmetric with respect to
an axis when the line AB is bisected at right angles by the axis.
If the points of a curve can be arranged in pairs which are
symmetric with respect to an axis or a point, then the curve
itself is said to be symmetric with respect to thai axis or point.
RULE I. // the equation of a locus remains unchanged in form
when in it y is replaced by — y, then the locus is symmetric with
respect to the axis of x.
For, if (x, y} can be replaced by (x, — y} throughout the
equation without affecting the locus, then if (a, 6) is on the
* The intercepts of a curve on the axis of x are the abscissas of the points of inter-
section of the curve and the z-axis. The intercepts on the j/-axis are the ordinates of
the points of intersection of the curve and the j/-axis.
62 MATHEMATICS [III, §53
locus, (a, — 6) is also on the locus, and the points of the locus
occur in pairs symmetric with respect to the axis of x.
We can also prove the following rules.
RULE II. // the equation of a locus remains unchanged in
form when in it x is replaced by — x, then the locus is symmetric
with respect to the y-axis.
RULE III. // the equation of a locus remains unchanged in
form when in it x and y are replaced by — x and — y, then the
locus is symmetric with respect to the origin.
54. Points of Intersection. If two curves whose equations
are given intersect, the coordinates of each point of intersection
must satisfy both equations when substituted in them for
x and y. In algebra it is shown that all values satisfying two
equations in two unknowns may be found by regarding these
equations as simultaneous in the unknowns and solving. Hence
the rule to find the points of intersection of two curves whose
equations are given.
Consider the -equations as simultaneous in the coordinates, and
solve for x and y.
Arrange the real solutions in corresponding pairs. These will
be the coordinates of all of the points of intersection.
EXERCISES
Plot the loci of the f ollowing equations :
1. 2z - 3y - 6 = 0. 12. 4z2 - ?/2 = 0.
2. 4z - Qy - 6 = 0. 13. 6z2 + 5xy - 6y* = 0.
3. 6z - 9y + 36 = 0. 14. x2 + yz = 4.
4. 2x + 3y + 5 = 0. 15. x2 - y2 = 4.
5. 3x - 2y - 12 = 0. 16. x2 + y2 = 25.
6. 5z + 2y - 4 = 0. 17. (x - 8)2 + (y - 4)2 = 25.
7. y = 7x - 3. 18. (x - 4)2 + (y - 2)2 = 5.
8. 2y - x = 2. 19. 4(x + !) = (?/- 2)2.
9. 2x + 9y + 13 = 0. 20. 10y = (x + I)2.
10. (x - 4)(y + 3) = 0. 21. y = x3 - 4z2 - 4x + 16.
11. (x2 - 4)(y - 2) = 0.
Ill, §55] GRAPHIC REPRESENTATION 63
22. y = x, xz, x3, x4, •••, xn. What points are common to these curves?
23. if = x, x2, x3, r». 24. y = (x - 1), (x - I)2, (x - I)3.
25. y = (x - l)(x - 2)(x - 3). 26. y = (x - l)(x - 2)2.
27. y = (x - 2)'. 28. rf = (x - l)(x - 2)(x - 3).
29. y"- = (x - l)(x - 2)2. 30. y2 = (x - 2)3.
31. y -— £-. 32. y =
x - 1 x + 1
33- !f---- 34. y-
x2 + 1 x2 + 1
^ ~ 3>
35. y = ~ - . 36. y =
(x - 2)(x - 4) (x - 2)(x + 4)
Find the points of intersection of the following curves :
I2x + y = 5, [x-y = 2,
' \x + 2y = l. 38' t 2* -3y = l.
+ y' = 18, f x2 + y2 = 18,
40' -3*.
Ans. (3, 3), (- f, - V). Ans. (3, 3), (3, - 3).
f 3x2 + 4y2 = 48, f 3x2 - 4?/2 = 11,
' '
x - y + 1 = 0. ' 1 4x = 3y2.
Ans. (2, 3), (- V, - -V)- ^«s. (3, 2), (3, - 2).
43. IX1J = 2> 44.
1 y2 = 4x. x2 + r/2 - 5.
45 f xy = x + y + 1, 46 f xy = x + ?/ + 1,
(xy =
\4x -
?/ = x - 1. I 4x - 3j/ + 1 = 0.
Ans. (3, 2). Ans. (2, 3), (- J, - J).
47. Find the length of the common chord of the two circles x2 -f- yz
= 4x and x2 + y2 = 4(x + y- 1). Ans. 2^3.
48. In what respects are the loci of the following equations sym-
metric?
(o) y = x2. (e) ?/2 = x2. (i) x3 — y3 — x — y = 0.
(6) y2 =x. (/)r/2 =x*. (T) XT/ = a.
(c) y = x3. (g) y = Xs - x. (fc) ax2 + by2 = 1.
(d) y2 = x3. (A) y = x4 - x2. (I) ax2 + 26xr/ + ct/2 = 1.
55. Straight Line Parallel to an Axis. Suppose a point
moves about on a piece of coordinate paper in such a way that
it is always two units to the right of the axis of y. It would
64 MATHEMATICS [III, §55
evidently be on the line A B that is parallel to the y-axis and
at a distance of two units to the right of
OF. Every point of the line AB has an
abscissa of two (x = 2), and every point
whose abscissa is two lies on the line AB.
For this reason we say that the equation
x = 2
I 2 A
FIG. 17
represents the line AB or is the equation of the line AB.
More generally, the equation
x = a,
where a is any real number, represents a straight line parallel
to the y-axis and at a distance a from it. Similarly, the equation
y = b represents a line parallel to the z-axis.
56. Straight Line through the Origin. Suppose a point
moves about on a piece of coordinate paper in such a way that
its distance from the x-axis, represented by y, is always equal
to m times its distance from the ?/-axis, represented by x. The
equation of the locus of the point is
y = mx.
This is the equation of a straight line through the origin. The
points of this line have the property that the ratio y/x of their
coordinates is the same number m, wherever on this line the
point is taken. Moreover for any point Q, not on this line,
the ratio y/x must evidently be different from m. The number
m is catted the slope of the line.
57. Proportional Quantities. Whenever two quantities y
and x vary in such a manner that their ratio y/x is always
constant, say m, they are said to be proportional. The constant
m is called the factor of proportionality. Many instances occur
HI, §57]
GRAPHIC REPRESENTATION
65
in the applied sciences of two quantities related in this manner.
It is often said that one quantity varies as the other. Thus
Hooke's law states that the elon-
gation E of a stretched wire
or spring varies as the tension t;
that is, E = kt, where A; is a con-
stant. For a given wire, when E
was expressed in thousandths of
an inch and t in pounds, the fol-
lowing relation was found:
E = .8«
Thus when t = 10, E = 8 and when t = 5, E = 4.
5
FIG. 18
10 T
EXERCISES
Draw the lines
1. x = 1, - 1, 0, 2, 3, - 2, - 3, - 4, 4. ^
2. y = 1, - 1, 0, 2, 3, - 2, - 3, - 4, 4.
3. What is the locus of a point if x > 3? x = 3? x < 3?
4. What is the locus of a point if2<x<3? 2 < x < 3? 2<x
< 3? 2 < x < 3?
5. What is the locus of a point if 2 < x < 3 and 1 < y < 2?
6. What is the locus of a point if x2 + y* < 16 and x > 2?
7. What is the locus of a point if 9 < x2 + y2 < 16?
8. A stand-pipe is filled at the rate of 150 gallons per hour. What is
the amount A of water in the stand-pipe h hours after filling begins?
9. A man saves $50 each month and deposits it in a bank. What
is the amount A which he has in the bank after t months?
10. A railroad track has a rise of 1 ft. in 20. Give its equation
and plot.
11. The extension E in feet of a spiral spring due to a tension I of
1 lb., was 1 inch. What is the relation connecting E and <? (Use
Hooke's law.)
6
66
MATHEMATICS
[III, §58
58. Slope of a Straight Line. The slope, m, of the line
passing through two points PI(XI, yi), P2(z2, 2/2), Fig. 19, is
given by the formula
(8)
m =
PiR
59. Equation of a Line through
two Points. Let the two given
points be PI(XI, yi), Pz(x2, y2).
Let P(x, y) be any other point on
the line joining PiP2. Draw PiRS
parallel to the z-axis. Draw P\M\,
P2M2, PM, parallel to the y-axis.
FIG. 19
Then since the triangles PiSP and PiRP2 are similar, we have
SP
PiS
RPj
PiR'
y -
?/2 -
X — Xi X-
which may be written in the form
(9) V* ~
The equation of a straight line with a given slope m and
passing through a given point (x\, y\) is seen from the last
equation to be
(10) y - yi = m(x - jci).
In particular if the y-intercept is given as b, the equation of
the straight line having the given intercept and with slope m is
which reduces to
(11)
y — b = m(x — 0)
y = mx + b.
HI, §61]
GRAPHIC REPRESENTATION
67
This last equation is called the slope form of the equation of
the straight line.
If both intercepts are given, say Z -intercept = a, ^/-intercept
= b, we can find the equation of the line by means of the
equation for a line through two given points. We have
which reduces to
(12)
This is called the intercept form of the equation of the straight
line.
60. Parallel Lines. Con-
sider two parallel lines PiRi
and PzRz- Draw R2Ri and PzPi
parallel to the ?/-axis, and RiSi,
RzSz parallel to the x-axis.
Then since the triangles RiSiPi
and RiSiPz are equal,
and
i = SZPZ.
FIG. 20
Hence,
That is the slopes of any two non-vertical parallel lines are equal.
61. Perpendicular Lines. Consider two perpendicular lines
LI and Z/2 intersecting at P\(x\, t/i). Let PI(XI + a, y\ + &) be
a second point on L\\ then since the given lines are perpen-
dicular, the point Qi(x\ — b, yi + a) lies on L2 as shown by
construction in the figure. Then the slope of L\ is mi = b/a,
by the definition of slope, § 58; and the slope of L2 is m2 =*
68
MATHEMATICS
[HI, §61
— (a/6), for the same reason. It follows that we have
(13) mim-j = — 1.
This proves the theorem:
// two non-vertical lines are
perpendicular, then the prod-
uct of their slopes is — 1.
The converse is also true:
// the product of the slopes
of two lines is — 1, then
they are perpendicular. The
proof, which is suggested
by Fig. 21, is left to the
student.
62. General Equation of the First Degree. The equation
V
FIG. 21
(14)
Ax + By + C = 0,
where A, B, C are constants, is called the general equation of
the first degree in x and y because every equation of the first
degree may be reduced to that form. For any values what-
soever of A, B, and C, provided A and B are not both zero,
the general equation of the first degree represents a straight
line.
EXERCISES
1. Find the slope of the line joining the points
(a) (1, 3) and (2, 7). (6) (2, 7) and (- 4, - 4).
(c) ( A/3, V2) and (- >/2, >/3). (d) (a + b, c + a) and (c + a, 6+c).
2. Prove by means of slopes that (- 4, - 2), (2, 0), (8, 6), (2, 4)
are the vertices of a parallelogram.
3. Prove by means of slopes that (0, - 2), (4, 2), (0, 6), (- 4, 2)
are the vertices of a rectangle.
4. What are the equations of the sides of the figures in Exs. 2
and 3.
Ill, §62] GRAPHIC REPRESENTATION 69
5. Find the intercepts and the slope of each of the following
lines:
(a) 2x + 3y = 6. (6) x - 2y + 5 = 0.
(c) 3z - y + 3 = 0. (d) 5x + 2y - 6 = 0.
(e) 7x - 4y - 28 = 0. (/) 3y - 2x = 8.
6. Find the equations of the lines satisfying the following conditions:
(a) passing through (—3, 1) and slope = 2.
(6) having the ^-intercept = 3, y-intercept = — 2.
(c) slope = — 3, x intercept = 4.
(d) x intercept = — 3, y intercept = — 4.
(e) passing through the point (2, 3) and with slope = — 2.
7. Find the points of intersection of
(a) x - 7y + 25 = 0, z2 + y2 = 25.
(b) 2x2 + 3y2 = 35, 3z2 - 4y = 0.
(c) x2 + y = 7, y* - x = 7.
(d) y = x + 5, 9z2 + 16y2 = 144.
8. Find the equations, and reduce them to the general form, of the
lines for which
(a) m = 2, b = - 3. (6) m = - 1/2, b = 3/2.
(c) m = 2/5, b = - 5/2. (d) m = 1, b = - 2.
(e) a = 3, 6 = 3. (/) a = 4, 6 = 2.
(0) a = - 3, 6 = - 3. (h) a = 4, & = - 2.
(t) a = - 3, b = 3. 0') a = 2, 6=4.
9. Write the equations of the lines passing through the points:
(a) (- 2, 3), (- 3, - 1). (b) (5, 2), (- 2, 4).
(c) (1, 4), (0, 0). (rf) (2, 0), (- 3, 0).
(e) (0, 2), (3, - 1). (/) (2, 3), (- 6, - 5).
10. Write the equations of the lines passing through the given
points and with the given slopes:
(a) (-2,3), m = 2. (6) (5,2), m = 1.
(c) (1, 4), m = i (d) (2, 0), m = - f.
(e) (0, 2), w = 0. (/) (3, - 2), m = - 2.
11. Write the equation of the line which shall pass through the
intersection of 2y + 2x + 2 = 0 and 3y — x — 8 = 0, and having a
slope = 4. Am. IQx — 4y + 51 =0.
70 MATHEMATICS [III, §62
12. What are the equations of the diagonals of the quadrilateral
the equations of whose sides are y — x + 1 = 0, y = — x + 2, y = 3x
+ 2, and y + 2x + 2 =0?
13. Required the equation of the line which passes through (2, — 1)
and is
(a) parallel to 2y - 3x - 5 = 0. Am. 2y - 3x + 8 = 0.
(6) perpendicular to 2y — 3x — 5 = 0. Ans. 2x + 3y — 1 =0.
14. Find the equations of the two straight lines passing through
the point (2, 3), the one parallel, the other perpendicular to the line
4x - 3y = 6. Ans. 4x - 3y + 1 = 0, 3x + 4y - 18 = 0.
15. Passing through (4, — 2), the one parallel, the other per-
pendicular to the line y = 2x + 4. Ans. y = 2x — 10, x + 2y = 0.
16. Passing through the point of intersection of 4x + y + 5 = 0
and 2x — 3y + 13 = 0, one parallel, the other perpendicular to the
line through the two points (3, 1) and (— 1, — 2).
Ans. 3x - 4y + 18 = 0, 4z + 3y - 1 =0.
17. Find the equation of the line joining the origin to the point of
intersection of 2z + 5r/ — 4 = 0 and 3x — 2y + 2 = 0.
Ans. y = — 8x.
18. Find the equation of the straight line passing through the
point of intersection of 2x + 5y — 4 = 0 and 2x — y + 1 =0 and
perpendicular to the line 5x — Wy = 17. Ans. 6x + 3y = 2.
19. Find the equations of the lines satisfying the following condi-
tions :
(a) through (2, 3), parallel to y = 7x + 3.
(6) through (4, — 1), perpendicular to 2x + 3y = 6.
(c) through (—2, - 1), parallel to 3y — 2x = 1.
(d) through (3, - 6), parallel to 2y + 4x = 7.
(e) through (— 1, — 1), perpendicular to x/2 + y/3 = 1.
(/) through (2, 2), perpendicular to y = — 3x -f 2.
20. Prove that the diagonals of a parallelogram bisect each other.
21. Prove that the diagonals of a rhombus bisect each other at right
angles.
22. Prove that the diagonals of a square are equal and bisect each
other at right angles.
23. A straight line makes an angle of 45° with the x-axis and its y
intercept = 2; what is its equation? Ans. y = x + 2.
Ill, §62]
GRAPHIC REPRESENTATION
71
24. The following data gives the height of a plant in inches on
different days.
Height
Day
0
0
28
40
33
60
36
80
40
100
52
120
62
140
66
160
Find the rate of growth after 60 days.
Find the rate of growth after 110 days.
[The rate of growth is the slope of the curve. The slope of a curve
at a given point is defined to be the slope of the tangent line drawn to
the curve at the given point. Draw the tangent with a ruler and
with the aid of the eye.] Ans. 7/10 in. per day; 0.55 in. per day.
CHAPTER IV
LOGARITHMS
63. Definitions and Preliminary Notions. In the equa-
tion
102 = 100,
three numbers are involved. By omitting each number in turn
there arise three different problems. If we omit the 100, we
have the familiar question in involution:
102 = ?.
If we omit the 10 we have the familiar question in evolution:
?2 = 100,
or, as it is usually written,
VlOO = ?.
If we omit the 2 we have the following question
10? = 100,
which we agree to write in the form,
\
logic 100 = ?
and we say that 2 = the logarithm of 100 to the base 10.
In general, if
(1) IP = N,
then x = the logarithm of N to the base b, and we write,
(2) x = logb N.
(1) and (2) are then simply two different ways of expressing
the same relation between b, x, and N. (1) is called the ex-
72
IV, §63] LOGARITHMS 73
ponential form. (2) is called the logarithmic form. Either of
the statements (1) or (2), implies the other. The exponent
in (1) is the logarithm in (2), a fact which may be emphasized
by writing
(3) (base)10*arithm = number.
For example, the following relations in exponential form:
32 = 9, 24 = 16, (1/2)3 = 1/8, a" = x,
are written respectively in the logarithmic form:
2 = logs 9, 4 = Iog2 16, 3 = logi/z 1/8, y = logo x.
We shall now give the following
DEFINITION OF A LOGARITHM. The power to which a given
number called the base must be raised to equal a second number is
called the logarithm of the second number.
EXERCISES
1. Write the following equations in logarithmic form:
(a) 9 = 32. (g) 7 = 71. _
(6) 64 = 43. (h) 25 = (Vs)4.
(c) 16 = 24. (i) 8 = (V2)«.
(d) 243 = 35. 0') 3 = (V3)2.
(c) 64 = 2". (fc) 3 - V9.
(/) 2401 = 74. (I) 4 = v/64.
2. Write the following equations in exponential form :
(a) log, 16 = 4. (g) logioO.l - - 1.
(b) Iog4 16 = 2. (A) logz 1/4 : 2.
(c) logio 1000 = 3. (t) Iog64 2 = 1/6.
(d) logs 729 = 5. (j) Iog2 1/8 = - 3.
(c) logs 625 = 4. (fc) logn 1 = 0.
(/) log» 1728 = 3. (0 loga o = l.
3. Find the numerical value of each of th following :
(a) Iog2 64. (e) Iog26 5.
(6) logio 0.001. (/) 3 Iog6 625 + logj 16.
(c) Iog27 3. (g) logi/2 4.
(d) logio 100 - | logo.i 100. (h) 5 logz 16-2 log« 625.
74 MATHEMATICS [IV, §64
64. Properties of Logarithms. Any positive number, ex-
cept 1, may be the base of a system of logarithms of all the real
positive numbers. In any such system,
1) The logarithm of 1 is zero.
For, 6° = 1, therefore logb 1=0.
2) The logarithm of the base itself is 1.
For, 61 = 6, therefore logb 6 = 1.
3) The logarithm of a product is the sum of the logarithms of
the factors.
For if logb M = k and logb -Y = I, then M — bk and N = bl,
MN = bk-bl = bk+l, whence
logb MN = k + I = logb M + logb N.
This can readily be extended to three or more factors.
4) The logarithm of a quotient is equal to the logarithm of the
dividend minus the logarithm of the divisor.
For,
' M =y
N ~ bl ''
therefore
logb jf = k - I = logb M - logb N.
5) The logarithm of the reciprocal of a number is the negative
of the logarithm of the number.
For on putting M — 1 under (4) above, we have
logb-r^ = logb 1 - logb N = - logb N,
since logb 1=0.
6) The logarithm of the pth power of a number is found by
multiplying the logarithm of the number by p.
For, N = bk and Np = (bk)p = bpk, whence
logb Np = pk = p logb N.
IV; §64] LOGARITHMS 75
7) The logarithm of the rth root of a number is found by dividing
the logarithm of the number by r.
For, N = bk and A/A7 = Nllr = (6*)1/r = bklr, whence
logs VAT = - =
N
r
EXERCISES
Express the logarithms of the following numbers in terms of the
logarithms of integers. In this book, when the base is omitted, 10 is
to be understood as the base.
352/3 17i/4 12-2
!• ]°g iQ2/3.Ai/2 • 2. log ,,-7,— . 3. log
132/3 . 6i/2 •
4. Prove that logs V81V729-9-2'3 = 31/18.
Express the logarithms of the following in terms of the logarithms
of prime numbers.
(63)1/4 88~1/2
(25)2(72)1/4 ' (75)3/4(12)2 *
7. log ^- — 3. 8. log (V2!V72V6).
9. Given log 2 = 0.3010, log 3 = 0.4771, log 7 = 0.8451, find the
logarithms of the following numbers.
(a) 6. (e) 32. (i) 420. (m) Vf/2.
(6) 14. (/)10.5. 0') 900. (n) A/504.
(c) 24. (g) 14?. (k) 35/48. (o) BVl3.5.
(d) 28. (h) 2.52. (I) 1/36. (p) >/294.
10. Express the logarithms of each of the following expressions to
the base a in terms of logo b, loga c, Iog0 d.
11. Prove that
= 2 loga (X + V-l).
12. If log 3 = 0.4771, what is the (a) log 30? (b) log 300? (c) log
3000? (d) log 30,000? What part of these logarithms is the same?
Why?
76 MATHEMATICS [IV, §65
65. Computation of Common Logarithms. While any
positive number except unity could be used as the base of a
system of logarithms, only two systems are in general use.
One, called the natural, or Napierian system is used in analytical
work and has the number e = 2.71828 + for its base. The
other, known as the common, or Briggs system is used for all
purposes involving merely numerical computations and has for
its base the number 10. Unless specifically stated to the
contrary the common system will be the one used throughout
this book.
In the following discussion of common logarithms, log x is
written as an abbreviation of logio x.
Every positive number has a common logarithm, and the
value of this logarithm may be obtained correct to as many
places of decimals as may be desired. Negative numbers and
zero have no real logarithms.
If we extract the square root of 10, the square root of the
result thus obtained, and so on, continuing the reckoning in
each case to the fifth decimal figure, ,we obtain the following
table :
1Q1/2
l0l/4
1Q1/8
1Q1/16
101/32
1Q1/64
= 3.16228,
= 1.77828,
= 1.33352,
= 1.15478,
= 1.07461,
= 1.03663,
1Q1/128 _
1Q1/256 _
101/512 =
1Q1/1024 _
1Q1/2048 _
1Q1/4096 =
1.01815,
1.00904,
1.00451,
1.00225,
1.00112,
1.00056,
and so on. The exponents 5, |, • • • on the left are the logarithms
of the corresponding numbers on the right.
By the aid of this table we may compute the common logar-
ithm of any number between 1 and 10, and hence of any positive
number.
IV, §66] LOGARITHMS 77
EXAMPLE. Find the common logarithm of 4.26.
Divide 4.26 by the next smaller number in the table, 3.16228. The
quotient is 1.34719. Hence 4.26 = 3.16228 X 1.34719. Divide
1.34719 by the next smaller number in the table, 1.33352. The quo-
tient is 1 0102. Hence 4.26 = 3.16228 X 1.33352 X 1.0102. Con-
tinue thus, always dividing the quotient last obtained by the next
smaller number in the table. We shall obtain by this method an ex-
pression for 4.26 in the form of a product:
4.26 = 3.16228 X 1.33352 X 1.00904 X • • •
= 101/2 X 101/8 X 101/256 X • • •
Therefore,
= .5000
+ .1250
+ .0039
= .6289
By referring to the table of logarithms at the end of the book we find
that, correct to four decimal places,
log 4.26 = .6294
Hence, by using only three terms in the above approximation we ob-
tain a result which is in error but 5 units in the fourth decimal place.
66. Characteristic and Mantissa. If two numbers are un-
equal, their logarithms are unequal in the same sense; that is if
a < b < c,
then
log a < log b < log c.
For example
log 100 < log 426 < log 1000,
that is,
2 < log 426 < 3.
78 MATHEMATICS [IV, §66
When the logarithm of a number is not an integer it may
be represented approximately by a decimal fraction correct to
any desired number of places; thus log 426 = 2.6294 to four
decimal places.
The integral part of the logarithm is called the characteristic
and the decimal part is called the mantissa. In log 426, the
characteristic is 2 and the mantissa is .6294. For convenience
in computing it is desirable to have the mantissa positive even
when the logarithm is a negative number. For example,
log \ = - 0.3010, but - 0.3010 = 9.6990 - 10, and we write
log \ = 9.6990 - 10,
in which the characteristic is 9 — 10 = — 1, but the mantissa
.6990 is positive.
It is convenient to write the logarithm of any number N in
the form
log # = M - A;- 10,
in which M is a positive number or zero and A; is a positive
integer or zero.
For example, log 426 = 2.6294, log 42.6 = 1.6294, log 4.26
= 0.6294, log 0.426 = 9.6294 - 10, log 0.0426 = 8.6294 - 10,
log 0.00426 = 7.6294 - 10.
Moving the decimal point n places to the right (left) in a number
increases (decreases) the characteristic of its common logarithm
by n, but does not affect its mantissa.
For this has the effect of multiplying (dividing) the number
by 10", and
log (N -10") = log N + log 10" = log N + n
and
log (N •*• 10") = log N - n.
Therefore, the mantissa of the common .logarithm of a number
is independent of the position of the decimal point. In other
IV, §66] LOGARITHMS 79
words, the common logarithms of two numbers which contain
the same sequence of figures differ only in their characteristics.
Hence, tables of logarithms of numbers contain only the man-
tissas and the computer must determine the characteristics
mentally. This can be done by the following simple rules.
RULE I. The characteristic of the common logarithm of any
number greater than 1, is one less than the number of digits before
the decimal point.
For if N is a number having n digits in the integral part (i. e.
before the decimal point), then
10n-i < N < 10n
and
n — 1 ^ log N < n;
therefore log N = (n — 1) -f (a decimal fraction) and its
characteristic is n — 1.
On the other hand if N is a decimal fraction (i. e., a positive
number less than 1), we may move the decimal point 10 places
to the right and apply Rule I., provided we subtract 10 from
the resulting logarithm. For example,
log 0.0006958 = log 6958000 - 10
and by Rule I. the characteristic is 6 — 10.
This process is easily seen to be equivalent to that specified in
RULE II. To find the characteristic of the common logarithm
of a number less than 1, subtract from 9 the number of ciphers
between the decimal point and the first significant figure. From
the number so obtained subtract 10.
. A very large number such as the distance in feet from the
earth to the sun, 490,000,000,000 (correct to two significant
figures), is conveniently written (on moving the decimal point
11 places to the left) in the form
4.9 X 1011
80 MATHEMATICS [IV, §66
and the characteristic of its common logarithm is 11. Similarly
a very small number such as 0.000,000,453,8 can be written
(on moving the decimal point 7 places to the right),
4.538 X 10-7
and the characteristic of its logarithm is — 7 = 3 — 10.
This form of expression is frequently used where only a few
significant figures are known to be correct, and if the decimal
point is placed after the first significant figure, the exponent of
10 is the characteristic of the logarithm of the number.
EXERCISES
Find the characteristics of the logarithms of the following numbers:
(1) 276.35 (5) 0.00072 (9) 73.187
(2) 0.0495 (6) 4589.5 (10) 8.421 X lO"26.
(3) 1.837 (7) 0.9372 (11) 7.268 X 101*.
(4) 6.3 X 10s. (8) 7.32 X 10~5. (12) 0.00008
67. Use of Tables. 1) The characteristic is not given in the
table of logarithms. It is to be found by the above two rules.
It should be written down first, and always expressed even
though it be zero, in order to avoid error due to forgetting it.
2) The mantissa of the common logarithms of numbers,
correct to four decimal places, are printed in Table I., at the
end of the book. For convenience in printing the decimal points
are omitted.
To find the mantissa of a number consisting of one, two, or
three digits (exclusive of ciphers at the beginning or end, and
the decimal point), look in the column marked N for the first
two digits and select the column headed by the third digit;
the mantissa will be found at the intersection of this row and
this column. For example, to find the mantissa of 456, we run
down the column headed N to 45 and then run across the page
IV, §68] LOGARITHMS 81
to the column headed 6 where we find the mantissa .6590; again,
the mantissa of 720 is found opposite 72 in the column headed 0,
and is 8573.
EXERCISES
Look up the following logarithms in Table I.
(1) log 276 = 2.4409 (11) log .00782
(2) log 8.64 = 0.9365 (12) log .0856
(3) log .829 = 9.9186 - 10. (13) log 20.
(4) log 7.34 X 105 = 5.8657 (14) log 8.5
(5) log 2.30 X 10-3 = 7.3617 - 10. (15) log 1870.
(6) log 24700 = 4.3927 (16) log 3.20 X 10~12.
(7) log 3.7 X 1012. .(17) log 5.47 X 1023.
(8) log 9. (18) log 7.58 X 10*.
(9) log 846000. (19) log 98.3
(10) log .000172 (20) log 3140000.
68. Interpolation. If there are more than three significant
figures in the given number, its mantissa is not printed in the
table; but it can be found approximately by the principle of
proportional parts: when a number is changed by an amount
which is very small in comparison with the number itself, the change
in the logarithm of the number is nearly proportional to the change
in the number itself.
For example, to find the logarithm of 37.68, we find from the
table,
Mantissa of 3760 = 5752,
Mantissa of 3770 = 5763.
The difference between these mantissas, called the tabular
difference, is 11. We note that an increase of 10 in 3760 pro-
duces an increase of 11 in its mantissa and we conclude that an
increase of 8 in 3760 (to bring it up to 3768, the given digits)
would produce an increase of .8 X 11 = 8.8 in the mantissa.
This number 8.8, called the correction, is to be added to the
7
82 MATHEMATICS [IV, §68
mantissa of 3760, but in using a four place table we retain only
four places in corrected mantissas, so here we add 9 (the integer
nearest to 8.8) ; thus,
log 37.60 = 1.5752
correction = 9
log 37.68 = 1.5761
Near the beginning of Table I. the tabular differences are so
large as to make this process of interpolation inconvenient and
in some instances unreliable. On this account there are printed
on the third and fourth pages of Table I., the mantissas of all
four figure numbers whose first digit is 1. By using these we
can avoid interpolation at the beginning of the table. Thus,
on the third page of the table we find,
log 103.2 = 2.0137,
but if we find it by interpolation on the first page,
log 103.2 = 2.0136
EXAMPLE 1. Find the logarithm of .003467. Opposite 34 in
column 6 find 5391; the tabular difference is 12; .7 X 12 = 8.4; the
mantissa is then 5391 + 8 = 5399; hence log .003467 = 7.5399 - 10
EXAMPLE 2. Find log 2.6582. Opposite 26 in column 5 find 4232;
the tabular difference is 17; .82 X 17 = 13.9; the mantissa is 4232
+ 14 = 4246; hence log 2.6582 = 0.4246.
69. Accuracy of Results. The accuracy of results obtained
by means of logarithms depends upon the number of decimal
places given in the tables that are used, and this accuracy has
reference to the significant figures counted from the left. In
general, a table will give trustworthy results to as many sig-
nificant figures, counted from the left, as there are decimal
places given in the logarithms. For example, four-place
logarithms would show no difference between 35492367 and
35490000.
IV, §70] LOGARITHMS 83
Neither a four-place nor a five-place table would be of any
use in financial computations where large sums are involved.
It would take a nine-place table to yield exact results if the
sums involved should reach a million dollars.
70. Reverse Reading of the Table. To find the number
when its logarithm is known. This is sometimes called finding
the antilogarithm. For this process we have the following rule.
RULE III. // the mantissa is found exactly in the table, the
first two figures of the corresponding number are found in the
column N of the same row, while the third figure of the number is
found at the top of the column in which the mantissa is found.
Place the decimal point so that the rules in § 66 are fulfilled.
EXAMPLE. Given log N = 1.7427; to find N.
We find the mantissa 7427 in the row which has 55 in column N.
The column in which 7427 is found has 3 at the top. Thus the sig-
nificant figures in the number are 553. Since the characteristic is 1 we
must have 2 figures to the left of the decimal point. Thus N = 55.3.
If the mantissa of the given logarithm is between two man-
tissas in the table, we may find the number whose logarithm
is given by the following
RULE IV. When the given mantissa is not found in the table,
write down three digits of the number corresponding to the mantissa
in the table next less than the given mantissa, determine a fourth
digit by dividing the actual difference by the tabular difference,
and locate the decimal point so that the rules for characteristics are
fulfilled.
EXAMPLE. Given log N = 0.4675; to find N.
The mantissa 4675 is not recorded in the table, but it lies between
the two adjacent mantissas 4669 and 4683. The mantissa 4669 corre-
sponds to the number 293. The tabular difference is 14. The actual
difference between 4669 and 4675 is 6. The number 4675 is 6/14 of
the interval from 4669 to 4683, and the corresponding number N is
84 MATHEMATICS [IV, §70
about 6/14 of the way from 293 to 294, or, reducing 6/14 to a decimal,
about .4 of a unit beyond 293. Hence the corresponding digits are
2934; hence TV = 2.934.
The work may be written down as follows:
log N = 0.4675
4669
14)60(4
N = 2.934
EXERCISES
Obtain the logarithm of each of the following numbers.
1. 3.1416 2. 1.732 3. 2.718
4. 1.414 5. 39.37 6. 0.4343
7. 3437 8. 0.0254 9. 0.9144
10. 0.003954 11. 0.016018 12. 0.0283
13. 7918. 14. 866500. 15. 92897000.
Find the antilogarithm of each of the following numbers.
16.
0.4563
17.
96378
- 10.
18.
5.3144
19.
1.7581
20.
8.2046
- 10.
21.
6.1126
22.
0.4971
23.
7.5971
- 10.
24.
4.9365
25.
4.6856 -
10.
26.
8.1530
- 10.
27.
8.6123
- 20.
28.
8.4048 -
10.
29.
8.4520
- 10.
30.
0.7318
- 20.
71. Cologarithms. The cologarithm of a number is the
logarithm of the reciprocal of the number. (Compare (5) § 64.)
Thus colog 425 = log — - = log 1 - log 425
4^o
= 0 - 2.6284
But since we always wish to have the mantissa of a logarithm
positive, we write 0 = 10 — 10, and subtract 2.6284 from this,
as follows:
log 1 = 10.0000 - 10
log 425 = 2.6284
colog 425 = 7.3716 - 10.
IV, §72] LOGARITHMS 85
In practice this is done mentally by beginning at the left not
omitting the characteristic, and subtracting each digit from 9,
except the last significant digit, which is subtracted from 10.
In the process of division subtracting the logarithm of a
number and adding its cologarithm are equivalent operations
since dividing by N is equivalent to multiplying by its reciprocal.
72. Computation by Logarithms. It should be kept in
mind that a logarithm is unchanged if at the same time any
given number is added to and subtracted from it. This is useful
in two cases:
First. When we wish to subtract a larger logarithm from a
smaller;
Second. When we wish to divide a logarithm by an integer.
EXAMPLE 1. Find the value of 27.4 -f- 652.
log 27.4= 1.4378
= 11.4378 - 10
log 652 = 2.8142
log x = 8.6236 - 10
x = 0.04304
EXAMPLE 2. Find the value of (0.0773)1/3.
log 0.0773 = 8.8882 - 10.
It is convenient to have, after division by 3, — 10 after the mantissa;
hence, before dividing we add 20.0000 - 20.
log 0.0773 = 28.8882 - 30 (divide by 3),
log * = 9.6294 - 10
x = 0.4250
EXAMPLE 3. Find the value of ' (42>6)
[" (42.6) (- 3.14) I'
02.4
We have no logarithms of negative numbers, but an inspection of
this problem shows that the result will be negative and numerically
86 MATHEMATICS [IV, §72
the same as though all the factors were positive; hence we proceed as
follows:
log 42.6 = 1.6294
log 3.14 = 0.4969
colog 62.4 = 8.2048 - 10 (add)
3)0.3311 (divide by 3)
log (- x) = 0.1104
- x = 1.290, whence x = - 1.290.
EXERCISES
Find approximate values of the following by aid of logarithms.
1. 231.6 X .0036. 2. 79 X 470 X 0.982.
3. 13750 X 8799000. 4. (- 9503) X (- 0.008657).
5. 0.0356 X (- 0.00049). 6. 9.238 X 0.9152
8075 . «
0.00542
n
24617
364.9'
0.04708 '
' -0.00054'
10.
67 X 9 X 0.462
11.
9097 X 5.408
. 12. (2.38S)5.
0.643 X 7095
- 225 X 593
X 0.8665
13.
(0.57)~4.
14.
(19/11)8.
15.
(1.014)25.
16.
A/67.54.
17.
A/- 0.3089.
18.
GV( - 9.718)3.
19.
85/4.
20.
(0.001 )2'3.
21.
(29^9r)3/2.
22.
(6f)3-4.
23.
(- 9306)3/7.
24.
(0.0067)2-5.
25.
vixVif.
26.
Vol
27.
(0. 00068) ~6/4.
(0.009)3/5 '
00
/ 854 X A/0.042
! on 3 |7"4 X 92"« X (0.01 )i/»
A/OOOl ' * (0.00026) s X 59681/3
30. V6A/0.5804 A/0.2405. 31. (6.89 X lO"22)16/17.
Ans. 1.21 X 10-20.
32. (5.67 X 10-18)9/11. 33. 8.4
M[(4.5 X lO-^lO6-58]1
Ans. 7.76 X lO"6. Ans. 5.51 X 107.
34. The amount a of a principal p at compound interest of rate r
for n years is given by the formula: a = p(l + r)n. Find the amount
of $486 in 5 years at five per cent, (r = .05) if the interest is com-
pounded annually. Ans. $620.27
IV, §72] LOGARITHMS 87
35. Find the amount of $384 in 40 years at four per cent., interest
compounded annually. , Ans. $1,843.59.
36. Find the simple interest on $6,237.43 for 7 years at six per cent.
Would the computation made with four-place logarithms, be sufficiently
accurate for commercial purposes? Explain. Ans. $2619.72.
37. The weight P in pounds which will crush a solid cylindrical cast-
iron column is given by the formula
,73.56
P = 98920^,,
where d is the diameter in inches and I the length in feet. What weight
will crush a cast-iron column 6 feet long and 4.3 inches in diameter?
[RiBTZ AND CRATHORNE] Ans. 834,200 Ibs.
The area A in acres, of a triangular piece of ground, whose sides are
a, b, c, rods, is given by the formula
_ Vs(s — a)(s — b)(s — c)
160
where s = %(a + 6 + c). Compute the areas, in acres, of the follow-
ing triangles:
38. a = 127.6,
39. a = 0.9,
40. a = 408,
41. a = 63.89,
b = 183.7,
b = 1.2,
b = 41,
6 = 138.24,
c = 201.3.
c = 1.5.
c = 401.
c = 121.15.
42. The percentage earning power, E, of an individual, in so far as
it depends upon the eyes is given by Magnus by the formula
E = c
where x takes one of the values 5, 7, or 10, C being the maximal central
visual acuity, VPi the visual field, A/Af the action of the extrinsic
muscles, Cj and C2 the central visual acuity of each eye, and A/P2 the
peripheric vision. Compute the value of # if C = 1, PI = 1, M = 1,
Ci = 1, Cs = 0.58, x = 10, P2 = 1. Ans. 97.2%
43. Compute E if d = 0.41, C2 = 0.25, x = 5, P2 = M = Pi = 1,
C = 0.41. Ans. 33.06%.
MATHEMATICS
[IV, §72
44. The percentage earning ability E, as dependent upon the eyes
is given by Magnus as
E =
where F — functional ability, V = necessary knowledge, K — the
ability to compete (demand for him), x has one of the values 5, 7, or 10.
Compute E for F = 0.78792, V = 1, x = 10, K = 0.39396.
Am. 71.78%.
45. Compute E f or F = 0.8254, x = 10, K = 0.4127, 7 = 1.
Ana. 75.52%.
46. When w grams of a substance is dissolved in v liters of water at t°
centigrade, the osmotic pressure, p, of the solution and the molecular
weight, M, of the solute are connected by the equation
pv = 0.082 (273 + t)w/M.
Compute the molecular weight of cane sugar from the data
(a) p = 12.06, v = 1, t = 22.62, w = 171.0 Ans. 343.7
(6) p = 24.42, » = 3,J = 23.56, w = 102.6 Ana. 340.5
Compute the osmotic pressure for glucose solution, given
(c) v = 1, t = 26.90, w = 72, M = 180.21 4ns. 9.824
(d) v = 2, t = 22,20, to = 360, M = 178.46 Ans. 24.36
73. The Slide Rule. The slide-rule is an instrument for
carrying out mechanically the operations of multiplication and
division. It is composed of two pieces, usually about the shape
of an ordinary ruler; one of the pieces (called the slide,) fits in
a groove in the other piece. Each piece is marked in divisions
FIG. 22
(Fig. 22), such that the distance from one end (e. g., A) is equal
to the logarithm of the number marked on it.
To multiply one number (e. g., 2.5) by another (e. g., 2) we
II, §73]
LOGARITHMS
89
set the point marked 1 on scale B opposite the point marked 2.5
on scale A (see Fig. 23). Then the product appears on scale A
FIG. 23
opposite the point 2 on scale B. Thus 5 on scale A lies oppo-
site 2 on scale B in Fig. 23. This follows from the fact that
log 2.5 + log 2 = log 5.
Likewise, if 1 on scale B is set opposite any number a on
scale A, then we find opposite any number 6 on scale B the
number ab on scale A.
Divisions can be performed by reversing this process. Thus
if 6 on scale B be set opposite c on scale A, the 1 on scale B
will be opposite c/b on scale A.
A little practice with such a slide-rule will make clear the
actual procedure in any case.
Scales C and D are made just twice the size of scales A and B.
It follows that any number on scale C, for example, is exactly
opposite the square of that number on scale A. This facilitates
the finding of squares and square roots, approximately.
Scales C and D may be used in place of scales A and B for
multiplication and division. Indeed, after some practice,
scales C and D will be preferred for this purpose.
More elaborate slide-rules, marked with several other scales,
are for sale by all supply stores. Descriptions of these and
full directions for their use will be found in special catalogs
issued bv instrument makers.
90 MATHEMATICS [IV, §73
A simple slide-rule can be bought at a moderate price. One
sufficient for temporary practice may be made by the student
by cutting out the large figure printed on one of the fly-leaves
of this book, and following the directions printed there.
The student should secure some form of slide-rule and he
should use it principally in checking answers found by other
processes.
As exercises the teacher may assign first very simple products
and quotients. When the operation of the slide-rule has been
mastered, the student may check the answers to the exercises
on p. 86.
CHAPTER V
TRIGONOMETRY
74. Introduction. The sides and angles of a plane triangle
are so related that any three given parts, provided at least one
of them is a side, determine the shape and the size of the triangle.
Geometry shows how, from three such parts, to construct the
triangle.
Trigonometry shows how to compute the unknown parts of a
triangle from the numerical values of the given parts.
Geometry shows in a general way that the sides and angles
of a triangle are mutually dependent. Trigonometry begins
by showing the exact nature of this dependence in the right
triangle, and for this purpose employs the ratios of the sides.
75. Definitions of Trigonometric Functions. Let A be any
acute angle. Place it on a pair of axes as in Fig. 24, with the
vertex at the origin, one side along
the ar-axis to the right, and the
other side in the first quadrant.
On this side choose any point M (ex-
cept 0) and drop M N perpendic-
ular to the or-axis. Let OM = r;
then by plane geometry,
x-
FIG. 24
r = v ar*
where x and y are the coordinates of the point M. The differ-
ent ratios of pairs of the three numbers x, y, and r, are designated
as follows
y _ ordinate _
r radius
x abscissa
(1)
(2)
(3)
r radius
y _ ordinate _
x abscissa
of angle A, written sin A,
= the cosine of angle A, written cos A,
tangent of angle
91
> written tan A.
92
MATHEMATICS
[V, § 75
The reciprocals of these rations are also used,
,., x abscissa
(4) ~ = :
y ordinate
= the cotangent of angle A, written ctn A,
(5) - = — - = secant of angle A, written sec A,
x abscissa
(6) - = — - = cosecant of angle A, written esc A.
y ordinate
These six ratios are called the trigonometric functions of the
angle A. They do not at all depend upon the choice of the
point M on the side of the angle but only upon the magnitude
of the angle itself.
For if we choose any two points M' and M " on the side of the
.
_
'
t
*.!
/
-/'
/
„
/
,
/
M
/
s
/
/
>
•\j
i
y
M
/
y
'
^"
•^
/
/
^
x
<"
>•
y
j
/
K
B
.
^
**
V
^
/
3
„
/
\
^
s
\
o
,/
X
A
A
0
^
^
~x
V
A'
\
X
'
X
FIG. 25
same angle A, and denote their coordinates by (xf, y'} and (x",
y") respectively, then by similar triangles,
~. = ^-- = sin A,
-- = %— = tan A, etc.
But if we take two points M' and M" at the same distance
r from 0 on the sides of two different angles A and B, then
y' v"
sin A = '— ^ '-*— = sin B,
r r
tan A =^-^^- =tanB,
x x
and similarly the other functions of A and B are unequal.
V, § 76]
TRIGONOMETRY
93
From these definitions we deduce the
following relations which are of fundamen-
tal importance in computing the unknown
parts of right triangles.
In any right triangle, having fixed atten-
tion on one of the acute angles,
side adjacent
FIG. 26
(7)
(8)
(9)
The side opposite = hypotenuse X sine.
also = side adjacent X tangent.
The side adjacent = hypotenuse X cosine.
also = side opposite X cotangent.
side opposite
The hypotenuse =
also =
sine
side adjacent
cosine
EXERCISES
Find the six functions of each of the acute angles in the right tri-
angle whose sides are :
1. 3, 4, 5. 2. 9, 40, 41. 3. 60, 91, 109.
4. 7, 24, 25. 5. 16, 63, 65. 6. 20, 99, 101.
7. 20, 21, 29. 8. 36, 77, 85. 9. 12, 35, 37.
10. 2n + 1, 2n(n + 1), 2n2 + 2n + 1. 11. 2n, n2 - 1, n2 + 1.
12. 2(n + 1), n(n + 2), n2 + 2n + 2. 13. a(62 - c2), 2abc, a(62 + c2).
76. Functions of Complementary Angles. Let A and B be
the acute angles in any right triangle. Then,
?/ *T
B sin A = cos B = - , cos A = sin B = — ,
r r
tan A = ctn # = -, ctn A = tan B = -
x y
x
FIG. 27
sec A = esc B = — , esc A = sec B = -
x y
94
MATHEMATICS
[V, §76
Since A -\- C = 90° (i. e., A and C are complementary), the
above results may be stated in compact form as follows:
A function of an acute angle is equal to the co-function of its
complementary acute angle.
77. Functions of 30°, 45°, 60°. On the sides of a right
angle lay off unit distances A B and AC and draw BC, forming
an isosceles right triangle, Fig. 28. The angles at B and C are
each 45°, and the hypotenuse BC is equal to V2~ (why?).
FIG. 28
\
AID B
FIG. 29
From the definitions,
sin 45° = cos 45° = I/ A/2 = A/2/2.
tan 45° = ctn 45° = 1.
sec 45° = esc 45° = A/2/1 = A/2.
Construct an equilateral triangle whose sides are 2 units long,
Fig. 29. Bisect one of its angles forming a right triangle ACD,
in which A = 60°, C = 30°, and the altitude CD is equal to A/3
(why?). Then from the definitions,
sin 60° = cos 30° = A/3/2.
cos 60° = sin 30° = 1/2.
tan 60° = ctn 30° = A/3.
ctn 60° = tan 30° = l/A/3.
sec 60° = esc 30° = 2.
esc 60° = sec 30° = 2/ A/3.
V, § 78]
TRIGONOMETRY
78. Eight Fundamental Relations. The following relations
hold for the trigonometric functions of any acute angle A,
(10) sin A esc A = 1, sine and cosecant are reciprocals ;
(11) cos A sec A = 1, cosine and secant are reciprocals ;
(12) tan A ctn A = 1, tangent and cotangent are reciprorocals ;
(13) tan A =
sin A
cos A
(15) sin2 A + cos2 A = 1 ;
(16) tan2 A + 1 = sec2 A •
(14) ctn A =
cos A
sin A '
(17) ctn2 A + I = esc2 A.
These eight identities are fundamental relations and should
be thoroughly learned by the student.
They may be proved as follows: (10), (11), (12) are direct
consequences of the definitions in § 75. To prove (13), we have
tan A = - , sin A = - , cos A = - ,
x r r
whence
sin A 11 x 11
-=--r-- = -= tan A.
cos A r r x
FIG. 30
Similarly,
cos A x . y x
- = - -:- - = - = ctn A.
sin A r r y
From Fig. 30,
(18) x2 + y2 = r2.
Dividing through by r2, we have
^+^-1,
whence cos2 A + sin2 A = 1.
Similarly, dividing (18) through by x2, and then by yz we prove
(16) and (17).
If the value of one function of an angle is known, the values
of all the others can be found by means of these relations.
96 MATHEMATICS [V, §78
EXAMPLE. Given sin A = 1/2. Then,
cos A = Vl — sin2 A = A/I = f V3,
and, by (13),
tan A = 1/V3 = iV§.
Since the other three functions are reciprocals of these three, we have
ctn A = V3, sec A = f V§, esc A = 2.
The values of these functions can also be found graphically by con-
structing a right triangle the ratio of whose
sides are such as to make the sine of one angle
equal to 1/2. This can evidently be done
by making the side opposite equal to 1 and
the hypotenuse equal to 2; then the side ad-
jacent is equal to >/3. (Why?) The other
functions can now be read directly from the
figure, using the definitions. Thus, tan A = side opposite -f- side ad-
jacent = l/A/3 = f A/3.
EXERCISES
1. Given sin 40° = cos 50°; express the other functions of 40° in
terms of functions of 50°.
2. The angles 45° + A and 45° — A are complementary; express
the functions of 45° + A in terms of the functions of 45° — A.
3. A and 90° — A are complementary; express the functions of
90° — A in terms of the functions of A.
4. Construct a right triangle, having given
(a) hypotenuse = 6, tangent of one angle = 3/2.
(b) cosine of one angle = 1/2, side opposite = 3.5.
(c) sine of one angle = 0.6, side adjacent = 2.
(d) cosecant of one angle = 4, side adjacent = 4.
(e) one angle = 45°, side adjacent = 20.
(/) one angle = 30°, side opposite = 25.
5. In Ex. 4, compute the remaining parts of each triangle.
6. Express each of the following as a function of the complementary
angle:
(a) sin 30°. (6) tan 89°. (c) esc 18° 10'. (d) ctn 82° 19'.
(e) cos 45°. (/)ctn!5°. (g) cos 37° 24'. (A) esc 54° 46'.
V, §80] TRIGONOMETRY 97
7. Express each of the following as a function of an angle less
than 45°:
(a) sin 60°. (6) tan 57°. (c) esc 69° 2'. (d) ctn 89° 59'.
(e) cos 75°. (/)ctn84°. (g) cos 85° 39'. (ft) esc 45° 13'.
8. Prove that if A is any acute angle,
(a) sin A -sec A = tan A. (6) sin A- ctn A = cos A.
(c) cos A -esc A = ctn A. (d) tan A- cos A — sin A.
(e) sin A-sec A-ctn A = 1. (/) cos A-csc A-tan A = 1.
(g) (sin A + cos A)2 — 1 + 2 sin A cos A,
(h) (sec A + tan A) (sec A — tan A) = 1.
(i) (1 + tan2 A) sin2 A = tan2 A.
0') (1 - sin2 A) esc2 A = ctn2 A.
(k) sin4 A — cos4 A = sin2 A — cos2 A.
(0 tan2 A cos2 A + cos2 A = 1.
(m) (sin A + cos A)2 + (sin A — cos A)2 = 2.
(n) sec A — cos A = sin A tan A.
(o) (sin2 A — cos2 A)2 = 1 — 4 sin2 A cos2 A.
(p) (1 - tan2 A)2 = sec4 A - 4 tan2 A.
9. Express the values of all the other functions of A hi terms of
(a) sin A, (6) cos A, (c) tan A, (d) ctn A, (e) sec A, (/) esc A.
79. Solution of Right Triangles. The values of the six
trigonometric ratios have been computed for all acute angles,
and recorded in convenient tables. They are given to four
decimal places in Table II, at the end of the book. These
tables, together with the definitions of the functions, enable
us to solve all cases of right triangles.
80. General Directions for Solving Right Triangles.
(1) Draw a diagram approximately to scale, indicating the
given parts. Mark the unknown parts by suitable letters, and
estimate their values.
(2) // one of the given parts is an acute angle, consider the
relation of the known parts to the one which it is desired to find,
and apply the appropriate one of formulas (7), (8), (9), p. 93.
(3) // two sides are given, and one of the angles is desired,
98
MATHEMATICS
[V, § 80
think of the definition of that function of the angle which em-
ploys the two given sides.
(4) Check the results. The larger side must be opposite the
larger angle, and the square of the hypotenuse must be equal to
the sum of the squares of the other two sides.
The following examples illustrate the process of solution.
EXAMPLE 1. Given the hypotenuse = 26, and
one angle = 43° 17'; find the two sides and the other
acute angle. Do not use logarithms.
Draw a figure ABC in which AC = 26, A = 43°
17' and denote the unknown parts by suitable let-
ters, x, y, and C. Find C as the complement of
A:
90° 00'
A = 43° 17'
C = 56° 43'
26
x B
FIG. 32
To find x note that it is adjacent to the given angle and that the hypo-
tenuse is given,
Then by (8) § 75
x = 26 cos 43° 17'
cos 43° 17' = 0.7280
26
Similarly by (7) §75
y = 26 sin 43° 17'
sin 43° 17' = 0.6856
26
4368
1456
x = 18.928
CHECK: tan A = y/x = 0.9418.
EXAMPLE 2. An acute angle
10' and the opposite side is 78.
Solve by means of logarithms.
By (8) § 75
x = 78 ctn 62° 10'
log 78 = 11.8921 - 10
log ctn 62° 10' = 9.7226 - 10
log x = 1.6147
x = 41.18
41136
13712
y = 17.8256
tan 43° 17' = 0.9418.
of a right triangle is 62°
Find the other parts.
By (9) § 75
r = 78/sin 62° 10'
log 78 = 11.8921 - 10
log sin 62° 10' = 9.9466 - 10
log r = 1.9455
r = 88.20
V, §81]
TRIGONOMETRY
99
CHECK:
r = 88.20
x = 41.18
r + x = 129.38
r - x = 47.02
log (r + x) = 2.1119
log (r - x) = 1.6723
3.7842
log 782 = 3.7842
EXAMPLE 3. The hypotenuse of a right triangle is 42.7 and one
side is 18.5. Find the other parts. To find one of the angles, as C,
note that the hypotenuse and side adjacent are known. Then
FIG. 34
ICC
cos C = ~ = 0.4332
C = 64° 19'.6
42?7 = 1823.29
18.52 = 342.25
x2 = 1481.04
x = 38.48
SOLUTION BY LOGARITHMS.
18.5
cosC =— .
log 18.5 = 1.2672
log 42.7 = 1.6304_
log cos C = 9.6368
C = 64° 19'
A = 90° - C = 25° 40'.4
CHECK: x = 18.5 ctn 25° 40'.4
= 18.5 X 2.0803 = 38.48
= 42.72 - 18.52
= 61.2 X 24.2
log 24.2 = 1.3838
log 61. 2 = 1.7868
log x2 = 3.1706
log x = 1.5853
x = 38.48
81. Graphical Solution. As shown in § 35, if the triangle be
drawn to scale, the unknown sides can be read off on the scale,
and the unknown angles on a protractor. The results so ob-
tained will be accurate enough to detect any large errors in the
computations.
100
MATHEMATICS
[V, §81
EXERCISES
Let A, B, C represent the three angles of any triangle and a, b, c
the sides opposite these angles.
1. Solve graphically the following triangles:
(a) a = 5, b = 4, c = 7. Ans. A = 44° 30', B = 34°, C = 101° 30'.
Ans. A = 22°, B = 60°, C = 98°.
Ans. A = 38°, B = 60°, C = 82°.
(6) a = 3, b = 7, c =
(c) a = 5, b = 7, c =
(d) a = 8, b = 7, B = 60°.
Ans. Ai = 82°, A2 = 98
(e) a = 3, b = 5, c = 7.
(/•) a = 7, A = 120°, b = 5.
(0) o = 42, b = 51, A = 55°.
Ans. Bi = 84°, B2 = 96°,
Ci = 38°, C2 = 22°, ci = 5, c2 = 3.
Ans. 4 = 22°, B = 38°, C = 120°.
Ans. £ = 38°, C = 22°, c = 3.
= 41°, C2 = 29°
2. Solve the following right triangles (C = 90°).
= 34, c2 = 25.
Required parts (Answers).
B = 60°,
B = 45°,
B = 30°,
A = 65°,
A = 50°,
A = 20°,
A = 45°,
A = 36° 52'
A = 4° 46',
B = 67°
5 = 53°,
A = 48°,
3. The width of the gable of a building is 32 ft. 9 in. The height of
the ridge of the roof above the plates is 14 ft. 6 in. Find the inclina-
tion of the roof, and the length of the rafters.
Ans. 41° 32', 21 ft. 10 in.
4. The steps of a stairway have a tread of 10 in. and a rise of 7 in. ;
at what angle is the stairway inclined to the floor? Ans. 35°.
5. The shadow of a tower 200 ft. high is 252.5 ft. long. What is the
angle of elevation of the sun?
Given parts.
(a)
A
= 30°,
a
= 12,
(6)
A
= 45°,
b
= 8,
(c)
A
= 60°,
c
= 20,
(d)
B
= 25°,
a
-72,
(^
B
= 40°,
b
= 33,
(!)
