of the University of California Los Angeles Form L 1 This book is DUE on the last date stamped below NOV 7 NOV 2 WOV 4 " RECD Form L-9-15m-8,'26 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 -x = o 4= 0, and every real number satisfies the equation -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 a mln by the equation . i a mln -\a m . For example, 8 2/3 = ^ = 4> ni In particular, a 1 '" = 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) m a r . 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 = a 1/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 a 6 and the two are reciprocals.* For example, 1 1 /I \~ 5/2 1 g-2/3 _ _!_ = L * 8 2/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, a 2 6c~ 3 _ 2- 1 d~ 1 e _ 2de~ l " a-^-'c 3 ~ 2c 3 d ' = (a? - z 2 )-i/ 2 , Va 2 x or 2 + 2x-*y~ l + 7/- 2 = - + + - . x z xy y 2 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) 9 s / 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?-8 4 . 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 ar 1 + y- 1 a" 1 - fe- 1 a;- 1 ?/ + xy ,.. 3 a-W -S-^tfb ffc x x- l y-*z~ 3 + x*y*z 3 2 a6- 1 +3a6 ' ar 3 ?/-^- 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. a b - a c = a b+c . EXAMPLES. 3 2 tr* = 3~ 2 . ( - |) B ( - f )- 3 = ( - f) 2 . (2)-l/3(|)l/2 = (!)l/ 6- g-l/3 . g2 = 8 5/3 . 11. a c b c = (ab) c . EXAMPLES. 2 3 5 3 = 10 3 . (- 3)- 2 (- 5)~ 2 = (15)~ 2 . (17)1/3(^)1/3 = 51/8, III. (a b ) c = a bc . EXAMPLES. (2 8 ) 2 = 2 6 . [(- |)- 2 ]- 3 = (- f) 6 . [(f) 1/3 l 6 = (I) 2 - I, 11] INTRODUCTION a b IV. r = a b ~ c . a c 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) a n -b n = 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 a n = b n , where n is a positive integer, then a = b. 11. Binomial Theorem. By multiplying out, we find the following equalities: (x + 7/) 2 = x 1 + 2#y + ?/ 2 , (x + y) 3 = x 3 + 3z 2 ?/ + 3z?/ 2 + y 3 , (x + 2/) 4 = a; 4 + 4:X 3 y + 6.r 2 7/ 2 + 4xy 3 + y 4 , (x + yY = x 5 + 5x 4 y + lOo; 3 ?/ 2 + Wx 2 y 3 + 5xy 4 + y 5 . 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-^-S 2 -j- 8 3/4 -8 1/12 . 2. 3 2/B -4 2 / B -5 2 / B 4- 15 2 / B -8 2 / 5 . 3. (3 2 -3 1/2 -5 6/2 ) 2 -i- (7 3 - 10 2 ). 4. (11 -3 2 + 7 4 ) 1 / 2 . C3/4 1 03/2 K (K4 _ 92.96^1/2 A _ 7 *"* D< 8 8 ' 12 ' 3 3 / 2 ' 40 2 " 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 -*Y l . 20 f^blY 3 ' 21 \ *V 2 / ' \8aV/ 22 / x + 2 \-i. / r &V \x*+x-2j \a-b) ' 24. 63a 4 x 5 -^ 9a 3 x 2 H- 3a 2 x. 25. (o + 6) (a + 6). ' 3a 2 6 2 28 30 2a 2 + 7ax + 3x 2 ^ 3a 2 + 7ax + 2x 2 2a + x a + 3x 31 a 2 - a - 20 a 2 - 2a - 15 a 2 - a a 2 33. x 2 - 5x + 4 x 2 - lOx +21 x 2 - 9x + 20 4. a + 3 v a 2 +a-2 . a 2 +3a+2 6 Multiply : 34. a 5/6 _ 35. a 1 / 2 + 2& 1 / 2 - 3c 1/2 by a 1/2 - 26 1/2 + 3c 1/2 . 36. x 4 / 3 + 2x + 3x 2 / 3 + 2x 1/3 + x by x 2 / 3 - 2X 1 / 3 + x. 37. Vx 3 - x 2 y - XT/ 2 + y 3 by Vx 3 + 3x 2 y + 3xy 2 + y 3 . 38. o 1/4 - 6 1/4 by a 3 / 4 + & 3/4 . 39. x 3 / 5 - ?/ 2/8 by x 2 / 5 + y 3 '*. 40. (a 1 / 2 ^/ 2 + c 1 / 2 ) 2 by (a 1 / 2 *) 1 / 2 - c 1/2 ) 2 . 41. (a- 1/2 - 3) 2 by (3O 1 / 2 + I) 2 . Divide : 42. x 8/2 + x 2 - 2X 1 / 2 + 1 by x + x 1 / 2 - 1. 43. x 3 + 27x - 9x 1/2 - 10 by x - 3x 1/2 + 5. 44. x-y- 6x 2 / 3 + 12X 1 / 3 - 8 by x 1 / 3 - y 1 ' 3 - 2. 45. a 6 / 2 - a 2 6 + a 3 2 c - oc + a 1/2 & - 1 by o 1/2 - 1. 46. a 2 + Sa 1 / 4 + 7 by a 1 / 2 + 2o 1/4 + 1. 10 MATHEMATICS [I, 11 Reduce each of the following to its simplest form : 47. Vl2.25zV. 48. v / 15.625a 6 6 9 . 49. V343a 10 6 25 . 50. V3a 2 6 - 2a 2 c. 51. V(a 3 + 53) ( 2 _|_ 52 _ a ty f 52 . Vx 4 ?/ 2 - 2xy + x 2 ?/ 4 . 53. VVl024. 54. 55. V27av / 27a6 4 . 57. \/1.35a 2 V6.25a ;! . 59. V|05 -2V605+V845. 60. *V 192 - 2v / 375 + #648 61. V72-V8-V50. 62. ^81 - 2\ y l92 + ^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+V5 x 5 -V2. \/3-V2 V2+V3 2-Vg 5+V2 74 V2 - \ 7 3 75 Vl89+3V20_ V3^_|_V48-V'50-v / 75 V84-V80 Expand each of the following expressions : 76. (2p 79. (1 - 82. 85. (1 + x 2 ) 2 . 88. (fc 2 +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 - X 2 ) 2 . 87. (1 +Vx) 2 . 89. (2t 2 + 5) 2 . 90. (a 2 + 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 + x 2 ) 3 . 101. (x+y- a) 2 . 102. (a 2 + 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)(z 2 - 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) x 2 - if + 2y = 1, (7) (2x ?/) 2 5x 2 = 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 i x+ 2_ 10 ? 3 ~2x~+~l { 2. Which of the numbers T V V7, 2 + V3, Vl4, 2 V3, are solutions of the equation x 2 + 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(x 2 y 2 ) = (x + y)(x 2 xy). (6) x(x 2 + y 2 ) = (x - y)(x 2 + xy). (c) x(x + 7) - (x + 3)(x + 4) + 12 = 0. (d) x(7 - x) + (3 - x)(4 - x) = 12. (c) 4x 2 + 7x + 2y = 0. (/) 4x 2 + 7x - 2y = 0. (g) x* = (x 2 + l)(x + l)(x - 1) + 1. (h) tf = (1 + x 2 )(l + x)(l -x) + 1. (i) (ax - b) 2 + (6x + a) 2 = (a 2 + 6 2 )(1 + x 2 ). 0") (ax - 6) 2 + (ax + &) 2 = (a 2 + 6 2 )(1 + x 2 ). (fc) (x - t/) 3 + (y - zY + (z- X? = 3(x - y)(y - 2) (a - x). (0 (x + 2/ + z) 3 - (x 3 + y 3 + s ) = 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 r 3 -4- "vr '-~--=x*-x + l and (x - l)(x - 2)(x - 3) = 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 x 2 1 _ 5x 6~ H ~ 6" is a solution of the equation x 3 + 6z = 5x*, which is formed by multiplying the first through by 6x; but they are not equivalent, since x = 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 3x 2 - 7x + 2 = x 2 - 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 x 2 - 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, 2x 2 + 2y 2 - 3x + y = 9, 3x 2 + 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 x 2 from the equations, - 2 5 5x 2 - 9x = 2, 2x 2 - 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 x 2 from the equations, - 2j|5x 2 + 9x = 2, 5||2z 2 + 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||3x 2 - 4x - 15y + 1 =0, - 3 || 2x 2 - 3x - 10y + 1 =0, the result is x - 1 = 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 y 2 ax lay + 2a 2 = 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 z 2 - 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 + 3 i ,_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; 15x 2 5x 8 w '2x~3 T 2x-3' 3x 2 + 6x + 4 (?) x - 1 = V3x - 5, x 2 - 5x + 6 = 0. = 5. (0 x = 2, x(x - 1) = 2(x - 1). 0') 2x = 1, Sx 3 - 12x 2 + 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 x 2 term from the equations x 2 - 2y* + 13x +2y = l, 3z 2 + 4?/ 2 - x + Qy = 3. 4. Eliminate y from the equations x 2 + 3xy - x + 1 = 0, 2x + y + 1 = 0. 5. Eliminate t from the equations , l+t*' 1+1? . 6. Eliminate t from the equations J 2 x = < 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 x 2 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, x z - 3% + 2, x+ A/3.T 3 + 3.4 + lx z , x* - 2x* + x - 5 , 3x, are examples of polynomials in x; y - 5 + 4y 2 , z 5 + ^ + A/2z 2 - \ , ^ + a 3 - 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 x 2 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 3x 2 + 0-x + 2, z 5 + 0-z 4 + 0-z 3 + 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 z 2 + 1, ?/ + y/2, z 2 - TTZ + V2, are polynomials, the expressions 3.1/2 + 1? y t _j_ 2/y, z 2 - 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= 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 = (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) ax 2 + 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 ax z + 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 ax 2 + bx + c =0 can be found by inspection. EXAMPLE 1. Solve 6x 2 + 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: 6z 2 + x - 15 = -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_ "" "^ 3x 2 i x = = - . Transposing and combining terms this reduces to & + & x = which factors by inspection into Z(- 2 7 Q X + *V) = whence xi = 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, 63x 2 - 14x = 3x 2 - 15x. EXERCISES Solve the following quadratic equations. 1. 2z 2 - 5x = 3. 2. 10x 2 + x = 2. 3. 6x 2 + 5x = 6. 4. 15x 2 - x = 6. 5. 6x 2 - 5 = 7x. 6. 28x 2 - 15 = x. 7. 135x 2 + 3x = 28. 8. 78x 2 - x = 2. 9. 3?/ 2 + y = 10. 10. 147/ 2 + y = 168. 11. Gy 2 + lly = 35. 12. 15?/ 2 + 4 = 16y. 13. 6a 2 + a = 5. 14. 2a + 3 = 8a 2 . 15. 9a(2a + 1) = 14. 16. 10(2a 2 - 3) + a = 0. 17. 3(2s 2 - 7) = 5s. 18. 15(2i 2 - 1) + 7t = 0. 19. p(12p - 7) = 10. 20. 5(3r 2 - 8) + r = 0. 26 MATHEMATICS [II, 25 21. A z 2 + 2x + 1 T 4 7 = 0. 22. v(y ~ 1} = 3(y + 1) - 2/2. 23. 2(2 - 1) = ;ft(6z - 1). 24. P + 3 - 1 = - 25. (1 - e 2 )z 2 - 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 ' x 2 - 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 ax 2 + bx + c 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 ax 2 + bx is to find a number d such that ax 2 + 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 6x 2 4x 1 =0. Transpose 1, and find the number to complete the square of 6x 2 - 4x by the above four steps: 1) xV6, 2) 2x>/6, 3) 2/V6, 4) 2/3; add this to both sides: 6x 2 - 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 x 2 a perfect square. In the above example we should have to solve the equivalent equation, 36z 2 - 24x + ( ) = 6. The number required to complete the square is 4, 36z 2 - 24z + 4 = 10. Whence 6z - 2 = VlO and Si -i- |ViO. EXERCISES Solve these equations by completing the square. 