LE MEN TART FUNCTIONS UC-NRLh IN MEMORIAM FLORIAN CAJORl ELEMENTARY FUNCTIONS AND APPLICATIONS BY ARTHUR SULLIVAN GALE, Ph.D. AND CHARLES WILLIAM WATKEYS, A.M. PROFESSORS OF MATHEMATICS IN THE UNIVERSITY OF ROCHESTER NEW YORK HENRY HOLT AND COMPANY 43 Copyright, 1920 BY Henry Holt and Company • < « • • CAJORl r PREFACE This book presents a coherent year's work in mathematics for college freshmen, consisting of a study of the elementary functions, algebraic and transcendental, and their applications to problems arising in various fields of knowledge. The treatment is confined to functions of one variable, with inci- dental exceptions, and complex values of the independent and dependent variables are excluded. The subject matter includes the essentials of plane trigonometry and topics from advanced algebra, analytic geometry, and calculus. The text is the result of experiments beginning in 1907-8. It has been used in the classroom since 1913-14, and each year extensive revisions have been made. Hence the content of the course, the order of topics, and the manner of presentation are based upon the experience of several years. The unity of the course is gained by an explicit analysis of the functions studied, which enables the student to compre- hend the purpose of the course as a whole and the nature of the investigation of properties of functions of a given type. This analysis consists of three parts: First. Relations between a given function and its graph (see pages 42 and 274). Most of these relations are considered in the first chapter so that at the start the student is made aware of a number of questions which will be investigated in studying a particular type of functions. Second. Relations between pairs of functions and their graphs (see page 152). These geometric transformations are introduced in connection with simple algebraic functions so that they are familiar tools by the time they are needed for the study of transcendental functions. 918206 iv PREFACE Third. Analogous properties of functions which have no immediate graphical interpretation. Several properties of x^ are grouoed together on page 153 in order to indicate further questions which should be investiagted in studying trans- cendental functions. Emphasis is also placed on characteristic properties which distinguish one class of functions from another. A very large group of freshmen taking mathematics do not continue the study of this subject in the following years, and the needs of these students have received primary considera- tion. For the general student, the interpretation of a graph, the fundamental concepts of the calculus, and the usefulness of mathematics are of fundamental importance. Fortunately these matters are also important for the student of mathe- matics, and experience with the text has shown that it is pos- sible in the second year satisfactorily to complete the usual first courses in analytic geometry and calculus. !,' To show the usefulness of mathematics, a wide range of problems which deal with matters of interest to the student have been introduced, although exercises which require con- siderable instruction in other subjects have been avoided. AppHcations to the solution of problems in mechanics, physics, chemistry, economics, and other subjects have been scattered throughout the course. The analysis of a problem in a field other than mathematics is usually more difficult for a freshman than the solution after the conditions and requirements have been stated in mathematical language. But from the broad standpoint in which mathematics appears as part of an educa- tional system, the training in such analysis is as important as the development of the mathematical processes to which the analysis leads. The obligation resting upon the teacher of mathematics to develop this power of analysis is increased by the proneness of other teachers to tread very lightly on the mathematical aspects of their own subjects, and it is quite possible that this inclination on the part of others is partly due to the failure of mathematicians to emphasize the appU- cations sufficiently. Simple appUed problems may furnish PREFACE V drill in mathematical technique, with added interest, and with but slightly increased difficulty. Simultaneously, they afford some training in analysis. Attention is called to the following features: 1. The chapter on the theory of measurements gives an outline of statistical methods which are used in many fields such as economics, biology, physics, education, etc. It is, perhaps, not too much to say that the average college graduate will find more use for this topic than for any included in the traditional freshman course. The treatment given here is intermediate between the books on statistics which presuppose very little mathematical theory and those which the mathe- matical prerequisites render unsuitable for the average college man. 2. Emphasis is placed upon the determination of a function from a given table of empirical data. In problems of this sort, which illustrate an important method of discovery in science, there is an element of general culture which is too often neg- lected in elementary courses in the various sciences on account of the mathematics involved. 3. Graphical methods of analyzing problems are used freely. 4. Graphical methods are used in problems which can be solved by straight Unes, and an algebraic solution is then obtained by finding the equation of certain Unes used in the graphical solution. 5. Use is made of the graphs of the trigonometric functions in tying together and affording the means of recalling many properties of these functions. This use of the graphs is merely a part of the general point of view of the course. 6. Trigonometric analysis, the most abstract topic included, is postponed until late in the course. 7. The introduction of a considerable amount of the ele- mentary portions of the calculus gives the general student a knowledge of the importance and utility of the fundamental ideas of derivative and integral. 8. The average rate of change of a function is introduced at the start, and it is used in studying the linear function. vi PREFACE The rate of change is introduced informally in connection with the quadratic function, while formal treatment is post- poned until later. The difficulties in grasping the concept of a derivative are thus separated, and time for thorough assimi- lation is afforded. 9. Integration is used to obtain the volumes of a pyramid, cone, and sphere. 10. Tables of squares, square roots etc., that is, tables of functions famiHar to the student, are used in advance of othisr tables. The table of logarithms is introduced as a general tool, and it is not regarded as something to be used primarily with the trigonometric functions. Use is made of various tables throughout a large part of the course, so that the student acquires faciUty in their use and in the selection of the most suitable table for a given computation. 11. The use of the slide rule, and of logarithmic and semi- logarithmic cross section paper, is explained in connection with the logarithmic function. 12. The functions a;" and 6^, which occur frequently in the appHcations of mathematics, are treated at some length. 13. An effort has been made to render expHcit the purpose of the various parts of the course. 14. Formal proofs of a number of theorems are omitted, and some are assumed without proof. The appeal to the intuition imderlying most of these assumptions is justified by the behef that the logical presentation of these theorems re- , quires a foundation too abstract for the general student, or too cumbersome for the purpose to be served. 15. The course includes more than a year's work so that teachers have an opportunity for choice of topics, and abundant naaterial is provided for selected sections which progress more rapidly than the average. 16. The course increases in difficulty with a corresponding increase in interest and gain in power on the part of the student. 17. A very wide range of problems, varying in difficulty, makes it possible for the instructor to emphasize different aspects of the subject, to select exercises suitable for students PREFACE vu of different abilities, and to assign different sets of exercises in different years. 18. Many of the exercises in the later chapters, while dealing with the newer topics, are constructed to afford review of principles presented earUer in the course, and to correlate various parts of the subject. We thank several friends for the inspiration of their interest and for their suggestions. We also thank the Trustees of The University of Rochester for making it possible for us to use the text in the classroom throughout our experimentation, in which we have participated equally. Combined courses are still to be regarded as in the experimental stage, but it is our conviction that they are fundamentally sound, and we shall feel well repaid if this volume contributes something of value to their development. Arthur Sullivan Gale Charles William Watkeys The IlNrvTERsiTY of Rochester May, 1920. CONTENTS PAai5 Some Principles of Algebra and Geometry xv CHAPTER I SECTION FUNCTIONS, EQUATIONS, AND GRAPHS 1. Comparison of the Reasoning in Natural Science and in Mathematics 1 2. Example of the Utihty of Mathematics in Science 2 3. Forms in which Data are Recorded 3 4. Variable. Function 4 5. Notation for a Function 8 6. Determination of the Function which Expresses the Func- tional Relation between Two Variables 8 7. Graphical Representation. Directed Lines 12 8. Rectangular Coordinates 14 9. Graph of a Function 16 10. Discussion of the Table of Values 21 11. Functions becoming Infinite. Asymptotes 25 12. Variation of a Function 28 13. Average Rate of Change of a Function 34 14. Classification of Functions 38 15. Summary 41 CHAPTER II LINEAR FUNCTIONS 16. Uniform Rate of Change 46 17. Characteristic Property of a Straight Line 49 18. Slope of a Straight Line 51 19. Graphical Solution of Problems Involving Functions which Change Uniformly 54 ix X CONTENTS 20. Graph of the Linear Function mx -{- b ... 56 21. Variation 61 22. Uniform Acceleration 63 23. Equation of a Straight Line 66 24. Application to the Solution of Problems 69 25. Remarks on Measurements 72 26. Possible Errors in Arithmetic Calculations. Abridged Mul- tipHcation and Division 74 27. Empirical Data Problems , . 78 CHAPTER III ALGEBRAIC FUNCTIONS 28. Introduction 87 29. Graph of x^ 87 30. Graphs of ax^ and af(x) 88 31. Translation of the Coordinate Axes 89 32. Instantaneous Velocity 93 33. Rate of Change. Slope of Tangent Line 94 34. Graph of the Quadratic Function ax^ + bx + c 98 35. Empirical Data Problems 104 36. The Function x" 107 37. Tables of Squares, Cubes, Square Roots, Cube Roots, and Reciprocals 107 38. Graph of x", w > 1 110 39. Graph of x", < w < 1. Graphs of Inverse Functions . . 113 40. Graph of x", n < 0. Graphs of Reciprocal Functions 117 41. Summary of Graph of x" 119 42. Interpolation 121 43. Variation 125 44. Empirical Data Problems 127 ax + b 45. The Linear Fractional Function — — : 131 ex + a 46. Integral Rational Functions 133 47. The Remainder Theorem 133 48. Synthetic Division 134 49. Graph of a Polynomial 136 50. Extent of the Tables 137 51. Solution of Equations. Rational Roots 140 62. Translation of the ^/-axis 144 CONTENTS xi 53. Horner's Method of Solution of Equations 147 54. Graph of the Function / (ax) 150 55. Related Functions and their Graphs 151 56. Some Operations of Algebra regarded as Properties of Func- tions 153 CHAPTER IV TEIGONOMETRIC FUNCTIONS 57. Introduction 158 58. Angles of any Magnitude 164 59. Trigonometric Functions of any Angle 165 60. Radians 170 61. Graphs of the Trigonometric Functions 172 62. Functions of Complementary Angles 177 63. Tables of Trigonometric Functions 178 64. Solution of Right Triangles 180 65. AppHcations 183 66. Parallelogram Law — Velocities, Accelerations, Forces 185 67. Conditions of Equilibrium of a Particle 187 68. Functions of n90° ^d 192 69. Apphcation to the Use of Tables 196 70. IncUnation and Slope of a Straight Line 199 71. Law of Sines 201 72. Law of Cosines 202 73. Solution of Oblique Triangles 203 74. Inverse Trigonometric Functions 209 CHAPTER V EXPONENTIAL AND LOGARITHMIC FUNCTIONS 75. Introduction 214 76. Graph of the Exponential Function 6*, 6 > 1 216 77. Properties of the Exponential Function 6*, 6 > 1 217 78. Computation by Means of an Exponential Function 219 79. The Logarithmic Function, the Inverse of the Exponential Function 221 80. Graph of the Logarithmic Function 221 81. Properties of the Logarithmic Function, log6 x,b > 1 222 xii CONTENTS 82. Common Logarithms 225 83. Computation by Means of Common Logarithms 229 84. Solution of Triangles 233 85. Exponential Equations 237 86. Compound Interest 241 87. Annuities 244 88. Graph of the exponential function kb"' 248 89. The Logarithmic Scale 249 90. Empirical Data Problems 256 CHAPTER VI DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 9L Introduction 264 92. Limits 265 93. Derivative of a Function 267 94. Fundamental Formulas for Differentiation 269 95. Derivative of a Polynomial 271 96. Corresponding Properties of a Function, its Graph and its Derivative 272 97. Velocity and Acceleration 276 98. Derivative of a Rational Function 278 99. Derivative of an Irrational Function 279 100. Equations of Tangent and Normal Lines 281 101. Problems in Maxima and Minima 285 102. Related Rates 289 103. Small Errors 291 104. Approximate value o{f{x-\- Ax) 296 CHAPTER VII INTEGRATION 105. Introduction 301 106. Area under a Curve 304 107. Motion in a Straight Line 308 108. Motion in a Plane , 311 109. Volume of a Right Prism 315 110. Volume of a Right Circular Cylinder 318 111. Volume of a Pyramid 320 112. Volume of a Solid of Revolution 323 CONTENTS xiii 113. Volume of a Cone of Revolution 325 1 14. Volume of a Sphere 327 115. Area of a Sphere 328 CHAPTER VIII PROPERTIES OF TRIGONOMETRIC FUNCTIONS Logarithmic Solution of Triangles, Cases III and IV 116. Introduction 332 117. Fundamental Trigonometric Relations 332 118. Trigonometric Equations 334 119. Trigonometric Identities 336 120. Functions of the Sum of Two Angles 338 121. Functions of the Difference of Two Angles 341 122. Functions of Twice an Angle, or the Functions of Any Angle in Terms of Half the Angle 342 123. Functions of Half an Angle, or Functions of Any Angle in Terms of Functions of Twice the Angle 343 124. Sum and Difference of the Sines or Cosines of Two Angles 345 125. Logarithmic Solution of Triangles, Case III 346 126. Logarithmic Solution of Triangles, Case IV 348 127. Miscellaneous Identities and Equations 351 128. Differentiation of Trigonometric Functions 352 129. Graph of the Function a sin (6x + c). Harmonic Curves . . 358 130. Empirical Data Problems 361 CHAPTER IX THEORY OF MEASUREMENT 131. Statistical Methods 365 132. Permutations 366 133. Combinations 367 134. The Binomial Expansion 370 135. Probabihty 373 136. Compound Events ' 375 137. Mortahty Tables 379 138. Frequency Distributions 382 139. Averages 388 xiv CONTENTS 140. Measures of Variability 398 141. Equation of the Frequency Curve Representing a Sym- metrical Distribution 404 142. The Probable Error 411 143. Least Squares 416 144. Correlation 420 SOME PRINCIPLES OF ALGEBRA AND GEOMETRY 1. Classification of the Numbers of Algebra. / Integral (1, 2, -3, etc.) [ I^ational ^ Fractional (f, -h, etc.) Real 1 _ _ [ Irrational (V2, V3, x, etc.) Complex (or imaginary) , which will not be considered in this course. 2. Laws of Addition and Multiplication. (a) Commutative laws: a + b = b + a. ah = ha. (b) Associative laws: a+ (b + c) = (a + h) +c. a{hc) = {ab)c. (c) Distributive law: a(b + c) = ah + ac. 3. Zero. (a) 0X0 = 0. (b) If o X 6 = 0, then a = or 6 = 0. (c) It Is impossible to divide by zero, for the quotient of a by zero, if it existed, would be a number q such that g x = a. But as g x = 0, by (a), we have a contradiction, and hence division by zero must be excluded. 4. Fractions. (a) The value of a fraction is unchanged if numerator and denomina- tor are divided by the same number not zero. This enables us to "cancel " a common factor of the numerator and denominator. (b) The value of a fraction is unchanged if numerate" and denominator are multiplied by the same number, that is, r = — r • This gives the rule: To simplify a given complex fraction, multiply numerator and denominator by the least common denominator of the fractions occurring in the numerator and denominator of the given frac- tion. Thus if we multiply numerator and denominator of the complex e .- a b . , . . bmx + amy fraction -r- by abm we get at once -r o* d '' ^ abcm + ahd c-\ m (c) Addition and subtraction: r+ 7= — Tj — * b d bd rvi SOME PRINCIPLES OF ALGEBRA AND GEOMETRY (d) Multiplication: r X 7 = r^- o a bd / \ T\' ' • CL c ad (e) Division: r ^ -7 = . — • a be (f) The reciprocal of a given number is a second number such that the product of the two is unity. Thus the reciprocal of a is -• 6. Rules for Signs. (a) (- a) (- b) = ab. (b) (- o) (6) = - ab. , . —a a ... —a a a , . /, x (^) Zb^b' ^"^^ T^ =6"~6* (e) a-(b-c)^a-b + c. 6. Factors. (a) a2 - 62 = (a + b) (a - 6). (b) o3 + 63 = (a + 6) (a2 - a6 + 62). (c) o' - 6» = (a - 6) (a2 + a6 + 62). (d) a« + 6« = (a + 6) (a^-i - a''-26 + a«-'62 - . . . + 6«-0, n odd. (e) a" - 6" = (a - 6) (a^-^ + a"-26 + a^'-^b- + . . . + 6»-i). (f) ofi+ (a + b)x + ab = (x + a) {x + 6). 7. Binomial Theorem. (a + 6)- - a- + na"-.6 + ^^^ a»- 6^ + "^""/''""^^ a- W + .-- + 6'». The coefficient of the (r + l)st term is >^ (^ - 1)(^ " 2) ••• (n - r + 1) Special cases: n = 2, (a + 6)2 •= a^ + 2ab + ¥. n = 3, (a + 6)3 = a' + 3a26 + 3a62 + b\ 8. Powers and Roots. (a) aO = l. (b) «-" = ^- (c) (a*")" = a»'"» - «/- (d) a« = VaP. (e) a'"-a'' = a"'+«. (f) ^ = a'«-». (g) Va VF = Vab. w 5=./? (i) To rationalize the denominator (or numerator) of a fraction con- taining a square root, multiply numerator and denominator by the quantity obtained by changing the sign of the radical in the denominator (or nu- merator). 9. Equations. (a) If equals be added to, or subtracted from, equals the results are equal. (b) If equals be multiplied or divided by equals, the results are equal. ELEMENTARY FUNCTIONS xvii Do not divide both sides of an equation by a quantity until the pos- sibihty of division by zero has been excluded. (c) Like powers, or roots, of equals are equal. (d) To simplify an equation containing fractions, multiply both sides by the lea^t common denominator of the fractions. (e) To simpHfy an equation containing square roots, transpose to one side all the terms except a single radical, and square both sides. (f) To solve the quadratic equation ax^ + bx + c =^ 0, factor the left- hand member, set each factor equal to zero, and solve for x. If the leflr-hand member cannot be easily factored, transform the equa- tion to the form and complete the square by adding to both sides the square of half the coefficient of x. Then extract the square root of both sides of the result- ing equation, and solve for r. (g) The roots of the equation , , » -b ^ V62 - 4ac ax^ + bx -\-c = are x = ^ The roots will be real and imequal, real and equal, or imaginary accord- ing as the discriminant b^ - 4ac is positive, zero, or negative. (h) Simultaneous equations in two variables x and y. If both equations are linear, to eliminate y, multiply each equation by the coefficient of y in the other, and subtract the results. If one equation is linear and the other quadratic, to eliminate one vari- able, solve the linear equation for x or y (whichever is easier) and sub- stitute in the quadratic equation. 10. Arithmetical Progression. In an arithmetical progression, a, a + d, a + 2dj a + 3d, . . . J each term is obtamed from the preceding by adding a constant quantity The nth term is I = a + (n - l)d. It The sum of n terms is S = j^ {a + I). The arithmetic mean between a and bis A = I (a + b). 11. Geometric Progression. In a geometric progression, a. ar, ar^, ar^, . . . , each term is obtained from the preceding by multiplying by a constant quantity. xviii SOME PRINCIPLES OF ALGEBRA AND GEOMETRY The nth term is I = ar"^'^. The sum of n terms is /^ ^ -_y__ ^ -^ - 4 5 2" ^ ^^ - it \ ^^ 1 Tr^-m IrU-Lr lOfl. 5ft. Fig. 6. then an increase of 1° would cause an expansion of 0.1/200, or 0.0005; hence for an increase of temperature of t degrees the expansion e would be e = 0.0005^. This equation expresses the generahzation in more compact form than the sentence above. In this illustration we have considered only a part of the table, a part for which the generah- zation is very simple. The determination of the mathematical expression of the generahzation from a table of values will be one of the objects of this course. The generalization from a mathematical point of view is considered in the following section. 