B
= 70°,
c
= 81,
(?)
a
= 6,
b
= 6,
(A)
a
= 3,
c
= 5,
(*)
b
= 12,
c
= 13,
0')
A
= 23°,
a
= 3.246,
(fc)
A
= 37°,
b
= 7.28,
(0
B
= 42°,
c
= 1021,
b
= 20.78,
c
= 24
a
= 8,
c
= 11.31
a
= 17.32,
6
= 10
b
= 33.57,
c
= 79.44
a
= 39.33,
c
= 51.34
a
= 27.70,
b
= 76.12
B
= 45°,
c
= 8.484
B
= 53° 8',
b
= 4
B
= 85° 14',
a
= i
b
= 7.647,
c
= 8.307
a
= 5.486,
c
= 9.116
a
= 758.7,
b
= 713.8
V, §82] TRIGONOMETRY 101
6. A cord is stretched around two wheels with radii of 7 feet and 1
foot respectively, and with their centers 12 feet apart. Find the
length of the cord. Ans. 12 V3 + 10w = 52.2 ft.
7. Two objects A, B in a rectangular field are separated by a thicket.
To determine the distance between them, the lines AC = 45 rods,
BC = 36 rods, are measured parallel to the sides of the field. Find
the distance AB. Ans. 57.63
8. One bank of a river is a bluff rising 75 ft. vertically above the
water. The angle of depression of the water's edge on the opposite
bank is 20° 27'. Find the width of the river. Ans. 201.1
9. A smokestack is secured by wires running from points on the
ground 35 ft. from its base to points 3 ft. from its top. These wires
are inclined at an angle of 40° to the ground, (a) What is the height
of the smokestack? (6) The length of the wires? (c) What is the
least number of wires necessary to secure the stack? If they are sym-
metrically placed, how far apart are their ground ends? (d) How
far are the lines joining their ground ends from the foot of the stack?
(e) From the top of the stack? (/) What angle do the wires make
with these lines? (<?) With each other? (h) What angle does the
plane of two wires make with the ground? (i) What angle does the
perpendicular from the foot of the stack on this plane make with the
ground? (j) What is its length? [DURFEE]
10. A tree stands on a horizontal plane. At one point in this plane
the angle of elevation of the top of the tree is 30°, at another point
100 feet nearer the base of the tree the angle of elevation of the top is 45°.
Find the height of the tree.
11. Find the length of a ladder required to reach the top of a building
50 ft. high from a point 20 ft. in front of the building. What angle
would the ladder in this position make with the ground?
82. General Angles. Rotation. Up to this point we have
defined and used the trigonometric functions of acute angles
only. Many problems require the consideration of obtuse angles
and others, particularly those concerned with the rotating parts
of machinery, involve angles greater than 180° or 360° even, and
it is necessary to distinguish between parts in the same or par-
allel planes which rotate in the same or in opposite directions.
102
MATHEMATICS
[V, §82
An angle may be thought of as being generated by the rota-
tion of one of its sides about the vertex; its first position is
called the initial side, its final
position the terminal side of
the angle. An angle gener-
ated by rotation opposite to
the motion of the hands of
a clock (counterclockwise) is
FIG. 35 said to be positive; an angle
generated by clockwise rotation is said to be negative. In draw-
ings a curved arrow may be used to show the direction of rota-
tion, the arrow head indicating the terminal side.
83. Trigonometric Functions of any Angle. Let $ = XOP
be any angle placed with its vertex at the origin and its initial
side along the positive z-axis. Let P be any point (except 0)
kY
J" a>
FIG. 36
on the terminal side and let x, y be its coordinates (positive,
negative, or zero depending upon the position of P in the plane) ;
let r be the distance from 0 to P (always positive). Then the
trigonometric functions of $ are defined as follows:
(19)
sin (f> = - ,
cos 0 = - .
The definitions (19) apply to all angles without exception.
(20)
tan </> =
sec d> = - .
x
V, §83]
TRIGONOMETRY
103
The definitions (20) apply to all angles except odd multiples of a
right angle; this exception is necessary because for all such
angles x is zero.
(21)
ctn <A = -
v
esc <6 = — .
y
The definitions (21) apply to all angles except even multiples
of a right angle; for all such angles y is zero.
These definitions apply of course to all acute angles and
give the same values as the definitions in § 75. These new
definitions are more general because they apply to angles to
which the former do not apply.
These ratios are independent of the choice of P on the terminal
side of the given angle. They depend upon the magnitude and
sign of the angle. For, if we choose a different point P' on the
terminal side of <£, we shall have
in magnitude and sign and this implies that
"-. = 2-i etc.
The signs of the trigonometric functions of an angle 0 depend
upon the quadrant of the plane in which
the terminal side of <f> falls when it is placed
on the axes. An angle <£ is said to be an
angle in the first quadrant when its ter-
minal side falls in that quadrant, and simi-
larly for the second, third, and fourth
quadrants. The signs of the sine and the
cosine of an angle in each of the quadrants
should be thoroughly learned. The accompanying diagram in-
dicates these signs.
FIG. 37
104
MATHEMATICS
[V, §83
The signs of the other functions are determined by noting that
tan <£ is positive when sin <£ and cos <j> have like signs and
negative when they have unlike signs; and that reciprocals
have like signs.
84. The Fundamental Rela-
tions. The fundamental identi-
ties (10) to (18) which were proved
for acute angles in § 78 are valid
for any angle whatever. The
proofs which are similar to those
already given are left to the student.
85. Quadrantal Angles. Let P be
a point on the terminal side of an
angle 0 at a distance r from the origin.
When (j> = 0°, P coincides with PI and its coordinates are
x = r and y = 0; then by § 83
sin 0° = - = 0,
r
tan 0° = - = 0.
x
cos 0° = -
r
1,
sec 0° = - = 1.
x
The angle 0° has no cotangent nor cosecant.
When 0 = 90°, P coincides with P2, x = 0, y = r; then
sin 90° = = 1,
r
ctn 90° = - = 0,
y
cos 90° = - = 0,
r
esc 90° = - = 1.
y
The angle 90° has no tangent nor secant.
When <j> = 180°, P coincides with P3, x = - r, y = 0; then
sin 180° = - = 0,
r
tan 180° = - = 0,
cos 180° = - = - 1,
r
sec 180° =
The angle 180° has no cotangent nor cosecant.
V, §86] TRIGONOMETRY 105
When 0 = 270°, P coincides with P4, x = 0, y = — r; then
sin 270° = - = - 1, cos 270° = - = 0,
r r
ctn 270° = - = 0, esc 270° = -=_!.
v y
The angle 270° has no tangent nor secant.
Often it is said that tan 90° = °o } but this does not mean
that 90° has a tangent; it means that as an angle 0 increases
from 0° to 90°, tan 0 increases without limit, and that before $
reaches 90°. Similar remarks apply to the statements ctn 0°
= oo , tan 270° = oo , etc.
86. Line Representations of the Trigonometric Func-
tions. The trigonometric functions denned in § 83 are abstract
numbers; each is the ratio of two lengths. They are not lengths
nor lines. They can however very conveniently be represented
by line segments in the sense that the number of length units in
the segment is equal to the magnitude of the function, and the sign
of the segment is the same as the sign of the function.
Let an angle 0 of any magnitude and sign be placed on the
axes, Fig. 39. With the origin as center and a radius one unit
length draw a circle cutting the positive z-axis at A, the positive
y-axis at B, and the terminal side of 0 at P. Draw tangents
to this circle at ^4. and at B and produce the terminal side in
one or both directions from 0 to cut these tangents in T and S
respectively. Draw PQ perpendicular to the z-axis. Then, if
we agree that QP shall be positive upward, OQ shall be positive
to the right, and that OT, or OS, shall be positive when it has
the same sense as OP and negative when it has the opposite
sense,
QP represents sin 0, OQ represents cos 0,
A T represents tan 0, AS represents ctn 0,
OT represents sec 0, OS represents esc 0.
106
MATHEMATICS
[V, §86
For, sin </» = QP/OP = the number of units of length in QP
since OP = unit length and sin $ and QP agree in sign from
quadrant to quadrant. Similarly the others may be proved.
\
FIG. 39
The student is warned against thinking or saying that " QP
is the sine of 0 "; say " The number of units in QP is sin (/> "
or, " QP represents sin 0."
87. Congruent Angles. Any angle formed by adding to or
subtracting from a given angle <f>, any multiple of 360° is said
to be congruent to 0; thus — 217° and 143° are congruent.
It is obvious from the definitions and from the line representa-
tions of the functions of an angle that two congruent angles
have equal functions. The functions of any angle formed by
adding to or subtracting from a given angle a multiple of 360° are
the same as the corresponding functions of the given angle.
v, §
TRIGONOMETRY
107
88. Trigonometric Equations. To solve the equation sin x
= 1/2 is to find all angles which satisfy it. We know that
x = 30° is a solution for sin 30° = 1/2; x = 150°, x = - 210°,
x = 750°, are also solutions. We can find all its solutions by
the following graphical method.
1) To solve the equation
sin x = s.
where s is a given number between
- 1 and + 1, draw a unit circle
center at the origin and on the
y-Sixis lay off OB = s (above 0 if
s > 0, below if s < 0) and through
B draw a parallel to the z-axis cut-
ting the circle in C and D, Then
the positive angles
FIG. 40
a = AOC and j8 = AOD
are solutions (and the only solutions between 0° and 360°) of
the given equation. Any angle congruent to a or to /3 is also a
solution, and there are no others. These results follow directly
from the line representations of the functions in § 86.
2) To solve the equation
cos x = c,
where c is a given number between — 1 and + 1, draw a unit
circle center at the origin, Fig. 41, and lay off on the z-axis
OB = c (to the right if c > 0, to the left if c < 0) and draw
through B a parallel to the 7/-axis cutting the circle in C and D.
Then the positive angles
a = AOC and 0 = AOD
are solutions (and the only solutions between 0° and 360°) of
the given equation. Any angle congruent to a or to /3 is also a
solution and there are no others.
108
MATHEMATICS
[V, §88
FIG. 41
3) To solve the equation
FIG. 42
tan x = t
where t is any given number whatever, draw a unit circle center
at the origin, and lay off on the tangent at A,
AB = t
and draw a line through 0 and B cutting the circle in C and D.
Then the positive angles
a = AOC, j8 - AOD
are solutions (and the only solutions between 0° and 360°) of
the given equation. Any angle congruent to a or to /3 is also
a solution, and there are no others.
Many other trigonometric equations can be reduced to one
of these three forms by the transformations given in § 78 and
hence can be solved by the above methods.
For example, the equation
is equivalent to
Again,
tan x = 3.
2 sin2 x — cos x = 1
can be reduced to the form
(cos x + l)(cos x — |) = 0
V, §89]
TRIGONOMETRY
109
,
\
0- \
f,l3
by replacing sin2 x by 1 — cos2 x, transposing all the terms to
the left side, and factoring.
8Q. Graphs of the Trigonometric Functions. The varia-
tion in the sine of a given angle as the angle increases from
0° to 360° may be exhibited graphically as follows.
Divide the circumference of a unit „ 1X &
circle into a convenient number of equal
arcs. In Fig. 43, the points of division
are marked 0, 1, 2,3, ••• 12. The
length of the circumference is approxi-
mately 6.3; lay this off on the z-axis
(Fig. 44) and divide it into the same
number of equal parts and number them
to correspond with the points of division on the circumference.
At each point of division on the a>axis lay off vertically the
line representation QP, of the sine of the angle whose terminal
side goes through the corresponding point of division on the
circle. Connect the ends of these perpendiculars by a smooth
curve. This is called the sine curve or the graph of sin x.
FIG. 43
*
A
23 4 5 6\
g\ io\ n\ Si2 is u is 16 x
FIG. 44
As the angle increases from 0° to 360°, P moves along the
circle successively through the points 0, 1, 2, 3, •••, 12, Q'
moves along the z-axis successively through the corresponding
points 0, 1, 2, 3, • • •, 12, and P' traces the sine curve.
The graphs of the other trigonometric functions, cos x, tan x,
110
MATHEMATICS
[V, §89
etc., are constructed in a similar manner by making use of their
line representations given in § 86.
If the angle increases beyond 360°, P makes a second revolu-
tion around the circle, and the values of all the trigonometric
functions repeat themselves in the same order and the graphs
from x = 6.3 to x = 12.6 will in all cases be a repetition of those
from x = 0 to x = 6.3. If P goes on indefinitely the graph
will be repeated as many times as P makes revolutions.
Functions which repeat themselves as the variable or argu-
ment increases are called periodic functions. The period is the
smallest amount of increase in the variable which produces the
repetition of the value of the function. Thus, sin a; is a peri-
odic function with a period of 360°, while the period of tan x
is 180°.
90. Functions of Negative Angles. Let AOC = </> be any
angle placed on the axes; and let AOC' be its negative, — <j>;
FIG. 45
lay off OP' = OP and draw PPf. Let x, y be the coordinates
of P and x', y' those of P'; let OP = r and OP' = r'. Then
no matter what the magnitude or sign of 4>,
y = - y ,
r = r'
V, §91]
TRIGONOMETRY
111
and by the definitions, § 83
sin ( — 0) = — = -- = — sin
>
cos (— $) =-7 = - = cos </»,
r r
tan (- <j>) = 7 = - - = - tan
ctn (—$)= — = -- = — ctn </>,
y y
sec (— <f>) =—,=- = sec
a; x
r r
esc ( — 0) = — = -- = — esc </>.
91. The Trigonometric Functions of 90° -f <t>. Let any
angle </> be placed on the axes; draw a circle, center at the origin,
with any convenient radius r, cutting the terminal side of 0 in P
and the terminal side of <f> + 90° in Q. Let the coordinates of P
be (a, &); then no matter in what quadrant P is, Q is in the
next quadrant and its coordinates are (—6, a), for the right
triangles OMP and QNO have the hypotenuse and an acute
angle of the one equal to the hypotenuse and an acute angle of
the other. Then by the definitions, § 83
C-6,a)\0
FIG. 46
MATHEMATICS
[V, §91
sin (90° + 0) = - = cos 0
r
cos (90° + 0) =
- b
= — sin 0,
tan (90° + 0) =
ctn (90° + 0) =
- 6
- 6
= — ctn 0,
= — tan 0,
sec (90° + 0) =
= — CSC 0,
csc (90° + 0) = - = sec 0.
a
These formulas hold for all angles.*
92. Functions of ± 9, 90° ± 6, 180° ± 0, 270° ± 6. If we
put for 0 in succession, - 6, 6, 90° - 6, 90° + 6, 180° - 0,
180° + 6, 270° - 9, 270° + 6, we obtain the values in the
following table, 6 being any angle.* By drawing diagrams the
results tabulated can be verified. The student is advised to
do this.
90°— e.
90°+0.
180°— e.
180° +0.
270"— 8.
270° +8.
360°— 9.
-e.
sin
cos 0
cos 9
sin 0
— sin 0
— cos 6
— cos 0
— sin 0
—sin 0
cos
sin 0
— sin 6
— cos 0
— cos 6
— sin 0
sin 0
cos 0
cos 0
tan
ctn 6
— ctn 0
— tan0
tan 0
ctn e
— ctn 0
— tan 0
— tan 0
ctn
tan 0
— tan 0
— ctn 6
ctn 6
tan 0
— tan0
— ctn 0
— ctn 0
sec
csc 0
— csc 0
— sec 0
— sec 0
— csc 0
csc 0
sec 0
sec 0
csc
sec B
sec 0
csc 0
— csc 6
— sec 5
— sec 0
— csc 0
— csc 0
If we inspect the table carefully, we find that it can be summed
up in the two rules that follow.
* Except that no angle whose terminal side falls on the y-axis has a tangent or
secant and no angle whose terminal side falls on the z-axis haa a cotangent or cosecant.
V, §93]
TRIGONOMETRY
113
1. Determine the sign by the quadrant in which the angle would
lie if 8 were acute; the result holds whether 6 is acute or not.
2. // 90° or 270° is involved, the function changes name to the
corresponding cof unction, while if 180° or 360° is involved the
function does not change name.
EXAMPLE 1. sin 177° = sin (180° - 3°) = + (rule 1) sin (rule 2) 3°.
EXAMPLE 2. cos 177° = cos (90° + 87°) = - (rule 1) sin (rule 2) 37°.
EXAMPLE 3. tan300° = tan (180° + 120°) = + (rule 1) tan (rule2) 120°.
93. Plotting Graphs from Tables. For many purposes,
such as the measurement of arcs and the speed of rotations, and
generally in the calculus and higher mathematics, angles are
measured in terms of a unit called the radian.
A radian is a positive angle such that when its vertex is placed
at the center of a circle the intercepted arc is equal in length
to the radius. This unit is thus a little less than one of the
angles of an equilateral triangle, 57°. 3 approximately. It is
easy to change from radians to degrees and vice versa, by
remembering that
(22) TT radians = 180 degrees.
Unless some other unit is expressly stated, it is always under-
stood that in graphs of the trigonometric functions the radian
is the unit angle and that 1 unit on the x-axis represents 1 radian.
These graphs can be constructed from a table of their values
such as Table III at the end of the book. Thus to plot the
graph of sin x, draw a pair of rectangular axes on squared paper
FIG. 47
114 MATHEMATICS [V, §93
and mark the points 1, 2, 3, • • • on the x-axis. These unit
lengths are divided by the rulings of the cross-section paper
into tenths. At each of these points of division on the x-axis
lay off parallel to the y-axis the sine of the angle from the table,
e. g., at 1 we plot AP = .84 = sin 1 (radian). The curve may
be extended beyond the first quadrant by the principles of § 92.
Similarly the graphs of cos x and tan x can be plotted from
their tabulated values.
EXERCISES
1. Express each of the following functions as functions of angles
less than 90°.
(a) sin 172°, (6) cos 100°, (c) tan 125°, (d) ctn 91°, (e) sec 110°,
(/) esc 260°, (g) sin 204°, (h) cos 359°, (i) tan 300°, (j) ctn 620°.
2. Express each of the preceding functions as functions of an angle
less than 45°.
3. Express each of the following functions in terms of the functions
of positive angles less than 45°.
(a) sin (-160°), (6) cos (- 30°), (c) esc 92° 25',
(d) sec 299° 45', (e) sin (- 52° 37'), (/) cos (- 196° 54'),
(g) tan 269° 15', (h) ctn 139° 17', (i) sec (- 140°),
(j) ctn (- 240°), (ft) esc (- 100°), (Z) sin (- 300°),
(m) cos 117° 17', (n) sin 143° 21' 16", (o) tan 317° 29' 31",
(p) ctn 90° 46' 12", (q) sec (- 135° 14' 11"), (r) cos (- 428°).
4. Simplify each of the following expressions.
(a) sin (90° + x) sin (180° + x) + cos (90° + x) cos (180° - x).
(6) cos (180° + x) cos (270° - y) - sin (180° + x) sin (270° - y).
(c) sin 420° cos 390° + cos (- 300°) sin (- 330°).
5. Prove each of the following relations,
(a) cos \(x - 270°) = + sin x/3.
(6) sec ( — x — 540°) = — sec x.
6. Verify each of the following equations,
(a) cos 570° sin 510° - sin 330° cos 390° = 0.
(6) cos (90° + a) cos (270° - a) - sin (180° - a) sin (360° - a)
= 2 sin2 a.
V, §93] TRIGONOMETRY 115
(c) 3 tan 210° + 2 tan 120° = - >/3.
(d) 5 sec2 135° - 6 ctn2 300° = 8.
(e) sin (90° + «) sin (180° + x) + cos (90° + x) cos (180° - x) = 0.
tan (90° + tt) + C8c2 (270° -«) = L+ ^c2 «.
7. Construct a table containing the functions of the eighths and
twelfths of 360°.
8. In each of the following equations find graphically the two solu-
tions which are between 0° and 360° and compute the values of the
other five functions of each of these angles.
(a) sin x = 3/5. (6) sin x = — 1/3. (c) cos x = — 1/3.
(d) ctnx = - 3. (e) sec x = - 5/3. (/) esc x = 13/5.
(0) esc x = — -^3. (h) tan x = — V?. (i) tan x = 2.5.
9. Verify each of the following equations.
(a) sin 90° + cos 180° = 0. (g) sec 270° + esc 0° = 0.
(6) sin 270° + cos 0° = 0. (h) sin 120° + sin 300° = 0.
(c) esc 90° + sec 180° = 0. (i) cos 150° + cos 330° = 0.
(d) esc 270° + sec 0° = 0. (j) tan 135° + tan 225° = 0.
(e) sin 0° + cos 270° = 0. (fc) ctn 315° + ctn 45° = 0.
(/) sin 180° + cos 90° = 0. (1) sin 120° + cos 210° = 0.
10. Find graphically another angle between 0° and 360° which has
the same
(a) sine as 140°, (6) sine as 220°, (c) cosine as 330°,
(d) tangent as 230°, (e) cotangent as 110°, (/) secant as 160°.
11. Find the values of 6 between 0° and 360° which satisfy the
following equations.
(a) sin 0 = sin 320°. (d) cos 0 = - cos 50°.
(6) tan 9 = tan 125°. (c) ctn 0 = - ctn 220°.
(c) sec 0 = sec 80°. (/) esc 0 = - esc 340°.
12. In what quadrant does an angle lie if sine and cosine are both
negative? if cosine and tangent are both negative? if cotangent is
positive and sine negative?
13. In finding cos x from the equation cos x = =*= Vl — sin2 x,
when must we choose the positive and when the negative sign ?
14. Plot the graphs of each of the following functions and determine
its period.
116
MATHEMATICS
[V, § 93
(a) cos x.
(d) sec x.
(g) cos (- x).
(6) tan x.
(e) esc x.
(K) sin (90° + x).
(c) ctn x.
(/) sin (- x).
(i) sin x — cos x.
15. Plot the graph of each of the following functions.
(a) x + sin x. (6) x2 + sin x. (c) sin x + cos x.
(d) x + cos x. (e) x — cos x. (/) x — 1 + sin x.
94. Sine and Cosine of the Sum of two Angles. Let
AOB = x, BOC = y, then AOC = x + y. With 0 as center
and a convenient radius r > 0, strike an arc cutting OC in P.
Drop PQ perpendicular to OB, also PR and QS perpendicular to
o R S
it o s
FIG. 48
CM. Through Q draw a parallel to OA cutting Pfl in T. Then
by (7), § 75,
r sin (x + y) = RP = SQ + TP.
Now by (7) and (8), § 75, we have
OQ = r cos y and <SQ = OQ sin x = r cos y sin x,
PQ - r sin i/ and TP = PQ cos x = r sin y cos 2.
Hence we may write
r sin (a: + y) = r cos y sin x + r sin y cos a:,
and
(23) sin (x + y) = sin x cos y + cos x sin y.
V, §95] TRIGONOMETRY 117
Similarly, we may write
r cos (x + y) = OR = OS - TQ.
Then as before,
OS = OQ cos x = r cos y cos x,
TQ = PQ sin x = r sin y sin x.
Hence we may write
r cos (x + T/) = r cos ?/ cos x — r sin y sin .r,
and
(24) cos (x + y) = cos x cos y — sin x sin y.
The above formulas, therefore, hold true for all acute angles
x and y. They are called the addition formulas.
It is readily proved that if x = a and y = /3 are any two
acute angles for which these formulas hold good they will hold
good for any two of the angles a, ft, a + 90°, a - 90°, /3 + 90°,
/3 — 90°. Therefore, since we have found that they hold good
for all acute angles, they hold good for all positive or negative
angles of any magnitude whatever.
The addition formulas may be translated into words as follows:
I. The sine of the sum of two angles is equal to the sine of the
first times the cosine of the second, plus the cosine of the first times
the sine of the second. '
II. The cosine of the sum of two angles is equal to the cosine of
the first times the cosine of the second minus the sine of the first
times the sine of the second.
95. Tangent of the Sum of two Angles. This can be de-
rived from the addition formulas as follows
sin (x + y) sin x cos y + cos x sin y
tan (x + y) = - — : — — : — .
cos (x + y) cos x cos y — sin x sin y
If we divide each term of the numerator and denominator of
118 MATHEMATICS [V, §95
the last fraction by cos x cos y, we have
sin x sin y
cosx cos y
tan (x + y) =
sin x sm
cos z cos y
that is
(25) «»(. + »)- *•" + «"'» .
1 — tan x tan y
This formula holds good for all angles such that z, y, and z + y
have tangents.
96. Functions of Twice an Angle. If we put z for y in
(23), (24), § 94, and (25), § 95, these formulas give
(26) sin 2z = 2 sin z cos z.
(27) cos 2z = cos2 z — sin2 z.
(28) = 2 cos2 z - 1.
(29) =1-2 sin2 z.
2 tan z
(30) tan2x =
1 — tan2 x
97. Functions of Half an Angle. The preceding formulas
are true for all values of x for which they have a meaning. Hence
we may replace x by any other quantity. If we write x/2 in
place of x in (28) and (29), § 96, and solve the resulting equa-
tions for sin (z/2) and cos (z/2), we find
, ._ — cos z
(31) sm \x = db
. /I + cos z
(32) cos \x = db ^ g '
Whence on dividing (31) by (32)
fl — cos z 1 — cos z sin z
(33) tan |z = ± x/ ,
1 cos z sin z 1 + cos z
V, §97] TRIGONOMETRY 119
The positive or 'negative sign is to be chosen according to the
quadrant in which z/2 lies.
EXERCISES
1. Putting 75° = 45° + 30°, find cos 75° and tan 75°.
2. ^Putting 15° = 45° + (- 30°), find sin 15°, cos 15°, and tan 15°.
3. 'Putting 15° = 60° + (- 45°), find sin 15°, cos 15°, and tan 15°.
4. Putting 90° = 60° + 30°, find sin 90° and cos 90°.
5. Show that sin (x — y) = sin x cos y — cos x sin y.
6. Show that cos (x — y) = cos x cos y + sin x sin y.
7. Putting 15° = 60° - 45°, find sin 15°.
8. Show that sin 3x = sin x(3 — 4 sin2 x) = sin x(4 cos2 x — 1).
9. Show that cos 3x = cos x(4 cos2 x — 3) = cos z(l — 4 sin2 x).
10. Find sin 4x; cos 4x; tan 4x.
11. Show that tan (45° + A) = ? + ^° AA .
1 — tan A
12. Show that
, . tan x — tan y
(a) tan (x — y) = — — ,
1 + tan x tan y '
, . ctn x ctn y — 1
Ctn(x + y) = ^nT + inT-
13. From the trigonometric ratios of 30°, find sin 60°, cos 60°, tan 60°.
14. Express sin 6,4, cos 6 A, tan GA in terms of functions of 3A.
15. Find sin 22|°, cos 22-J-0, and tan 22£°, from cos 45°.
16. Find sin 15°, cos 15°, and tan 15°, from cos 30°.
17. Find cos (x + y), having given sin x = 3/5 and sin y = 5/13,
x being positive acute, y being positive obtuse. Ans. — 63/65.
18. Verify the following:
(a) sin (60° + x) - sin (60° - x) = sin x.
(6) cos (30° + y) - cos (30° - y) = - sin y.
(c) cos (45° + x) + cos (45° — x) = V2 cos x.
(d) cos (Q + 45°) + sin (Q - 45°) = 0.
(e) sin (x + y) sin (x — y) = sin2 x — sin2 y.
. , , sin (x + y) tan x + tan y 2 tan x
(f ) ~ — 7 • (0) sin 2x =
^•* ' otr\ tfm 1*1 -for* />- *.,»!,! ™'
sin (x — y) tan x — tan y ' 1 + tan2 x '
,, , esc2 x sin $x
(h) sec 2x = — ^. (i) tanjx = — =— r-.
esc2 x — 2 1 + cos \x
(fi ctn ix - sin^ (H t«n ' A - l ~ C08 A
n*X 1-cos^x* sin A '
120 MATHEMATICS [V, §97
(I) 2 esc 2s = sec s esc s.
(m) tan (x + 45°) + ctn (x - 45°) = 0.
19. Prove each of the following identities,
(a) cos (A + B) cos (A - B) = cos2 A - sin2 B.
(6) sin (A + B) cos B — cos (A + B) sin B = sin A.
(c) sin (A + B) + cos (A - B) = (sin A + cos A) (sin B + cos B).
(d) cos4 A = f + 5 cos 2A + | cos 4A.
(e) sin4 A = f — 5 cos 2A + | cos 4A.
(/) sin2 A cos2 A = | - | cos 4A.
(g) sin2 .A cos4 A. = ^ + jj cos 2A — ^ cos 4A — ^ c°s 6A.
(/i) cos (x — y -\- z) = cos a; cos y cos 2 + cos x sin ?/ sin z
— sin x cos y sin z + sin x sin y cos z.
(i) cos £ sin (y — z) + cos y sin (z — x) + cos z sin (x — y) = 0.
0') sin A + sin 5 = 2 sin f(A + 5) cos f (A - B).
(fc) sin A - sin B = 2 cos i(^ + B} sin |(A - B).
(1) cos A + cosB =2 cos \(A + B) cos f(A - B).
(m) cos A — cos 5 = — 2 sin |(A + .B) sin %(A — B).
(n) sin A cos (B — C) — sin £ cos (A. — C) = sin (A. — JB) cos C.
(o) cos2 %<f>(l + tan |0)2 = 1 + sin 0.
(p) sin2 |x(ctn |x — I)2 = 1 — sin x.
, . 2 sec A . x . 2 sec A
(g) sec2 JA = - — : -. . (r) esc2 §A = s .
1 + sec A sec A — 1
98. Solution of Oblique Triangles. One of the chief uses
of trigonometry is to solve triangles. That is, having given
three parts of a triangle (sides and angles) at least one of which
must be a side, to find the others. In plane geometry it has
been shown how to construct a triangle, having given
CASE I. Two angles and one side.
CASE II. Two sides and the angle opposite one of them.
CASE III. Two sides and the included angle.
CASE IV. Three sides.
When the required triangle has been constructed by scale and
protractor the parts not given may be found by actual measure-
ment. The results obtained by such graphic methods are not,
however, sufficiently accurate for many practical purposes.
V, § 99] TRIGONOMETRY 121
Nevertheless, they are very useful as a check upon the com-
puted values of the unknown parts. Other checks are fur-
nished by the theorems of plane geometry that the sum of the
angles of any triangle is 180°, and that if two sides (angles) are
unequal the greater side (angle) lies opposite the greater angle
(side). The properties of isosceles triangles can also be used in
certain special cases.
The direction solve a triangle tacitly assumes that a sufficient
number of parts of an actual triangle are given. A proposed
problem may violate this assumption and there will be no
solution. Thus, there is no triangle whose sides are 14, 24,
and 40 ; likewise, there is no triangle of which two sides are 9
and 10 and the angle opposite the former is 64° 10'. Any tri-
angle which can be constructed can be solved.
Any oblique triangle can be divided into right triangles by a
perpendicular from a vertex upon the opposite side, and this
method when applied to the various cases leads to three laws,
called the law of sines, the law of cosines, and the law of tan-
gents, by means of which the unknown parts of any oblique
triangle can be computed. We proceed to prove these
laws.
99. Law of Sines. Any two sides of a triangle are to each
other as the sines of the opposite angles.
In any oblique triangle let a, b and c be the measures of the
lengths of the sides and A, B, and C the measures of the angles
opposite. Drop the perpendicular CD = p from the vertex
of angle C to the opposite side.
Two possible cases are shown in Figs. 49, 50. In either of
these figures,
p = b sin A.
In Fig. 49,
p = a sin B.
122
MATHEMATICS
[V, §99
In Fig. 50,
p = a sin (180° - B) = a sin B.
Therefore, whether the angles are all acute, or one is obtuse
a sin B = b sin A,
D B
FIG. 49
whence dividing first by sin A sin B, and second by 6 sin B,
(34)
sin A sin B '
or
sin A
sin B '
Similarly, by drawing perpendiculars from A and B to the
opposite sides, we obtain
be a c
Hence,
(35)
sin B sin C" sin A sin C '
a b c
sin A sin B sin C *
It is evident that a triangle may be solved by the aid of the
law of sines if two of the three known parts are a side and its
opposite angle. The case of two angles and the included side
being given, may also be brought under this head, since we
may find the third angle which lies opposite the given side.
100. Law of Cosines. In any triangle, the square of any
side is equal to the sum of the squares of the other two sides minus
twice the product of these two sides into the cosine of their included
angle.
V, § 100]
TRIGONOMETRY
123
Let ABC be any triangle. Drop a perpendicular BD from B
on AC or AC produced. Two possible cases are shown in
FIG. 51
Figs. 51, 52. Then we have either
or else
and
CD = b - AD (Fig. 51),
= 6 — c cos A,
CD = b + AD (Fig. 52)
= 6 + c cos (180° - A)
= 6 — c cos A,
p = c sin A (Fig. 51),
p = c sin (180° - A) = c sin A (Fig. 52).
Hence, in either figure, we may write
CD = b — c cos A and p = c sin A.
Again, in either figure,
a2 = CD2 + P2
= (b — c cos A)2 -f- (c sin A)-
= b2 - 2bc cos A + c2 (sin2 A + cos2 A)
= b- — 2bc cos A + c2
that is
(36)
COS
In like manner it may be proved that the law of cosines applies
to the side b or to the side c.
124 MATHEMATICS [V, § 100
These formulas may be used to find the angles of a triangle
when the three sides are given and also to find the third side
when two sides and the included angle are given.
101. Law of Tangents. The sum of any two sides of a tri-
angle is to their difference as the tangent of half the sum of their
opposite angles is to the tangent of half their differ&nce.
From the law of sines, we have
a sin A
b = sin B'
whence, by division and composition in proportion, we find
o + b _ sin A + sin B
a — b sin A — sin B '
Let x + y = A and x — y = B. Then we have
2x = A + B, and x = f (A + B),
2y = A - B, and y = %(A - B).
Hence, substituting in (37), we find
sin A + sin B sin (x + y) + sin (x — y)
sin A — sin B sin (x + y) — sin (x — y)
_ 2 sin x cos y _ tan x
2 cos x sin y tan y
_ tan |(A + B)
~ tan |(A - B} '
From (37) and the preceding result, we have
(38) ° + b = tan %(A + B)
a- b tan %(A - B) '
Since
tan \(A + B) = tan |(180° - (7) = tan (90° - £C) = ctn |C,
we may write the law of tangents in the form
(39) tan %(A - B) = ?—^-ctn \C.
a + o
V, §103] TRIGONOMETRY 125
As a check, (38) is the more convenient form, while for solving
triangles, (39) is preferred by some computers. If 6 > a, then
B > A. The formula is still true, but to avoid negative num-
bers the formula in this case should be written in the form
. b + a = tan %(B + A)
b - a ~ tan \(B - A) '
When two sides and the included angle are given, as a, 6, C,
the law of tangents may be employed in finding the two unknown
angles A and B.
102. Methods of Computation. The method to be used in
computing the unknown parts of a triangle depends on what
parts are given. In what follows triangles are classified ac-
cording to the given parts and the methods of computation are
stated and illustrated by examples.
103. Case I. Given two Angles and one Side. There is
always one and only one solution, provided the sum of the
given angles is less than 180°.
The third angle is found by subtracting the sum of the two
given angles from 180°. The unknown sides are found, suc-
cessively, by the law of sines.
EXAMPLE. In a triangle given two angles 38°
and 75° 43', and the side opposite the former
180; find the other parts.
Construct the triangle approximately to scale
and denote the unknown parts by suitable let-
ters as in Fig. 53. £ A
First compute the third angle C = 66° 17'. FIG. 53
To compute b use the law of sines,
_6_ sin 75° 43'
180 ™ sin 38° *
In any proportion imagine the means and the extremes to be paired
by lines crossing at the equal sign,
126 MATHEMATICS [V, §103
then the rule: Multiply the pair of knowns and divide by the known in
the other pair; or, Add the logarithms of the pair of knowns and the co-
logarithm of the known in the other pair.
FIRST METHOD: without logarithms.
sin 75° 43' = 0.9691
180
775280
9691
sin 38° = 0.6157)174.4380(283.3
12314
51298, etc.
whence 6 = 283.3.
SECOND METHOD: with logarithms.
log 180 = 2.2553
log sin 75° 43' = 9.9864 - 10
colog sin 38° = 0.2107
log b = 2.4524
18
15)60(4 b = 283.4.
Similarly we may compute c. Using logarithms, we find c = 267.7.
Not using logarithms, we find 267.6. The difference in the two answers
is due to the slight inaccuracy caused by our using only four decimal
places.
EXERCISES
1. Given two angles 43° and 67° and the included side 51; find the
other parts. Ans. 70°, 49.96, 37.02.
2. Given two angles 24° 14' and 43° 13' and the side opposite the
latter 240; find the other parts. Ans. 112° 33', 143.9, 323.8.
3. Solve the triangle ABC being given A = 17° 17', B = 102° 25',
and a = 36.84. Ans. C = 60° 18', c = 107.7, 6 = 121.1.
4. Solve the triangle LMN being given L = 28°, M = 51°, I = 6.3.
Ans. N = 101°, n = 13.17, m = 10.43.
V, §104]
TRIGONOMETRY
127
104. Case II. Given two Sides and the Angle opposite
one of Them. This case sometimes admits two solutions and
on this account is called the ambiguous case. The number of
solutions can be determined by constructing the triangle to
scale as follows.
To fix our ideas, let the given angle be A, the given opposite
side a, and the given adjacent side 6. Construct the given
angle A, and on one of its sides lay off AC = b, the given
adjacent side, and drop a perpendicular CP, of length p, from
C to the other side of the given angle A. With C as center
and with radius o, the given opposite side, strike an arc to
determine the vertex of the third angle B. Several possible
cases are shown in Fig. 54.
t. One Solution
S, One .Solution
FIG. 54
A study of these diagrams shows that there will be two
solutions when, and only when, the given angle is acute and the
length of the given opposite side is intermediate between the
lengths of the perpendicular and the given adjacent side; that is
A < 90° and p < a < b.
The two triangles to be solved are AB\C and AB2C. Since
128
MATHEMATICS
[V, §104
the triangle BiCB2 is isosceles, the obtuse angle BI (i. e., angle
ABiC) is the supplement of the acute angle B-2.
The following examples illustrate the method of computing
the unknown parts in Case II.
EXAMPLE 1. One angle
of a triangle is 34° 23', the
side opposite is 44.24 and
another side is 60.35; find
the other parts.
On constructing the tri-
j. B^~ p <B angle to scale as in Fig. 55,
pIG 55 it appears that there are two
solutions. This is verified by
computing p = 60.35 sin 34° 23'. Noting from the tables that sin 35° < .6,
it is evident that p < 40.
Let us solve first the triangle AB2C, the angle B2 being acute. By
the law of sines,
60.35 sin
44.24 sin 34° 23'
B2 = 50° 23'
log 60.35 = 1.7807
s sin 34° 23' = 9.7518 - 10
colog 44.24 = 8.3542 - 10
log sin 52 = 9.8867 - 10
64
10)30(3
Then find C2 (i. e., angle ACBJ = 95° 14'. To find c2 (i. e., side
use the law of sines again,
c2 sin 95° 14'
44.24 sin 34° 23'
c2 =78.02
log 44.24 = 1.6458
log sin 95° 14' = 9.9982 - 10
colog sin 34° 23' = 0.2482
log c2 = 1.8922
21
6)10(2
To solve the triangle AB&, we first find BI = 129° 37' being the
supplement of 52, and then the third angle Ci = 16° 00'. To find Ci
(i. e., the side ABi) use the law of sines,
V, §104]
TRIGONOMETRY
129
C]
sin 16°
44.24 ein 34° 23'
d = 21.59
CHECK.
c2 = 78.02
ci = 21.59
log 44.24 = 1.6458
log sin 16° = 9.4403 - 10
colog sin 34° 23' = 0.2482
log ci = 1.3343
= 2(44.24 cos 50° 23')
c2 -
= 56.43
log 2 = 0.3010
log 44.24 = 1.6458
log cos 50° 23' = 9.8046 - 10
log £i£2 = 1.7514
= 56.41
EXAMPLE 2. One angle of a triangle is 34° 23', the side opposite is
60.35 and another side is 44.24. Solve.
There is only one solution
as shown by constructing.
44.24 sing
60.35 ~ sin 34° 23' '
whence B = 24° 27' and the
third angle C = 121° 10'.
c
FIG. 56
sin 121° 10'
60.35 sin 34° 23' '
whence c = 91.46
EXERCISES
1. Two sides of a triangle are 17.16 and 14.15 and the angle opposite
the latter is 42°. Find the other parts.
Ans. 125° 46', 12° 14', 4.483, or 54° 14', 83° 46', 21.02
2. In the triangle AGK, A = 31° 14', a = 54, g = 48.6. Find the
other parts. Ans. 27° 49', 120° 57', 89.3
3. A 50 ft. chord of a circle subtends an angle of 100° at the center.
A triangle is to be inscribed in the larger segment having one side
40 ft. long. How long is the third side? How many solutions?
Ans. 65.22
4. If the triangle of Ex. 3 is to have one side 60 ft. long, how many
solutions? How long is the third side. Ans. 18.88 or 58.25
10
130
MATHEMATICS
[V, §104
105. Case III. Given two Sides and the included Angle.
There is always one and only one solution. The third side
can be found by the law of cosines and if the angles are not
required, this is a convenient method of solution, especially if
the given sides are not large.
EXAMPLE 1. Two sides of a triangle are 2.1 and 3.5 and the in-
cluded angle is 53° 8'. Find the third side.
x2 = 27P + iTB2 - 2 (2.1) (3.5) cos 53° 8'
= 4.41 + 12.25 - 14.7 X 0.6000 = 7.84,
whence x = 2.8.
If the other two angles as well as the third side are required,
the two angles should be found by the
law of tangents and then the third side
can be found by the law of sines.
Both these computations can be made
by logarithms.
EXAMPLE 2. In the triangle ARK,
a = 23.45, r = 18.44, and K = 81° 50'.
Find the other parts.
By the law of tangents,
a+r tan
+ R)
_
a — r tan 5 (A — R) '
The actual computation may be arranged as follows.
a = 23.45
r = 18.44
a + r = 41.89
a - r = 5.01
180° 00'
K = 81° 50'
A + R = 98° 10'
1(A + R) = 49° 5'
41.89
tan 49° 5'
5.01 tan \(A-R)
log 5.01 = 0.6998
log tan 49° 5' = 0.0621
colog 41.89 = 8.3779 - 10
log tan $(A - R) = 9.1398 - 10
i(A - R) = 7° 51'
HA + B) = 49° 5'
A = 56° 56'
R = 41° 14'
V, §106] TRIGONOMETRY 131
CHECK.
23.45 = sin 56° 56'
18.44 sin 41° 14'
log 23.45 = 1.3701 log 18.44 = 1.2658
log sin 41° 14' = 9.8190 - 10 log sin 56° 56' = 9.9233 - 10
1.1891 1.1891
To compute k use the law of sines,
k sin 81° 50'
23.45 ~ sin 56° 56' '
whence k =27.70
EXERCISES
1. In the triangle ABC given a = 52.8, b = 25.2, C = 124° 34';
find the other parts. Ans. 38° 15', 17° 11', 70.2
2. Given I = 131, m = 72, N = 39° 46', find n, L, M.
Ans. 88.57, 108° 54', 31° 20'.
3. Given u = 604, v = 291, W = 106° 19', find U, V, w.
Ans. 51° 32', 22° 9', 740.4
4. To find the distance between two objects A and B, separated by
a swamp, a station C is selected so that CA = 300 ft., CB = 277 ft.,
and angle ACB = 65° 47', can be measured. Compute AB.
Ans. 313.9
5. Two sides of a parallelogram are 23.47 and 62.38 and one angle
is 71° 30'. Find its diagonals. Ans. 59.27 and 73.29
106. Case IV. Given the three Sides. There is one and
only one solution, provided no side is grea-
ter than the sum of the other two.
The angles can be computed, in succes-
sion, by the law of cosines.
EXAMPLE 1. The sides of a triangle are 5, 7,
8. Find the angles.
49 = 25 + 64 - 2 X 5 X 8 cos A,
132 MATHEMATICS [V, §106
whence cos A = |, A = 60°.
25 = 49 + 64 - 2 X 7 X 8 cos B,
cos B = -B = 0.7857, B = 38° 13'.
64 = 25 + 49 - 2 X 5 X 7 cos C,
cos C = | = 0.1429, C = 81° 47'.
CHECK. 60° + 38° 13' + 81° 47' = 180° 00'.
The law of cosines is not adapted to logarithms but can be
transformed as follows. The three sides of a triangle ABC,
being given, then
a2 = &2 + c2 - 2bc cos A,
whence
62 + c2 - a2
(41) cosA=--2bT~'
To adapt this to logarithmic computation, subtract each
member from unity
&2 + c2 - a2 2bc - b2 - c2 + a2
1 — cos A = 1 —
2bc
a2 - (b - c)2
2bc
Hence we have
(42) 2sin4A = l-cosvl=(a + fe-
If we now set a + & + c = 2s, we have
a + & — c = 2(s — c),
a - 6 + c = 2(« - 6).
Substituting these values in (42) we find
(43) sm
Similarly,
- a)(s - c) -rfir (s - a)(g - 6)
, sin' ^o = — - - -. —
ac ab
V, §106] TRIGONOMETRY 133
Again, adding each member of (41) to unity,
52 + C2 _ Q2 (6 + c)2 - a2
1 + cos A = 1 + - -^- -2£- -
_ (b + c + a)(& + c - a)
26c
Therefore,
2 cos* |A = 1 + cos A = 2'( V" a) ,
oc
whence
(44) cos* 4A = '-^^ .
Similarly,
, ID «(* - 6) , ir< s(s - c)
cos2 §5 — - — , cos2 \C = -- - — .
ac ab
Dividing sin2 %A by cos2 \A, we have, by (43) and (44).
tan2 %A = (s — b)(s - c)/s(s - a)
= (s - a)(s - b)(s - c)/s(s - a)2.
It follows that
1 KS -a)(s -b)(s-c)
(45) tan \A = - — \/— — .
s — a \ s
If we now set
(46) r = V(* - a)(s - b)(s - c)/s,
the equation (45) becomes
(47) tan \A = — ^— .
s — a
Similarly,
7" *
tan %B = - tan
s — b s — c
It will be shown in § 107 that r is the radius of the circle in-
scribed in the given triangle.
134
MATHEMATICS
[V, § 106
EXAMPLE. The sides of a triangle are 77, 123, 130. Find the
angles.
Us - a)(s - b)(s -c) log (s - a) = 1.9445
= V s log (s - 6) = 1.6232
log (s - c) = 1.5441
colog s = 7.7825 - 10
tan \A =
s — a
a = 77
b = 123
c = 130
2)2.8943
2s = 330
s = 165
s - a = 88
s - b = 42
s - c = 35
CHECK 165
logr = 1.4472
log tan \A = 9.5027 - 10
log tan \E = 9.8240 - 10
log tan \C = 9.9031 - 10
\A = 17° 39'
\B = 33° 42'
\C = 38° 40'
CHECK 90° 01'
Therefore A = 35° 18', B = 67° 24', C = 77° 20'.
The sum of the half angles should check within 3'.
107. Area of a Triangle. It is shown in plane geometry
that the area of a triangle is equal to
one half the product of any side and
the perpendicular from the opposite ver-
tex upon that side.
If two sides and their included angle
are given, say b, c, and A, then
p = 6 sin A
FIG. 59
and
(48) Area = \bc sin A,
whence, the area of a triangle is equal to one half the product of
any two sides and the sine of their included angle.
If the three sides are given, a formula for the area can be
deduced from (48) as follows. From (26), § 96, we have
sin A = 2 sin |A cos %A
9 Vs(.9 — a)(s — b)(s — c)
~bc~
V, § 107] TRIGONOMETRY
by (43) and (44), § 106. It follows that
(49)
135
Area = Vs(s — a)(s — V)(s — c),
in which s denotes one half the perimeter.
Let r be the radius of the inscribed circle of the triangle
whose sides are a, b, c. Then since the area of the triangle
FIG. 60
ABC is equal to the sum of the areas of the triangles AOB,
BOG, CO A, we have,
(50) Area = \cr + \ar + \br = rs.
Equating (49) and (50), and dividing through by s,
(51) r ••
— c)
which proves that the r of § 106 is in fact the radius of the in-
scribed circle.
EXERCISES
1. Solve each of the following triangles.
(a) a = 50, A = 65°, B = 40°.
Ans. C = 75°, 6 = 35.46, c = 53.29
(b) a = 30, b = 54, C = 46°.
Ans. A = 33° 6', B = 100° 54', c = 39.56
(c) a = 872.5, b = 632.7, C = 80°.
Ans. A = 60° 36', B = 39° 24', c = 986.2
136
MATHEMATICS
[V, § 107
(d) a = 120, b = 80, B = 35° 18'.
(a) A
= 21° 30',
(b) A
= 62° 15',
(c) A
= 53° 25',
(d) a
= 30,
(e) a
= 25.8,
(/) a
= 37,
(fiO a
= 25.3,
(h) a
= 42,
(») a
= 3,
(j) a
= 640,
(fc) a
= .0428,
(0 a
= 12,
(m) a
= 6.02,
(a) C
= 83° 30',
(&) c
= 69°,
(c) C
= 56°,
(d) Ci
= 125° 14',
C2
= 14° 46',
(e) c
= 30.57
(/) No solution
(gr) No solution
(A) B
= 56°,
(0 A
= 111° 44',
(?) A
= 51° 58',
(/c) A
= 30° 58',
(0 A
= 32° 10',
(m) A
= 47° 24',
Ans. A = 60°, C =
84° 42',
c = 137.9
39, C = 72° 15'.
Ans. A = 51° 15', B =
56° 30',
c = 95.24
following triangles.
Given parts.
B = 75°,
a =
31.24
B = 48° 45'
6 =
402.3
B = 70° 35',
c =
6.031
6 = 50,
A =
20°.
b = 40,
A =
40° 10'.
b = 25,
A =
37°.
6 = 54,
A =
28°.
b = 42,
A -
56°.
b =2,
C =
30°.
6 - 800,
C =
48° 10'.
c = .0832,
B =
58° 30'.
6 = 16,
c =
22.
& = 4.82,
c =
8.12
Answers : Required parts
b = 82.32,
c =
84.68 '
a = 473.4,
c —
499.4
a = 5.841,
6 =
6.861
Bl = 34° 46',
Ci =
71.63
B2 = 145° 14',
C2 =
1.577
C = 50°,
B =
90°.
C = 68°,
c =
46.97
B = 38° 16',
c —
2.403
B = 79° 52',
c =
605.4
C = 90° 32',
b =
.0709
B = 45° 12',
/-Y
102° 38'.
B = 36° 8',
C =
96° 24'.
V, § 107]
TRIGONOMETRY
137
3. Find the areas of each of the following triangles.
(a) Given a = 40,
(6) Given a = 502,
(c) Given a = 27.2,
(d) Given a = 38,
FIG. 61
b = 13, c = 37. Ans. Area = 240.
b = 62, c = 484. Ans. Area = 14,590.
b = 32.8, C = 65° 30'. Ans. Area = 406.
c = 61.2, 5 = 6° 56'. Ans. Area = 1,078.
4. Venus is nearer to the Sun than the Earth. Assume that the
orbit of Venus is a circle with the Sun at its center. The distance from
the Earth to the Sun is 92.9 millions of miles. What is the distance
from Venus to the Sun if the greatest angular distance of Venus from
the sun as seen from the Earth is 46° 20'? Ans. 67,200,000 mi.
5. On a clear day, twilight ceases when the sun
has reached a position 18° below the horizon
(HAS = 18°) . Find the height AE of the atmos-
phere which is sufficiently dense to reflect the
sun's rays. Take OC = 4,000 miles. The result
must be diminished by 20% to allow for re-
fraction. [MORITZ] . Ans. 40 miles.
6. The mean distances of the Earth and Mars from the sun are 92.9
and 141.5 millions of miles respectively. How far is Mars from the
Earth when its angular distance from the sun is 28° 10' ?
Ans. 21,280,000 mi.
7. From two points on the same meridian, the
zenith distances of the moon are 35° 25' and
40° 11'. The difference in latitude between the
points of observation is 74° 26'. Find the dis-
tance of the moon from the earth, assuming the
FIG. 62 radius of the earth as 3,959 miles. [MORITZ]
Ans. 239,000 miles, approximately.
8. A search light 20 feet above the edge of a tank is directed to a
point on the surface of the water 40 feet from the edge. If the tank
is 15 feet deep how far will be the illuminated spot on the floor of the
tank from the edge, the index of refraction being 4/3? Ans. 62.5 ft.
9. A man whose eye is 6 feet above the edge of a tank 10 feet deep sees
a coin in a direction making an angle of 34° with the surface of the
water. If the index of refraction is 4/3, how far is the coin from the
side of the tank? Ans. 16.83 ft.
138 MATHEMATICS [V, § 107
10. Three forces of 12, 16, and 22 pounds in equilibrium can be
represented by the 3 sides of a triangle taken in order. Find the
angles which they make with each other.
Am. 77° 22', 134° 48', 147° 50'.
11. A sharpshooter and an enemy are 220 feet apart and on the
same side of a street 100 feet wide. Both are concealed by buildings.
A bullet striking a building on the opposite side of the street at an angle
x is deflected from the building at an angle y so that 3 sin a; = 4 sin y.
Find x so that the sharpshooter may be able to hit the enemy.
Ans. 40° 6'.
12. A ship is going 15 miles per hour. How far to the side of a target
1 mile distant must the gunner aim if the shot travels 2000 ft. per
second and the shot is fired when directly opposite?
Ans. 0° 38' or 58 ft.
13. An aeroplane is observed from the base and from the top of a
tower 40 feet high. The angles of elevation are found to be 10° 40'
and 9° 50'. Find the distance from the base to the plane and the
height of the plane. Ans. 2713 ft., 502.4 ft.
14. To determine the distance of a hostile fort A from a place B, a
line BC and the angles ABC and BCA were measured and found to be
1006.6 yd., 44°, and 70°, respectively. Find the distance AB.
Ans. 1,036 yd.
15. In order to find the distance between two objects, A and B,
separated by a pond, a station C was chosen, and the distance CA
= 426 yd., CB = 322.4 yd., together with the angle ACB = 68° 42',
were measured. Find the distance from A to B. Ans. 430.9 yd.
16. A surveyor wished to find the distance of an inaccessible point
0 from each of two points A and B, but had no instrument with which
to measure angles. He measured A A' = 150 ft. in a straight line with
OA, and BE' — 250 ft. in a straight line with OB. He then measured
AB = 279.5 ft., BA' = 315.8 ft,, A'B' = 498.7 ft. From these
measurements find each of the distances AO and BO.
Ans. 152.3 ft., 319.7 ft.
17. Two stations, A and B, on opposite sides of a mountain, are both
visible from a third station C. The distance AC = 11.5 mi., BC = 9.4
mi., and angle ACB = 59° 30'. Find the distance between A and B.
Ans. 10.5 mi.
CHAPTER VI
LAND SURVEYING
108. The Surveyor's Function. Land surveying consists
in measuring distances and angles and marking corners and lines
upon the ground, and in recording these measurements in field
notes from which a map can be drawn and the area computed.
The original survey of a tract of land having been made and
recorded, a surveyor may subsequently be called upon to find
the corners, to relocate them if lost, to retrace the old boundaries,
and to renew the corner posts and monuments if decayed or
destroyed. This is called a resurvey.
A surveyor may make a resurvey of a tract of land in order
to divide it by new lines and to
map and compute the areas of
the subdivisions.
109. Instruments. Distances
on the ground are measured with
the chain or tape. The land sur-
veyor's chain is 66 feet (4 rods)
long and is divided into 100 links
each 7.92 inches long. The steel
tape is usually 100 feet long, sub-
divided to hundredths of a foot.
Angles, horizontal or vertical,
are usually measured with the
transit. This is an instrument
mounted on a tripod, and composed of the following parts: (a)
the telescope provided with cross hairs to determine the line of
sight, a sensitive spirit level, and a graduated circle on which the
139
FIG. 63
' 140 MATHEMATICS [VI, § 109
angular turn of the telescope in the vertical plane is read; (6)
the alidade, carrying the telescope, provided with spirit levels
to bring its base into the horizontal plane and a large gradu-
ated circle on which is read the angular turn of the telescope in
measuring horizontal angles; and (c) the magnetic compass.
110. Bearing of Lines. The direction of a line on the ground
may be given by its bearing; this is the angle between the line
and the meridian through one end of it. For example, a line
bearing N 26° E is one which makes an angle of 26° on the east
side of north; one bearing S 85° W makes an angle of 85° on the
west side of south. The bearing of a line which is run by the
transit is read off on the compass circle but is subject to a cor-
rection depending upon the time and place since the magnetic
needle does not point due north at all times and places.
111. Government Surveys. In government surveys of the
public lands, a north and south line called a principal meridian
is first accurately laid out and marked by permanent monuments.
From a convenient point on the principal meridian a base line
is run east and west and carefully marked. North and south
lines, called range lines, are then run from points six miles apart
on the base line. Then township lines six miles apart are run
east and west from the principal meridian.
The land is thus divided into townships six miles square. A
tier of townships running north and south is called a range.
Ranges are numbered consecutively east and west from the
principal meridian. Townships are numbered north and south
from the base line.
In deeds and records a township is located, not by the county,
but as " Township No. — north (or south) of a certain base line
and in range No. — east (or west) of a certain principal me-
ridian. Townships are divided into thirty-six sections each
one mile square containing 640 acres, and are numbered from
VI, § 111]
LAND SURVEYING
141
1 to 36 as shown in Fig. 64. The sections are often subdivided
into halves, quarters, eighths, etc., as illustrated in Fig. 65.
6
S
4
3
2
1
7
8
9
10
11
12
18
17
16
15
14
13
19
20
21
22
23
24
30
29
28
27
26
25
31
32
33
34
35
36
Iff A.