1. 4z 2 + 3z = 9. 2. 25z 2 - 14z + 1 = 0. 3. 50z 2 + 12x = x 2 - i 4. (x 2 + 1) V3 = 4z. 5. 12z 2 + 5x = 1. 6. 6^ 2 + 1 = 6y. 7. 32 2 = 13(2 - 1). 8. x + 2 = llar(l - x). 9. 12< 2 - 4(o + b)t + a6 = 0. 10. 2(?/ 2 + c 2 ) = 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, 4a 2 z 2 + 4a6x + 6 2 = - 4ac + 6 2 , extracting the square roots, we have 2ax + b = V6 2 - 4oc, whence we find - & =fc Vfe 2 - 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 3x 2 + 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. 2x 2 + 3x = 4. 2. x 2 = 220 + 9x. 3. 5x 2 + 3x = 3. 4. 5x 2 + 5x + 1 = 0. 5. 15?/ 2 = 86y + 64. 6. 5z 2 = 80 + 21. 7. a 2 + a = 3. 8. p 2 + 3p = 40. 9. I 2 + 3a 2 = 4o< - 1. 10. 5m 2 + 21m + 4=0. Solve the following equations by any method and test all results in the given equation. 11. (2x - 3) 2 = 8x. 12. x 2 - 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. 3x 2 + (9o - l)x - 3a = 0. 18. x 2 - 2ax + a 2 - fc 2 = 0. 19. c 2 x 2 + c(a - 6)x - afc = 0. 20. x 2 - 4ax + 4a 2 - 6 2 = 0. 21. x 2 - 6acx + a 2 (9c 2 - 46 2 ) = 0. 22. (a 2 - 6 2 )x 2 - 2 (a 2 + 6 2 )x + a 2 - ft 2 = 0. II, 28] REVIEW OF EQUATIONS 29 Solve for y in terms of x. 23. x 2 + 12xy + 9y 2 + 3 = 0. 24. x 2 - 4xy - 4y* + x = 0. 25. llx 2 + 3Qxy + 25y* = 3. 26. 8x 2 - 12x?/ + 4y 2 = x + 1. 27. Gx 2 - XT/ - 2y 2 = 0. 28. 21x 2 = xy + lOy 2 . 29. 30x 2 + 150* = 43xy. 30. 12x 2 + 41xy + 35y 2 = 0. 31. 2x 2 + 3xy - 2y 2 + x + 7y - 3 = 0. 32. 3x 2 + lOxy + 8?/ 2 + 4x + 2y - 15 = 0. 33. 10x 2 + 7xy + r/ 2 - x - 2?/ - 3 = 0. 34. 12x 2 = 4xy + 2ly 2 + 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, x 4 13z 2 + 36 = is not a quadratic in x but it is a quadratic in x 2 ] again if the terms of x* 6x 3 + 7x 2 + Qx = 8 be grouped in the form - 2(x 2 - 3z) = 8 it is seen to be a quadratic in (x 2 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.x 4 - 13x 2 + 36 = 0. 2. x + Vx + 6 = 14. Ans. 10. 3. 2x 2 + 3 Vx 2 - 2x + 6 = 4x + 15. Ans. - 1 and 3. = 2. Ans. and 1. x + Vl -x 2 5. s Vx + 18 = 5 3 Vx~ 2 . Ans. 8 and - 729/125. 6. x 4 - 6X 3 + 7x 2 + 6x = 8. Ans. - 1, 1, 2, 4. 29. Imaginary Roots. . There are quadratics which are not satisfied by any real number. For example, z 2 = 4, x 2 + 2x + 2 = 0. This is because the square of every real number (except 0) is positive. If we attempt to solve the equation x 2 + 2x + 2 = 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 = 2 2 i 2 = - 4; (- 3i) 2 = (- 3) 2 i 2 = - 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 i 3 , + 1 for i 4 , etc. We can now say that every quadratic equation can be solved. The solutions of the equation x 2 - 2x + 2 = 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. x z + <ox + 13 = 0. -4ns. 2 i. Ans. 3 2i. 3. 36x 2 - 36x + 13 = 0. 4. 2x 2 + 2x + 1 = 0. Ans. 1/2 t'/3. -4ns. 1/2 t/2. 32 MATHEMATICS [II, 29 5. x 2 + 4 = 0. 6. x 2 + x + 1 = 0. Ans. 2i. Ans. - 1/2 i V3/2. 7. z 2 - 2z + 3 = 0. 8. x 2 - |x + 1 = 0. 9. x 2 - 2x VJj + 7 = 0. 10. 2z 2 - 2x + 5 = 0. 11. x 2 + 3x + 2.5 = 0. 12. 49z 2 - 56x + 19 = 0. 30. The Sum and Product of the Roots. The two roots of the quadratic equation ax 2 + bx + c = are by (14), 27, - 6 + V& 2 - 4oc - 6 - Vb 2 - 4oc zi = and x 2 = - . 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, 36z 2 - 36z + 13 = the sum of the roots is 1, and their product is 13/36. Again in the equation, my z 4ay + 4a6 = 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 ax 2 + bx + c = 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 10z 2 - llz - 6 = 0. 32. Number of Roots. Conversely, it is readily shown that if r and s are roots of the quadratic equation, ox 2 + bx + c = then the left member can be factored in the form, (15) a(x - r)(x - s) = and this shows that no quadratic can have more than two roots. Some quadratics have only one root; for example 4z 2 + 9 = 12x is satisfied only by x = 3/2. If 6 2 4ct = 0, then the polynomial ax 2 + bx + c is a perfect square and the equation ox 2 + bx + c = has only one root, and conversely. For, if 6 2 4oc = 0, then c = 6 2 /4a and this is precisely the number necessary to complete the square of ox 2 + bx. Also if 6 2 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 6 2 4ac > 0, then the quadratic equation ax* + bx + c = has two real roots; but if 6 2 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. 4z 2 - 12x + 9 = 0. Here b 2 4ac = 144 144 = 0, the left member is a perfect square and the equation has only one root. EXAMPLE 2. 3x 2 - 5x + 2 = 0. Compute b 2 4ac = 25 24 = + 1, which shows that the equa- tion has two real roots. EXAMPLE 3. x 2 + x + I = 0. Here b 2 4oc = 3, which shows that the equation has imaginary roots. If a, b, and c, are rational numbers, then the roots of the equation ax 2 + bx + c = are rational if b 2 4ac is a perfect square, i. e., the square of a rational number: in particular if a, b, and c, are integers and if b 2 4ac is a perfect square the left member of the equation can be factored by inspection. EXAMPLE 4. 2x 2 - x - 6 = 0. Here b 2 4ac = 1 + 48 = 49; the left member factors into (2x + 3)(x - 2) = 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. () a 2 , /S 2 , 7 2 - (0) a - 0, - 7, 7 - (P) "A 07, 7- 2. Determine the nature of the roots of the following equations. (a) 3x 2 - 4.r - 1 =0. (6) 5x 2 + 6z + 1 = 0. (c) 2x 2 + x - 6 = 0. (d) x* - 2x - 1 = 0. (e) 5x 2 - 6x + 5 = 0. (/ ) x 2 - 6 V3x - 5 = 0. (g) x 2 + x + 1 = 0. (h) 13x 2 - 6 >/3x + 7 = C. (1) 3x 2 + 2x + 1 = 0. 0') 2 * 2 - 16x + 9 = 0. (*) 5x 2 - 12x - 8 = 0. (0 6x 2 + 4x - 5 = 0. II, 33] REVIEW OF EQUATIONS 35 (m) 5x + 7 = (3x + 2)(x - 1). (n) 5(x 2 + x + 1) = 1 - 16x. (o) 2x(x - 3) = 7(3x + 2). (p) 7(x 2 + 5x + 3) = x(l - x). (9) 3x(x + 1) = (3 - x)(3 + x). (r) 3x 2 = 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 = 6 2 + c 2 . (6) (x + a) 2 = 86 2 . (c) a (ax 2 + 26x - a) = b(bx 2 - lax - b). (d) 7/ 2 = 2a(y -b) + 2b(y - a). (e) (a + 6 - c)?/ 2 - 2cy = (a + b + c). (/) (a + b - c)x 2 + 4(o + b)x + (a + b + c) = 0. (0) (b + c - 2a)x 2 + (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 = a 2 - 8a + 15. (d) x 2 - 15 = 2a(x - 4). (e) 3(x 2 + Sax - a) = x. (/) 9x 2 + 6(0 - 4)x + a 2 = 0. (g) 3ax(x - 1) = ax - 2. (h) (a + l)x 2 -(a + 2)x +fa = 0. (t) (4a 2 + 3)x 2 + 8a(3 - 2a)x + 4(4a 2 - 12a - 3) = 0. 6. Find a value of k such that the sum of the roots of the equation x 2 3(k + l)x + Qk = 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 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 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 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 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 4 2 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. A r cb. 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 P 2 be the end points of a given segment in the plane. PI and P 2 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 PiP 2 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 SP 2 is 7/2 - y\ ; also pj% = !\s* + isp?. Therefore if we let d represent the required distance PiP2, (1) d = (x 2 - xtf + (y z - 7/0 2 . 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 - x 2 Y + (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 = 5 2 + 12 2 = 13. 46. Ratio of Division. Let PI and P 2 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 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, PiP 2 = x 2 - Xi ; HI, 47] GRAPHIC REPRESENTATION 55 hence the abscissa x of P must satisfy the condition (3) X X 2 - whence solving for x, (4) x = Xi + \(x 2 - Xi). If the segments PiP and PiP 2 have the same sense, the division ratio is positive and P and P 2 lie on the same side of P\. If the segments PiP and PiP 2 are oppositely directed, then the division ratio is negative and P and P 2 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 PiP 2 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 P 2 (o; 2 , y 2 ) be the two given points, X the given ratio, and P(x, y) the required point. Draw PiQi, PQ, P 2 Q 2 parallel to the y-axis, and PiRi, PR, P 2 R 2 parallel to the z-axis. Then Q and R divide Q\Q 2 and R\Ri, respectively, in the ratio X. Now as OQ\ = Xi, OQ 2 = x 2 , OQ = x, it follows from (4) 46 that (5) x = Xi + \(x 2 - Xi). In the same way we find (6) y = yi + K(y 2 - yi). Thus, the coordinates x, y of P are expressed in terms of the coordinates of PI and P 2 and the division ratio X. 56 MATHEMATICS [III, 48 48. Middle Point. If P be the middle point of PiP 2 , 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 x 2 4y + 4 = 0. If P is always 3 units from the origin, then x 2 + y z = 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 ^x 2 + (y a) 2 . Hence the condition expressed in the problem gives y = Vz 2 -|- (y a) 2 . This simplifies to x 2 + 2ay = a 2 . 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 x 2 + 2ay = a 2 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? + b 2 - c 2 - d 2 }. 4. Find the equation of the locus of a point whose distance from the point ( 3, 4) is always equal to 5. .4ns. x 2 + y 2 + 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 x 2 = 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. . . 1 2 3 4 5 6 . 