4. Variable. Function. The following table gives the lengths of an iron bar suspended from one end when carrjdng different loads. Load in lbs. Length in in. 0, 500, 1000, 1500, 2000 30, 30.67, 30.91, 31.23, 31.52 FUNCTIONS, EQUATIONS, AND GRAPHS 5 Any one of the numbers in each of these sets of numbers is conveniently represented by a single symbol, as x, y, t, etc. Thus X may be taken to represent one of the numbers 0, 500, 1000, 1500, 2000 and any other number that might be included in the half of the table giving loads, while y may represent the cor- responding number giving lengths in the second part of the table. The symbols x and y are called variables, in accordance with the Definition. A variable is a symbol for any one of a set of numbers. In the experiment giving rise to the above table, the load was changed arbitrarily by the experimenter and the length of the bar for the chosen load measured. In consequence of this order of measurement the variable representing the load is called the independent variable and the one representing the length of the bar the dependent variable. It is customary to denote the independent variable by x and the dependent variable by y. If the experiment were repeated, under the same conditions, it would be found that for a specified load the length of the bar would be the same, that is, there is a law connecting the load and the length of the bar. This relation between the vari- ables is expressed by saying that the length, y, is a function of the load, x, in accordance with the Definition. A function is a variable so related to another variable (called the independent variable) that for every ad- missible value of the independent variable, one or more values of the function are determined. The function is also called the dependent variable. The idea of a function arises wherever there is a relation be- tween magnitudes which are changing, and it underlies all magnitude relations which mankind has discovered. Example 1. The algebraic expression 2x + 3 is a variable whose value is determined whenever a definite value is assigned to x. If x be given the value 1, then 2x + 3 has the value of 5, and if x has the value 2, 2x + 3 haa the value 7. Hence 2a; + 3 is a function of x. 6 ELEMENTARY FUNCTIONS Example 2. A theorem from physics relating to a falling body states that if a heavy object be dropped, the distance it falls in t seconds is 16^^. This distance is a function of the variable t; for if t be given a definite positive value, the distance is determined. li t = 2 seconds, the distance fallen is 64 feet; if < = 3 seconds, the distance is 144 feet. Negative values of t are meaningless and hence not admissible. Example 3. The formula for the area of a circle is A = irr^. If the radius r be given a definite value the area is determined. Hence the area A is a function of r. The symbol tt represents the number 3.14159 . . . which remains the same for any pair of corresponding values of r and A, and hence is called a constant in accordance with the Definition. A constant is a symbol for a particular number. It has the same value throughout a discussion. Example 4. A simple equation which occurs frequently in practice is y = mx, where x represents the independent variable, y the function, and m is a constant representing a fixed value as x and y vary. If X represents the number of pairs of shoes of a certain kind sold during a limited time by a dealer, and y the amount of the sales, then m represents the price per pair which remains fixed, for the time considered, while x and 2/ vary. Example 5. An equation in two variables establishes a functional re- lation between the variables. For if a value be given to either, the cor- responding value or values of the other may be found by substituting the given value of one variable and solving for the other. Either variable may be regarded as a function of the other, and the form of the function may be found by solving for one variable in terms of the other. Thus, if the equation ^y-x^^O be solved for y and then for x we have a? /- y = j and a; = * 2Vy- The given equation defines t/ as a function of x to be the function x^/4, and X as a function of y to be the function ±2V2/- Example 6. Some of the elements entering into the cost of a suit of clothes are the supply of cloth, the supply of labor, rent, style, etc. As these elements vary the cost of the suit will vary, so that the cost of the suit is a function of a number of variables. Considering one of the independent variables at a time, a part of the cost of the suit may be expressed as a function of this variable, e.g., the supply of cloth. FUNCTIONS, EQUATIONS, AND GRAPHS 7 The law of supply and demand from economics states that the price of an article increases or diminishes as the supply diminishes or in- creases. If X represents the supply of cloth and y the price of the suit of clothes, then it is assumed in economics that the functional relation may be ex- pressed by the equation m where m is a constant which can be determined in any concrete case. In this course we shall confine ourselves to the study of functions of one variable. Example 7. The temperature at a given place is a function of the time. For at a given time the temperature must have a definite value. But this function is so little understood that the Weather Bureau can only approxi- mate the value for any future time, and that, indeed, only for times in the hnmediate future. The data of such departments of human knowledge as physics, astronomy, and engineering are so complete that many of the functions arising there can be identified and studied by mathe- matical methods. In other subjects, for example, chemistry and economics, the data have only recently been made suf- ficiently complete to warrant an increasing use of mathematics. But there still remains a countless number of functions which mankind has been unable to represent by a mathematical expression. EXERCISES In the following exercises give the reason for the statement that one variable is a function of another. 1. Mention three variables which are functions of the side of an equi- lateral triangle of varying size. 2. A train goes from one station to another at a variable rate. Mention two variables of which the rate is a function. 3. What are some of the variables of which the cost of erecting an office building is a function? 4. Mention some variables involved in heating water in a pan on a gas stove. Which are independent variables? Which are functions? 5. Find the functions of x defined by the following equations and tabu- late three pairs of values. 8 ELEMENTARY FUNCTIONS (a) 2/ - x2 - 3x - 2 = 0. (b) x^ - 3?/2 + 2^/ + 4 = 0. (c) 3xy + Qx -9y + ^ = 0. (d) 2x^ + ^xy + 4y^ + Qx - Sy + 7 = 5. Notation for a Function. It is convenient to represent a function of the variable x by the symbol f{x) which is read " function of x," *^ the function / of a;," or merely ^' f of a;." The various parts of the symbol are to be regarded as forming a single compound symbol, never as separate symbols meaning the product of two numbers / and x. This symbol is used to denote either any function or a par- ticular function such as fix) =x^ + x-l. Similar symbols convenient for distinguishing different functions are F(^), g(.x), W, etc. An advantage of this notation is that the value of a function f(x) for any value of x, say x ^ a, may be suggestively represented hy f{a). For example, if f{x) = x2 + a; - 1, /(a) = a^ + a - 1, /(2) = 2^ + 2 - 1 = 5, /(-I) = (-1)2 + (-1) - 1 = - l,/(0) = - 1, f{-x) = {-xY + i-x) - 1 = a:2 - X - 1, etc. This notation also enables us to state certain theorems in a more compact form. EXERCISES 1. If /(x) = x3 _ xS find /(I), /( - 3), /(O), /(- X). 2. If F{x) = l/x2, find F(2), F{- 1), F{a), F{- x). 3. If = 100 and r = 1.05. Substituting these values in the above formula, we have for the required function A = 100(1.05)'-!. Check. Substituting the values 1, 2, 3, for t in this function we get the values 100 105 and 110.25 respectively for A, which results agree with the table. EXERCISES 1. In each of the following tables let z represent any one of the numbers in the first column (value of the independent variable), and find a function of X which will represent the corresponding number in the second colunm. Add two additional pairs of values of x and the function to each table. (a) X (b) X fix) (c) X 1 fix) 1 (d) x 1 fix) 1 1 1 1 1 27 3 4 8 2 8 4 2 64 4 9 27 3 27 9 3 125 5 16 64 (e) X FTJNCTIONS, fix) (f) X EQUATIONS, AND GRAPHS fix) (g) a: /(x) (h) X 11 1 8 27 64 1 4 2 9 4 16 6 3 1 6 2 9 3 3 1 5 2 7 3 9 4 2 4 8 16 2. Find the perimeter of an equilateral triangle as a function of the altitude x. 3. Express the area of a square as a function of (a) the side, (b) the diagonal. 4. Express the area of a regular hexagon as a function of the side. 5. A regular octagon is formed by cutting off isosceles right triangles from the corners of a square. Express the area as a function of the side of one of the right triangles cut off. ■6. Express the side of a regular decagon as a function of the radius of the circumscribed circle. 7. A Norman window consists of a square surmounted by a semi- circle. Find the area as a function of the side of the square x. 8. (a) A man walks from a certain town toward a second at the rate of three miles an hour. Express the distance traveled as a function of the time. (b) A second man starts at the same time from the second town, which is 10 miles from the first, and travels at the rate of 4 miles an hour toward the first town. Express his distance from the first town as a function of the time. (c) A third man starts from the same town as the first man but two hours later and travels at the rate of 3^ miles an hour in the same direction. Express his distance from town as a function of the time elapsed since the first man started. How far will each man be from the first town 4 hours after the fiirst man starts? When will each man be 20 miles from the first town? 9. A man starts from a town 15 miles directly west of a city and travels east at the rate of 4 miles an hour. A second man starts from the same town at the same time and travels west at the rate of 3 miles an hour. Express the distance of each from the city after t hours as a function of t. How far from the city will each be in 3 hours? When will they be 25 miles apart? 10. A ball starting from rest rolls down an inclined plane 4 feet in the first second, 8 feet in the next, 12 feet in the next, etc. Express the dis- tance it rolls in any second t as a function of t. How far will it roll in the 8th second? 11. With the data of the preceding problem and the formula for the smn of an arithmetic progression find the total distance rolled as a function of the time t. When will the ball have rolled 108 feet? \ 12 ELEMENTAKY FITNCTIONS 12. A body starting from rest falls 16 feet in the first second, 48 feet in the next, 80 feet in the next, etc. Find the distance fallen in any second as a function of the time of falling. Find the total distance fallen as a function of the time. 13. $100 is placed at simple interest at 4 per cent. Express the amount at the end of t years as a function of t. 14. If the interest in the preceding problem is compounded annually, express the amount at the end of t years as a function of t. 15. The cross-section of a gutter pipe is in the form of an isosceles trapezoid. The lower base and the inclined sides are each 3 inches long. Find the area as a function of the width across the top. 16. A rectangle is inscribed in a circle 4 inches in diameter. Express the area as a function of one of the sides. 7. Graphical Representation. Directed Lines. The func- tional relation can be represented by a geometrical figure which fui-nishes a valuable method for studying the properties of the function, since the whole of the relation is placed before the mind at once. The system of coordinates devised by Rene Descartes (1596-1650), which is developed in the following section, is the basis of the representation. This system of coordinates rests on the theory of directed lines. Let XX' be any line, and let the direction from X' to X be called positive, from X to X' negative. These words are used instead of such terms as north and south, to the right and left, up and down, backward and forward. A line upon which a positive direction has been fixed is called a directed line. The positive direction is commonly indicated by an arrow-head. If A and B be two points on a directed line, the symbol AB is used to denote either: (1) The Hne drawn from ^4 to B, or (2) The real number whose numerical value is the number of times the unit of length is contained in the line, and whose sign is positive or negative according as the direction from A to B is positive or negative. Thus, in the figure, AB and A'B' denote certain lines. x'\ — \ — f— HH — hH — \—\ — I— HH — I— f-x A B B' A' Fig. 8. ^ FUNCTIONS, EQUATIONS, AND GRAPHS 13 They also denote the numbers AB = S and A'B' = — 2, provided the unit of length is a quarter of an inch. It follows that if A and B are points on a directed line BA =-. -AB (1) I B Definition. No matter what the position of three points A, B, C, on SL directed Une may be, the sum oi AB and BC is defined to be AB + BC '= AC (2) If ^5 is thought of as a motion from A to B, and BC as a motion from B to C, the sum gives a motion from A to C. The sum gives the distance from A to C in both magnitude and direction, but not necessarily the total number of units passed over in going from A to B and then from B to C. That this definition agrees with elementary algebra, when AB, BC, and AC are regarded as numbers, is seen in the fol- lowing illustrations, of which the first agrees with arithmetic. x i I ^ I I I ? I y I l>x XI I -t I ^ I I I I f I | . x AB+ BC = 4^ + 2 = G= AC AB+BC=? y i f I [ f i I I [ ? t> X X'\ I ? I 1 I AB+BC=? AB + BC=-S+i-'i) = -7 = AC X' l I ? I I I ^ I f I I |. X X^ l ? I I I I ? I I ^ I |> X AB+BC = ! AB + BC=1 Fig. 9. An especially important use of directed lines is the following: Let be any point on a directed line X' X, and let points be laid off on each side of at a unit's distance from each other. Then with every point P on X'X is associated a real number OP, and conversely, with every real number is associated a point on the line. X'- I ol I I I I I -^x Fig. 10. In the figure, the numbers associated with the points A, B, C, D are respectively OA = 3, OB - - 3, OC ^ 5, OD = -4.5. u ELEMENTARY FUNCTIONS Using this association of points and numbers, if Pi and Pi are two points on the Hne, and xi and X2 are the numbers asso- ciated with them, we have from (2) OPl+PlP2 = OP2, ^1 JC' I I I I I I r ■^x Fig. 11. whence Pi Pa = OP2 - OPi = X2 - xi (3) This difference of the values of the x^s is denoted by Ax, so that Ax =PiP2=X2-Xi (4) Notice that Ax is a single symbol (never the product of two numbers A and x), and that its value is obtained by subtract- ing the value of x corresponding to the first point from that corresponding to the second. EXERCISES 1. Illustrate (3) by numerical examples for the six possible relative positions of 0, Pi, and P2. Find Ax for each case. 2. Show that Ax is positive or negative according as Pi lies to the left or right of Pi, using the definition that a 'b _ u Fig. 17. 24 ELEMENTARY FUNCTIONS The following theorems, whose proofs are left as exercises, follow from the facts that the points {- x, - y) and (x, - y) are symmetrical to the point (a;, y) with respect to the origin and the X-axis respectively. Theorem 2A. The graph of Theorem 2B. The graph of fix) is symmetrical with respect an equation is symmetrical with to the origin if respect to the origin if the equa- f{—x) = —fix), Hon obtained by replacing x by —X and yby —y, and simplify- ing, is identical with the given equation. Theorem SA. The graph of Theorem SB. The graph of f(x) is symmetrical with respect an equation is symmetrical with to the x-axis if its values occur in respect to the x-axis if the pairs which are numerically equal equation obtained by replacing but opposite in sign. y by — y, and simplifying, is identical with the given equation. We shall use the phrase to discuss the table of values of a func- tion to mean that the Symmetry, Values to be excluded. Intercepts, and Asjnuptotes (see next section) are to be determined before building the table of values. For the last three considerations, solve the equation for y in terms of x and for x in terms of y. But if the intercepts are desired in- dependently, they may be found by setting either variable equal to zero and solving for the other. EXERCISES 1. Does/(- x) always equal either ±/(a;)? 2. Discuss the table of values (omitting asymptotes) and plot the graph of each of the functions and equations. (a) x^ - 2. (b) xy9. (c) 3-x2. (d) y = X*. (e) 2/' - 4x + 2 = 0. (f) 2/=x» + 2. (g) x^ - 9x. (h) 2/2 + 6x = 0. (i) y*-{-4x^0. FUNCTIONS, EQUATIONS, AND GRAPHS 25 3. Discuss the table of values (omitting asymptotes) and plot the graph of (a) a:2+ 2/2 = 16. (b) x^ + y^-Qx = 0. (c) x^ + y^ + 4x = 0. (d) ix'- + 2/2 = 16. (e) a:2_2/2 = 16. (f) a;2 - 1/2 _ 6a: = 0. (g) 4X2 ^ Qy2 ^ 36. (h) 9x2 + ^2 + 36^ ^ 0. (i) y =x^-Ax\ 4. If J{x) is any one of the functions whose graphs are given below, determine whether or not J{- x) = */(x), find the value of /(O), and the values of x for which /(x) is zero, and for which it is imaginary. Fig. 18. 11. Functions becoming Infinite. As3miptotes. Continuing the discussion in the preceding section, in the following ex- ample we shall need the Definition. It is said that a function becomes infinite as x approaches a if the numerical value of the function can be made larger than any positive number, however large, by giv- ing X a value sufficiently near to a. 26 ELEMENTARY FUNCTIONS Example. Build a table of values and plot the graph of the function v-~ (1) Symmetry. The tests for symmetry show that the graph is not symmetrical with respect to either axis or the origin. Solving (3) for x we get x = ^±^- (2) y \ J Values excluded. As no radicals occur in (1) and (2), no imaginary values are encountered. Hence no values of x or y need be excluded on this account, and the graph runs indef- initely up and down, and to the right and left. But the values x = 4 and t/ = must be excluded as it is impossible to divide by zero (see 3 (c), page xv). Intercepts. Setting a; = in (1), the intercept on the ?/-axis is y = - h But as 2/ = is an excluded value, the curve does not cut the x-axis. Asymptotes. To determine the form of the graph between the points corresponding to a; = 3 and x = 5, the table includes a number of pairs of values near the excluded value a; = 4. 0, 1, 2, 3, 3.5, 3.8, 3.9, 4, 4.1, 4.2, 4.5, 5, 6, 7, 8 4 4' 4' ^' "^' "^' ~^' "^^' "' ^^' ^' ^' ^' h i I The nimaerical value of the function, according to the table, increases as x gets nearer to 4, and it is readily seen that by giving x values sufficiently near 4 the numerical value of ;; can be made larger than any given positive number, a; — 4 however large. Thus to make ^ numerically greater than X "~ ^ 10, let X have any value between 3.9 and 4.1 except the value 4; to make it greater than 100, let x have any value between 3.99 and 4.01 except the value 4; etc. Hence the function -z ^^^omes infinite as x approaches 4. This fact is indicated in the table by the symbols (4, »). FUNCTIONS, EQUATIONS, AND GRAPHS 27 In the figure it is seen that as the graph gets nearer the line perpendicular to the x-axis at the point for which a; = 4, it recedes farther and farther from the x-axis. Such a Hne is called an asymptote in accordance with the Definition. An asymptote of a curve is a straight line such that the distance from a point on the curve to the line approaches zero as the point recedes in- definitely from the origin along the curve. Similar reasoning establishes the fact that the x-axis is also an asymptote, corresponding to the excluded value y = 0. The example furnishes an illustration of the general prin- ciple: If a function becomes infinite CLS X approaches a, the line per- pendicular to the X-axis at x = a is an asymptote of the graph. We shall consider asymptotes of the graph of an equation or function only when they are parallel to one of the coordinate axes. Asymptotes parallel to the y-axis may be found by solving the equation for y; the solution is a function of x, and to each value of X for which this function becomes infinite there corresponds a vertical asymptote. Horizontal asymptotes may be found in a similar manner by solving the equation for x. An algebraic function becomes infinite for real, finite values of the variable only if the variable is contained in the denominator, and if the denominator is zero for one or more real values of the variable. EXERCISES 1. Discuss the table of values and plot the graph of 1 > ^ 4 Vi ■■"T- 1 1 ' 1 1 1 1 \ ^ 1 1 V, - , . o ] I ^ — _ ™i>x" — ... 1 . \l \ 1 — -€ 1 Fig. 19. (d) - (b) x^ + 1 (f) 2xy - 2» + 3y » 0. 28 ELEMENTARY FUNCTIONS (g) 2xy + Zx-by = 0. (h) x^ - 2a:y + 4 = 0. (i) a:^ - 4a:?/ + 6 = 0. (m) x^ -^y + 2x = 0. (n) a;^^/ + 2^ - 3a: = 0. (o) x'y - ^y - 3x = 0. 2. If a gas is kept at the same temperature, the product of the pressure and the volume is constant. Discuss the table of values and plot the pressure as a function of the volume, assuming a numerical value for the constant. 