NE 1
s.j-x.w.j
160 A.
80 A.
t
I
2 ,
s.w.4
S~E. i
4
160 A.
80 A.
FIG. 64
FIG. 65
The first principal meridian runs north from the junction of
the Ohio and Big Miami rivers on the boundary between Ohio
and Indiana. The second coincides with 86° 28' of longitude
west of Greenwich running north from the Ohio river near the
towns of English, Bedford, Lebanon, Culver, Walkerton, and
Warwick, Indiana. The surveys in Indiana (with the exception
of certain lands in the southeast corner) are governed by this
second principal meridian and a base line in latitude 38° 28' 20"
crossing this meridian about 5 miles south of Paoli, in Orange
County.* Thus a certain parcel of land is described in the
Indiana records as " E \ of NW \ of Section nineteen (19),
Township twenty-three (23) N, Range four (4) W."
The surveys extending east from one meridian will not gener-
ally close with those extending west from the preceding meridian;
the same is true of the ranges of townships extending north
* The first six principal meridians are designated by number ; some twenty -odd others
by name. E. g., the Mount Diablo meiidian, 120° 54' 48" W, which governs sur-
veys in California and Nevada. The first six base lines are neither numbered nor
named but all subsequent ones are named. The locations of all the principal merid-
ians and base lines is given in the Manual of Instructions for the Survey of the Public Lands
issued from timo to time by the GENERAL LAND OFFICE, Washington. D. C. For de-
tails and a historical sketch see also, PENCE AND KETCHUM, Surveying Manual.
142 MATHEMATICS [VI, § 111
and south from the base lines. These circumstances and the
presence of rivers and lakes give rise to fractional townships and
sections.
112. Corners. In an original survey one of the most im-
portant of the surveyor's duties is the marking of corners in
such a manner as to perpetuate their location as long as possible.
The Manual of Instructions (see 1894 edition, p. 44) says, " If
the corners be not perpetuated in a permanent and workman-
like manner, the principal object of the surveying operations
will not have been attained."
The Instructions prescribe in detail the kind of monument and
the mark to be put upon it to establish each of the various kinds
of corners that are located in the government surveys. Wooden
posts and stakes, stones, trees, and mounds of earth are used.
Witness trees or witness points are trees or other objects located
near the corner, suitably marked and described in the field notes
to make easy a subsequent relocation of the corner.
If called upon to make a resurvey of land that was originally
laid out under the direction of the General Land Office, the
surveyor will do well to make a careful study of the instructions
concerning corners that were in force when the original survey
was made, as the practice has varied somewhat from time to
time.
113. Judicial Functions of Surveyors.* Many years have
elapsed since the greater part of the government surveys were
made and in many cases the original corner marks have entirely
disappeared. The first settlers and original owners often failed
to fix their lines accurately while the monuments remained, and
the subsequent owners have no first hand knowledge of their
location. When in such cases a surveyor is called upon to
* This topic is based upon a paper of the same title by Justice Cooley of the Michi-
gan Supreme Court, published in the Michigan Engineer's Annual for 1880-81.
pp. 18-25.
VI, § 114] LAND SURVEYING
1*
make a resurvey, it is his duty to find if possible where the
original corners and boundary lines were, and not at all where
they ought to have been. However erroneous the original sur-
vey may have been, the monuments that were set must never-^
theless govern, for the parties concerned have bought with refer-
ence to these monuments and are entitled only to what is
contained within the original lines.
If the original monument and all the witness trees and other
identification marks mentioned in the field notes of the original
survey have disappeared, the corner is lost and it is the duty of
the surveyor to relocate it in the light of all the evidence in the
case, including the testimony of persons familiar with the
premises, existing fences, ditches, etc., at the point where this
evidence shows it most probably was.
In making a resurvey the surveyor has no authority to
settle disputed points ; if the disputing parties do not agree
to accept his decision, the question must be settled in the
courts. In a controversy between adjacent owners over the
location of corners and division lines, it is well established in law
that a supposed boundary line long accepted and acquiesced
in by both parties is better evidence of where the real line should
be than any survey made after the original monuments have
disappeared. It is common belief that boundary lines do not
become fixed by acquiescence in less than 21 years, but there is
no particular time that must elapse to establish boundary lines
between private owners where it appears that they have ac-
cepted a particular line as their boundary and all concerned
have claimed and occupied up to it.
114. Measuring on Level Ground. The line to be meas-
ured is first ranged out and marked with range poles or its
direction is determined by the line of sight of the transit
set on the line. The leader sticks a pin at the starting point,
takes ten in his hand and steps forward on the line dragging the
44 MATHEMATICS [VI, § 114
chain behind him. At a signal from the follower, given just
before the full length has been drawn out, he turns, aligns, and
levels the chain, stretches it to the proper tension, and, while
the follower holds the rear end at the starting point, sticks a pin
at the forward end on the line determined by the follower and
a range pole or by the transitman. At a signal from the
leader the follower pulls his pin and both move forward on the
line another chain's length and set the next pin. This process
is repeated until the leader has set his tenth pin, when the
follower goes forward, counting his pins as he goes and, if there
are ten, hands them to the leader who also counts them. The
count of tallies is kept by both. When the end of the line is
reached the follower walks forward and reads the fraction of
the chain at the pin and notes the number of pins in his hand to
determine the distance from the last tally point which is re-
corded in the field notes.
115. Measuring on Slopes. The horizontal distance which
is required can be found on slopes by leveling the chain and
plumbing down from the end
off ground. On steep slopes
only a part of the chain can
be used as at A and B in Fig.
66. The part used should be
a multiple of ten links and
great caution must be used
by both leader and follower to avoid mistakes and confusion in
the count of pins.
116. Offsets. In case measurements cannot be made on the
desired line on account of a fence, hedge, pond or other obstacle,
a perpendicular to the line, called an offset, is measured, suf-
ficiently long to avoid the obstruction and the measurements are
VI, § 117]
LAND SURVEYING
145
made on an auxiliary line parallel to the required line. Stakes
may then be set on the required line by offsets from the auxiliary
line. See Fig. 67.
c D
FIG. 68
FIG. 67
From any point on a line a right angle (or any other required
angle) can be laid off with the tran-
sit. An angle of 90° or 60° can be
laid off in a clear space with chain or
tape and pins as shown in Fig. 68,
from the facts that (1) a triangle
whose sides are to each other as 3 :
4 : 5 has a right angle opposite the
longest side; and (2) an equilateral triangle has three 60° angles.
117. Passing Obstacles. An obstacle in the line can be
passed and the line prolonged beyond it by means of perpen-
dicular offsets as shown in Fig. 67, if the nature of the locality
makes it convenient.
The same result can be accomplished by a triangle as shown
in Fig. 69. The angle HAB, the distance AB, the angle KBC,
are measured; then the distance BC
and the angle MOD are computed;
the distance BC is measured off and
the point C is located and the angle
at C is turned off and the direction
CD established; AC is computed
and the point D is taken a whole
The angles at A and B and the
FIG. 69
number of chains from A
11
146 MATHEMATICS [VI, §117
distance AB are arbitrary and may be taken so as to avoid
difficulties of the surroundings. If the circumstances permit
the angle HAB may be made 60°, and angle KEG = 120°;
then the triangle ABC will be equilateral and computations will
be avoided.
1 18. Random Lines. When it is desired to mark out a long
line, such as AB, Fig. 70, whose end points are established but
A^ k T& Ts Tt T5 B
FIG. 70
are invisible each from the other, a line AC, called a random line,
is run as nearly in the direction of A B as can be determined
and stakes Si, S2, 83, etc., are set at regular measured distances.
On coming out near B a perpendicular is let fall from B to AC
precisely locating the point C. The lengths of the offsets
SiTi, $2^2, $3^3, etc., all perpendicular to AC, can be computed
and on retracing CA, stakes can be set at T$, Tt, T3, etc., on
the desired line AB. For example, if the stakes on AC are 12
chains apart, if S$C = 6.46 chains, and if BC = 54 links, then
the offset, in links, at any stake S, is found by multiplying its
distance AS, in chains, from A, by the ratio 54/66.46 = 0.8125.
Thus the offset S4T4 = 48 X 0.8125 = 39. It is left to the
student to show that A B is longer than AC by less than 1/4 a
link and that the stakes on AB are practically 12 chains apart.
119. Computing Areas. If the boundaries of a tract are all
straight lines, i. e., if its perimeter is a polygon, the area can be
computed by dividing it into triangles, or into rectangles and
triangles, provided enough measurements are made so that the
required dimensions of each part are known or can be com-
VI, §121]
LAND SURVEYING
147
puted. It is customary to measure more lines on the ground
than is theoretically necessary in order to check the computa-
tions. These extra measurements are called proof lines in the
field notes.
120. Irregular Areas by Offsets. When one side of a field
is not straight as occurs if the boundary is a stream or curved
road, a line may be run cutting off the
irregular part and leaving the remain-
der of the field in a shape whose area
is easily computed; as AD in Fig. 71.
Stakes are set at regular measured
intervals on AD and the offsets AB,
SiTi, SzTz, SzTs, etc., are measured.
The area can be approximated by considering each of the strips
to be a trapezoid. On computing and adding we are led to the
following rule.
RULE: From the sum of all the offsets subtract half the sum of
the extreme ones and multiply the remainder by the common
distance between them.
121. Areas by Rectangular Coordinates. If the irregular
side of the field is a broken
line or if the nature of the
place makes it inconvenient
to measure the offsets at regu-
lar intervals the area can be
found by measuring the rec-
tangular coordinates of the
points A, B, C, D, E, Fig. 72, referred to the axes OX and OY.
Let the coordinates of A, B, C, • • • be (0, y0), (zi, T/I), (x2, 7/2),
• • • respectively. Then the sum of the areas of the trapezoids is
FIG. 72
(1)
2/2)
148
MATHEMATICS
[VI, § 121
where n is the number of trapezoids. On combining terms
this reduces to
(2) |[{zi(ffe - 2/2) + £2(2/1 - 2/3) + • • •
+ £n_l(2/n-2 - 2/n)} + £n(2/n-l + 2/n)]-
Hence we have the following rule.
RULE: From each ordinate subtract the second succeeding
ordinate and multiply the remainder by the abscissa of the inter-
mediate point; also multiply the sum of the last two ordinates by
the last abscissa; and divide the
algebraic sum of the products by
two.
If the coordinates of the ver-
tices of a closed polygon are
known its area can be computed
as follows. Consider the con-
vex pentagon shown in Fig. 73.
The area included may be found
by adding the trapezoids under
the sides ED and DC and subtracting those under the other
three sides; this gives
(3) |[(z4 - £5X2/4 + 2/5) + (£3 - £4X2/3 + 2/4)
— (£3 — £2X2/3 + 2/2) — (£2 — £1X2/2 + 2/0
- (xi - x5)(yi + 2/5)].,
Combining like terms, we find that this reduces to either
(4) $[xi(yt - 2/5) + £2(2/3 - 2/i) + £3(2/4 - 2/2)
+ £4(2/5 - 2/s) + £5(2/1 - 2/4)],
TJI ~o
or
(5) ?[yi(x6 — £2) + yz(x\ — £3) + 2/3(^2 — £4)
— £5)
— £1)].
VI, § 121]
LAND SURVEYING
149
These formulas are easily extended to convex polygons of
any number of sides and prove the following rule.
Multiply each abscissa by the difference of its adjacent ordinatcs,
always making the subtractions in the same sense around the perim-
eter, and take one-half the algebraic sum of the products.
The result will be the same (except as to sign) if in this rule
the words abscissa and ordinate be interchanged.
EXERCISES
1. Find the area of a field in the form of a right triangle.
(a) Base = 31.28 ch., Altitude = 16.25 ch. Ans. 25.42 A.
(6) Base = 28.46 ch., Altitude = 38.65 ch. Ans. 55.00 A.
2. Find the area of a triangular field,
(«) whose three sides are 24.50, 10.40, and 21.70 ch.
(6) having two sides 35.60, 23.70 ch., and their included angle 42° 30'.
Ans. (a) 11.27 A. (6) 28.50 A.
3. How many acres in a rectangular field whose dimensions are
17.44 and 32.65 ch. Ans. 56.94 A.
4. One side of a 200 acre rectangular field is 33.60 chains. Find the
other side. Ans. 62.50 ch.
5. What is the length of one side of a square field which contains
36 acres? Ans. 18.97 ch.
6. The diagonals of a four-sided field measure 21.40 and 24.50 ch.,
and they cross at an angle of 74° 40'. Find the area. Ans. 25.28 A.
7. One diagonal of a quadrangle runs N. 36° 20' E. 22.40 ch., and the
other S. 69° 30' E. Find its area. Ans. 25.22 A.
8. Find the areas of the fields whose boundaries are given.
Sta-
tion.
Bearing.
Distance.
A
North
9.14 ch.
B
S. 73° 25' E.
8.27
C
S. 28° 15' E.
10.04
D
N. 80° 45' W.
12.84J
Sta-
tion.
Bearing.
Distance.
P
West
19.66 ch.
Q
North
13.77
R
N. 64° 15' E.
16.66
S
S. 12° 30' E.
21.51
Ans. (a) 8.74 A. (6) 30.97 A.
150
MATHEMATICS
[VI, § 121
9. Find the areas of the fields whose boundaries are given,
(a) (6)
Sta-
tion.
Bearing.
Distance.
A
N. 25° 30' E.
10.50 ch.
B
N. 76° 50' E.
7.00
C
S. 19° 30' E.
7.92
D
S. 53° 34' W.
11.90
E
N. 64° 30' W.
4.20
Sta-
tion.
Bearing.
Distance.
1. .
N. 12° 46' W.
6.80 ch.
2. .
N. 49° 10' E.
2.40
3. .
S. 40° 50' E.
6.00
4. .
S. 10° 30' W.
4.00
5. .
N. 85° 50' W
4.52|
Ana. (a) 10.09 A. (6) 3.30 A.
10. The coordinates, in chains, of the vertices of a broken line are:
Vertex.
A.
B.
c.
D.
E.
F.
X
0.00
2.95
1.10
0.60
2.20
1.80
y
10.00
8.12
7.25
5.00
4.50
0.00
Find the area included by the broken line and the axes. Ans. 2.36 A.
11. The coordinates, in chains, of the corners of a field are:
Vertex.
1.
2.
3.
4.
5.
6.
X
0.00
7.00
12.50
18.00
15.00
10.00
y
6.00
12.00
20.00
15.00
8.25
0.00
Make a plot and find the area.
Ans. 16.175 A.
12. Starting on the bank of a river a line is run across a bend 20.00
ch., to the bank again. Offsets are measured every two chains as fol-
lows: 1.61, 2.27, 1.96, 4.23, 3.70, 2.92, 3.26, 2.50, 1.25 ch. Make a
plot of the land between the line and the river and find the area.
Ans. 4.74 A.
13. Find the measurements so as to run a line from the vertex A
of a triangle ABC to a point D on the side BC = 8.75 ch., so as to cut
off 2/5 of the area next to B. Ans. BD = 3.50 ch.
14. Find the measurements so as to run a line through a point E
on BC of the triangle of Ex. 13, parallel to AB so as to cut off 2/5 of
the area in the trapezoid. Ans. CE = 6.78 ch.
VI, § 121]
LAND SURVEYING
151
15. Two lines meet at P. PA bears S. 65° 30' E., PB bears N. 78°
15' E. Determine measurements to run a line perpendicular to PA
so as to cut off five acres. Ans. Base = lOVctn 36° 15' = 11.68 ch.
16. A triangular field contains 6 A. Show how to find on the plot
a point inside the triangle from which lines drawn to the vertices will
divide it into three triangular fields of 1, 2, and 3 A., and so that the
smallest and largest shall be adja-
cent respectively to the smallest and
largest sides of the field.
17. If the bases of a trapezoid
are a and b, a < 6, and the slant
sides are c and d, as in Fig. 74, de- FIG. 74
termine measurements to run a line
parallel to the bases to cut off, adjacent to the shorter base a, a frac-
tion /, of the whole area.
Ans. x = Va2 + /(b2 - a2), y = c
b -a'
b -a'
18. Given a = 20, b — 30, c = 54.40 ch., determine x and y to cut
off \ the area, Fig. 74. Ans. x = 23.80, y = 20.69 ch.
19. In a four sided field ABCD, AB runs S. 8.40 ch., BC, E. 9.24 ch.,
and CD, N. 5.68 ch.
(a) Run a north and south line so as to divide it into two parts whose
areas shall be to each other as 2 : 3 with the smaller on the east.
(b) Run a north and south line so as to cut off 3 A. on the west.
Ans. (a) 4.14 ch. from the east; (6) 5.40 ch. from the east.
20. A tract of land A BCD, lies
between two converging streets as
shown in Fig. 75. AB = 1980 ft.,
AC = 590 ft., BD = 1380 ft. De-
termine the measurements for run-
ning lines PQ, RS, etc., perpen-
dicular to AB, so as to divide the
tract into ten lots of equal area.
[HINT. Use the method of Ex. 17
to find PQ and AP. Or otherwise,
1
^^
S
Q
c
A
P
R
FIG. 75
find the tangent of the angle between the streets AB and CD ; find the
MATHEMATICS
[VI, § 121
Corner.
Bearing.
Distance.
1. .
N. 75° 30' W.
N. 5° 15' E.
S. 68° 10' E.
S. 23° E.
30.08 ch.
2 ...
3
4
Area = 139.84 acres
area of CAPQ in terms of x( = AP) ; this leads to a quadratic equation
in x. Find the positive root.]
Ans. AP = 300.11 ft. PQ = 709.74 ft. Area CAPQ = 4.477 A.
21. From the notes in Ex. 8 (6), make a plot and (a) run a line from S
to a point M on PQ so as to divide the field into two parts of equal areas,
(6) run a line from R to a point N on SP so as to cut off 10 acres in the
triangle.
22. From the accompany-
ing notes from a farm survey
compute the lengths of the
first, second, and fourth sides.
[HiNT. Produce the second
and fourth sides to form a
triangle.] Ans. 51.38, 36.56, 40.16.
23. Suppose the lengths of the first and fourth sides of the field
in Ex. 9 (a) are unknown. Compute them from the other data if the
area is 10.094 acres.
24. It is desired to mark out and measure a line PQ. A random
line PR is run and stakes are set on it every 100 ft. The perpendicular
from Q upon PR is 48.82 ft. long and meets it at R, 22.18 ft. beyond the
42nd stake. Compute the offsets for setting the stakes over on PQ,
their distance apart, and the length of PQ.
25. To prolong a line AB past an obstacle 0, a right turn 40° is
made at B, 400 ft. is measured to C, and a left turn of 116° is made.
Compute the distance to D on AB produced through O and the right
turn which must be made at D. How far from D should hundred foot
stakes be resumed?
CHAPTER VII
STATICS
122. Statics. Statics treats of bodies at rest and of bodies
whose motion is not changing in direction or in speed. A body
whose motion is not changing is said to be in equilibrium. The
chief problem of statics is to find the conditions of equilibrium.
123. Mass. The weight, W, of a body is not constant. For
instance a body weighs less on a mountain top than at sea level.
Also the acceleration, g, due to gravity is not constant. It
likewise is less on a mountain top than at sea level. An increase
in acceleration is accompanied by a proportional increase in
weight. But the ratio W/g is constant. The constant number
represented by this ratio is called the mass of the body. A unit
of mass is the gram, and is 1/1000 of the mass of a certain piece
of platinum which is preserved at Paris. Another unit of mass
is the avoirdupois pound. One thousand grams is equal to
2.20462125 Ibs. The mass of any body is then the number ex-
pressing the ratio of its weight to the weight of a unit of mass.
The weight is to be determined by means of a spring balance.
124. Momentum. When a given mass is in motion, we
require to know not only the magnitude of the mass, but also
its velocity. The product of the mass of a body and its velocity
is called its momentum.
125. Force. If a body possesses a certain amount of momen-
tum, it is impossible for it to alter its motion in any manner
unless acted upon by some other body which pushes or pulls it.
Force is that which tends to produce a change of motion in a
body on which it acts. This change of motion is proportional to
the force and takes place in the direction of the straight line in
153
154 MATHEMATICS [VII, § 125
which the force acts. Thus, to increase the speed of an auto-
mobile, the driving force must be increased. The greater the
force, the greater the rate of increase in the speed.
This illustrates the fact that forces are of different magnitudes.
If a motionless croquet ball is struck, its subsequent motion
depends upon the direction of the stroke. This illustrates the
fact that forces have different lines of action. If a billiard ball
is struck, the motion of the ball depends upon the point at which
the cue struck the ball. This illustrates the fact that we must
take into account the point of application of the force.
A force is said to be completely determined if we know (a) its
magnitude; (6) its line of action; (c) its direction along the
line of action; (d} its point of application.
In practice forces are never applied at a point. The force is
applied over an area such as the pressure of a thumb on the
head of a tack or the pressure of a book on a table. A force
may act throughout an entire volume as is the case with at-
traction. These forces are called distributed forces. In prac-
tice we often consider the forces which applied at a point would
produce the same effect as the given distributed forces. Such
forces are termed concentrated forces.
126. Unit of Force. The unit of force is sometimes taken
as the weight of a unit mass. This unit of force is not constant.
It changes both with altitude and with latitude. These changes
are small but for scientific purposes cannot be neglected. To
obtain a constant unit it is sufficient to make the following
definition :
The unit of force is the weight of a unit of mass at a fixed place,
say at London, Paris, or Washington.
127. Graphic Representation of Forces. A force P is com-
pletely determined if we know its magnitude, its line of action,
its direction along this line, and its point of application. It
VII, 128J STATICS 155
follows that a force can be completely represented by anything
which possesses these attributes. It can, for example, be repre-
sented by a directed segment of a
straight line. For we may let any
point 0, Fig. 76, represent the point
of application. From 0 draw any
line segment OA the number of units
in whose length is the same as the number of units in the given
force. The length of the segment represents then the magnitude
of the force. The line of which OA is a part represents the line
of action of the force. We can represent the direction along the
line by an arrowhead placed on OA.
128. Composition of Forces. — Parallelogram of Forces.
If two or more forces act in the same straight line and in the
same direction, their resultant, or sum, is obtained by adding
the numbers representing the magnitudes of the forces.
If the forces act in the same straight line but in opposite
directions, the resultant is equal to their difference, that is to
their algebraic sum.
When the forces do not act in the same straight line the
total or resultant force is found by means of a rule called the
parallelogram of forces: If two forces not in the same straight
line are represented in direction and in magnitude by two adjacent
sides of a parallelogram, the single force which would produce the
same effect as the two given forces is represented in direction and
in magnitude by that diagonal of the parallelogram which passes
through the same vertex as the two given forces.
In Fig. 77, the forces FI and F2 are represented by the lines
AB and AC, respectively. Their resultant R is represented by
AD. The magnitude of the resultant is given by the equation
(1) R = VFS + F22 + 2FiF2 cos 0,
where 6 stands for the angle BAC.
156 MATHEMATICS [VII, § 128
It will be noted that BD, being parallel and equal to AC,
represents the magnitude and the direction (but not line of
FIG. 77
action) of the force F2. If we let a equal the angle BAD, we
have, from the triangle BAD
sine sin a'
whence
F2 sin 9
(3) sin a = ^—,
The direction a of the resultant force may be found from this
equation. Thus R is completely determined.
When 6 = 90°, equation (1) reduces to
(4) R = fi* + Ft.
We also have
F2 Fl
(5) sin a = — , cos a = — .
ti T
Two forces which have a given force for their resultant are
called the components of this force. Thus FI and F2 are com-
ponents of R. The process of finding the resultant of any
number of forces is known as the composition of forces. The
process of finding the components of a given force is called the
resolution of forces. Two systems of forces acting on a particle
and having the same resultant are said to be equivalent.
VII, §129]
STATICS
157
FIG. 7&
129. Rectangular Components of a Force. Frequently,
it is desired to resolve a force
into components which are, re-
spectively, parallel and perpen-
dicular to a given line. Such
components are called rectan-
gular components. In this case —
the magnitudes of the compon-
ents may be found by the solu-
tion of equations (5), or directly from a figure. See Fig. 78.
Thus we find
(6) Fi = R cos a, F2 = R sin a.
These formulas give FI and F2 as the rectangular components
of R.
Similarly the component of any given force along any given
line is equal to the magnitude of the force multiplied by the
cosine of the angle between the line and the force.
EXERCISES
1. Given F! = 48.7, F2 = 69.8, 6 = 65° 20 , find R and a.
2. Given FI = 20.3, F2 = 60.2, 0 = 135° 10'; find R and a.
3. Given FI = 60.3, F2 = 30.2, 0 =90°, find R and «.
4. Given F! = 26.7, F2 = 45 7, 0 = 60°; find R and a.
5. R = 140, a = 15°; find Fi and Fi.
Ans. Fi = 135.2, F, = 36.2
6. R = 125, a = 25°; find Fi and FI.
Ans. Fi = 113.3, F, = 52.8
7. R = 325, a = 35°; find Fi and F,.
Ans. F! = 266.2, F, = 186 4
8. R = 600, a = 55°; find Fi and F2.
Ans. F, = 344.1, F, = 491.5
9. A particle is acted upon by two forces, of 8 and 10 pounds re-
spectively, making an angle of 30° with each other. Find the mag-
nitude of the resultant. Ana. 17.39
158 MATHEMATICS [VII, § 129
10. A boat is being towed by two ropes making an angle of 60° with
each other. The pull on one rope is SCO pounds, the pull on the other
is 300 pounds. In what direction will the boat tend to move? What
single force would produce the same result? [MILLER-LILLY]
Ans. 21° 47' with force of 500 Ibs.; 700 Ibs.
11. Let a raft move in a straight line down stream with a uniform
speed of 2 feet per second; suppose a man upon the raft walks at a
uniform speed of 4 feet per second in a direction making an angle of 60°
with the direction of movement of the raft. Find the speed and
direction qf_the man relative to the earth.
Ans. V28 ft. per sec. at an angle of 40° 54' with direction of raft.
12. A river one mile wide flows at a rate of 2.3 miles per hour. A
man, who in still water can row 4.2 miles per hour, desires to cross to a
point directly opposite. Find in what direction he must row and how
long he will be in crossing.
Ans. Upstream at an angle of 56° 48' with direction of stream; 17
minutes approximately.
13. A man in a house observes rain drops falling with a speed of
32 feet per second. The direction of descent makes an angle of 30°
with the vertical. Find the velocity of the wind.
Ans. 18.5 ft. per sec.
14. A motor boat points directly across a river which flows at the
rate of 3.5 miles per hour; the boat has a speed in still water of 10 miles
per hour. Find the speed of the boat and the direction of its motion.
Ans. 10.59 mi. per hr., 70° 43' with direction of stream.
15. From a railway train going 40 mi. per hour a bullet is fired
2,000 ft. per second at an angle of 65° with the track ahead. Find its
speed and direction.
130. Triangle of Forces. It will be seen at once on re-
ferring to Fig. 77 that the sum or
resultant of the two forces FI and F2
could be obtained more easily by
7g drawing a triangle ABD, as in Fig. 79;
when applied to find the resultant of
two forces the triangle ABD is called the triangle of forces.
Referring again to Fig. 79, it is evident that if a force equal
VII, §130] STATICS 159
and opposite to the resultant R were applied at A, this force
and the forces FI and Fz would balance, and the point A would
be in equilibrium. Another way of stating the proposition
would be as follows.
// three concurrent forces are in equilibrium, their magnitudes
arc proportional to the three sides of a triangle whose sides, taken
in order, are parallel to the directions of the given forces. Con-
rcrwly, if the magnitudes of three concurrent forces are propor-
tional to the three sides of a triangle and their directions are paral-
lel to the sides taken in order, these forces will be in equilibrium.
EXERCISES
1. Draw a triangle ABC whose sides BC, CA, AB are 7, 9, 11 units
long. If ABC is a triangle for three forces in equilibrium at a point P,
and if the force corresponding to the side BC is a force of 21 Ibs., show
in a diagram how the forces act, and find the magnitude of the other
two forces. Ans. 27, 33.
2. Draw two lines AB and AC containing an angle of 120°, and sup-
pose a force of 7 units to act from A to B and a force of 10 units from
A to C. Find by construction the resultant of the forces, and the
number of degrees in the angle its direction makes with AB.
Aris. V79; 77°, approximately.
3. Draw an equilateral triangle ABC, and produce BC to D, making
CD equal to BC. Suppose that BD is a rod (without weight) kept at
rest by forces acting along the lines AB, AC, AD. Given that the
force acting at B is one of 10 units acting from A to B, find by con-
struction (or otherwise) the other two forces, and specify them com-
pletely.
4. Find the resultant of two velocities of 9 and 7 ft. per second
acting at a point at an angle of 120°. Ans. ^G7.
5. Find the magnitude and direction of the resultant of two velocities
of 5 and 4 ft. per second acting at a point at an angle of 45°.
Ans. 8.32; 19° 52'.
6. A certain clothes line which is capable of withstanding a pull of
300 pounds, is attached to the ends A and B of two posts 40 feet apart,
160
MATHEMATICS
[VII, § 130
A and B being in the same horizontal line. When the rope is held
taut by a weight W, attached to the middle point, C, of the line, C is
four feet below the horizontal line AB. Find the weight of the heaviest
boy it will support without breaking. [MILLER-LILLY]
Ans. 117.7, Ibs.
7. A street lamp weighing 100 pounds is supported by means of a
pulley which runs smoothly on a cable supported at A and B, on oppo-
site sides of the street. If A is 10 feet above B, and the street 60 feet
wide, and the cable 75 feet long, find the point on the cable where the
pulley rests, and the tension in the cable. [MILLER-LILLY]
8. A particle of weight W lies on a smooth plane which makes an
angle a with the horizon. Show that P = W sin a, R — W cos a,
where P is the force acting along the plane to keep the particle from
slipping and R is the reaction of the plane.
131. The Simple Crane. One of the most useful applica-
tions of the triangle of forces is the case of an ordinary crane.
It has a fixed upright member AB called the crane post, a member
AC called the jib, and a tie-rod BC, A weight W suspended
FIG. 80
rigidly at C is kept in position by three forces in equilibrium.
These forces are (a) the weight W, (b) the pull in the tie-rod,
and (c) the thrust in the jib. To determine their magnitudes
construct to scale a force triangle EFG. Draw EF parallel to
the line of action of the weight W and equal to W in magni-
tude. From F draw F G parallel to the jib and from E draw
VII, § 131] STATICS 161
EG parallel to the tie-rod. The lengths of EG and FG to the
same scale on which EF was drawn represent the thrust in the
jib and the pull in the tie-rod. The directions of the forces
acting along the tie-rod and jib are given by following around
the triangle in order from E to F to G to E.
When a crane is used to raise or lower a weight, the weight
is held by a rope passing over a pulley at C. The tension of the
rope must now be taken into account.
Suppose a chain or rope supporting the weight is made to
pass over a pulley at C, and is then led on to a drum at A round
which the rope or chain is coiled. The pull in the rope and
tie-rod together is the same as before and is represented by EG.
The tension in the rope is the same on each side of the pulley.
Therefore if we mark off on EG a distance HE equal to EF,
this distance will represent the pull in the rope, thus leaving
GH to represent the pull in the tie-rod.
EXERCISES
Find the pull in the tie-rod and the thrust in the jib of a crane when
the dimensions and weight are as given below. (Weight suspended
rigidly at C.)
1. AB = 10, BC = 24, AC = 31, W = 12 tons.
2. AB =6, BC = 12, AC = 16, W = 6 tons.
3. AB = 15, BC = 50, AC = 45, W = 5 tons.
4. AB = 9, BC = 16, AC = 21, W = 4 tons.
5. The jib of a crane is subjected to a compressive force equal to
the weight of 24 tons, the suspended load being 10 tons. If the in-
clination of the jib to the horizontal is 60°, find the tension in the tie-
rod. Ans. 16.1 tons.
6. In a crane the pull in the tie-rod inclined at an angle of 60° to
the vertical is 18 tons. If the weight lifted be 8 tons, find the thrust
in the jib. Ans. 23.06 tons.
7. In exercises 1-4, find the forces acting in each member of the
crane when the load is suspended, but not rigidly, at the jib head, for
12
162
MATHEMATICS
[VII, § 131
the two cases when the rope passes from the pulleys to the drum (a) par-
allel to the tie-rod, (6) parallel to the jib.
8. The jib of a crane is subjected to a compressive force equal to the
weight of 4000 Ibs., the suspended load being 2000 Ibs. If the inclina-
tion of the jib to the horizontal is 45°, find the tension in the tie-rod.
9. In a crane the pull in the tie-rod inclined 45° to the vertical is
1000 Ibs. Find the thrust in the jib if the weight is 2000 Ibs.
10. In Ex. 9 find the thrust in the jib if the weight is 1000 Ibs.
11. The thrust in the jib inclined 60° to the vertical is 1800 Ibs.
The load is 900 Ibs. Find the tension in the tie-rod.
132. Polygon of Forces. The resultant of three or more
concurrent forces lying in the same plane may be found by
repeated applications of the triangle of forces.
Let a particle at 0 be acted upon by any number of forces,
Fi, F2, • • • ; to be definite, say Fi, F2, F3, F4. To find their
resultant proceed as follows. For the forces FI and F2 con-
struct the triangle of forces OAB (Fig. 81). Then OB is the
FIG. 81
resultant of FI and F2. For the forces OB and F3 construct
the triangle of forces OBC. The sum is given by OC. In a
similar manner combine OC and F4. The resultant is R = OD.
The construction of the lines OB and OC is unnecessary and
should be omitted. The figure OABCDO is called the polygon
of forces. OD, the closing side, is called the resultant. It will
be noticed that the arrows on the vectors representing the
VII, §133]
STATICS
163
given forces all run in the same sense around the polygon,
while the arrow of the resultant runs in the opposite sense.
If any number of forces acting at a point can be represented
by the sides of a closed polygon taken in order, the point is in
equilibrium and the resultant is zero.
From the above discussion we obtain the following rule for
finding the resultant of any number of forces.
From any point 0 draw a line OA to represent in magnitude
and direction the force F\. From the extremity A draw a linc~ AB
to represent in magnitude and direction the force F%. Continue
thix process for each of the given system of forces. Then the line
which it is necessary to draw from 0 to close the polygon represent*
the resultant in magnitude and direction.
133. Resultant of Several Concurrent Forces. Analytic
Formula. Let any number of forces Fi, F2, •••, lying in the
same plane, act on a particle at 0. To fix the ideas, suppose
there are three forces. With 0 as origin refer the forces to a
pair of coordinate axes, OX and OY (Fig. 82). Resolve each
force into two components, one along OX and one along OY.
The components of F! will be OA and OB; of Fz, OC and OE;
164 MATHEMATICS [VII, § 133
of FS, OD and OF. If a\, a^, a3 represents the angles which
FI, FZ, FS make respectively with the axis 0 X, we have :
Xt = OA = FI cos ai, F! = OB = Fl sin ah
Xz = OE = Fz cos «2, F2 = OC = Fz sin a2,
Z3 = 0Z> = F3 cos as, Y3 = OF = F3 sin a3.
If a component acts upward or toward the right we will assume
it to be positive ; if downward or toward the left, negative.
The given system of forces is equivalent to another set con-
sisting of the rectangular components of the forces of the given
system. Let us use the letters X and Y to represent the sum
of these components along the x-axis and the 7/-axis, respectively.
Then
(7)
X = FI COS «i + Fz COS «2 + FZ COS «3
= the sum of all the horizontal components.
Y = FI sin «i + FZ sin «2 + F3 sin a3
= the sum of all the vertical components
The two forces X and Y acting at right angles to each other
are equivalent to the given system of forces. The single force
R which is the resultant of X and Y is also the resultant of the
given system of forces. We have
(8) R = X2 + P.
The resultant R is always thought of as being positive. We
now have the magnitude of the resultant force. To find the
line of action we have
Y
(9) tana = — ,
Ji.
where a is the angle between the positive direction of the x-axis
and the positive direction of the resultant R.
VII, § 134] STATICS 165
To find the direction along the line of action the two following
equations are used :
Y IT
(10) sin a = — , cos a = — .
It is obvious that equations (10) determine both the line of
action and the direction along that line.
EXERCISES
1. If four forces of 5, 6, 8, and 11 units make angles of 30°, 120°,
225°, and 300° respectively, with a fixed horizontal line, find the mag-
nitude and the direction of the resultant. Ans. 7.39 ; — 81° 6'.
2. Forces P, 2P, 3P, and 4P act along the sides of a square taken
in order. Find the magnitude, the direction, and the line of action of
the resultant.
Ana. 2V2P, - 45° with line of force of 4P, through (- 2a, - 4a)
where side of square is 4o and origin of coordinates is intersection of
3P, 4P.
3. A particle is acted on by five coplanar forces ; a force of 5 Ibs.
acting horizontally to the right, and forces of 1, 2, 3, 4 Ibs. making
angles of 45°, 60°, 225°, and 300° respectively with the 5-lb. force.
Find the magnitude and the direction of the resultant.
Ans. R = 7.31, 0 = 334° 28'.
4. Find the resultant of the following concurrent, coplanar forces :
(a) (14, 45°), (6, 120°), (5, 240°).
(6) (2, 0°), (3, 50°), (4, 150°), (5, 240°).
(c) (2, - 30°), (3, 90°), (4, 135°), (5, 225°).
(d) (5, - 30°), (6, 270°), (4, 120°), (3, 135°).
134. Resultant of Parallel Forces. Let FI and F2 be two
parallel forces acting in the same direction and with their
points of application at the points A and B, Fig. 83. At A
and B apply two equal and opposite forces, AS and BT, whose
line of action coincides with AB. These will balance and will
not change the effect of the other forces. Find the resultant
AD of AS and FI, and the resultant BE of B T and F2, by con-
structing the parallelograms of forces. Then by constructing a
166
MATHEMATICS
[VII, §134
parallelogram of forces at 0, the intersection of AD and BE
produced, we may find their resultant OR, which is evidently
the resultant of FI and F2. Draw MK parallel to AB. Then
E
since OM is equal to AD in magnitude and in direction and MR
is equal to BE in magnitude and direction, it follows that the
triangles OMK and ADFi are equal, and the triangles MKR
and BTE are equal. Hence the resultant OR is equal to
FI + F2, and its line of action is a line through the point 0
parallel to the lines of action of FI and F2.
Let C be the intersection of AB and OR. Then from the
pairs of similar triangles OCA and AFiD, and OCB and BF2E,
we have
AC AS BC _BT _ AS
oc~J\ oc " FT ~ 77 '
Hence
Fi BC
(ii) F = IF"
r 2 ^10
A similar proof can be given for the case of unequal parallel
forces acting in opposite directions. Both results may be
combined into the following theorem.
The resultant of any two parallel forces, acting in the same
direction, or of two unequal forces acting in opposite directions,
VII, §136] STATICS 167
is parallel to the forces and equal to their algebraic sum and cuts
a line joining their points of application into segments, the lengths
of which are inversely proportional to the magnitudes of the forces.
135. Moment of a Force. The moment of a force with re-
spect to a point, called the center of moments, is the product of
the magnitude of the force and the perpendicular distance, called
the arm, from the point to the line of action of the force.
Geometrically the moment of a force is represented by twice
the area of a triangle whose base is the line representing the given
force and whose vertex is the center of moments.
The moment of a force in a given plane with respect to a line
perpendicular to that plane is the moment of the force with
respect to the foot of that perpendicular. The line is called the
axis of moments.
Moments are positive or negative according as they tend to
produce counter clockwise or clockwise rotation about the axis
of moments.
136. Composition of Moments. The algebraic sum of the
moments of any two forces with respect to any point of their plane
is equal to the moment of their resultant with respect to the same
point.
There are two cases.
CASE 1. When the lines of action of
the forces are not parallel.
PROOF. Let OP, OQ be two forces
acting at 0, and OR their resultant; and
let A be any point in the plane about
which moments are to be taken. Join AO, AP, AQ, and AR.
Then
Area &AOQ = Area &APR + Area ARPO,*
*By convention areas are positive or negative according as their boundaries are
travel sed in counterclockwise or clockwise direction.
168 MATHEMATICS [VII, § 136
since they have equal bases OQ and PR, and the altitude of
AAOQ is equal to the sum of the altitudes of APR and RPO.
Area AAOR = Area AAOP + Area &APR + Area ARPO,
for obvious reasons it follows that
Area AAOR = Area LAOP + Area AAOQ.
Therefore the moment of OR about A is equal to the sum
of the moments of OP and OQ about A.
Frequently it is easier to determine the moment of a force
by computing the sum of the moments of its components than
to determine it directly.
CASE II. When the lines of action of the forces are parallel.
We exclude the case in which the forces are equal and opposite.
Suppose that two forces P and Q act on the body at the
points A and B, Fig. 85. From any point 0, draw OACB per-
pendicular to the lines of action of the forces. Let OA = p,
AC = x. Then by § 134, CB = Px/Q. Taking moments about
0 we find
moment of P = P • p, moment of Q = Q(p + x +
moment of P + moment of Q = Pp -\-Qp-\-Qx-\-Px
= moment of R.
If P and Q are in opposite directions the proof is similar to the
above and is left to the student. The proof in case P and Q
are equal but opposite in direction is given in the following
section.
f
It
r
-
FIG. 86
6
A
FIG. 85
0 B
137. Couples. A system of two parallel forces, equal irt
VII, §138] STATICS 169
magnitude and opposite in direction, is called a couple. The
perpendicular distance between the lines of action of the forces
is called the arm of the couple; and the plane containing the
forces is called the plane of the couple.
The moment of a couple is the algebraic sum of the moments
of its forces about any axis perpendicular to its plane and is
equal to the product of either force and the length of the arm. For,
let 0 be any axis, perpendicular to the plane of the couple, and
OA and OB, the moment arms of the forces with respect to 0.
Taking moments about 0, we have
F-OB - F-'OA = F-AB.
The sign of the couple is plus if it tends to turn with clock-
wise rotation, and minus if it tends to turn with counter-clock-
wise rotation.
138. Conditions of Equilibrium.
(a) Concurrent coplanar forces. In order that the forces of
a system may balance each other, the resultant must be equal
to zero, that is
(12) R = V(SZ)2 + (S7)2 = 0.
Hence we have also
(13) SX = 0, and 2Y = 0.
The algebraic sum of the moments of the forces (written SAf)
about any point is equal to the moment of the resultant. If
the forces are in equilibrium, R = 0; therefore
(14) SM = 0.
These conditions are used in the second method of Ex. 1, below.
(6) System of parallel forces. If the algebraic sum of a sys-
tem of parallel forces is not zero, the resultant is a single force
and the system is not in equilibrium. Hence a necessary con-
170
MATHEMATICS
[VII, § 138
dition for equilibrium is that
= 0,
where F represents the magnitude of a force. 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. Hence a necessary
condition for equilibrium is that
(15) 2F • x = 0,
where x is the moment arm of the force F.
EXERCISES
BALANCED SYSTEMS OF FORCES ACTING THROUGH THE SAME POINT
1. A triangular frame ABC (Fig. 87) carries a load of 1000 Ibs. at A.
Find the stresses in the members AB and AC.
1000
FIG. 87
FIG. 88
SOLUTION. We have in this problem a balanced system of forces
acting through the point A, namely, the load of 1000 Ibs. and the forces
FI and F2 in the members AC and AB. Both AC and AB are subjected
to a compression. Hence both members exert a thrust in the direction
indictated by the arrows. The problem is to determine the magnitude
of two unknown forces in a balanced system of three forces, the direc-
tions of the forces being known. This problem may be solved in any
one of the three following ways.
FIRST METHOD. (Triangle of Forces.) The forces may be repre-
VII, § 138]
STATICS
171
sented by the sides of a triangle taken in order, Fig. 88. If the figure
is drawn to scale the magnitudes of the unknown forces F\ and Ft may
be obtained directly from the figure by measurement.
woo
FIG. 89
1000
FIG. 90
If the lengths of all of the members of the frame ABC are known or
can be computed, we can obtain the magnitudes of FI and Ft by pro-
portion, since the triangle ABC and the force triangle are similar.
In this particular problem we observe that the force triangle is right-
angled and one acute angle is 60°. Hence
F! = 1000 sin 60° = 866 Ibs., F2 = 1000 cos 60° = 500 Ibs.
SECOND METHOD. (Resolution of Forces.) Refer the forces to a
system of coordinate axes, Fig. 88, and use the conditions (13) of equi-
librium. We have
SX = F2 cos 30° - F! cos 60° = 0,
2F = Ft sin 30° + Fl sin 60° - 1000= 0.
The solution of these equations gives,
Fl = 866 Ibs., Ft = 500 Ibs.
THIRD METHOD. (Moments.) The sum of the moments of all the
forces about any arbitrarily chosen point leads to one equation contain-
ing the unknowns. If we take the sum of the moments of all the forces
about as many arbitrary points as there are unknowns then we will have
as many equations as unknowns. The solution of these equations gives
the magnitudes of the unknown forces. It is often advantageous to
choose for the points about which moments are taken, points on the lines
of action of the unknown forces, one on each line.
Taking moments about B we find
172
MATHEMATICS
[VII, § 138
whence
SAf = 8V3F, - 1000 X 12 = 0,
Fi = 866 Ibs.
Taking moments about C we find
SM = 1000 X 4 - 8F2 = 0,
whence
F2 = 500 Ibs.
2. Find the stresses in the members AB and AC, of the triangular
frame ABC, Fig. 91, the load at A being
1000 Ibs.
[HiNT. Use the triangle of forces.]
Ans. AB, 739.1 Ibs.; AC, 922.2
3. S Ive Ex. 2 (a) by using the
method of resoluti n of forces ; (6) by
the method of moments.
4. Assuming that the frame in Ex.
2 is supported by a vertical force at B, find the magnitude of the force
and the stress in BC.
5. A crane is loaded with 3000 Ibs. at C. Determine the stresses
in the boom CD, the tie BC, the mast BD
and the stay AB, Fig. 92.
[HiNT. Use the triangle of forces.]
Ans. CD, 6250 Ibs. (compression) ; BC,
4250 Ibs. (tension) ; AB, 5858 Ibs. (tension) ;
BD, 2500 Ibs. (compression).
6. Solve Ex. 5, using the method of reso-
lution of forces.
7. Find the horizontal and vertical components of the supporting
forces at A and D, Ex. 5.
8. Find the stresses in the members of the
crane in Ex. 5, when the boom makes an an-
gle of 15° with the horizontal.
9. What is the smallest force F which will
prevent a weight of 150 Ibs. from slipping
down the incline represented in Fig. 93 if friction is neglected?
Ans. 212.2 Ibs.
FIG.
VII, § 138] STATICS 173
10. Let F = 150 Ibs. (Fig. 93) and let the weight also be 150 Ibs
What will be the largest angle between the inclined plane and the hori
zontal at which the weight will not slip ? Ans. 30°.
11. Experiments indicate that a horse exerts a pull on his traces
equal to about one-tenth of his weight, when the working day does not
exceed 10 hours. The draft of a certain wagon is due to (a) axle
friction = 5 Ibs. per 2000 Ib. load ; (6) gradient or hills ; (c) rolling draft
depending on height of wheel, width of tire, condition of road-bed, etc.
How large a load can a team of horses each weighing 1000 Ibs. pull
up a 10% grade if the rolling draft is zero. (A 10% grade is a rise of
10 feet for each 100 feet measured horizontally along the roadway.)
Ans. 1961 Ibs.
FIG. 94
12. What extra pull must a horse exert on his traces (assumed
horizontal) if on a level road the wheel, 4 feet in diameter, strikes a
stone 2 inches high, the load being 1000 Ibs. Ans. 436 Ibs.
13. A carriage wheel whose weight is W and whose radius is r rests
on a level road. Show that any horizontal force acting through the
center of the wheel greater than
r — h
will pull it over an obstacle whose height is h.
14. In Ex. 13, let P = 100 Ibs., W = 1000 Ibs., r = 2 feet. Find h.
Ans. 0.126 in.
15. A 50 Ib. boy swings on the middle of a clothes line which is 50 feet
long. The lowest point is 2 feet below either end. Find the tension
in the rope. Ana. 625 Ibs.
16. A wire 90 feet long carries a weight of 25 Ibs. at each of its trisec-
tion points. When the wire is taut each weight is 5 feet below the hori-
zontal line connecting the points of support. Find the tension in each
segment of the wire. Ans. 150 ; 147.9 ; 150 Ibs.
174
MATHEMATICS
[VII, §138
17. Steam in the cylinder of an engine exerts a pressure of 20,000
pounds on the piston-head. The guides N, Fig. 95, are smooth. What
N
1
; i
N
FIG. 95
is the thrust in the connecting rod when it makes an angle of 20° with
the horizontal? What is the pressure on the guides N ? [MILLER-LILLY]
PARALLEL FORCES ACTING IN THE SAME PLANE
18. Determine the resultant R of each of the following systems of
parallel forces.
50 20 30
Y~—4— -»!<— -
6'— ->f -5-' —
20
40
80
(a) FIG. 96
10
60
(6) FIG. 97
70
500
U-3./4- +•__ ... „.•
800
(c) FIG. 98
19. Let AB (Fig. 99) represent a beam carrying the weights indi-
cated and supported by the vertical forces FI and F2. Find FI and F2.
1000 2000
>*-»* — 4+ — 4*
F0= 2500
FIG. 99
20. The system of parallel forces in Fig. 100 is in equilibrium. Find
the magnitudes and directions of the unknown forces FI and Ft.
4P F,
FIG. 100
VII, § 138]
STATICS
175
21. If a horse exerts a pull on his traces equal to one-tenth of his
weight, where should the single-tree for each of two horses weighing
1200 and 1600 Ibs., respectively, be fastened to a double-tree in order
that each horse shall do his proper share of the work ?
22. The center clevis pin A, of a double-tree is a inches in front of the
mid-point B, of the line connecting the end clevis pins C and D, which
are b inches apart. If one horse is pulling c inches ahead of the other
what fraction of the load L is each horse pulling, Fig. 101 ?
1 _ ac
2
Ans.
2
- c2
O
FIG. 101
23. Find what fractional part of the load each horse is pulling if
a = 2, when
(a) b = 41, c = 9. (b) b = 39, c = 15.
(c) b = 34, c = 16. (d) 6 = 52, c = 20.
(e) b = 37, c = 12. (/) b = 50, c = 14.
(jr) b = 61, c = 11. (h) b = 36, c = 4.
24. In Ex. 22, if the evener makes an angle 0 with the tongue, what
fractional part of the load is pulled by each horse ?
Ans. +
\ - \ tan 9.
176 MATHEMATICS [VII, § 133
25. In Ex. 24 put a — 2, b — 40. Plot a curve using values of 9 as
abscissas and values of the load pulled by one horse as ordinates. What
can you say about the part of the load pulled by this horse as 0 increases ?
26. In each of the cases of Ex. 23 find the pounds of pull exerted by
each horse if the total pull on the load is 362.88 Ibs.
27. The middle clevis pin A of a three-horse evener is a inches in front
of the point B of the line connecting the end clevis pins C and D. The
end clevis pins are b and 26 inches from the point B. Find what frac-
tional part of the load is borne by the horse on the longer end when it is
c inches behind the other horses.
28. Find what fractional part of the load the horse on the long end is
pulling if a = 2, when
(a) b = 24, c = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
(b) b =25, c = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
(c) b = 26, c = 2, 4, 6, 8, 10, 12, 14.
29. In Ex. 27, if the evener makes an angle 0 with the tongue, what
fractional part of the load is pulled by the horse on the long end.
Ans. - + --tau6.
3 3b
30. In Ex. 29 put a = 2, b = 25. Plot a curve using values of 0
as abscissas and fractional parts of the load pulled by the horse on the
long end as ordinates. Discuss the problem.
31. A steel rail 60 ft. long weighs 1595 Ibs. Where must a fulcrum
be placed so that a 180 Ib. man at one end can raise 4 tons at the other?
Ans. 6 ft.
CHAPTER VIII
SMALL ERRORS
139. Errors of Observation. Suppose that we measure the
length of a building and record the result. Such a record is
called a reading or an observation. Suppose that we measure
the same length and record the reading on each of several suc-
cessive days. On comparison it is likely we shall find that
they do not exactly agree. What then is the true length?
Whatever the actual length may be the difference between it
and any observation of it is called an error of observation.
Suppose that we measure the length of a building with a tape
whose smallest division is one foot. If the length is not a whole
number of feet, we estimate by the eye the fraction of a foot left
over. This estimate will almost certainly be in error. If we
measure the same length with a tape divided to eighths of an inch,
the end of the building may coincide with a division of the tape
or we may have to estimate the fraction of an eighth. Subse-
quent readings are not likely to agree exactly with the first, and
even if they do all agree we cannot be sure that we have the true
length. Inattention and lack of precision of the observer, in-
experience in using the measuring instrument, or the use of an
instrument which is defective or out of adjustment, all tend
to introduce errors. It is important to keep in mind that such
errors are always present, in greater or less degree, in every set
of observations.
If a is the recorded reading of a measurement of an unknown
quantity u, a measure of the error in this reading is a positive
number m, such that u lies between a — m and a + m. The
actual error may be very much less than its measure m. For
example if a rod of (unknown) length / be measured with a scale
177
178 MATHEMATICS [VIII, 139
divided to tenths of an inch and the reading is 47.8, it is fairly
certain that 47.7 < / < 47.9, and we write I = 47.8 ±0.1.
It is evident that any number will be in error if it is derived by
computation from other numbers which are inexact. Approxi-
mations are used in computations not only for recorded meas-
urements but also in the case of irrational numbers, such as
surds, most logarithms, trigonometric functions, TT, etc. We
have 3-place, 5-place, 7-place, 10-place tables in order to secure
the degree of accuracy desired in the computed result. In what
follows it is shown how to find a measure of the error in a number
computed by some of the simpler processes of arithmetic from
given numbers the measures of whose errors are known.
140. Error in a Sum. Suppose that in measuring two quan-
tities whose actual (and unknown) values are u and v, we make
errors Aw and Av respectively, and record the readings a and b.
Then u = a ± Aw, v = b ± A» and their sum lies between
a + b — (Aw + A0) and a + b + (Aw + Au).
Whence, w + -» = a + 6± (Aw + Aa).
That is, the error in the sum of two readings is measured by the
sum of their errors.
This result is readily extended to the sum of more than two
readings. The error in the difference of two readings is never
greater than the sum of their errors, though it may be greater
than their difference.
EXAMPLE. Find the sum and difference of 46.8 ± 0.65 and 12.4 ±
0.15. Here the readings are 4.68, 12.4 and the measures of their errors
are 0.65, 0.15 respectively. The measure of the error of their sum is
065 + 0.00 = 0.80 ;
whence (46.8 ± 0.65) + (12.4 ± 0.15) = 59.2 ± 0.8
and (46.8 ± 0.65) - (12.4 ± 0.15) = 34.4 ± 0.8
141. Error in a Product. With the same notation as above,
the product uv lies between
ah — (aA0 + feAw + Aw • At)) and ab + (aAw + 6Aw -f Aw • Ac),
VIII, § 142] SMALL ERRORS 179
whence, neglecting the small term Aw • A», we have approxi-
mately,
uv = ab ± (aAv + 6 Aw).
That is, a measure of the error in the product of two readings is the
first times the error of the second plus the second times the error of
the first.
142. Error in a Fraction. The quotient of u divided by v
evidently lies between
a — Au i a + AM
6 + A« b — At)'
that is between
a aAfl + bAu j n . aAy +
b b(b + At)) 6 b(b - At>)
whence a measure of the error in the fraction is
aAt) -f feAw aAa -f &Aw
— — — - -- , approximately.
b(b — At)) fr2
That is, a measure of the error in the .quotient of two readings is a
measure of their product divided by the square of the divisor.
EXAMPLE. Find the product and quotient of 12.4 ± 0.15 and
46.8 ± 0.65. By § 141, a measure of the error in the product is (12.4)
(0.65) + (46. 8) (0.15) = 15.08 and the error in their quotient is meas-
ured by 15.08/(46.8)2 = 0.0069.
Whence, (12.4 ± 0.15) (46.8 ± 0.65) = 580.32 ± 15.08
and (12.4 ± 0.15)/(46.8 ± 0.65) = 0.265 ± 0.0069
EXERCISES
Make each of the following computations and state the result so as to
show a measure of the error in it.
1. (123 ± 0.2) ± (241 ± 0.1). 2. (222 ± 0.5) ± (111 ± 0.4).
3. (217 ± 0.2J(117 ± 0.3). 4. (1267 ± 0.5)(1342 ± 0.4).
5. (163 ± 0.2)/(25 ± 0.5). 6. (732 ± 0.3)/(21 ± 0.4).
7. In Ex. 3 and 4 compute the term Au • Ay neglected.
8. In Ex. 5 and 6 compute " ~ A>1 and a + A';\ Find the differ-
0 + Ay b — Aj
180
MATHEMATICS
[VIII, § 142
ence between the error thus computed and those computed in exercises
5 and 6 and consider the influence of this difference upon the quo-
tient.
9. A line is measured with a chain (100 links each 1 ft. long). After-
wards, it is found that the chain is one foot too long. If the measured
length was 10.36 chains, what is its true length if the error is assumed
to be distributed through the chain? Ans. 10.4636 chains.
10. A line is measured with a 100-ft. tape and found to be 723.36
feet long. The tape is afterwards found to be 0.02 of a foot short.
What is the true length of the line? Ans. 723.22 ft.
11. A certain steel tape is of standard length at 62° F. A tape will
expand or contract sixty-five ten millionths of its length for each
Fahrenheit degree change of temperature. A line is measured when
the temperature of the tape is approximately 80° and found to be
323.56 feet long. What is its true length? Is it necessary to know
the nominal or standard length of the tape to solve this problem?
Ans. 323.52 ft.
12. What change in temperature is necessary to change a 100-foot
tape by 0.01 of a foot, or 1 in 10,000? Ans. 15°.38
13. A certain 100-foot steel tape, standard length at 62° F., is used
to measure from the monuments (Fig. 102) to the point A, in a line
X Monument
so'-;
7th
St.
1
2
3
4
5
6
8th
St.
Monument X
FIG. 102
between lots 2 and 3 extended, when the temperature is 40° F. As-
suming that the map distances are correct, what lengths must be
measured from 7th street and 8th street monuments respectively to
locate the point A, the monuments being in the center lines of the
streets? Ans. 160.02; 280.04
Show that if x, y, and z are small that
14. (1 + z)(l + y} is nearly equal to 1 + x + y.