7 - 1 - 2 - 3 y. . - .75 - 1 - .75 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 x 2 + y 2 = Qx + 2y. Solving the given equation for y, we have y = 1 Vl + 6z - x 2 , and we tabulate solutions as follows. X. ... 1 2 3 4 5 6 7 - 1 - 2 y. . . . - 1.45 2 - 2.16 - 2 - 1.45 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 x 2 < 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 -y 2 -, 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. 4z 2 - ?/ 2 = 0. 2. 4z - Qy - 6 = 0. 13. 6z 2 + 5xy - 6y* = 0. 3. 6z - 9y + 36 = 0. 14. x 2 + y z = 4. 4. 2x + 3y + 5 = 0. 15. x 2 - y 2 = 4. 5. 3x - 2y - 12 = 0. 16. x 2 + y 2 = 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 = x 3 - 4z 2 - 4x + 16. 11. (x 2 - 4)(y - 2) = 0. Ill, 55] GRAPHIC REPRESENTATION 63 22. y = x, x z , x 3 , x 4 , , x n . What points are common to these curves? 23. if = x, x 2 , x 3 , 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. y 2 = (x - 2) 3 . 31. y - -. 32. y = x - 1 x + 1 33- !f---- 34. y- x 2 + 1 x 2 + 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 x 2 + y 2 = 18, 40 ' -3*. Ans. (3, 3), (- f, - V). Ans. (3, 3), (3, - 3). f 3x 2 + 4y 2 = 48, f 3x 2 - 4?/ 2 = 11, ' ' x - y + 1 = 0. ' 1 4x = 3y 2 . Ans. (2, 3), (- V, - -V)- ^s. (3, 2), (3, - 2). 43. I X1J = 2 > 44. 1 y 2 = 4x. x 2 + 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 x 2 -f- y z = 4x and x 2 + y 2 = 4(x + y- 1). Ans. 2^3. 48. In what respects are the loci of the following equations sym- metric? (o) y = x 2 . (e) ?/ 2 = x 2 . (i) x 3 y 3 x y = 0. (6) y 2 =x. (/)r/ 2 =x*. (T) XT/ = a. (c) y = x 3 . (g) y = X s - x. (fc) ax 2 + by 2 = 1. (d) y 2 = x 3 . (A) y = x 4 - x 2 . (I) ax 2 + 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 x 2 + y* < 16 and x > 2? 7. What is the locus of a point if 9 < x 2 + y 2 < 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), P 2 (z 2 , 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(x 2 , y 2 ). Let P(x, y) be any other point on the line joining PiP 2 . Draw PiRS parallel to the z-axis. Draw P\M\, P 2 M 2 , PM, parallel to the y-axis. FIG. 19 Then since the triangles PiSP and PiRP 2 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 P z Rz- Draw R 2 Ri 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 = S Z P Z . 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 L 2 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 L 2 is m 2 =* 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, z 2 + y 2 = 25. (b) 2x 2 + 3y 2 = 35, 3z 2 - 4y = 0. (c) x 2 + y = 7, y* - x = 7. (d) y = x + 5, 9z 2 + 16y 2 = 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 = 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 = 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 = 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 = 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 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 10 2 = 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: 10 2 = ?. 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 = log b 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, 2 4 = 16, (1/2) 3 = 1/8, a" = x, are written respectively in the logarithmic form: 2 = logs 9, 4 = Iog 2 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 = 3 2 . (g) 7 = 7 1 . _ (6) 64 = 4 3 . (h) 25 = (Vs) 4 . (c) 16 = 2 4 . (i) 8 = (V2). (d) 243 = 3 5 . 0') 3 = (V3) 2 . (c) 64 = 2". (fc) 3 - V9. (/) 2401 = 7 4 . (I) 4 = v/64. 2. Write the following equations in exponential form : (a) log, 16 = 4. (g) logioO.l - - 1. (b) Iog 4 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) Iog 2 64. (e) Iog 26 5. (6) logio 0.001. (/) 3 Iog 6 625 + 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, 6 1 = 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 b k and N = b l , MN = b k -b l = b k+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 ~ b l '' 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 = b k and N p = (b k ) p = b pk , whence logb N p = 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 = b k and A/A 7 = N llr = (6*) 1/r = b klr , 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 17 i/4 12 -2 ! ] g iQ2/3.Ai/2 2. log ,,- 7 , . 3. log 13 2/ 3 . 6 i/ 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 ! V7 2 V6). 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) B Vl3.5. (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, log a c, Iog 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 l l/4 1Q1/8 1Q1/16 10 1/32 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 = 10 1 / 2 X 10 1/8 X 10 1 / 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 10 n-i < N < 10 n 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 10 11 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 10 1 *. (4) 6.3 X 10 s . (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 10 5 = 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 10 12 . .(17) log 5.47 X 10 23 . (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 = - 2.6284 But since we always wish to have the mantissa of a logarithm positive, we write = 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. G V( - 9.718) 3 . 19. 8 5/4 . 20. (0.001 ) 2 ' 3 . 21. (29^ 9 r ) 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 5968 1 / 3 30. V 6 A/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-^lO 6 - 58 ] 1 Ans. 7.76 X lO" 6 . Ans. 5.51 X 10 7 . 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 C 2 the central visual acuity of each eye, and A/P 2 the peripheric vision. Compute the value of # if C = 1, PI = 1, M = 1, Ci = 1, C s = 0.58, x = 10, P 2 = 1. Ans. 97.2% 43. Compute E if d = 0.41, C 2 = 0.25, x = 5, P 2 = 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 > wr i tte n 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 ^ ^ ~x V A' \ X ' X FIG. 25 same angle A, and denote their coordinates by (x f , 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 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), 2n 2 + 2n + 1. 11. 2n, n 2 - 1, n 2 + 1. 12. 2(n + 1), n(n + 2), n 2 + 2n + 2. 13. a(6 2 - c 2 ), 2abc, a(6 2 + c 2 ). 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) sin 2 A + cos 2 A = 1 ; (16) tan 2 A + 1 = sec 2 A (14) ctn A = cos A sin A ' (17) ctn 2 A + I = esc 2 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) x 2 + y 2 = r 2 . Dividing through by r 2 , we have ^+^-1, whence cos 2 A + sin 2 A = 1. Similarly, dividing (18) through by x 2 , and then by y z 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 sin 2 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 + tan 2 A) sin 2 A = tan 2 A. 0') (1 - sin 2 A) esc 2 A = ctn 2 A. (k) sin 4 A cos 4 A = sin 2 A cos 2 A. (0 tan 2 A cos 2 A + cos 2 A = 1. (m) (sin A + cos A) 2 + (sin A cos A) 2 = 2. (n) sec A cos A = sin A tan A. (o) (sin 2 A cos 2 A) 2 = 1 4 sin 2 A cos 2 A. (p) (1 - tan 2 A) 2 = sec 4 A - 4 tan 2 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 78 2 = 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 18.5 2 = 342.25 x 2 = 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 = 42.7 2 - 18.5 2 = 61.2 X 24.2 log 24.2 = 1.3838 log 61. 2 = 1.7868 log x 2 = 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, A 2 = 98 (e) a = 3, b = 5, c = 7. (/) a = 7, A = 120, b = 5. (0) o = 42, b = 51, A = 55. Ans. Bi = 84, B 2 = 96, Ci = 38, C 2 = 22, ci = 5, c 2 = 3. Ans. 4 = 22, B = 38, C = 120. Ans. = 38, C = 22, c = 3. = 41, C 2 = 29 2. Solve the following right triangles (C = 90). = 34, c 2 = 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 to P (always positive). Then the trigonometric functions of $ are defined as follows: (19) sin (f> = - , cos = - . 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 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 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, r tan = - = 0. x cos = - r 1, sec = - = 1. x The angle has no cotangent nor cosecant. When = 90, P coincides with P 2 , 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 P 3 , 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 = 270, P coincides with P 4 , 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 increases from to 90, tan increases without limit, and that before $ reaches 90. Similar remarks apply to the statements ctn = 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 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 at P. Draw tangents to this circle at ^4. and at B and produce the terminal side in one or both directions from 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 "; 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 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 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 = AOD are solutions (and the only solutions between 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 and B cutting the circle in C and D. Then the positive angles a = AOC, j8 - AOD are solutions (and the only solutions between 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 sin 2 x cos x = 1 can be reduced to the form (cos x + l)(cos x |) = V, 89] TRIGONOMETRY 109 , \ 0- \ f,l3 by replacing sin 2 x by 1 cos 2 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 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 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 = 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 PP f . 