12. Variation of a Function. Under this heading we con- sider primarily three things. First. Sign of the function. If x increases through a given interval, the sign of the function will be positive if, and only if, the graph hes above the a:-axis; and the sign will be negative if, and only if, the graph hes below the a:-axis. For the ordinate representing a value of the function lies above or below the X-axis according as the function has a positive or negative value. Second. Changes of the function. If x increases through a given interval, the value of the function increases if, and only if, the graph rises as it runs to the right; and the value de- creases if, and only if, the graph falls as»it runs to the right. For the ordinate representing a value of the function increases or decreases according as the curve rises or falls. The motion to the right corresponds to increasing values of x. Third. Average rate of change of the function. This will be considered in the following section. In order to state in what intervals the function is positive or negative, or is increasing or decreasing, it is necessary to de- termine the values of x bounding these intervals. These values of X, the corresponding values of the function, and the points on the graph representing them, are the remaining important elements of the variation of the function. They are: First: The real zeros of the function, represented by the in- tercepts of the graph on the a;-axis. Second: The values of x for which the function becomes in- finite, which give rise to the vertical asymptotes. Third: The maximum and minimum values of the function, which we proceed to define. FUNCTIONS, EQUATIONS, AND GRAPHS 29 Definition. A function f{x) is said to have a maximum value for X = a if /(a) is greater than all other values of f(x) when X is near x = a. The point on the graph whose coordi- nates are x = a and y = f(a) is called a maximum point If X increases through x = a, the value of f(x) will increase to the value /(a), and then decrease. And the maximum point will be higher than the other points on the graph nearby. Definition. A function f{x) is said to have a minimum valu£ for X = a if /(a) is less than all other values of f{x) when x is near x = a. The point [a, /(a)] on the graph is called a minimum point. If X increases through x = a, the value of f{x) will decrease to the value /(a), and then increase. The minimum point will be lower than other points on the graph nearby. Example 1. Discuss and plot the graph of the function 2x'^ - 3a; - 9. Let y = 2x2 - 3x - 9, (1) To solve for x in terms of y we first write (1) in the form 2x^ -Sx -9 -y '0. Then by the formula for solving a quadratic 3 ^V9 + 72 + 8y 3 VSl + Sy ,_. ^ = 4 = 4 " 4 (2} Symmetry. The tests for symmetry show that the graph is not sym- metrical with respect to either axis or the origin. Valites excluded. From (1), no values of x need be excluded, so that the graph extends indefinitely to the right and left. Equation (2) shows that we must exclude all values of y for which 81 + 82/ < 0. Hence we must exclude y < - 10^, so that no part of the curve Hes more than lOi units below the x-axis. Intercepts. Setting y = in (2), the intercepts on the x-axis, or the zeros of the function, are found to be 3 9 . . o 4 4 Setting X = in (1), the intercept on the y-a.xis is y = -9. Asymptotes. There are no vertical or horizontal asymptotes. We now proceed to build a table of values and draw the graph. The variation of the function is readily discussed in connection with the graph. 30 ELEMENTARY FUNCTIONS Zeros of the function. These are represented by the intercepts on the rc-axis, OA and OB. The curve crosses the axis at A and B. Sign of the function. The graph is above the X-axis to the left of the point A and to the right of B. Hence 2x2 - 3a; - 9 > if ^ ^ _ j ^^ ^^ ^ ^ ^ The graph is below the x-axis between A and B, and hence 2x2 - 3x - 9 < if - 1.5 < X < 3. The last set of symbols is read "x is greater than — 1.5 and less than 3." Maximum and minimum values. We saw above that all values of y less than - lOg must be ex- cluded. Hence y = - lOl is the smallest value of the functions. Substituting this value in (2), the corresponding value of x is seen to be f. Hence the function has the minimum value - 10| when X = f . This value of the function is represented Fig. 20. ^y *^® ordinate ED, and the point Z>(|, - 10|) is a minimum point. This function does not have a maximum value. Changes of the function. Since the graph falls to the right at the left of the point D, 2x2 - 3x - 9 is decreasing as x increases if x < f . And since the curve rises to the right of D, 2x2 - 3x - 9 is increasing as x increases if x > f . The variation may also be stated as follows, but care must be exercised to see that none of the important elements are omitted. For a numerically large negative value of x the function is positive. As X increases to- 1.5, /(x) decreases to zero. It then becomes nega- tive, and continues to decrease until it assumes its minimum value of - lOff when x = f . As x increases from | to 3, the function is negative and increases to zero. As x increases beyond x = 3, /(x) is positive and increasing. y \ \ \ \ „ 1 —1 *1 E B •^ r 1 3 .i 1 \ 1 \ \ 7l \ '1 j \ 1 ^ <1 1 D In this example, the function does not change sign unless x increases througli a zero of the function. But a function may f FUNCTIONS, EQUATIONS, AND GRAPHS 31 change sign if x increases through a value for which the function becomes infinite. Thus j, whose graph is given on page 27, changes sign, from negative to positive, if x increases through the value 4. The intercepts on the a:-axis and the vertical asymptotes should therefore be determined before considering the sign of a function. In the example just cited, the vertical asymptote does not separate intervals in which the function increases or decreases, but it may do so. For example, the function l/x"^ becomes in- finite if X approaches zero, and hence the y-axis is a vertical asymptote. As x increases, this function increases if x is negative, and decreases if x is positive. This function also shows that a vertical asymptote need not separate intervals in which the function has opposite signs, for 1/x^ is positive for all real values of x. A good order for considering the elements of the variation of a function is as follows: Zeros of the function, Function becomes infinite, Sign of the function, Maxima and minima. Changes of the function. As the zeros and asymptotes are taken up in discussing the ible of values, only the last three are new ideas. These properties of a function may be determined approxi- mately by inspection of the graph of the function. This pro- cedure is especially useful if the table of values or the graph be given and the functional relation itself is unknown. The accuracy of the results will depend upon the choice of units on the axes, the care with which the graph is drawn, and the closeness with which it is read. 32 ELEMENTARY FUNCTIONS Example 2. A thermograph is an instrument which records the tem- perature continuously by means of such a curve as in the figure. Discuss the variation of the temperature as a function of the time. E "1 id /" "m / \ — / \ -30 / ^ ^ / "^ \ 20- / N 1 A, f > / -10° — > I 1 / ^l N ^ aJ 1 i -0_ N B \>1 A .M M 3 A \m t A M. f/^ M. S 1 Too % 31 jtf 6P M. 9 P M. ^f'^l \ !^ J A \ ft u / A ^ — — — X 1 Fig. 21. Zeros of the function. The graph cuts the time axis at B and D, hence the temperature was zero at 3 a.m. and at 8:30 a.m. Sign of the function. The graph is above the time axis from A io B and from D to F and below from B to D. Hence the temperature was above zero from 12 Mid. to 3 a.m. and from 8.30 a.m. to 12 Mid. the next night, and below zero from 3 a.m. to 8:30 a.m. Maximum and minimum values. At C the ordinates cease to decrease and begin to increase. Hence, the temperature had a minimum value of about - 14° at 6 a.m. At E the ordinates cease to increase and begin to decrease. Hence, the temperature had a maximum value of about + 45" at 3 p.m. Changes of the function. The graph rises from C to E and falls from A to C and from E to F. Hence, the temperature increased from 6 a.m. to 3 P.M. and decreased from 12 Mid. to 6 a.m. and from 3 p.m. to 12 Mid. If it is not desired to treat each topic separately, the results might be stated as follows: The temperature at midnight was 10°. It decreased until it became zero at 3 a.m. and continued to decrease until 6 a.m., when it reached a minimum value of about - 14°. It then increased, becoming zero at 8:30 A.M., until 3 p.m., when its maximum value was about 45°. From that time on it decreased to about 20° at midnight. The discussion of this function is continued in the next section. FUNCTIONS, EQUATIONS, AND GRAPHS 33 EXERCISES 1. Discuss the table of values, plot the graph, and determine the varia- tion of each of the functions: (a) y =x/2. (b) x^. (c) 2 - x\ (d) y = 16a: - xK (e) xy +6y = 4. (f) x2-42/ + 8 = 0. (g) x" -2y + Ax^Q. (h) 2xy -x -3y = 0. (i) x^ -2xy-\-l=^ 0. (J) x^ -5x -4y = 0. (k) a;2 -4xy + 2 = 0. (1) x^ -xy + S^O. (m) xy = Q. (n) xh/ = 12. (o) a; + 2/» = 0. 2. The surplus and shortage of railroad cars in thousands for the months of the years 1915 and 1916 were as follows. Plot the graph and discuss the function. 1915. 180, 279, 321, 327, 291, 299, 275, 265, 185, 78, 26, 38. 1916. 46, 21, - 20, 3, 33, 57, 52, 9, -19, ^60, -114, -107. 3. The following are the data for the surplus reserves of New York banks in millions for the months of the years indicated. Plot the graph and discuss the function. The Federal Reserve System was inaugurated November 16, 1914. 1914. 29, 32, 20, 17, 38, 40, 15, -29, -28, -0.4, 72, 117. 1915. 121, 135, 131, 156, 168, 185, 159, 177, 197, 179, 168, 155. 4. One side of a rectangle whose perimeter is 12 inches is x. Find the area as a function of x. Construct the graph of the function, discussing the table of values, and find the value of x if the area is a maximum. What is the maximum area? 5. A farmer wishes to fence off a poultry yard whose area is to be 6 square rods. If one dimension is x, express the perimeter (the amount of fencing needed) as a function of x. Discuss the table of values and plot the graph of the function. What will the dimensions be to require the least amount of fencing? How much should he purchase? 6. There are a number of diseases with continued fever in which the course of the temperature is sufficiently characteristic to furnish the diagnosis. Croupous pneumonia is one of these. (See Fig. 22.) Discuss the three functions. The zero for temperature would be the normal temperature 98°.4. By comparison of the three functions state in words some of the symptoms of the disease. 7. The data for the temperature curves of measles and scarlet fever 34 are given in the following tables, following graph. ELEMENTARY FUNCTIONS Compare the graphs with the J)igeaae 1 2 S ■f 5 e 7 8 9 10 11 12 70 m 1 60 UO / sh uA A A ir y V V / \A h\^ — 50 120 • r\ ••— ' i^'7 ./^ ^ ^m "A V v^ ./ V i \h Jfi 100 1 -I V^ t SO 80 i! \\ sN '7' jl X^/ 'rn 20 60 i V - V— 103^ 102" lOf 10(P 99'* 98'* 97'* Temperature.,-.^^ BeapiratiotL-.-^ Pulse Temperature, pulse, and respiration tables in croupous pneumonia. Fig. 22. Day of fever. 1 2 3 4 56 7 8 T. (Measles) a.m. 99^.4 101°.2 101°.2 101°.8 103^.4 102'*.8 99° 98^.4 P.M. 103°.8 ior.8 102^8 104°.8 105^8 103^4 99^8 normal T. (Scarlet / 98°.4 105^2 fever) ' ' \ 103°.8 104''.8 104° 103°.6 102°.6 101°.8 100°.8 99°.98 P.M. 103°.4 105°.8 105°.2 104°.4 103°.4 103°.2 101°.4 99°.4 Discuss these functions. How do the temperature curves distinguish pneumonia, measles, and scarlet fever, one from another? 13. Average Rate of Change of a Function. The average rate of change of temperature during a given period of time is a famiUar idea. For instance, in Example 2 of the preceding section, the temperature rose from — 14° to 45° between 6 a.m. and 3 P.M., a total rise of 59° in 9 hours. Dividing 59 by 9, we see that the average rate of change in this interval of time was about 6.5 degrees per hour. On the graph, C K represents the change in time from 6 a.m. to 3 P.M., and KE the corresponding change in temperature. Hence the average rate of change of temperature in this in- terval, 6.5 degrees per hour, is represented by the ratio KE/C K. The graph has its most abrupt rise from about the point G to FUNCTIONS, EQUATIONS, AND GRAPHS 35 Ax X y ^y ^y/^x 3 12 Mid. 3 a.m. 10 -10 -3.3 9 6 a.m. 3 P.M. -14 +45 +59 +6.5 2 8 a.m. 10 A.M. -6 +19 +25 +12.5 the point H, which indicates that the average rate of change of temperature was greatest from about 8 a.m. to 10 a.m. This average rate of change is represented by the ratio JH/GJ, Let y denote the temperature at any time x, and Ai/ the change in temperature during an inter- val of time Ao;. Since the aver- age rate of change is the ratio of the change of temperature to the corresponding change in time it is expressed by A?//Aa:. The computation of the average rate of change for several different intervals may be effected con- veniently in tabular form, the values of y for the given values of X being obtained from the graph. The idea of the average rate of change of a function of the time has been extended to include a function of any variable. This generalization is given in the Definition. The average rate of change of a function of Xy for a particular change in x, is the ratio of the corresponding change in the function to the change in x. If y denotes the function, and Ai/ the change in y due to a change of Ax in x, then the average rate of change of y with re- spect to X is symbolized by Ay /Ax. If X increases from one given value to another, the average rate of change of y may be found as in the illustration above, except that the values of y would be computed from the given function instead of being read from the graph. The method of finding a general expression for the average rate of change for any interval is illustrated in the Example. Find the average rate of change of the function y = Ix^ for the intervals of x from 1 to 3, from 2 to 4, and from 2 to 6. Find also the average rate of change for any interval. The details of the computation for the three given intervals are given b the table on page 36. To find the average rate of change of the function for any interval, we start with any pair of corresponding values x and y, and let Ay be the change in y produced by a change of Ax in a; . Then x + Ax and y + Ay are cor- 36 ELEMENTARY FUNCTIONS responding values of the independent variable and the function, anc hence these values satisfy the given equation. Therefore y + Ay = l{x + Ax)^, or y + Ay = lx^ + ^x Ax + ^Ax% But y = Ix^. Subtracting, Ay = ^x Ax + ^Ax^. This gives the change in y due to a change of Ax in X. Dividing by Ax, the requirec average rate of change for any interval Ax is Ax 2""^^^' ^^' The average rate for any particular interval may be obtained from thu result by substitution. For example: For the first given interval, from 1 to 3, x starts with the value x = 1 and increases by Ax = 2. Substituting these values in (1), we get Ay 1.1 Ax x 1 3 2 4 2 6 y 1 21 1 4 1 9 Ay Ay/Ax 2 2 1 2 3 Ih 4 8 2 Ax 2^1 + 4 X2 = l. For the interval from 2 to 4, x Ay 1 2, and Ax = 2. Hence Ax 2^2 + ^x2 = 1-. 2, and Ax =» 4, so that ^x2 + |x4 = 2. desirable And for the interval from 2 to 6, x Ay Ax As these results agree with those obtained above, we have check on the correctness of the entire procedure. The graphical representation of the average rate of change is important, The corresponding pairs of values (x, y) and (x + Ax, y + Ay) are repre- sented by two points P and Q on the graph, so that X = OM, y = MP, x + Ax = ON, y + Ay ^ NQ. Hence Ax = MN = PR, and Ay = RQ. Then Ay /Ax is represented by the ratio RQ/PR. That is, the average rate of change of a func- tion is represented graphically by the ratio of the difference of the ordinates of two points on the graph to the difference of their abscissas. This interpretation holds for any two points P and Q on the curve. For no matter what the relative positions of the points may be, we have, by the definition on page 13, OM + MN ' ON and X ^^ -1 1 2 3 4 5 6 V <. 1 y 1 I \ f 1 1 1 / y / p / R N y 1, -I -J J 1 ii I ' ' Fio. NR + RQ^ NQ. 23. FUNCTIONS, EQUATIONS, AND GRAPHS 37 EXERCISES 1. Find the average rate of change of each of the functions below if X changes from to 1; from 1 to 3; from 1 to 4. What can be said of these average rates? Find the average rate for any interval, draw the graph, and interpret the average rate graphically (a) 2x-3. (b) 2x. (c) -x +a. (d) \x + 2. (e) - ix + 3. (f) - 2a: + 5. 2. Find the average rate of change of each of the following functions for the intervals from 1 to 3; from 1 to - 1; from to 3; for any interval. Check the results. Plot the graph and interpret the average rates. (a) - x2 + 16. (b) x2 - 2x. (c) x2 - 2a; + 1. (d) a;2 + 2. (e) x^ - 5x + 6. (f) a:^. * 3. For each of the following functions, discuss the table of values, plot the graph, and determine the variation, including the average rate of change for any interval. (a) 4 - x2. (b) x2 - 4a;. (c) 4a; - x^. (d) x2 - 4a; + 4. (e) x^ - 3a; + 2. (f) 7? - 3x2. 4. Discuss the function represented by the following temperature graph. Find the greatest and the least average rate for the three-hour intervals indicated on the time axis. so" ^ V / \ -: \ N I \ \ r \ i^ -/ \ ». ±^.^± s 6 PM. P.M. tl2M. 8 A.M. 6 A.M. 9 A.M. 12 Noon 3 P.M. Fig. 24. 6. The daily variation of temperature for a person in normal condition 18 given in the following table. Plot the graph and discuss the function. For what hour of the day is the average rate the greatest? The least? Hour 6 A.M. Temperature 97''.6 2 99° 7 98M 3 99M 8 98°.4 4 99°.2 9 10 11 98°.7 99° 98°.9 5 6 7 99°.3 99°.2 99° 12 m. 1p.m. 98°.8 98°.9 8 9 98°.8 98°.7 10 98°.4 11 98°.2 12 98° 1 A.M. 2 97°.6 97°.6. The temperature from 2 a.m. to 6 a.m. is stationary. fe 38 ELEMENTARY FUNCTIONS 14. Classification of Functions. In discussing functions it is convenient to separate them into classes according to the properties they possess or the character of the operations that are involved in calculating them. The following scheme indicates the important divisions and subdivisions of the functions which we shall study in this course. Linear 1. Algebraic Functions Rational Irrational Integral Quadratic Cubic Biquadratic Polynomial Fractional Elementary Exponential Logarithmic 2. Transcen- "^ Trigonometric dental I Inverse Trigonometric Functions Other transcendental functions studied in higher mathe- ^ matics. An algebraic function * is a function whose value may be computed when that of x is given by the application, a finite number of times, of the operations of algebra, namely, addi- tion, subtraction, multiplication, division, involution, and evolution. The following are examples of algebraic functions: 2x + 3, ax'^ + bx + c, Vx^ - 9, ?^^- X — fj In the work of this course only real num bers will be used. Hence, for us, such an algebraic function as Vl6 — x^ is defined only for values of x such that -4 = a; = + 4 (read ** x is greater * This definition is sufficiently general for the purposes of this course. The definition used in higher mathematics, which is more inclusive is as follows: Given a polynomial in y, fiy) = (W + flilT"* + fW^ + . . . + an-\y + an, whose coefficients oo, ai, 02, • • • On-i, fln are polynomials in x, tnen any solution of the equation f{y) = for 2/ in terms of a; is called an algebraic function. . ffi FUNCTIONS, EQUATIONS/AND GRAPHS 39 than or equal to - 4, and less than or equal to + 4 ")> since for all other values, as a; = 5, the value of the function is im- aginary. An integral rational function (integral function or polynomial) is a function whose value, for a given value of x, may be found by the operations of addition, subtraction, and multipUcation applied a finite number of times. The following is the general form: 2/ = ax" + 6x"-i -\- . . . + kx -\- 1 Polynomials are classified according to the degree of the highest power of x occurring. Linear function ax + b. Quadratic function ax^ -\-bx + c. Cubic function ox^ + bx"^ -\- ex -{■ d. Biquadratic function ax* + 6x^ 4- cx^ -\- dx + e. A rational function is a function whose value, for a given value of x, may be found by the four rational operations of addition, subtraction, multiplication, and division, appUed a finite num- ber of times. If division involving x in the divisor occurs in the computation, the function may be expressed as the quotient of two polynomials, and it is then called a fractional function. The general form is '^ ^ ba^ + 6ix"»-i + . . . + bm-ix + bm An irrational function is a function which involves tne extrac- tion of roots in addition to the four rational operations. In contrast with algebraic functions, all other functions are called transcendental. This term is merely a synonym for non-algebraic. Inverse functions. If y be given as a function of x by the relation 2/ = x2 - 1, X is also a function of y obtained by solving this equation for z, namely, X = =t Vy -h 1. 