15. (1 + x)l(\ + y) is nearly equal to 1 + x — y.
VIII, § 143] SMALL ERRORS 181
16. (1 + x)(l + ?/)(l + z) is nearly equal to 1 + x + y + z.
17. Show that (1 + 0.03) (1 - 0.05) = 0.98 nearly.
18. Compute (1.04)(1.06)(0.95). Ans. 1.05
19. Compute (a) 1.03/1.02; (6) (1.03)(1.02).
Ans. (a) 1.01; (6) 1.05
20. Compute (a) (1.03) (0.98); (6) 1.03/0.98.
Ans. (a) 1.01; (b) 1.05
21. Draw a figure (rectangle) to represent (4.03) (9.02) and indi-
cate 4 X 0.02; 9 X 0.03; 0.03 X 0.02; 4X9.
22. Show that the error in abc due to errors Aa, Ab, Ac in a, b, and c
respectively, is be • Aa + ac • Ab + ab • Ac.
23. Compute 2.01 X 4.02 X 3.02 Draw a figure (parallelepiped)
to represent this product and indicate 3 X 4 X .01; 2 X 3 X .02;
2 X 4 X .02; .01 X .02 X .02; 2X4X3. Ans. 24.4
143. Data derived from Measurements. The preceding
results apply immediately to the case in which numbers ob-
tained by measurement are stated without any accompanying
indication of the probable error.
In such cases it is understood that the given figures are all
reliable, i. e., that we stop writing decimal places as soon as
they are doubtful. The last figure written down should be as
accurate as is possible. Then the error will surely not be
more than 5 in the next place past the last one actually written.
Thus, if a certain length is reported to be 2.54 ft., we would
understand that the true length is not more than 2.545 ft., and
not less than 2.535 ft. For if the true length is more than 2.545
ft., it should be given as 2.55 ft.; and so on.
It may happen that the last figure written down is 0. This
means that that place is reliable. Thus, to say that a given
length is 2.4 ft. means that the true length is between 2.35 ft.
and 2.45 ft. But to say that a given length is 2.40 ft. means
that the true length is between 2.395 ft. and 2.405 ft.
In computations based upon numbers obtained by measure-
ment, these facts must be kept in mind, and the result of any
182 MATHEMATICS [VIII, §143
calculation should not be stated to more decimal places than
are known to be reliable.
EXAMPLE 1. Find the area of a rectangle whose sides are found,
by actual measurement, to be 2.54 ft. and 6.24 ft., respectively.
Since the error in writing 2.54 ft. may be as great as .005, we must
write for the length of this side (2.54 ± .005) ft. Likewise, we must
write for the other side (6.24 ± .005) ft. Hence, by the rule of § 141,
the error in the product may be as large as
2.54 X .005 + 6.24 X .005,
that is .043. Hence we are not justified in expressing the answer to
more than one decimal place; although
2.54 X 6.24 = 15.8496,
we must sacrifice all the figures past 15.8, and write
2.54 X 6.24 = 15.8 ± .1
since the true answer may be as large as 15.894 Even the figure 8
in the first decimal place is not reliable, since the true area may be
nearer 15.9 than 15.8 sq. ft.
EXERCISES
1. Assuming that the numbers stated below are the results of
measurements, and that each of them is stated to the nearest figure
in the last place, find the required answer and state it so that it also is
correct to the nearest figure in the last place you give, or else to within
a stated limit of possible error.
(a) 2.74 -f 3.48 + 11.25 + 7.34 Ans. 24.8
(6) 3.25 - 7.348 + 4.26 - 6.1 Ans. 20.9 ± .1
(c) 6.27 X 3.14 (g) 61.54 X 45.2 + 14.81
(d) 26.5 X 11.4 (A) 8.26 -=- 2.14
(e) 7.32 X 5.4 (i) 43.7 + 5.4
(/) 36.4 X 2.78 0') (6-42 X 2.35) -?- 4.5
2. The sides of a rectangle are measured, and are found to be 4 ft.
6.3 in. by 3 ft. 5.4 in. Express correctly the area of the rectangle.
3. The three sides of a rectangular block are measured and are
found to be 7.4 in. by 3.6 in. by 4.7 in. Express the volume.
VIII, § 146] SMALL ERRORS 183
4. Suppose that the dimensions of a bin are measured roughly to
the nearest foot, and that they are 8 ft. by 4 ft. by 3 ft. How large
may the volume actually be? How small may it be?
Ans. 118.1 cu. ft., 65.6 cu. ft.
5. The floor of a room is found by measurement to be 22 ft. X 15 ft.,
each dimension being to the nearest foot. How should the area be
stated? Ans. 330 ± 18 sq. ft., or 300 sq. ft.
6. If, in Ex. 5, the height of the room is 9 ft. to within the nearest
foot, express the volume of the room.
144. Error in a Square. If a is an observed value of an un-
known quantity u, then it follows directly from § 141 that a
measure of the error in w2 is approximately
aAw + aAw — 2aAw, and we write
w2 = a2 ± 2aAw.
That is, a measure of the error in the square of a reading is twice
the reading times its error.
145. Error in a Square Root. With the same notation as
above, u = a ± Aw is nearly equal to
— 2
"la"'
since the last term is small. This is a perfect square and hence
the positive square root of w is approximately
Va±-^=.
2V a'
That is, a measure of the error in the positive square root of a read-
ing is equal to its error divided by twice its square root.
EXAMPLE. Find Vl25 ± 0.5 A measure of the error is
0.5/2(11.18) = .022 and Vl25 ± 0.5 = 11.18 ± 0.022
Again V2400 = V2401 - 1 = 49 - & = 49 - 0.0102
146. Errors in Trigonometric Functions. Suppose a ex-
pressed in radians is an observed value of an unknown angle a.
184 MATHEMATICS [VIII, 146
Then a = a ± Aa and by § 94,
sin a = sin (a ± Aa) = sin a cos Aa ± cos a sin Aa.
Now if Aa is small, cos Aa is nearly equal to 1, and sin Aa is nearly
equal to Aa. Whence we have, approximately,
sin a •= sin a ± cos a • Aa,
and the smaller Aa is, the better the approximation. Hence,
a measure of the error in the sine of an angle is the error in the
reading (expressed in radians) multiplied by the cosine of the
reading.
Similarly we can show that a measure of the error in the cosine
of an angle is the error in the reading multiplied by the sine of the
reading.
By means of these results and the principles of § 142 we can
readily find a measure of the error in the other trigonometric
functions. For example
_ sin a _ sin (a ± Aa)
tan a — — - ~
cos a cos (a ± Aa)
and by § 142, a measure of the error in tan a is
Aa(sin2 a + cos2 a)/cos2 a = sec2 a • Aa
EXAMPLE, sin (36° 40' ± 5') = sin 36° 40' ± .00145 cos 36° 40'
= .5972 ± .0012
cos (36° 40' ± 5') = cos 36° 40' ± .00145 sin 36° 40' = .8021 ± .0009
tan (36° 40' ± 5') = tan 36° 40' ± .00145 sec* 36° 40' = .7445 ± .0023
147. Computation of Error from Tables. This will be illus-
trated by an example. To find sin (36° 40' ± 10') we look in
a table of sines and find sin 36° 50' = .5995, sin 36° 40' = .5972,
sin 36° 30' = .5948 ; the difference between the first and second
is .0023 and that between the second and third is .0024. Choos-
ing the larger we write sin (36° 40' ± 10') = .5972 ± .0024.
This method applies to tables of logarithms, squares, square
roots, etc., in fact to any tables giving the values of a function
VIII, § 148]
MATHEMATICS
185
In practical
4
FIG. 103
for regularly spaced values of the argument. For example, a
measure of the error in log u = log (o ± Aw) is the greater of
the differences log (a + Aw) — log a and log a — log (a — Aw).
Thus to find log (17.4 ± 0.7) we look up in the table log 16.7 =
1.2227, log 17.4 = 1.2405, log 18.1 = 1.2577. The larger
difference is 0.0178 and we write
log (17.4 ± 0.7) = 1.2405 ± 0.0178
148. Errors in Computed Parts of Triangles,
applications, e.g. in surveying,
the given parts of triangles are
subject to errors of measure-
ment and consequently the com-
puted parts are also in error.
Suppose the base AB of the tri-
angle ABC in Fig. 103 is 23.4 ±
0.02, the side AC = 15.6 ± 0.04, and the angle A = 32° 30' ±
10'. Then the altitude
CD = (15.6 ± 0.04) sin (32° 30' ± 10')
= (15.6 ± 0.04) (0.5373 ± 0.0025) §§ 146, 147
= 8.382 ± 0.060 § 141
Again, the area is given by the following computation.
Area = £(23.4 ± 0.02) (8.382 ± 0.06) = 98.069 ± 0.786.
Similarly a measure of the error in any computed part of a
triangle may be found by the foregoing principles of this chapter.
EXERCISES
Calculate the error and the per cent, error of the square in each of
the following numbers. Where no estimate of the error is expressed
the error is supposed to be not greater than 5 in the next place past
the last one written (§ 143).
1. a = 76 ± 0.1 4. a = 432 ± 0.03
2. a = 101 ± 0.4 5. a = 2.46
3. a = 32 ± 0.04 6. a = 13.4
186 MATHEMATICS [VIII, §148
Find the error and the per cent, error in the square root of each of the
following:
7. 121 ± 0.4 11. 216 ± 0.03
8. 169 ± 0.5 12. 165 ± 0.2
9. 144 ± 0.02 13. 43.7
10. 625 ± 0.01 14. 6.45
15. Show that the error of the cube of a ± Aa is ± 3a2-Aa. Hence
find a correct expression for the volume of a cube of side 2.6 ft.
16. Show that the error of the fourth power of a ± Aa is ± 4o3-Aa.
17. Show that the error of the cube root of a ± Aa is Aa/3a2/3.
18. Find by the use of the tables and by use of the results of Ex. 17
the error in the cube root of (a) 1728 ±2; (6) 15625 ±1; (c) 343 ± 0.2
Ans. .005; .0006; .014
19. By applying twice the formula for the error of the .square root
of a ± Aa, show that the error of the fourth root of a ± Aa is Aa/4a.
Find the error in the fourth root of 256 ± 1. Ans. 0.001
20. Find the error by both methods of sin a for each of the following:
(a) 26° ± 10'. (6) 45° ± 15'.
(c) 80° ± 30'. (d) 10° db 10'.
Ans. .0026; .0031; .0015; .0028
21. Find the error of (a) cos a; (6) tan «; (c) ctn a; (d) sec a; (e)
esc a due to an error Aa in a.
22. Find by the use of the tables the error of (a) cos (26° db 25') ;
(6) tan (20° ± 3'); (c) ctn (70° ± 20'); (d) sec (24° ± 10'); (e) esc (46°
± 10').
Ans. (a) .0032; (b) .0009; (c) .0066; (d) .0014; (e) .0039
23. Find the error of the area of the triangle for each of the following :
(a) a = 120 db 0.3 rod, 6 = 144 ± 0.2 rod, y = 47° ± 10'.
(6) a = 80 ± 0.1 rod, b = 160 ± 0.5 rod, y = 89° ± 30'.
(c) a = 40 ± 0.5 rod, 6 = 60 ± 0.3 rod, y = 45° d= 10'.
(d) a = 32 ± 0.4 rod, b = 146 ± 0.8 rod, y = 26° ±5'.
24. If A, B, C denote the angles and a, b, c the sides opposite in a
plane triangle and if a, A, B are known by measurement, then
b = a sin B/sin A.
VIII, §148] SMALL ERRORS 187
Show that the error, called the partial error in b due to a (written A06),
in the computed value of b due to an error Aa in measuring a is, approxi-
mately,
A0& = sin B • esc A • Aa.
Likewise show that
AAb = — a-sin B-csc A-ctn A-&A, and Agb = a cos B-csc A-&B,
and that the total error is, approximately,
A6 = Aa& + AAb + AB&.
Note that A and B are to be expressed in radian measure.
25. The measured parts of a triangle and their probable errors are
a = 100 ± 0.01 ft.; A = 100° ± 1'; B = 40° ± 1'.
Show that the partial errors in the side b are
AQb = ± 0.007 ft.; AAb = ± 0.003 ft.; ABb = ± 0.023 ft.
If these should all combine with like signs, the maximum total error
would be A6 = ± 0.033 ft.
26. If a = 100 ft., B = 40°, A = 80°, and each is subject to an error
of 1 %, find the per cent, of error in b.
27. Find the partial and total errors in angle B, when
a = 100 db 0.01 ft., b = 159 ± 0.01 ft., A = 30° ± 10'.
28. The radius of the base and the altitude of a right circular cone
being measured to 1%, what is the possible per cent, of error in the
volume? Ans. 3%.
29. The formula for index of refraction is m = sin i/sin r, where i
denotes the angle of incidence, and r the angle of refraction. If i = 50°
and r = 40°, each subject to an error of 1%, what is m, and what its
actual and percentage error?
30. Water is flowing through a pipe of length L ft., and diameter
D ft., under a head of // ft. The flow in cubic feet per minute, is
Q = 2356 J- *
IL + 30D
If L = 1000, D = 2, and H = 100, determine the change in Q due to
an increase of 1% in H; in L; in D.
31. The formula for the area of a triangle hi terms of its three sides
188 MATHEMATICS [VIII, §148
is A = Vs(s - a)(s - b)(s — c) where s = \(a + b + c). A tri-
angular field is measured with a chain that is afterwards found to be
one link too long. The sides as measured are 6 chains, 4 chains, and
3 chains respectively. What is the computed area, and what is the
true area?
32. Show that the erroneous area of a field, determined from measure-
ments with an erroneous tape, will be to the true area as the square of
the nominal length of the tape is to the square of its true length.
33. An irregular field is measured with a chain three links short.
The area is found to be 36.472 acres. What is the true area?
34. The acceleration of gravity as determined by an Atwood's
machine is given by the formula: g = 2s /I2. Find approximately the
error due to small errors in observing s and t.
Ans. A-sg = 2As/P; Atg = - 4s/l3.
35. A right circular cylinder has an altitude 12 ft. and the radius
of its base is 3 ft. Find the change in its volume (a) by increasing
the altitude by 0.1 ft., and (6) the radius by 0.01 ft. (c) By increasing
each simultaneously. Ans. (a) 2.83; (6) 3.02; (c) 5.85
36. The period of a simple pendulum is
.2, jr.
\<7
9
Show that AT IT = %Al/l — %Ag/g and hence a small positive error
of k per cent, in observing I will increase the computed time by k/2%,
and a small positive error of k'% m the value of g will decrease the
computed time by k'/2 per cent.
37. Let Wi denote the weight of a body in air, and w-i its weight in
water; then the formula
o —
Wi — Wz
gives the specific gravity of a body which sinks in water. If
wi = 16.5 ± 0.01, w>2 = 12.3 ± 0.02,
find the error in S due to the error in w\; due to the error in w2; the
total error in S; the relative error AS/S.
38. The specific gravity S of a floating body is given by the expression
S= ^— >
VIII, §148] SMALL ERRORS 189
where Wi is the weight of the body in air, w2 is the weight of a sinker
in water, and w3 is the weight in water of the body with sinker attached.
Determine the specific gravity of a body and the probable error if
wi = 16.5 ± 0.01
wz = 182.2 ± 0.03
wa = 176.5 ± 0.02 [RIETZ AND CRATHORNE]
39. To determine the contents of a silo I measure the inside diameter
and height in feet and inches and find D = 8 ft. 2 in., h = 21 ft. 6 in.
Find the error in the computed contents if there are errors AD = ± 0.4
in., A/i = 0.3 in. in the measured dimensions. Ans. 2.22 cu. ft.
40. My neighbor wants to buy the wheat from one of my bins.
The measurements are: length = 12 feet; width = 6 feet; depth of
wheat in bin = 8 ft. I make a mistake however of 1 /4 inch in measur-
ing each 2 feet of linear measure. Find the error of contents in cubic
inches. Find the error in bushels if 2150.4 cu. in. make 1 bushel.
A more accurate value is 2150.42 Find the error due to using 2150.4
instead of 2150.42 Find the error if 2150 is used.
41. I decide to sell to a neighbor by measurement my corn in the
crib. I measure with a yard stick placing my thumb to mark the
end of the yard and holding my thumb in place proceed to measure
beyond it thus making an error of 1/2 inch. My measurements are
length = 30 ft. 3 in.; width = 11 ft. 9 in.; height 13 ft. 6 in. Find the
error in cubic inches due to my method of measuring.
42. The quantity of water in cubic feet per second flowing through
a rectangular weir is given by the formula.
Q =* 3.33 [L - 2h]hw,
where h is the depth in feet of water over the sill of the weir, and L
the length in feet of the sill. Find Q and the error hi Q if L = 26 ± 0.1,
h = 1.6 ± 0.02
CHAPTER IX
CONIC SECTIONS
149. Derivation. The circle, the ellipse, the parabola, and
the hyperbola, are curves which can be cut out of a right circular
conical surface by planes passing through it in various directions.
For this reason, they are called also conic sections. Being
plane curves, however, they can be defined and studied as the
locus of a point moving in a plane under certain conditions.
150. The Circle. The circle is the locus of a point moving at
a fixed distance r from a fixed
point C.
The fixed distance r is called
the radius; the fixed point C is
called the center.
EQUATION OF THE CIRCLE.
Given the center, C(x0, y0) and
the radius, r, of a circle, to de-
duce its equation.
Let P(x, y) be any point on the locus (Fig. 104). Then by
(D§ 45,
CP = V(x — z0)2 + (y — 2/o)2,
and by the definition of the circle CP = r. Hence, squaring
and equating the two values of CP , we find
(1) (x - x0)2 + (y - y0Y = r\
Conversely, let Q(x\, y\) be any point which satisfies (1); i. e.,
~7
FIG. 104
- x0)2 + (t/i - y0)» =
190
IX, §152] CONIC SECTIONS 191
whence
(*i - so)2 + (</! - yoy = r,
but this says that CQ = r, and therefore Q is on the circle.
Therefore (1) is the equation of the circle.
If the center is at the origin, x0 = yo = 0, and the equation
reduces to
(2) x2 + y2 = r2.
151. Equation of the Second Degree. The most general
equation of the second degree in x and y is of the form
(3) a.-c2 + bxy + cy2 + dx + cy + f = 0,
in which the coefficients are real numbers and a, 6, c, are not
all zero. The equation of the circle which we have obtained
is of this form and has always 6 = 0 and a = c. Conversely,
the special equation of the second degree
(4) ox2 + ay2 + dx + ey + f = 0.
is the equation of a circle or of no locus. To show this we
have only to complete the square of the terms in x and of the
terms in y. This process will reduce it to the form of (1) § 150,
as is shown in the next paragraph.
152. Determination of Center and Radius. When the
equation of a circle is given, the center and radius can be found
by transposing the constant term to the right and completing
the square of the terms in x and also of the terms in y.
EXAMPLE 1. Find the center and radius of the circle
x2 + y* - 3x - 2y - 3 = 0.
To reduce this equation to the form (1) we complete the squares as
follows:
(z2 - 3x + ) + (y2 - 2y + ) = 3,
(z2 - 3x + |) + & - 2y + 1) = 3 + | + 1,
(x - f)2 + (y - I)2 = (f)2
192 MATHEMATICS [IX, § 152
Comparing this with the standard equation (1), we see that the center
is at (3/2, 1) and r = 5/2.
EXAMPLE 2. Examine the equation
9x2 + 9?/2 - 6x + 12y + 6 = 0,
We complete the squares as follows :
& + y2 - lx + f y + f = 0,
X* - \x + $ + 2/2 + f y + f = - f + i + f,
(s ~ i)2 + (2/ + §)2 = ~ i
But since the square of a real number is positive (or zero), this shows
that there are no points in the plane which satisfy the given equation.
Therefore it has no locus.
EXAMPLE 3. Examine the equation
225x2 + 225?/2 - 270x - 300t/ + 181 = 0.
We complete the squares as follows:
x2 + y2 - fx - fa + iH = 0,
x2 - fx + A + 2/2 - f !/ + I = - Mi + & + *,
(x - f )2 + (y - I)2 = o.
This shows that the given equation is satisfied by the point (3/5, 2/3)
and by no other point in the plane. This case may be looked upon as
the limiting case of a circle whose center is at (3/5, 2/3), and whose
radius is zero.
EXERCISES
1. Write the equation of the circle determined by each of the follow-
ing conditions.
(0) Center (2, 4), radius = 3. (6) Center (— 1, 3), radius = 5.
(c) Center (-2, -3), radius = 3. (d) Center (3, - 2), diameter = 7.
(e) Center (a, a), diameter a. (f) Center (r, 0), radius = r.
(g) Center (4, 6) passes through the point (0, 3).
(h) Abscissa of center = 1, passes through the points (0, — 1), (0, 7).
(1) The segment from (1, — 3) to (7, 5) is a diameter.
0) Center is on the line x = y, tangent to x-axis at (— 6, 0).
IX, § 152] CONIC SECTIONS 193
2. Write the equation of a circle of radius 6 when the origin is (a) at
the highest point of the circle ; (6) at the lowest point ; (c) at the left-
most point ; (d) at the rightmost point ; (e) when the origin divides
the horizontal diameter from left to right in the ratio 1/3.
3. Determine which of the following equations represent circles;
find the center and the radius in each case.
(a) x2 + y2 = 4x. (6) x2 + y2 = 6y.
(c) x2 + 8y = 4x - y2. (d) 3z2 + 3y2 = 14y.
(e) x* + y2 + 4x + 7 = 0. (/) x2 + y2 + 3x + 5y = 0.
(0) x2 + y2 = 2(y + 4). (h) x2 + y2 = 4(x - 2).
(1) x2 + y2 - 4x - 6y + 9 = 0.
0') *2 + y2 + 101 = 87y - 20x.
(A;) 2z2 + 27/2 + 15y = 12x + 7.
(1) 9x2 + 9y2 + 6y = 24x + 47.
(TO) 16.x2 + 167/2 = 24x + 40y - 34.
(n) 49x2 + 49y2 + 28x - 2% + 9 = 0.
(o) 4a(ax2 + 6x - by) + b2 + 4o(ay2 - ex - cy) + c2 = 0.
4. Show that if the coefficients of x2 and y2 in the equation of a
circle are each + 1, the coordinates of the center can be found by
taking negative one-half the coefficient of x and negative one-half the coef-
ficient of y.
For example, the center of the circle
x2 + y2 - 5x + 4y - 3 = 0
is (5/2, - 2).
5. Find the coordinates of the center of each of the following circles,
by the process of Ex. 4.
(a) x2 + y2 - 4x - 6y + 9 = 0. (d) x2 + y2 - 2x + 4y + 1 = 0.
(6) x2 + y2 + 6x + 4y + 9 = 0. (e) x2 + y2 - 3x + 5y + 3 = 0.
(c) x2 + ?/ - 4y = 0. (/) 2x2 + 2y2 + 4x - 6y + 1 =0.
6. The value of the polynomial P = x2 + y2 — 2x — 4y + 3 at any
point of the xy-plane is found by substituting the coordinates of the
point for r and y in P. Thus at (3, 2), P = 2. Show that all points
at which P is positive lie outside a certain circle, and all points at
which P is negative lie inside the same circle. With respect to this
circle, where are the points (0, 1), (1, 2), (2, 3), (4, 5), (0, 3), (1, 4),
(2, 2)?
14
194
MATHEMATICS
[IX, § 153
153. Translation of Axes. Given a pair of axes OX and
0 Y, a curve C, and its equation in terms of the coordinates
x = OA and y = AP. (Fig. 105.) Move the origin to the
point 0' whose coordinates
referred to the old axes are
(h, k} and draw new axes
O'X' and O'Yf parallel to the
x old axes. The curve is not
moved or changed but the
Y
-»- coordinates of all its points
are changed, and its equation
is changed.
4-°-
FIG. 105
From the figure we see that
and
x = x' + h
y = y' + k.
These equations are true no matter which way nor how far the
origin is moved if the new axes are parallel to the old ones.
These values substituted in the old equation of the curve,
give the new equation. Hence, to find the new equation,
substitute in the old equation, in the place of x, the new x plus
the abscissa of the new origin and in the place of y, the new y plus
the ordinate of the new origin.
EXAMPLE. Translate the origin to the point (1, — 2) on the circle
3x2 + 3?/2 - 5x + 2y = 6.
The new equation is
3(x' + I)2 + 3(y' - 2)' - 5(z' + 1) + 2(y' - 2) = 6,
and this reduces to
3x'2 + 3y'2 + x' - 1<V = 0.
IX, §155]
CONIC SECTIONS
195
154. Parabola. The parabola is the locus of a point which
moves so as to be always equidistant from a fixed point F and a
fixed line L.
The fixed point F is called the focus. The fixed line L is
called the directrix.
155. Equation of the Parabola. Let F be the focus and
RS the directrix of a parabola. (Fig. 106.) Draw FD per-
pendicular to the directrix. The
midpoint 0 between D and F
is on the parabola. Take 0 for
the origin, OF for the re-axis,
and take OY parallel to the
directrix for ?/-axis. Let the
distance DO = OF = p. Then
the coordinates of the focus
are (p, o). Let P(x, y) be any
point on the parabola. By
definition, FP = NP; but
FIG. 106
and
whence
FP = (z - p)2 + y2,
NP = x + p,
Squaring this, we find
(5)
p)2 + 2/2 = x + p.
= ipx.
We have now proved that every point on the parabola satis-
fies the equation (5). It follows that the parabola has no
points on the left of the y-axis, for negative values of x cannot
satisfy the equation (5).
Conversely, let PI(XI, y\) be a point which satisfies (5); then
?/i2 = 4pzi, and (x\ — p)2 = (xi — p)2,
196
MATHEMATICS
[IX, § 155
whence, adding, we have
(xi - p)2 + yS = (xi + p)2,
that is
FP? = N\P?.
Therefore PI is on the parabola. This completes the proof
that (5) is the equation of the parabola.
The parabola is symmetric with respect to the line through
its focus perpendicular to its directrix. This line is called the
axis of the parabola. The point where the parabola crosses
its axis is called its vertex. The chord through the focus
perpendicular to the axis of the parabola is called its latus
rectum. Let the student show that the length of the latus
rectum is 4p.
The parabola y2 = 4px crosses every horizontal line exactly
once, and every vertical line to the right of the 7/-axis twice,
once above and once below the z-axis. The farther the vertical
line is to the right, the farther from the z-axis does the curve
cut it.
By analogy to (5) it is evident that the equations of the
parabolas shown in Figs. 107, 108, 109 are, respectively,
FIG. 107
FIG. 109
(6) t/2 = - 4pz, (7) x* = 4py, (8) z2 = - 4py.
The position of each of these curves should be related to its
equation as follows: yz = 4px is a parabola tangent to the y-axis
at the origin, having its focus on the x-axis to the right. The
student should make similar statements concerning equations
(6), (7), and (8).
IX, §156] CONIC SECTIONS 197
156. Vertex not at the Origin. Each of the equations
(9) (y - k)2 = db 4p(z - h),
(10) (x - h)* = ± 4p(y - k)
represents a parabola whose vertex is at (h, k) and whose axis is
either horizontal (equation (9)) or vertical (equation (10)). For,
on translating the axes to this point they reduce to one of
the types (5), (6), (7), or (8) considered above.
In particular, the equation
(11) y = ax*+bx + c (fl+0)
represents a parabola whose axis is vertical. It is concave up
or down according as a is positive or negative, and the vertex,
focus, and directrix can be found by completing the square of
the terms in x and reducing it to the form (10).
EXAMPLE 1. Locate the parabola y = 2x2 — 8x + 5. Transposing,
2x2 - 8x = y - 5;
dividing by 2,
x2 — 4x = \y — \ ;
adding 4,
x2 - 4x + 4 = \y + f ;
Hence the vertex is the point (2, — 3), and p = |. The parabola is
concave upwards; its focus is | above the vertex, and its directrix is £
below the vertex.
EXAMPLE 2. Examine the equation y = — 2x2 + 4x. We may
write successively the equations
x2-2x=-|y, x2-2x + l =
Hence the vertex is at the point (1, 2), and p = |. The parabola is
concave downwards, its focus is £ below the vertex, and its directrix
is j above the vertex.
Similarly, the equation x = ay2 -\-by-\-c can be reduced to
the type (9) by completing the square of the terms in y, and
from this a sketch of the parabola can be made.
198 MATHEMATICS [IX, § 156
EXERCISES
1. Sketch each of the following parabolas, write the equation of its
directrix, and the coordinates of its focus and vertex:
(a) y* = Sx. (d) %2 = 3z. (g) (x + 3)2 = 5(3 - y).
(b) x* = Gy. (e) 2y* = 25z. (h) x2 = I0(y + 1).
(c) y2 = - 3z. (/) (y - 2)2 = 8(s - 5). (i) (y + 4)2 = - 6z.
2. Sketch each of the following parabolas, and find the coordinates
of the vertex and focus and the equations of the directrix and axis.
(a) y2 - 2y - 4x + 6 = 0. (b) y* + ±y - 6x = 0.
(c) x2 + 4z + 6y - 8 = 0. (d) a;2 - x + y = 0.
(e) 4z2 - 12x + 3y - 2 = 0. (/) 3y2 + 6?/ - 7x - 10 = 0.
3. Sketch the parabolas with the following lines and points as direc-
trices and foci, respectively; and find their equations.
(a) x - 3 = 0, (6, - 3). (b) x = 0, (- 2, - 2).
(c) y + 4 = 0, (- 2, 0). (d) y - 26 = 0, (0, 0).
4. Find the parabolas with the following points as vertices and foci,
respectively.
(a) (0, 0), (2, 0). (6) (1, 1), (3, 1).
(c) (- 2, - 2), (- 4, - 2). (d) (3, 2), (3, 6).
5. Find the parabola with vertex at the origin and axis parallel to
the x-axis, and passing through the point :
(4,1); (2,3); (1,1); (-1,2); (2, - 4); (- 2, - 5).
6. The cable of a suspension bridge assumes the shape of a parabola
if the weight of the suspended roadbed (together with that of the cables)
is uniformly distributed horizontally. Suppose the towers of a bridge
240 ft. long are 60 ft. high and the lowest point of the cables is 20 ft.
above the roadway. Find the vertical distances from the roadway to
the cables at intervals of 20 ft.
7. An arch in the form of a parabolic curve is 29 ft. across the
bottom and the highest point is 8 ft. above the horizontal. What is
the length of a beam placed horizontally across it, 4 ft. from the top?
8. A parabolic reflector is 8 inches across and 8 inches deep. How
far is the focus from the vertex? Ans. 2 in.
IX, §157]
CONIC SECTIONS
199
157. Ellipse. An ellipse is the locus of a point which moves
so that the sum of its distances from two fixed points is constant.
The fixed points F and F' (Fig. 110) are called the foci. Let
the constant distance be 2a;
this cannot be less than F'F.
If it is just equal to F'F the
locus is evidently the seg-
ment F'F. Hence we assume
that 2a > F'F. Take the
x-axis through the foci, and
FIG. 110
the origin midway between
them. Then for all positions of the moving point P, we have
0
(12)
F'P + FP = 2a.
One position of P is a certain point A on the z-axis to the
right of F, and by (12),
and
F'A + FA = 2a
OA = $(F'A + FA) = a.
Similarly the point A' to the left of F' such that A'O = a, is a
point on the ellipse. The points A and A' are called the vertices.
The segment A' A is called the major axis of the ellipse.
Another position of P is a point B on the ?/-axis above 0 and
OB is denoted by b. By (12), we have
F'B + FB = 2a,
and since B is on the perpendicular bisector of F'F,
F'B = FB = a.
Similarly, the point B' below 0 such that B'O = b, is a point on
the ellipse. The distance B'B is called the minor axis. The
200 MATHEMATICS [IX, § 157
intersection of the major and minor axes is called the center of
the ellipse.
The rectangle formed by drawing lines perpendicular to the
major and minor axes at their extremities is called the rectangle
on the axes.
Let a denote the acute angle OFB. Then cos a is called the
eccentricity of the ellipse, and is denoted by e. It is evident
that e = OF JO A. Hence, from the right triangle OFB, we
have
62
(13) 0 < e < I and - = sin2 a = 1 - e2.
a2
Since OF = ae the coordinates of the foci F and F' are (ae, o)
and (— ae, o), respectively.
Then for all positions of the moving point P, by (12), we have
(14) V(z + ae}2 + y2 + V(z - ae)2 + y2 = la.
Transposing the second radical, squaring, and reducing, we find
(15) V(z - ae)2 + y2 = FP = a - ex,
which is the right-hand focal radius.
Similarly, on transposing the first radical in (14), we obtain
the equation
(16) (a + ae)2 + i/ = F'P = a + ex,
which is the left-hand focal radius. Squaring either (15) or (16)
and reducing, we find
(17) (1 - e2)x2 + y2 = a2(l - e2),
whence, by (13),
-j.2 n/2
a* + V = L
We have now proved that every point on the ellipse satisfies
IX, §157]
CONIC SECTIONS
201
(18). It can be proved, conversely, that every point which
satisfies (18) is on the ellipse. Hence we may state the fol-
lowing theorem.
The equation of the ellipse whose semi-major axis is a, whose
semi-minor axis is b, whose center is at the origin, and whose foci
are on the x-axis, is
(19)
W
The numbers a, b, e, are positive, a > 6, e < 1, &2/a2 = 1 — e2.
The coordinates of the foci are (ae, o) and (— ae, 0). The focal
distances of any point on the ellipse are a — ex and a + ex,
respectively.
The equation shows that the curve is symmetric with respect
to the x-axis and also with respect to the ?/-axis. It follows
that the curve is symmetric with respect to the origin. It is
only necessary to plot that part of the curve which lies in the
first quadrant to determine the shape of the whole curve, which
is as shown in Fig. 111.
FIG. 112
The ellipse can be drawn by the continuous motion of a pencil
point by means of a pair of tacks set at the foci and a loop of
string around them as shown in Fig. 112. This1 is the best
method of tracing an ellipse on a drawing board. It can be
used to lay out an ellipse of any desired size on the ground.
Let the student show that the length of the loop of string is
2o(l + e).
202
MATHEMATICS
[IX, §158
158. Auxiliary Circle. A comparison of the equation of the
ellipse (19) with that of the circle
(20)
shows that any ordinate of the ellipse is to the corresponding
ordinate of the circle as b is to a. The
diameter of this circle (20) is the
major axis of the ellipse. For this
reason, the circle (20) is called the
major auxiliary circle, or simply the
auxiliary circle. The points where
any ordinate cuts the ellipse and the
auxiliary circle are called correspond-
ing points.
FIG. 113
159. Area of an Ellipse. Since the horizontal dimensions
of the ellipse and its auxiliary circle are the same, and since
their vertical dimensions are in the ratio b : a, we have
(21)
Area of ellipse 6
Area of auxiliary circle a
Hence, since the area of the circle is known to be iraz, the area,
of an ellipse whose semi-axes are a and b is irab.
160. Projection. If a circle of
radius a be drawn on a plane making
an angle a with the horizontal plane,
then the vertical projection of this
circle on the horizontal plane is an
ellipse whose semi-major axis is a and
whose semi-minor axis is a cos a,
since its ordinates are to the corre-
sponding ordinates of the circle as a cos a is to a.
FIG. 114
IX, §160] CONIC SECTIONS 203
EXAMPLE 1. Reduce the equation of the ellipse 3x2 + 4y2 = 48 to
standard form; find a, b, and c, the coordinates of the foci, the focal
distances to the point (2, 3), and the area of the ellipse.
Dividing through by 48, we find
Then, by comparison with (19), we have o2 = 16 and b2 = 12, whence
a = 4 and b = 2V3. From (13) we find e = £; hence ae = 2. It
follows that the foci are (—2, 0) and (2, 0). The right-hand focal
distance to (2, 3) is a — ex = 3 and the left-hand focal distance is
a + ex = 5. The area is irab = 87rA/3 = 43.53 +
EXAMPLE 2. Reduce the equation 15x2 + 28y2 = 12.
Dividing by 12, we have
_
4 "3 '
or
Hence, by comparison with (19), we have o = §VH and b = f V21.
EXERCISES
1. Find the semi-axes, the eccentricity, locate the foci, and find the
focal distances to any point (x, y) on the curve; construct the rectangle
on the axes, and sketch the curve:
(a) 4x2 + 9i/2 = 36. (b) x* + 25y* = 100.
(c) 9x2 + 25y2 = 225. (d) 9x2 + 16?/2 = 144.
(e) x2 + 2y2 = 4. (/) 6x2 + 9?/2 = 20.
2. In each of the following cases find the values of a, b, e, if they
are not given. Locate the foci, and write the equation of the ellipse.
Construct the rectangle on the axes and sketch the curve.
(a) o = 10, b = 6. (g) b = 2^, e = 1/2.
(6) a = 10, 6=8. (h) a = 5, e = 2/3.
(c) o = 5, b = 3. (t) a = 6, e = 0.
(d) a = 13, e = 12/13. (j) b = 8, e = 3/5.
(e) a = 7, e = 5/7. (Jfc) 6 = 12, e = 5/13.
(/) a = 10, e = 3/5. (06=2, e = 1/3.
204 MATHEMATICS [IX, §160
3. Find the area of each of the ellipses in Ex. 1.
4. Show that any oblique plane section of a circular cylinder is an
ellipse.
5. Find the semi-axes and the area of the section formed by cutting
off a log 14 inches in diameter by a plane making an angle of 60° with
its length.
6. Design a flashing (sheet metal collar) for a four inch soil pipe
projecting vertically through a roof whose pitch is 1/3.
7. A circular window in the south wall of a building is 4 ft. in diam-
eter. Light from the sun passes through the window and falls on the
floor. Find the area of the bright spot at noon, when the angle of
elevation of the sun is (a) 60°, (6) 45°, (c) 30°.
8. An ellipse whose semi-axes are 10 and 9 is in a horizontal position.
Through what angle must it be rotated about its minor axis hi order
that its projection on a horizontal plane shall be a circle.
Ans. 25° 50'.
161. Hyperbola. A hyperbola is the locus of a point which
moves so that the difference of its distances from two fixed points is
constant.
The fixed points are called the foci. Other terms are defined
in a manner analogous to those for the ellipse.
By an analysis similar to that given in § 157 for the ellipse,
it can be shown that the equation of the hyperbola whose semi-
transverse axis is a, whose semi-conjugate axis is 6, whose
center is at the origin and whose foci are on the cc-axis, is
a* V
The curve consists of two branches and is symmetric with
respect to both axes and with respect to the origin, as shown in
Fig. 115. The quantities a, b, and e (= sec a), are positive,
a = b, e > 1, 62/a2 = e2 — 1; the coordinates of the foci are
(ae, o) and ( — ae, o) ; the focal distances to a point on the
right branch are ex — a and ex + a, and to a point on the
left branch, the negatives of these. The diagonals OC and OC",
IX, §162]
CONIC SECTIONS
205
of the rectangle on the axes are called the asymptotes of the
hyperbola, and the curve approaches nearer and nearer to
FIG. 115
them as the moving point recedes from the vertices,
equations of the asymptotes are
The
(23)
and
y- --*.
162. Rectangular or Equilateral Hyperbola. If the semi-
axes of a hyperbola are equal, b = a, its equation reduces to
the form
(24) x2 - y2 = a2.
The rectangle on the axes is a square, the eccentricity is sec 45°
= V2, and the asymptotes are the two perpendicular lines
y = x and y = — x. This is called a rectangular or equilateral
hyperbola. It plays a role among hyperbolas analogous to that
played by the circle among ellipses.
The product of the distances of any point on an equilateral
hyperbola to its asymptotes is constant. For the distance to
the asymptote y = x is (x — y) cos 45°, and the distance to
the asymptote y = .— x is (x + y) cos 45°; hence the product
of these distances is a2 cos2 45° = £a2.
206
MATHEMATICS
[IX, §162
It follows from this property that if the asymptotes of an
equilateral hyperbola be taken for coordinate axes the equation
of the curve will be
(25) xy = a positive constant,
when the branches are in the first and
third quadrants, as shown in Fig. 116;
and the equation will be
(26) xy = a negative constant,
when the branches of the curve are in
the second and fourth quadrants.
FIG. 116
EXAMPLE. Reduce the equation of the hyperbola lQxz — 9?/2 = 144
to standard form.
Dividing by 144, we find
i2 _ yL =
9 16
Hence, by comparison with (22), we have a = 3, b = 4. From b?/a2
= e2 — 1 we find e = 5/3.
It follows that the coordinates of the foci are (5, 0) and (— 5, 0).
The focal distances to a point on the right branch are
ex — a = l(5x — 9) and ex + a = |(5a; + 9).
For example to the point (6, 4V3) they are 7 and 13. The equations
of the asymptotes are
\
rf
A'
y = fx and y = — fz.
To sketch the curve, lay off OA = 3, OB
= 4, Fig. 117, construct the rectangle on the
axes, locate the foci by circumscribing a circle
about this rectangle. Sketch in the curve
free hand in four parts beginning each time
at a vertex, using the asymptotes as guides,
the curve approaching them in distance and direction.
FIG. 117
IX, §162] CONIC SECTIONS 207
EXERCISES
1. Find the semi-axes, the eccentricity, the coordinates of the foci,
the focal distances to the point indicated, the equations of the asymp-
totes; construct the rectangle on the axes and the asymptotes, and
sketch each of the following hyperbolas.
(a) 4x2 - 9?/2 = 36, (Vl3, 4/3).
(6) 4x2-7/2 -8, (-3/2, 1).
(c) 3x2 -i/2 = 9, (3, -3>/2).
(d) 3x2 - 4y2 = 1, (- V?; V5).
(e) 144x2 - 25y2 = 3600, (10, - 12 V3).
(/) 9x2 - 16y2 = 576, (12, 31/5).
(g) 25x2 - y2 = 100, (- V29, 25).
(h) 225x2 - 647/2 = 14400, (17, 28 J).
(i) x2 - y2 = 9, (- 5, 4).
0') x2 - ?/2 = 400, (101, 99).
2. Plot on the same axes the curves xy = c, for c = 1, 4, 6, — 1,
- 4, - 6.
3. Find the equation of the locus of a point which moves so that
the difference of its distances from the two points (1,1) and (— 1, — 1)
is constant and equal to 2.
4. Find the locus as in Ex. 3, when the foci are (a, a) and ( — a, — a)
and the constant is 2a.
5. Find the locus of a point where two sounds emitted simultaneously
at intervals one second apart at two points 2,000 ft. apart are heard at
the same time, the speed of sound in air being 1,090 ft. per second.
6. On a level plain the crack of a rifle and the thud of the bullet
on the target are heard at the same instant. The hearer must be on a
certain curve; find its equation. (Take the origin midway between the
marksman and the target.)
7. By translation of the axes (§ 153) find the equation of the ellipse
.(a) whose foci are (— 4, 2) and (0, 2), and whose eccentricity is 5.
Ans. 3x2 + 4y2 + 12x - IQy = 20.
(6) whose vertices are (—2, 2) and (4, 2), and which passes through
the point (1,4). Ans. 4x2 + 9y2 - 8x - 36?/ + 4 = 0.
(c) whose semi-axes are 5 and 3, whose right-hand focus is at (4, — 4),
and whose left-hand vertex at (— 5, — 4).
Ans. 9x2 + 25r/2 + 200y + 175 = 0.
208 MATHEMATICS [IX, §162
[HINT. Start with the equation of the same curve when its center is
at the origin.]
8. By the method of Ex. 7, find the equation of the hyperbola
(a) whose vertices are (—2,2) and (4, 2), and whose eccentricity is 5/3.
Ans. 16x2 - 9?/2 - 32x + 36y = 164.
(6) whose semiminor axis is 15, whose left-hand vertex is at (— 15, 3)
and whose right-hand focus is at (10, 3).
Ans. 225x2 - 64y2 + 3150x + 384y = 3951.
(c) which passes through the origin and whose asymptotes are the
lines x = 2 and y = 1. Ans. xy = x + y.
163. Intersection of Loci. If a point lies on a curve, its
coordinates must satisfy the equation of that curve. Con-
versely, any pair of values of x and y which satisfy an equation
determines a point on the locus of that equation. If the same
pair of values of x and y satisfies two equations, it locates a
point which is common to the two curves, i. e., a point of inter-
section. Hence, to find the points of intersection of two curves,
solve their equations simultaneously to find all their common
solutions.
EXAMPLE 1. Find the intersections of the line 3x — y = 5 and the
ellipse 4x2 + 9?/2 = 25.
Solving the first equation for y = 3x — 5, substituting this in the
second and reducing, we have
17x2 - 54x + 40 = 0.
We can factor this quadratic by inspection :
(17x - 20) (x - 2) = 0,
whence
xi = 20/17 and x2 = 2.
Substituting these values in the equation 3x — y — 5, gives 7/1 = (—25/17)
j/s =1. Therefore the points of intersection are (20/17, — 25/17), and
(2, 1).
Let the student plot the curves on the same axes and verify these
results.
IX, §163] CONIC SECTIONS 209
EXAMPLE 2. Where does the parabola
3y = x2 - 5x + 12
intersect the ellipse
4x2 + 3y2 = 48?
Substituting the value of y from the first equation in the second
and reducing, we get
x4 - lOx3 + 61x2 - 120x = 0.
Factoring this equation, we have
x(x -3)(x2 - 7x + 40) = 0
and we see by inspection that Xi = 0 and x2 = 3 are roots. The quad-
ratic x2 — 7x + 40 has imaginary roots.
Substituting these values of x in the first given equation we find
?/i=4 and 7/2 = 2. Hence the points of intersection are (0, 4) and (3, 2)-
EXERCISES
Find the points of intersection of the following pairs of curves.
1. x2 + Qxy + 9j/2 =4, 4x + 3y = 12.
Ans. (14/3, - 20/9), (10/3, - 4/9).
2. x2 - if =0, 3x - 2y = 4. Ans. (4, 4), (4/5, - 4/5).
3. y2+x=0, 2y + x = 0. Ans. (0, 0), (- 4, 2).
4. x2 — 5y = 0, x — y = 1.
Ans. x = i(5 ± V5), y = |(3 ± V5).
5. 2x + 3y = 5, 4x2 + 9(/2 + 16x - 18^ -11=0.
Ans. (1,1), (-2,3).
6. x - y + 1 = 0, (x + 2)2 - 4y = 0. Ans. (0, 1).
7. y - 2x = 0, x2 + y* - x + 3y = 0.
Am. (0,0), (- 1, - 2).
8. x - 2y + 4 = 0, 5x2 - 4y2 + 20 = 0. Ans. (I, 2£).
9. y = 2x - 3, 4y2 = (x + 3)(2x - 3)«.
Ans. (3/2, 0), (1, - 1).
10. 4i/2 = x2(x + 1), y2 = x(x + I)2. Ans. (0, 0), (- 1, 0).
11. 2x2 - 3i/2 = - 58, 3x2 + t/2 = 111.
Ans. (5, 6), (- 5, 6), (5, - 6), (-5, - 6)
12. x2 = 4ay, y = 8n3/(x2 + 4oJ). Ans. (± 2a, a).
ITu x2 + y2 = 2, x2 + 7/2 - 6x - 6y + 10 = 0. Ans. (1, 1).
15
210
MATHEMATICS
[IX, §154
164. Straight Line and Conic. The equations of the circle,
parabola, ellipse, and hyperbola, are all of the second degree in
x and y. Conversely, it can be shown that every such equa-
tion represents a conic section, if it represents any curve at all.
Given a straight line and a circle we know that one of three
things will happen, 1) there may be two intersections, 2) there
may be no intersection, or 3) there may be only one point in
common and then the line is a tangent. The same three cases
occur with the intersections of a straight line with any conic
section.*
When we solve simultaneously the equation of a straight line
with the equation of a conic, we may begin by substituting the
value of y from the first equation in the second. The result is
a quadratic equation in x. This quadratic equation (§§ 32, 33),
(27) Ax2 + Ex + C = 0
will have 1) two real roots when B2 — 4 AC > 0, or 2) no real
roots when Bz — 4 AC < 0, or 3) one real root when B2 — 4AC
= 0. These algebraic cases cor-
respond exactly to the geometric
cases enumerated above.
EXAMPLE. Of the three parallel
lines 8z - 9y = 20, 8z - Qy = 30,
and 8x — 9y = 25, the first cuts
the ellipse 4x2 + 9?/2 = 25 in two
points (5/2, 0) and (7/10, - 8/5),
the second does not intersect it at
all, and the third intersects it at
(2, — 1) only, i. e. it is tangent at
that point.
The resulting quadratics are, respectively,
20x2 - 64x + 35 = 0, B2 - 4AC = 1296,
. 20x2 - 96x + 135 = 0, B2 - 4AC = - 1584,
x2 - 4x + 4 = 0, B* - 4AC = 0.
* There is one exception to this rule: any line parallel to the axis of a parabola h&t>
one and only one point in common with the curve, but no such line is a tangent to
the parabola.
IX, §165] CONIC SECTIONS 211
1. Show that one of the three lines 4x + 25 = Wy, 4x + 27 = lOy,
4x + 21 = lOy, intersects the parabola y2 = 4x in two points, another
is tangent, and the third does not intersect it at all.
2. Determine whether the following given lines are tangent, secant,
or do not meet, the corresponding given conic.
(a) x + y + 1 = 0, x2 = 4y.
(6) x - 2y + 20 = 0, x2 + y* = 16.
(c) 2x + 3y =8, y2 = 4x.
(d) x + 2y = 5, x2 + y2 = x + 2y.
(e) 2x = 3y, 4x2 - 3y2 + 8x = 16.
(/)x + 7/ = 8, 4x2+?/2 = 16x.
3. Find the points in which the circle x2 + y2 = 45 is cut by the lines
(a) 2x - y = 15, (6) 2x - y = 0, (c) 2x - y = - 15.
Ans. (a) (6, - 3), (6) (3, 6), (- 3, - 6), (c) (- 6, 3).
4. Find the points in which the circle x2 + yz — 6x — Qy + 10 = 0
is cut by the lines (a) x + y = 2, (6) x + y = 6, (c) x + y = 10,
(d) x - y = 0.
Ans. (a) (1, 1), (6) (1, 5), (5, 1), (c) (5, 5), (d) (1, 1), (5, 5).
5. Find the points in which the parabola 3y = 2x2 — 8x + 6 is cut
by the lines (a) 4x + 3y = 4, (6) 4x + 3y = 6, (c) 4x + 3y = 12.
Ans. (a) (1, 0), (6) (0, 2), (2, - f), (c) (3, 0), (- 1, 5|).
6. Find the points in which the ellipse 3x2 + 4t/2 = 48 is cut by the
lines (a) x + 2y = 0, (6) x + 2y = 4, (c) x + 2y = 8, (d) x + 2y = -4,
(e) x + 2y = -_8.
Ans. (a) (2\3, - V3), (-2^3, V3), (b) (4, 0), (- 2, 3), (c) (2, 3),
(rf) (-4,0), (2, -3), (e) (-2, -3).
165. Tangent and Normal. Focal Properties. The
equation of the tangent to a conic can be found by the principles
of § 164 if the slope of the tangent is known, or if the coordinates
of one point on the tangent are known. This given point may
be the point of contact or some other point through which the
tangent is to pass.
The perpendicular to the tangent at the point of contact is
called the normal to the conic at that point. When the slope
212 MATHEMATICS [IX, §165
of the tangent is known or can be found, the equation of the
normal can be written by the principles of §§59 and 61.
EXAMPLE 1. Find the equation of the tangent to the parabola
yz = 24x which is perpendicular to the line x + 3y + 1 =0.
By (13) § 61, the slope of the required tangent is 3, and by (11) § 59,
y = 3x + b is parallel to it no matter what value b has. Proceeding
to find the points where this line intersects the parabola we are led to
the quadratic equation,
9z2 + 6(6 - 4)z + 62 = 0.
By § 32, this quadratic will have only one root and the line will be
tangent to the parabola, if
36(6 - 4)2 - 3662 = 0.
This gives 6=2; whence, the equation of the required tangent is
y = 3x + 2.
EXAMPLE 2. Find the equation of the tangent and of the normal
to the ellipse 3z2 + 4y2 = 48 at the point (2, 3).
We' first verify that the given point is in fact on the ellipse. Then
by (10) § 59, y — 3 = m (x — 2) is the equation of a line through (2, 3)
no matter what value m has. Solving this simultaneously with the
equation of the ellipse we get the quadratic equation,
(4m2 + 3)z2 + 8m(3 - 2m)x + 4(4m2 - 12m - 3) = 0.
This equation will have only one root and the line will be tangent to
the ellipse if (§32),
64m2 (3 - 2m)2 - 16 (4m2 + 3) (4m2 - 12m - 3) =0,
that is if,
4m2 + 4m + 1 =0,
whence m = — \ and the equation of the required tangent is
y - 3 = - \(x - 2), or x + 2y = 8,
and the equation of the normal (whose slope by (13) § 61 is 2) is
y - 3 = 2(x - 2), or 2x - y = 1.
The normal at any point .P on a parabola bisects the angle
between the focal radius FP, and the line through P parallel
to the axis of the curve.
IX, §165]
CONIC SECTIONS
213
We learn in Physics that light is reflected by a mirror in such
a way that the angle of incidence is equal to the angle of reflection.
Hence, a ray of light emanating from a source at the focus and
FIG. 119
Fia. 120
striking the parabola at any point, will be reflected parallel to
the axis. This is the principle of parabolic reflectors which
are extensively used for head lights. It is easily seen that if
the light be moved slightly beyond the focus, the reflected rays
will tend to illuminate the axis.
The normal at any point of an ellipse bisects the angle between
the focal radii to that point, Fig. 120. It follows that rays
of light, or sound, emanating from one focus F, will after re-
flection by the ellipse, converge at the other focus F". Hence
the name focus. This is the principle of whispering galleries.
EXERCISES.
1. Find the equations of the tangents and normals to the following
curves at the points indicated:
(a) y2 = 8x, (2, 4), (6) x2 - y2 = 64, (10, 6),
(c) x2 + 3J/2 = 21, (3, - 2), (d) 28?/2 = 27x, (2J, 1J).
2. Find the equations of the two tangents which can be drawn to
the parabola y* + 8x = 0 from the point (2, 1) and verify that they
are perpendicular.
3. Find the equations of the tangent and normal to the circle x2 + y*
- 6x - Gy + 10 = 0 at the point (1, 1).
Ans. x + y = 2,x — y = 0.
214 MATHEMATICS [IX, § 165
4. Find the equations of the tangents from the point (9, 3) to the
circle x2 + y2 — 45. Ans. 2x — y = 15, x + 2y = 15.
5. Find the equations of the tangent and normal to the circle xz + y2
= 6z + 2y at the point (2, 4).
Ans. x - 3?/ + 10 = 0, 3x + y = 10.
6. Find the equations of the tangent and normal to the hyperbola
xy = 6 at the' point (2, 3). Ans. 3x + 2y = 12, 2x - 3y + 5 = 0.
7. Find the tangent to the parabola y2 = 12x which makes an
angle of 60° with the z-axis.
8. Find the tangent to the parabola y2 = 6x which makes an angle
of 45° with the re-axis.
9. Find the equations of the tangents to the circle
(a) x2 + y2 = 4 parallel to 2x + 3y + 1 =0,
(b) x2 + y2 = 16 parallel to 3x - 2y + 2 = 0.
10. Find the tangents to the ellipse Qx2 + 16?/2 = 144 which make
an angle of 30° with the x-axis.
11. Find the equations of the tangents to the following conies which
satisfy the condition indicated.
(a) y2 = 4x, slope = 1/2. (/) x2 + y2 = 25, at (4, - 3).
(6) x2 + y2 = 16, slope = - 4/3. (g) x2 + 4?/2 = 8, at (- 2, 1).
(c) 9z2 + l&y2 = 144, slope = - 1/4. (h) x2 - y2 = 16, at (- 5, 3).
(d) x2 = 4y, passing through (0, - 1). (i) 2y2 - x2 = 4, at (2, - 2).
(e) x2 = Sy, passing through (0, 2). 0') 2/2 = %x> at (2» 16)-
12. Determine the condition for tangency of the following pairs of
curves.
(a) x2 — y2 = a2, y — kx. Ans. k = ± 1.
(6) x2 + y2 = r2, 4y - 3x = 4Je. Ans. 16/b2 = 25r2.
(c) 4x2 + y2 - 4x - 8 = 0, y = 2x + k. Ans. k2 + 2k - 17 = 0.
(d) xy + x - 6 = 0, x = ky + 5. Ans. k2 + 14fc + 25 = 0.
13. A parabolic reflector is 12 inches across and 8 inches deep.
Where is the focus?
14. The ground plan of an auditorium is elliptic in shape. The
extreme length is 2,725 ft. and the width is 2,180 ft. By what path
will a sound made at one focus arrive first at the other focus, i. e., directly
or by reflection from the walls? How much sooner if sound travels
1,090 ft. per second?
IX, §166]
CONIC SECTIONS
215
166. Intersection of Conies. Simultaneous Quadratics.
Two conies intersect, in general, in four points. Since their
equations are of the second degree in x and y, this corresponds
to the fact that two quadratics in x and y have, in general, four
solutions. In some cases these solutions are not all real, or
there may be less than four so that
the conies represented intersect in
less than four points.
As shown in Fig. 121a, the hyper-
bola x2 — y- = 5 intersects the ellipse
x2 + 4?/2 = 25 in the four points (3, 2),
(- 3, 2), (-3, - 2), and (3, - 2).
The parabola 4x2 = Qy cuts the same
ellipse only in (3, 2) and (— 3, 2), as
shown in Fig. 1216.
FIG. 121
Certain types of these equations
can be solved by elementary methods. The most important
cases will now be explained.
CASE I. When all the terms (except the constant terms) are
of the second degree in x and y.
Eliminate the constant terms and factor the result into two
linear factors.
X2 - 7/2 = 5,
x2 + 4?/2 = 25.