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 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 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 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 cos 9 sin sin cos 6 cos sin sin cos sin sin 6 cos cos 6 sin sin cos cos tan ctn 6 ctn tan0 tan ctn e ctn tan tan ctn tan tan ctn 6 ctn 6 tan tan0 ctn ctn sec csc csc sec sec csc csc sec sec csc sec B sec csc csc 6 sec 5 sec csc csc 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 sin 2 a. V, 93] TRIGONOMETRY 115 (c) 3 tan 210 + 2 tan 120 = - >/3. (d) 5 sec 2 135 - 6 ctn 2 300 = 8. (e) sin (90 + ) sin (180 + x) + cos (90 + x) cos (180 - x) = 0. tan (90 + tt) + C8c2 (270 -) = L+ ^c 2 . 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 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. (6) sin 270 + cos = 0. (h) sin 120 + sin 300 = 0. (c) esc 90 + sec 180 = 0. (i) cos 150 + cos 330 = 0. (d) esc 270 + sec = 0. (j) tan 135 + tan 225 = 0. (e) sin + 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 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 and 360 which satisfy the following equations. (a) sin = sin 320. (d) cos = - cos 50. (6) tan 9 = tan 125. (c) ctn = - ctn 220. (c) sec = sec 80. (/) esc = - 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 sin 2 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) x 2 + 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 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 = cos 2 z sin 2 z. (28) = 2 cos 2 z - 1. (29) =1-2 sin 2 z. 2 tan z (30) tan2x = 1 tan 2 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 sin 2 x) = sin x(4 cos 2 x 1). 9. Show that cos 3x = cos x(4 cos 2 x 3) = cos z(l 4 sin 2 x). 10. Find sin 4x; cos 4x; tan 4x. 11. Show that tan (45 + A) = ? + ^ A A . 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- , 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) = sin 2 x sin 2 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 + tan 2 x ' ,, , esc 2 x sin $x (h) sec 2x = ^. (i) tanjx = = r-. esc 2 x 2 1 + cos \x (fi ctn ix - sin ^ (H tn ' 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) = cos 2 A - sin 2 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) cos 4 A = f + 5 cos 2A + | cos 4A. (e) sin 4 A = f 5 cos 2A + | cos 4A. (/) sin 2 A cos 2 A = | - | cos 4A. (g) sin 2 .A cos 4 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) cos 2 %<f>(l + tan |0) 2 = 1 + sin 0. (p) sin 2 |x(ctn |x I) 2 = 1 sin x. , . 2 sec A . x . 2 sec A (g) sec 2 JA = - : -. . (r) esc 2 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, a 2 = CD 2 + P 2 = (b c cos A) 2 -f- (c sin A)- = b 2 - 2bc cos A + c 2 (sin 2 A + cos 2 A) = b- 2bc cos A + c 2 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 AB 2 C. Since 128 MATHEMATICS [V, 104 the triangle BiCB 2 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, p IG 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 AB 2 C, the angle B 2 being acute. By the law of sines, 60.35 sin 44.24 sin 34 23' B 2 = 50 23' log 60.35 = 1.7807 s sin 34 23' = 9.7518 - 10 colog 44.24 = 8.3542 - 10 log sin 5 2 = 9.8867 - 10 64 10)30(3 Then find C 2 (i. e., angle ACBJ = 95 14'. To find c 2 (i. e., side use the law of sines again, c 2 sin 95 14' 44.24 sin 34 23' c 2 =78.02 log 44.24 = 1.6458 log sin 95 14' = 9.9982 - 10 colog sin 34 23' = 0.2482 log c 2 = 1.8922 21 6)10(2 To solve the triangle AB&, we first find BI = 129 37' being the supplement of 5 2 , 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. c 2 = 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') c 2 - = 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. x 2 = 27P + iTB 2 - 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 a 2 = & 2 + c 2 - 2bc cos A, whence 6 2 + c 2 - a 2 (41) cosA =--2bT~' To adapt this to logarithmic computation, subtract each member from unity & 2 + c 2 - a 2 2bc - b 2 - c 2 + a 2 1 cos A = 1 2bc a 2 - (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) -rfi r (s - a)(g - 6) , sin' ^o = - - -. ac ab V, 106] TRIGONOMETRY 133 Again, adding each member of (41) to unity, 52 + C 2 _ Q 2 (6 + c)2 - a 2 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) cos 2 5 - , cos 2 \C = -- - . ac ab Dividing sin 2 %A by cos 2 \A, we have, by (43) and (44). tan 2 %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', C 2 = 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 B l = 34 46', Ci = 71.63 B 2 = 145 14', C 2 = 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. 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 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 & T s T t T 5 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, S 2 , 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, $3^3, etc., all perpendicular to AC, can be computed and on retracing CA, stakes can be set at T$, T t , T 3 , 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 S 4 T 4 = 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, y ), (zi, T/I), (x 2 , 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) |[(z 4 - 5X2/4 + 2/5) + (3 - 4X2/3 + 2/4) (3 2X2/3 + 2/2) (2 1X2/2 + 2/0 - (xi - x 5 )(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(x 6 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 = Va 2 + /(b 2 - a 2 ), 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 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 F 2 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 + F 2 2 + 2FiF 2 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 F 2 . If we let a equal the angle BAD, we have, from the triangle BAD sine sin a' whence F 2 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 F 2 F l (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 F 2 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, F 2 = R sin a. These formulas give FI and F 2 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, F 2 = 69.8, 6 = 65 20 , find R and a. 2. Given FI = 20.3, F 2 = 60.2, = 135 10'; find R and a. 3. Given FI = 60.3, F 2 = 30.2, =90, find R and . 4. Given F! = 26.7, F 2 = 45 7, = 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 F 2 . 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 F 2 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 F z 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 be acted upon by any number of forces, Fi, F 2 , ; to be definite, say Fi, F 2 , F 3 , F 4 . To find their resultant proceed as follows. For the forces FI and F 2 con- struct the triangle of forces OAB (Fig. 81). Then OB is the FIG. 81 resultant of FI and F 2 . For the forces OB and F 3 construct the triangle of forces OBC. The sum is given by OC. In a similar manner combine OC and F 4 . 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 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 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, F 2 , , lying in the same plane, act on a particle at 0. To fix the ideas, suppose there are three forces. With 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 F z , OC and OE; 164 MATHEMATICS [VII, 133 of FS, OD and OF. If a\, a^, a 3 represents the angles which FI, FZ, FS make respectively with the axis X, we have : Xt = OA = FI cos ai, F! = OB = F l sin a h Xz = OE = F z cos 2 , F 2 = OC = Fz sin a 2 , Z 3 = 0Z> = F 3 cos a s , Y 3 = OF = F 3 sin a 3 . 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 + F 3 sin a 3 = 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 = X 2 + 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, = 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 F 2 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 F 2 , 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 F 2 . 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 + F 2 , and its line of action is a line through the point parallel to the lines of action of FI and F 2 . Let C be the intersection of AB and OR. Then from the pairs of similar triangles OCA and AFiD, and OCB and BF 2 E, 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 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 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 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 F 2 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., F 2 = 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 = F 2 cos 30 - F! cos 60 = 0, 2F = Ft sin 30 + F l sin 60 - 1000= 0. The solution of these equations gives, F l = 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 - 8F 2 = 0, whence F 2 = 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 F 2 . Find FI and F 2 . 1000 2000 >*-* 4+ 4* F = 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 F t . 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 - c 2 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 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 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 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 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)) fr 2 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- + 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 w 2 is approximately aAw + aAw 2aAw, and we write w 2 = a 2 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(sin 2 a + cos 2 a)/cos 2 a = sec 2 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 3a 2 -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 4o 3 -Aa. 17. Show that the error of the cube root of a Aa is Aa/3a 2/3 . 