40 ELEMENTARY FUNCTIONS The two functions x^ - 1 and =«= Vy + 1 are called inverse functions. The independent variable in the first is, as usual, x, but in the second it is y. In order to write the inverse function with X as the independent varia ble we must replace y by x. Hence the inverse of x^ - 1 is =*= Va; + 1. Definition. If y be set equal tof(x), the equation solved for X in terms of y, and y replaced by x in the final result, then the function of x so obtained is called the inverse oi f{x). The same result is obtained by interchanging x and y in the equation y = /(x), and then solving for y in terms of x. EXERCISES 1. What kind of a function is (a) 3x + 2? (b) x^ -4x + 1? (c) a^ -2x^+ 7? 4x -2 (d)^^? (e) Vx»-4? 2. Give a numerical example of a (a) quadratic function. (b) fractional function, (c) cubic function. (d) irrational function. 3. Find the inverse of each of the following functions; classify eacb function and its inverse. (a) x3 + 3. (b) Vx2 - 4. (c) \/^^- (d) vT+x+ VFT^. (e) y =- y/x + Vx. 4. Find the inverse functions defined by the following equations and classify them: (a) xy + 2y -X -3 = 0. (b) x^ + 2xy + 3y^ + I = 0. (c) xy" "k. (d) x^ + y^ = a'. 15. Summary. Suppose that the data of a law of a science can be recorded in a table of values of two variables, or, by some mechanical device, in the form of a curve. The generalization which expresses the law connecting the two variables is a func- I Ip FUNCTIONS, EQUATIONS, AND GRAPHS 41 tional relation (Sections 1-4). If this relation can be expressed in mathematical symbols, then mathematics becomes, to a considerable extent, the language of that science. An important problem, therefore, is the determination of an expression in mathematical symbols for a function which is given otherwise (Section 6; see also 5 and 6 below). Function — table of values — graph. These are merely dif- ferent manifestations of the same concept. A functional re- lation expresses a law connecting each pair of numbers of two sets; the table gives particular pairs of these numbers; and the graph affords a geometric representation of them. The fundamental problems arising in connection with this concept are the determinations of any two of these manifesta- tions from the third. They are six in number. 1. Given a function, to build the table of values. This is an easy process for simple algebraic functions. 2. Given a table of values, to plot the graph. This is also easily done if the table is sufficiently extensive. 3. Given a function, to draw the graph. This is done by means of 1 and 2 (Section 9). 4. Given a graph, to construct a table of values. Pairs of values may be read, approximately, directly from the graph. 5 and 6. Given a table of values, to find the function. This is precisely the problem which confronts the scientist in seeking an unknown law, and it is by far the most difficult of these problems. Methods of solution will be considered in the follow- ing chapters (see Sections 25, 44, and 90). In Sections 10-13 we have considered properties of a function, its table, and its graph, which are important in the study of any function. The correspondence between these properties of a function and its graph may be exhibited compactly as follows : PROPERTY OF GRAPH PROPERTY OF FUNCTION Length of an ordinate. A value of the function. Graph symmetrical respect y-axis. J{- x) = j{x). Graph symmetrical respect origin. f{-x) = - f(x). Vertical lines do not cut graph. Excluded values of x. Function imaginary or infinite. 42 ELEMENTARY FUNCTIONS Intercepts on the a:-axis. Intercepts on the y-axis. Vertical asymptote. Horizontal asymptote. Graph above a;-axis. - Graph below x-axis. Ordinate of maximum point. Ordinate of minimum point. Graph rises to the right. Graph falls to the right. Ratio of difference of ordinates to difference of abscissas. Zeros of function, i.e., j{x) = 0. Values of /(O). Function becomes infinite. Value of function as x becomes in- finite. Function is positive, i.e., /(x) >0. Function is negative, i.e., j{x) <0. Maximum value of function. Minimum value of function. /(x)increases as x increases. /(x) decreases as x increases. Value of Ay /Ax. Average rate of change of function. Interpretation of a graph. The properties of the graph may be determined by inspection, if only one has in mind what to look for, and the corresponding properties of the function may then be stated. It is true that the properties of the graph are proved by first establishing the properties of the function. But after the graph is drawn, this interpretation of the graph affords a comprehensive point of view of many properties of the function and furnishes a simple means of stating any par- ticular property. As we come to study any one class of functions defined in Section 14, we shall take up the properties listed above, and in addition the characteristic properties which distinguish that class of functions from others. A typical graph for each class of functions should he fixed in mind, as it enables one to tie to- gether and recall quickly the characteristics of a function as soon as it is classified. This is of great importance in analyz- ing problems. From time to time we shall add other properties of graphs and functions to the list above. We shall also take up the relations between certain pairs of functions and their graphs, which enable us to obtain the graph of one function from that of the other. Thus the graph of f{x) + k may be obtained from that of f(x) (see Exercise 3, page 19). These two sets of relations constitute the framework of the entire course. FUNCTIONS, EQUATIONS, AND GRAPHS 43 We turn next to the detailed study of particular classes of functions, beginning with the linear function, which occurs frequently in the applications of mathematics. MISCELLANEOUS EXERCISES 1. State as many properties as possible of the functions whose graphs are given below. * 1 y. \ (a) I ^ ^^»^:^ -* ^7 { V ^ \ ^ \ T 5^ -i ± it 'Ml -X ' t - V^^V ' ^ \^y « Ky ^ V- 1 / t '^(c) t t t t _^ 1"!^ 7 1 _r 1 / I V (d) / S •"^ ^^ s / X __^ T ' T 4^ X /i /I (.) ^^ .^fc -^XoZX ^ i / ' i /^' -U ^ ^ ^ X it Jl IL " ^ (c) r 7 f _.Z . 2c y ■ <« / \ r ^ V y . \ / X V' (f) / \ ^ -_ 1 X u (i) —1 ^ ^ ' / / *x / Fig. 25. 2. Discuss the table of values, plot the graph, and determine the vari- ation, including the average rate of change, of: (a) |x - 3. (b) x2 - 42/ - 3 = 0. (c) x^ - 16x + 3y = 0. 3. Discuss the table of values, plot the graph, and determine the varia- tion, omitting the average rate of change, of: (a) 2/2 + 2x - 9 = 0. (h) x^ + y^ -4x = 12. (c) x^ + ^y^ - ix = 16. (d) x2 - 2x2/ + 5 = 0. (e) x" + 4c2/ + 2 - 0. (f ) 3x2/ + 4x - 22/ - 0. I 44 ELEMENTARY FUNCTIONS (g) y^ = x«. a) 2/^(4 -x) (h) x^ -4y -1 x2(4 + x). (k) x3 + &r - 16 ■ 0. (i) x^y - 42/ - X = 0. 0. (1) xy^ + x -4y = 0. 4. For a given abscissa, the ordinate of a point on the graph of y =f(x) + g{x) is the sum of the ordinates of the points with the same abscissa on the graphs of /{x) and g{x). Hence the graph of y may be obtained by drawing the graphs of f{x) and g{x) and then adding the ordi- nates of points with the same abscissa. Using this method, which is called the addition of ordinates, construct the graphs of the following functions. (a) y = x^ + ^x. Solution. Draw the graphs of x2 and of \x. If MP and MQ are the ordinates of points on these graphs with the same abscissa OM^ then MR = MP + MQ = MP+ PR is the ordinate of a point on the graph of the given function. Several points on the graph may be obtained in this way, and a smooth curve drawn through them. Notice that if OM is negative, so also is MQ, so that MR we have PP4 ^ QP2 .PzR PiQ' and hence The ratio of the difference of the ordinates of any two points on a straight line to the difference of the abscissas, Ay /Ax, is the same no matter what two points on the line he chosen. This fact has been illustrated in the preceding section. Conversely, a line is a straight line if the ratio of Ay /Ax is always the same for any two points on the line. Let Pi and P2 be two definite points and P3 any third point on the given line which is to be proved straight. That the given line is straight will be proved by showing that P3 hes on the straight hne determined by Pi and P2. Draw the hne through Pi parallel to the x-axis cutting the ordinates of P2 and P3 at Q and R respectively. Since, by hypothesis, the values of Ay /Ax computed for Pi and P2 and for Pi and P3 are equal, it follows that QP2 _ PP3 PiQ PiR' y' ^^^"^3- -"t^i'i 1 1 1 1 ^ Ml 4 Ms » 50 ELEMENTARY FUNCTIONS Hence the triangles P1QP2 and PiRPz are similar (why?), and therefore Z QP1P2 = Z RP1P3. Then P1P3 coincides with P1P2, and hence P3 hes on the straight hne P1P2. Definition. The con- stant value of the ratio of the difference of the ordi- nates of any two points on a straight Une to the differ- ence of the abscissas is called Fig 30. the slope of the line. If the slope is denoted by m, this definition may be expressed in the symbolic form Ay For any graph, the ratio of the difference of the ordinates to the difference of the abscissas, Ay /Ax, represents the average rate of change of a function. Hence the facts proved above may be stated as the Theorem. If y is a function of x, the graph of y is a straight line if, and only if, the average rate of change of y with respect to X is constant. To plot the graph of a function we usually express the func- tional relation in the form of an equation, build the table of values, and plot the curve. This process is unnecessary if it Ls known that the average rate of change of the function is constant, for the theorem just proved shows that the graph is a straight line. To draw the graph it is sufficient to obtain two pairs of values, plot the points representing them, and draw the straight line through these points. Example. Commercial alcohol, 95% pure, is poured into a bottle. Construct a graph showing the amount of alcohol in the bottle as a func- tion of the amount of liquid in the bottle. Interpret the slope. Plot the amount of liquid, I, on the horizontal axis, and the amount of alcohol, a, on the vertical axis. If Ao is the amount of alcohol in Ai units of the liquid, we have An = 0.95Ai, since 95 % of the liquid is alcohol. Hence Ao/Ai = 0.95, so that the rate of change of a with respect to I is LINEAR FUNCTIONS 51 constant. The graph is therefore a straight line. If there is no liquid in the bottle, there is no alcohol, and hence the origin (0,0) is on the line; and if the bottle contains 10 units of liquid, there are 9.5 units of alcohol in the bottle. Hence (10, 9.5) is a second point on the line, and these two points are sufficient to determine the graph. The slope represents the rate of change of a with respect to I, Aa/Al = 0.95, which gives the percentage of alcohol in the liquid. In plotting the graph by using the point (10, 9.5), it is assumed that the unit of volume is some such unit as an ounce or cubic centimeter. If a bottleful were chosen as the unit, which is convenient in many problems, it would be better to use the point (1, 0.95) and choose the unit on the Z-axis as large as convenient. For then I = 1 would mean that the bottle was full. 18. Slope of a Straight Line. " Slope of a line " (defini- tion, Section 17) is the term used technically by mathematicians for what might be called the measure of steepness of the hne. It may represent many things. Thus in the example in Section 2/f r ~^ (10,^^.5), / c / / / y^ / —5 A / —3 / / / / i , O _ ., 1 _ — \ t'?'! Fig. 31. y F ^ ^y^ Ay Pr ^ ^ Q ^ Ax ^ fj ik U ^ Fig. 32. 16, As/Ai is the slope of the Hne, and hence the velocity v is represented by the slope. If the slope m is computed from two points Pi (xi, yi) and ^2(0:2, 2/2) on the hne, we have, in either figure, m = ^ - ^^^ - ^^^^ ~ ^^^^ - y^-y^ _ yi -^2 Ax PiQ OM2 - OMi X2 - xi xi - X2 In finding Ay and Ax it is essential that both coordinates of Pi be sub- tracted from those of P2, or vice versa. 52 ELEMENTARY FUNCTIONS Theorem 1. A line runs up to the right or dovm to the right according as its slope is positive or negative. For two points Pi and P2 may be chosen on the hne so that Pi lies to the left of P2, whence Aa; is positive. Then Ay will be positive or negative according as P2 is above or below Pi. Hence m = Ay /Ax will be positive or negative according as P2 is above or below Pi, that is, according as the hne runs up to the right or down to the right. The precise relation between the value of m and the direc- tion of the hne as determined by the angle the hne makes with the X-axis involves a transcendental function, and will be con- sidered in a later chapter (see also Exercises 3 and 4 below). Theorem 2. If two lines are parallel their slopes are equal, and conversely. The triangles P1QP2 and Pi'Q'P^ are similar, since their sides are parallel by pairs. Hence QP2_qPl PiQ Pi'Q" that is, the slopes of the two Hnes are equal. The proof of the converse is left as an exercise. Construction. To construct a line through a given point Pi whose slope is a given positive fraction a/b, take Q b units to the right of Pi, and P2 a units above Q; then P1P2 is the required line. Yov we have eP2 a Fig. 33. m = PiQ'h If the slope is negative, take P2 below Q. If the slope is an integer, it may be regarded as a fraction with unit denominator. If two points are close together, the hne through them can- not be drawn as accurately as if they were farther apart. Hence, in this construction, it is sometimes desirable to take PiQ equal to some multiple of b and QP2 equal to the same multiple of a. LINEAR FUNCTIONS 53 EXERCISES 1. Plot the following pairs of points and the lines determined by them. Find Ax, Ay, and the slope m, and indicate the graphical significance of each of these quantities. (a) (1, 2) and (4, 5). (b) (1, 2) and (4, 1). (c) (1, 2) and (- 2, 0). (d) (1, 2) and (3, 4). (e) (3, - 2) and (5, 2). (f) (2, 5) and ( 4, 2). 2. Construct the lines through the points indicated with the given slope. (a) P(2, 3), m == i (b) P(3, - 1), m = - ^ (c) P(0, 4), m = - |. (d) P(3, 0), m = 2. (e) P(- 2, 4), m = f . (f) P(5, 2), m = 0. 3. Show that of two lines through the same point the one which makes the greater acute angle with the a;-axis has the larger slope numerically. 4. Show that the slope of a line parallel to the x-axis is zero. 6. If a telegram containing 10 words or less can be sent to a certain place for S5^, and a 24-word message for 63ff, construct a graph showing the cost of a message containing any number of words, and determine the charge for each additional word above 10. What represents this charge? 6. Construct a graph to show the cost of any number of eggs at 40^ a dozen. Determine from the graph the number of eggs which can be purchased for 70)!f. What does the slope represent? 7. Construct a graph to show the amount of silver in any amount of an alloy containing 25% of silver. Determine from the graph how much silver there is in 20 pounds of the alloy. What does the slope represent? 8. A pound of an alloy contains 3 parts of silver and 5 of copper. Con- : struct a graph to show the amount of silver in any amount of the alloy. Determine from the graph how much of the alloy contains 4 1 pounds of i silver, and how much silver there is in 10 pounds of the alloy. Interpret j the slope. j 9. Ten ounces of alcohol 95 % pure are poured into a bottle and then I 5 ounces of water are added. Construct a graph showing the amount of I alcohol in the bottle during the process as a function of the amount of I liquid. I 10. Solve Exercise 9 if 5 ounces of 50 % alcohol are added instead of 5 I ounces of water. I 11. Is the slope of the line joining (3, 6) to (6, 1^) the same as the slope of [the line joining (3, 6) to ( - 3, -6)? What can be said of the three I points? 12. Are the points (6, 1), (- 2, - 3), and (0, - 2) on a straight line? 13. Show that the points (2, 1), (3, 7), (5, 3), and (0, 5) are the vertices of a parallelogram. 14. Find the value of b if the line determined by (2, h) and (- 1, 3) ia parallel to the line joining (-4, - 5) and (7, - 2). 54 ELEMENTARY FUNCTIONS 15. Two boys start from the same point and run in the same direction, one at the rate of 15 feet per second, the other at the rate of 20 feet per second. Construct the graphs showing the distances from the starting point to each boy at any time. What do the slopes represent? De- termine from the graphs how far apart the boys are after 3 seconds. Solve the same problem if the boys run in opposite directions. 16. Water is admitted into a tank through two pipes at the rates of 3 and 5 gallons per second. On the same axes, plot graphs showing the amount which has entered through each pipe at any time, and also the total amount. What do the slopes represent? 17. If the cost of setting type for a circular is 50 cents, and if the cost of paper and printing is half a cent a copy, construct a graph showing the cost of any number of copies. From it determine the cost of 500 copies. What does the slope represent? Hint: The cost of setting the t3rpe may be regarded as the cost of zero copies. 18. If $100 is deposited in a bank to draw simple interest at 6 %, con- struct a graph to show the amount at any time. What does the slope represent? The intercept on the " amount axis " ? Determine graphically how long it would take the principal to double itself. What is the relation between the amount and the time? 19. One of the horizontal boards in a flight of stairs is called a tread, and one of the vertical boards a riser. How can the steepness of the stairs be expressed in terms of the widths of the treads and risers? j 20. Show that the average rate of change of a function other than aj linear function is represented by the slope of a secant Une of the graph, j 19. Graphical Solution of Problems Involving Functions which Change Uniformly. Graphical methods are used freely by engineers and others because of their relative simplicity. A graphical solution of a problem is also often used as a simple means of checking the accuracy of some other solution. A method of ob- taining approximate solutions of many problems by means of graphs is illustrated in the foUo^j ing problems: Example 1. A druggist has a 50' solution of a certain disinfectant an^ also a 10 % solution. How much of the former must be added to 3 pint of the latter to obtain a 20 % solution? d 1 /" ';.. L V 7 /_ A\ / /I i / y ^ / / / g / _^ / eL^^ "^ / y ^ ^' / ^- ^ D H 1 3 — ' T Fig. 34. LINEAR FUNCTIONS 55 The graph representing the amount of disinfectant, d, in a quantity, I, of the 50 % solution is the straight line OA through the origin whose slope is 0.5. Similarly, the graphs for 10% and 20% solutions are re- spectively OB and OC whose slopes are 0.1 and 0.2. On the Z-axis take OD = 3, and draw the ordinate DE to the line OB. Then DE represents the amount of disinfectant in 3 pints of the 10% solution. Through E draw EF parallel to OA. Then the ordinate of a point on EF represents the amount of disinfectant in the mixture as some of the 50 % solution is added to the 3 pints of the 10 % solution. If DE cuts OC at G, and if GH be drawn perpendicular to the Z-axis, then in OH pints of the combined mixture there are HG pints of disinfec- tant; and since G lies on OC, the strength of the combined mixture is 20 %. Hence we must add DH, or 1 pint approximately, of the 50 % solution. Example 2. A freight train goes from ^4 to £ at the rate of 15 miles per hour. Four hours after it starts, an express train leaves A and moves at the rate of 45 miles per hour. The express arrives at B half an hour ahead of the freight. How far is it from A to B and how long does it take each train to go? Let abscissas denote time in hours measured from the time of departure of the freight train, and ordinates distances from A. The graph representing the distance of the freight at any time is the straight line OC through the origin with the slope 15, and that of the express is the line DE through the point D(4, 0) with the slope 45. At B the express is half an hour ahead of the freight, and hence we must determine a point on OC half a unit to the right of DE. To do this, take F on the <-axis so that DF = |. Draw the line through F parallel to DE, and let it meet OC at G. If the line through G parallel to the Z-axis cuts DE at H, then HG = |. Hence, the required distance AB is represented by the ordinates IH = JG = 100 miles approximately. The time of the freight is represented by OJ = 6.7 hours, approximately, and that of the express by DI = 2.2 hours, approximately. The line FG may be interpreted as the graph of a train leaving half an hour later than the express and moving at the same rate of 45 miles an hour. By the conditions of the problem, such a train would overtake the freight at B. ^^ Wf¥^ Z W -^0 -^d. L y^Li -60 X / / -.^ tt~~~ -^ tt -«. ^/ ri "" - ^ Tl "" ^ .£Lji 1 1 " ^^±_l-J_±Zl Fig. 35. 56 ELEMENTARY FUNCTIONS EXERCISES Solve each of the following exercises graphically. Use a sharp, hard pencil, and choose the units on each axis so that the figure will be as large as convenient. 