Multiplying the first equation by 5 and subtracting, we -have
4x2 - 9y2 = 0,
whence
(2x - 3y)(2x + 3y) = 0.
Now solving simultaneously the two pairs of equations
\ 2x — 3y = 0. \ 2x + By = 0.
We find that the solutions of (a) are (3, 2) and ( — 3, — 2) ; and those
of (6) are (3, — 2) and (— 3, 2). It is easy to verify that these are all
solutions of the given equations by actual substitution.
EXAMPLE 1
•{
216 MATHEMATICS [IX, §166
f x2 + 3xy = 28,
EXAMPLE 2. •< '
t 4?/2 + xy = 8.
Eliminating the absolute terms, we have
2X2 - Xy - 282/2 = 0,
whence
(2x + 7y)(x - 4y) = 0.
This gives the two pairs of simultaneous equations
(«\ S 42/2 + W = 8' ™ / 4^2 + sy = 8,
W I 2x + 7y = 0, W 1 x - 4y = 0.
The solutions are therefore (14, - 4), (- 14, 4), (4, 1), (- 4, - 1).
Verify each of these by actual substitution.
SPECIAL METHOD, CASE I. If there is no term in xy the
equations can be solved as linear equations considering z2
and i/2 as the unknowns.
r x2
EXAMPLE. •!
t x2 •
- y1 = 5,
+ 42/2 = 25.
Eliminate x2 and solve for y2. This gives y2 = 4, whence y = ± 2.
Then eliminate y2 and solve for x2. This gives x2 = 9, whence x = ± 3.
Verify that (3, 2), (- 3, 2), (-3, - 2), (3, - 2), are all solutions of
the given equations.
CASE II. When the equations are symmetric in x and y;
i. e., when the interchange of x and y leaves the equations
unchanged.
/• 2 _i_ q.2 "1 O
EXAMPLE. -\
L xy = 6.
Substituting s + t = x and s — t = y in the given equations, we find
2s2 + 2£> = 13,
S2 _ ft - Qf
Solving these equations, we have
s = ±5/2, t = ±1/2.
Hence the values of x and y are x = ± 3 or ± 2, y = ± 3, or ± 2.
Testing these values in the given equations we verify that (3, 2), (2, 3)
(—2, — 3), (—3, — 2), are solutions.
IX, § 166] CONIC SECTIONS 217
EXERCISES
Solve the following pairs of simultaneous equations.
J 4x2 + 4xy - y2 = 7x - y, | x2 - y2 + 16 = 0,
i 4x + 3y = 1. 1 (x + I)2 = (y + I)2.
f 3x2 + 4y2 = 48, J 5x2 + 7y2 = 225,
1 y2 = 3(1 - x). \ 2x + 3y = 9.
f 4x2 + 3xy = 10, f a? + y2 = 153>
I 3y2 + 4xy = 20. 1 xy = 36.
J 2x2 + 2xt/ + 5y2 = 40, f x2 + y2 - x - y = 204,
1 x2 -
f 2x2 + xy + 37/2 = 12, x2 - 2y* + 1 = 0,
y2 + 2x - 2y = 0. xy + x + y = 129.
,
1 2x + y = 0.
y = 0. 2x2 - 3y2 - 23 = 0.
3x2 + 4?/2 = 48, j 4x2 + 6xi/ + 4y2 = 46,
\ X2 + ^ = 34.
x2 + 2xy = 407, f x2 + 2?/2 = 123,
7/2 + 2xy = 455. 1 y2 + 2x2 = 99.
, ,
= 2x + 5y. 1 2x + y = 3.
f x(3x + y) = j/(r/ + 3), f 3x(x - 4) = y - 5,
1 (3x - y)(x - 2y + 3) = 0. 1 2x + y = 30.
2 + 4j/2 = 48,
10 \ 20
V + y2 = 58 U + 3(x + l) =0.
f x2 + xy + y2 = 7, f 2x(2x - 3) = 184,
1 x2 - xy + y2 = 19. 1 9y(2x + y) = - 135.
f x2 - 3xy + y2 = 1, „. f x2 + xy + y2 = 7,
1 (x+y+2)(2x-2/ + l) =0, 1 y2 - x2 = 5.
' x2 + 3xy + y2 - 4x - 2y - 1 = 0,
0O«
15x + 4y - 1 = 0.
26' I x + 2xy + y - 17 = 0.
CHAPTER X
VARIATION
167. Function and Variables. One of the most common
scientific problems is to investigate the causes or effects of
certain changes. The change or variation of one quantity in
the problem is produced or caused by changes in other variable
quantities and is said to depend upon, or be a function of these
variables. Thus the growth of a plant depends on the amount
of certain constituents in the soil, upon the temperature and
humidity of the soil and of the atmosphere, upon the intensity
of the light, and doubtless upon several other variables. The
volume of gas contained in an elastic bag depends on the pressure
and the temperature. The circumference of a circle depends
only on the radius.
To study the effect of any one variable upon a function of
two or more variables, we try to arrange conditions so that
all the other variables of the problem shall remain constant,
while this one varies. Thus we keep the temperature of a gas
constant to find the effect on the volume of a change of the
pressure. To study the effect of carbonate of lime on the
growth of alfalfa, we arrange a series of plats of soil so that
they shall have all the other constituents the same, and all
be subject to the same conditions of light, heat, and moisture,
but differ from plat to plat by known amounts of pulverized
limestone.
The precise form of the relation between a function and its
variables is often very complicated and difficult or impossible
to obtain. Often, the best that can be done is to record the
results of experiments and to study these records to deduce
218
X, §170] VARIATION 219
general effects. Such results are called empirical. This is
especially true of the so-called applications of science to the
processes of nature.
168. Direct Variation. One of the simplest relations that
can exist between two variables is called direct variation.
When the ratio of two variables is constant, each is said to
vary directly as the other.
The statement that y varies directly as x or simply y varies
as x, is written
which means that the ratio y/x is constant and implies the
equation
y = kx,
where k is called the constant of variation.
The circumference of a circle varies as the radius; i. e., C °c r,
or C = kr. The constant of variation is known to be k = 2ir
— 6f , approximately.
169. Inverse Variation. When the product of two variables
is constant, each is said to vary inversely as the other. If y
varies inversely as x, then
xy = k, or y = k I - ) ,
\ * /
whence y varies directly as 1/x, the reciprocal of x.
EXAMPLE. The volume v, of a gas kept at constant temperature,
varies inversely as the pressure p; i. e.
k k
pv = k, or v = - , or p = - .
p v
170. Joint Variation. When a function z depends upon two
variables x and y, in such a manner that z varies as the product
xy, i. e., z = kxy, then z is said to vary jointly as x and y. Thus,
the area of a rectangle varies jointly as the length and the
220
MATHEMATICS
[IX, § 170
breadth. This definition may be extended to functions of three
or more variables. A function /, depending upon several vari-
ables x, y, • - • , z, is said to vary jointly as x and y, • • • , and z,
when it varies as their product, i. e., / = kx-y • - - z. Thus,
simple interest varies jointly as the principal, and the rate,
and the time.
It is evident that if one variable z depends on two other
variables x and y, and if z varies as x when y is constant, and z
varies as y when x is constant, then z varies jointly as x and y
when x and y vary simultaneously. Thus, the area of a triangle
varies as the altitude when the base is constant and varies as
the base when the altitude is constant; therefore the area varies
jointly as the base and the altitude.
This principle is readily extended to functions of three or
more variables. Thus, simple interest varies as the principal
when rate and time are constant, as the rate when principal
and time are constant, and as the time when principal and rate
are constant; therefore simple interest varies jointly as the prin-
cipal, the rate, and the time.
171. Graphic Representation. When y varies directly as
x, the graph of the relation, y = kx, which connects them is
1
X
x*
X
X
S
X
^
^
X
x
f
n-"*1
•^
^
?
o
f*
x
.Y
^
1
2
?
J
5
1
1
FIG. 122
a straight line through the origin whose slope is k. The position
of this line is fixed and the value of k can be determined if we
X, § 171]
VARIATION
221
know one other point on the line, i. e., one pair of simultaneous
values of x and y; and values of y corresponding to any given
values of x, can be read directly from the graph. Then k is the
difference of two values of y divided by the difference of the
corresponding values of x (§ 58).
EXAMPLE. Given that y varies as x and that y = 1 when x = 2.
Plotting the point (2, 1) and drawing the line OP we have the graph
of the relation between x and y. From this we read off y = | when
x = 1, y = 3i when x = 6|, etc. Fig. 122.
When y varies inversely as x, the graph of their relation
xy = k is a rectangular hyperbola asymptotic to the x and y
axes. Here again one point is sufficient to determine k and fix
the curve.
EXAMPLE. Given that volume v, varies inversely as pressure p,
and that v = 12 when p = 3.
Then pv — k, 3-12 = k, pv = 36. The graph of this is shown in
10
10
FIG. 123
Fig. 123 for positive values of p and v. From this we can read off
t; = 6 when p = 6, v = 4 when p = 9, etc.*
* When z varies jointly as x and y, the graph of the relation z = kxy, in three
dimensions, is a surface called a hyperbolic paraboloid with which the student is not
yet familiar.
24
222 MATHEMATICS [X, § 172
172. Determination of the Constant. By substituting in
an equation of variation a set of simultaneous values of the
variables, the constant of variation can be determined.
EXAMPLE 1. Given, y varies as x and y = 8 when x = 10. We
may write y = kx, as in § 168. Substituting x = 10 and y = 8, we
find 8 = ft- 10. From this equation, we can find k by dividing both
sides by 10. This gives k = 4/5. Hence we have y = (4/5)x.
From this equation, the value of y corresponding to any given
value of x can be found. Thus, y = If when x = 2.
EXAMPLE 2. A light is 24 inches above the cen-
ter of a table. The illumination I at any point P of
the surface of the table varies directly as the cosine
of the angle of incidence, i, of the ray LP, and also
•p ^24 inversely as the square of the distance LP = x to the
light. If the illumination at C is 10, what is it at
any point P of a circle of radius 18 inches about C?
SOLUTION. The illumination 7 at any point is
T _ , cos i
:^>?~'
but x = 24 sec i and therefore
i*
-r A/ . .
7 =mco#t.
Since 7 = 10 when i = 0, k = 5760, and hence
7 = 10 cos' i.
Now when CP = 18, cos i = 4/5, and 7 at P is equal to 5.12
EXERCISES
1. Write equations equivalent to each of the following statements;
determine the constant of variation and construct the graph.
(a) y varies as x; y = 7 when x = 3.
(6) y is proportional to x; y = 3 when x = 2|.
(c) y varies inversely as x; y = 1J when x = 1-|.
(d) v varies inversely as p] v =3 when p — 2.
2. Write equations equivalent to each of the following statements
and find the value asked for in each case.
X, § 172] VARIATION 223
(a) y varies as x2 ; y = 81 when x = 3 ; find y when x = 51.
(b) y varies as sin x ; y — 2 when x = 30° ; find y when x = 150°.
(c) u varies inversely as v ; u = 8 when v = 2 ; find u when v = 6.
(d) z varies jointly as x and y;z = Q when x = 2, y = 7 ; find 2 when
x = 4, y = 6.
(e) y varies directly as r and inversely as s ; y = 16 when r = 10, s = 8 ;
find y when r = 7, s = 12.
(/) M varies jointly as x, and y2, and z"1 ; tt = 6 when x = 2, y = 3,
2 = 4; find w when x = 10, ?/ = 15, 2 = 25.
(0) 2 varies directly as x and inversely as y2 ; z = 2 when x = 17, y = 3 ;
find x when 2 = 6, y = 4.
3. Express each of the following by means of an equation.
(a) The volume of a cone varies directly as the height when the radius
of the base is constant.
(b) The volume of a cone varies directly as the square of the radius of
the base when the height is constant.
(c) The number of calories of heat produced when a moving body is
stopped varies jointly as the mass and the square of the velocity.
(d) The squares of the periods of the planets vary directly as the cubes
of their mean distances from the sun.
4. With the statement of Ex. 3 (c) find the heat generated by a mass
of 8 kilograms striking the sun with a velocity of 500 miles per second
if a body weighing one kilogram and moving with a velocity of 380
miles per second on striking the sun produces 45,000,000 calories of heat.
5. The simple interest due on P dollars varies jointly as the amount
P, the rate, and the time. If $1000 yields $30 interest in six months
find the interest on $1200 for eight months at 7%.
6. The amount of heat received by a given planet varies inversely
as the square of its distance from the sun and directly as the square of its
radius.
(a) What is the effect of doubling the distance?
(b) Mercury has a diameter of 3000 miles and is 36 million miles
from the Sun. The Earth has a diameter of 8000 miles and is 93 million
miles from the Sun. Compare the amounts of heat they receive.
7. With the statement in Ex. 3(d), taking the distance of the Earth
from the Sun as the unit and the period of the Earth as the unit of time,
224 MATHEMATICS [X, § 172
find the period of Neptune whose distance from the Sun is known to be
30 units. Ans. 165 yrs.
8. The amount of heat received on a surface of given size varies in-
versely as the distance from the source. One body is twice as far as
another from the source. Compare the amounts of heat received.
9. The resistance offered to a rifle bullet varies directly as the square
of the velocity. Discuss the effect of doubling the velocity.
10. The maximum load P that a rectangular beam supported at one
end will hold without breaking varies directly as the breadth, the square
of the depth and inversely as the length. A beam 4" X 2" X 10'
supports 300 pounds. What load will the same beam support when
placed on edge?
11. The deflection y in a rectangular beam supported at the ends and
loaded in the middle varies directly as the cube of the length, inversely
as the breadth, and inversely as the cube of the depth. A beam 6 inches
wide, 8 inches deep, 15 feet long, supporting 1000 Ibs., has a deflection
of \ inch at the middle. Find the deflection in a beam 4 inches wide,
4 inches deep, 10 feet long, supporting 800 Ibs.
12. With the data of Ex. 10, find the load which a beam 4 inches wide,
6 inches deep, and 16 feet long will support.
13. With the data of Ex. 10, find the longest beam 4 inches wide and
4 inches deep which will support 100 Ibs.
14. With the data of Ex. 10, find the least depth of a beam 12 feet
long and 4 inches wide that will support 400 Ibs.
15. With the data of Ex. 10, find the least breadth of a beam 12 feet
long and 4 inches deep that will support 500 Ibs.
16. Evaporation from a surface varies directly as its area.
(a) Of two square vats the side of one is 10 times that of the other.
What is the ratio of evaporation?
(5) Of two circular vats one evaporates 10 times as fast as the other.
Compare their radii.
17. The distance traversed by a falling body varies directly as the
square of the time. If a body falls 144 feet in 3 seconds, how far will
it fall in 5 seconds?
18. The area of a triangle varies jointly as the length of the base b
and the altitude a. Write the law if the area is 12 square inches when
a = 6 inches and b = 4 inches.
X, §172] VARIATION 225
19. Similar figures vary in area as the squares of their like dimensions.
A new grindstone is 48 inches in diameter. How large is it in diameter
when one-fourth of it is ground away?
20. A circular silo has a diameter of a feet. What must be the
diameter of a circular silo of the same height to hold 4 times as much?
21. What is the effect on the area of a regular hexagon if the length
of each side of the hexagon is doubled.
22. Similar solids vary in volume as the cubes of their like dimen-
sions. A water pail that is 10 inches across the top holds 12 quarts.
Find the volume of a similar pail that is 12 inches across the top.
23. Using the rectangular pack, 432 apples 2 inches in diameter can
be put in a box 12 X 12 X 24. How many 3 inch apples can be packed
in the same box? How many 4 inch apples? Ans. 128; 54.
24. If a lever with a weight at each end is balanced on a fulcrum,
the distances of the two weights from the fulcrum are inversely propor-
tional to the weights. If 2 men of weights 160 Ibs. and 190 Ibs. respect-
ively are balanced on the ends of a 10 foot stick, what is the length
from the fulcrum to each end? Ans. 4^ ft.; 5f ft.
25. A wire rope 1 inch in diameter will lift 10,000 Ibs. What will
one f inches in diameter lift? Ans. 1,406 Ibs.
26. Two persons of the same build are similar in shape; their weights
should vary as the cube of their heights. A man 5| ft. tall weighs
150 Ibs. Find the weight of a man of the same build and 6 feet tall.
Ans. 194.74 Ibs.
27. A man 5 ft. 5 in. tall weighs 140 Ibs., and one 6 ft. 2 in. tall weighs
216 Ibs. Which is of the stouter build?
28. The size of a stone carried by a swiftly flowing stream varies as
the 6th power of the speed of the water. If the speed of a stream is
doubled, what effect does it have on its carrying power? What effect
if trebled?
16
CHAPTER XI
EMPIRICAL EQUATIONS
173. Empirical Formulas. In practice, the relations be-
tween quantities are usually not known in advance, but are to
be found, if possible, from pairs of numerical values of the
quantities discovered from experiment.
In order to determine the relation between these quantities
it is useful to first plot the corresponding pairs of values upon
cross-section paper, and draw a smooth curve through the
plotted points. If the curve so drawn resembles closely one
of the following types of curves:
(1) y = ax + b (straight line),
(2) y = a + bx + ex2 (parabola),
(3) x = a + by + c?/2 (parabola),
(4) y = kxn (parabolic in form),
(5) xy = c (hyperbola),
(6) y = c!0kx (exponential curve),
we assume that the relation connecting the quantities is the
corresponding equation of the above set and it remains to
determine the constants of the equation.
If the plotted data does not fit any of the type curves men-
tioned above, a general method of procedure is to assume an
equation of the type
(7) y = OQ + a\x + a2z2 + • • • + anxn (nth degree curve) .
The coefficients OQ, a\, a%, • • • , an can be found from any n + 1
pairs of values of x and y.
Since the measurements made in any experiment are liable
226
XI, § 174]
EMPIRICAL EQUATIONS
227
to be in error, errors will occur in the computed values of the
coefficients. The curve represented by the final equation will
not in general pass through the points representing the ob-
served data. Some of these points will be on one side and
some on the other. All will be near the curve.
174. Computation of the Coefficients in the Assumed
Formula. In case the plotted points appear to be upon a
straight line, a parabola, or a curve of the nth degree, the
corresponding equation is assumed and we proceed to determine
the coefficients by a method which is illustrated in the following
example.
EXAMPLE 1. Let the observed values of x and y be
X
43
85
127
169
V . .
17
33
49
65
Plotting this data, the points will be seen to lie roughly on a straight
line. Hence we assume a relation of the form
y = ax + b.
y
•10
o
40
120
160
80
FIG. 125
In this equation replace x and y by their observed values. In this way
228 MATHEMATICS [XI, § 174
we obtain four equations connecting a and b :
43a + b = 17,
85a + b = 33,
127a + b = 49,
169a + b = 65.
Two equations are necessary and sufficient for the determination of the
two unknowns a and b. In general if we have more equations than
unknowns the equations are not consistent. That is, the values of
a and b as determined from the first two equations are not the same
as those obtained from the last two, or from the second and third, etc.
Our problem then is to derive from the given set two equations such
that the values of a and b obtained therefrom when used as coefficients
in the assumed equation will give us a straight line which fits closely
the points plotted from the observed data. There are in common use a
number of ways of doing this.
FIRST METHOD. Multiply each equation in turn by the coefficient
of a in that equation and add. This gives one equation containing a
and b. Multiply each equation in turn by the coefficient of b in that
equation and add. This gives a second equation containing a and b.
Using the data in (8) above we find in this way the following equations :
53764a + 4246 = 20744,
424a + 46 = 164.
The solution of these equations for a and 6 gives
(10) a = 0.39, 6 = - 0.34
Substituting these values of a and 6 in the assumed equation, we find
(11) y = 0.39* - 0.34.
SECOND METHOD. When on plotting it is clear that a straight line
is the best fitting curve, draw a straight line among the points so that
about half are above and half below. The y coordinate of the inter-
section of this line with the y-axis can then be read directly from the
graph and gives the value of 6 in the equation y = ax + b. Measure
the angle a that this line makes with the z-axis and then a = tan a.
In ca*se different scales are used on the two axes select two points
(xi, yi) (x2, 2/2) on the line, then
(12) 0=W2-H-Wl.
Xt — Xi
XI, §-174]
EMPIRICAL EQUATIONS
229
THIRD METHOD. Suppose the best fitting curve is a straight line,
i.e. that the equation should be of the form
y = ax + b.
Use for a and b the values obtained on solving the first and last of equa-
tions (8). The straight line so found actually passes through the first
and last points.
If the points are so distributed that one of the forms (2) or (3) § 173
should be used, proceed to find a, b, and c by using the first, middle, and
last points only.
If an equation of degree n [(7), § 173], i.e. an equation of the form
y = QO 4- o,\x + a2x2 + ••• + anxn
should be assumed, use n + 1 points evenly distributed along the
curve. This method gives us always the same number of equations as
there are unknown coefficients to be determined.
FOURTH METHOD. If it is known that the curve is a straight line
through the origin then
y = kx.
Substitute the observed pairs of values of x and y in this equation,
add the resulting equations and solve for k. See § 172.
EXERCISES
1. In the following example a series of observed values of y and x
are given. The variables are known to be connected by a relation
of the form
y = ax + b.
Ans. a = 0.498, b = 0.96
Find a and b.
11. .
6
10.8
16.1
20.6
26
X
10
20
30
40
50
2. The following table gives the density 8 of liquid ammonia at vari-
ous degrees centigrade. Find a relation of the form
5 = at + b.
i.e. determine the values of a and b.
t
0
5
10
15
s
.6364
.6298
.6230
.6160
Ans. d = 0.6364 - 0.0014 t
230
MATHEMATICS
[XI, § 174
3. The following table gives the specific heat s of hot liquid ammonia
at various degrees Fahrenheit. Find a relation of the form s = at + b.
t . .
5
10
15
20
25
s
1 090
1 084
1 078
1.072
1 066
Ans. s = 1.096 - 0.0012i
4. In an experiment to determine the coefficient of friction between
two surfaces (oak) the following values of F were required to give steady
motion to a load W. Plot F and W on squared paper, and find M where
M = F/W. [CASTLE] Ans. M = 3.302
F
5
10
15
20
25
30
35
40
W
2
3
6}
74
10V
111
2
2
5. In the following examples a series of values of x and y are given.
In each case the variables are connected by an equation of the form
y = ax + b. Find a and b.
(a)
(b)
(c)
(d)
Ans. a = 0.33, b = 0.7
In the two following sets of data plot the values of E (Electromotive
force) and R (Resistance), and determine an equation of the form
E = aR + b.
11. .
5
7.8
11.1
14.2
17
X
9
18
27
36
45
Ans.
a = 0.337
, b = 1.9
y
2
3.1
4
5.2
6.2
x
4
8
12
16
20
Ans. a
= 0.2625,
6 = 0.95
y . .
5
6.1
8.2
10
12.1
X . .
1
2
3
4
5
Ans.
a = 1.81,
b = 2.85
y
4
7
11
14
17
X
10
20
30
40
50
XI, § 174]
EMPIRICAL EQUATIONS
231
E. .
0
5
1
1 5
2
9 5
3
3
5
4
45
5
R
7
5
18
28
38
49
t
>9
f
58
8(1
90
100
E..
R....
1
3
4
4
(
[.5
28
6
42
7.75
56
9.5
70
1]
&
I
I
IS
g
.5
8
13.5
112
15
126
6. A wire under tension is found by experiment to stretch an amount
I, in thousandths of an inch, under a tension T, in pounds, as follows:
T. .
10
15
20
25
30
1
8
12.5
155
20
23
Find a relation of the form I = kT (Hooke's law) which best represents
these results.
7. In an experiment with a Weston differential pulley block, the
effort E, in pounds, required to raise a load W, in pounds, was found
to be as follows:
W. .
10
90
30
40
50
60
70
80
90
100
E
•H
4f
61
74
9
10|
m
13f
15
16i
Find a relation of the form E = aW + b.
8. If 6 denotes the melting point (Centigrade) of an alloy of lead
and zinc containing x per cent, of lead, it is found that
x
40
50
60
70
80
90
e
186
205
226
250
276
304
Find a relation of the form 0 = a + bx + ex2.
9. The readings of a standard gas-meter S and those of a meter T
being tested on the same pipe line were found to be
S .
3,000
3 510
4022
4 533
T
0
500
1,000
1,500
Find a formula of the type T = aS + b which best represents these
data. What is the meaning of a? of fe?
232
MATHEMATICS
[XI, §174
10. An alloy of tin and lead containing x per cent, of lead melts at
the temperature 0 (Fahrenheit) given by the values
X
25
50
75
e
482
370
356
Determine a formula of the type 6 = a + bx + ex2.
11. A restaurant keeper finds that if he has G guests a day his total
daily expenditure is E dollars, and his total daily receipts are R dollars.
The following numbers are averages, obtained from the books
G. .
210
270
320
360
E
16.7
19.4
21.6
23.4
R
15.8
21.2
26.4
29.8
Find the simple algebraic laws which seem to connect E and R with
G. [R = mG; E = aG + b.] What are the meanings of m, a, and fe?
Below what value of G does the business cease to be profitable?
12. The following statistics (taken from Bulletin 110, part 1 of the
Bureau of Animal Industry, U. S. Dept. of Agriculture) give the changes
in average egg production between 1899 and 1907:
Year.
Birds
Competing
per Year.
Eggs Laid.
Actual
Average
Production.
Added to
Actual
Average.
Modified
Average
Due to
Abnormal
Conditions.
1899-1900
70
9,545
136.36
0
136.36
1900- 01
85
12,192
143.44
0
143.44
01- 02
48
7,468
155.58
0
155.58
2- 3
147
19,906
135.42
23.73
159.15
3- 4
254
29,947
117.90
11.24
129.14
4- 5
283
37,943
134.07
0
134.07
5- 6
178
24,827
140.14
13.95
154.09
6- 7..
187
21.175
113.24
29.53
142.77
With the actual and modified averages in hand we may inquire:
what has been the general trend of the mean annual egg production
during the period covered by the investigation? The clearest answer
to this question may be obtained by plotting the figures in the fourth
and sixth columns of the above table, and then striking through each
XI, §175]
EMPIRICAL EQUATIONS
233
of the two zigzag lines so obtained the best fitting straight line, as
determined by the method of least squares. The equations of the
two straight lines are as follows:
actual averages:
modified averages:
y = 148.48 - 3.10z,
y = 144.13 + 0.043z.
In these equations y represents the mean annual egg production and
x the year. The origin for x is at 1898-99. Verify these two equations.
13. The following table, taken from the same bulletin, gives the
percentage of the flocks laying (a) less than 45 eggs, and (6) 195 or more
eggs in a year.
Annual Egg Production.
1899-
1900.
1900-
•81.
•01-'02.
•02-
'03.
•03-
'04.
•04-'05.
'05-
•06.
•06-
•07.
Less than 45 in % . .
4.29
1.18
0
1.36
6.70
7.07
0.56
4.81
195 or more in % ...
4.29
10.60
18.75
6.12
0.79
12.71
5.06
0
Plot this data, using years for abscissa and percentages for ordinates,
making two curves and find by the method of least squares the best
fitting lines.
Poor layers: y = 1.795 + 0.3225x.
Good layers: y = 11.639 - 0.966x.
Interpret the sign of the coefficient of x in each equation, and give the
meaning of the constant term in each equation.
175. Substitution. If on plotting the given values of x
and y the plotted points are seen to be approximately on a
branch of a rectangular hyperbola with vertical and horizontal
asymptotes we assume a relation of the form
(14) (x - a)(y -b) =c,
where (a, 6) are the coordinates of the intersection of the asymp-
totes, and proceed to determine a, b, and c.
In many of the cases in which this form appears both a and 6
are zero and the equation (14) becomes
y = c/x.
234 MATHEMATICS [XI, §175
In some cases a is zero and equation (14) becomes
y = b + c/x.
There are many curves which resemble closely the curve
given by equation (14), but whose equation is somewhat differ-
ent. In order to determine whether (14) is the best equation
to represent the plotted data, obtain from the figure an approxi-
mate value of a. In many cases a = 0. Make the substitution
l/(x — a) = u and plot the new points (u, y). If these are
approximately upon a straight line then
y — b + cu*
and equation (14), in one of its forms, is the proper relation to
assume.
If on plotting the observed values of x and y the plotted points
appear to be on a parabola with axis parallel to one of the axes
and vertex on that axis then call that axis the ?/-axis and assume
(15) y = a + bx2. '
The determination of the coefficients a and b can be reduced
to that of finding the coefficients in the linear form
y = a + bu,
where u = xz. As a check that (15) is the correct form to as-
sume plot pairs of values of u and y. If these points appear to
be on a straight line then equation (15) is the correct form to
assume.
EXAMPLE. The distance s, in feet, passed over by a falling body in t
seconds is found by experiment to be
s
0
0
5
.5
16
1
35
1.5
65
2
t
Find a law connecting s and t.
*This is sometimes called the reciprocal curve.
XI, §175]
EMPIRICAL EQUATIONS
235
Upon plotting this data, the points are seen to fall on a parabola with
vertex upward and at the origin. This suggests that we assume the
relation of the form
s = aP.
As a check on this assumption we plot the points («, s) given in the
following table:
5
.5
.25
16
1
1
35
1.5
2.25
65
2
4
These points are approximately
upon a straight line s = au. The de-
termination of a by the method of least
squares gives a = 16.9, whence
s = 16.9<2.
EXERCISES
10
In
1. The following data on the relation
of temperature to insect life gives the
number of days at a given temperature
to complete a given stage of develop-
ment and is taken from Technical Bul-
letin, No. 7, Dec. 1913 of the New
Hampshire College Ag. Exp. Station,
each case the plotted points are on a curve of the type
FIG. 126
y — b = c/x (x = days, y = temperature).
The term developmental zero is used to designate that point at
which an insect may be kept, theoretically at least, without change for
an indefinite period. The developmental zero for the insect and stage
approximates the point where the reciprocal curve (calculated from the
time factor) intersects the temperature axis. (6 = developmental
zero.)
For each set, plot the data, and the reciprocal curve; find the
developmental zero, and obtain an equation of the form y — b = c/x
connecting the data.
236 MATHEMATICS [XI, §175
(a) Malacosoma americana, pupal stage. Developmental zero = 11°C.
y .
32.4
32
26.1
20
16
X
9.7
10
13.2
22.5
54
(b) Tenebrio molitor, incubation of eggs. Developmental zero = 9.5° C.
31.1
26.6
21
11 6
X
6
7.4
12.1
57
(c) Leptinotarsa decemlineaia, incubation of eggs. Developmental zero
= 6°.
32.2
26.7
18.6
X
3
3.9
6.3
(d) Toxoptera graminum, birth to death. Developmental zero = 5°
V •
32.5
26.5
21
15.5
10
x
10
12
20
30
58
(e) Incubation period of eggs of codling moth. Developmental zero
= 6°.
y
28
25
22
18
16
15
X . ...
4.5
6
7
9
12
16
(/) Toxoptera graminum, birth to maturity. Developmental zero = 5°.
11
26'.5
21
15.5
10
x
6.5
9
15
32
Ans. y — 5 =
150 (nearly)
2. Determine a relation of the form y = a + fez2 that best represents
the values.
X
1
2
3
4
5
6
11. •
14.1
25.2
44.7
71.4
105.6
147.9
3. The pressure p, measured in centimeters of mercury, and the
volume v, measured in cubic centimeters, of a gas kept at constant
temperature, were found to be as follows.
XI, §176]
EMPIRICAL EQUATIONS
237
V
145
155
165
178
191
V • •
117.2
109.4
102.4
95
88.6
Determine a relation of the form pv = k.
4. Find a formula of the type u = kv2 that best represents the
following values.
u
3.9
15.1
34.5
61.2
95.5
137.7
187.4
V
1
2
3
4
5
6
7
5. If a body slides down an inclined plane, the distance «, in feet,
that it moves is connected with the time t, in seconds, after it starts
by an equation of the form s = kP. Find the best value of k con-
sistent with the following data.
s
2.6
10.1
23
40.8
63.7
t
1
2
3
4
5
Am. k = 2.556.
6. Find approximately the relation between s and t from the fol-
lowing data.
s
3.1
13
30.6
50.1
79.5
116.4
/
.5
1
1.5
2
2.5
3
176. Logarithmic Plotting. In case the plotted points (x, y}
appear to lie on one of the parabolic or hyperbolic curves of the
family
(16) y = bxm
there is a distinct advantage in taking the logarithm (base 10)
of both sides:
(17) log y = m log x + log 6,
and then substitute
(18)
X for log x, Y for log y, B for log b
238
MATHEMATICS
[XI, §176
so that the equation (17) becomes,
(19) F = mX + 5.
If the values of x and y are tabulated in columns, and their
logarithms X and Y are looked up and written in parallel
columns opposite, then the points (X, Y) should lie on a straight
line to justify the assumption of equation (16). And if they do
lie fairly on a line, its slope and y-intercept determine the constants
m and b of equation (16). This can often be done graphically
from the drawing with sufficient accuracy, but if greater ac-
curacy is required they can be determined from the data by
least squares.
EXAMPLE.
X.
y-
A'= log z.
r=iogi/.
2
4
8
16
6.000
24.60
70.80
338.8
0.3010
0.6020
0.9030
1.2040
0.7782
1.3909
1.8500
2.5299
FIG. 127
and these values in equation (16) give,
y = 1.574X1-"*
The points (X, F) lie nearly on
a line BD, Fig. 127. Graphically,
we scale off from the figure,
B = the t/-intercept OB = 0.2,
CD
m — the slope = -^
-DO
-47-188
-25-1'88
By least squares, putting the
data into equation (19), we find
B = 0.1970 = log 6;
hence
b = 1.574,
m = 1.914,
XI, §177]
EMPIRICAL EQUATIONS
239
In case the quantities x and y are connected by a relation of
the form
(20) y = cWkx,
it is advantageous to compute Y = log y and plot x and Y.
If these new values when plotted appear to be on a straight line
we write
(21) F = kx + log c
and determine k and log c by the method of least squares.
177. Logarithmic Paper. Paper, called logarithmic paper,
may be bought that is ruled in lines whose distances, horizontally
and vertically, from a point 0 are proportional to the logarithms
of the numbers 1, 2, 3, etc.
Such paper may be used instead of actually looking up the
logarithms in a table. For if the given values be plotted on this
new paper, the resulting "figure is identically the same as that
obtained by plotting the logarithms of the given values on ordi-
nary squared paper.
The use of logarithmic paper is however not essential; it is
merely convenient when one has a large number of problems
of this type to solve.
EXERCISES
1. A strong rubber band stretched under a pull of p kg. shows an
elongation of E cm. The following values were found in an experiment:
p
05
1 0
1.5
?n
?5
30
3 5
40
45
5.0
E
0 1
03
Of>
0.9
1 3
1 7
?,?,
2.7
33
3.9
Find a relation of the form E = kpn. Ans. E = .Sp1-'
2. The amount of water A, in cu. ft., that will flow per minute
through 100 feet of pipe of diameter d, in inches, with an initial pressure
of 50 Ibs. per sq. in., is as follows:
d
]
1.5
2
3
4
6
A
4.88
13.43
27.50
75.13
152.51
409.54
Find a relation of the form A = kdn.
Ans. A = 4.88eP-473
240
MATHEMATICS
[XI, §177
3. In testing a gas engine corresponding values of the pressure p,
measured in Ibs. per sq. ft., and the volume v, in cubic feet, were obtained
as follows:
V
7.14
7.73
859
J> • •
54.6
50.7
45.9
Find a relation of the form p = kvn. Ans. p = 387.6#~-938
4. Find a relation between p and v from the following data:
v
6.27
534
3 15
V • •
20.54
25.79
54.25
Ans. pvlM = 273.5
5. The intercollegiate track records for foot-races are as follows,
where d means the distance run, and t the record time:
d
100 yds.
220 yds.
440 yds.
860 yds
1 mi
2 mi
t
0:094
0:214
0:48
l-54f
4-151
9-241
Find a relation of the form t = kdn. What should be the record time
for a race of 1,320 yds.?
6. In each of the following sets of data find a relation of the form
y = kxn connecting the quantities.
(a)
(6)
V
1
2
3
4
5
v . .
137.4
62.6
39.6
286
226
u
12.9
17.1
23 1
285
30
v
63.0
27
13 8
85
6 9
(d)
e
§2°
212°
390°
K<7(
1°
7
50°
1100°
c
f
5.09
2.69
2.90
9 C
IS
s
09
3 28
X . .
1 ^
2 5
3 5
4 5
[
5
6 L
7 5
8 5
V. .
3(1
5
3 92
4 65
5 30
5 82
6 '
10
6 85
7 25
Ans. y = 2.5x1/2.
XI, § 177]
EMPIRICAL EQUATIONS
241
7. Draw each of the following curves:
(a) y = x1/2. (6) y = 2x2.
(c) y = 2s1/2. (d) y = 3X3/2.
(e) y = 8x~3/2. (/) y = 1.5x2'3.
(g) y = 9.2X-2/3. (h) y = log x2/3.
(i) 2/ = 10. tf) y = 2-10*2.
(fc) y = 10*/2. (Q y = 10*+2.
8. Find an empirical equation connecting the x and y values given
in the following tables.
(a)
x
0.2
0.4
0.6
0.8
•u . .
3.18
3.96
5.00
6.30
Ans. y = 2.51(10"*).
(d)
x
0.2
0.4
0.6
0.8
y .
5.8
4.4
3.4
2.6
X
0
14.4
28.4
42.2
y . .
180
24
3
0.7
X
0
41.4
83.6
126.2
y
180
92
46
22
9. Given age in years and diameter in inches of a tree If feet from
the ground as follows.
Age
19
58
114
140
181
229
Diameter
3
7
13.2
17.9
24.5
33
Plot the data and determine a relation of the form y = kxn.
10. Given age in years and height in feet of a tree as follows :
Age
13
34.4
50.5
218
247
Height
13.4
27.5
38.4
72.5
73
Plot the data and determine a relation of the form y = kxn.
11. Following are vapor pressures, in mm. of mercury, of methyl
alcohol at various temperatures:
17
242
MATHEMATICS
[XI, § 177
t
6
13
21
30
40
' 42
64
100
160
260
Represent these by an empirical formula.
12. The safe load W in tons of 2000 Ibs. for a beam 4 inches wide when
the distance between the supports is 12 feet is given by
W = KD\
where D is the depth in inches. Find K from the following table :
D. .
10
12
14
16
18
W
1.85
2.67
3.63
4.74
6.00
13. Plot a curve from the following data, find its equation, and esti-
mate the price of 36-inch pipe.
Diameter of Sewer
Pipe . . .
8
10
19!
14
16
18
20
22
24
Price in i per linear
ft
?6
7,1
30
36
50
68
93
125
150
14. Plot a curve from the following data, find its equation, and
estimate the pressure for a velocity of 110 miles per hour. The pressure
is given in pounds per square foot of cross section of the first car in a
train of ten, and the velocity in miles per hour.
V. . . .
p —
32
.97
37
1.35
43
1.80
48
2.25
55
3.32
64
4.18
68
4.83
83
6.75
88
7.72
91
8.37
95
9.01
CHAPTER XII
THE PROGRESSIONS
178. Arithmetic Progression. A sequence of numbers in
which each term differs from the preceding one by the same
number is called an arithmetic progression (denoted by A. P.).
The common difference is that number which must be added
to any term to obtain the next one.
To determine whether or not a given sequence is an arith-
metic progression we find and compare the successive differences
of consecutive terms. Thus
3, 10, 17,24,31, •••
is an A. P. in which the common difference is 7.
5,8, 11, 15, 18, •••
is not an A. P.
179. Notation. The following symbols are commonly used
to denote five important numbers, called elements, which are
considered in connection with arithmetic progressions.
a or ai = the first term
n = the number of terms
I or an = the last or rith term
d = the common difference
5 or sn = the sum of the first n terms
180. Formulas. If the terms of an arithmetic progression
are written down and numbered as follows,
Terms : a, a + d, a + 2rf, a + 3d,- • • •
Number of term : 1, 2 , 3 , 4 , •••
243
244 MATHEMATICS [XII, § 180
we observe that the coefficient of d in each term is one less than
the number of the term. Hence for the last or nth term we have
(1) I = a + (n- i)d
We may write the progression in which I is the last term as
follows:
a, a + d, a + 2d, • • • , I — Id, I — d, I.
The sum of an arithmetic progression is found by adding the
n terms together:
s = a + (a + d) + (a + 2d) + • • • + (I - 2d) + (I - d) + I.
Inverting the order of the terms
s = I + (I - d} + (I - 2d) + • • • + (a + 2d) + (a + d) + a.
By addition of corresponding terms, we have
2s = (a + 1) + (a + Z) + (a + Z) + • • • + (a + 1) + (a + I)
= n(a + 0-
EXAMPLE. Find the sum of an arithmetic progression of six terms
whose first term is 4 and whose common difference is 2.
Since n = 6, we have I = 4 + 5-2 = 14. Hence s = |(4 -}- 14)
= 54.
Given any three of the elements a, n, I, d, s, either of the other
two can be found by substituting in (1) or (2) and solving. If
n is to be found, the given elements must be such that the
formula will be satisfied by a positive integral value of n.
EXAMPLE. Given d = 5, Z = f, s=— -1/; find a and n. Sub-
stituting in (1) and (2), we have
3- i-lfci i\ 15-1
W 2 ~ a + 2 (n ~ l)' ~ 2" ~ 2
XII, §181] THE PROGRESSIONS 245
Eliminating a,
n* - 7n - 30 = 0.
Solving for n,
n = 10 or - 3.
The value n = — 3 is inadmissible. Substituting n = 10 in (3), we
obtain a = — 3. Hence n = 10, a = — 3, and the arithmetic pro-
gression is - 3, - 2\, - 2, - 1|, - 1, - i, 0, i, 1, H.
181. Arithmetic Means. The terms of an arithmetic progres-
sion between the first and last terms are called arithmetic means.
Between any two numbers as many arithmetic means as desired
can be inserted. To do this we can use equation (1) to compute
the common difference d, for a and / are known and n is two more
than the number of terms to be inserted. Then the required
means are. a + d, a + 2d, etc.
The problem of inserting one arithmetic mean between two
numbers is the same as the problem of finding the average of
two numbers. If m is the average of o and b, then
and a, m, b form an arithmetic progression. For this reason
m is called the arithmetic mean of a and 6.
EXAMPLE. Insert 4 arithmetic means between 7 and 20. Here
a = 7, I = 20, n = 6. Substituting these values in (1), we have
20 = 7 + 5-d, whence d = 2|. Hence, the required means are 9$,
12i, 14f, 17|.
EXERCISES
Determine which of the following suites of numbers form arithmetic
progressions.
1. 1, 7, 9, 12, ». 2. x, x\ 3x, —
3. 5, 8, 11, 14, ••• 4. a - 26, a, a + 26, •••
5. 3, 7, 11, 15, ••• 6. 4, 2, 0, - 2, •••
7. 2, 4, 6, 9, - 8. 5, 3, 1, - 1, •••
246 MATHEMATICS [XII § 181
Find I and s for the following progressions :
9. - 2, - 6, - 10, ••• to 17 terms.
10. 3, 10, 17, ••• to 50 terms.
11. 5, 7.5, 10, ••• to 36 terms.
12. 2, |, V°> 4, ••• to 48 terms.
13. Solve formula (1) for a, n, and d in turn.
14. Solve formula (2) for a, n, and I in turn.
15. Given n = 20, a = 1, d = 7 ; find I and s.
16. Given n = 1000, I = 500, d = £ ; find a and s.
17. Given n = 16, a = 2, Z = 3 ; find d and s.
18. Given a = 2, I = 3, s = 100 ; find n and d.
19. Given n = 9, a = 1, s = 37; find d and Z.
20. Given a = 4, d = 0.1, Z = 8; find n and s.
21. Given n = 10, d = 0.2, s = 78 ; find a and Z.
22. Given n = 12, I = - 3, s = 140 ; find a and d.
23. Given d = 3, I! = 22, s = 87 ; find a and n.
24. Given a = 8, d = 8, s = 80 ; find I and n.
25. Insert 3 arithmetic means between 1 and 17.
26. Insert 4 arithmetic means between 2 and 18.
27. Insert 5 arithmetic means between 3 and 38.
28. Insert 6 arithmetic means between 4 and 6.
29. Eight stakes are to be set at equal distances between the two cor-
ners of a 60 ft. lot. How far apart must they be? Ans. 6 ft. 8 in.
30. I desire to close up one side of crib 12 feet 4 inches high, with 6
inch boards. I have just 21 boards. I desire to leave a 1 inch crack
at top and bottom. How far apart must I place the boards to have
them equally spaced? Ans. 1 inch.
31. At the end of each year for 10 years a man invests $200 on which
he collects annual interest at 6%. Find the total interest received.
Ans. $540.
32. The population of a certain town has made a net gain of the same
number of people each year for the last 30 years. In 1893 it was 1523 ;
in 1906 it was 2212. What was it in 1890 ? in 1902 ? in 1916 ? Predict
the population for 1925.
33. What will it cost to erect the steel work of a 20 story building at
$3000 for the first story and $250 more for each succeeding story than
for the one below? Ans. $107500.
XII, § 183] THE PROGRESSIONS 247
34. I drop a rock over a cliff 400 ft high. How long before I hear it
strike bottom if it falls 16 ft. the 1st second, 48 ft. the 2d second, 80 ft.
the 3d second, etc., and sound travels 1090 ft. per second in air?
Ans. 5f sec. nearly.
35. A ball rolling down an incline goes 2 ft. the first second and 6 ft.,
10 ft., 14 ft., respectively in the next three seconds, starting from rest.
How far will it roll in 15 seconds? Ans. 450 ft.
36. A clock strikes the hours and also 1, 2, 3, 8, respectively, at the
quarter hours. How many strokes does it make in a day ? Ans. 422.
37. A farmer is building a fence along one side of a quarter section.
The post holes are dug one rod apart and the posts are piled at the first.
How far will he walk to distribute them one at a time and return to set
the first one? Ans. 20| miles.
38. Find the sum of all multiples of 7 less than 1000. Ans. 71071.
39. Find two numbers whose arithmetic mean is 11 and the arith-
metic mean of their squares is 157.
40. Show that if an A. P. has an odd number of terms the middle term
is the arithmetic mean of the first and last.
41. If the sum of any number of terms of the A. P. 8, 16, 24, ••• be
increased by 1, the result is a perfect square.
182. Geometric Progression. A sequence of numbers in
which each term may be found by multiplying the preceding
term by the same number is called a geometric progression
(denoted by G. P.). The constant multiplier is called the
common ratio. Thus
3, 15, 75, 375, •••
is a G. P. in which the common ratio is 5.
The elements of a geometric progression are the first term a
or oi, the number of terms n, the last or nth term / or an, and the
sum s or sn of the first n terms.
183. Formulas. If the terms of a geometric progression be
written down and numbered as follows,
Term : a, ar, ar2, ar3, • • •
Number of term: 1, 2, 3, 4 , •••
248 MATHEMATICS XII, § 183
we see that the exponent of r in each term is one less than the
number of the term. Hence for the nth or last term we have
(4) I = ar"-1
The sum of the first n terms of the preceding geometric pro-
gression is
s = a + ar + ar2 + ••• + arn~l
Multiplying both sides by r,
sr = ar + ar2 + ar3 + • • • + arn
By subtraction, we have
sr — s = arn — a.
Solving the last equation for s, we get
r - 1 1 - r
From (4) we obtain rl = arn. Hence (5) may also be written
a - rl
(6)
1 - r
The two fundamental formulas (4) and (6) contain the five
elements a, I, n, r, s, any two of which may be found if the
other three are given.
EXAMPLE 1. Find s if a = 1, n = 7, r = 4.
Substituting these values in (5), we get
47 - 1 16384 - 1
S = T— ^ -3- =5461.
184. Geometric Means. If three positive numbers are in
geometric progression the middle one is said to be the geo-
metric mean of the other two. It is easy to see that the geo-
metric mean of two numbers is the square root of their product.
Thus 3 is the geometric mean of 2| and 4.
If several numbers are in geometric progression all the inter-
XII, § 184] THE PROGRESSIONS 249
mediate terms are said to be geometric means between the first
and last terms. We can insert as many geometric means as we
wish between any two positive numbers. To do this we use
equation (4), § 183, to compute r; a, I, and n being known.
Then the desired means are ar, ar2, ar3, etc.
EXAMPLE. Insert three geometric means between 4 and 16. Since
16 is to be the 5th term we have a = 4. ar4 = 16, whence r4 = 4 and
r = V2 ; hence the five terms are 4, 4V2, 8, 8 V2, 16.
EXERCISES
Which of the following sets of numbers form geometric progressions ?
1. 3, - 6, 12, - 24, ••• 2. 4, 6, 9, 13.5, •••
3. 7, 18, 40, ••• 4. 8, 12, 18, 26, •••
5. a, 2a, 3a, 4a, ••• 6. a, a?, a3, •••
7. V3 - 1, V2. V3 + 1, - 8. 8, 4, 2, 1, •••
9. a, - a2, a3, - a«, ••• 10. \/2, 2, 2^2, 4, —
11. \/2, V6, 3^2, ••• 12. 9, 3, 1, i —
13. Solve formula (4) for a, n, and r in turn.
14. Solve formula (6) for a, I, and r in turn.
15. Given a = 2, r = 3, n = 12 ; find I and s.
16. Given a = 3, r = 5, n = 10 ; find I and s.
17. Given a = 4, n = 6, s = 252 ; find I and r.
18. I = 486, a = 2, n = 6 ; find r and s.
19. Given a = 15, r = 3, I = 3645 ; find n and s.
20. Given n = 5, r = \, I = 512 ; find a and s.
21. Insert two geometric means between 2 and 128.
22. Insert 3 geometric means between 2 and 162.
23. Insert 2 geometric means between "N/2 and 108.
24. What is the geometric mean between a/6 and 6/a?
25. Find the 6th term and the sum of the series 2, 4, 8, •••.
26. It takes 32 nails to shoe a horse. A blacksmith agrees to drive
them as follows : 2 cents for the first, 4 cents for the second, 8 cents for
the third, etc. What is the total cost? Ans. $85,899,345.90
27. Find the amount of $500 in 5 years at 6% compounded annually j
compounded semiannually. Ans. $669.10; $672.45
250 MATHEMATICS [XII, § 184
28. In how many years will $100 amount to $200, interest at 8%
compounded annually ? In how many years with interest at 6 % com-
pounded annually?
Ans. 9 years approximately ; 12 years approximately.
29. A man promises to pay $10,000 at the end of 5 yr. What amount
must be invested each year at 6 % compound interest so that at the end
of the time the debt can be paid?
30. A premium of $104 is paid to an insurance company each year
for 10 years.
What is the value of these amounts at the end of the time if accumu-
lated at 3% compound interest?
31. A premium of $91 is paid to an insurance company each year for
10 years.
What is the value of these amounts at the end of the time if accu-
mulated at 3% compound interest?
What is the value if accumulated at 4% compound interest?
32. An insurance company agrees to pay me $1000 a year for 10
years, or an equivalent cash sum to myself or heirs at the end of the
period.
Compute the equivalent cash sum if money is worth 6% compound
interest.
33. A father invests $100 each year for a newborn son, beginning
when he is one year old.
If money is worth 4% compounded annually, what sum is due the
son on his twenty-first birthday ?
What does he receive on his twenty-first birthday if the amounts in-
vested bear 5% compound interest?
34. A potato cuts into 4 parts for planting, each piece produces
5 good sized potatoes, 80 of which make a bushel. If I plant each
year all that I raised the preceding year, how many bushels of potatoes
will I have at the end of the fifth year? How much are they worth
at $4.00 per bu. ? Ans. $160,000.
35. One kernel of corn planted produces a stalk with 2 ears with
16 rows each, 50 kernels to the row. Suppose 100 ears make a bushel
and that I plant each year one-half of all that I raised the preceding
year and that one-half of the kernels grew and produced. How many
XII, §184]
THE PROGRESSIONS
251
bushels would I have at the end of the fifth year? (Assume two kernels
planted the first year.)
36. I have one sow. Let us suppose that the average litter of pigs
is 6, sexes equally distributed, and that I keep all of the sows each
year but sell all the others. How many sows in the sixth generation?
How many pigs will have been sold after I have disposed of 1 /2 of the
last or 5th litter? Am. 243; 363.
37. The common housefly matures and incubates a new litter every
3 weeks. There are approximately 200 to a litter evenly distributed
as to sex. What will be the number of descendents of one female fly
in 12 weeks? Ans. 2 X 108.
38. Grasshoppers hatch yearly a brood of 100 evenly distributed
as to sex. Assuming that none are destroyed, what will be the number
of descendants of one female grasshopper at the end of 5 years? 6 years?
39. The apple aphis matures and incubates in 10 days. The progeny,
all females, are 5 in number. The female propagates 5 each day for
30 days. What will be the number of descendants of one female at
the end of 30 days?
40. If the population doubled every 40 years, how many descend-
ants would one person have after 800 years? Ans. 1,048,576.
41. Find the amount of money that could profitably be expended
for an overcoat which lasts 5 years provided it saved an annual doctor
bill of $5, money being worth 6% compound interest.
42. The effective heritage contributed by each generation and by
each separate ancestor according to the law of ancestral heredity as
stated by Galton is shown in the following table from Davenport.
Generation Back-
ward.
Eflective Contribu-
tion of Each Gen-
eration.
Number of Ancestors
Involved.
Effective Contribution
of Each Ancestor.
1
1/2
2
1/4
2
1/4
4
1/16
3
1/8
8
1/64
4
1/16
16
1/256
5
1/32
32
1/1024
Compute the effective contribution of the last 20 generations. The
number of ancestors involved in the 20th generation backward and
the total number of ancestors involved. The effective contribution of
each ancestor in the 20th generation backward.
252 MATHEMATICS [XII, § 185
185. Infinite Geometric Series. A geometric progression can
be extended to as many terms as we please, since on multiplying
any term by the common ratio we obtain the next one. Any
series which has no last term and can be indefinitely extended is
called an infinite series.
Suppose the terms of a geometric series are all positive. If
we begin at the first and add term after term the sum always
increases. If r > 1, this sum becomes infinite, i.e., if we choose
a positive number N no matter how large it is possible to add
terms enough that the sum will exceed N. If however r < 1,
the case is quite different. The sum does not become infinite ;
it converges to a limit, i.e., it is possible to find a number L such
that the sum will exceed any number whatever less than L, but
it will never reach L. For example the sum obtained by adding
terms of the geometric series
1 +i+i+^ + •••
in which r = -|, will never reach 1.5, but terms enough can be
added to make the sum exceed any number less than 1.5. If,
e.g., we wish to make the sum greater than 1.49, five terms are
sufficient.
A geometric series in which r < 1 is called a decreasing geo-
metric series. The limit to which the sum of the first n terms of
a decreasing geometric series converges is a/(\ — r), i.e., the first
term divided by one minus the ratio.
For by (5) § 183,
8 = a(l - r") = a a ^ ^
1 - r 1 - r I - r
Now as we add more and more terms, the n in this formula gets
larger and larger, a and r remain fixed. Since r < 1, it follows
that r2 < r, r3 < r2, etc., and rn converges to zero when n is
taken larger and larger. Therefore the second term on the
right converges to zero, and sn converges to a/(l — r). This
XII, § 185] THE PROGRESSIONS 253
limit is sometimes called the " sum " (although strictly it is not
a sum) of the infinite decreasing geometric series
a + ar + or2 + • • •,
and we write
(7) s=^—' r<1-
EXAMPLE. The repeating decimal .666 ••• can be written thus
.6 + .06 + .006 + •••. It is therefore an infinite geometric series
whose first term is .6 and whose common ratio is .1. Hence
.6 _2
fin 3-
EXERCISES
Find the sum of the following infinite series :
1. 1+0.5 +0.25 + -. 6. 1+I + H + --
2. 1 -0.5 + 0.25 -0.125 + ••-. 7. 3 + f + T35 + •••.
3. l + i + i + — . 8. 100 + 1 +0.01 +•••.
4. 1 - i + I - iV + •••• 9. 3 + 0.3 + 0.03 + ••-.
5.. 1 + f + | + ••-. 10. 0.23 + 0.023 + 0.0023 + -.
Find the value of the following repeating decimals :
11. .1111 -. 17. .00032525 •••.
12. .2222 •••. 18. .1234512345 •••.
13. .252252-. 19. 20.2020—.
14. 1.2424 •••. 20. 5.312312 ••-.
15. 2.53131 •••. 21. 6.4141 -.
16. 2.3452345 ••-. 22. 3.214214 —.
CHAPTER XIII
ANNUITIES*
186. Definitions. Suppose you take out a life insurance
policy on which you agree to pay a premium of $100 at the end
of each year for 10 years. Such an annual payment of money
for a stated time is termed an annuity. Instead of paying $100
a year you may prefer to pay $24 at the end of every three
months or $206 at the end of every two years. In any case the
stated amount paid at the end of equal intervals of time is called
an annuity.
Suppose the stated sums are not paid when due and that after
the lapse of say 5 years you desire to pay off your indebtedness
with interest compounded. The sum due is called the amount
of the annuity for the five years.
Suppose you buy a house and agree to pay $1000 at the end
of each year for 4 years. This is an annuity. An equivalent
cash price at the time of sale is called the present value of the
annuity.
187. Notation. The letter r stands for the rate of interest,
e.g. 6 ; the letter f ( = r/100) stands for the annual interest on one
dollar, e.g. .06.
The symbol S^ stands for the amount of an annuity of one
dollar paid at the end of each year for n years.
are indebted for many ideas, methods, and exercises.
254
XIII, § 189] ANNUITIES 255
The symbol S^ stands for the amount of an annuity of one
dollar paid at the end of each pth part of a year for n years.
The symbol a^\ stands for the present value of one dollar paid
at the end of each year for n years.
The symbol a^ stands for the present value of an annuity of
one dollar paid at the end of each pth part of a year for n years.
188. Amount of an Annuity. It is sufficient to consider an
annuity of one dollar since the amount for any other sum will
be proportional to this.