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 A 6), in the computed value of b due to an error Aa in measuring a is, approxi- mately, A & = sin B esc A Aa. Likewise show that A A b = a-sin B-csc A-ctn A-&A, and Agb = a cos B-csc A-&B, and that the total error is, approximately, A6 = A a & + A A b + 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 A Q b = 0.007 ft.; A A b = 0.003 ft.; A B b = 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 /I 2 . Find approximately the error due to small errors in observing s and t. Ans. A- s g = 2As/P; A t g = - 4s/l 3 . 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 w 2 ; 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, w 2 is the weight of a sinker in water, and w 3 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 w z = 182.2 0.03 w a = 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(x , y ) 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 z ) 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 - x ) 2 + (y - y Y = r\ Conversely, let Q(x\, y\) be any point which satisfies (1); i. e., ~7 FIG. 104 - x ) 2 + (t/i - y ) = 190 IX, 152] CONIC SECTIONS 191 whence (*i - so) 2 + (</! - y o y = 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, x = yo = 0, and the equation reduces to (2) x 2 + y 2 = r 2 . 151. Equation of the Second Degree. The most general equation of the second degree in x and y is of the form (3) a.-c 2 + bxy + cy 2 + 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 = and a = c. Conversely, the special equation of the second degree (4) ox 2 + ay 2 + 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 x 2 + y* - 3x - 2y - 3 = 0. To reduce this equation to the form (1) we complete the squares as follows: (z 2 - 3x + ) + (y 2 - 2y + ) = 3, (z 2 - 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 9x 2 + 9?/ 2 - 6x + 12y + 6 = 0, We complete the squares as follows : & + y 2 - 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 225x 2 + 225?/ 2 - 270x - 300t/ + 181 = 0. We complete the squares as follows: x 2 + y 2 - fx - fa + iH = 0, x 2 - 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) x 2 + y 2 = 4x. (6) x 2 + y 2 = 6y. (c) x 2 + 8y = 4x - y 2 . (d) 3z 2 + 3y 2 = 14y. (e) x* + y 2 + 4x + 7 = 0. (/) x 2 + y 2 + 3x + 5y = 0. (0) x 2 + y 2 = 2(y + 4). (h) x 2 + y 2 = 4(x - 2). (1) x 2 + y 2 - 4x - 6y + 9 = 0. 0') * 2 + y 2 + 101 = 87y - 20x. (A;) 2z 2 + 27/ 2 + 15y = 12x + 7. (1) 9x 2 + 9y 2 + 6y = 24x + 47. (TO) 16.x 2 + 167/ 2 = 24x + 40y - 34. (n) 49x 2 + 49y 2 + 28x - 2% + 9 = 0. (o) 4a(ax 2 + 6x - by) + b 2 + 4o(ay 2 - ex - cy) + c 2 = 0. 4. Show that if the coefficients of x 2 and y 2 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 x 2 + y 2 - 5x + 4y - 3 = is (5/2, - 2). 5. Find the coordinates of the center of each of the following circles, by the process of Ex. 4. (a) x 2 + y 2 - 4x - 6y + 9 = 0. (d) x 2 + y 2 - 2x + 4y + 1 = 0. (6) x 2 + y 2 + 6x + 4y + 9 = 0. (e) x 2 + y 2 - 3x + 5y + 3 = 0. (c) x 2 + ?/ - 4y = 0. (/) 2x 2 + 2y 2 + 4x - 6y + 1 =0. 6. The value of the polynomial P = x 2 + y 2 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 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 3x 2 + 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 between D and F is on the parabola. Take 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 + y 2 , 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 ?/i 2 = 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 y 2 = 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) z 2 = - 4py. The position of each of these curves should be related to its equation as follows: y z = 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 = 2x 2 8x + 5. Transposing, 2x 2 - 8x = y - 5; dividing by 2, x 2 4x = \y \ ; adding 4, x 2 - 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 = 2x 2 + 4x. We may write successively the equations x 2 -2x=-|y, x 2 -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 = ay 2 -\-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) x 2 = I0(y + 1). (c) y 2 = - 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) y 2 - 2y - 4x + 6 = 0. (b) y* + y - 6x = 0. (c) x 2 + 4z + 6y - 8 = 0. (d) a; 2 - x + y = 0. (e) 4z 2 - 12x + 3y - 2 = 0. (/) 3y 2 + 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 (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 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 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 6 2 (13) < e < I and - = sin 2 a = 1 - e 2 . a 2 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 + y 2 + V(z - ae) 2 + y 2 = la. Transposing the second radical, squaring, and reducing, we find (15) V(z - ae) 2 + y 2 = 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 - e 2 )x 2 + y 2 = a 2 (l - e 2 ), 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 /a 2 = 1 e 2 . 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. This 1 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 ira z , 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 3x 2 + 4y 2 = 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 o 2 = 16 and b 2 = 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 15x 2 + 28y 2 = 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) 4x 2 + 9i/ 2 = 36. (b) x* + 25y* = 100. (c) 9x 2 + 25y 2 = 225. (d) 9x 2 + 16?/ 2 = 144. (e) x 2 + 2y 2 = 4. (/) 6x 2 + 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, 6 2 /a 2 = e 2 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) x 2 - y 2 = a 2 . 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 a 2 cos 2 45 = a 2 . 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 lQx z 9?/ 2 = 144 to standard form. Dividing by 144, we find i 2 _ yL = 9 16 Hence, by comparison with (22), we have a = 3, b = 4. From b?/a 2 = e 2 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) 4x 2 - 9?/ 2 = 36, (Vl3, 4/3). (6) 4x 2 -7/ 2 -8, (-3/2, 1). (c) 3x 2 -i/ 2 = 9, (3, -3>/2). (d) 3x 2 - 4y 2 = 1, (- V?; V5). (e) 144x 2 - 25y 2 = 3600, (10, - 12 V3). (/) 9x 2 - 16y 2 = 576, (12, 31/5). (g) 25x 2 - y 2 = 100, (- V29, 25). (h) 225x 2 - 647/ 2 = 14400, (17, 28 J). (i) x 2 - y 2 = 9, (- 5, 4). 0') x 2 - ?/ 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. 3x 2 + 4y 2 + 12x - IQy = 20. (6) whose vertices are (2, 2) and (4, 2), and which passes through the point (1,4). Ans. 4x 2 + 9y 2 - 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. 9x 2 + 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. 16x 2 - 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. 225x 2 - 64y 2 + 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 4x 2 + 9?/ 2 = 25. Solving the first equation for y = 3x 5, substituting this in the second and reducing, we have 17x 2 - 54x + 40 = 0. We can factor this quadratic by inspection : (17x - 20) (x - 2) = 0, whence xi = 20/17 and x 2 = 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 = x 2 - 5x + 12 intersect the ellipse 4x 2 + 3y 2 = 48? Substituting the value of y from the first equation in the second and reducing, we get x 4 - lOx 3 + 61x 2 - 120x = 0. Factoring this equation, we have x(x -3)(x 2 - 7x + 40) = and we see by inspection that Xi = and x 2 = 3 are roots. The quad- ratic x 2 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. x 2 + Qxy + 9j/ 2 =4, 4x + 3y = 12. Ans. (14/3, - 20/9), (10/3, - 4/9). 2. x 2 - if =0, 3x - 2y = 4. Ans. (4, 4), (4/5, - 4/5). 3. y 2 +x=0, 2y + x = 0. Ans. (0, 0), (- 4, 2). 4. x 2 5y = 0, x y = 1. Ans. x = i(5 V5), y = |(3 V5). 5. 2x + 3y = 5, 4x 2 + 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, x 2 + y* - x + 3y = 0. Am. (0,0), (- 1, - 2). 8. x - 2y + 4 = 0, 5x 2 - 4y 2 + 20 = 0. Ans. (I, 2). 9. y = 2x - 3, 4y 2 = (x + 3)(2x - 3). Ans. (3/2, 0), (1, - 1). 10. 4i/ 2 = x 2 (x + 1), y 2 = x(x + I) 2 . Ans. (0, 0), (- 1, 0). 11. 2x 2 - 3i/ 2 = - 58, 3x 2 + t/ 2 = 111. Ans. (5, 6), (- 5, 6), (5, - 6), (-5, - 6) 12. x 2 = 4ay, y = 8n 3 /(x 2 + 4o J ). Ans. ( 2a, a). ITu x 2 + y 2 = 2, x 2 + 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) Ax 2 + Ex + C = will have 1) two real roots when B 2 4 AC > 0, or 2) no real roots when B z 4 AC < 0, or 3) one real root when B 2 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 4x 2 + 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, 20x 2 - 64x + 35 = 0, B 2 - 4AC = 1296, . 20x 2 - 96x + 135 = 0, B 2 - 4AC = - 1584, x 2 - 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 y 2 = 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, x 2 = 4y. (6) x - 2y + 20 = 0, x 2 + y* = 16. (c) 2x + 3y =8, y 2 = 4x. (d) x + 2y = 5, x 2 + y 2 = x + 2y. (e) 2x = 3y, 4x 2 - 3y 2 + 8x = 16. (/)x + 7/ = 8, 4x 2 +?/ 2 = 16x. 3. Find the points in which the circle x 2 + y 2 = 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 x 2 + y z 6x Qy + 10 = 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 = 2x 2 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 3x 2 + 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 y z = 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, 9z 2 + 6(6 - 4)z + 6 2 = 0. By 32, this quadratic will have only one root and the line will be tangent to the parabola, if 36(6 - 4) 2 - 366 2 = 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 3z 2 + 4y 2 = 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, (4m 2 + 3)z 2 + 8m(3 - 2m)x + 4(4m 2 - 12m - 3) = 0. This equation will have only one root and the line will be tangent to the ellipse if (32), 64m 2 (3 - 2m) 2 - 16 (4m 2 + 3) (4m 2 - 12m - 3) =0, that is if, 4m 2 + 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) y 2 = 8x, (2, 4), (6) x 2 - y 2 = 64, (10, 6), (c) x 2 + 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 = from the point (2, 1) and verify that they are perpendicular. 