1. Two trains start from the same place at the same time and run in the same direction at the rates of 30 and 50 miles per hour respectively. How far apart will they be at the end of two hours? When will they be 100 miles apart? 2. Solve Exercise 2 if the trains run in opposite directions. 3. For printing business cards one firm charges 75^5 for setting the type, and a apiece for the cards and printing. Another firm makes no charge for typesetting, but charges f ^ apiece. Which firm will print more cheaply 100 cards? 500 cards? Which will do the more printing for $1.25? For $2.00? How many cards will both print at the same cost, and what will the cost be? 4. A man rides horseback at the rate of 8 miles an hour for an hour and a half, and then walks back at the rate of 4 miles an hour. Another man starts from the same place at the same time and walks in the same direction at the rate of 3 miles an hour. When and where will they meet? 5. How much water must be added to 10 gallons of alcohol, 95 % pure, to reduce it to a 66| % solution? 6. A farmer sells three and a half dozen eggs to a grocer at 25}* a dozen, and takes his pay in rice at 8ff a poimd. How many pounds of rice will he receive? 7. At what time are the hands of a clock together between three and four o'clock? When are they opposite? Hint: The position of either hand may be determined by the number of minute spaces it has passed over since leaving "12." Plot this number as a function of the number of minutes after three o'clock. 8. Solve Exercise 7 for eight and nine o'clock. 9. The earth revolves around the sun in 52 weeks, and the planet Mars in 98 weeks. How often is the earth between Mars and the sun? 20. Graph of the Linear Function mx + b. Let the function be denoted by 2/, so that y = mx + b, (1) ' Let us first determine the average rate of change of y with respect to x. Let the independent variable change from an^ value re to X + Arc, and suppose that the dependent variable LINEAR FUNCTIONS 57 changes from y to y + Ay. Then the corresponding values X + Ax and y + Ay satisfy (1), so that y + Ay = m{x + Ax), (2) Subtracting (1) from (2), Ay = m Ax, and dividing by Ax, Ay /Ax = m. Hence the average rate of change of y with respect to x is constant, and the graph of (1) is therefore a straight Une : (Theorem, Section 17). I If the graph is a straight Hne the value of Ay /Ax is the slope I of the Une (Definition, Section 17), and hence the slope of the I graph of (1) is m. The constant term b is the intercept on the ?/-axis, for if a; = 0, then from (1) y = b. The hne is thus determined by the point (0, 6) and the slope m (Construction, Section 18). We may therefore express the fundamental theorem of this chapter as follows: Theorem 1. The graph of the linear function mx -{• b, or of the equation y = mx + b, is the straight line whose slope is m, i and whose intercept on the y-axis is b. If 6 = 0, the hne passes through the origin. Hence the important Corollary 1. The graph of the function mx, or of the equation y = mx, is the straight line through the origin whose slope is m. Corollary 2. The graphs of the equxitions y = mx + b and y = mx + b' are parallel lines. An equation of the first degree in x and y is called linear, and may be reduced to the form Ax -h By + C = 0. (3) Theorem 2. The graph of any linear equation (3) is a straight lin£. 58 ELEMENTARY FUNCTIONS If the equation contains y, so that B 9^ (read " B is not equal to zero")> we may divide by B and solve for y, obtaining the linear function ^ q whose graph is the straight Hne whose slope is m = — A/B, and whose intercept on the y-axis is 6 = - C/B (Theorem 1). li B = 0, equation (3) has the form Ax + C = 0, whence, if ^ ^ ^^ a; = - CM. The graph of this equation is a straight line parallel to the ?/-axis. For all points whose abscissas have the same value, -C/A, no matter what the ordinates may be, are equidistant from the ^/-axis. Hence the graph of (3) is always a straight line. Corollary 1. The graph of an equation of the form x ^ a con- stant is a straight line parallel to the y-axis, and that of y = a constant is a line parallel to the x-axis. The proof of the first statement appears explicitly in the preceding proof. Is that of the second included in the proof | (see Exercise 4, Section 18)? | Corollary 2. // a linear equation be solved for y, then the] coefficient of x is the slope of the graph of the equation, and the constant term is the intercept on the y-axis. In constructing the graph of a linear equation it is necessary to plot hut two points. The intercepts (Section 10) usually fur- nish the two most convenient points. The line may be drawn j more accurately by choosing points some distance apart thanj by taking them close together. The graph of y = mx may he drawn by plotting but one point and joining it to the origin. EXERCISES 1. Construct the graphs of the following equations: (a) 2/ = 2x + 5. (h) y=-2x + 5. (c) yx + Z. id) y= -x + 7. (e) y = 3a;. (f) ?/ = 2a; - 4. (g) a; - 2j/ + 6 - 0. (h) 2a; + Sy - 12 = 0. (i) 2x + 3y + 12 = G) 1/ = |ar. (k) y = ix + 2. (1) z/ = - ^a; + 5. LINEAR FUNCTIONS 59 2. In experiments on the temperature at various depths in a mine, the temperature t (in degrees Centigrade) was found to be connected with the depth d (in feet) by the equation ^ = 20 + 0.01 d. Construct the graph. What is represented by the ordinate and abscissa of a point on the graph? by the intercept on the f-axis? by the slope? 3. In experiments on a pulley block the pull P, in pounds, required to lift a weight of W pounds was found to be P = 0.3 W + 0.5. Construct the graph. What does the slope represent? The intercept on the P-axis? What is the value of the rate of increase of P with respect to Wl 4. The readings, 2\ of a gas meter being tested, were found in com- Darison with those of a standard meter, S, and the two sets of readings satisfied the equation T = 300 + 1.2 S. Draw the graph. What do the ibscissa and ordinate of a point on it represent? The slope? The in- ■ercept on the 7'-axi3? Has the consumer whose meter is being tested )een getting more or less gas than the meter indicated? I 5. Plot the graph of the equation s = -3^ + 12. Describe the motion I or the first 6 hours if s represents the distance of a man from town, in i niles, at the time t, in hours. Interpret the slope and the intercepts on i )oth axes. i 6. The ordinate of a point on one of the graphs in the figure represents s he distance s (in feet) of a moving body from a certain station at the time i (in seconds) repre- jented by the abscissa. ! )escribe the motion epresented by each of jhe graphs (a), (6), (c), I d), from t = -Q to ; -12. I Solution for (a). From he graph, s = 7 when « - 6. Hence the body Uarts at a point 7 feet i*om the station on the I ositive side of it. Ast I icreases, s decreases, j nee the graph is falling 3 it runs to the right jhanges of the function). Hence the body is approaching the station. Then t = 0, the distance from the station is s = 4 (interpretation of the itercept on the s-axis). The body passes through the station when « 8, for then s = (interpretation of the intercept on the i-axis). Four jconds later, that is, when t = 12, the body has reached a point 2 feet syond the station in the negative direction; for when t = 12, s = - 2. The rate of change of s with respect to t, namely, the velocity, is con- 8^ V ^ c^ ^ ^ "V K ^ ^ U) X ^ 2 < *^ ^ ^-' ^ N .^ ^ ' > ^ <^ t*.. (. k. K ^. ^ -> .-^ "^ / s 'i -i "•^ X ■^^ \ . (d) ^ ^ -^ ^ ^ (c) _ _J Fig. 36. 60 ELEMENTARY FUNCTIONS Btant since the graph is a straight line, points (0, 4) and (8, 0), we get _ 4f 0-4 "~ At~ S-0' Computing its value from tl Solving the equation for TF, we get TT = ^ D + t- The negative sign indicates that the total change in s in any interval of time A^ is negative, and hence the body is moving in the negative direction at the uniform rate of half a foot per second. 7. In the following equations, s denotes the number of feet, measured along the road, at the time t (in seconds), from a tree to a steam roller. Construct the graph of each equation, and from it describe the motion of the steam roller from f = - 5, to i = 10. (a) s = 2t + S. {h) s= -^t + S. (c) s = - 2< - 4. 8. Construct and interpret the graph of each of the equations below, in which W denotes the part of a piece of work that a man can do in D days. (a) 20W - D - 5 = 0. ^ " Hence the graph is the straight line whose slope is ^V and whose intercept on the TF-axis is \. Since W = 1 indicates that the piece of work has been completed, the graph need be plotted only from W = OtoW = 1. The intercept on the TF-axis indicates that one fourth of the job has been completed when we begin to consider the problem, and the slope represente the rate at which the man works, namely, ^V o^ the piece of work per day. When TF = 1, the graph shows that D = 15 (which may be verified from the equation), that is, it will take the man 15 days tc complete the work. If we care to interpret negative values of D, the graph shows that wher TF = 0, D = - 5. Hence if we regard the whole piece of work as performec by the man in question, he started 5 days ago. (b) 12TF - D - 6 = 0. (c) lOTF - 3D = 0. (d) lOTF - 2D + 4 = 0. (e) 30TF - 5D + 10 = 0. 9. Mention some function of the form mx arising in daily Ufe, and die cuse it graphically. Fig. 37. fl LINEAR FUNCTIONS 61 10. Mention some function of the form mx + b arising in daily life, and discuss it graphically. 11. Plot the graphs of the equations below, using a single set of co- ordinate axes for each case. Each figure will contain several lines cor- responding to the values of the constants indicated. (a) y = mx for w = |, |, 1, 2, 4. (h) y = 2x + b for 5 = - 3, - 2, 0, 2, 5. (c) y = mx + 2 for m = |, 1, 3, - 3, - 1, - J. 12. Exhibit in tabular form as many as possible of the corresponding properties of a straight line and the function fix) = mx + b oi which the line is the graph (see Section 15). Indicate special properties of this function which distinguish it from all other functions. 21. Variation. We have already had a number of illustra- tions of the important relation y = mx, in which one variable is a constant times the other. For example, the price paid for eggs is the price per dozen times the number of dozen bought; if a man walks at a uniform rate, the distance is the rate times the time; etc. Two technical terms are used in this connection. Definition. A variable is said to vary as, or to be propor- tional to, a second variable if the first is equal to a constant times the second. SymboHcally, y varies as x, or y is propor- tional to X, ii y = mx, m being a constant. This use of the word proportional is justified by the fact that • two values of y are proportional to the corresponding values of ) X. For if t/i = mxi, and y^ = mx^ then by dividing the first equation by the second, we obtain th^ proportion ^ = — . I 2/2 X2 j The phrase " variation of a function " used in Section 12, I should not be confused with " variation " as a heading under i which we discuss proportional variables. A more general use j of the latter term will be taken up in Chapter III. I The mass of a body is the amount of matter in it. If the ! nature of the material in the body is the same throughout, the I mass varies as the volume. That is, if M denotes the mass and V the volume, M = mV, where m is a constant called the density of the substance of which the body is composed. The value of m is the mass of a unit volume. 62 ELEMENTARY FUNCTIONS Example. If the mass of an aluminum object is 13, and its volume is 5, find the mass of an aluminum object whose volume is 8.6, and the volume of an object whose mass is 17.4. Let M denote the mass and V the volume of any aluminum object. Then since M varies as V, M = mV. (1) Graphical solution. The graph is a straight line through the origin (Corollary 1 to Theorem 1, Sec- tion 20). It also passes through the point A (5, 13), since M = 13 when V = 5. It is therefore the line OA in the figure. From the graph, if 7= 8.6, the value of M is CD = 22.4; and if M = 17.4, the value of F is OG = 7. These are the required values. The slope of the line, computed from and A, is M' ■ -22 D/ ^ ■■■ r/ E — — ■'fl J -L 13 -^ (i^ r -10 '\ ! / / 1 ^ ?! 1 1 ClfV Fig. 38. m = AM BA 13 AV OB ^ = ^ = 2.6. Since -\^ is the mass of a unit volume of aluminum, or the density, the slope represents the density. Algebraic solution. Substituting the given values M = 13 and V = 5 in (1), we obtain 13 = 5m, whence the density m = ^^- = 2.6. Substituting the value of m in (1), we have the relation between the mass and volume of any aluminum object, (2) Then if F = 8.6, And if M = 17.4, whence M = 2.6F. Af = 2.6x8.6 = 22.36. 17.4 = 2.6F, V = 17.4/2.6 = 6.69. The graph has the advantage of exhibiting the relation to the eye, and if it is constructed carefully, it is sufficiently accurate for the purpose of a " ready reckoner/' It is very useful when a number of values of either variable are desired and great ac- curacy is not essential, and in any case it furnishes a valuable check on the accuracy of the computation. When we are given the fact that a variable y is proportional to a second variable x, the general form of the law connecting X and 2/ is 2/ = mx. ' The exact form of the law involves a par- ticular value of the constant m. This value of m may be found from a pair of corresponding values of x and y, other than (0, 0), as in the algebraic solution of the example above LINEAR FUNCTIONS 63 22. Uniform Acceleration. The motion of a body is said to be accelerated if the velocity is variable. It is said to be uni- formly accelerated if the change in velocity is proportional to the change in time. The rate of change of velocity is called acceleration, and it is measured by the number of feet per second by which the velocity changes in each second. A freely falhng body moves with a uniformly accelerated motion, if the resistance of the air is neglected. The accelera- tion, which is called the acceleration due to gravity and which is denoted by g, is approximately 32 feet per second per second. If the positive direction is downward, g = 32, but if the positive direction is upward, ^ = — 32. Since the rate of change of velocity of a uniformly accelerated body is constant, v is a linear function of t, and the graph is a straight line. The slope of the hne represents the acceleration, and the intercept on the !;-axis represents the velocity when t = 0, which is called the initial velocity. EXERCISES 1. Hooke's law says that the amount of stretching in a stretched elastic string is proportional to the tension. If a 2-lb. weight will stretch a string 3 feet, what tension arises when the string is stretched 1 foot? Draw the graph. 2. Prove graphically, that if y varies as x, then any two values of y are proportional to the corresponding values of x. 3. Is the definition near the end of Section 16 equivalent to the state- ment following? A variable y changes uniformly with respect to x if the change in y is proportional to the change in x producing it. 4. The mass of a body varies as its volume. On the same axes con- struct the graphs showing the relation between mass and volume for bodies composed of the following substances whose densities are given: Lithium, 0.6; alcohol, 0.8; India rubber, 1.0; magnesium, 1.7; diamond, 3.5; silver, 10.5. How are the various densities represented? 5. Wihelmy's law for chemical reactions states that the amount of chemical change in a given time is proportional to the quantity of re- acting substance in the system. Construct the graph. 6. The velocity acquired in t seconds by a body falling freely from rest is given by the equation v = d2t. Plot the graph and from it determine how fast the body would be falling after 4 seconds. What does the slope represent? 64 ELEMENTARY FUNCTIONS 7. If a ball is dropped from a high building, how fast will it be moving at the end of one second? at the end of 2 seconds? at the end of 4 seconds? If a ball is thrown vertically upward, how will its velocity be affected in any second? If it is thrown up with an initial velocity of 96 feet per second, how long will it rise? 8. A train starting from rest acquires a velocity of 50 feet per second in 15 seconds. Find the average acceleration. 9. An automobile is moving at the rate of 40 feet per second, when the power is shut ofif and the brakes applied. If it moves thereafter with a uniform acceleration of - 5 feet per second per second, how long before it will stop? 10. Name two quantities arising in every-day experiences which are proportional, and give the value of the constant involved in the relation between them. 11. The pressure of a liquid is proportional to the depth. If the pres- sure per square inch on the suit of a diver is 5.2 lbs. for a depth of 12 ft. how deep can he go safely if 78 lbs. per square inch is an allowable pressure? Construct the graph. 12. For small changes in altitude, atmospheric pressure varies as the altitude. If the change in the reading of a barometer is 0.1 of a unit for each 90 ft. ascent, construct the graph of this relation and determine the difference in readings of two places with a difference in altitude of 1000 ft. 13. In a spring balance the extension of the spring is proportional to the weight. If a weight of 2 lbs. lengthens the spring 1 inch, construct a graph and determine the extension of the spring for weights of 5, 8, and 17 pounds. 14. Thestockof a corporation yields an annual dividend of 5%. What is the relation between the total amount a stockholder receives in divi- dends and the time the stock is held? If the graph of this relation is drawn, what does the slope of the line represent? What is the value of the slope if the par value of the stock held is $10,000? 15. The rent charged to each department of a store is proportional to the percentage of the total floor space which the department occupies. If the store pays a rent of $12,000, construct a graph to show the rent charged to a department of any size, using as large a scale as convenient. From the graph, determine the rent charged to each department if the various departments occupy 17, 25, 10, 8, 5, 12, and 23 per cent of the total floor space. 16. An automobile moving at the rate of 22 feet per second (15 miles per hour) begins to coast down a hill. If its velocity after t seconds ia given by V = \Qt + 22, find from the graph of v how fast it would be going after 5 seconds, and how long it would take it to acquire a velocity of 60 miles per hour. What is the acceleration? LINEAR FUNCTIONS 65 17. If a ball is thrown vertically upward with a velocity of 100 feet per second, its velocity after t seconds is given by the equation v = - 32< + 100. Plot the graph, and describe the motion. What is represented by the intercept on the v-axis? the intercept on the <-axis? the slope? 18. The ordinate of a point on one of the graphs in the figure repre- sents the velocity y of a moving body at the time t represented by the abscissa. Describe the motion of a body whose velocity is repre- sented by each of the graphs, from < = - 6 to < = 12. Solution for (a). From the graph, the intercept on the i-axis is f = 6, when y = 0, so that the body is not moving at this instant. The intercept on the f-axis is y = - 2, which gives the velocity when < = 0. The line is below the <-axis if < < 6, so that during this time v is negative and the body is moving in the opposite of a certain direction which has been assumed as positive, i.e., in the negative direction. The line is above the <-axis if i > 6, so that v is positive and the body is moving in the positive direction. The slope of the line, computed from the points (6, 0) and (0, - 2) is the , ^. Ay - 2 - 1 ., acceleration m = ^ = — — — - = - ft. per sec. per sec. V. Vj V, (c) ^ V, V V 4 b) — - _^ ^ _,9 K '^ ^ — " .^ — ■ — ' 'v. "v. ^ V ^ ^ -.' _.: I i>- c ^ V, L$ ,^ ^ ^ "^ ^ ".^ ^ (d) *^ -J Fig. 39. Since the line rises to the right, the velocity constantly increases, which agrees with the fact that the acceleration is positive. From the graph, when t= - 6, y = - 4, and when i = 12, y = 2. Discarding the technical terms positive and negative, the motion may be described as follows : At the start the body is moving in a certain direc- tion at the rate of 4 feet per second. During the next 12 seconds it slows down uniformly and comes to rest for an instant, and then begins to move in the opposite direction with a constantly increasing speed. After 6 seconds it has acquired a velocity of 2 feet per second. The velocity changes by | of a foot per second in each second of the motion. Notice that where the body is at any time, and in particular where it comes to rest, cannot be determined from the graph of the velocity. 19. The velocity of an automobile, in feet per second, at the time <, 66 ELEMENTARY FUNCTIONS in seconds, is given by one of the following equations. Construct the graph, and describe the motion, from t = to t = Q. (a) V = U. (b) y = 3< - 6. (c) « = O.U - 2. (d) w = - 2i + 5. 