The first payment of one dollar made at the end of the first
year will bear interest for n — 1 years, and at the end of the
period the amount due will be (1 + t)™"1. The second payment
will bear interest for n — 2 years and will increase to (1 + i)n~2.
The next to the last payment will bear interest for one year and
will increase to 1 + i. The last payment will be one dollar and
it will bear no interest. The total amount S^, due at the end of
n years is therefore
1 + (1 + i} + (1 + t)2 + ». + (1 + t)"-2 + (1 + i}"-1.
In this geometric progression the first term is 1, the last term
is (1 + i)n~S and the ratio is 1 + i. Substituting these values
in the formula (5) § 183 for the sum of a geometric progression,
we find
189. Partial Payments. Suppose that the payments instead
of being made at the end of each year are made at the end of
each pth part of a year for n years. Consider an annuity of one
dollar.
The payment to be made at each payment period is l/p. The
first payment will bear interest for n — \/p years. The second
payment will bear interest for n — 2/p years, and so on. The
next to the last payment will bear interest for l/p years. The
256 MATHEMATICS [XIII, § 189
last payment will bear no interest. The total amount due is
then
i + 1 (i + o* + - a + o* + - • + - a + *r* .
p P P P
i
In this geometric progression the common ratio is (1 + i}v ,
and by (5), § 183, the sum of the terms is
As shown in § 145 for the square root, the pth root of 1 + i
is nearly equal to 1 + i/p. In fact it is customary in comput-
ing the amount of one dollar at interest compounded p times a
year, to use 1 + i/p instead of Vl + i- See § 217. If this ap-
proximate value be used in formula (2), the right member
reduces to
^
which shows that S^ is approximately equal to S^.
EXERCISES
1. Find the amount of an annuity of $200 for 10 years at 3% ; 4% ;
5% ; 6% ; 8%. Ans. For 3% $2292.78
2. The semiannual premium on an insurance policy is $50. Find
the amount of this annuity for 10 years at 4%. Ans. $606.37
3. The quarterly premium on a policy is $62.10. Find the amount
of this annuity for 10 years at 3%. Ans. $719.11
4. The annual rent of a house is $480. Find the amount of this
annuity for 20 years at 6%. Find the amount if the rent is paid
monthly. Ans. $17657.08
5. A man saves and at the end of each year for 40 years deposits $100
in a savings bank which pays 4% compounded annually. Find the
amount. Ans. $9502.55
6. A man saves $500 a year and invests savings and interest in bonds
yielding 6%. What will his accumulations amount to in 10, 15, 20,
30 years? Ans. $6590.40
XIII, § 190] ANNUITIES 257
190. Given the Amount of an Annuity to find the Annuity.
Let the annual payment be x. The first payment made one
year from the beginning of the term of the annuity will bear in-
terest for n — 1 years and will increase to x(l + i)n~l. Like-
wise, the second will increase to x(l + i}n~~, the third to
(1 + z')"~3> and so on, while the last payment x will bear no
interest. If the sum of the amounts due at the end of n years
is $1, we have
x[(\ + t)"-1 + (1 + i)"-2 + •" + (1 + i) + i] = 1.
The expression within the square brackets is a geometric pro-
gression of n terms with ratio (1 + i} ; hence, by (5), § 183, we
have
or
'"(TT^"1'
which gives the annuity whose amount after n years is $1. This
formula for x may be written symbolically in the form
(4) x=-f.
S*i
EXERCISES
1. In 10 years a man desires to be worth $30,000. What sum must
he set aside yearly to realize that amount if money is worth 8% ?
2. An auto truck costing $2000 lasts 5 years. What sum must be
set aside annually at 6% to replace the truck when worn out?
3. An automobile costs $1500 and lasts 5 years. What is the equiva-
lent annual expenditure, money worth 6%?
4. A city decides to pave some of its streets. For this purpose bonds,
bearing 6% interest, to the amount of $50,000 are issued. The bonds
are due in 10 years. What sum must be collected yearly in taxes and
invested at 6% to pay off the bonds when due?
258 MATHEMATICS [XIII, § 191
191. Present Value of an Annuity. The present value of
one dollar due in one year is (1 + *)-1,
one dollar due in two years is (1 + i)~2>
one dollar due in n years is (1 + i}~n.
The present value of one dollar paid at the end of each year
for n years will then be
(1 + i)~l + (1+ i)-2 + - + (1 + ^
The sum of this geometric progression is the present value
sought. Hence the present value, a$j-,, of an annuity of $1 is
(5) a . - (1 + ^ - (1 + fl"1
(1 + i)-1 - 1
Multiplying numerator and denominator by 1 + i we find
EXERCISES
1. A man buys a farm, agreeing to pay $1500 cash and $1500 at the
end of each year for three years. What would be the equivalent cash
value of the farm if money is worth 6%?
2. A man buys a farm, agreeing to pay $2000 cash and $2000 at the
end of each year for ten years. What would be the equivalent cash value
of the farm if money is worth 6%?
3. A contractor performs a piece of work for a city and takes bonds
in payment. The bonds do not bear interest, and are payable in 10
equal annual installments of $2000, the first payment to be made one
year from date. Money being worth 6%, payable annually, what is
the cash value of the bonds on the date of issue ?
4. Prove that the present value of one dollar paid at the end of each
pth part of a year for n years is
1
1 +
(1 + i)"
and show that this is approximately equal to a^. See § 189.
XIII, § 192] ANNUITIES 259
5. A man contracts to buy a house paying $200 every three months
for 8 years. Find the equivalent cash price, money being worth 6%.
6. Find the cash value of semiannual payments of $500 for 5 years,
money being worth 6%.
192. Cost of an Annuity. A man desires to provide for his
family, in event of his death, an annuity of $5000 a year for 20
years. What amount must he set aside in his will to provide
for this annuity, assuming that money is worth 6%.
The cost of an annuity of one dollar per year for n years is
^/fi» §§ 187, 191. Whence the cost C, of an annuity of P dollars
per year for n years is
(7)
i
From this we compute that the man should set aside in his will
about $57350.
EXERCISES
1. What will be the cost of an annuity of $500 a year for 10 years,
money being worth 4%? Ans. $4055
2. A man agrees to pay $700 a year for 5 years for a house. What is
the cash value of the house, money being worth 6%. Ans. $2948.66
3. A man agrees to pay $700 a year for 20 years for a farm. What
is the cash value of the farm, money being worth 5%? Ans. $8723.55
4. A man 70 years old has $3000. His expectation of life being 8
years, what annuity can an insurance company offer him, money being
worth 4% ?
5. A man with $10,000 pays it into a life insurance company which
agrees to pay him or his heirs a stated sum each year for 20 years.
What is the yearly payment, money being worth 4%?
6. A man buys a house for $4000. What annual payment will can-
cel the debt in 5 years, money being worth 6%? Ans. $949.60
7. How long will it take a man to accumulate $100,000, by saving
$1000 a year and investing it at 6%. Ans. 33 yrs.
8. A man inherits $20,000 which is invested at 4%. If $1000 a
year is spent, how long will the inheritance last. Ans. 41 yrs.
260 MATHEMATICS [XIII, § 193
193. Perpetuities. In the previous problems treated in this
chapter the payments continued over a fixed number of years
and then stopped. The annual amount expended for repairs
on a gravel road does not stop at the end of a given period, but
continues forever. Such payments constitute an endless an-
nuity, which is called a perpetuity. Other examples are the
annual repairs on a house, taxes, annual wage for a flag man,
annual pay of a section gang. The amount of an annuity would
evidently increase indefinitely as time went on. The present
value of a perpetuity, however, has a definite meaning. The
present value of a perpetuity is a sum which put at interest at
the given rate will produce the specified annual income forever.
Denote by V the present value of the perpetuity and by P the
annual payment. Then
(8) F • i = P.
If the payments are made every n years instead of yearly,
the present value of the perpetuity is denoted by Vn ; its value
will be
(9) Vn = P[(l + i)- + (1 + i)-2n + - + (1 + i~)~pn + •••].
This is an infinite geometric progression whose first term is
P(l + i)-" and whose ratio is (1 + t)"". Hence, by (7), § 185,
the present value of the perpetuity is
(1 + i)-» P
(10) Vn = P
1 - (1 + i)-» (1 + i)w - 1
EXERCISES
1. What is the present cash value of a perpetual income of $1200 per
year, money being worth 6% ? Ans. $20,000.
2. How much money must be invested at 6% to provide for an in-
definite number of yearly renewals of an article costing $24?
3. How much money must be invested at 4% to provide for the pur-
chase every 4 years of a $1000 truck?
XIII, § 193] ANNUITIES 261
4. What is the cash value of a farm that yields an average annual
profit of $2400, money being worth 6%?
5. The life of a certain farming implement costing $100 is 6 yrs.
Find what sum must be set aside to provide for an indefinite number of
renewals, money being worth 4%.
6. The life of a University building costing $100,000 is 100 years.
A man desires to will the University enough money to erect the building
and to provide for an indefinite number of renewals. How much must
he leave the institution?
CHAPTER XIV
AVERAGES *
194. Meaning of an Average. In referring to a group of
individuals, a detailed statement of the height of each would
take considerable time, when large numbers are involved. In
comparing two or more groups, such a mass of detail might fail
to leave a definite impression as to their relative heights. What
is needed is a single number, between that of the shortest and
that of the tallest, which is representative of the group with
respect to the character measured. Such an intermediate
number is called an average.
The idea of an average is in use in everyday affairs. We
hear mentioned frequently such expressions as the average rain-
fall, the average weight of a bunch of hogs, the average yield
of wheat per acre for a county or state, the average wage, the
average length of ears of corn, the average increase in popula-
tion, etc. Often these expressions are used with only an indefi-
nite idea as to what is really meant.
In this Chapter we shall discuss some of the averages in com-
mon use, and we shall explain the circumstances under which
each is to be used.
195. Arithmetic Average. The arithmetic average is the
* The authors of this book are indebted for many ideas in this Chapter and for some
of its methods to an Appendix by H. L. RIETZ to E. DAVENPORT, Principles of Breeding,
Ginn and Co. Some use has been made also of ZIZEK, Statistical Averages, Henry Holt
and Co. ; PEARSON, Grammar of Science; BOWLEY, Elements of Statistics; and SECRIST,
Introduction to Statistical Methods, Macmillan/
262
XIV, § 196] AVERAGES 263
number obtained by dividing the sum of the measurements taken
by the number of those measurements :
/1N .,, ,. sum of all measurements
arithmetic average = — — .
number of measurements
Thus, if we measure seven ears of corn and find their lengths
to be 6, 7, 8, 9, 10, 11, 12 inches, the arithmetic average of their
lengths is 9 inches. Again, the arithmetic average of 6, 7, 8, 12,
12 is 9. This example shows that the arithmetic average gives
no indication of the distribution of the items and that there
may be no item whose measurement coincides with the average.
However, it is influenced by each of the items, and it is easily
understood and computed. It should seldom be used except in
conjunction with other forms of averages. When used alone it
should be for descriptive purposes only.
196. Weighted Arithmetic Average. In measuring the given
items it frequently happens that there are
n\ items with the same measurement /i,
HZ items with the same measurement 1%,
nt items with the same measurement l^.
Then the weighted arithmetic average is given by the formula
(2) weighted arithmetic average = ni/1 + n'2/2 + -+"***.
ni + w2 + "• + nk
In the simple case mentioned above, the weighted arithmetic
average gives the same result as the arithmetic average. Its
chief advantage is that it facilitates computations. For
example the average length of the ears of corn whose individual
lengths are 6, 7, 8, 12, 12 can be found as follows :
average length = 1X6 + 1X7 + 1X8 + 2X12 =
1+1 +1+2
There may be other reasons, however, for counting one item
264 MATHEMATICS [XIV, § 196
several times. Thus, in measurements, an item that is known
to be particularly trustworthy may be counted doubly or triply.
In such cases, the weighted average differs from the arithmetic
average.
197. The Median. If we arrange the numbers representing
the measurements of the items in order of magnitude, the
middle number is called the median. Thus, the median length
of the ears of corn whose lengths are 6, 7, 8, 12, 12 inches is 8
inches. In case there are an even number of items the median
is midway between the two middle terms. Thus if the lengths
of four ears of corn are 6, 7, 9, 10 inches, the median length is
8 inches. There is no ear of this length among those measured.
The median is often used because it is so easily found. Like
the arithmetic mean, it gives no indication of the distribution.
It can be used even when a numerical measure is not attached
to the various items. For example, ears of corn can be ar-
ranged in order of length without knowing the numerical length
of any ear ; clerks can be ranked in order of excellence ; shades
of gray may be arranged with respect to darkness of color ; etc.
The median is the central one of a group and is unaffected by the
relative order of the other members of the group. Thus it is
used when the primary interest is in the central members.
198. The Mode. In measuring the items of a given set it
may happen that some one measurement occurs more frequently
than any other. This measurement is called the mode. Thus,
the modal length of six ears of corn whose lengths are 6, 7, 8, 12,
12, 13 inches is 12 inches. A set of measurements may have
more than one mode. Thus in a given factory there might be
few men who received $2 per day, a large number who received
$3, a small number who received $4, and a large number who
received $5, while few received more than $5. There would then
be two modes for wages, namely $3, and $5.
XIV, § 199] AVERAGES 265
If a curve be plotted using measurements as abscissas and the
number of items corresponding to each frequency as ordinates,
the mode corresponds to the maximum ordinate or ordinates.
(See § 225.)
Unlike the arithmetic average, and the median, the mode is
always the value of one individual measurement. Extreme
measurements have no effect upon it.
In measuring heights of men we might place all those over
4.5 and under 5.5 feet at 5 feet. For this distribution the mode
would necessarily fall at one of the integers. If we arrange
the heights in three-inch intervals the mode might not appear
as an integer, although it would be near the mode first obtained.
Thus it is seen that the mode depends upon the grouping of the
measurements.
The existence of a mode shows the existence of a type. It is
the mode that we have in mind when we speak of the average
height of a three-year-old apple tree, the average price of land, or
the average interest rate.
199. The Geometric Average. The geometric mean of two
positive numbers has been defined in § 184. By analogy we
may define the geometric average of n positive numbers as the
nth root of their product.
If a growing tree doubles its diameter in 20 years what is its
annual percentage rate of increase ? It is not 5%, for an increase
of 5% a year would give the following diameters at the end of the
1st, 2d, 3d, . . ., 20th year
which would give a final diameter greater than 2.6d. Evi-
dently what is wanted is a rate r such that
/ f \20
M -i--L- \ — 2
I J. | — ,
v 100;
whence r = 100(V2 — 1) = 3.53+. Hence an annual increase
266 MATHEMATICS [XIV, § 200
of about 3|% will double anything in 20 years. The geometric
average is used in many practical affairs. Knowing the average
rate of growth of a city in the past the geometric average is used
to predict its future growth. When a new school building is
being designed, for example, it should be made large enough to
meet the future growth of the community as shown by this
geometric average.
200. Conclusion. Given a set of items numerically measured
or not, we should first determine whether or not the data is such
as to warrant any kind of an average. Then the decision
whether one or another kind of average is to be employed de-
pends upon the use to which the result is to be put. If the data
is not complete, the arithmetic average cannot be used. If we
desire to characterize a type in such a case, we may find the
mode, for which the data need not be complete.
Frequently it is best to make use of more than one kind of
average in describing a distribution. It must be remembered that
any average at best conveys only a general notion and never con-
tains as much information as the detailed items which it repre-
sents.
EXERCISES
1. From the heights of the members of your class, find each of the
following kinds of average height : (a) arithmetic, (6) median, (c) mode.
2. Determine in the following cases which average is meant : mean
daily temperature ; average student ; average price of butter ; average
of a flock with respect to egg production ; average salary for all of the
teachers of a state ; average number of bushels of corn per acre for a
state or nation ; normal rainfall ; average number of pigs per litter ;
average number of hours of sunshine per day ; average speed of train
between two stops ; average wind velocity ; mean annual rainfall ; aver-
age sized apple ; average price of oranges when arranged according to
sizes ; average date of the last killing frost in the spring ; average price
of land per acre in a given locality ; average gain in weight per day of
a hog.
XIV, § 200] AVERAGES 267
3. What kind of an average is meant in each of the following cases :
one fly lays on an average 120 eggs; 63% of the food of bobolinks is
insects ; every sparrow on the farm eats j oz. of weed seed every day ;
the average gas bill is $2 per month ; the average price received for lots
in a subdivision was $800; repairs, taxes, and insurance on a house
average $100 per year ; the average amount of material for a dress pat-
tern is 8 yards, 36 inches wide ; a college graduate earns on an average
$1125 a year, while the average yearly earnings of a day laborer, who
has no more than completed the elementary school, is $475.
4. Suppose that we consider 5 millionaires and 1000 persons who are
in poverty. Find the arithmetic average, the median, and the mode of
the wealth of this group. Which best portrays conditions?
5. In the Christian Herald for March 10, 1915, p. 237, it is stated that :
"The average salary of ministers of all denominations is $663. The
few large salaries bring up the average." Which average is used here?
Is it the best to portray conditions? Is the result too high or too low
to represent conditions properly?
6. Compute for the members of your family the mean age, and arith-
metic average. Is there a mode?
7. On a given street ascertain the number of houses per block for 5
blocks. Find the arithmetic average and the median. Is there a mode ?
8. On a given business street ascertain the number of stories of each
business house for one block. Find the arithmetic average and the
median. Is there a mode?
9. Proceed as in Ex. 8 for a residence street. Is there a mode ?
10. In 4 years the number of motorists killed at railroad crossings
doubled. Find the annual rate of increase, using the geometric average.
Ans. 19%.
11. If in the last 20 years the number of deaths in the U. S. due to
consumption has increased 50%, find the annual rate of increase, using
the geometric average. Ans. 2%.
12. Land increased in value from $40 to $150 per acre from 1890 to
1915. What was the average yearly increase?
13. Find the average (arithmetic) word, sentence, and paragraph
length, of some one of the writings of Longfellow, Holmes, Whittier,
Poe ; of some short story ; of some newspaper article.
14. The total of the future years which will be lived by 100,000
268 MATHEMATICS [XIV, § 200
persons born on the same day are 5,023,371. If the total number of
-years to be lived is divided by the number of persons the quotient will
be the average number of future years to be lived by each person.
What kind of an average is this ? What average age does it give ?
15. Out of 100,000 males born alive on the same date about one-half,
namely 50,435, attain age 59. This is then an average age attained.
What kind of an average is it?
CHAPTER XV
PERMUTATIONS AND COMBINATIONS
201. Introduction. In how many ways can I make a selec-
tion of two men to do a day's work if there are 3 men available
for the forenoon and 4 for the afternoon? Having hired one
man for the forenoon, I can hire any one of 4 for the afternoon,
and since this is true for each of the three, there are 3 X 4 = 12
ways of making the selection. This reasoning is general ; that
is, it does not depend upon the special properties of the numbers
3 and 4. Hence we see that if there are p ways of doing a first
act, and if corresponding to each of these p ways there are q ways
of doing a second act, then there are pq ways of doing the sequence
of two acts in that order.
It is evident also that this principle applies to a sequence of
more than two acts and we may say,
If there are p ways of doing a first act; and if after this has
been done in any one of these p ways there are q ways of doing a
second act; etc.; and if after all but the last of the sequence have
been done there are r ways of doing the last act, then all the acts of
the sequence can be done in the given order in pq ••• r ways.
EXERCISES
1. With 4 acids and 6 bases, how many salts can a student make?
2. A ranchman has 5 teams, 4 drivers, and 3 wagons. In how many
ways can he make up one outfit?
3. There are 6 routes from Chicago to Seattle, 4 from Seattle to Port-
land, 3 from Portland to San Francisco. How many ways are there of
going from Chicago to San Francisco via Seattle and Portland?
269
270 MATHEMATICS [XV, § 202
202. Combinations and Permutations. A group of things
selected from a larger group is called a combination. The
things which constitute the group are called elements. Two
combinations are alike if each contain all the elements of the
other irrespective of the order in which they appear. Two
combinations are different if either contains at least one element
not in the other.
A permutation of the elements of a group or combination, or
simply a permutation, is any arrangement of these elements.
Two permutations are alike if, and only if, they have the same
elements in the same order. Thus, eat, tea, and ate are the same
combination of three letters a, e, t ; but they are different per-
mutations of these three letters.
203. Number of Permutations. The number of permuta-
tions of three elements taken all at a time is 6, as may be seen by
writing them down and counting them :
abc, acb, bac, bca, cab, cba.
The number of permutations of 4 elements taken 2 at a time is
12. Thus,
ab, ac, ad ; ba, be, bd ;
ca, cb, cd ; da, db, dc.
If the number of elements is large the process of counting is
tedious. It is possible to derive general formulas for the num-
ber of permutations of any number of elements by which the
number can be easily computed.
204. Permutations of n Things. A rule for the number of
permutations of n things taken all at a time is easily deduced
by means of the principle of § 201 . We have n elements and n
places to fill. We may think of a row of cells numbered from
1 to n.
XV, § 205] PERMUTATIONS AND COMBINATIONS 271
1
2
3
4
n
The first cell can be filled in n different ways and after it has
been filled the second cell can be filled in n — 1 ways. There-
fore the first two can be filled in n(n — 1) ways. When they
have been filled in any one of these possible ways the third cell
can be filled in (n — 2) ways. Therefore the first three cells
2an be filled in n(n — l}(n — 2) ways. Continuing thus we
see that the first k cells (k < n) can be filled in n(n — l}(n — 2)
••• (n — k + 1) ways, and that all the n cells can be filled
in n(n — l}(n — 2) -"2 • 1 ways. This product of all the
natural numbers from 1 to n is called factorial n, and is denoted
by n ! or \n. Thus, 2 ! = 2, 3 ! = 6, 4 ! = 24, 10 ! = 3,628,800.
Therefore,
The number of permutations of n things taken all at a time is
factorial n.
For example, 4 horses can be hitched up in 24 ways ; 10 cows
can be put into 10 stanchions in 3,628,800 ways.
By the same reasoning the number of permutations of n
things k at a time (k ^ n) is the number of ways that k cells
can be filled from n things. The symbol nPt is used to denote
this number. Then, as shown above,
(1) nPk = n(n - l)(n - 2) ••• (n - k + 1)
To remember this formula, note that the first factor is n and the
number of factors is k. Thus B^3 = 5 • 4 • 3 = 60. The
number of ways in which 4 stanchions can be filled out of a herd
of 10 cows is 10P4 = 10 • 9 • 8 • 7 = 5040. In this notation
we should write for the number of permutations of n things
all at a time
(2) nPn = n\
205. Repeated Elements. The above reasoning assumes that
the elements are all distinct. If some of the n elements are alike,
272 MATHEMATICS [XV, § 205
the number of distinguishable permutations is less than n \
For example, the number of distinct permutations that can be
made out of the 7 letters of the word reserve is not 7 ! The
number of permutations of the 7 characters ri, ei, s, 62, r2, v, 63
is indeed 7 ! ; but when the subscripts are dropped the permuta-
tions TI e\ s e^rzv 63 and r^ £2 s e3 r\ v e\ become identical.
Let x be the number of different permutations of the letters
of the word reserve. For each of these x there will be 2 ! per-
mutations of the characters r\ e s e r<z v e and for each of these
x • 2 ! there will be 3 ! permutations of the characters r\ c\-
s cz rz v 63, making x • 2 ! 3 ! in all. It follows that
x • 2 I 3 ! = 7 ! and x = -^~
2!3!
This reasoning can be extended to show that the number of
distinguishable permutations of n elements of which p are alike,
q others are alike, etc., •••, r others are alike, is equal to
(3)
n !
p ! q I ••• r I
EXERCISES
1. How many 3-letter words can be formed from the letters a, p, <?
How many 2-letter words ? How many of each are used in the English
language ?
2. How many different 2-digit numbers can be made from the ten
digits 0, 1, 2, •••, 9 ? How many if repetitions are allowed ? How many
of these are used?
3. Find the number of permutations of the letters in each of the fol-
lowing words : (a) degree, (6) natural, (c) Indiana, (d) Mississippi,
(e) Connecticut, (/) Kansas, (g) Pennsylvania, (h) Philadelphia,
(i) Onondaga, (j) Cincinnati.
4. In how many ways can a pack of 52 cards be dealt into four piles
of 13 each?
5. With 15 players available, in how many ways can the coach fill
the various positions on a baseball team?
XV, § 206] PERMUTATIONS AND COMBINATIONS 273
6. How many different signals of two flags, each one above the other,
can be made with five different colored flags ?
7. How many different sounds can be made by plucking the five
strings of a banjo one or more at a time?
8. How many football signals can be given with four numbers, no
repetitions being allowed?
9. In how many ways can four fields be cropped with corn, oats, wheat,
and clover, one field to each?
10. A seed store offers 12 varieties of garden seeds. My garden
has 8 rows. In how many ways can I plant one row of each variety
selected?
11. In how many ways can a gardener plant 2 rows of lettuce, 3 of
onions, 3 of beans, 4 of potatoes, if his garden has 12 rows?
12. How large a vocabulary could be formed with 9 letters, no repe-
titions being allowed? How many with ten? How many with
twenty-six? (There are about 100,000 words in Webster's dictionary.
The average man has a vocabulary of less than 5000 words.)
206. Combination of n Things k at a Time. The symbol
nCk or (2) is used to denote the number of different combinations
(§ 202) that can be made from n elements taken k at a time.
A combination of k elements can be arranged into k ! permuta-
tions of these elements. That is, there are k I times as many
permutations as there are combinations of k elements taken all
at a time. Whence
nPk = k\nCk.
Making use of the value of nPt, (1), § 203, and solving for nCk
we have,
(M r _"(" ~l)(n -2)--(n -fc + 1)
1.2.3-*
To remember this formula note that the first factor of the nu-
merator is n, and that there are k factors in the numerator and
k in the denominator.
Another useful form of this result is obtained by multiplying
274 MATHEMATICS [XV, § 206
both numerator and denominator of (4) by (n — k}(n — k — 1)
(n -k - 2) ••• 2 • 1. This gives
(5) nCk = '1-TW
k i(n — k) I
We note that the interchange of Jc and n — k leaves (5) un-
altered and hence conclude that
(6) n^n-k = nCk-
This is what we should expect when we think that the numbers
of ways that k things can be selected from a group of n must be
the same as the number of ways that n — k can be rejected.
EXERCISES
1. From a pack of 52 cards how many different hands can be dealt?
2. How many combinations of 5 can be drawn from 42 dominoes?
3. How many different tennis teams can be made up from 6 players
(a) singles ; (6) doubles ?
4. How many straight lines can be drawn through 8 points, no three
of which lie on a straight line ? How many circles ?
5. How many diagonals has a convex polygon of n vertices?
Ans. nCi. — n.
6. Two varieties of corn are planted near each other. How many
varieties will be harvested? Ans. zCz + 2.
7. If four varieties of oats are sown near each other, how many varie-
ties will be harvested? Ans. 4^2 + 4.
8. A starts with two kinds of pure-bred chickens. How many kinds
will he have at the end of the third hatching if all stock is sold when
one year old? Ans. «Cz + 6.
9. In how many ways can 15 gifts be made to 3 persons, 5 to each?
Ans. isCe • i0C5.
10. In how many ways can 15 gifts be made to 3 persons, 4 to A,
5 to B, 6 to C? Ans. 630,630.
11. Given (a) nC2 = 45; (6) BC2 = 190; (c) nC2 = 105; find n.
12. In how many different ways can 500 ears of corn be selected from
505 ears?
13. Compute: (a) loooCW; (&) mCm; (c) 10002^10000.
CHAPTER XVI
THE BINOMIAL EXPANSION— LAWS OF
HEREDITY
207. Product of n Binomial Factors. If the indicated mul-
tiplications are performed and terms containing like powers of x
are collected,
(1) (x + Oi)(a; + a2)(.r + a,)(x + o4) • • • (x + an)
= Xn + CiX—1 + CZXn~* + C3Xn~3 + • • • + Cn-iX + Cn
in which the coefficients have the following values:
Ci = 01 + a2 + «3 + • • • + an.
The number of these terms is n.
C2 = Oi02 + • • • aian + aza3 +'•••+ a3a4 + • • • + an-i«n.
The number of these terms is the number of combinations
that can be made from n a's, 2 at a time, i. e., nC2.
Cs = aia2a3 + aia2a4 + • • • + a2a3a4 + • * • + an_2an_iOn.
The number of these terms is the number of combinations
that can be made from n a's, 3 at a time, i. e., nC3.
d = Oia2a3a4 + aia2a3a6 + • • • + an_3on_2an_ian.
The number of these terms is nCt.
Cr = aiO2o3 • • • ar +
The number of these terms is «Cr.
Cn = aidzds • ' • an, and consists of one term.
275
276 MATHEMATICS [XVI, § 208
If now each of the a's be replaced by y, it is evident that,
Ci = ny, Cz = nC2y*, C3 = nC3y3,
r - r if r - nn
\sr — n^ry , > ^n — y j
and therefore
(2) (x + y)n = xn + nxn~ly + nC2x"-2z/2 + nC3xn~3y3 + • • •
+ nCrxn~ryr + ••••+ nxyn~l + yn.
This is known as the binomial expansion, or binomial formula.
208. Binomial Theorem. If x and y are any real (or imagin-
ary) numbers and if n is a positive integer, then the binomial
formula (2) is valid. The following observations will be of
value.
(1) The exponent of x in the first term is 1 and decreases by
1 in each succeeding term.
(2) The exponent of y in the second term is 1 and increases
by 1 in each succeeding term.
(3) The coefficient of the first term is 1, that of the second
term is n. The coefficient of any term can be found from the
next preceding term by multiplying the coefficient by the exponent
of x and dividing by one more than the exponent of y.
(4) The (r + l)th term is nCrxn-ryr, i. e.,
n(n - l)(w - 2) • • • (n - r + 1)
__i 1J ' : ' ' ~n— r,,r
r!
The coefficient of this (r + l)th term is the product of the
first r factors of factorial n, divided by factorial r.
(5) The sum of the exponents of x and y in any term is n.
(6) The number of terms is n -\- 1.
To prove the rule in statement (3) apply it to the (r + l)th
term,
n . ~n—r*,r
n^r X y .
XVI, § 210] THE BINOMIAL EXPANSION
277
It gives
n — r n(n —
r + 1 r!
n(n - l)(n - 2) • • • (n - r)
— 2) • • • (n — r + 1) n — r
' r +1
(r + l)l
but this is precisely nCV+i, which was to be proved.
209. Binomial Coefficients. The coefficients in the bi-
nomial expansion are called binomial coefficients. Their values
are given in the following table for a few values of n. This
table is called Pascal's triangle.
TABLE OP BINOMIAL COEFFICIENTS, nCr. — PASCAL'S TRIANGLE
r=0
r = l
r=2
r=3
r=4
r=5
r=6
r=7
r -8
r=9
r=lO
r=ll
n= 1
1
1
n= 2
1
2
1
n= 3
1
3
3
1
n= 4
1
4
6
4
1
n= 5
1
5
10
10
5
1
n= 6
1
6
15
20
15
6
1
n= 7
1
7
21
35
35
21
7
1
n= 8
1
8
28
56
70
56
28
8
1
n= 9
1
9
36
84
126
126
84
36
9
1
n = 10
1
10
45
120
210
252
210
120
45
10
1
n = ll
1
11
55
165
330
462
462
330
165
55
11
1
etc.
etc.
etc.
NOTE. If any number in the table be added to the one on
its right, the sum is the number under the latter.
210. Sum of Binomial Coefficients. A great many uses for
binomial coefficients and a great many relations among them
have been discovered. Two of these are as follows.
(1) The sum of the binomial coefficients of order n is 2n. We
verify from the above table that
1 + 1- 21; 1+2 + 1= -22; 1 + 3 + 3 + 1 = 23; etc.
To prove it for any value of n, put x = 1 and y = 1, in the
278 MATHEMATICS [XVI, § 211
binomial formula:
(1 + l)ra = 1 + nCl + nC2 + • • • + »Cn-l + nCn
which proves the statement.
Transposing 1, we have
«Ci + nC2 + nC3 + • ' • + nCn = 2n -I
i. e., the total number of combinations of n things taken 1,2, 3,
" ' , n, at a time is 2n — 1.
(2) The sum of the odd numbered coefficients is equal to the
sum of the even numbered ones and each is 2""1.
We verify from the table, that
1 = 1, 1 + 1=2, 1+3 = 3 + 1,
1+6 + 1=4 + 4, etc.
To prove it for any value of n, put x — 1, y = — 1, in the bi-
nomial formula:
(1 - 1)" = 1 - nd + nC2 - nC3 + nC4 - • • • ± nCn
whence
1 + nC'2 + nC* + • • • = nC\ + nCs + nCg + ' • • .
211. Use of the Binomial Theorem. In expanding a bi-
nomial with a given numerical exponent, the student is urged
to find thq successive coefficients by using the statement (3) § 208,
and not by substitution in a formula. This is illustrated in the
following examples.
EXAMPLE 1. Expand (2z — 3?/)5.
(2x - Si/)5 = (2x)» + 5(2a;)«(- 3?/)1 + 10(2z)3(- Sy)2
+ 10(2x)2(- 3?/)3 + 5(2z)(- ZyY + (- 3y)fi.
XVI, § 211] THE BINOMIAL EXPANSION 279
The coefficients are computed mentally as follows,
the 3d coefficient from the 2d term : 5X4/2 = 10,
the 4th " " " 3d term : 10 X 3/3 = 10,
the 5th " " " 4th term : 10 X 2/4 = 5,
the 6th " " " 5th term : 5X1/5 = 1.
Simplifying the terms, we have
(2x - 3?/)5 = 32z8 - 240x4t/ + 720z32/2 - 2160z2?/3 + 810zy4 - 243t/5.
EXAMPLE 2. Expand (3 — |)6.
(3 - |)6 = 3« + 6(3)6(- i)i + 15(3)<(- W + 20(3)«(- i)3
+ 15(3)2(- |)< + 6(3)'(- I)5 + (- £)6.
The coefficients are computed as follows:
6X5/2 = 15, 15 X 4/3 = 20, 20 X 3/4 = 15, etc.
Simplifying, we have
+
729. 729.
303.75 67.5
8.4375 0.5625
0-015625 797.0625
1041.203125
797.0625
(2|)6 = 244.140625
EXAMPLE 3. Expand (a + b + c)3.
[(a + 6) + c]3 = (a + 6)3 + 3 (a + 6)2c + 3 (a + 6)c2 + c3
= a3 + 3a26 + 3afe2 + b3 + 3 (a2 + 2ab + 62)c
+ 3(o + b)c2 + c3
= a3 + 63 + c3 + 3a26 + 3a2c + 362c + 362a + 3c2a
+ 3c26 + 6o6c.
EXERCISES
Expand the following expressions by the binomial theorem.
1. (x + 3)5. 2. (y - 4)«. 3. (2 - a:)4.
4. (2z + 3y)3. 5. (3x - 4y)3. 6. (3o + x2)6.
7. (x*+ y*)*. 8. (or1 + 2ay~1)*. 9. (a"1 - x~2)*.
280 MATHEMATICS [XVI, § 212
10. (a2 - b2)8.
11. (3a2b + 2C3)8.
12. (1 + x)10.
(<!• 9 V0
l+f) .
14. (2x - £)9.
15. (5 + i)8-
16. (4.9)3.
17. (1.01)B.
18. (0.99)«.
19. (1.9)5.
20. (1.02)4.
21. (15/8)7.
22. Expand (1 + i)5 and (2 - f)5 and check results.
23. Prove that any binomial coefficient, counted from the first, is
equal to the same numbered one, counted from the last.
212. Selected Terms. To select a particular term in the
expansion of a binomial without computing the preceding terms,
we can use the formula for the (r + l)th term, namely,
the first r terms of n !
nCrxn^ryr = - xn^yr.
r\
(x \20
9 ~~ 2y ) .
Here r + 1 = 10, r = 9, n = 20, and the required term is
20- 19- 18- 17- 16- 15- 14- 13- 12 /x\" _ _ 41990xlly9
9-8-7-6.5-4. 3-2-1 \2 ) (
(r- 1 V3
\x + - j which contains x2.
The (r + l)th term is iSCfr(x1/2)13-r(x-1)r = nCr-x«3-3r^2, whence r
must be 3 and the 4th term is required. It is
EXERCISES
1. Find the 4th term of (4a - 6)12.
2. Find the llth term of (2x - y)17.
3. Find the 6th term of (xVy + yVx)9.
4. Find the middle term of (x + 3?/)8.
(x 2 V°
- + - j which does not contain x.
fx y2\12
6. Find the term of I *s. I which contains neither x nor
XVI, §213] THE BINOMIAL EXPANSION 281
213. The Binomial Series. The binomial theorem and the
symbols nCr for the number of combinations of n things taken r
at a time, have no meaning except when n and r are positive
integers. On the other hand we know that such expressions as
(1 + i)*/2, (2 + 5)-2, (32 + 3)1/*, (1 - O.I)-1/2,
have perfectly definite meanings; e. g., (2 + 5)~2 = 1/49.
If we should expand a binomial whose exponent is not a
positive integer by the binomial theorem (that is form the
coefficients and exponents by the same rules as though the
exponent were a positive integer), we should get a non-termi-
nating series of terms. For example,
(32 + 3)1/5 = 321/5
+ Tfy(32)-""(3)« ---- .
Now it is shown in advanced courses in mathematics, that
this binomial series is actually valid, provided the numerical value
of the first term of the binomial is greater than the numerical value
of the second term. It is then valid, in the sense that if we
begin at the first and add term after term, the more terms we
take the nearer the sum approaches to the true value sought
and that, by taking terms enough, the sum which we are com-
puting will approximate the true value as nearly as we please.
EXAMPLE. Find VlO by the binomial series.
VlO = (8 + 2)1/3 = 2(1 + i)1/3
Whence computing, we have
+
1.0000- • •
.0833- • •
.0010- • •
.0000- • •
0.0069- • •
.0002- ••
.0000- ••
1.0843---
.0071 • • •
1.0772---
2
1.0843---
0.0071
2.1544- ••
= VIo.
282 MATHEMATICS [XVI, § 214
The student should note carefully that while the binomial
series for (1 + £)1/3 is valid, that for (-£ + 1)1/3 is not.
EXERCISES
Expand the following in binomial series and simplify five terms.
1 (1 4- x)1^2 2 (1 -r- x)"1/2 3 (1 x)"1/3
4. (0.98)1/3. 5. (1.02)1/2. 6. (0.99)1/2.
7. V96. 8. v/30. 9. v'tKJ. 10. -\/33~.
11. \/15. 12. v/65. 13. \/732.
14.
V1025.
15. ^2400.
16
. ^125.
17.
flVimv thnt ,
11 1 ,«2 1
1 . 3^
,4 , 1
• 3-
5
** +
...
T 2* ~T
2- 4
' 2
• 4-
6
18.
01, 1
_1_ 1 _L
1 • 4 .
' 1 x
• 4-
7
C3 +
...
VI - x
1 + 3X ~T
3- 6
1 3
• 6-
9
19.
Show that (1 —a;)"2
= 1 + 2x +
3x2 +
4x3 -i
.....
20.
Ohow that 1
-1 *x+-
• 3r2
1-
3- £
i«
' + '
"•.
Vl +x
214. Mendel's Law.* An Austrian monk by the name of
Mendel planted some sweet peas of different colors in the garden
of the monastery. These blossomed and produced seed. This
seed was gathered and planted the following year. The flowers
produced the second summer contained all of the colors of the
first summer, but other colors were present. By observing and
counting the number of flowers of each color Mendel discovered
the law which bears his name. In its simplest form it may be
explained as follows.
Suppose a bed of sweet peas with blossoms half of which are
red and half of which are white. Fertilization of the flowers
by wind and insects will take place without selection. That is,
pollen from a white flower is equally likely to fertilize a red or a
white flower. If pollen from a white flower fertilizes a white
* The following articles (§§ 214-216) are based largely upon Chapter XIV of E.
DAVENPORT, Principles of Breeding, Ginn and Co. Much additional information may
be found there.
XVI, § 215]
LAWS OF HEREDITY
283
flower the seed produced is of pure stock and will produce pure
white flowers the following year. Such flowers let us denote by
W2. If pollen from a white flower fertilizes a red flower, or vice
versa, the seed produced will be mixed stock and the following
year will show its mixed character by producing flowers which are
neither red nor white but some intermediate shade. Such
flowers let us denote by RW. The symbol R2 is now self-explan-
atory. On counting the flowers which are pure white, mixed,
and red, we would discover their numbers to be approximately
in the ratio 1:2: 1. These are the coefficients in the expansion
of (R + W)2. This is what one might have expected beforehand,
as is seen from the adjoined table. Observe that there are twice
as many flowers of mixed color as of either of the pure colors.
Color of fertilizing flower.
Color of flower fertilized.
R
W
R...
R?
RW
RW
W2
w
Result of mixing : R2 + 2RW + W2.
215. Successive Generations. Let R* denote the result of
fertilizing R2 with R2 ; RSW denote the result of fertilizing R2
with RW, and so on. Then the results of indiscriminate fertili-
zation of the flowers will be shown in the second generation, but
in the third year, as given in the following table.
Color of fertilizing
flower and its relative
numbers.
Color of flower fertilized and their relative numbers.
fi»
2RW
W*
&..
R4
2R3W
R*W*
2R3W
4R?W2
2RW3
/PTP
2RW3
W*
2RW
JP
Result of mixing : R* + 4RW + 6RW2 + 4RW3 + W*
Observe that the result in the second generation of mixing is
the binomial expansion of (R + W)4.
284 MATHEMATICS [XVI, §216
Similarly we can show that the result in the third generation
of mixing is given by (R + W)8, and so on.
216. Mixing of Three Colors. Make a table, as above, but
for three colors. Suppose the third color to be blue (B). Then
a complete expression for the effect, in the first generation after
mixing, is the following :
(R + W + BY = R2 + W2 + B2 + 2RW + 2RB + 2WB.
In case the ratio of the number of white flowers to red flowers
is as 2 to 3 then the result in the first generation after mixing
is as follows :
(2W + 3fl)2 = 4PF2 + 12WR + 9fl2.
Mendel's law of heredity, as illustrated above by the dis-
tribution of color in the successive generations of plants, applies
to other transmissible characters in both plants and animals.
That this distribution follows the mathematical laws of the
binomial formula is due to the fact that each individual plant
or animal inherits the characteristics of two parents, and hence
the number two and its mathematical properties have their
analogies in the laws of biology.
EXERCISES
1. Plot a few graphs, using binomial coefficients as ordinates and
the number of the corresponding term as abscissas.
2. How many varieties of sweet peas are produced by sowing in the
same bed three different strains (a) first year ; (6) second year.
Ans. (a) 6; (6) 14.
3. A farmer buys two different kinds of thoroughbred chickens but
allows them to mix freely. How many different kinds of chickens will
he have at the end of (a) the first, (6) the second, (c) the third year of
hatching? Ans. (a) 3, (6) 5, (c) 9.
4. Four different varieties of wheat are planted side by side. How
many different varieties will be harvested? Ans. 10.
5. Plot graphs as indicated in Ex. 1 for the results of Ex. 3.
XVI, § 216] LAWS OF HEREDITY 285
6. What varieties and in what proportion are obtained by freely
mixing the first and second generations?
7. I plant 8 sweet pea seeds — 4 red, 4 white. Each seed produces
16 flowers — each flower matures 2 seeds which germinate and grow
the following season. Find the total number of flowers, the proportion
and number of the different kinds of flowers, in the (a) first, (6) second,
and (c) third generations.
CHAPTER XVII
THE COMPOUND INTEREST LAW
217. Compound Interest. Suppose one dollar to be loaned
at compound interest at r% per annum payable annually. The
interest i, due at the end of the first year, is r/100. The amount
due is 1 + i- If interest is payable semiannually the amount
due at the end of the first half year is 1 + i/2* If the interest
is payable quarterly the amount due at the end of the first quar-
ter is 1 + i/4.
In general terms if the interest is payable p times a year at
r% per annum compound, the amounts due on a principal of
one dollar at the end of the 1st, 2d, • • •, pth period are respec-
tively,
and the amounts due at the end of the 1st, 2d, •••, nth years are
respectively,
(i\p
1+*)
pJ \ p \ p
The amount A at the end of n years at r% per annum payable
p times a year on a principal of P dollars is given by the formula
* The amount of one dollar for n years compound interest at r% payable annually
is (1 + i)n. If a settlement is made between two interest dates there is some divergence
of practice in computing the interest for the fractional part of a year. The amount of
one dollar for the pth part of a year by analogy to (1 + t)w would be (1 +t) l/p = Vl + i,
but 1 + - is often used instead. When, however, by the terms of the note the interest
is payable p times a year, and is to be compounded, it is clear that the amounts due at
the end of 1, 2, •••, n periods are
+LY ..,
p/
286
XVII, § 218] THE COMPOUND INTEREST LAW 287
(,) A - P(
218. Continuous Compounding. The larger p is the shorter
the interval between the successive interest paying dates. As
p increases without bound this interval approaches zero ; i.e.
we can take p large enough to make this interval as small as we
please. In the limit interest is said to be compounded contin-
uously. While this state is never realized in financial affairs
it is closely approximated. For example, large retail stores
sell goods over the counter very nearly continuously and con-
tinuously replenish their stock.
Let us see what form equation (1) takes when p becomes
infinite. Put x for i/p which approaches zero when p becomes
infinite. Then (1) becomes
in 1
(2) A = P(l + x}* = P[(l + x)x]in.
Now it is shown in books on the Calculus that as x approaches
i
zero, the quantity (1 + x)x converges to a certain number be-
tween 2 and 3. This number is the base of the natural or
Napierian system of logarithms and is usually denoted by e.
To five decimal places c = 2.71828. It can be shown that the
following steps are justifiable, although the proof will not be
given here. By the Binomial Formula,
x
As x approaches zero the terms on the right converge respectively
to the terms of the series
288 MATHEMATICS [XVII, J 218
If we begin at the first and add the terms of this series, the
more terms we add the nearer the sum comes to e. „ 00000 0
The sum of the first ten terms is 2.71828, as is shown
u.ouuuu u
7
7
in the adjoining computation. ~
Then we conclude that as x approaches zero
i
0.04166
(1 + x)x converges to e.
Returning now to equation (2) we see that as p 0.00138 9
becomes infinite and x approaches zero, A converges 0.00019 9
to Pein. Hence we say that when interest is com- 0.00002
pounded continuously, the amount of P dollars at 0-00000 2
r% per annum for n years is given by the equation 2.71828
(3) A = Pein,
in which i = r/100 is the simple interest on one dollar for one
year. This equation is said to represent the compound interest
law.
Scientific investigations reveal many examples of quantities
whose rate of increase (or decrease) varies as the magnitude of
the quantity itself. For example, the number of bacteria in a
favorable medium, or the growth of an organic body by cell
multiplication ; again the rate of decrease in atmospheric pres-
sure in ascending a mountain is proportional to the pressure,
and the rate of change in the volume of a gas expanding against
resistance varies as the volume. The proverbial phrases, the
rich grow richer, the poor poorer; nothing succeeds like success;
a stitch in time saves nine; are expressions in popular language
which show a recognition of this law in crude form.*
In general terms if y and x are two varying quantities such
that the rate of change in y (as regards a change in x) is known to
vary directly as y itself, then they are connected by an equation
of the form
* See DAVIS, The Calculus, § 81.
XVII, § 218] THE COMPOUND INTEREST LAW 289
(4) y =
in which c and k are constants.
EXAMPLE. Suppose that atmospheric pressure at the earth's surface
is 15 Ibs. per square inch and that it is 10 Ibs. per square inch at a height
of 12,000 ft. If now it be assumed that the rate of decrease in the pres-
sure is proportional to the pressure, we have from equation (4)
p = cetrt.
Substituting p = 15 when & = 0, we find c = 15; then substituting
p = 10, h = 12,000, c = 15, we find
12000 '
and these values of c and k give
12000
p =
by means of which the pressure at any height h can be computed.
This example illustrates the method of solving similar problems which
fall under the compound interest law. We assume an equation of the
form of (4) and determine the constants c and k by substituting in known
pairs of values of x and y. Having determined the constants we insert
them in the assumed formula which is then in form to give the value
of y corresponding to any value whatever of x.
EXERCISES
1. Do you see any relation between the growth of plants, or the
increase in population, and the compound interest law? Is the relation
exact? What circumstances tend to limit its application?
2. Is there any relation between your ability to acquire knowledge
and to think clearly and the compound interest law?
3. The population of the state of Washington was 349,400 in 1890 and
in 1900 it was 518,100. Assume the relation P = ceT, where P =
population, T = time in years after 1890, and predict the population
for 1910.
4. Using the data of Ex. 3, find the average annual rate of increase
from 1890 to 1900. Assuming the same average rate to be maintained
for the next 10 years, predict the population for 19^0.
290 MATHEMATICS [XVII, § 218
5. When heated, a metal rod increases in length according to the
compound interest law. If a rod is 40 ft. long at 0° C., and 40.8 ft. long
at 100° C., find (a) its length at 300° C; (&) at what. temperature its
length will be 41 ft. 6 in. Ans. (a) 42.448 ; (&) 185°.8
6. The rate of increase in the tension of a belt is proportional to the
tension as the distance changes from the point where the belt leaves the
driven pulley. If the tension = 24 Ibs. at the driven pulley, and 32 Ibs.
ten feet away, what is it six feet away? fAns. 28.52
7. Assuming that the rate of increase in the number of bacteria in a
given quantity of milk varies as the number present, if there are 10,000
at 6 A.M., 60,000 at 9 A.M., how many will there be at 2 P.M.? At
3 P.M.? At 6 P.M.? Ans. 2 P.M., 1,188,700.
8. In the process of inversion of raw sugar, the rate of change is pro-
portional to the amount of raw sugar remaining. If after 10 hours 1000
Ibs. of raw sugar has been reduced to 800 Ibs., how much raw sugar will
remain at the end of 24 hours? Ans. 586 Ibs.
CHAPTER XVIII
PROBABILITY
219. Definition of Probability. // an event can happen in h
ways, and fail in f ways, the total number of ways in which the
event can happen and fail is h +/. Then h/(h +/) is said to
be the probability that the event will happen, andf/(h-\-f) is said
to be the probability that the event will fail.
For example, suppose we have a box containing 4 red marbles
and 5 white ones. Let us determine the chance of drawing a
red marble the first time. This event can happen in 4 ways, and
fail in 5 ways, while the total number of ways in which the
event can happen and fail is nine. Then by the preceding defi-
nition the probability of drawing a red marble is 4/9, and the
probability of not drawing a red marble is 5/9. Observe that
one of these things is certain to happen. The measure of this
certainty is the sum of the probabilities of the separate events. This
sum is 1 . Hence, if p is the probability that an event will happen,
the probability q that it will not happen is 1 — p.
220. Statistical Probability. In a throw of a penny, before
the event takes place, there is no reason to suspect that heads
are more likely to turn up than tails. In a throw of a die any one
of the six faces is equally likely to turn up and this probability
does not depend upon the particular die used. The probability
of a man's making a safe hit in a game of baseball, and that of
not making a safe hit are not equal. Here the individuality
of the batter enters and before the event takes place, if the batter
291
292 MATHEMATICS [XVIII, § 220
is unknown, we have nothing on which to make an estimate.
If the batter is known, our estimate is based on his past perform-
ance and this, unlike a throw of dice, depends upon the particu-
lar individual at bat. If out of the last 60 times at bat, he has
made a safe hit 20 times, then we say that the probability of his
making a safe hit this time at bat is 1/3.
Again what is the probability that a man aged 70 will die
within the next year? Clearly this depends upon the individual,
his present state of health, his habits, etc. In this case, how-
ever, we can construct a measure of his probability of dying
which is independent of these personal elements. From the
American Experience Mortality Table (see Tables, p. 329), we
find that out of 38,569 persons living at age 70, within the year
2,391 die. Hence the probability that a man aged 70 will die
within the year is 2,391 -4- 38,569.
To derive the probability of an event from statistical data divide
the number of cases h in which the event happened by the total num-
ber n of cases observed.
221. Expectation. If p is the probability that a man will
win a certain sum s of money, then the product sp is called the
value of his expectation.
Thus the value of a lottery ticket in which the prize is $25
and in which there are 500 tickets is $25 X 1/500, or 30 cents.
EXERCISES
1. According to the mortality table (p. 329) it appears that of
100,000 persons at the age of 10, only 5,485 reach the age of 85. What is
the probability that a child aged 10 will reach the age of 85?
2. On 200 of 240 school days a student has had a grade of 90. What
is the probability that his grade will be 90 on the 241st day?
3. The weather bureau predicts rain for to-day. What is the prob-
ability that it will rain, if on the average 90 out of every 100 predictions
are correct?
XVIII, § 223] PROBABILITY 293
/ 4. Compute the probability of throwing with 2 dice a sum of (a)
seven, (6) eight, (c) nine, (d) ten, (e) eleven.
Ans. (a) i; (6) &; (c) *; (d) & ; (e) &.
5. Find the probability in drawing a card from a pack that it be (a) an
ace, (6) a spade, (c) a face card, (d) not a face card.
Ans. (a) A; (6) i; (c) &; (d) £.
6. Find the expectation of a man who is to win $300 if he holds one
ticket out of a total of 1000 tickets. Ans. 30 cents.
222. Mutually Exclusive Events . Two events are said to be
mutually exclusive if the occurrence of one of them precludes the
occurrence of the other. For example, in a race between A, B,
and C, if A wins, B and C do not win.
// the probabilities of the mutually exclusive events E\, E%, •••,
En are p\, pz, •••, pn, then the probability that some one will occur
is the sum of the probabilities of the separate events.
The meaning will be made clear by means of the following
illustration. A bag contains 3 red, 4 white, and 5 blue balls.
What is the probability that in a first draw we obtain a red or
a white ball? There are 12 balls in all and 7 cases are favorable,
namely 3 red and 4 white balls. Then from the definition of
probability the chance of drawing a red ball or a white ball is
7/12. But the probability of drawing a red ball is 3/12 and that
of drawing a white ball is 4/12 and (3/12) + (4/12) = 7/12.
223. Dependent Events. Events are said to be dependent
if the occurrence of one influences the occurrence of the other.
// the probability of a first event is p\ ; and if after this has happened
the probability of a second event is p^; etc., ••• ; and if after all
those have happened the probability of an nth event is pn ; then 'the
probability that all of the events will happen in the given order is
Pi, P2 "• pn.
For, if the first event can happen in hi ways and can fail in /i
ways ; and if after this has happened the second can happen in h%
ways and can fail in fa ways ; etc., ••• ; and if after these have hap-
294 MATHEMATICS [XVIII, § 223
pened the nth event can happen in hn ways and can fail in /„
ways ; then they can all happen and fail in (hi + fi)(hz + f%) ••-
(hn + /») ways. Now all the events can happen together in the
given order in hi A2 ••• hn ways. Then by the definition of prob-
ability the chance that all of the dependent events will take
place in the given order is
hi hz "- hn _ h\ hi hn
' "
(hi +fi)(h2 +/2) -. (hn +/„) hi +/! A2+/2 +/„
= PlP* '" Pn-
Thus the problem of drawing 2 red balls in succession from a
bag containing 3 red and 2 black balls is (3/5) X (2/4) = 3/10.
For after drawing one red ball and not replacing it the probability
of drawing a red ball the second time is 2/4.
224. Independent Events. Events are said to be independent
when the occurrence of any one of them has nothing to do with
the occurrence of the others.
The probability that all of a set of independent events will take place
is the product of the probabilities of the independent simple events.
This follows as a corollary from the theorem of § 223.
Thus the probability of throwing a deuce twice in succession
is (1/6) X (1/6) = 1/36.
EXERCISES
1. If the batting average of Tyrus Cobb is 0,400 what is the chance
that in any single time at bat he will make a safe hit ?
2. What is the probability of holding 4 aces in a game of whist?
Ans. 1/270,725.
3. Suppose I enter 2 horses for a race and that the probabilities of
their winning are respectively | and j. What is the probability that
one or the other will win the race? Ans. 3/4.
4. Does Ex. 3 teach us anything with respect to diversified farming?
Discuss the probability of crop failure of a single crop as compared with
that of two or more different crops.
5. Three men A, B, C go duck hunting. A has a record of one bird
XVIII, § 223] PROBABILITY 295
out of two, B gets two out of three, C gets three out of four. What is
the probability that they kill a duck at which all shoot at once ?
Ans. 23/24.
6. What is the chance of drawing a white and red ball in the order
named from a bag containing 5 white and 6 red balls? Ans. 3/11.
7. In a certain zone in times of war 23 out of 5000 ships are sunk by
submarine in one week. What is the chance that a single vessel will
cross the zone safely? What is the chance that all of 4 vessels which
enter the zone at the same time will cross in safety ? What is the chance
that of these 4 exactly 3 will cross in safety ? That at least 3 will cross
in safety?
8. In certain branches of the army service 2% of the men are killed
each year. Three brothers enlist in this branch of the service for a
period of two years. Compute the probability that (a) all will survive,
(b) exactly 2 will survive, (c) at least 2 will survive, (d) exactly one will
survive, (e) at least one will survive, (/) none will survive.
9. At the time of marriage the probabilities that a husband and wife
will each live 50 years are \ and j respectively. Compute the probabil-
ity that (a) both will be alive, (b) both dead, (c) husband alive and wife
dead, (d) wife alive husband dead.
10. From the American Experience Table of Mortality (Tables,
p. 329) compute your chances of living 1, 10, 20, 30, 40, 50 years.
11. From the American Experience Table of Mortality (Tables,
p. 329) find that age to which you now have an even chance of living.
12. Find from the same table that age to which a person aged 20 has
an even chance of living. Ans. 66+.
13. Three horses are entered for a race. The published odds are 5 : 4
for A ; 3:2 against B ; 4:3 against C. Is it possible to place bets in
such a way that I win some money no matter which horse wins ?
Ans. Yes.