3. Find the equations of the tangent and normal to the circle x 2 + y* - 6x - Gy + 10 = 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 x 2 + y 2 45. Ans. 2x y = 15, x + 2y = 15. 5. Find the equations of the tangent and normal to the circle x z + y 2 = 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 y 2 = 12x which makes an angle of 60 with the z-axis. 8. Find the tangent to the parabola y 2 = 6x which makes an angle of 45 with the re-axis. 9. Find the equations of the tangents to the circle (a) x 2 + y 2 = 4 parallel to 2x + 3y + 1 =0, (b) x 2 + y 2 = 16 parallel to 3x - 2y + 2 = 0. 10. Find the tangents to the ellipse Qx 2 + 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) y 2 = 4x, slope = 1/2. (/) x 2 + y 2 = 25, at (4, - 3). (6) x 2 + y 2 = 16, slope = - 4/3. (g) x 2 + 4?/ 2 = 8, at (- 2, 1). (c) 9z 2 + l&y 2 = 144, slope = - 1/4. (h) x 2 - y 2 = 16, at (- 5, 3). (d) x 2 = 4y, passing through (0, - 1). (i) 2y 2 - x 2 = 4, at (2, - 2). (e) x 2 = 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) x 2 y 2 = a 2 , y kx. Ans. k = 1. (6) x 2 + y 2 = r 2 , 4y - 3x = 4Je. Ans. 16/b 2 = 25r 2 . (c) 4x 2 + y 2 - 4x - 8 = 0, y = 2x + k. Ans. k 2 + 2k - 17 = 0. (d) xy + x - 6 = 0, x = ky + 5. Ans. k 2 + 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 x 2 y- = 5 intersects the ellipse x 2 + 4?/ 2 = 25 in the four points (3, 2), (- 3, 2), (-3, - 2), and (3, - 2). The parabola 4x 2 = 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. X 2 - 7/ 2 = 5, x 2 + 4?/ 2 = 25. Multiplying the first equation by 5 and subtracting, we -have 4x 2 - 9y 2 = 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 x 2 + 3xy = 28, EXAMPLE 2. < ' t 4?/ 2 + xy = 8. Eliminating the absolute terms, we have 2 X 2 - X y - 282/2 = 0, whence (2x + 7y)(x - 4y) = 0. This gives the two pairs of simultaneous equations (\ S 4 2/ 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 z 2 and i/ 2 as the unknowns. r x 2 EXAMPLE. ! t x 2 - y 1 = 5, + 42/ 2 = 25. Eliminate x 2 and solve for y 2 . This gives y 2 = 4, whence y = 2. Then eliminate y 2 and solve for x 2 . This gives x 2 = 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 2s 2 + 2> = 13, S 2 _ ft - Q f 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 4x 2 + 4xy - y 2 = 7x - y, | x 2 - y 2 + 16 = 0, i 4x + 3y = 1. 1 (x + I) 2 = (y + I) 2 . f 3x 2 + 4y 2 = 48, J 5x 2 + 7y 2 = 225, 1 y 2 = 3(1 - x). \ 2x + 3y = 9. f 4x 2 + 3xy = 10, f a? + y 2 = 153> I 3y 2 + 4xy = 20. 1 xy = 36. J 2x 2 + 2xt/ + 5y 2 = 40, f x 2 + y 2 - x - y = 204, 1 x 2 - f 2x 2 + xy + 37/ 2 = 12, x 2 - 2y* + 1 = 0, y 2 + 2x - 2y = 0. xy + x + y = 129. , 1 2x + y = 0. y = 0. 2x 2 - 3y 2 - 23 = 0. 3x 2 + 4?/ 2 = 48, j 4x 2 + 6xi/ + 4y 2 = 46, \ X 2 + ^ = 34. x 2 + 2xy = 407, f x 2 + 2?/ 2 = 123, 7/ 2 + 2xy = 455. 1 y 2 + 2x 2 = 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 + y 2 = 58 U + 3(x + l) =0. f x 2 + xy + y 2 = 7, f 2x(2x - 3) = 184, 1 x 2 - xy + y 2 = 19. 1 9y(2x + y) = - 135. f x 2 - 3xy + y 2 = 1, . f x 2 + xy + y 2 = 7, 1 (x+y+2)(2x- 2/ + l) =0, 1 y 2 - x 2 = 5. ' x 2 + 3xy + y 2 - 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 = m co#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 x 2 ; 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 y 2 , 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 y 2 ; 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 + ex 2 (parabola), (3) x = a + by + c?/ 2 (parabola), (4) y = kx n (parabolic in form), (5) xy = c (hyperbola), (6) y = c!0 kx (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 + a 2 z 2 + + a n x n (nth degree curve) . The coefficients OQ, a\, a%, , a n 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) (x 2 , 2/2) on the line, then (12) =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 + a 2 x 2 + + a n x n 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 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. . 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 = a + bx + ex 2 . 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 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 (Fahrenheit) given by the values X 25 50 75 e 482 370 356 Determine a formula of the type 6 = a + bx + ex 2 . 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 9,545 136.36 136.36 1900- 01 85 12,192 143.44 143.44 01- 02 48 7,468 155.58 155.58 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 37,943 134.07 134.07 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 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 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 + bx 2 . ' 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 = x z . 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 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 = 11C. 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 + fez 2 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 = kv 2 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 = bx m 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=io g i/. 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 = 1.574X 1 -"* 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 = cW kx , 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 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 1.5 ?n ?5 30 3 5 40 45 5.0 E 1 03 Of> 0.9 1 3 1 7 ?,?, 2.7 33 3.9 Find a relation of the form E = kp n . Ans. E = .Sp 1 -' 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 = kd n . 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 = kv n . 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. pv lM = 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 = kd n . 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 = kx n 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 = 2.5x 1/2 . XI, 177] EMPIRICAL EQUATIONS 241 7. Draw each of the following curves: (a) y = x 1 / 2 . (6) y = 2x 2 . (c) y = 2s 1 / 2 . (d) y = 3X 3 / 2 . (e) y = 8x~ 3 / 2 . (/) y = 1.5x 2 ' 3 . (g) y = 9.2X- 2 / 3 . (h) y = log x 2/3 . (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 14.4 28.4 42.2 y . . 180 24 3 0.7 X 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 = kx n . 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 = kx n . 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 a n = the last or rith term d = the common difference 5 or s n = 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 a n , and the sum s or s n of the first n terms. 183. Formulas. If the terms of a geometric progression be written down and numbered as follows, Term : a, ar, ar 2 , ar 3 , 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 + ar 2 + + ar n ~ l Multiplying both sides by r, sr = ar + ar 2 + ar 3 + + ar n By subtraction, we have sr s = ar n a. Solving the last equation for s, we get r - 1 1 - r From (4) we obtain rl = ar n . 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 4 7 - 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, ar 2 , ar 3 , etc. EXAMPLE. Insert three geometric means between 4 and 16. Since 16 is to be the 5th term we have a = 4. ar 4 = 16, whence r 4 = 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?, a 3 , 7. V3 - 1, V2. V3 + 1, - 8. 8, 4, 2, 1, 9. a, - a 2 , a 3 , - 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 10 8 . 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 r 2 < r, r 3 < r 2 , etc., and r n converges to zero when n is taken larger and larger. Therefore the second term on the right converges to zero, and s n 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 + or 2 + , 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 + T 3 5 + . 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 V n ; its value will be (9) V n = 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) V n = 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%, n t 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 + w 2 + " + n k 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 n P t is used to denote this number. Then, as shown above, (1) n P k = 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 10 P 4 = 10 9 8 7 = 5040. In this notation we should write for the number of permutations of n things all at a time (2) n P n = 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, r 2 , v, 63 is indeed 7 ! ; but when the subscripts are dropped the permuta- tions TI e\ s e^rzv 63 and r^ 2 s e 3 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 r z 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 n Ck 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 n P k = k\ n C k . Making use of the value of n P t , (1), 203, and solving for n Ck 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) n C k = ' 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 = n Ck- 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. n Ci. 