20. Describe the motion of a ball rolling on the side of a hill if its veloc- ity is given by one of the following equations, by interpreting the graph. What is the acceleration in each case? (a) V = 10/. (b) t; = 8f + 5. (c) y = 6« - 8. (d) y = - 4i + 12. (e) y = - 6i + 12. (f) t; = 3^-12. 21. Construct the graph of the equation 2/ = - 2x + 4. From it de- scribe the motion of a body (a) if y represents the distance of a body from a certain station at the time x] (b) if y represents the velocity of a body at the time x. 23. Equation of a Straight Line. We have seen that the graph oi y = mx + ?> is the straight line whose slope is m and whose intercept on the 2/-axis is h. Conversely, if a hne is given whose slope is, say, 2 and whose intercept on the 2/-axis is, say, 3, then the equation of which it is the graph is ?/ = 2x + 3; for the graph of this equation is known to be the given hne. The equation of which a given line or curve is the graph is called the equation of ihedine or curve. To find the equation of a hne whose slope and intercept on the 2/-axis are given, substitute the given values of m and 6 in the equation y = mx + h. (1) This is called the slope-intercept form of the equation of a straight line. To find the equation of a line determined by its slope and any point on it we apply the Theorem. The equation of the line whose slope is m and which passes through the point Pi{xi, yi) is y -yi = m(x - xi). (2) Let P(x, y) be any point on the given line. Since the value of Ay /Ax for any two points on a straight hne is constant (page 49), and equal to m (definition page 50), we have LINEAR FUNCTIONS 2/ -2/1 67 and hence JSx X — Xi 2/ - 2/1 = m{x = w, xr). This is called the point-slope form of the equation of a straight Une. To ^nd the equation of a straight line determined by two points^ find the slope and then apply (2). Example 1. Find the equation of the line determined by the points A (1, 4) and B (2, 3). The slope oi AB'm 3-4 , Substituting xi = 1, t/i = 4, and m = - 1 in (2) we get 2/ - 4 = - l(x - 1), or X + 1/ - 5 = 0. Fig. 40. Check. The coordinates of A and of B satisfy this equation. Example 2. Find the equation of the hne through the point A{b, 2) which is parallel to the Hne x - 3?/ + 6 = 0. Solving the given equation for y we get K 1 " \ A (1,4^ ^ \ \ B(3,S -1 \ - \ . 1 1 '! t1 i/ ,- Fig. 44. W 10, 20, 30. 40 W 3.1, 5.8, 8.1, 11.2' the unit of weight being the gram. Determine approximately the re- lation between W and W. Fig. 45. Plot the points whose coordi- nates are the pairs of values of W and W, using values of W as abscissas. The four points obtained lie very nearly on a straight line through the origin, and hence we assume that the graph of the relation is a straight line through the origin. 80 ELEMENTARY FUNCTIONS (That the graph ought to pass through the origin is clear, for if TT = 0, so also must W = 0.) The required relation must therefore be of the form W = mW, where m is the slope of the line. The slopes of the lines joining the origin to each of the four points are respectively 3.1 -0 10-0 0.31; 5^ 20 0.29; 8J. 30 = 0-27; ^-^ = 0.28. From these values it appears that m may have any value between 0.27 and 0.31, and the line W = mW would be a fair approximation to the graph required. We shall choose as a good value of m the average of the four values of m, namely, m = ^ (0.31 + 0.29 + 0.27 + 0.28) = —■ 0.29. In dividing 1.15 by 4, the second decimal figure is 8, but the quotient Is nearer to 0.29 than to 0.28; hence the former value is chosen. Hence the relation desired is represented approximately by the equation W = 0.29Tr. (1) The accuracy with which this equation represents the given data may be determined by constructing the table adjoined. The first two columns give the observed values of W and W, the third, the values of W computed by means of equation (1) from the observed values of W] the fourth, the error in the com- puted values of W, which are obtained by subtracting the second column from the third; and the fifth, the percentage of error. The percentage of error is found by dividing the error by the observed value of W\ Thus 0.2/3.1 =0.064, or 6.4%. w 10 W 0.29Tf Error %of error 3.1 2.9 -0.2 -6.4 20 5.8 5.8 0.0 30 8.1 8.7 + 0.6 + 7.4 40 11.2 11.6 + 0.4 + 3.5 I Note the part played by mathematics in this illustration the scientific method. The given data are obtained by ohservi tion. The principles of graphic representation enable us t put the data in a form which makes reasonable the hypothesis that the graph of the law under investigation is a straight line, and that the law is represented by a linear equation. By d&- ductive processes we determine the numerical values of the coefficients of the equation, and the accuracy of the represen- tation of the given data by the equation found. The verification LINEAR FUNCTIONS 81 would consist, in part, of repeating the experiment a number of times, varying the values of Wy and seeing if the law re- mained approximately the same. But a more satisfactory verification would be to deduce from this law some other law which could be verified by an experiment of a different sort. This will be done in a later section. The law obtained in Example 1 can be stated in a more con- venient form. The force tending to make the block slide is equal to W, because the tension of the string is the same at all points. This force acts in the direction of the surfaces in contact, and when motion is about to begin, it is numerically equal to the force of friction which is preventing the block from moving. W is the pressure on the board acting perpen- dicularly to the surfaces in contact. Hence the result obtained may be stated as follows, using the language of Section 21 : The friction of wood on wood varies as the pressure perpen- dicular to the surfaces in contact. If different kinds of wood, or other substances, be used in the experiment, the value of m obtained would not be the same. But extensive experiments have shown that we are reasonably justified in stating the law: When motion is about to take place, the friction between two surfaces varies as the pressure perpendicular to the surfaces. Hence if F denotes the force of friction and P the perpen- dicular pressure F = mP. The constant m is called the coefficient of friction, and it is equal to the ratio of the friction to the pressure perpendicular to the surfaces in contact. The coefficient of friction for the block and board in Example 1 is 0.29. Example 2. In an experiment with a block and tackle, the pull P necessary to raise a weight W, both measured in pounds, was found for W HOP, 200. 300. 400, 1^"^ "^ *^ ^'"/^ "[^ *° '^' t^"^'^" -p-— ^y^ --^ -^ -— p Fmd approximately the equation ex- ' ' ' ' pressing P as a function of W. Plotting the points representing the pairs of numbers in the table, using values of W as abscissas, it is seen that they he very nearly on a 82 ELEMENTARY FUNCTIONS straight line. This line ought not to pass through the origin, because if ' W = 0, B, certain force P is necessary to raise the lower block. Hence P is approximately a linear function of W of the form P=.mW + b. A good value of m is the average of the slopes of the lines determined 1 by each pair of points. Denoting the points by A, B, C, D respectively, the slope of the line | ABis 59 - 37 m 200 - 100 = 0.22. In like manner we find the value of m for each of the lines determined by two of the points A, B, C, D. These values are given in the table following. Line | AB, AC, AD, BC, BD, CD 0.22, 0.25, 0.25, 0.28, 0.24 400 jy 0.26, The average value of m is found to be m = 0.25. Hence the func- tion required has the form, approximately, P = 0.25W + b. (2) n the graph of (2) passes through the point A, the coordi- nates of A must satisfy (2), and hence 37 = 0.25 X 100 + b, whence 6 = 12. In like manner, we find the value of b on the assumption that each of the points B, C, D, lies on the graph of (2). The results are given in the table. Point I A, B, C, D b I 12, 9, 12, 11 The average of these values of 6 is 6 = 11. Substituting this value in (2), the function required is, approximately, P = 0.25Tr + 11. The accuracy with which this equation represents the observed data is shown by the table below. The first two columns give the observed values of W and P. The third column gives the values of P computed Fig. 46. LINEAR FUNCTIONS 83 from the observed values of W by means of (3). The fourth gives the error in the computed value of P as compared with the observed value, while the last column gives the percentage of error, which is obtained from the second and fourth columns. w Observed P Computed P Error Per cent of error 100 37 36 - 1 -2.7 200 59 61 + 2 + 3.3 300 87 86 - 1 -1.1 400 111 111 property of a linear function is illustrated in the table, which gives X X y 3 1 1 5 1 3 9 1 4 11 Ay values of x, y, Ax, and Ay for the function y = 2x +S. The values of x being such that the successive values of Ax are equal, it appears that the successive values of Ay are also equal. That this is always the case follows from the fact that the value of Ay /Ax is al- ways the same for points on a straight line. Now suppose, as in the examples above, the points representing a given table of values appear to He on a straight line. To test the accuracy of this assumpn tion, find the successive values of Ax and Ay. If the values of Ax are equal, and if those of Ay are nearly equal, we are justified in believing that y is indeed a linear function of x. The labor involved in computing an average may be lessened by means of the following rule: Assume a number x which appears to be a reason- able "guess " for the average desired. Subtract x from each of the given numbers, find the average of the differences, and add it to x, paying due regard to signs throughout. The process is illustrated for the average of the numbers in the first column, the value of x being chosen as 85. The differences obtained by sub- tracting 85 from each of the numbers are given in the second and third columns. Their, average is obtained by adding them and dividing by 6, the number of numbers to be averaged, and is found to be - 2.5. This must be added to 85, which gives 82.5 as the average. The advantage of the method lies in the fact that it involves only relatively small numbers, which decreases the probability of error, and also enables one to handle simple cases mentally. 85 70 -15 80 - 5 90 + 5 75 -10 95 + 10 + 15 -30 - 30 6| - 15 - 2.5 85 + (2.5) = = 82.5 84 ELEMENTARY FUNCTIONS The proof of the correctness of the rule, for 4 numbers a, 6, c, d, follows from the fact, which is readily verified, that the equation a + b + c + d _ (a -x) + (b -x) + (c - x) + (d - x) 4 is true no matter what the value of x may be. EXERCISES 1. In an experiment on friction of metal on metal, values of W and W (see Example 1) were found as in the following table. Find the equation connecting W and W. What is the value of the coefficient of friction? W I 10, 20, 30, 40, W I 1.9, 4.2, 6.1, 7.8, 2. In an experiment on the friction of leather on metal, values of W and W (see Example 1) were determined as in the table. Find the equa- W I 10, 20, 30, 40, tion connecting W and W. What is TF' I 6.2, 11.7, 17.9, 24.3, the coefficient of friction? 3. The table gives corresponding values of P and W (see Example 2) PT I 1 00 200 ^00 400 obtained in an experiment on a block -^- .-, o — ^Fr~i — iA>7 a — 1^1 I and tackle. Find the relation be- P 41.2, 74.6, 107.6, 141.5, , d j w ' tween P and W. 4. In an experiment on a system of pulleys in which all but one were Tfl200, 400, 600, 1000, movable the values of Tf and Pin ' — — — the table were found. Fmd the P 37.3, 61.8, 87.1, 137.2, :. , ^^ , J — ' equation of the relation. 5. The table gives the length, in inches, of an iron rod at different tem- n 10 20 30 40 Peratures, in degrees r 40, 40.0(;46, 40.0lk 40.ol45 40.0193, ^^^f^^f^' ^md ^he ' ' length I at any tem- perature t. (Assume that the length when < = 0° is exact.) 6. The table gives the length in centimeters of a brass rod at various temperatures in degrees Centi- r-kTT^rb — ^^^^TTi na Aa o a aUa g^ade. Find the length of the M 36.405, 36.411, 36.416,3 6.424, . . no r. .- j j +u ' , ' ' ' rod at Centigrade, and the rate of change of the length of the rod with respect to the temperature. 7. Find the volume of alcohol at any temperature from the table* t I 10, 20, 40, 60, V I 101.1, 101.9, 103.9, 106.0, 8. Determine the density (see page 61) of brass from the table, which 5, 10, 15, 20, gives the mass m of pieces of brass m I 41.2, 83.6, 124.1, 166.8, with the volume e;. LINEAR FUNCTIONS 85 9. In order to graduate a spring balance, the extension e in inches was measured for different weights w^ w I 2, 5y 7, 10^ as in the table. Find to hundredths T"! 0.52, 1.31, 1.84, 2.62, of an inch the extension due to one pound. 10. The pressure p, per square foot, at various depths d, in feet, under water is given in the table. Find the d I 20, 40, 60, 80, pressure at a depth of one foot, and the p~| 1240, 2500, 3760, 4980, rate of change of the pressure with re- spect to the depth. 11. Barometer readings h were made simultaneously at various heights 6 I 29.00, 28.88, 28.79, 28.45, ^^' |f /Z^*' ^^"7". ^ ,^^7 P^^^.*' ^ — j^ Qg-^ 2or"^ — 495~^ ^^^^ *^® ^^^ ^^"^^^ *^® change m ' ' ' ' altitude as a function of the change in barometer for small changes in altitude. MISCELLANEOUS EXERCISES 1. Construct the line through the point (3, 2) whose slope is - \, and find its equation. 2. Construct the line determined by the points (2, 1) and (4, 5), and find its equation. 3. Construct the graph of the equation y = 2t + Z. Discuss and in- terpret the graph if y represents the distance from a station to a moving body at the time t; if y represents the velocity of a moving body at the time t. 4. The volume of a cake of ice is proportional to the volume of the water from which the ice was obtained. If 55 cubic centimeters of ice are made by freezing 50 cubic centimeters of water, how much water will be required to make 500 cubic centimeters of ice? Interpret the solution graphically. 6. A train is moving at the rate of 35 miles an hour when the brakes \ are applied. Six seconds later the train's rate is 17 miles an hour. Con- struct the graph of the velocity as a function of the time, and find its ; equation. How long after the brakes are applied will the train come to I rest? Check the result graphically. i 6. A 12-pound specimen of copper ore lost 2.5 pounds when weighed in I' water. How much copper did it contain if 8.9 pounds of copper and 2.7 pounds of the material with which the copper was combined each lost 1 pound when weighed in water? ! 7. How many cubic centimeters of cork, density 0.3, must be combined with 75 cubic centimeters of steel, density 0.9, in order that the combined I mass will just float in water? Hint: The density of the combination must equal that of water, which is unity. 86 ELEMENTARY FUNCTIONS 8. Specific gravity is the ratio of the density of a body to that of another body taken as a standard. Water which is taken as the standard has a density of unity in the metric system. Hence the mass of a substance is equal to the specific gravity times the volume. How much water must be added to 25 cubic centimeters of concentrated hydrochloric acid, specific gravity 1.20, to reduce the specific gravity to 1.12? 9. How much water must be added to 30 cubic centimeters of ammonia, specific gravity 0.90, to raise the specific gravity to 0.96? 10. It is desired to reduce the specific gravities of quantities of sul- phuric acid, specific gravity 1.84, and of nitric acid, specific gravity 1.42. How much water must be added to 80 cubic centimeters of each to reduce the specific gravities to 1.18 and 1.20 respectively? 11. What mass of hydrogen (molecular weight 2) will be displaced by 10 grams of zinc (molecular weight 65) acted upon by hydrochloric acid if the amounts exchanged are proportional to the molecular weights? 12. Find the money value of pure gold in a $20 gold piece, if one ounce of gold is worth $20.66 and the coin weighs 516 grains and is ^^ pure. 13. A tank can be filled by one pipe in 3 hours, and emptied by a second in 2 hours, and by a third in 4 hours. How long will it take to empty the tank when it is full if all the pipes are opened? 14. The report of a gun was heard in 3 seconds at a place 3189 feet distant, toward which the wind was blowing; and in 2 seconds at a place 2074 feet distant, from which the wind was blowing. Find the velocity of sound and of the wind. 15. A railway passenger observed that a train moving in the opposite direction passes him in 2 seconds, but when moving in the same direction it passed him in 30 seconds. Compare the rates of the trains. 16. Two cities are 39 miles apart. If A leaves one city-two hours earlier than B leaves the other, they meet two and a half hours after B starts. Had B started at the same time as A, they would have met three hours after they started. How many miles an hour does each walk? 17. The accuracy with which one can weigh an object on a balance de- pends on the number of spaces on the scale which the pointer moves when a small weight is added. The W I 5, 25, 50. 100, table gives the deflection, D, of the D I 1.99, 1.72, 1.76, 0.95, pointer with a weight W in the pan when one milligram is added to W. Determine D, approximately, as a function of W. CHAPTER III ALGEBRAIC FUNCTIONS 28. Introduction. In this chapter we shall study the proper- ties (Sections 10 to 13) of certain types of algebraic functions (Section 14). Among these types are the quadratic function, ax^ + bx + c, which occurs in the solution of quadratic equations, and the function ax^. The latter is an integral rational function if n is a positive integer, a rational fractional function if n is a negative integer, and an irrational function if n is a fraction. Other types are the linear fractional function, {ax -\- h)/{cx + d), and polynomials, which are studied in order to obtain a method for solving equations of higher degree than the second, an extension of the solution of Unear and quadratic equations. These functions find frequent appUcation in many fields. For example, the quadratic function appears in the theory of falling bodies, and among the laws which can be represented by a function of the type ax" are Newton's law of gravitation, and Boyle's law connecting the pressure and volume of a gas. In the study of quadratic functions we shall proceed from special cases to the general case by methods which are impor- tant in other connections. 29. Graph of jc^. Since (- x)^ = x^, the graph is symmetrical with respect to the y-axis (Theorem lA, page 23). Hence the ^/-axis is called an axis of symmetry. only positive values of x. 87 T 4 1~ H . ^ t 4 I i X t ^ 1 f 4 X-^ t \] / ik^ _i -'-'-'" ' ' '" 0, FiQ. 47. 1> 2, 1, 0, 1, 4, 9, The table of values need include 88 ELEMENTARY FUNCTIONS It is readily seen that the intercepts are x = and y = 0, Hence the curve passes through the origin, but does not cut the axes elsewhere. Since x^ is positive if x is real, no part of the curve lies helow the X-axis. As X increases, so also does x^, and hence the curve broadens out as it rises. If x > 1, x^ increases more rapidly than x, so that, to the right of x = 1, the curve rises more rapidly than it broadens out. The origin is a minimum point. 30. Graphs of ax^ and af(x). Consider the graph of the function 2x\ The ordinate of any point on the graph is twice that of the point with the same ab- scissa on the graph of x^. Hence the graph of 2x^ may be plotted by doubling the ordinates of a number of points on the graph of x^, and drawing a smooth curve through the points so obtained. In like manner, the graph of Sx^ may be obtained by trebling the ordinates of points on the graph of x^, of Jx^ by bisecting them, etc. The ordinates of points on the graph of — 2x'^ are numerically twice those of points on the graph of x^, but as they are negative, the points on the graph lie below the x-axis. The graphs of 2x'^ and - 2x^ are symmetrical to each other with respect to the x-axis. i From the method of constructing these graphs we see that The y-axis is an axis of symmetry of the graph of ax^. The curve runs up or down, and the origin is a minimum or maximum point, according as a is positive or negative. If a is positive, the graph of ax^ rises more or less rapidly than that of x^, for which a = 1, according as a is greater or less than unity. The larger the value of a, the more rapidly the curve rises, and the smaller the value of a is, the less rapidly it rises. Any one of these curves is called a parabola. The axis of symmetry of a parabola is called its axis. TI-; in -W-W ^4t_ Ifr ^u Cy _^B /// Sa,* /// \^>f ^-'-^ 771^ SV ttt-l m i/lrHJ rit ' iffi Fig. 48. ALGEBRAIC FUNCTIONS 89 I ^KThe reasoning employed above may be used to prove the ^^Theorem. The graph of af{x) may be obtained by multiplying by a the ordinates of points on the graph of f{x). Corresponding points on the two graphs lie on the same or opposite sides of the X-axis according as a is positive or negative. If a = - 1 the graph is symmetrical to that of f(x) with re- spect to the a;-axis. This theorem is the second one we have considered belonging to the set of relations between pairs of functions and their graphs (see the last paragraph but one on page 43). 