14. Suppose n horses entered for a race, and let the published odds
be (a — 1) to 1 against the first ; (6 — 1) to 1 against the second, (c — 1)
to 1 against the third and so on. A man bets (a — l)/a to I/a against
the first; (b — l)/6 to 1/6 against the second, etc. Show that whatever
horse wins his gains are represented algebraically by the formula
f5+; +
296
MATHEMATICS
[XVIII, § 225
225. Frequency Distribution Curves.* A sample of 400 oats
plants were taken from an experimental plot and measured as to
height in centimeters with the following results : f
Height, H
45
50
50
55
55
60
60
65
65
70
70
75
75
80
80
85
85
90
90
95
Frequencies, F
2
9
21
34
97
m
89
?4
0
1
Let us plot this data with heights as abscissas and frequencies
as ordinates. Construct rectangles, with bases on the horizontal
rr1
^,
/
\
/
v
\
/
80
/
/
V
/
-
\
\
-GO
\
/
j
—10
/
/
\
\
i
j
/
'
/
^
.-
""
\
5
s i
'i
45
5
0
5
5
6
0
f
5
7
0
'
5
E
0
f
0
<J
0
9
5
/*•
FIG. 128
axis. Let the width of the base in each case be 5 units, which
agrees with the grouping of the measurements as to height.
* In the remainder of this Chapter (§§ 225-231), the authors are indebted for many
ideas to E. DAVENPORT, Principles of Breeding [Chapter XII and Appendix (H. L.
HIETZ)]. Other books containing similar matter are JOHNSON, Theory of Errors and
Method of Least Squares: WRIGHT AND HAYFORD, Adjustments of Observations; MERRI-
MAN, Textbook of Least Squares; WELD, Theory of Errors and Least Sqiiares; etc.
t MEMOIR No. 3, CORNELL UNIVERSITY AGRICULTURAL EXPERIMENT STATION,
Variation and Correlation of Oats, by H. H. LOVE and C. E. LEIGHTY, Aug., 1914.
XVIII, § 226] PROBABILITY 297
Let the height of the individual rectangles be representative of
the frequency for the corresponding heights of plants, as shown
in Fig. 128.
The upper parts of these rectangles form an irregular curve
made up of segments of straight lines. A smoother curve is
obtained by connecting the middle points of the upper bases of
these rectangles by segments of straight lines as shown by the
dotted line in Fig. 128. Instead of the dotted line we may draw
a smooth curve as near as possible to the middle points of the
upper bases. Any curve drawn as nearly as possible through
a series of plotted points representing a distribution with respect
to a given character is called a frequency distribution curve.
Such curves are useful in presenting to the eye some of the
features of a distribution. The type of character most fre-
quent is represented by the mode (§ 198), which is the value
of the abscissa corresponding to the highest point of the curve.
The median measurement of the group (§ 197) is represented by
the abscissa of that ordinate on either side of which there are
equal areas under the curve. The arithmetic average (§ 195)
is the abscissa of the center of gravity of the area under the
curve.
Frequency distribution curves are plotted for a great variety
of things, such as frequency distribution of people with respect
to height, weight, or age ; grains of wheat with respect to weight ;
alfalfa with respect to duration of bloom in days ; cherry trees
with respect to earliness of bloom ; pigs with respect to size of
litter; diphtheria with respect to time of year; women with
respect to age of marriage ; etc.
226. Probability Curve. If a large number of measurements
are made upon the same item, they will not in general agree.
Let us plot as abscissas the measurements observed and as ordi-
nates their relative frequencies. In most cases, the positive
298
MATHEMATICS
[XVIII, § 226
and negative errors are equally likely to occur, and small errors
are more numerous than large ones. The frequency curve for
the observed data would then have its highest point at the true
value of the measured magnitude, would be symmetric about an
ordinate through this highest point, and would rapidly approach
the axis of abscissas both to the right and left of this maximum
ordinate. If we take the vertical through the highest point as an
axis of y, then abscissas will represent errors of observation and
ordinates will represent frequency of error.
The curve so drawn is well represented by the equation
(1)
y =
in which cr is what we shall call the standard deviation, e =
2.71828 ••• the base of Napierian logarithms, n the number of
observations, x the error of a reading, y the probability of an
error x. This curve is called the probability curve or curve
of error.
While the theoretical curve (1) is symmetric, the curves ob-
tained by plotting the results of statistical study are often
not symmetric. However the formulas developed in this chapter
for the symmetric case can be used for approximate results in the
non-symmetric cases.
227. Standard D e viation . It is not enough to know the value
XVIII, § 227] PROBABILITY 299
of the arithmetic average or the mode. It is important to have
a measure of the tendency to deviate from the average or from
the mode.
The general theory will be explained by means of the data of
§ 225, which represents the measurements of the heights of 400
oat plants. From this data the average height of oat plant is
70.8 centimeters. Compute the deviation, D, of these plants
from their average height. Multiply the square of each devia-
tion by its corresponding frequency and add the results. We
get 19,320. Divide by the sum, 400, of the frequencies. The
quotient is 48.3. We next extract the square root since the de-
viations have all been squared in the above calculations. We
get 6.95~, and this is called the standard deviation.
In general, to find the standard deviation,
Compute the deviation of each frequency from the arithmetic
average. Multiply the square of each deviation by its corresponding
frequency and add the results. Divide by the sum of thefrequenci.es.
Extract the square root.
This rule is symbolized in the following formula :
(2) „
The curve A in Fig. 129 represents the distribution when a is
small, and the curve B represents the distribution when a is large.
For example, the two sets of numbers 7, 7, 8, 8, 8, 8, 9, 9 and
5, 6, 7, 8, 8, 9, 10, 11 have the same arithmetic mean. The
second set, however, shows a greater tendency to vary from
the arithmetic average (type) than does the first. This greater
tendency to vary is shown by the larger value for cr for the
second set. The values of a- are 0.706 and 1.87 respectively.
Again, suppose two men are shooting at a mark, and that we
compute the standard deviation for each. The man for whom <r
is smallest is said to be the more consistent shot.
300 MATHEMATICS [XVIII, § 228
228. Coefficient of Variability. A comparison of the standard
deviations of two different groups conveys little information
as to their respective tendencies to deviate from the arithmetic
average. This is due to two causes : (1) the measurements may
be in different units, as centimeters and grams, (2) one average
may be much larger than the other, for example the average
height of a group of men would be larger than the average length
of ears of corn. We need then a measure of variability which
is independent of the units used and takes into account the
relative magnitudes of the means. Such a measure is the
coefficient of variability, which is denoted by C and is determined
by the formula,
/Ox r _ Standard deviation _ <r
\y) v — "7 — r~j : •
Arithmetic average ra
For example, the coefficient of variability in height of the 400
oat plants considered in § 225 is 6.95/70.8, or approximately
10%.
229. Probable Error of a Single Measurement. Any indi-
vidual measurement is likely to be in error. This error is ap-
proximately the difference between this measurement and the
arithmetic average of all the measurements. Compute these
errors for all the measurements, some positive, some negative.
Give them all positive signs and arrange them in order of magni-
tude. The median of this list is called the probable error of a
single measurement of the set and is denoted by Es. It is
shown in the theory of probability that
(4) Es = 0.67450-.
230. Probable Error in the Arithmetic Average. Take a
sample of 500 ears of corn from a crib. Compute the arithmetic
average of their lengths. We use this to represent the mean
length of all the ears in the crib. Quite likely it differs from their
true arithmetic average. We now find by means of equation
XVIII, §231] PROBABILITY 301
(5) below, a number Em, called the probable error in the arith-
metic average. This is a number such that it is equally likely
whether or not the computed arithmetic average of the 500 ears
selected lies between ra — Em and m + Em, where m denotes the
(unknown) true arithmetic average for all the ears in the crib.
In other words if a very large number of persons take a sample
of ears and each computes an average length, then, in a sufficiently
large number of cases, one half of these averages will be within
the limits set and one half will be without.
In treatises on probability it is shown that
P E, 0.6745o-
\P) &m - ~j= = — T=— •
Vn Vn
This formula shows that in order to double the precision of the
computed arithmetic average it is necessary to take four times as
many observations.
231. Probable Error in the Standard Deviation. Compute
the standard deviation, § 227, of the lengths of 500 ears of corn
from a crib. This will differ slightly from the true standard
deviation <r, of the lengths of all the ears in the crib. Next find,
by means of equation (6) below, the probable error Eay of the
standard deviation. Then for a sufficiently large number of
samples from the crib, the computed standard deviations will
fall one half within the limits a- — E* and a- + Ey, and one half
without. The formula for the probable error in the standard
deviation is
ttrt F .- Em _ 0.6745Q-
\P) &v — — j=— .
V2 V2n
EXERCISES
1. Compute E,, Em, E« for the data in § 225.
2. Compute <r, C, E,, Em, E<r for the following sets of measurements,
(a) 5, 6, 7, 8, 8, 9, 10, 11 ; (6) 5, 5, 5, 7, 9, 10, 11, 12.
(c) 1, 6, 8, 8, 8, 8, 10, 15 ; (d) 51, 56, 58, 58, 58, 58, 60, 65.
302
MATHEMATICS
[XVIII, § 231
3. Compute <r, C, E,, Em, E* for the following distribution of oat
plants with respect to height in centimeters [LOVE-LEIGHTY].
Height
60
65
70
75
80
85
90
Frequency ....
2
11
45
140
122
73
7
(a)-
(6)
Height
60
65
70
75
80
85
90
95
Frequency. . . .
11
36
60
94
99
102
68
18
4. Compute from the following data the mode, the mean, the coef-
ficient of variability, the standard deviation, the probable error in the
mean, and the probable error in the standard deviation.
Lbs. of butter fat . .
No. of cows
400
1
375
?,
350
4
325
5
300
7
275
6
250
5
225
2
200
1
Draw the distribution curve.
5. The following table is taken from BULLETIN 110, PART 1, Bureau
of Animal Husbandry, U. S. Dept. of Agriculture on "A BIOMETRICAL
STUDY OF EGG PRODUCTION IN THE DOMESTIC FOWL" and shows the
frequency distribution for hens in first-year egg production.
Annual Egg
Production^
A
£*
IS
M
60
7?
If
90
ITJ?
m
120
134
m
ill
i-n
180
±y?
195
25'J
fl 0
23
IIS
1902-03
1903-04
7
2
5
5
1
10
5
10
8
?0
17
?4
18
W
17
5?
26
37
17
W
18
16
9
8
2
2
6
1
1905-06
r
?,
4
q
13
?5
?4
?,?,
3?!
17
90
q
1906-07
(a)
2
2
5
5
q
16
30
39
?6
?1
1Q
1?
1
00
10
8
8
15
29
32
48
39
36
25
18
6
5
2
From this data compute for each year the mean, the median, and the
mode for egg production. Compute ff, C, E<r, Em, E,. Draw the dis-
tribution curve.
6. From Table I at the end of Chapter XIX compute for each weight
(length) the mean, the median, and the mode for length (weight).
Compute <r, C, E<r, Em, E, of weight (length) for each length (weight).
7. For Table II (p. 312) follow the directions as given in Ex. 6 for
Table I, reading however number of kernels instead of weight.
XVIII, § 231] PROBABILITY 303
8. For Table III (p. 312) follow the directions as given in Ex. 6
for Table I, reading yield and number of culms in place of weight
and length.
9. For Table IV (p. 313) follow the directions as given in Ex. 6.
Read height of mid-parent and height of adult children in place of
weight and length.
CHAPTER XIX
CORRELATION*
232. Meaning of Correlation. Whenever two quantities
are so related that an increase in one of them produces or is ac-
companied by an increase in the other and the greater the in-
crease in the one the greater the increase in the other, these
quantities are said to be correlated positively. If an increase
in one produces, or is accompanied by, a decrease in the other,
they are said to be correlated negatively. If a change in one is
not accompanied by any change in the other, there is no corre-
lation, and the quantities are said to be unrelated. Perfect
positive correlation is represented by the number + 1, perfect
negative correlation by — 1, no correlation by zero. There
is perfect positive correlation between the area of a rectangular
field and its length, the extension of a spiral spring and the sus-
pended load. There is perfect negative correlation between
the pressure and volume of a perfect gas. No relation exists
between the price of coal and the length of ears of corn.
There are quantities, common in everyday life, such that a
change in one is not accompanied by a proportionate change
in the other, but a given change in one is always accompanied by
some change in the other. Such quantities are still said to be
correlated. The degree of relationship may be anywhere be-
tween complete independence and complete dependence, that is
* Throughout this Chapter, the authors have consulted the following books, and are
indebted to them for ideas: E. DAVENPORT, Principles of Breeding (Chap. XIII);
ZIZEK, Statistical Averages; SECRIST, Introduction to Statistical Methods; PEAKSON,
Grammar of Science; BOWLEV, Elements of Statistics.
304
XIX, § 232]
CORRELATION
305
between zero and + 1 or between zero and — 1. For example
we may mention the effect of potato prices on acreage, and vice
versa.
^Ve desire a numerical measure for this correlation. Any
adequate expression must be such that it becomes zero when
there is no correlation, — 1 when there is perfect negative corre-
lation, + 1 for perfect positive correlation, and which is always
between — 1 and + 1. Yule has proposed a formula which
satisfies these conditions. Arrange the observed data with refer-
ence to the two quantities in question as in the following dia-
gram :
x present.
x absent.
y present
U
V
y absent
T
S
Then a measure m of the correlation existing is given by the
equation
(1)
If either r or v is zero
If either u or s is zero
If us = rv
us + rv
m = + 1.
m = — 1.
m = 0.
EXERCISES
1. Compute from the following table the degree of effectiveness of
vaccination against diphtheria :
Recoveries.
Deaths.
Vaccinated .
2843
106
Not vaccinated
254
225
2. Compute from the following table the correlation between prohi-
bition and the arrests per day in a given city for one year :
306
MATHEMATICS
[XIX, § 232
Days with more than 20 arrests.
Less than 20.
Wet
281
84
Dry
142
223
3. Compute the correlation between use of fertilizer and yield of
potatoes in bushels per acre when the results from fifty plats are as
follows :
Yield over 100 bushels.)
Under 100 bushels.
Fertilizer.
47
3
No fertilizer
14
36
Ans. 0.95
This high value of correlation is considered evidence of some connec-
tion between use of fertilizer and yield.
233. Correlation Table. Let it be proposed to find the de-
gree of correlation, if any, between the lengths of ears of corn and
their weight, between their lengths and number of rows of
kernels, between length and circumference, between length and
yield per acre, between length of head of wheat and yield per
acre, between height of wheat and yield per acre. The problem
is now more complex. Let us take for example a given number
of ears of corn and examine them as to weight in ounces and
length in inches. The measurements may be tabulated as shown
in the accompanying table. Each column is a frequency dis-
tribution of lengths for a constant weight. Each row is a fre-
quency distribution of weights for a constant length. The
distribution of the ears of length 8 inches with respect to weight
is 3, 7, 19, 25, 17, 22, 17, 3, 1.
It is to be noticed that the table extends across the enclosing
rectangle from the upper left-hand corner to the lower right-
hand corner. Whenever data tabulated with respect to two
measurable characters show this skew arrangement, correlation
exists. In the accompanying table weights increase from left
XIX, §234]
CORRELATION
307
to right and lengths increase as we move downward. We
have in this case positive correlation. An extension of the array
from the upper right-hand corner to the lower left would have
indicated negative correlation.
234. Coefficient of Correlation. The method of obtain-
ing the correlation coefficient may be explained in connection
CORRELATION BETWEEN WEIGHT AND LENGTH OF EAR *
Weight of Ear in Ounces.
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
is
19
20
21
3
1
2
1
3.5
4
1
4
3
5
5
1
o5
4.5
6
5
4
1
QJ
£
5
2
4
7
2
4
o
a
5.5
2
9
15
14
8
4
1
s
6
1
2
12
16
13
13
6
1
a
«iH
6.5
1
6
11
26
11
8
6
1
03
7
1
2
2
12
18
12
12
11
4
1
H
7.5
1
2
4
20
12
13
21
11
6
6
1
1
•s
8
3
7
19
25
17
22
17
3
1
,D
8.5
1
1
12
9
23
30
26
26
5
1
1
9
1
7
10
23
35
26
24
12
1
2
1
a
CJ
9.5
1
4
14
19
29
17
10
1
3
1
1
^
10
1
1
3
8
18
10
(I
4
2
10.5
2
3
6
7
2
5
1
11
1
1
2
1
11.5
]
with the above table. Find the arithmetic mean of each char-
acter involved — in this case mean length of ear, MI, and mean
weight of ears, Mw. Find the deviation DI of ear length from
mean length, and the deviation Dw of weight from mean weight,
for each ear tabulated. For each ear tabulated find the product
of DI and Dw and then add all of these products. This sum we
will indicate by 3DiDw. Find in the usual way the standard
deviation of length of ears, <TI, and the standard deviation of
weight of ears, <rw. Then the coefficient of correlation, r, is
* E. DAVENPORT, Principles of Breeding, p. 458.
308
MATHEMATICS
[XIX, § 234
given by the formula
(2)
where n is the number of things observed, in this case the total
number of ears.
A convenient arrangement for computing DI for each ear length and
Dw for each ear weight is shown in the table below.
The row labeled 6.5 inches (table § 233), gives the frequency distri-
bution of ears with respect to weight. There is one ear of weight
4 oz., 6 ears of weight 5 oz., 11 ears of weight 6 oz., 26 ears of weight
7 oz., etc.; a total of 70 ears, fi, of length 6.5 inches.
fiVi = 1X4 + 6X5 + 11X6 + 26 X7 + 11X8 + 8X9
+ 6 X 10 + 1 X 11 = 455.0
The mean length of ear is obtained by adding the numbers in the
column headed fiVi and dividing this sum by the total number,
n = 993, of ears.
CORRELATION OF WEIGHT TO LENGTH OF EARS OF CORN
Length,
ji
SiVi
DI
Weight,
t
fvVw
Dw
DiDv
Inches.
Ounces.
3
4
12.0
- 4.8
2
4
8
-8.7
143.0
3.5
5
17.5
- 4.3
3
22
66
-7.7
156.9
4
14
56.0
- 3.8
4
27
108
-6.7
394.4
4.5
16
72.0
- 3.3
5
50
250
-5.7
347.2
5
19
95.0
- 2.8
6
47
282
-4.7
297.6
5.5
53
291.5
- 2.3
7
71
497
-3.7
618.9
6
64
384.0
- 1.8
8
75
600
-2.7
465.8
6.5
70
455.0
- 1.3
9
71
639
-1.7
306.8
7
75
525.0
- 0.8
10
75
750
-0.7
110.8
7.5
98
735.0
- 0.3
11
88
968
0.3
14.9
8
114
912.0
0.2
12
107
1,284
1.3
1.4
8.5
134
1,139.0
0.7
13
114
1,482
2.3
129.6
9
142
1,278.0
1.2
14
112
1,568
3.3
466.3.
9.5
100
950.0
1.7
15
65
975
4.3
564.4
10
53
530.0
2.2
16
37
592
5.3
431.0
10.5
26
273.0
2.7
17
8
136
6.3
364.0
11
5
55.0
3.2
18
13
234
7.3
107.2
11.5
1
11.5
3.7
19
4
76
8.3
27.0
993
7,791.5
20
2
40
9.3
21
1
21
10.3
M
7791.5
5
993
10,576
1 993 ™
Mw
10,576
= 10.65
993
* E. DAVENPORT, Principles of Breeding, p. 461.
XIX, § 235] CORRELATION 309
All of the symbols used have been defined with the exception of the
following : <r/ is the standard deviation of length ; /„, is the number (fre-
quency) of ears of same weight w ; Vi stands for the value of length of
ears with given frequency ; Vw represents the value of weight of ears
with given frequency. This gives MI = 7.85. In the row labeled 6.5
and in the column headed DI we write the difference between this mean
length 7.85 and the length 6.5. This gives the number — 1.3 of the col-
umn headed D/. The number 306.8 in the last column is obtained as
follows :
(- 1.3)[1(- 6.7) + 6(- 5.7) + 11(- 4.7) + 26(- 3.7)
+ 11(- 2.7) + 8(- 1.7) + 6(- 0.7) + 1(0.3)] = 306.8
That is, the ear of weight 4 oz. deviates from the mean weight by 6.7 oz.,
the 6 ears of weight 5 oz. deviate from the mean weight by 5.7 oz.,
the 11 ears of weight 6 oz. deviate from the mean weight by 4.7 oz.,
etc.
The number 306.8 represents the sum of the products of the cor-
responding length and weight deviations for every individual in the
horizontal row to which the number belongs. To find the correlation
coefficient add the numbers in the column headed DiDw, obtaining in
this case 4947.2.
Divide this number 4947.2 by n X <TI X «•„. In this case n = 993,
and <TI, ffw have been computed to be 1.57 and 3.63 respectively.
This gives the correlation coefficient
- - 4947'2 = 0.87
993(1.57) (3.63)
235. The Regression Curve. For each recorded height (see
table, § 233) compute the arithmetic average of length of ears.
Thus the ears of weight 4 oz. have an average length of 5.1 inches.
The ears of weight 5 oz. have an average length of 5.46 inches,
etc. Plot a curve using for abscissas the weights, and for
ordinates the computed average lengths. The curve so plotted
is called a regression curve. In many cases this curve is a
straight line. It can be shown that the straight line which best
represents the plotted data is given by the equation
310 MATHEMATICS [XIX, § 235
(2) Mt = r^lw.
aw
Another regression curve can be plotted for the same data,
using lengths as ordinates and mean weights for abscissas. This
curve does not in general coincide with the first. Its equation is
Mw = r^l.
&1
By means of these curves the mean value of one character can be
read off when a fixed value is given to the other character.
EXERCISES
1 . Find, for the correlation table in § 233 :
(a) the regression of weight relative to length ;
(b) regression of length relative to weight.
Ans. (a) 2.03 (6) 0.38
2. Find the equation of the line of regression in both cases of Ex. 1.
3. Plot the line of regression in Ex. 2 from the equation found there
and then again plot the line from the data as suggested in § 235.
4. From Table II, p. 312, which gives the correlation of height of oat
plants with the average number per plant of kernels per culm, compute
the mean height, the mean number of kernels per culm, the standard
deviation with respect to height, the standard deviation with respect to
number of kernels per culm, the correlation coefficient, and the regres-
sion coefficients.
5. Examine Table IV, p. 313, which gives the number of children of
various statures born of 205 mid-parents of various statures. From
this table compute :
Mp = mean height of mid-parents,
Me — mean height of adult children,
ffp = standard deviation of height of mid-parents,
ffe = standard deviation of height of adult children,
r = the correlation coefficient, and both regression coefficients.
6. For Ex. 4 plot the lines of regression (a) from their equations,
(6) from the data directly.
7. For Ex. 5 plot the lines of regression (a) from their equations,
(6) from the data directly.
XIX, §235]
CORRELATION
311
8. From the following table find a measure of the effectiveness of
vaccination against smallpox.
Recoveries.
Deaths.
Total.
Vaccinated
3,951
200
4,151
Not vaccinated
278
274
552
Total
4,229
474
4,703
9. Construct a correlation table from your own observations on
length and breadth of leaves, (a) Use 30 classes for length, (fe) Use
15 classes for length, thus making the class interval twice as large.
Compute in each case the correlation coefficient.
10. From Table I, below, which gives the conslation of lengths and
weights of ears of corn, compute the mean length, the mean weight,
the standard deviation with respect to length, the standard deviation
with respect to weight, the correlation coefficient, and both regression
coefficients.
11. The same as Ex. 10 after writing number of kernels in place of
weight, using Table II, p. 312, in place of Table I.
I. CORRELATION OP LENGTH AND WEIGHT OF EARS OF CORN
Length
In Inches.
Weight In Ounces.
2
3
4
5
c
7
8
9
10
11
12
13
14
15
16
17
is
3.0
1
1
1
3.5
1
2
2
1
4.0
2
3
5
4
1
4.5
4
5
6
2
1
5.0
4
7
8
6
4
1
5.5
3
9
12
13
8
3
1
6.0
1
5
10
15
12
9
5
2
6.5
2
6
12
26
14
10
5
3
1
7.0
1
3
4
14
18
15
10
7
2
1
7.5
1
2
6
13
17
19
13
9
6
4
2
8.0
2
7
10
13
19
7
6
2
1
8.5
1
3
9
14
25
17
8
5
1
9.0
1
4
7
19
25
16
11
3
9.5
2
3
8
18
20
15
6
1
10.0
1
3
9
18
13
7
5
•2
10.5
2
3
7
5
4
1
11.0
1
2
3
2
312
MATHEMATICS
[XIX, § 235
II. CORRELATION OP AVERAGE HEIGHT OF OAT PLANTS IN CENTI-
METERS AND AVERAGE NUMBER OF KERNELS PER CULM PER
PLANT. [LOVE-LEIGHTY.] r = 0.73.
Number of Kernels.
Height.
30
40;
50
GO
70
80
90
100
110
120
40
50
60
70
80
90
100
110
120
130
55-60
1
1
60-65
4
7
65-70
7
22
9
6
1
70-75
1
13
30
59
32
5
75-80
2
16
40
38
23
3
80-85
1
12
26
23
9
2
85-90
3
2
2
III. CORRELATION OF NUMBER OF CULMS PER OAT PLANT AND
TOTAL YIELD OF PLANT IN GRAMS. [LOVE-LEIGHTY.]
r = 0.712
Yield
Number of Culms per Plant.
2
3
4
5
6
7
0-1
3
28
18
1
19
66
42
7
3
20
58
59
26
1
1
7
11
14
4
1
1
3
2
3
1
1
1-2
2-3
3-4
4-5 .
5-6
6-7
7-8 .
8-9
XIX, §235]
CORRELATION
313
IV. CORRELATION OF HEIGHTS OF ADULT CHILDREN AND PARENTS
DATA FOR CHILDREN OF 205 MID-PARENTS* OF VARIOUS STATURES
Heights of
Mid-parents.
Heights of Adult Children In Inches.
Above.
73.2
72.2
71.2
70.2
69.2
68.2
67.2
66.2
65.2
64.2
63.2
62.2
Below.
Above
3
1
72.5
4
2
7
2
1
2
1
71.5
2
2
9
4
10
5
3
4
3
1
70.5
3
3
4
7
14
18
12
3
1
1
1
1
69.5
5
4
11
20
25
33
20
27
17
4
16
1
68.5
3
4
18
21
48
34
31
25
16
11
7
1
67.5
4
11
19
38
28
38
36
15
14
5
3
66.5
4
13
14
17
17
2
5
3
3
65.5
1
2
5
7
7
11
11
7
5
9
1
64.5
5
5
1
4
4
1
1
Below
1
1
2
2
1
4
2
1
* Height of mid-parent is the mean height of the two parents.
[GALTON-DAVENPORT]
GREEK ALPHABET
LETTERS NAMES
A a Alpha
H,
B 0 Beta
90
T 7 Gamma
It
A 6 Delta
KK
E € Epsilon
AX
Z f Zeta
MM
LETTERS NAMES
LETTERS NAMES
N v
Nu
TT Tau
H£
Xi
T v Upsilon
Oo
Omicron
€> 0 Phi
UTT
Pi
XX Chi
Pp
Rho
¥ \ff Psi
So-s
Sigma
Q w Omega
314
FOUR PLACE TABLES
PAGES
I. LOGARITHMS OF NUMBERS 316-319
II. VALUES AND LOGARITHMS OF TRIGONOMETRIC
FUNCTIONS 320-324
III. RADIAN MEASURE — TRIGONOMETRIC FUNCTIONS 325
IV. SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS 326
V. IMPORTANT CONSTANTS 327
VI. DEGREES TO RADIANS 327
VII. COMPOUND INTEREST 328
VIII. AMERICAN EXPERIENCE MORTALITY TABLE. . . 32!)
IX. HEIGHTS AND WEIGHTS OF MEN 330
EXPLANATION OF TABLE II . . 331-333
315
Table I. Logarithms of Numbers
N.
0
1
2
3
4
5
6
7
8 9
Prop. Parts
0
0000
3010
4771
6021
6990
7782
8451
9031
9542
22
21
1
0000
0414
0792
1139
1461
1761
2041
2304
2553
2788
l
2.2
2.1
2
3010
3222
3424
3617
3802
3979
4150
4314
4472
4624
2
4.4
4.2
3
4771"
4914
5051
5185
5315
5441
5563
5682
5798
5911
3
4
6.6
8.8
6.3
8.4
5
11.0
10.5
4
6021
6128
6232
6335
6435
6532
6628
6721
6812
6902
6
13.2
12.6
5
6990
7076
7160
7243
7324
7404
7482
7559
7634
7709
7
15.4
14.7
6
7782
7853
7924
7993
8062
8129
8195
8261
8325
8388
8
9
17.6
19.8
16.8
18.9
7
8451
8513
8573
8633
8692
8751
8808
8865
8921
8976
20
19
8
9031
9085
9138
9191
9243
9294
9345
9395
9445
9494
1
2.0
1.9
9
9542
9590
9638
9685
9731
9777
9823
9868
9912
9956
2
3
4.0
6 0
3.8
5 7
10
0000
0043
"0086
0128
0170"
0212
0253"
0294
0334
0374
4
8.0
7^6
11
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
5
10.0
9.5
12
0792
0828
0864
0899
0934
0969
1004
1038
1072
1106
6
7
12.0
14 0
11.4
13 3
13
1139
1173
1206
1239
1271
1303
1335
1367
1399
1430
8
ie!o
15.2
9 118.0
17.1
14
1461
1492
1523
1553
1584
1614
1644
1673
1703
1732
IB
17
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
1
lo
1.8
i /
1 7
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
2
3.6
3.4
3
5.4
5.1
17
2304
2330
2355
2380
2405
2430
2455
2480
2504
2529
4
7.2
6.8
18
2553
2577
2601
2625
2648
2672
2695
2718
2742
2765
5
9.0
8.5
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
6
7
10.8
12.6
10.2
11.9
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
8
14^4
13.6
21
3222
3243
3263
3284
3304
3324
3345
3365
3385
3404
9
16.2
15.3
22
3424
3444
3464
3483
3502
3522
3541
3560
3579
3598
16
15
23
3617
3636
3655
3674
3692
3711
3729
3747
3766
3784
1
1.6
1.5
2
3.2
3.0
24
3802
3820
3838
3856
3874
3892
3909
3927
3945
3962
3
4.8
4.5
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
4
5
6.4
8.0
6.0
7.5
26
4150
4166
4183
4200
4216
4232
4249
4265
4281
4298
6
9^6
9^0
7
11.2
10.5
27
4314
4330
4346
4362
4378
4393
4409
4425
4440
4456
8
12.8 12.0
28
4472
4487
4502
4518
4533
4548
4564
4579
4594
4609
9
14.4 13.5
29
4624
4639
4654
4669
4683
4698
4713
4728
4742
4757
14
13
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
1
1.4
O o
1.3
31
4914
4928
4942
4955
4969
4983
4997
5011
5024
5038
3
— .O
4.2
2.6
3.9
32
5051
5065
5079
5092
5105
5119
5132
5145
5159
5172
4
5.6
5.2
33
5185
5198
5211
5224
5237
5250
5263
5276
5289
5302
5
7.0
6.5
6
8.4
78
34
5315
5328
5340
5353
5366
5378
5391
5403
5416
5428
7
9.8
9.1
35
5441
5453
5465
5478
5490
5502
5514
5527
5539
5551
8
9
11.2
12.6
10.4
11.7
36
5563
5575
5587
5599
5611
5623
5635
5647
5658
5670
12
11
37
5682
5694
5705
5717
5729
5740
5752
5763
5775
5786
1
1 2
1.1
38
5798
5809
5821
5832
5843
5855
5866
5877
5888
5899
2
3
2.4
3.6
2.2
3.3
39
5911
5922
5933
5944
5955
5966
5977
5988
5999
6010
4
4^8
4.4
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
5
6.0
7 A
5.5
ft ft
41
6128
6138
6149
6160 ; 6170
6180
6191
6201
6212
6222
7
.«
8.4
D.n
7.7
42
6232
6243
6253
6263 6274
6284
6294
6304
6314
6325
8
9.6
8.8
43
6335
6345
6355
6365
6375
6385
6395
6405
6415
6425
9
10.8
9.9
9
8
44
6435
6444
6454
6464
6474
6484
6493
6503
6513
6522
1
0.9
0.8
45
6532
6542
6551
6561 6571
6580
6590
6599
6609
6618
2
1.8
1.6
46
6628
6637
6646
6656 6665
6675
6684
6693
6702
6712
3
2.7
2.4
4
3.6
3.2
47
6721
6730
6739
6749
6758
6767
6776 6785
6794
6803
5
4.5
4.0
48
6812
6821
6830
6839
6848
6857
6866 6875
6884
6893
6
7
5.4
6.3
4.8
5.6
49
6902
6911
6920
6928
6937
6946
6955 6964
6972
6981
8
7^2
6^4
50
6990
6998
7007
"7016"
7024
7033
7042 7050
7059
7067
9 8.1
7.2
N.
0
1
2
3
4
5
6
7
8
9
316
Table I. Logarithms of Numbers
N.
0
1
2
3
4
5
6 | 7
8
9
Prop. Parts
50
6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
51
7076
7084
7093
7101
7110
7118
7126
7135
7143
7152
9
52
7160
7168
7177
7185
7193
7202
7210
7218
7226
7235
2
1 8
53
7243
7251
7259
7267
7275
7284
7292
7300
7308
7316
3
2.7
4
3.6
54
7324
7332
7340
7348
7356
7364
7372
7380
7388
7396
5
4.5
55
7404
7412
7419
7427
7435
7443
7451
7459
7466
7474
6
5.4
56
7482
7490
7497
7505
7513
7520
7528
7536
7543
7551
7
8
6.3
7.2
9
8.1
57
7559
7566
7574
7582
7589
7597
7604
7612
7619
7627
58
7634
7642
7649
7657
7664
7672
7679
7686
7694
7701
59
7709
7716
7723
7731
7738
7745
7752
7760 7767
7774
8
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
1
0.8
61
7853
7860
7868
7875
7882
7889
7896
7903
7910
7917
3
2.4
62
7924
7931
7938
7945
7952
7959
7966
7973
7980
7987
4
3.2
63
7993
8000
8007
8014
8021
8028
8035
8041
8048
8055
5
4.0
6
4.8
64
8062
8069
8075
8082
8089
8096
8102
8109
8116
8122
7
5.0
65
8129
8136
8142
8149
8156
8162
8169
8176
8182
8189
8
g
7 9
66
8195
8202
8209
8215
8222
8228
8235
8241
8248
8254
67
8261
8267
8274
8280
8287
8293
8299
8306
8312
8319
7
68
8325
8331
8338
8344
8351
8357
8363 8370 8376
8382
1
0.7
69
8388
8395
8401
8407
8414
8420
8426 8432 8439
8445
2
1.4
70
8451
8457
8463
8470
8476
84 S2
8488 8494
8500
8506
3
4
2.1
2.8
71
8513
8519
8525
8531
8537
85-13
8549
8555
8561
8567
5
3.5
72
8573
8579
8585
8591
8597
8603
8609
8615
8621
8627
6
4.2
73
8633
8639
8645
8651
8657
8663
8669
8675
8681
8686
7
4.9
8
5.6
74
8692
8698
8704
8710
8716
8722
8727
8733
8739
8745
9
6.3
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
76
8808
8814
8820
8825
8831
8837
8842
8848
8854
8859
6
77
8865
8871
8876
8882
8887
8893
8899
8904
8910
8915
1
2
0.6
1.2
78
8921
8927
8932
8938
8943
8949
8954
8960
8965
8971
3
1.8
79
8976
8982
8987
8993
8998
9004
9009
9015
9020
9025
4
2.4
80
!K.)31
9036 9042
9047
9053
905X
9063
9069
9074
9079
6
3.6
81
9085
9090
9096
9101
9106
9112
9117
9122
9128
9133
7
4.2
82
9138
9143
9149
9154
9159
9165
9170
9175
9180
9186
8
4.8
83
9191
9196
9201
9206
9212
9217
9222
9227
9232
9238
9
5.4
84
9243
9248
9253
9258
9263
9269
9274
9279
9284
9289
85
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
86
9345
9350
9355
9360
9365
9370
9375
9380
9385
9390
2
1.0
3
1 5
87
9395
9400
9405
9410
9415
9420
9425
9430
9435
9440
4
2.0
88
9445
9450
9455
9460
9465
9469
9474
9479 , 9484
9489
5
2.5
89
9494
9499
'.).-,( )4
9509
9513
9518
9523
9528
9533
053S
e
3.0
90
9542
9547
9552
9557
9562
9566
9571
9576
95S1
9586
8
4.0
91
9590
9595
9600
9605
9609
9614
9619
!)<>•_> i
9<>L>S
9633
9
4.5
92
9638
9643
9647
9652
9657
9661
9666
9671
9675
9680
93
9685
9689
9694
9699
9703
9708
9713
9717
9722
9727
4
94
9731
9736
9741
9745
9750
9754
9759
9763
9768
9773
1
04
95
9777
9782
9786
9791 | 9795
9800
9805
9809 9814 9818
3
1 2
96
9823
9827
9832
9836
9841
9845
9850
9854
9859
9863
4
1.6
5
2.0
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
G
2.4
98
9912
9917
9921
9926
9930
9934
9939
9943
9948
9952
7
2.8
99
9956
9961
<)'.)< if,
9969
9974
0!>7S
9983 ; 9987 ; 9991
9996
8
9
3.2
3.6
100
0000
0004
0009
0013
0017
0022
0026 ; 0030
0035
0039
N.
0
1
2
3
4
5
6
7
8
9
317
Table I. Logarithms of Numbers
No.
0
1
2
3
4
5
6
7
8
9
Prop. Parts
100
0000
0004
0009
0013
0017
0022
0026
0030
0035
0039
101
0043
0048
0052
0056
0060
0065
0069
0073
0077
0082
102
0086
0090
0095
0099
0103
0107
0111
0116 ,0120
0124
103
0128
0133
0137
0141
0145
0149
0154
0158
0162
0166
104
0170
0175
0179
0183
0187
0191
0195
0199
0204
0208
105
0212
0216
0220
0224
0228
0233
0237
0241
0245
0249
106
0253
0257
0261
0265
0269
0273
0278
0282
0286
0290
5
1
0.5
107
0294
0298
0302
0306
0310
0314
0318
0322
0326
0330
2
1.0
108
0334
0338
0342
0346
0350
0354
0358
0362
0366
0370
3
1.5
109
0374
0378
0382
0386
0390
0394
0398
0402
0406
0410
5
2.5
110
0414
0418
0422
0426
0430
0434
0438
0441
0445
0449
6
3.0
111
0453
0457
"0461
04~65"
0469"
0473"
0477
0481
0484
0488
7
g
3.5
•1 0
112
0492
0496
0500
0504
0508
0512
0515
0519
0523
0527
9
4.5
113
0531
0535
0538
0542
0546
0550
0554
0558
0561
0565
114
0569
0573
0577
0580
0584
0588
0592
0596
0599
0603
115
0607
0611
0615
0618
0622
0626
0630
0633
0637
0641
116
0645
0648
0652
0656
0660
0663
0667
0671
0674
0678
4
117
0682
0686
0689
0693
0697
0700
0704
0708
0711
0715
1
0.4
118
0719
0722
0726
0730
0734
0737
0741
0745
0748
0752
2
0.8
119
0755
0759
0763
0766
0770
0774
0777
0781
0785
0788
3
1.2
120
0792
0795
0799
0803
0806
0810
0813
0817
0821
0824
5
2.0
121
0828
0831
0835
0839
0842
0846
0849
0853
0856
0860
6
2.4
122
0864
0867
0871
0874
0878
0881
0885
0888
0892
0896
7
2.8
123
0899
0903
0906
0910
0913
0917
0920
0924
0927
0931
8
9
3.2
3.6
124
0934
0938
0941
0945
0948
0952
0955
0959
0962
0966
125
0969
0973
0976
0980
0983
0986
0990
0993
0997
1000
126
1004
1007
1011
1014
1017
1021
1024
1028
1031
1035
127
1038
1041
1045
1048
1052
1055
1059
1062
1065
1069
128
1072
1075
1079
1082
1086
1089
1092
1096
1099
1103
3
129
1106
1109
1113
1116
1119
1123
1126
1129
1133
1136
2
0.6
130
1139
1143
1146
1149
1153
1156
1159
1163
1166
1169
3
0.9
131
1173
1176
1179
1183
1186
1189
1193
1196
1199
1202
5
1 5
132
1206
1209
1212
1216
1219
1222
1225
1229
1232
1235
6
1.8
133
1239
1242
1245
1248
1252
1255
1258
1261
1265
1268
7
2.1
8
2.4
134
1271
1274
1278
1281
1284
1287
1290
1294
1297
1300
9
2.7
135
1303
1307
1310
1313
1316
1319
1323
1326
1329
1332
136
1335
1339
1342
1345
1348
1351
1355
1358
1361
1364
137
1367
1370
1374
1377
1380
1383
1386
1389
1392
1396
138
1399
1402
1405
1408
1411
1414
1418
1421
1424
1427
139
1430
1433
1436
1440
1443
1446
1449
1452
1455
1458
2
140
1*461
1464
1467
1471
1474
1477
1480
1483
1486
1489
o
0.4
141
1492
1495
1498
1501
1504
1508
1511
1514
1517
1520
3
0.6
142
1523
1526
1529
1532
1535
1538
1541
1544
1547
1550
4
0.8
143
1553
1556
1559
1562
1565
1569
1572
1575
1578
1581
5
6
1.0
1.2
7
1 4
144
1584
1587
1590
1593
1596
1599
1602
1605
1608
1611
8
1.6
145
1614
1617
1620
1623
1626
1629
1632
1635
1638
1641
9
l.S
146
1644
1647
1649
1652
1655
1658
1661
1664
1667
1670
147
1673
1676
1679
1682
1685
1688
1691
1694
1697
1700
148
1703
1706
1708
1711
1714
1717
1720
1723
1726
1729
149
1732
1735
1738
1741
1744
1746
1749
1752
1755
1758
150
1761
1764
1767
1770
1772
1775
1778
1781
1784
1787
No.
0
1
2
3
4
5
6
7
8
9
318
Table I. Logarithms of Numbers
N.
0
1
2
3
4
5
6
7
8
9
Prop. Parts
150
1761
1764
1767
1770
1772
1775
1778
1781
1784
1787
151
1790
1793
1796
1798
1801
1804
1807
1810
1813
1816
152
1818
1821
1824
1827
1830
1833
1836
1838
1841
1844
153
1847
1850
1853
1855
1858
1861
1864
1867
1870
1872
154
1875
1878
1881
1884
1886
1889
1892
1895
1898
1901
155
1903
1906
1909
1912
1915
1917
1920
1923
1926
1928
156
1931
1934
1937
1940
1942
1945
1948
1951
1953
1956
3
1 0.3
157
1959
1962
1965
1967
1970
1973
1976
1978
1981
1984
2 0.6
158
1987
1989
1992
1995
1998
2000
2003
2006
2009
2011
3 0.9
159
2014
2017
2019
2022
2025
2028
2030
2033
2036
2038
4 1.2
5 15
160
2041
2044
2047
2049
2052
2055
2057
2060
2063
20(10
6 1.8
161
2068
2071
2074
2076
2079
2082
2084
2087
2090
2092
7 2.1
162
2095
2098
2101
2103
2106
2109
2111
2114
2117
2119
9 2.7
163
2122
2125
2127
2130
2133
2135
2138
2140
2143
2146
164
2148
2151
2154
2156
2159
2162
2164
2167
2170
2172
165
2175
2177
2180
2183
2185
2188
2191
2193
2196
2198
166
2201
2204
2206
2209
2212
2214
2217
2219
2222
2225
167
2227
2230
2232
2235
2238
2240
2243
2245
2248
2251
168
2253
2256
2258
2261
2263
2266
2269
2271
2274
2276
169
2279
2281
2284
2287
2289
2292
2294
2297
2299
2302
170
2304
2307
2310 2312
2315
2317
2320
2322
2:525
2327
171
2330
2333
2335
2338
2340
2343
2345
2348
2350
2353
172
2355
2358
2360
2363
2365
2368
2370
2373
2375
2378
173
2380
2383
2385
2388
2390
2393
2395
2398
2400
2403
174
2405
2408
2410
2413
2415
2418
2420
2423
2425
2428
175
2430
2433
2435
2438
2440
2443
2445
2448
2450
2453
176
2455
2458
2460
2463
2465
2467
2470
2472
2475
2477
2
1 0.2
177
2480
2482
2485
2487
2490
2492
2494
2497
2499
2502
2 0.4
178
2504
2507
2509
2512
2514
2516
2519
2521
2524
2526
3 0.6
179
2529
2531
2533
2.-,:;(>
2538
2541
2543
2545
2548
2550
5 1.0
180
2553
2555
2558
2560
2562
2565
2567
2570
2572
2574
6 1.2
181
2577
2579
2582
2584
2586
2589
2591
2594
2596
2598
7 1.4
8 16
182
2601
2603
2605
2608
2610
2613
2615
2617
2620
2622
9 1.8
183
2625
2627
2629
2632
2634
2636
2639
2641
2643
2646
184
2648
2651
2653
2655
2658
2660
2662
2665
2667
2669
185
2672
2674
2676
2679
2681
2683
2686
2688
2690
2693
186
2695
2697
2700
2702
2704
2707
2709
2711
2714
2716
187
2718
2721
2723
2725
2728
2730
2732
2735
2737
2739
188
2742
2744
2746
2749
2751
2753
2755
2758
2760
2762
189
2765
2767
2769
2772
2774
2776
2778
2781
2783
2785
190
27SH
2790
2792
2794
2797
2799
2801
2804
2806
2808
191
2810
2813
2815
2817
2819
2822
2824
2S2(>
2S2S
2831
192
2833
2835
2838
2840
2842
2844
2847
2849
2851
2853
193
2856
2858
2860
2862
2865
2867
2869
2871
2874
2876
194
2878
2880
2882
2885
2887
2889
2891
2894
2896
2898
195
2900
2903
2905
2907
2909
2911
2914
2916
2918
2920
196
2923
2925
2927
2929
2931
2934
2936
2938
2940
2942
197
2945
2947
2949
2951
2953
2956
2958
2960
2962
2964
198
2967
2969
2971
2973
2975
2978
2980
2982
2984
2986
199
2989
2991
2993
2995
2997
21MHI
3002
3004
3006
3008
200
3010
3012
3015
3017
3019
3021
3023 3025
3028
3030
N.
0
1
2
3
4
5
6
7
8
9
,519
Table II. Values and Logarithms of Trigonometric Functions
[Characteristics of Logarithms omitted — determine by the usual rule from the value]
RADIANS
DEGREES
SINE
TANGENT
COTANGENT
COSINE
Value Log10
Value Lo<r10
Value Logio
Value Log10
.0000
0°00'
0000
0000
1.0000 .0000
90° 00'
1 .5708
!(X)29
10
.0029 .4637
.0029 .4637
343.77 .5363
1.0000 .0000
50
1^5679
.0058
20
.0058 .7(548
.0058 .7648
171.89 .2352
1.0000 .0000
40
1.5650
.0087
30
.0087 .9408
.0087 .9409
114.59 .0591
1.0000 .0000
30
1.5621
.0116
40
.0116 .0658
.0116 .0658
85.940 .9342
.9999 .0000
20
1.5592
.0145
50
.0145 .1627
.0145 .1627
68.750 .8373
.9999 .0000
10
1.5563
.0175
1°00'
.0175 .2419
.0175 .2419
57.290 .7581
.9998 .9999
89° 00'
1.5533
.0204
10
.0204 .3088
.0204 .3089
49.104 .6911
.9998 .9999
50
1.5504
.0233
20
.0233 .3668
.0233 .3(569
42.964 .6331
.9997 .9999
40
1.5475
.02(52
30
.0262 .4179
.0262 .4181
38.188 .5819
.9997 .9999
30
1.5446
.021)1
40
.0291 .4637
.0291 .4638
34.368 .5362
.9996 .9998
20
1.5417
.0320
50
.0320 .5050
.0320 .5053
31.242 .4947
.9995 .9998
10
1.5388
.0349
2° 00'
.0349 .5428
.0349 .5431
28.636 .4569
.9994 .9997
8 8° 00'
1.5359
.0378
10
.0378 .5776
.0378 .5779
2(5.432 .4221
.9993 .9997
50
1.5330
.0407
20
.0407 .6097
.0407 .6101
24.542 .3899
.9992 .9996
40
1.5301
.0436
30
.0436 .6397
.0437 .6401
22.904 .3595)
.9990 .9996
30
1.5272
.0465
40
.0465 .6677
.0466 .6682
21.470 .3318
.9989 .9995
20
1.5243
.0495
50
.0494 .6940
.0495 .6945
20.206 .3055
.9988 .9995
10
1.5213
.0524
3° 00'
.0523 .7188
.0524 .7194
19.081 .2806
.9986 .9994
87° 00'
1.5184
.0553
10
.0552 .7423
.0553 .7429
18.075 .2571
.9985 .C993
50
1.5155
.0582
20
.0581 .7645
.0582 .7652
17.169 .2348
.9983 .9993
40
1.5126
.0611
30
.0610 .7857
.0612 .7865
16.350 .2135
.9981 .9992
30
1.5097
.0640
40
.0640 .8059
.0(541 .8067
15.605 .1933
.9980 .9991
20
1.5068
.0669
50
.0669 .8251
.0670 .8261
14.924 .1739
.9978 .9990
10
1.5039
.0698
4° 00'
.0698 .8436
.0699 .8446
14.301 .1554
.9976 .9989
86° 00'
1.5010
.0727
10
.0727 .8613
.0729 .8624
13.727 .1376
.9974 .9989
50
1.4981
.0756
20
.0756 .8783
.0758 .8795
13.197 .1205
.9971 .9988
40
1.4952
.0785
30
.0785 .8946
.0787 .8960
12.706 .1040
.9969 .9987
30
1.4923
.0814
40
.0814 .9104
.0816 .9118
12.251 .0882
.9967 .9986
20
1.4893
.0844
50
.0843 .9256
.0846 .9272
11.826 .0728
.9964 .9985
10
1.4864
.0873
5° 00'
.0872 .9403
.0875 .9420
11.430 .0580
.9962 .9983
85° 00'
1.4835
.0902
10
.0901 .9545
.0904 .9563
11.059 .0437
.9959 .9982
50
1.4806
.0931
20
.0929 .9682
.0934 .9701
10.712 .0299
.9957 .9981
40
1.4777
.09(50
30
.0958 .9816
.0963 .9836
10.385 .0164
.9954 .9980
30
1.4748
.0989
40
.0987 .9945
.0992 .95X56
10.078 .0034
.9951 .9979
20
1.4719
.1018
50
.1016 .0070
.1022 .0093
9.7882 .9907
.9948 .9977
10
1.4690
.1047
6° 00'
.1045 .0192
.1051 .0216
9.5144 .9784
'.9945 .9976
84° 00'
1.4661
.1076
10
.1074 .0311
.1080 .0336
9.'_'5.r)3 .9664
.9942 .9975
50
1.4632
.1105
20
.1103 .0426
.1110 .0453
9.0098 .9547
.9939 .9973
40
1.4(503
.1134
30
.1132 .0539
.1139 .05(57
8.7769 .94a3
.9936 .9972
30
1.4573
.1164
40
.1161 .0648
.1169 .0678
8.5555 .9322
.9932 .9971
20
1.4544
.1193
50
.1190 .0755
.1198 .0786
8.3450 .9214
.9929 .9969
10
1.4515
.1222
7° 00'
.1219 .0859
.1228 .0891
8.1443 .9109
.9925 .9968
83° 00'
1.4486
.1251
10
.1248 .0961
.1257 .0995
7.9530 .9005
.9922 .9966
50
1.4457
.1280
20
.1276 .10(50
.1287 .109(5
7.7704 .8904
.9918 .9964
40
1.4428
.1309
30
.1305 .1157
.1317 .1194
7.5958 .8806
.9914 .9963
30
1.4399
.1338
40
.1334 .1252
.1346 .1291
7.4287 .8709
.9911 .9961
20
1.4370
.1367
50
.13(53 .1345
.1376 .1385
7.2(587 .8615
.9907 .9959
10
1.4341
.1396
8° 00'
.1392 .1436
.1405 .1478
7.1154 .8522
.9903 .9958
82° 00'
1.4312
.1425
10
.1421 .1525
.1435 .15(59
6.9682 .8431
.9899 .9956
50
1.4283
.1454
20
.1449 .1612
.1465 .1(558
6.8269 .8342
.9894 .9954
40
1.4254
.1484
30
.1478 .1697
.1495 .1745
6.6912 .8255
.9890 .9952
30
1.42-24
.1513
40
.1507 .1781
.1524 .1831
6.5606 .8169
.9886 .9950
20
1.4106
.1542
50
.1536 .1863
.1554 .1915
6.4348 .8085
.9881 .9948
10
1.41(56
.1571
9° 00'
.1564 .1943
.1584 .1997
6.3138 .8003
.9877 .9946
81° 00'
1.41 ."7
Value Log10
Value Lojrln
Value Log10
Value Log10
DEGREES
RADIANS
COSINE
COTANGENT
TANGENT
SINE
320
Table II. Values and Logarithms of Trigonometric Functions
[Characteristics of Logarithms omitted —determine by the usual rule from the value]
UADIANS
DEGREES
SINE
TANGEN-T
COTANGENT
COSINE
Value Log10
Value Loi^m
Value L<>£10
Value Logla
.1571
9° 00'
.1504 .1943
.1584 .1997
6.3138 .8003
.9877 .9946
81° 00'
1.4137
.1600
10
.1593 .2022
.1014 .2078
0.1970 .7922
.9872 .9944
50
1.4108
.1629
20
.1622 .2100
.1044 .21.78
6.0844 .7842
.98(58 .9942
40
1.4079
.1658
30
.1050 .2170
.1673 .22:50
5.9758 .77(54
.9863 .9940
30
1.4050
.1687
40
.1679 .22.-. 1
.1703 .2313
5.8708 .7(587
.9858 .9938
20
1 .4021
.1710
50
.1708 .2324
.1733 .2389
5.7094 .7611
.9853 .99:1(5
10
1.3992
.1745
10° 00'
.1736 .2397
.1763 .2463
5.6713 .7537
.9848 .9934
80° 00'
1.39(53
.1774
10
.1765 .2468
.1793 .25: JO
5.5704 .7464
.9843 .99: :i
50
1.3934
.1804
20
.1794 .25:38
.1823 .2(509
5.4845 .7391
.9838 .9! 129
40
1.3904
.1833
30
.1822 .2606
.1853 .2(580
f).:;9.-)5 .7320
.9833 .9927
30
1.3875
.1862
40
.1851 .2674
.1883 .2750
5.3093 .7250
.9827 .9924
20
1.384fi
.1891
50
.1880 .2740
.1914 .2819
5.2257 .7181
.9822 .9922
10
1.3817
.1920
11°00'
.1908 .2806
.1944 .2887
5.1446 .7113
.9816 .9919
79° 00'
1.3788
.1949
10
.1937 .2870
.1974 .2!r,:;
5.0658 .7047
.9811 .9917
50
1.3759
.1978
20
.1965 .2934
.2004 .3020
4.9894 .6980
.9S05 .9914
40
1.3730
5007
30
.1994 .2997
.2035 .3085
4.9152 .(5915
.9799 .9912
30
1.3701
.2036
40
.2022 .3058
.20(55 .3149
4.8430 .6851
.9793 .9909
20
1.3672
.2005
50
.2051 .3119
.2095 .3212
4.7729 .6788
.9787 .9907
10
1.3643
.2094
12° 00'
.2079 .3179
.2126 .3275
4.7046 .6725
.9781 .9904
78° 00'
1.3614
.'21--':$
10
.2108 .:52:;s
.2150 .3336
4.6382 .6664
.9775 .9901
50
1.3584
5153
20
.2136 .3290
.2186 .3397
4.5736 .6(503
.9769 .9899
40
1.3555
.2182
30
.2164 .3353
.2217 .3458
4.5107 .6542
.9763 .9896
30
1.3526
.2211
40
.2193 .3410
.2247 .3517
4.4494 .0483
.9757 .9893
20
1.3497
.21-40
50
.2221 .3466
.2278 .357(5
4.3897 .6424
.9750 .9890
10
1.3468
5269
13° 00'
.2250 .3521
.2309 .3634
4.3315 .6366
.9744 .9887
77° 00'
1.3439
5298
10
.2278 .3575
.2339 .3091
4.2747 .0:509
.97:57 .9884
50
1.3410
.2327
20
.2306 .3629
.2370 .3748
4.2193 .02.72
.9730 .'.issi
40
1.3381
.2356
30
.2334 .30H2
.2401 .3804
4.1053 .6196
.9724 .9878
30
1.3352
5380
40
.2:503 .3734
.2432 .3859
4.1120 .6141
.9717 .9S7.-
20
i .:;:;23
.2414
50
.2391 .3786
.2462 .3914
4.0(511 .6086
.9710 .9872
10
1.3294
.2443
14° 00'
.2419 .3837
.2493 .3908
4.0108 .6032
.9703 .98(59
76° 00'
1.32(55
5473
10
.2447 .3887
.2524 .4021
3.9(517 .5979
.9090 .'.'sro
00
L.323B
5502
20
.2470 .3937
.2555 .4074
3.9130 .5926
.<iOs9 .9863
40
1.3200
.2331
30
.2504 .3986
.2586 .4127
3.81 5G7 .5873
.9681 .9S59
30
1.3177
5560
40
.2532 .4035
.2617 .4178
3.8208 .5822
.9674 .98.-.0
20
1.31 48
.2589
50
.2560 .4083
.2648 .4230
3.7700 .5770
.9007 .9853
10
1.3119
.2618
15°00'
.2588 .4130
.2679 .4281
3.7321 .5719
.9(559 .9849
75 00'
1.3090
.2647
10
.2616 .4177
.2711 .4331
3.6891 .5(569
.9052 .'.MI;
no
1.3061
5676
20
.2641 .4223
.2712 .4381
3.0470 .5019
.9(544 .984:5
40
1.3032
5708
30
.2672 .420!)