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 i C 5 . 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) n C 2 = 45; (6) B C 2 = 190; (c) n C 2 = 105; find n. 12. In how many different ways can 500 ears of corn be selected from 505 ears? 13. Compute: (a) loooCW; (&) mC m ; (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; + a 2 )(.r + a,)(x + o 4 ) (x + a n ) = X n + CiX 1 + C Z X n ~* + C 3 X n ~ 3 + + Cn-iX + C n in which the coefficients have the following values: Ci = 01 + a 2 + 3 + + a n . The number of these terms is n. C 2 = Oi0 2 + aia n + a z a 3 +'+ a 3 a 4 + + a n -in. The number of these terms is the number of combinations that can be made from n a's, 2 at a time, i. e., n C 2 . Cs = aia 2 a 3 + aia 2 a 4 + + a 2 a 3 a 4 + * + a n _ 2 a n _iO n . The number of these terms is the number of combinations that can be made from n a's, 3 at a time, i. e., n C 3 . d = Oia 2 a 3 a 4 + aia 2 a 3 a 6 + + a n _ 3 o n _ 2 a n _ia n . The number of these terms is n Ct. C r = aiO 2 o 3 a r + The number of these terms is C r . C n = aidzds ' a n , 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, C z = n C 2 y*, C 3 = n C 3 y 3 , r - r if r - n n \sr n^ry , > ^n y j and therefore (2) (x + y) n = x n + nx n ~ l y + n C 2 x"- 2 z/ 2 + n C 3 x n ~ 3 y 3 + + n C r x n ~ r y r + + nxy n ~ l + y n . 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 n C r x n - r y r , 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 . ~nr*,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 n CV+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, n C r . 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 2 n . We verify from the above table that 1 + 1- 2 1 ; 1+2 + 1= -2 2 ; 1 + 3 + 3 + 1 = 2 3 ; 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 + n Cl + n C 2 + + Cn-l + n C n which proves the statement. Transposing 1, we have Ci + n C 2 + n C 3 + ' + n C n = 2 n -I i. e., the total number of combinations of n things taken 1,2, 3, " ' , n, at a time is 2 n 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 - n d + n C 2 - n C 3 + n C 4 - n C n whence 1 + nC'2 + nC* + = n C\ + n Cs + 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 = 32z 8 - 240x 4 t/ + 720z 3 2/ 2 - 2160z 2 ?/ 3 + 810zy 4 - 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) 2 c + 3 (a + 6)c 2 + c 3 = a 3 + 3a 2 6 + 3afe 2 + b 3 + 3 (a 2 + 2ab + 6 2 )c + 3(o + b)c 2 + c 3 = a 3 + 6 3 + c 3 + 3a 2 6 + 3a 2 c + 36 2 c + 36 2 a + 3c 2 a + 3c 2 6 + 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 + x 2 ) 6 . 7. (x*+ y*)*. 8. (or 1 + 2ay~ 1 )*. 9. (a" 1 - x~ 2 )*. 280 MATHEMATICS [XVI, 212 10. (a 2 - b 2 ) 8 . 11. (3a 2 b + 2C 3 ) 8 . 12. (1 + x) 10 . (<! 9 V 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 ! n C r x n ^ r y r = - x n ^y r . 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 V 3 \x + - j which contains x 2 . The (r + l)th term is i S C f r (x 1 / 2 ) 13 - r (x- 1 ) r = n C r -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. f x 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 n C r 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 = 32 1/5 + 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 C 3 + ... VI - x 1 + 3 X ~T 3- 6 1 3 6- 9 19. Show that (1 a;)" 2 = 1 + 2x + 3x 2 + 4x 3 -i ..... 20. O how that 1 -1 *x+- 3 r2 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 W 2 . 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 R 2 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 W 2 w Result of mixing : R 2 + 2RW + W 2 . 215. Successive Generations. Let R* denote the result of fertilizing R 2 with R 2 ; R S W denote the result of fertilizing R 2 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* &.. R 4 2R 3 W R*W* 2R 3 W 4R?W 2 2RW 3 /PTP 2RW 3 W* 2RW JP Result of mixing : R* + 4RW + 6RW 2 + 4RW 3 + 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 = R 2 + W 2 + B 2 + 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 = 4PF 2 + 12WR + 9fl 2 . 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 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 Pe in . 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 = Pe in , 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 = ce trt . 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 = ce T , 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 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%, , E n are p\, pz, , p n , 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 p n ; then 'the probability that all of the events will happen in the given order is Pi, P2 " p n . 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 h n ways and can fail in / ways ; then they can all happen and fail in (hi + fi)(hz + f%) - (h n + /) ways. Now all the events can happen together in the given order in hi A 2 h n 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 "- h n _ h\ hi h n ' " (hi +fi)(h 2 +/ 2 ) -. (h n +/) hi +/! A 2 +/ 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 f 5+; + 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 1 Let us plot this data with heights as abscissas and frequencies as ordinates. Construct rectangles, with bases on the horizontal rr 1 ^, / \ / v \ / 80 / / V / - \ \ -GO \ / j 10 / / \ \ i j / ' / ^ .- "" \ 5 s i 'i 45 5 5 5 6 f 5 7 ' 5 E f <J 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, / O x 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 E s . It is shown in the theory of probability that (4) E s = 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 E m , 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 E m and m + E m , 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 E ay 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- + E y , and one half without. The formula for the probable error in the standard deviation is ttrt F .- E m _ 0.6745Q- \P) &v j= . V2 V2n EXERCISES 1. Compute E,, E m , E for the data in 225. 2. Compute <r, C, E,, E m , 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,, E m , 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 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, E m , 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, E m , 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, M w . Find the deviation DI of ear length from mean length, and the deviation D w of weight from mean weight, for each ear tabulated. For each ear tabulated find the product of DI and D w and then add all of these products. This sum we will indicate by 3DiD w . Find in the usual way the standard deviation of length of ears, <TI, and the standard deviation of weight of ears, <r w . 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 D w 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 D w DiD v 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 M w 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 ; V w 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 DiD w , obtaining in this case 4947.2. Divide this number 4947.2 by n X <TI X . In this case n = 993, and <TI, ff w 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) M t = r^lw. a w 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 M w = 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 : M p = mean height of mid-parents, M e mean height of adult children, ff p = standard deviation of height of mid-parents, ff e = 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 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 > Phi UTT Pi X X 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. 1 2 3 4 5 6 7 8 9 Prop. Parts 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 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 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. 1 2 3 4 5 6 7 8 9 316 Table I. Logarithms of Numbers N. 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. 1 2 3 4 5 6 7 8 9 317 Table I. Logarithms of Numbers No. 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 1 112 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. 1 2 3 4 5 6 7 8 9 318 Table I. Logarithms of Numbers N. 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. 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 Log 10 Value Lo<r 10 Value Logio Value Log 10 .0000 000' 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 100' .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 Log 10 Value Lojr ln Value Log 10 Value Log 10 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 Log 10 Value Loi^m Value L<> 10 Value Log la .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 1100' .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 1500' .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<; 73 C 00' 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 Log 10 Value Log 10 Value Log- 10 Value Lopr 10 .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 Log 10 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 Log 10 Value Log 10 Value Log 10 Value Log 10 .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. ,:; 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 Log 10 Value Log 10 Value Lop 10 Value Log 10 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 Log le Value Log t .