31. Translation of the Coordinate Axes. Consider a system of coordinate axes with the origin O, and a second system, parallel to the first, with origin O'. Replacing the first system by the second is called translating the axes. By the equations for translating the axes we ^' Pry mean equations which express the coordinates of a point re- ferred to the first, or old, axes in terms of the coordinates of the same point referred to the second, or new, axes. Let the old and new coordi- nates of any point P be respec- tively {x, y) and {x\ y'), and let the old coordinates of the new origin, 0', be Qi, k). From the figure we readily obtain Theorem 1. The equations for translating the axes are V' Ei X' Fig. 49. X = x' -\-hf\ y = y' + hi (1) 1^ where Qi, k) are the coordinxites of the new origin. Suppose that the graph of an equation in x and y has been plotted. To find the equation of this graph referred to new axes parallel to the old axes we substitute in the given equation the values of x and y given by (1). The following examples illustrate the utility of the theorem. 90 ELEMENTARY FUNCTIONS Example 1. Given the equation 2/ = x2 - 6x + 13, (2) translate the axes so that the new origin is the point (3, 4) and find the equation in the new coordinates which has the same graph as (2). Plot the graph, and state its most noteworthy properties. By Theorem 1 we have M ' ; ^^ 7 t t 3 ' k iZ ^ '0 T ^ ei' -^ a; = x' + 3, 2/ = ?/' + 4. Substituting these values in (2), we have y' + 4.= {x' + 3)2 - 6(x' + 3) + 13. Removing the parentheses, y' + 4 = x'2 + 6x' + 9 - 6x' - 18 + 13, or (3) (4) (5) Fig. 50. The graph of (5), plotted on the new axes, is the same as the graph of (2), referred to the old axes. But the graph of (5) is a parabola which is easily plotted on the new axes. It is shown in the figure. This curve is also the graph of (2) when plotted on the old axes. From it we see that the axis of symmetry of the parab- ola, the 2/'-axis, is the line x = 3, and that the minimum point is the new origin (3, 4). In this example it turned out that the equation obtained by translating the axes was much simpler than the given equation. The question arises : If an equation is given, how can we de- termine how to move the axes so as to obtain a simpler equation with the same graph? The method of doing this is illustrated in Example 2. Simplify the equation 2/ = 2x2 - &c + 11 (6) by translating the axes, and construct the graph. First solution. Substituting in (6) the values of x and y given by (1), we get ?/' + fc « 2{x' + A)2 - 8(x' + /i) + 11 = 2x'2 + 4/ix' H- 2/i2 _ 8x' - 8^1 -h 11, whence y' = 2x"' + {Ah - 8)x' + 2/i2 - 8/i + 11 - fc. (7)'i This equation will be very simple if the coeflficient of x' and the constant term are zero, that is, if 4A-8 = 0.| (gy I and 2^2 - 8A + 11 - fc y.\ ^J ' ^ 1 J 'K g 3 i s .^J X ALGEBRAIC FUNCTIONS 91 Solving these equations for h and k, we get h = 2, \ /qx ,and /c = 2x22-8x2 + ll = 3./. ^""^ If tiiese values are substituted in (7) we obtain y' = 2x'\ (10) Hence, if the axes be translated so that the origin is moved to the point v2, 3), equation (6) assumes the simpler form (10). Plotting the graph of (10) on the new axes we get the curve in the figure. Second solution. The form of the given equa- ^iQ- SI. tion (6) may be changed as follows: 2/ = 2x2 - 8x + 11 = 2(x2 - 4a;) + 11 = 2(x2 - 4a; + 4 - 4) + 11 - 2(a;2 - 4x + 4) - 8 + 11 = 2(x - 2)2 + 3, whence y -Z = 2{x - 2)^. It is seen by inspection of this equation that if we set X = x' + 2, "I and y = y' + S, ) we will obtain the simpler equation y' = 2x'2, which is identical with (10). Equations (11; are the equations for translating the axes so that the origin is moved to the point (2, 3) (by Theorem 1). Let the graph of any function y = f{x) be given. In order to move the x-axis up k units, leaving the 2/-axis unchanged, k we set y = y' +k, obtaining y' -\-k = f{x), or ■ y'=f(x)-k. (12) ; The graph of (12) is identical with the given graph, if the j equation is plotted on the new axes. Now suppose that the graph of (12) is plotted on the original axes. It may be obtained by moving the given graph down k I units (see the Theorem in Exercise 3, page 19). Hence the I graph of (12) may be interpreted in two ways, which are es- sentially identical since the effect of moving the a;-axis up k (11) 92 ELEMENTARY FUNCTIONS units, and erasing the original x-axis, is the same as moving the graph down k units and erasing the original curve. In Uke manner, if we set x = x' + h, and leave y imchanged, we get y=f{x' + h), (13) The graph of this equation is identical with the graph of /(x), if plotted on the original x-axis and a new ?/-axis h units to the right of the old. But the effect of moving the 2/-axis h units to the right and erasing the old 2/-axis gives the same figure as moving the curve h units to the left and erasing the old curve. Hence the graph of (13) referred to the original axes may be ob- tained by moving the graph of f{x) h units to the left. A B Fig. 62. In plotting (13) on the original axes it is convenient to write X in place of x'. We thus obtain Theorem 2. The graph of f{x + h) may he obtained by moving the graph of f(x) h units to the left. The motion will be to the right if h is negative. This theorem, for which we find application in later chap- ters, should be associated with the last paragraph but one on page 43. EXERCISES 1. Plot on the same axes the graphs of: (a) x2, 3x2, |2;2, - ^x\ (b) x^ - 4x, 2(x2 - 4x), Ux^ - 4x). (c) x^ + 3, 2(x2 + 3), Ux^ + S). (d) (X - 2)2, Six - 2)2, lix - 2)2. 2. Find the value of o if the graph of y - ox* passes through the point (2, 3), and construct the graph. 1 ALGEBRAIC FUNCTIONS 93 I 3. Show that the points (2, 1), (3, 2J), and (4, 4) lie on one of the parab- ! olas ax^. 1 4. Show that one of the curves ax^ passes through any point \ Pi(xi, 2/i). [ 6. Find the average rate of change of y with respect to x, for the inter- I vals from x = to x = 1, from a; = 1 to a; = 2, from a; = 2 to a; = 3, and for I any interval if (a) y = x^\ (b) 2/ = 2a;2; (c) y = ^x^; (d) y = ax^. 6. Translate the axes to the new origin indicated, and construct both pairs of axes and the graph: (a) y = x^ + S, O'(0, 3). (b) y = x^ - 4x, 0'(2, - 4). (c) 2/ = - x^ 0'(2, - 4). (d) y = 1/x, 0'(3, 2). 7. Using Theorem 2, Section 31, construct on the same axes the graphs of x\ (x + 2)2, and (x - 3)2. 8. Construct the graphs of x^ and x^ + 8x + 16 on the same axes. 9. On the same axes construct the graphs oi y = x^, y = 2x2, y = (x + 2)^, y = a;2 + 2. 10. Simplify the following equations by translating the axes. In each case, construct both pairs of axes and the graph. Determine the ! axis of symmetry and the maximum or minimum point in the given co- ordinates. (a) x2 + 6x + 4. (b) 4x - x\ (c) 2x2 _ 16a; ^ 35. (d) Sx^ - 12x + 5. (e) - x2 + 2x + 5. (f) - 2x2 _ i2x - 20. 32. Instantaneous Velocity. If a ball is dropped, its velocity ! changes continually. An approximate value of what we mean when we speak of its velocity at some given instant is given by I the average velocity in an interval of time A^ beginning at I' that instant. The smaller the value of A^ is, the more ac- ■ curate is the approximation. A precise notion of the velocity ■ at an instant is given by the Definition. The velocity of a body at an instant, or its in- ; stantaneous velocity, is the limit * of the average velocity in an i interval At beginning at that instant, as the interval At ai> j preaches zero. The computation of an instantaneous velocity is illustrated in the following example. * The limit of a variable is a constant such that the numerical value of i the difference between the variable and the constant becomes and r&- j mains less than any assigned positive number, however small. 94 ELEMENTARY FUNCTIONS Example. If a ball is dropped, its distance from the starting point at any time <, in seconds, is given by s = lQt\ (1) Find the velocity at any time. The distance s at any time t is given by (1). If the distance changes by As in the next A^ seconds, the distance at the time t + At will be s + As. These corresponding values of the distance and time must satisfy (1), and hence s + As = 16(i + At)^ = 16f2 + 32f At + IQAtK (2) Subtracting (1) from (2) we obtain As = S2t At + lQAt\ As and dividing by At, ^ = 32< + IQAt. (3) This is the average velocity during an interval of At seconds beginning at the time t. By the definition above, the velocity at the time t is the limit of (3) as At approaches zero. Denoting it by y, we get v^S2t, (4) since IQAt approaches zero when At approaches zero. 33. Rate of Change. Slope of Tangent Line. A generaliza- tion of the idea of instantaneous velocity, which may be de- scribed as the rate of change of the distance s with respect to the time t, is given in the DEFmiTiON. The rate of change of y with respect to x for a given value of x is the limit, as Ax approaches zero, of the average rate of change of y with respect to x in the interval froi X to X + Ax. If 1/ is a function of x, y-Ax), (1] the rate of change of y with respect to x may be determined follows : Substitute x + Ax for x and y + Ay for y in (1). The result, 2/ + A2/=/(x + Ax), (2) is true because Ay is the change in y due to a change of Ax in X, and hence x + Ax and y -{- Ay are corresponding values oi the independent variable and the function. ALGEBRAIC FUNCTIONS 95 Fig. 53. Subtract (1) from (2), which gives Ay, and divide by Ax, which gives Ay /Ax, the average rate of change of y in the interval Ax. Then find the limit of the average rate of change as Ax approaches zero. To interpret this process graphi- cally we shall need the Definition. If P is a given point on a curve and Q any other point on it, the line tangent to the curve at P is the limiting position of the secant PQ as Q moves along the curve and approaches P. The points P(x, y) and Q{x -\- Ax, y + Ay) are on the graph of (1), and Ay /Ax is the slope of the secant PQ. As Ax ap- proaches zero, Ay /Ax approaches the rate of change of y; also, Q approaches P, the secant PQ approaches the tangent PT, and the slope of PQ approaches that of PT. Hence, The rate of change of y with respect to x is represented by the dope of a line tangent to the graph of y. The tangent to the graph of 2/ at a maximum or minimum point is horizontal, and hence its slope is zero, and hence, also the rate of change of y. This fact affords a general method of finding maximum and minimum points. The slope of the tangent line and the rate of change of the function should be added to the table of properties of graphs and functions on page 42. Example 1. Find the rate of change of y = ax\ (3) and discuss its meaning for the graph. Replacing x and y by x + Ax and y + Ay respectively, we get y + Ay = aix + Ax)^ = ax^ + 2ax Ax + a Ax^. btracting (3) from (4), Ay = 2ttx Ax + aAx\ Ay (4) Hviding by Aa;, Ax = 2ax + a Ax, (5) ich is the average rate of change of y in any interval Ax. 96 ELEMENTARY FUNCTIONS Letting Ax approach zero, the limit of (5) is m = 2ax, (6) the required rate of change, or the slope of the tangent line at any point. If X = 0, we have m = 0, so that the parabola which is the graph of (3) is tangent to the x-axis at the origin. For a given value of x, the larger the value of a, the larger also is the slope of the tangent line. This justifies the statement made in Section 80 that " the larger the value of a the more rapidly the curve rises." If a = 1, so that y = x^, we have m = 2x. Hence the slope of the tangent line is less or greater than 1 according as x is less or greater than |. There- fore, to the right of x = ^, the curve rises more rapidly than it broadens out. (Compare with the statement in Section 29). For any given value of a, the larger the value of x the larger is the value of m. Hence the curve rises more and more rapidly as it runs to the right. Example 2. Find the slope of the line tangent at any point to the graph of y = x^ - 4x. (7) Find the slope at the point for which x = 3, construct the line, and find its equation. Find also the coordinates of the maximum or minimum points. The table of values and the graph are readily constructed. Replacing x by x + Ax and y by ?/ + Ay in (7) we get 2/ = (x + Ax)2 - 4(x + Ax) or ?/ = x2 + 2x Ax + Ax2 - 4x - 4Ax. (8) Subtracting (7) from (8), At/ = 2x Ax + Ax2 - 4Ax. Dividing by Ax, Ax l^i T k it ^ t 4 V t^ A tt J 'k - ■10 19 3 / 5 6 7 b X 1 , f X J .3 -L . S2 " ^x Fig. 64. 4. (9) Passing to the limit as Ax approaches zero, the slope of the tangent line at any point is jr I - 1, 0, 1, 2, 3, 4, 5, y I ^5, 0, - 3, - 4, - 3, 0, 6, 2x (10) At the point A (3, - 3) the slope of the tangent line is therefore m = 2x3-4 = 2, and the line may b« drawn by the construction on page 5% ALGEBRAIC FUNCTIONS 97 Using the point-elope form of the equation of a straight Hne (page 66), the equation of the line tangent at A is I 2/ + 3 = 2(x-3), or 2/ = 2x - 9. As a check, notice that the intercept on the ^-axis, - 9, agrees with the line as constructed. At a maximum or minimum point the tangent line is horizontal, and ' hence the slope m given by (10) must be zero. Hence I 2x - 4 = 0, ' whence x = 2. Substituting this value in (7), the minimum value of y is 2/ = - 4. Hence the minimum point is the point D(2, - 4). EXERCISES 1. A ball is rolled down an inclined plane. Its distance from the start- I ing point is given by s = ^t^ + 4L Find the velocity at any time; at the I instant ^ = 3. Plot the graphs of s and v on the same axes, and from them describe the motion. 2. Prove that a line tangent to a circle, as defined in Section 33, is per- pendicular to the radius drawn to the point of contact, using the form of reasoning of elementary geometry. 3. The distance from a fixed station to a moving body is given by s = f2 _ 3^^ Find the velocity at any time, and show that the acceleration is constant. Plot the graphs of s and v on the same axes, and describe the motion from t = to t =" 5. 4. Find the slope of the line tangent at any point to the graph of each I of the functions below. Find the slope of the line tangent at the point for I which X = 2, construct the line, and find its equation. Find the coordi- \ nates of the maximum or minimum points. (a) 4 - a;2. (b) y = 3x - x\ (c) y = x^ - 1. (d) x^ -4y + 2x = 0. i 6 Tabulate the values of the slope of the line tangent to the graph of j r* at the points for which a: = — 3, — 2, — 1, 0, 1, 2, 3. What can be said I of the value of m as a; increases? Does the value of m, the slope of the ; tangent line, always increase as x increases if the graph is concave up- i ward? Is the converse true? If a curve is concave downward, how does the slope of the tangent line : change as x increases? Is the converse true? 1 6. Find the maximum and minimum points oi y = x^ - x, and then plot [the graph. i 7. Find the equation of the line tangent to the graph of x^ at the point for which x = 1; for which x = 2; for which x = 4. Find the intercept 98 ELEMENTARY FUNCTIONS of each line on the y-axis. How does the intercept compare with the ordinate of the point of contact? 8. Prove that if two lines are perpendicular the slope of one is the negative reciprocal of the slope of the other. Definition. The line perpendicular to a line tangent to a given curve at the point of tangency is called a normal to the curve. The slope of the normal at any point may be found from that of the tangent by the preceding exercise. 9. Find the equations of the tangent and normal to the graph ol y = x^ 4- 2x at the point (1, 3). Construct the figure. 10. Find the equation of the line normal to the graph of x^ at the poini for which x = 1; for which x = 3; for which x = 5. Find the intercept' of each line on the 2/-axis. How does the intercept compare with the or- dinate of the point at which the normal cuts the curve? 11. Find the acceleration of a moving body if its velocity is given by V = 2t^ - 5t. Plot the graphs of v and o on the same axes and discuss the variation of both functions. 12. If the position of a moving body is given by the equation s = t^ - % find the velocity and acceleration. Plot the graphs of s, v, and a on the same axes. 13. Find the acceleration of a body if its position is given by s = 4:(^ - fi. 14. If a ball is dropped the velocity after the ball has fallen s feet is given by v^ = 2gs. Find the rate of change of s with respect to v. 34. Graph of the Quadratic Function ax^ + bx + c. Denote the function by y so that y = ax^ + hx + c. (1) In order to determine the form of the graph we shall first show that it always has a single maximmn or minimum point, and then translate the axes so as to have this point for the new origin. To find the maximum or minimum point we need first the slope of the tangent line at any point. In (1) replace x by X + Ax and y by y + Ay, This gives y + Ay = a{x + Ax)^ + h(x + Ax) + c, or y + Ay ^ ax"^ + 2ax Ax + Ax^ -\-hx + bAx + c. (2) Subtracting (1) from (2), Ay ==2axAx-{-Ax^ + b Ax, J ALGEBRAIC FUNCTIONS 99 and dividing by Ax, ■jr = 2ax -f- Ax + 6. Ax Passing to the limit as Ax approaches zero, the slope of the tangent line at any point is m = 2ax-\- b. (3) The tangent hne will be horizontal if m = 0, that is, if 2ax + b = 0, whence .= -A. (4) Substituting this value in (1) the corresponding value of y is (J) 4,ac — b'^\ - — , — J j is either a maximum or a mini- mum point. To translate the axes so that O' is the new origin we set , b , 4:ac - b^ Substituting these values in (1), , iac -V / „ bx' 6n , / , 6 \ , = ax'2-6x'+f^ + 5x'-|+c, 4a or y' = cix'\ (6j The graph of this equation is a parabola (Section 30) con- gruent to the graph of ax^. Its axis of symmetry is vertical, and the curve runs up or down according as a is positive or negative. 100 ELEMENTARY FUNCTIONS Then since the graph of (1), plotted on the old axes, is identi- cal with that of (6), plotted on the new axes, we have the Theorem. The graph of the quadratic function ax^ + hx -{- c is a paraholaj with vertical axis, which is congruent to the graph of ax^. It runs up or down according as the coefficient of x^ is positive or negative. Example. If a ball is thrown vertically upward with a velocity of 80 feet per second, its height s, in feet, after t seconds, is given by the equation s = - 16^2 + 80t. (7) Find when the ball will be highest, how high it will rise, and compare the time of rising with that of falling. Construct the graph. At the highest point the velocity of the ball is zero, and we therefore seek first the velocity at any time. Replacing thy t + At and shy s + As we get s + As = - m{t + Aty + 80(< + At), or s + As = - 16^2 _ 32^ A« - IQAt^ + 80^ + 80A^ (8) Subtracting (7) from (8), As = - 32t At - 16Ai2 + SOAt, and dividing by A^, ^ = - 32f - IQAt + 80. Passing to the limit as At approaches zero, the velocity at any time is v= -S2t + 80, f (9) since the limit of the average velocity As/ At is the instantaneous velocity. At the highest point reached by the ball v = 0, and hence < = |f = | seconds. Substituting this value in (7) the maximum value of s is or: K s=-16x'=r + 80x^=-100 + 200 = 100 feet. 4 2 To translate the axes to the maximum point set f = r + I, s - s' + 100. j Substituting in (7), s' + 100 = - 16(/' + f)' + 80(r + f), = - le^'* - m' - 100 + sot' + 200, whence s' - - 16f'*. (10)1 The graph of this equation plotted on the new axes is identical with that of (7) plotted on the old axes. The figure shows the graph, which is readily plotted on the new axes. It passes through the old origin sincej ALGEBRAIC FUNCTIONS 101 Fig. 65. « = when t = 0. From the symmetry of the graph wi^hresp^cito tte'^ «'-axis the other intercept on the i-axis h t = 5, so tliat the ball returns to the hand after 5 seconds. Hence the time of rising is equal to the time of falling. The figure also shows the graph of V [equation (9)]. As y is a linear function of t, the slope - 32 gives the acceleration, which is con- stant. A simultaneous dis- cussion of the variation of 8 and V describes the motion fully. At the start s = and V = 80. As t increases, s increases but the velocity decreases, until when t - 5 the ball reaches its maxi- mum height of 100 feet and the velocity becomes zero. The ball then begins to fall, since the velocity changes sign, and its height decreases to zero at < = 5, when the velocity is - 80, i.e., 80 feet per second downward. The position of the parabola which is the graph of a quadratic function with reference to the x-axis is determined by the dis- criminant (9(/) page xvii) of the function, for the discriminant determines the nature of the roots of the equation ax^ + 6x 4- c = 0, which are identical with the zeros of the function and which are represented by the intercepts on the x-axis. If the discriminant 6^ — 4ac is positive the roots are real and distinct, if the discriminant is zero the roots are equal, and if the discriminant is negative the roots are imaginary. It follows that the graph of a quadratic function cuts the a:-axis twice if the discriminant is positive. If the discrimi- nant is zero, the graph has but one point in common with the rc-axis, and hence the graph is tangent to the x-axis. The point of tangency is the maximum or minimum point. If the dis- criminant is negative, the curve does not meet the a;-axis at all. 102 ELEMENTARY FUNCTIONS . A method of using the zeros of a quadratic function in con- structing the graph of the function is illustrated in the Fig. 66. Alternative solution for the example above. The graph of equation (7 a parabola running downward, since s is a quadratic function of t which the coefficient of f^ is negative. To find the intercepts on the <-axis set s = 0, which gives- 16^2 ^80< = 0. Divid- ing by - 16, and factoring, t{t - 5) = 0, whence t = or 5. Hence the origin and the point A (5, 0) are on the graph. These intercepts give the time of starting and when the ball returns to the hand. The axis of symmetry is vertical. Hence it must bisect OA, and it is therefore the line a: = f = 2.5. Since the maximum point lies on the axis of symmetry, the ball reaches its greatest height when t = f seconds. Substituting this value in (7), the maximum height is - 100 + 200 = 100 feet. Hence the point M(2.5, 100; is the maximum point. In order to plot the curve from O to M we need only to compute the values of s for ^ = 1 and 2. The other half of the curve is obtained by means of the symmetry. EXERCISES 1. The Theorem, Section 34, shows that the graph of y = 6a: - x^ is a parabola, with vertical axis, running downward. Construct the graph: (a) By translating the axes, as in Section 31. (b) By finding the maximum point as in Section 33 and translating the axes. (c) By finding the intercepts on the a;-axis, determining the axis of symmetry from the intercepts, and the maximum value of y by means o: the axis of symmetry. (d) By finding the values of y to be excluded, from which the maximum value of y may be found, and determining the axis of symmetry by means of the maximum value. In (c) and (d) it will be necessary to obtain a few pairs of values of a>' ALGEBRAIC FUNCTIONS 103 and y. The most desirable values of x to be assumed may be determined after finding the coordinates of the maximum point. 