.2773 .4430
30059 .5570
.9(5:50
30
1.3003
.2734
40
.2700 .4314
.2805 .417!)
3.5050 .5521
.902S .'.is:,-;
20
1.2974
.2763
50
.2728 .4:»9
.2836 .4527
3.5261 .5473
.9621 .9S32
10
15948
5793
16° 00'
.2756 .4403
.2S07 .4575
3.4874 .5125
.9613 .9-S2S
74° 00'
1.2918
5822
10
.2784 .4447
.2*99 .4(522
3.4495 ..T.J7S
.9605 .9*2".
60
.2851
20
.2812 .4491
.29:51 .4(509
3.1124 .5331
.959(5 .9821
40
1.2857
.2880
30
.2840 .4533
.2902 .471(5
3.3759 .52S4
.!i:,ss .9817
30
1.2828
.2909
40
.2868 .4571!
.2994 .4762
3.3402 ..72: IS
.95SO .9814
20
1.2799
.2938
50
.2896 .4618
.3026 .4808
3.3052 .5192
.9572 .9810
10
1.2770
.2967
17° 00'
.2924 .4659
.3057 .48.-,:?
3.2709 .r,117
.9503 .9SO<;
73C00'
1.2741
5996
10
,2!i:,2 .4700
.3089 .4S9S
3.2371 .5102
.'.).-,.-,.-, .9M>2
50
1.2712
.3028
20
.297!) .4741
..-5121 .4943
3.2041 .r,0.-,7
.9.140 .9798
40
1.2683
.3054
30
.3007 .4781
.3153 .4987
3.1 7 10 ..Vi 115
.9537 .97d!
30
1.2664
.3083
40
.:;(»:•,;, .4821
.3185 .5(): ;i
3.1:197 .4909
.9528 .9790
20
1.2025
.3113
50
.3062 .4861
.3217 .5075
3.1084 .4925
.9520 .978(5
10
1.2595
.3142
18° 00'
.3090 .4900
.:-.2i9 ..-11*
3.0777 .4882
.9511 .97S2
72° 00'
1.2500
Value Logu
V III Ui- I.l'lTi,,
Value Loffio
Value I."L',n
DEGREKB
UADIANS
Oomra
COTANGENT
|'AXI;ENT
SINK
321
Table II. Values and Logarithms of Trigonometric Functions
[Characteristics of Logarithms oniitii-d — ilctcnnim' !>y the usual rule IVom tlic valucj
RADIANS
DEGREES
SINK
TANGENT
COTANGENT
COSINE
Value Log10
Value Log10
Value Log-10
Value Lopr10
.3142
18° 00'
.3090 .4900
.3249 .5118
3.0777 .4882
.9511 .9782
72° 00'
1.2500
.3171
10
.3118 .4939
.3281 .5101
3.0475 .4839
.9502 .9778
50
1.2537
.3200
20
.3145 .4977
.3314 .5203
3.0178 .4797
.9492 .9774
40
1.2508
.3221 i
30
.3173 .5015
.3340 .5245
2.9887 .4755
.9483 .9770
30
1.2479
.3258
40
.3201 .5052
.3378 .5287
2.9(500 .4713
.9474 .9705
20
1.2450
.3287
50
.3228 .5090
.3411 .5329
2.9319 .4671
.9465 .97(51
10
1.2421
.3316
19° 00'
.3250 .5120
.3443 .5370
2.9042 .4030
.9455 .9757
71° 00'
1.2392
.3345
10
.3283 .5103
.3476 .5411
2.8770 .4589
.944(5 .9752
50
1.2363
.3374
20
.3311 .5199
.3508 .5451
2.8502 .4549
.943(5 .9748
40
1.2334
.3403
30
.3338 .5235
.3541 .5491
2.8239 .4509
.9420 .9743
30
1.23! '5
.3432
40
.33<55 .5270
.3574 .5531
2.7980 .4409
.9417 .9739
20
1.2275
.341 W
50
.3393 .5300
.3007 .5571
2.7725 .4429
.9407 .9734
10
1.22-tfi
.3401
20° 00'
.3420 .5341
.3(540 .5011
2.7475 .4389
.9397 .9730
70° 00'
1 .2217
5520
10
.3448 .5375
.3073 .5050
2.7228 .4350
.9387 .9725
50
1.2188
.3649
20
.3475 .5409
.370(5 .5089
2.6985 .4311
.9377 .9721
40
1.2150
.3578
30
.3502 .5443
.3739 .5727
2.0740 .4273
.9367 .9716
30
1.2130
.3007
40
.3529 .5477
.3772 .5766
2.0511 .4234
.9350 .9711
20
1.2101
.3030
50
.3557 .5510
.3805 .5804
2.6279 .419(5
.9340 .9700
10
1.2072
.3005
21° 00'
..3584 .5543
.3839 .5842
2.6051 .4158
.9336 ,9702
69° 00'
1.2043
.3694
10
.3011 .5570
.3872 .5879
2.5826 .4121
.9325 .9097
50
1.2014
.3723
20
.3(538 .5009
.3900 .5917
2.5(505 .4083
.9315 .9092
40
1.1985
.3752
30
.3I505 .5041
.3939 .5954
2.5386 .4046
.9304 .9087
30
1.1950
.3782
40
.3092 .5(573
.3973 .5991
2.5172 .4009
.9293 .9(582
20
1.1920
.3811
50
.3719 .5704
.4000 .0028
2.4960 .3972
.9283 .1)077
10
1.1897
.3840
22° 00'
.3746 .5730
.4040 .6064
2.4751 .3936
.9272 .9(572
68° 00'
1.1808
.3860
10
.3773 .5707
.4074 .6100
2.4545 .3900
.9201 .1X507
50
1.1839
.3808
20
.3800 .5798
.4108 .0130
2.4342 .38(54
.9250 .9001
40
1.1810
.3027
30
.3827 .5828
.4142 .6172
2.4142 .3828
.9239 .1)050
30
1.1781
.35)66
40
.3854 .5859
.4176 .6208
2.3945 .3792
.9228 .9651
20
1.1752
.3983
50
.3881 .5889
.4210 .6243
2.3750 .3757
.9216 .9046
10
1.1723
.4014
23° 00'
.3907 .5919
.4245 .6279
2.3559 .3721
.9205 .9640
67° 00'
1.1094
.4043
10
.3934 .5948
.4279 .6314
2.3309 .3086
.9194 .9(535
50
I.IK;.-,
.4072
20
.3901 .5978
.4314 .6348
2.3183 .3652
.9182 .9629
40
1.10.30
.4102
30
.3987 .0007
.4348 .6383
2.2998 .3(517
.9171 .9024
30
1.1000
.4131
40
.4014 .0036
.4383 .6417
2.2817 .3583
.9159 .9018
20
1.1577
.41(50
50
.4041 .0005
.4417 .6452
2.2637 .3548
.W147 .9013
10
1.1548
.4189
24° 00'
.4067 .0093
.4452 .6486
2.24(50 .3514
.9135 .9007
66° 00'
1.1519
.4218
10
.4094 .6121
.4487 .0520
2.2286 .3480
.9124 .9602
50
1.1400
.4247
20
.4120 .6149
.4522 .(5553
2.2113 .3447
.9112 .9596
40
1.1401
.4270
30
.4147 .0177
.4557 .6587
2.1943 .3413
.9100 .9590
30
1.1432
.4300
40
.4173 .6205
.4592 .6620
2.1775 .3380
.9088 .9584
20
1.1403
.4334
50
.4200 .6232
.4628 .0654
2.1609 .3346
.9075 .9579
10
1.1374
.4303
25° 00'
.4220 .0259
.4063 .6687
2.1445 .3313
.9063 .9573
65° 00'
1.1346
.4392
10
.4253 .6280
.4699 .0720
2.1283 .3280
.9051 .95(57
50
1.1310
.4422
20
.4279 .6313
.4734 .0752
2.1123 .3248
.9038 .95(51
40
1.128H
.4451
30
.4305 .6340
.4770 .0785
2.09(55 .3215
.9020 .9555
30
1.1257
.4480
40
.4331 .6:366
.4800 .0817
2.0809 .3183
.9013 .9549
20
1.1228
.4501)
50
.4358 .0392
.4841 .0850
2.0(555 .3150
.9001 .9543
10
1.1190
.4538
26° 00'
.4384 .6418
.4877 .0882
2.0503 .3118
.8988 .9537
64° 00'
1.1170
.4507
10
.4410 .6444
.4913 .0914
2.0353 .3080
.8975 .95:50
50
1.1141
.4590
20
.4436 .6470
.4950 .0940
2.0204 .3054
.89(52 .9524
40
1.1112
.4(525
30
.4402 .6495
.4986 .15977
2.0057 .3023
.8949 .9518
30
1.1083
.4054
40
.4488 .6521
.5022 .7009
1.9912 .21)91
.8930 .9512
20
1.1054
.4(583
50
.4514 .6546
.5059 .7040
1.9768 .29(50
.8923 .9505
10
1.1025
.4712
27° 00'
.4540 .6570
.5095 .7072
1.9026 .2928
.8910 .9499
63° 00'
1.0996
Value Logjo
Value Login
Value Log10
Value Logjo
DEGREES
RADIANS
COSINE
COTANGENT
TANGENT
SINE
322
Table II. Values and Logarithms of Trigonometric Functions
[Characteristics of Logarithms omitted — determine by the usual rule from the value]
RADIANS
DEGREEK
SlXE
TANGENT
COTANGENT
COSINE
Value Log10
Value Log10
Value Log10
Value Log10
.4712
27° 00'
.4540 .6570
.5095 .7072
1.9(526 .2928
.8910 .9499
63° 00'
1.0996
.4741
10
.45(56 .6595
.5132 .7103
1.9486 .2897
.8897 .9492
50
1.09(5(5
.4771
20
.4592 .6620
.5169 .7134
1.9347 .28(56
.8884 .948(5
40
1.0937
.4800
30
.4617 .6644
.5206 .7165
1.9210 .2835
.8870 .9479
30
1.0908
.4829
40
.4643 .66(58
.5243 .7196
1.9074 .2804
.8857 .9473
20
1.0879
.4858
50
.4669 .6692
.5280 .7226
1.8940 .2774
.8843 .9466
10
1.0850
.4887
28° 00'
.4695 .6716
.5317 .7257
1.8807 .2743
.8829 .9459
62° 00'
1.0821
.4916
10
.4720 .(5740
.5354 .7287
1.8676 .2713
.8816 .9453
50
1.0792
.4945
20
.4746 .6763
.5392 .7317
1.8546 .2683
.8802 .944(5
40
1.07(53
.4974
30
.4772 .6787
.5430 .7348
1.8418 .2652
.8788 .9439
30
1.0734
.5003
40
.4797 .6810
.5467 .7378
1.8291 .2(522
.8774 .9432
20
1.0705
.5032
50
.4823 .6833
.5505 .7408
1.8165 .2592
.8760 .9425
10
1.0676
.5061
29° 00'
.4848 .6856
.5543 .7438
1.8040 .2562
.8746 .9418
61° 00'
1.0647
.5091
10
.4874 .6878
.5581 .7467
1.7917 .2533
.8752 .9411
50
1. 0(517
.5120
20
.4899 .6901
.5619 .7497
1.7796 .2503
.8718 .9404
40
1.05S8
.5149
30
.4924 .6923
.5658 .7526
1.7675 .2474
.8704 .9397
30
1.0559
.5178
40
.4950 .6946
.6696 .7556
1.7556 .2444
.8689 .9390
20
1.0530
.5207
50
.4975 .6968
.5735 .7585
1.7437 .2415
.8675 .9383
10
1.0501
.5236
30° 00'
.5000 .6990
.5774 .7614
1.7321 .2386
.8660 .9375
60° 00'
1.0472
.5265
10
.5025 .7012
.5812 .7644
1.7205 .2356
.8646 .9368
50
1.0443
.521)4
20
.5050 .7033
.5851 .7673
1.7090 .2327
.8631 .9361
40
1.0414
.5323
30
.5075 .7055
.5890 .7701
1.6977 i2299
.8(516 .9353
30
1.0385
.6352
40
.5100 .7076
.5930 .7730
1.6864 .2270
.8601 .9346
20
1.0356
.5381
50
.5125 .7097
.5969 .7759
1.6753 .2241
.8587 .9338
10
1.0327
.5411
31° 00'
.5150 .7118
.6009 .7788
1.6643 .2212
.8572 .9331
59° 00'
1.0297
.5440
10
.5175 .7139
.6048 .781(5
1.6534 .2184
-S557 .9323
50
1.0268
.5469
20
.5200 .7160
.6088 .7845
1.6426 .2155
.8542 .9:;i5
40
1.0239
.5498
30
.5225 .7181
.6128 .7873
1.6319 .2127
.8526 .9308
30
1.0210
.5527
40
.5250 .7201
.6168 .7902
1.6212 .2098
.8511 .9:500
20
1.0181
.5556
50
.5275 .7222
.6208 .7930
1.6107 .2070
.8496 .9292
10
1.0152
.5585
32° 00'
.5299 .7242
.6249 .7958
1.6003 .2042
.8480 .9284
58° 00'
1.0123
.5(514
10
.5324 .7262
.6289 .798(5
1.5900 .2014
.84(55 .927(5
50
1.0094
.5643
20
.5348 .7282
.6330 .8014
1.5798 .19S6
.8450 .9268
40
1.0065
.5(572
30
.5373 .7302
.6371 .8042
1.5697 .1958
.8434 .9260
30
1.003(5
.5701
40
.5398 .7322
.6412 .8070
1.55! )7 .1930
.8418 .9252
20
1.0007
.5730
50
.5422 .7342
.6453 .8097
1.5497 .1903
.8403 .9244
10
.9977
.5760
33° 00'
.5446 .7361
.6494 .8125
1.5399 .1875
.8387 .92:!6
57° 00'
.9948
0789
10
.5471 .7380
.(55.'56 .8153
1.5301 .1847
.8371 .9228
50
.9919
.5818
20
.5495 .7400
.6577 .8180
1.5204 .1820
.8.'555 .9219
40
.9890
.5847
30
.5519 .7419
.(',619 .8208
1.5108 .1792
.8339 .9211
30
.98(51
.5876
40
.5544 .743S
.6(561 .8235
1.5013 .17C.5
..s: !23 .9203
20
.9632
.5905
50
.5568 .7457
.(5703 .8263
1.4919 .1737
.8307 .9194
10
.9803
.5934
34° 00'
.5592 .7476
.6745 .8290
1.4826 .1710
.8290 .918(5
56° 00'
.9774
.5963
10
.5(516 .7494
.6787 .8317
1.4733 .1683
.8274 .9177
50
.9745
.5992
20
.5640 .7513
.68:50 .8344
1.4641 .1656
.8258:. 91(19
40
.9716
.6021
30
.5(5(54 .7531
.6873 .8371
1.4650 .1629
.8241 .91(50
30
.'.H1S7
.6050
40
.5<;ss .7550
.6916 .8398
1.44(50 .1602
.-S-J25 .9151
20
.9657
.6080
60
.5712 .7568
.6959 .8425
1.4370 .1575
.8208 .9142
10
.9628
.6109
35° 00'
.5736 .7586
.7002 .8452
1.4281 .1548
.8192 .9134
55° 00'
.9599
.6138
10
.57(50 .7(504
.Tdlil .8479
1.4193 .1521
.8175 .9125
50
.9570
.6167
20
.5783 .7622
.7089 .8506
1.4106 .1494
.8158 .9116
40
.9541
.6196
30
.5S07 .7640
.7133 .S5.0,:;
1.4019 .14(57
.8141 .9107
30
.9512
.6225
40
.5831 .7(157
.7177 .8559
1.3934 .1441
.8124 .9098
20
.9483
.6254
5i >
.5854 .7675
.7221 .8586
1.3848 .1414
.8107 .9089
10
.9454
.6283
36° 00'
.5878 .7692
.7265 .8613
1.37(54 .1387
.8090 .9080
54° 00'
.9425
Value Log10
Value Log10
Value Lop10
Value Log10
D EG BEES
RADIANS
COSINB
COTANGENT
TANGENT
SINE
323
Table n. Values and Logarithms of Trigonometric Functions
[Characteristics of Logarithms omitted — determine by the usual rule from the value]
KAIHA N
DEGREE
SINK
TANGENT
COTANGENT
COSINE
Value Logj
Value LOJBTJ
Value Logle
Value Logt
.6283
36° 00
.5878 .7692
.7265 .8613
1.3764 .1387
.8090 .9080
54° 00
.9425
.6312
10
.5901 .7710
.7310 .8639
1.3680 .1361
.8073 .9070
50
.9396
.6341
20
.5925 .7727
.7355 .8666
1.3597 .1334
.8056 .9061
40
.9367
.6370
30
.5948 .7744
.7400 .8692
1.3514 .1:308
.8039 .91152
30
.9338
.6400
40
.5972 .7761
.7445 .8718
1.3432 .1282
.8021 .9042
20
.9308
.6429
50
.5995 .7778
.7490 .8745
1.3351 .1255
.8004 .9033
10
.9279
.6458
37° 00
.6018 .7795
.7536 .8771
1.3270 .1229
.7986 .9023
53° 00
.9250
.6487
10
.6041 .7811
.7581 .8797
1.3190 .1203
.7969 .9014
50
.9221
.6516
20
.6065 .7828
.7627 .8824
1.3111 .1176
.7951 .9004
40
.9192
.6545
30
.6088 .7844
.7673 .8850
1.3032 .1150
.7934 .8995
30
.9163
.6574
40
.6111 .7861
.7720 .8876
1.2954 .1124
.7916 .8985
20
.9i;34
.6603
50
.6134 .7877
.7766 .8902
1.2876 .1098
.7898 .8975
10
.9105
.6632
38° 00'
.6157 .7893
.7813 .8928
1.2799 .1072
.7880 .8965
52° 00
.8076
.6661
10
.6180 .7910
.7860 .8954
1.2723 .1046
.7862 .8955
50
.9047
.6690
20
.6202 .7926
.7907 .8980
1.2647 .1020
.7844 .8945
40
.9018
.6720
30
.6225 .7941
.7954 .9006
1.2572 .0994
.7826 .8935
30
.8988
.6749
40
.6248 .7957
.8002 .9032
1.2497 .09(58
.7808 .8925
20
.8959
.6778
50
.6271 .7973
.8050 .9058
1.2423 .0942
.7790 .8915
10
.8930
.6807
39° 00'
.6293 .7989
.8098 .9084
1.2349 .0916
.7771 .8905
51° 00
.8901
.6836
10
.6316 .8004
.8146 .9110
1.2276 .0890
.7753 .8895
50
.8872
.6865
20
.6338 .8020
.8195 .9135
1.2203 .0865
.7735 .8884
40
.8843
.6894
30
.6361 .80:35
.8243 .9161
1.2131 .0839
.7716 .8874
30
.8814
.6923
40
.6383 .8050
.8292 .9187
1.205!) .0813
.7698 .8864
20
.8785
.6952
50
.6406 .8066
.8342 .9212
1.1988 .0788
.7679 .8853
10
.8756
.6981
40° 00'
.6428 .8081
.8391 .9238
1.1918 .0762
.7660 .8843
50° 00
.8727
.7010
10
.6450 .8096
.8441 .9264
1.1847 .0736
.7642 .8832
60
.8698
.7039
20
.6472 .8111
.8491 .9289
1.1778 .0711
.7623 .8821
40
.8668
.7069
30
.6494 .8125
.8541 .9315
1.1708 .0685
.7604 .8810
30
.8639
.7098
40
.6517 .8140
.8591 .9341
1.1640 .0659
.7585 .8800
20
.8610
.7127
50
.6539 .8155
.8642 .9366
1.1571 .0634
.7566 .8789
10
.8581
.7156
41° 00'
.6561 .8169
.8693 .9392
1.1504 .0608
.7547 .8778
49° 00
.8552
.7185
10
.6583 .8184
.8744 .9417
1.1436 .0583
.7528 .8767
50
.8523
.7214
20
.6604 .8198
.8796 .9443
1.1369 .0557
.7509 .8756
40
.841)4
.7243
30
.6626 .8213
.8847 .9468
1.1303 .0532
.7490 .8745
30
.8465
.7272
40
.6648 .8227
.8899 .9494
1.1237 .0506
.7470 .8733
20
.8436
.7301
50
.6670 .8241
.8952 .9519
1.1171 .0481
.7451 .8722
10
.8407
.7330
42° 00'
.6691 .8255
.9004 .9544
1.1106 .0456
.7431 .8711
48° 00'
.8378
.7359
10
.6713 .8269
.9057 .9570
1.1041 .0430
.7412 .86<)9
50
.8348
.7389
20
.6734 .8283
.9110 .9595
1.0977 .0405
.7392 .8688
40
.8319
.7418
30
.6756 .8297
.9163 .9621
1.0913 .0379
.7373 .8676
30
.8290
.7447
40
.6777 .8311
.9217 .9646
1.0850 .0354
.7353 .8665
20
.8261
.7476
50
.6799 .8324
.9271 .9671
1.0786 .0329
.7333 .8653
10
.8232
.7505
43° 00'
.6820 .8338
.9325 .9697
1.0724 .0303
.7314 .8641
47° 00'
.8203
.7534
10
.6841 .8351
.9380 .9722
1.0661 .0278
.7294 .8629
50
.8174
.7563
20
.6862 .8365
.9435 .9747
1.059!) .0253
.7274 .8618
40
.8145
.7592
30
.6884 .8378
.9490 .9772
1.0538 .0228
.7254 .8606
30
.8116
.7621
40
.6905 .8391
.9545 .9798
1.0477 .0202
.7234 .8594
20
.8087
.7650
50
.6926 .8405
.9601 .9823
1.0416 .0177
.7214 .8582
10
.8058
.7679
44° 00'
.6947 .8418
.9657 .9848
1.0355 .0152
.7193 .8569
46° 00'
.8029
.7709
10
.6967 .8431
.9713 .9874
1.0295 .0126
.7173 .8557
50
.7999
.7738
20
.6988 .8444
.9770 .9899
1.0235 .0101
.7153 .8545
40
.7970
.7767
30
.7009 .8457
.9827 .9924
1.0176 .0076
.7133 .8532
30
.7941
.7796
40
.7030 .8469
.9884 .9949
1.0117 .0051
.7112 .8520
20
.7912
.7825
50
.7050 .8482
.9942 .9975
1.0058 .0025
.7092 .8507
10
.7883
.7854
45° 00'
.7071 .8495
1.0000 .0000
1.0000 .0000
.7071 .8495
45° 00
.7854
Value Log10
Value Login
Value Log]0
Value Log10
)EGREES
vADIANS
COSINE
COTANGENT
TANGENT
SINE
324
Table III. Radian Measure — Trigonometric Functions
Had.
Deg. Mln.
sin.
cos.
tan.
Rad.
Deg. Mln.
sin.
cos.
tan
0.0
0 0
0
1
0
3.2
183 20.8
-.058
-.998
.058
0.1
5 43.8
.100
.995
.100
3.3
189 4.6
-.158
-.987
.161
0.2
11 27.5
.199
.980
.203
3.4
194 48.3
-.255
-.967
.264
0.3
17 11.3
.296
.955
.309
3.5
200 32.1
-.351
-.936
.375
0.4
22 55.1
.389
.921
.423
3.6
206 15.9
-.443
-.897
.493
0.5
28 38.9
.479
.878
.546
3.7
211 59.7
-.530
-.848
.625
0.6
34 22.6
.565
.825
.684
3.8
217 43.4
-.612
-.791
.774
0.7
40 6.4
.644
.765
.842
3.9
223 27.2
-.688
-.726
.947
0.8
45 50.2
.717
.697
1.030
4.0
229 11.0
-.757
-.654
1.158
0.9
51 34.0
.783
.622
1.260
4.1
234 54.8
-.818
-.575
1.424
1.0
57 17.7
.841
.540
1.557
4.2
240 38.5
-.872
-.490
1.778
1.1
63 1.5
.891
.454
1.965
4.3
246 22.3
-.916
-.401
2.286
1.2
68 45.3
.932
.362
2.572
4.4
252 6.1
-.952
-.307
3.096
1.3
74 29.1
.964
.267
3.602
4.5
257 49.9
-.978
-.211
4.638
1.4
80 12.8
.985
.170
5.798
4.6
263 33.6
-.994
-.112
8.859
1.5
85 56.6
.997
.071
14.101
4.7
269 17.4
-1.00
-.012
80.713
1.6
91 40.4
1.000
-.029
-34.233
4.8
275 1.2
-.996
.088
-11.385
1.7
97 24.2
.992
-.129
- 7.700
4.9
280 45.0
-.982
.187
- 5.267
1.8
103 7.9
.974
-.227
- 4.286
5.0
286 28.6
-.959
.284
- 3.381
1.9
108 51.7
.946
-.323
- 2.927
5.1
292 12.5
-.926
.378
- 2.449
2.0
114 35.5
.909
-.416
- 2.185
5.2
297 56.3
-.883
.469
- 1.885
2.1
120 19.3
.863
-.505
- 1.710
5.3
303 40.1
-.832
.554
- 1.501
2.2
126 3.0
.808
-.588
- 1.374
5.4
309 23.8
-.773
.635
- 1.217
2.3
131 46.8
.746
-.666
- 1.119
5.5
315 7.6
-.706
.709
- .996
2.4
137 30.6
.675
-.737
.917
5.6
320 51.4
-.631
.776
- .814
2.5
143 14.4
.598
-.801
.747
5.7
326 35.2
-.551
.835
- .600
2.6
148 58.1
.516
-.857
- .602
5.8
332 18.9
-.465
.886
- .525
2.7
154 41.9
.427
-.904
- .473
5.9
338 2.7
-.374
.927
- .403
2.8
160 25.7
.335
-.942
- .356
6.0
343 46.5
-.279
.960
- .291
2.9
166 9.5
.239
-.971
- .246
6.1
349 30.3
-.182
.983
- .185
3.0
171 53.2
.141
-.990
.143
6.2
355 14.0
-.083
.997
- .083
3.1
177 37.0
.042
-.999
.042
6.3
360 57.8
+.017
1.000
+ .017
325
Table IV. Squares and Cubes Square Roots and Cube Roots
No.
SQUARE
C'UUE
SQUARE
UOOT
CUBE
ItOOT
No.
SQUARE
CUBE
SQUARE
KOOT
ClT.E
HOOT
1
1
1
1.000
1.000
51
2,601
132,651
7.141
3.708
2
4
8
1.414
1.260
52
2,704
140,608
7.211
3.733
3
9
27
1.732
1.442
53
2,809
148,877
7.280
3.75(5
4
16
64
2.000
1.587
54
2,916
157,464
7.348
3.780
5
25
125
2.236
1.710
55
3,025
166,375
7.416
3.803
6
36
216
2.449
1.817
56
3,136
175,616
7.483
3.826
7
49
343
2.646
1.913
57
3,249
185,193
7.550
3.84!)
8
64
512
2.828
2.000
58
3,364
195,112
7.616
3.871
9
81
729
3.000
2.080
59
3,481
205,379
7.681
3.893
10
100
1,000
3.162
2.154
60
3,600
216,000
7.746
3.915
11
121
1,331
3.317
2.224
61
3,721
226,981
7.810
3.936
12
144
1,728
3.464
2.289
62
3,844
238,328
7.874
3.968
13
109
2,197
3.606
2.351
63
3,969
250,047
7.937
3.979
14
196
2,744
3.742
2.410
64
4,096
262,144
8.000
4.000
15
225
3,375
3.873
2.466
65
4,225
274,625
8.062
4.021
16
256
4,096
4.000
2.520
66
4,356
287,496
8.124
4.041
17
.289
4,913
4.123
2.571
67
4,489
300,763
8.185
4.062
18
324
5,832
4.243
2.621
68
4,624
314,432
8.246
4.082
19
361
6,859
4.359
2.668
69
4,761
328,509
8.307
4.102
20
400
8,000
4.472
2.714
70
4,900
343,000
8.367
4.121
21
441
9,261
4.583
2.759
71
5,041
357,911
8.426
4.141
22
484
10,648
4.690
2.802
72
5,184
373,248
8.485
4.160
23
529
12,167
4.796
2.844
73
5,329
389,017
8.544
4.17H
24
576
13,824
4.899
2.884
74
5,476
405,224
8.602
4.198
25
625
15,625
5.000
2.924
75
5,625
421,875
8.660
4.217
26
676
17,576
5.099
2.962
76
5,776
438,!>76
8.718
4.236
27
729
19,683
5.196
3.000
77
5,929
456,533
8.775
4.254
28
784
21,952
5.292
3.037
78
6,084
474,552
8.832
4.273
29
841
24,389
5.385
3.072
79
6,241
493,039
8.888
4.291
30
900
27,000
5.477
3.107
80
6,400
512,000
8.944
4.:50!>
31
961
29,791
5.568
3.141
81
6,561
531,441
9.000
4. :','21
32
1,024
32,768
5.657
3.175
82
6,724
551,368
9.055
4.:U4
33
1,089
35,937
5.745
3.208
83
6,889
571,787
9.110
4.3(52
34
1,156
39,304
5.831
3.240
84
7,056
592,704
9.165
4.380
35
1,225
42,875
5.916
3.271
85
7,225
614,125
9.220
4.397
36
1,2'W
46,656
6.000
3.302
86
7,396
636,056
9.274
4.414
37
1,3(59
50,653
6.083
3.332
87
7,569
658,503
9.327
4.431
38
1,444
54,872
6.164
3.362
88
7,744
681,472
9.381
4.448
39
1,521
59,319
6.245
3.391
89
7,921
704,969
9.434
4.4(;r,
40
1,600
64,000
6.325
3.420
90
8,100
729,000
9.487
4.481
41
1,681
68,921
6.403
3.448
91
8,281
753,671
9.588
4.49S
42
1,764
74,088
6.481
3.476
92
8,464
778,688
9.592
4.614
43
1,849
79,507
6.557
3.503
93
8,649
804,a57
9.644
4.631
44
1,<>:!6
85,184
6.633
3.530
94
8,836
HlJu,.-^
9.698
4.547
45
2,025
91,125
6.708
3.557
95
9,025
857,370
!>.747
4.56.",
46
2,116
97,336
6.782
3.583
96
9,216
884,736
9.798
4.579
47
2,20!»
103,823
6.856
3.609
97
9,409
912,673
9.849
4.595
48
2,304
110,592
6. 928
3.634
98
9,604
941,192
9.899
4.610
49
2.401
117,649
7.000
3.<>59
99
9,801
970,299
9.950
4.626
50
2,500
125,000
7.071
3.684
100
10,000
1,000,000
10.000
4.1142
For a more complete table, see THE MACMILLAN TABLES, pp. 94-111.
326
Table V. Logarithms of Important Constants
Ar = NUMBER
VALUE OF JV
LOGjo ff
f
3.14159265
0.49714987
1 + *
0.31830989
9.50285013
X*
9.86960440
0.99429975
V^r
1.77245385
0.24857494
e = Napierian Base
2.71828183
0.43429448
M = logio e
0.43429448
9.63778431
1 -H M = loge 10
2.30258509
0.36221569
! 180 -i- r = degrees in 1 radian
57.2957795
1.75812262
IT -=- 180 = radians in 1°
0.01745329
8.24187738
IT -T- 10800 = radians in 1'
0.0002908882
6.4637261
x H- 648000 = radians in 1"
0.000004848136811095
4.68557487
sin 1"
0.000004848136811076
4.68557487
tan 1"
0.000004848136811152
4.68557487
centimeters in 1 ft.
30.480
1.4840158
feet in 1 cm.
0.032808
8.5159842
inches in 1 m.
39.37
1.5951654
pounds in 1 kg.
2.20462
0.3433340
kilograms in 1 Ib.
0.453593
9.6566660
g
32.16 ft. /sec. /sec.
1.5073160
= 981 cm. /sec. /sec.
2.9916690
weight of 1 cu. ft. of water
62.425 Ib. (max. density)
1.7953586
weight of 1 cu. ft. of air
0.0807 Ib. (at 32° F.)
8.9068735
cu. in. in 1 (U. S.) gallon
231.
2.3636120
ft. Ib. per sec. in 1 H. P.
550.
2.7403627
kg. m. per sec. in 1 H. P.
76.0404
1.8810445
watts in 1 H. P.
745.957
2.8727135
Table VI. Degrees to Radians
1°
.01745
10°
.17453
100°
1.74533
6'
.00175
6'
.00003
2°
.03491
20°
.34907
110°
1.91986
7'
.00204
7'
.00003
3°
.05236
30°
.52360
120°
2.09440
8'
.00233
8'
.00004
4°
.06981
40°
.69813
130°
2.26893
9'
.00262
9'
.00004
5°
.08727
50°
.87266
140°
2.44346
10'
.00291
10'
.00005
6°
.10472
60°
1.04720
150°
2.61799
20'
.005S2
20'
.00010
7°
.12217
70°
1.22173
160°
2.79253
30'
.00873
30'
.00015
8°
.13963
80°
1.39626
170°
2.96706
40'
.01164
40'
.00019
9°
.15708
90°
1.57080
180°
3.14159
50'
.01454
50'
.00024
327
Table VII. Compound Interest Table
Amount of One Dollar Principal with Compound Interest at Various Rates.
a
•<
H
P
2} Per
Cent.
3 Per
Cent.
3J Per
Cent.
4 Per
Cent.
4.^ Per
Cent.
5 Per
Cent.
5J Per
Cent.
6 Per
Cent.
6£ Per
Cent.
7 Per
Cent.
8 Per
Cent.
1
$1.025
$1.030
$1.035
$1.040
$1.045
$1.050
$1.055
$1.060
$1.065
$1.070
$1.800
2
1.051
1.061
1.071
1.082
1.092
1.103
1.113
1.124
1.134
1.145
1.166
3
1.077
1.093
1.109
1.125
1.141
1.158
1.174
1.191
1.208
1.225
1.260
4
1.104
1.126
1.148
1.170
1.193
1.216
1.239
1.262
1.286
1.311
1.360
5
1.131
1.159
1.188
'1.217
1.246
1.276
1.307
1.338
1.370
1.403
1.469
6
1.160
1.194
1.229
1.265
1.302
1.340
1.379
1.419
1.459
1.501
1.587
7
1.189
1.230
1.272
1.316
1.361
1.407
1.455
1.504
1.554
1.606
1.714
8
1.218
1.267 1.317
1.369
1.422
1.477
1.535
1.594
1.655
1.718
1.851
9
1.249
1.305 1 1.363
1.423
1.486
1.551
1.619
1.689
1.763
1.838
1.999
10
1.280
1.344
1.411
1.480
1.553
1.629
1.708
1.791
1.877
1.967
2.159
11
1.312
1.384
1.460
1.539
1.623
1.710
1.802
1.898
1.999
2.105
2.332
12
1.345
1.426
1.511
1.601
1.696
1.796
1.901
2.012
2.129
2.252
2.518
13
1.379
1.469
1.564
1.6G5
1.772
1.886
2.006
2.133
2.267
2.410
2.720
14
1.413
1.513
1.619
1.732
1.852
1.980
2.116
2.261
2.415
2.579
2.937
15
1.448
1.558
1.675
1.801
1.935
2.079
2.232
2.397
2.572
2.759
3.172
16
1.485
1.605
1.734
1.873
2.022
2.183
2.355
2.540
2.739
2.952
3.426
17
1.522
1.653
1.795
1.948
2.113
2.292
2.485
2.693
2.917
3.159
3.700
18
1.560
1.702
1.857
2.026
2.208
2.407
2.621
2.854
3.107
3.380
3.996
19
1.599
1.754
1.923
2.107
2.308
2.527
2.766
3.026
3.309
3.617
4.316
20
1.639
1.806
1.990
2.191
2.412
2.653
2.918
3.207
3.524
3.870
4.661
21
1.680
1.860
2.059
2.279
2.520
2.786
3.078
3.400
3.753
4.141
5.034
22
1.722
1.916
2.132
2.370
2.634
2.925
3.248
.3.604
3.997
4.430
5.437
23
1.765
1.974
2.206
2.465
2.752
3.072
3.426
3.820
4.256
4.741
5.871
24
1.809
2.033
2.283
2.563
2.876
3.225
3.615
4.049
4.533
5.072
6.341
25
1.854
2.094
2.363
2.666
3.005
3.386
3.813
4.292
4.828
5.427
6.848
26
1.900
2.157
2.446
2.772
3.141
3.556
4.023
4.549
5.142
5.807
7.396
27
1.948
2.221
2.532
2.883
3.282
3.733
4.244
4.822
5.476
6.214
7.988
28
1.996
2.288
2.620
2.999
3.430
3.920
4.478
5.112
5.832
6.649
8.627
29
2.046
2.357
2.712
3.119
3.584
4.116
4.724
5.418
6.211
7.114
9.317
30
2.098
2.427
2.807
3.243
3.745
4.322
4.984
5.743
6.614
7.612
10.063
31
2.150
2.500
2.905
3.373
3.914
4.538
5.258
6.088
7.044
8.145
10.868
32
2.204
2.575
3.007
3.508
4.090
4.765
5.547
6.453
7.502
8.715
11.737
33
2.259
2.652
3.112
3.648
4.274
5.003
5.852
6.841
7.990
9.325
12.676
34
2.315
2.732
3.221
3.794
4.466
5.253
6.174
7.251
8.509
9.978
13.690
35
2.373
2.814
3.334
3.946
4.667
5.516
6.514
7.686
9.062
10.677
14.785
36
2.433
2.898
3.450
4.104
4.877
5.792
6.872
8.147
9.651
11.424
15.968
37
2.493
2.985
3.571
4.268
5.097
6.081
7.250
8.636
10.279
12.224
17.246
38
2.556
3.075
3.696
4.439
5.326
6.385
7.649
9.154
10.947
13.079
18.625
39
2.620
3.167
3.825
4.616
5.566
6.705
8.069
9.704
11.658
13.995
20.115
40
2.685
3.262
3.959
4.801
5.816
7.040
8.513
10.286
12.416
14.974
21.725
41
2.752
3.360
4.098
4.993
6.078
7.392
8.982
10.903
13.223
16.023
23.462
42
2.821
3.461
4.241
5.193
6.352
7.762
9.476
11.557
14.083
17.144
25.339
43
2.892
3.565
4.390
5.400
6.637
8.150
9.997
12.250
14.998
18.344
27.367
44
2.964
3.671
4.543
5.617
6.936
8.557
10.547
12.985
15.973
19.628
29.556
45
3.038
3.782
4.702
5.841
7.248
8.985
11.127
13.765
17.011
21.002
31.920
46
3.114
3.895
4.867
6.075
7.574
9.434
11.739
14.590
18.117
22.473
34.474
47
3.192
4.012
5.037
6.318
7.915
9.906 12.384
15.466
19.294
24.046
37.232
48
3.271
4.132
5.214
6.571
8.271
10.401 13.065
16.394
20.549
25.729
40.211
49
3.353
4.256
5.396
6.833
8.644
10.921
13.784
17.378
21.884
27.530
43.427
50
3.437
4.384
5.585
7.107
9.033
11.467
14.542
18.420
23.307
29.457
46.902
328
Table VIII. American Experience Mortality Table
Based on 100,000 living at age 10.
At
Number
At
Number
At
Number
At
Number
Age.
Surviving.
Deaths.
Age.
Surviving.
Deaths.
Age.
Surviving.
Deaths.
Age.
Surviving.
Deaths.
10
100,000
749
35
81,822
732
60
57,917
1,546
85
5,485
1,292
11
99,251
746
36
81,090
737
61
56,371
1,628
86
4,193
1,114
12
98,505
743
37
80,353
742
62
54,743
1,713
87
3,079
933
13
97,762
740
38
79,611
749
63
53,030
1,800
88
2,146
744
14
97,022
737
39
78,862
756
64
51,230
1,889
89
1,402
555
IS
96,285
735
40
78,106
765
65
49,341
1,980
90
847
385
16
95,550
732
41
77,341
774
66
47,361
2,070
91
462
246
17
94,818
729
42
76,567
785
67
45,291
2,158
92
216
137
18
94,089
727
43
75,782
797
68
43,133
2,243
93
79
58
19
93,362
725
44
74,985
812
69
40,890
2,321
94
21
18
20
92,637
723
45
74,173
828
70
38,569
2,391
95
3
3
21
91,914
722
46
73,345
848
71
36,178
2,448
22
91,192
721
47
72,497
870
72
33,730
2,487
23
90,471
720
48
71,627
896
73
31,243
2,505
24
89,751
719
49
70,731
927
74
28,738
2,501
25
89,032
718
50
69,804
962
75
26,237
2,476
26
88,314
718
51
68,842
1,001
76
23,761
2,431
27
87,596
718
52
67,841
1,044
77
21,330
2,369
28
86,878
718
53
66,797
1,091
78
18,961
2,291
29
86,160
719
54
65,706
1,143
79
16,670
2,196
30
85,441
720
55
64,563
1,199
80
14,474
2,091
31
84,721
721
56
63,364
1,260
81
12,383
1,964
32
84,000
723
57
62,104
1,325
82
10,419
1,816
33
83,277
726
58
60,779
1,394
83
8,603
1,648
34
82,551
729
59
59,385
1,468
84
6,955
1,470
329
Table IX. Heights and Weights of Men
Light-face figures are 20 per cent, under and over the average.
AGES.
S
£
01
I
•*
n
i
O
n
i
IO
w
105
131
157
1
OJ
I
••#
107
134
161
s
s
107
134
161
§
I
10
i<
|
cq
OJ
<N
I
C*
•*
«
A
123
154
185
a
3
00
5
i
OJ
2
•*
129
161
193
50-54
o
B
1
10
130
163
196
Ft.
5
In.
0
96
120
144
100
125
150
102
128
154
106
133
160
107
134
161
Ft.
5
In.
8
117
146
175
121
151
181
126
157
188
128
160
192
130
163
196
1
98
122
146
101
126
151
103
129
155
105
131
157
107 109
134 136
161 163
109 109
136 136
163 163
9
120 124 127
150155 159
180186191
130 132
162 165
194 198
133
166
199
134
167
200
134
168
202
2
3
4
5
99
124
149
102
127
152
102
128
154
105
131
157
105
131
157
106
133
160
109
136
163
109
136
163
110
138
166
113
141
169
110
138
166
113
141
169
110
138
166
113
141
169
10
11
123
154
185
127
159
191
127 131
159 164
191 197
134
167
200
136 137
170 171
204205
138
172
206
142
177
212
138
173
208
142
178
214
107
134
161
111
139
167
131
164
197
135
169
203
138
173
208
140 142
175 177
210212
105 108 110
131 135 138
157:162 166
112
140
168
114
143
172
115 116
144 145
173 174
116
145
174
6
0
132 136 140 143
165 170 175 179
198 204 210 215
144 146' 146 146
180 183 182 183
216220218220
107
134
161
110
138
166
113
141
169
114
143
172
118
147
176
117
146
175
118
147
176
119
149
179
119
149
179
1
136142
170 177
204;212
145 148
181 185
217 222
149 151
186 189
223 227
150 151
188 189
226 227
6
110114
138 142
166170
iir,
145
174
120
150
180
124
155
186
121
151
181
122
153
184
122
153
184
2
141 147
176 184
211 221
150
188
226
154 155! 157
192 194 196
230:233 235
166
194
233
155
194
233
7
114118
142 147
170 176
i
120
150
180
122
152
182
125
156
187
126
158
190
126
158
190
3
145
181
217
152
190
228
156
195
234
160
200
240
1621163
203204
244,245i
161
201
241
158
198
238
330
FOUR PLACE TABLES 33 J
EXPLANATION OF TABLE II*
VALUES AND LOGARITHMS OF TRIGONOMETRIC FUNCTIONS
1. DIRECT READING OF THE VALUES. This table gives the sines,
cosines, tangents and cotangents of the angles from 0° to 45°; and by
a simple device, indicated by the printing, the values of these functions
for angles from 45° to 90° may be read directly from the same table.
For angles less than 45° read down the page, the degrees and minutes
being found on the left; for angles greater than 45° read up the page
the degrees and minutes being found on the right.
To find a function of an angle (such as 15° 27', for example) we
employ the process of interpolation. To illustrate, let us find tan 15°
27'. In the table we find tan 15° 20' = .2742 and tan 15° 30' = .2773;
we know that tan 15° 27' lies between these two numbers. The process
of interpolation depends on the assumption that between 15° 20' and
15° 30' the tangent of the angle varies directly as the angle; while this
assumption is not strictly true, it gives an approximation sufficiently
accurate for a four-place table. Thus we should assume that tan
15° 25' is halfway between .2742 and .2773. We may state the problem
as follows: An increase of 10' in the angle increases the tangent .0031;
assuming that the tangent varies as the angle, an increase of 7' in the
angle will increase the tangent by .7 X .0031 = .00217. Retaining only
four places we write this .0022. Hence
tan 15° 27' = .2742 + .0022 = .2764.
The difference between two successive values in the table is called
the tabular difference (.0031 above). The proportional part of the
tabular difference which is used is called the correction (.0022 above),
and is found by multiplying the tabular difference by the appropriate
fraction (.7 above).
Example 1. Find sin 63° 52'.
\Vefind
sin 63° 50' = .8975.
tabular difference = .0013 (subtracted mentally from the table).
correction = .2 X .0013 = .0003 (to be added).
Hence,
sin 63° 52' = .8978.
* The use of Table I. is explained on pages 80-86 of the text.
332 MATHEMATICS
Example 2. Find tan 37° 44'.
tan 37° 40' = .7720
tabular difference = .0046
correction = .4 X .0046 = .0018.
Hence,
tan 37° 44' = .7738.
Example 3. Find cos 65° 24'.
cos 65° 20' = .4173
tabular difference = 26; .4 X 26 = 10
(to be subtracted because the cosine decreases as the angle increases).
Hence
cos 65° 24' = .4163.
Example 4. Find ctn 32° 18'.
ctn 32° 10' = 1.5900
tabular difference = 102; .8 X 102 = 82 (to be subtracted).
Hence,
ctn 32° 18' = 1.5818.
Rule. To find a trigonometric function of an angle by interpolation:
select the angle in the table which is next smaller than the given angle, and
read its sine (cosine, tangent, or cotangent as the case may be) and the
tabular difference. Compute the correction as the proper proportional
part of the tabular difference. In case of sines or tangents ADD the cor-
rection: in case of cosines or cotangents, SUBTRACT it.
2. REVERSE READINGS. Interpolation is also used in finding the
angle when one of its functions is given.
Example 1. Given sin x = .3294, to find x.
Looking in the table we find the sine which is next less than the given
sine to be .3283, and this belongs to 19° 10'. Subtract the value of the
sine selected from the given sine to obtain the actual difference = .0011;
note that the tabular difference = .0028. We may state the problem
as follows: an increase of .0028 in the function increases the angle 10';
then aa increase of .0011 in the function will increase the angle 11/28
of 10 = 4 (to be added). Hence x = 19° 14'.
Example 2. Given cos x = .2900, to find x.
The cosine in the table next less than this is .2896 and belongs to
73° 10'; the tabular difference is 28; the actual difference is 4; correction
= 4/28 of 10 = 1 (to be subtracted). Hence x = 73° 9'.
FOUR PLACE TABLES 333
Rule. To find an angle when one of its trigonometric functions is
given: select from the table the same named function which is next less than
the given function, noting the corresponding angle and the tabular differ-
ence: compute the actual difference (between the selected value of the func-
tion and the given value), divide it by the tabular difference, and multiply
the result by 10; this gives the correction which is to be added if the given
function is sine or tangent, and to be subtracted if the given function is
cosine or cotangent.
3. THE LOGARITHMS OF THE TRIGONOMETRIC FUNCTIONS. If it is
required to find log sin 63° 52', the most obvious way is to find sin 63° 52'
= .8978, and then to find in Table I, log .8978 = 9.9532 - 10, but this
involves consulting two tables. To avoid the necessity of doing this,
Table II gives the logarithms of the sines, cosines, tangents, and co-
tangents. The student should note that the sines and cosines of all
acute angles, the tangents of all acute angles less than 45° and the
cotangents of all acute angles greater than 45° are proper fractions,
and their logarithms end with — 10, which is not printed in the table,
but which should be written down whenever such a logarithm is used.
Example 1. Find log sin 58° 24'.
In the row having 58° 20' on the right and in the column having
sine at the bottom find log sin 58° 20' = 9.9300 - 10; the tabular differ-
ence is 8; correction = .4 X 8 = 3 (to be added). Hence
log sin 58° 24' = 9.9303 - 10.
(In case of sine and tangent add the correction.)
Example 2. Find log cos 48° 38'.
log cos 48° 30' = 9.8213 - 10, tabular difference 15;
.8 X 15 = 12 (subtract) therefore log cos 48° 38' = 9.8201 - 10.
(In case of cosine and cotangent, subtract the correction.)
Example 3. Given log tan x = 0.0263, to find x.
The log tan in Table II next less than the given one is 0.0253 and
belongs to 46° 40'; actual difference is 10; tabular difference is 25;
correction = 10/25 of 10 = 4. Hence x = 46° 44'.
Example 4. Given log cos x = 9.9726 — 10, to find x.
The logarithmic cosine next less than the given one is 9.9725 — 10
and belongs to 20° 10'; actual difference = 1; tabular difference = 5;
correction = 1/5 X 10 = 2 (subtract). Hence x = 20° 8'.
INDEX
Abscissa, 38
Addition formulas, 116
Angles, 91, 94, 101, 104, 106
trigonometric functions of,
91, 102
Annuity, amount of, 254, 2551
cost of, 259
present value of, 254, 258
Area, by offsets, 147
by rectangular coordinates,
147
of ellipse, 202
of triangle, 134
Associative law, 3
Asymptote, 205, 206
Auxiliary circle, 202
Average, 262
arithmetic, 262, 263
weighted arithmetic, 263
geometric, 265
Axis, 38, 61, 196, 199
Bearing, 140
Binomial coefficients, 8, 277
series, 281
theorem, 7, 276, 278
Characteristic, 77
Circle, 190
auxiliary, 202
Coefficients, 23
binomial, 8, 277
variability, 300
Cologarithm, 84
Combinations, 270, 273
Commutative law, 3
Components, of a force, 156
rectangular, 157
Compound interest, 255, 256,
286
Conic sections, 190
Coordinates, 38
Corners, 142
Correlation, 304
coefficient of, 307
Correlation, measure of, 305
table, 306
Cosecant, 92, 103
Cosines, 91, 102
law of, 122
Cotangent, 92, 103
Couple, 168
Crane, 160
Degree, 23
Deviation, standard, 298
Diagrams, 41, 47, 49, 50, 51
Directrix, 195
Distance between two points, 53
Distributive law, 3
Division, point of, 55
ratio of, 54
Eccentricity, of ellipse, 200
of hyperbola, 204
Elimination, 17
Ellipse, 190, 199
area of, 202
Equation, of a circle, 190
of a curve, 58
of an ellipse, 201
of a hyperbola, 204, 205,
206
of a parabola, 195, 197
of a straight line, 63, 64, 66,
67
Equations, definition of, 11
conditional, 13
empirical, 226
equivalent, 14
general, 68, 191
having given roots, 32
in quadratic form, 29
linear, 18, 24, 68
quadratic, 24
simultaneous, 16
trigonometric, 107
transformation of, 14
Equilibrium, conditions of, 169
Error, 177, 181
335
336
INDEX
Error, curve of, 298
in a fraction, 179
in parts of a triangle, 185
in a product, 178
in a square, 183
in a square root, 183
in a sum, 178
in trigonometric functions,
183
probable, 300
Evolution, 4
Expectation, 292
Exponents, 4, 5, 6, 23
Focal properties, 211, 213
Focus, 195, 199, 201, 204
Forces, components of, 156
concentrated, 154
distributed, 154
graphical representation of,
154
moments of, 167
parallelogram of, 155
resolution of, 156
Frequency distribution curves,
296
Functions, 218
of complementary angles, 93
of half an angle, 118
of negative angles, 110
of twice an angle, 118
periodic, 110
trigonometric, 91, 102, 105,
109
Fundamental relations, 95, 104
Graphical solution, 36, 99, 154,
158, 220, 228, 238
Graphs, 41, 113, 154, 220
Hyperbola, 190, 204
equilateral or rectangular,
205
Identities, 12
Imaginary numbers, 30
Intercepts, 61
Interpolation, 81, 331
Intersection, points of, 62
of conies, 215
of loci, 62, 208
Involution, 3
Irrational numbers, 2
Latus rectum, 196
Law, associative, 3
commutative, 3
distributive, 3
of cosines, 122
of exponents, 6
of sines, 121
of tangents, 124
Lines, base, 140
bearing of, 140
parallel, 63, 67
perpendicular, 67
random, 146
range, 140
slope of, 66
through the origin, 64
through two points, 66
township, 140
Locus, of a point, 57
of an equation, 58, 59
Logarithmic paper, 239
plotting, 237
Logarithms, Briggs, 76
computation by, 87
computation of, 76
definition of, 72
Napierian, 76
properties of, 74
Mantissa, 77
Mass, 153
Mean, arithmetic, 245, 262, 263
geometric, 249, 265
Measurement, 1, 181
on level ground, 143
on slopes, 144
of force, 154
Median, 264
Mendel's law, 282
Middle point, 56
Mode, 264
Moments, of force, 167
center of, 167
composition of, 167
Momentum, 153
Normal, 211
Number, 2, 3, 30, 31
INDEX
337
Oblique triangles, solution of,
120, 125
Offsets, 144
Ordinate, 38
Origin, 38
Parabola, 190, 195
Parallelogram of forces, 155
Periodic functions, 110
Permutations, 270
Perpetuity, 260
Point, of division, 55
of intersection, 62
Polygon of forces, 162
Polynomial, 22
Principal meridian, 140
Probability, 291
curve, 297, 298
Probable error, in a single
measurement, 300
of arithmetic average, 300
of standard deviation, 301
Progression, arithmetic, 243
geometric, 247, 252
Proportional quantities, 64, 219,
220
Quadrantal angles, 104
Quadratic equation, 24
kind of roots, 30, 33
number of roots, 33
solution of, 26, 27
sum and product of roots,
32
Radian, 113
Ratio of division, 54
Rational number, 2
Rectangular components, 157
coordinates, 38
Rectangular hyperbola, 205, 206
Regression curve, 309
Resolution of forces, 156
Resultant, 155
of concurrent forces, 163
of parallel forces, 165
Right triangles, solution of, 97
Root, 12
Scales, 36
Secant, 92, 102
Series, binomial, 281
infinite geometric, 252
Sines, 91, 102
law of, 121
Slide rule, 88
Slope, 66
Statistical data, 40
Substitution, 12, 233
Symmetry, 61
Tabular difference, 81, 331
Tangents, 91, 102, 211
law of, 124
Translation of axes, 194
Triangle, oblique, 120, 125
of forces, 158
right, 97
Trigonometric functions, of an
acute angle, 91
of any angle, 102
graphs of, 109
line representation of, 105
Variable, 218
Variation, 219
direct, 219
inverse, 219
joint, 219
constant of, 219, 222
Printed in the United States of America.
TRIGONOMETRY
BY
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Edited by EARLE RAYMOND HEDRICK, Professor of Mathematics
in the University of Missouri
Plane and Solid Geometry, doth, izmo, 319 pp., $125
Plane Geometry, cloth, I2mo, 213 pp., $ .80
Solid Geometry, doth, I2mo, 106 pp., $ .80
STRONG POINTS
I. The authors and the editor are well qualified by training and experi-
ence to prepare a textbook on Geometry.
II. As treated in this book, geometry functions in the thought of the
pupil. It means something because its practical applications are shown.
III. The logical as well as the practical side of the subject is emphasized.
IV. The arrangement of material is pedagogical.
V. Basal theorems are printed in black-face type.
VI. The book conforms to the recommendations of the National Com-
mittee on the Teaching of Geometry.
VII. Typography and binding are excellent. The latter is the reenforced
tape binding that is characteristic of Macmillan textbooks.
"Geometry is likely to remain primarily a cultural, rather than an informa-
tion subject,"" say the authors in the preface. " But the intimate connection
of geometry with human activities is evident upon every hand, and constitutes
fully as much an integral part of the subject as does its older logical and
scholastic aspect." This connection with human activities, this application
of geometry to real human needs, is emphasized in a great variety of problems
and constructions, so that theory and application are inseparably connected
throughout the book.
These illustrations and the many others contained in the book will be seen
to cover a wider range than is usual, even in books that emphasize practical
applications to a questionable extent. This results in a better appreciation
of the significance of the subject on the part of the student, in that he gains a
truer conception of the wide scope of its application.
The logical as well as the practical side of the subject is emphasized.
Definitions, arrangement, and method of treatment are logical. The defi-
nitions are particularly simple, clear, and accurate. The traditional manner
of presentation in a logical system is preserved, with due regard for practical
applications. Proofs, both foimal and informal, are strictly logical.
THE MACMTLLAN COMPANY
Publishers 64-66 Fifth Avenue New York
SLIDE-RULE
r
(1) (*) (3)
I la'
LLL.I
DIRECTIONS
A reasonably accurate slide-rule
may be made by the student, for
temporary practice, as follows.
Take three strips of heavy stiff
cardboard 1".3 wide by 6" long;
these are shown in cross-section in
(1), (2), (3) above. On (3)
paste or glue the adjoining cut
of the slide rule. Then cut strips
(2) and (3) accurately along the
lines marked. Paste or glue the
pieces together as shown in (4)
and (5). Then (5) forms the
slide of the slide-rule, and it will
fit in the groove in (4) if the work
has been carefully done. Trim
off the ends as shown in the large
cut.
m o
UNIVERSITY OF CALIFORNIA LIBRARY
Los Angeles
This book is DUE on the last date stamped below.
RHTD LD-UR0
FEB161971
16 1971
Form L9-Series 444
UC SOUTHERN REGIONAL LIBRARY FACILITY
A 000933189 3
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