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 Log 10 Value Log in Value Log ]0 Value Log 10 )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 1 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 A r = 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 = log e 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. 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 6 132 136 140 143 165 170 175 179 198 204 210 215 144 146' 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 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, 255 1 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 ALFRED MONROE KENYON PROFESSOR OF MATHEMATICS, PURDUE UNIVERSITY AND LOUIS INGOLD ASSISTANT PROFESSOR OF MATHEMATICS, THE UNIVERSITY OF MISSOURI Edited by EARLE RAYMOND HEDRICK Trigonometry, flexible cloth, pocket size, long I2mo (xi- s rij2 pp.) with Complete Tables (xviii + 124 pp.), $t-5O Trigonometry (xi-\- 132 pp.) with Brief Tables (xviii -{- 12 pp.), $1.20 Macmillan Logarithmic and Trigonometric Tables, flexible cloth, pocket size, long izmo (xviii + 124 pp.) , $60 FROM THE PREFACE The book contains a minimum of purely theoretical matter. Its entire organization is intended to give a clear view of the meaning and the imme- diate usefulness of Trigonometry. The proofs, however, are in a form that will not require essential revision in the courses that follow. . . . The number of exercises is very large, and the traditional monotony is broken by illustrations from a variety of topics. Here, as well as in the text, the attempt is often made to lead the student to think for himself by giving suggestions rather than completed solutions or demonstrations. The text proper is short; what is there gained in space is used to make the tables very complete and usable. Attention is called particularly to the com- plete and handily arranged table of squares, square roots, cubes, etc. ; by its use the Pythagorean theorem and the Cosine Law become practicable for actual computation. The use of the slide rule and of four-place tables is encouraged for problems that do not demand extreme accuracy. Only a few fundamental definitions and relations in Trigonometry need be memorized; these are here emphasized. The great body of principles and processes depends upon these fundamentals; these are presented in this book, as they should be retained, rather by emphasizing and dwelling upon that dependence. Otherwise, the subject can have no real educational value, nor indeed any permanent practical value. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New Tork ELEMENTARY MATHEMATICAL ANALYSIS BY JOHN WESLEY YOUNG PROFESSOR OF MATHEMATICS IN DARTMOUTH COLLEGE AND FRANK MILLETT MORGAN ASSISTANT PROFESSOR OF MATHEMATICS IN DARTMOUTH COLLEGE Edited by EARLE RAYMOND HEDRICK, Professor of Mathematics in the University of Missouri Cloth, i2tno, 542 pp., $2.60 A textbook for the freshman year in colleges, universities, and technical schools, giving a unified treatment of the essentials of trigonometry, college algebra, and analytic geometry, and intro- ducing the student to the fundamental conceptions of calculus. The various subjects are unified by the great centralizing theme of functionality so that each subject, without losing its fundamental character, is shown clearly in its relationship to the others, and to mathematics as a whole. More emphasis is placed on insight and understanding of fundamental conceptions and modes of thought ; less emphasis on algebraic technique and facility of manipulation. Due recog- nition is given to the cultural motive for the study of mathe- matics and to the disciplinary value. The text presupposes only the usual entrance requirements in elementary algebra and plane geometry. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New Tcrk ANALYTIC GEOMETRY BY ALEXANDER ZIWET PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF MICHIGAN AND LOUIS ALLEN HOPKINS INSTRUCTOR IN MATHEMATICS IN THE UNIVERSITY OF MICHIGAN Edited by EARLE RAYMOND HEDRICK Flexible cloth. 111., izmo, viii + j6g pp., $fj6o Combines with analytic geometry a number of topics, tradi- tionally treated in college algebra, that depend upon or are closely associated with geometric representation. If the stu- dent's preparation in elementary algebra has been good, this book contains sufficient algebraic material to enable him to omit the usual course in College Algebra without essential harm. On the other hand, the book is so arranged that, for those students who have a college course in algebra, the alge- braic sections may either be omitted entirely or used only for review. The book contains a great number of fundamental applications and problems. Statistics and elementary laws of Physics are introduced early, even before the usual formulas for straight lines. Polynomials and other simple explicit func- tions are dealt with before the more complicated implicit equa- tions, with the exception of the circle, which is treated early. The representation of functions is made more prominent than the study of the geometric properties of special curves. Purely geometric topics are not neglected. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New Tork Analytic Geometry and Principles of Algebra BY ALEXANDER ZIWET PROFESSOR OF MATHEMATICS, THE UNIVERSITY OF MICHIGAN AND LOUIS ALLEN HOPKINS INSTRUCTOR IN MATHEMATICS, THE UNIVERSITY OF MICHIGAN Edited by EARLE RAYMOND HEDRICK Cloth, viii + 369 pp., appendix, answers, index, I2mo, $i-7j This work combines with analytic geometry a number of topics traditionally treated in college algebra that depend upon or are closely associated with geometric sensation. Through this combination it becomes possible to show the student more directly the meaning and the usefulness of these subjects. The idea of coordinates is so simple that it might (and perhaps should) be explained at the very beginning of the study of algebra and geometry. Real analytic geometry, however, begins only when the equation in two variables is interpreted as defining a locus. This idea must be introduced very gradu- ally, as it is difficult for the beginner to grasp. The familiar loci, straight line and circle, are therefore treated at great length. In the chapters on the conic sections only the most essential properties of these curves are given in the text ; thus, poles and polars are discussed only in connection with the circle. The treatment of solid analytic geometry follows the more usual lines. But, in view of the application to mechanics, the idea of the vector is given some prominence; and the representation of a function of two variables by contour lines as well as by a surface in space is explained and illustrated by practical examples. The exercises have been selected with great care in order not only to fur- nish sufficient material for practice in algebraic work but also to stimulate independent thinking and to point out the applications of the theory to con- crete problems. The number of exercises is sufficient to allow the instructor to make a choice. To reduce the course presented in this book to about half its extent, the parts of the text in small type, the chapters on solid analytic geometry, and the more difficult problems throughout may be omitted. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York THE CALCULUS BY ELLERY WILLIAMS DAVIS PROFESSOR OF MATHEMATICS, THE UNIVERSITY OF NEBRASKA Assisted by WILLIAM CHARLES BRENKE, Associate Professoi ol Mathematics, the University of Nebraska Edited by EARLE RAYMOND HEDRICK Cloth, semi-flexible, xxi + 3&3 PP- + Tables (63), szmo, $2.10 Edition De Luxe, flexible leather binding, India paper, $2.30 This book presents as many and as varied applications of the Calculus as it is possible to do without venturing into technical fields whose subject matter is itself unknown and incomprehensible to the student, and without abandoning an orderly presentation of fundamental principles. The same general tendency has led to the treatment of topics with a view toward bringing out their essential usefulness. Rigorous forms of demonstra- tion are not insisted upon, especially where the precisely rigorous proofs would be beyond the present grasp of the student. Rather the stress is laid upon the student's certain comprehension of that which is done, and his con- yiction that the results obtained are both reasonable and useful. At the same time, an effort has been made to avoid those grosser errors and actual misstatements of fact which have often offended the teacher in texts otherwise attractive and teachable. Purely destructive criticism and abandonment of coherent arrangement are just as dangerous as ultra-conservatism. This book attempts to preserve the essential features of the Calculus, to give the student a thorough training in mathematical reasoning, to create in him a sure mathematical imagination, and to meet fairly the reasonable demand for enlivening and enriching the subject through applications at the expense of purely formal work that con- tains no essential principle. THE MACMILLAN COMPANY Publisher! 64-66 Fifth Avenue Hew Tork GEOMETRY BY WALTER BURTON FORD JUNIOR PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF MICHIGAN AND CHARLES AMMERMAN THE WILLIAM McKiNLEY HIGH SCHOOL, ST. Louis 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