2. Construct the graph of each of the functions below. Find the axis of symmetry, and the maximum or minimum point. Find the zeros of each function, and state how they are represented graphically. (a) x^ + 2x- 8. (b) 2x2 + 3x - 2. (c) - x^ + 3x. (d) - x2 - X + 2. (e) x2 - 4x -i- 4. (f) - x^ - 2x - 1. (g) x^ + X + 1. (h) 4x2 + 2x - 1. (i) _ 6x2 + X + 5. (j) x2 - X + 1. (k) 5x2 + 2x + 1. (1) - 3x2 4. 2x - 2. 3. Without solving the following equations, determine whether the roots are real or imaginary, equal or unequal. From this, what can be said of the graph of the quadratic function in the left-hand member? (a) x2 + 3x - 5 = 0. (b) 2x2 - 3x + 4 = (c) 4x2 + 12x + 9 = 0. (d) x2 - |x + 3 « 0. (e) 4x2 - X + 1 = 0. (f) x«-7x-2 = 0. 4. A special quadratic function is one in which one or more of the co- efficients are zero. Discuss and plot the graph of the special fimctions below, and state the special properties of the graph which distinguish it from the general case. (a) x2 - 4. (b) x2 - 4x. (c) x2/3. (d) 9 - x2. (e) 3x - x2. (f) _ 2x2. (g) ax2 + c. (h) ax2 + 6x. (i) axK 5 If /(x) = ax2 + 6x + c, what must be true of the coefficients if /(- x) = fix)? 6. If a ball is thrown vertically upward with a velocity of 50 feet per second, its height after t seconds is given by the equation s = - 16^2 + 50^. Construct the graph. Find when the ball will be highest, how high it will rise, and when it wiHYeturn to the hand. 7. Construct the graphs of x and x2 from x = to x = 1. What is the numerical difference of the ordinates of points on these graphs with the same abscissa? Construct the graph of this difference, and find the value of X for which it is greatest. What is the greatest difference? 8. A Norman window consists of a rectangle (width 2x and height y) surmounted by a semi-circle. If the perimeter of the window is 16 feet, find the area in terms of the half width x, and construct its graph. Find the dimensions which make the area (and hence the amount of light ad- mitted j a maximum. 104 ELEMENTARY FUNCTIONS 9. A clifif 40 feet high overhangs a river. A man on the cliff throws a Btone vertically upward with a velocity of 30 ft. per second. If the height of the stone above the cliff after t seconds is given by s = - 16f2 + 30t, find how high the stone rises, and when it will strike the water. 10. If the perimeter of a rectangle is 8 feet, find the area as a fmiction of one of the dimensions. Construct the graph of the area, and find the dimensions and area of the largest rectangle of this sort. 11. Solve Exercise 10 if the perimeter is any constant p. [The parcel post regulations require that the sum of the length and girth (greatest perimeter of a section at right angles to the length) of a package shall not exceed 6 feet. By means of this Exercise, what can be said of the shape of the largest rectanglar box which can be mailed?] 12. The equation of the path of a ball thrown into the air at an angle of 60° with the horizon with a velocity of 32 feet per second is y = i.7x - xym. Construct the graph, find the greatest height attained, and where the ball will hit the ground. 13. If a body is projected vertically upward with an initial velocity of V feet per second, its height s after t seconds Is given by the equation s= - 16^2 ^ vt. Find how long the body will rise and its maximum height. Prove that the time of rising equals the time of falling. 14. A farmer estimates that if he digs his potatoes now he will have 100 bushels worth $1.25 a bushel; but that if he waits, the crop will increase 16 bushels a week, while the price will drop Sff a week. Find the value of his crop as a function of the time in weeks, and draw the graph. When should he dig to get the greatest cash returns? 15. The amount of wheat obtained per acre depends on the intensity of cultivation. A farmer finds that he can cultivate 15 acres with suf- ficient intensity so that the return will be 30 bushels per acre; 20 acres BO that the return will be 25 bushels per acre; 25 acres so that the return will be 20 bushels per acre; etc. Find the law giving the total return as a function of the number of acres under cultivation, and plot the graph. What is the best size acreage for him to cultivate in order to get the lari est gross returns? 35. Empirical Data Problems. The table of values of linear function is such that if the successive values of Ax equal so also are the successive values of Ay. Any quadratic function has a somewhat analogous property, which is illustrated in the table for y = 2x^ + Sx + 4. Tb ALGEBRAIC FUNCTIONS 105 Ax X y t^y 4 1 5 1 9 1 9 2 18 1 13 3 31 1 17 4 48 4^_ 4 4 4 values of Ax being equal, those of Ai/, sometimes called the jird differences of 2/, are not equal. But the successive differences of Ay, called the second differences of y and denoted by A^y, are equal. Every quadratic function pos- sesses this property, that if the values of x are such that the successive values of Ax are equal, then the su^ccessive valves of the second differences oj y, A^y, are equxil. This property enables us to tell whether a given table of values may be represented approximately by a quadratic function, and also to determine the coefficients of the function. Example. Find a quadratic function which represents approximately the law connecting the values of x and y given in the table. Inspection of the values of x shows that the successive values of Ax are equal. Computing the successive values of Ay and A^y, it is found that the latter are nearly equal. Hence the values of y resemble those of a quadratic function. Now let y = ax^ + hx-\-c (1) be the required function. The second table gives the values of y, Ay, Ahf computed from (1) for the values of x in the given table. If the given values of y were exactly the values of some quadratic func- tion, then for the proper •values of a, h, c, all the numbers of one table would equal the corresponding numbers in the other. In particular, the average val- ues of y, Ay, and Ahj found in the second table would equal those in the first. As it is, these averages are equal for values of a, h, c, for which (1) represents the required law approximately. X y Ay Ah/ 2 3.1 8.1 4 11.2 15.7 7.6 6 26.9 24.1 8.4 8 51.0 32.4 8.3 10 83.4 5 1 175.6 35.12 4 180.3 3 1 24.3 20.08 8.1 X y Ay AV 2 4a + 26+ c 12a + 26 4 16a + 45+ c 20a + 26 8a 6 36a + 6&+ c 28a + 26 8a 8 64o+ 86+ c 36a + 26 Sa 10 100a + 106+ c 5 1 220a + 306 + 5c 4 1 96a + 86 31 24a 44a + 66+ c 24a + 26 8a 106 ELEMENTARY FUNCTIONS Equating the average values of Ahj, Sa = 8.1, whence a = 1.01. Equating the average values of Ay, 24a + 26 = 20.08. Substituting the value of a, and solving for 6, we get 6 = - 2.08. Equating the average values of y, 44a -f 66 + c = 35.12. Substituting the values of a and 6, and solving for c, we get c = 3.16. Substituting the values of a, 6, c, in (1), _x Observed y Computed y \ Error % error the required approxima- 2 3.1 3.04 -0.06 -1.9 tion of the law is 4 11.2 11.00 -0.20 -1.8 2/ = 1.01x2 -2.08X+ 3.16 6 26.9 27.04 +0.14 +0.5 The accuracy with 8 51.0 51.16 +0.16+0.3 which this relation ex- 10 83.4 83.36 -0.04 -0.05 presses the law is indi- cated in the table. If it is desired to find the equation of a parabola with verti- cal axis which passes through, or near, several points whose coordinates are given, the method used in the example may be employed even though the values of Ax, and hence also those of A^y, are not equal. At least three points must be given. EXERCISES 1. Find the equation of the parabola with vertical axis which passes through the points (1, 0) ,(3, 10), (5, 28). Construct the graph and check the result. 2. In sinking a deep mine, new material as well as labor must be in- vested each day, and the required depth cannot be determined in advance with accuracy. A company which has set aside $100,000 for the cost of sinking a shaft finds the total capital invested as the work increases as rr.. . ., 1 n 1 o o given in the table. Find Time m months i n i o o Investment in thousands | 4, 12, 26.1, 46.1, 0, 1, 2, the law giving the invest- ment as a function of the time. If the work continues to progress according to the same law, when must the work be completed if the cost is not to exceed the amount set aside? 3. A contractor agreed to build a breakwater for $125,000. After spend- ing $33,000 as indicated in the table, he threw up the job without receiv- ing any pay, because he estimated that the cost would increase according to the same law, and that it would require six months more to complete the work. How much would he have lost if he had finished the breakwater? Time in months 0, 1, 2, Investment in thousands I 5, 6.1, 11.2, 20.2, 33.1 4. An electric conductor gives out a definite amount of current in every mile of its length. Let x be the distance of any point in miles from the ALGEBRAIC FUNCTIONS 107 end of the line remote from the generator, and y the voltage there. As- Buming that the law is of the form y = ax^ + 5, determine the law from the table. What is the voltage at a point 2.5 miles from the end? Hint: The value of a may be de- : I 0, 1, 2, 3, 4 termined by the general method; ^1 200, 212.5, 250, 312.5, 400 after findmg a, substitute its value in y = ax^ -\-h. Then a value of 6 may be determined by substituting any pair of values of x and y. A better value of h is obtained by determining its value for each pair of values of X and y and averaging the results (compare the method of finding h in Section 27 after m is determined). 36. The Function x^» Among the functions represented by a;" which may be obtained by assigning a numerical value to n may be mentioned the following: x^, for n = 2. x^j for n = 3. xi = =fc ^s/x, for n = J. x^ = -s/x, for n = \. x~^ = 1/x, f or n = — 1. If no numerical value is assigned to n, the sjnnbol x^ may represent either the totality of all functions obtained by assign- ing a numerical value to n, or a particular, but unassigned, one of these functions. The function x", sometimes called the power function, is con- sidered in the following sections. 37. Tables of Squares, Cubes, Square Roots, Cube Roots, and Reciprocals. These tables are extensive tables of values of the function a;" for n = 2, 3, |, J, and - 1. They are labor- saving devices, for from them we can find, for example, the square of a number without the labor of multiplying it by it- self. We shall use Huntington's Four Place Tables, Unabridged Edition. Tables of sqicares and cubes. The square of a number n, between 1 and 10, may be found from the table on page 2. In order to economize room, the table is not arranged in two long columns of rows. Instead, the first digits in n are given in the border on the left and the last digit in the border at the 108 ELEMENTARY FUNCTIONS top of the table. To find the square of 1.56, look in the row in which 1.5 stands on the left, under " n," and in the column headed " 6." Here we find 2.434, which is the square of 1.56 to four figures. The exact square, obtained by multiplication, is 2.4336, which is nearer 2.434, as given in the table, than to 2.433. The lack of exactness in the tables is usually im- material, for in most of the applications of mathematics three or four figure accuracy is all that is desired^ and in many it is all that can be attained. Thus if 1.56 is the side of a square, ob- tained by measurement, the area is 1.56^ = 2.43. Only three figures are retained in the area since the product cannot con- tain more significant figures than the factors (Section 26). Foi^ such an example the table of squares is more accurate than' necessary. To find the squares of numbers greater than 10, or less than 1, we use, respectively, the relations (10n)2,= 1007i2 and {n/lOy = n^lOO. Since multiplication by 10 or 100 shifts the decimal point one or two places to the right, respectively, while division shifts it to the left, we have the rule given at the top of the table: " Moving the decimal point one place in n is equivalent to moving it two places in n^." Repeated application of this rule enables us to find the square of any number. Thus to find 247^, we find from the table 2.472, = 6.10, whence, applying the rule twice, 247^ = 61,000. Sinularly, 0.247^ = 0.0610. The table of cubes, on page 4, is very similar to the table of squares. The rule at the top of the table for shifting the decimal follows from the relations: {lOnY = lOOOn' and (n/ioy = nVlOOO. Tables of square roots and cube roots. The table of square roots on page 3 is separated into two parts which give the square roots of numbers from 0.1 to 1 and from 1 to 10. The reason for this lies in the, rule for shifting the decimal point. ALGEBRAIC FUNCTIONS 109 Since VlOOn = lOVn and X/Tno 10" we have the rule at the top of the table: *' Moving the decimal point two places in n is equivalent to moving it one place in V^" If n is greater than 10 or less than 0.1, by moving the point two places at a time, to the left or right, respectively, we will ultimately obtain a number in which the decimal point either precedes or follows the first significant figure. In the former case the square root is found in the upper part of the table, and in the latter case in the lower part. The square root of the original number is then found by re-shifting the decimal point in accordance with the rule. For example, to find the square root of 35,700, we shift the point two places to the left twice in succession, obtaining 3.57. From the lower part of the table we obtain V3.57 = 1.889. Then applying the rule twice, shifting the point to the right, \/35700 = 188.9. To find the square root of 0.0024, we shift the point two places to the right, getting 0.24. From the upper part of the table, VO^ = 0.490, whence, by the rule, VO.0024 = 0.0490. Notice that the zero following the nine is retained in order to show the value of the radical to three figures. The table of cube roots on page 5 is divided into three parts because in moving the decimal point in any number three places at a time (see rule at the top of the table), we ultimately obtain a number in which the point either follows the first significant figure, precedes it immediately, or is followed by a single cipher. Table of reciprocals. The reciprocal of n, 1/n, differs from n^, n^, Vn, \/n, in that as n increases the reciprocal decreases. For a reason which will appear in Section 42, it is desirable to have all the tables so arranged that the numbers in the body of the table increase as we read down the page, and from left to right. The table of reciprocals on page 7 is so arranged. The values of n, given in the border of the table, increase as we read up and to the left. 110 ELEMENTARY FUNCTIONS The rule for shifting the decimal point at the top of the table follows from the fact that the product of n and its re- ciprocal is imity. Hence if one is multipHed by 10, the other must be divided by 10. To find 1/436, for example, we read up the table, on the right, until we come to 4.3, then over to the left until we are in the col- umn headed 6, where we find 0.2294, which is the value of 1/4.36. Then by the rule for shifting the point, 1/436 = 0.002294. In using any one of these tables the position of the decimal point may frequently be determined by inspection, and it is well to check the result obtained from the table. EXERCISES 1. Find 3.242, ^0.47, V4.72, 7.43, ^o.0235, ^0.24, ^1.84, 1/24. 2. Find the squares of 25.4, 0.86, 3540, 0.0043. 3. Find the square roots of 59, 590, 4300, 0.000382. 4. Find the cubes of 54, 0.317, 53200, 0.0000371. 6 Find the cube roots of 1540, 470, 18.3, 0.0048, 0.0000259. 6. Find the reciprocals of 23.4, 0.4 78, 532 , 0.0074. 7. Find the value of 0.07S 0.06^, V'14.23, 1/234^, ^0.02472. 8. Find the hypothenuse of a right triangle whose sides are 61 and 74. 9. Find the mean proportional between 6 and 34. 38. Graph of jc", n>l. It will be seen in this section, and those following, that a part of the graph of a:" hes in the first quadrant, and that the remaining part may be found by means of the symmetry of the curve. Hence particular attention should be made to fix in mind the various forms of the part of the graph in the first quadrant. The general appearance of this part of the graph varies according as n>l, 0l. In this case the graph of y = x^ (1) is tangent to the x-axis at the origin (0, 0), rises to the right and passes through the point (1, 1), at which the slope of the tangent Hne is n. That it passes through these points is seen by substituting their coordinates in (1). To find the slope of the fine tangent to the graph at any point, rhen w is a positive integer, replace a; by x + Aa: and yhyy + Ay (1), which gives y + Ay = (x + Ax)". Expanding the right hand member by the binomial theorem, n(n — 1) + Ai/ = X" + nx"-^Ax + Subtracting (1) from (2), n(n — 1) ^n-2^^2 _|. + Ax" (2) X«-2Ax2 + Ay = nx"~^Ax + Dividing by Ax, • +Ax". + Ax»-^ Passing to the limit as Ax approaches zero, the slope of the ]ent line at any point is m = nx"-i. (3) This proof assumes that n is a posi- tive integer, but in Chapter VI it will be shown that the result holds for fractional and negative values of n. At the origin x = 0, so that, by (3), m = if ri >1. Hence the tan- gent at the origin is the x-axis. At the point (1, 1) the slope of the ,' tangent line is m = n. Hence the larger t the value of n the steeper the curve is at this point, from which it follows that it must be flatter near the origin. If n is a positive integer greater than unity, the part of the graph of X" in the first quadrant is then very much like the graph of x2. 112 ELEMENTARY FUNCTIONS If n is a positive fraction greater than unity the part of the graph in the first quadrant has the same general appearance. To see this, let r, s, and t be three values of n such that rl, always resembles one of these curves. Thus the graph of x*^ is very much like a parabola, but it is flatter near the origin and steeper elsewhere. The a;-axis is tangent to the graph of a^ at the origin. This tangent differs from any we have encountered hitherto in that it crosses the curve at the point of tangency. The graph of x^ is remarkable in that it has a sharp point a+ the origin. It differs from other curves we have studied in detail in that vertical lines to the right of the 2/-axis cut it ALGEBRAIC FUNCTIONS 113 9 {y,^) P{x*y) are twice, corresponding to the fact that x' = ±Va^ has two values for each positive value of x. 39. Graph of x", 0l? Ifp/?<1? 6. Plot the graph of x^, find the inverse function, and plot its graph. 7. Would it be easy to build a table of values for a:*? Construct its graph by first finding the inverse function. 8. Plot the graphs of the following functions, in each case finding the inverse function and its graph. (a) 2/ = a;2 - 2. (b) 2/ = x^ + 2a;. (c) y = - x^ + 4x - 4. 9. Using Huntington's Tables, construct the graphs, on the same axes, of X, x^, x^, x% x^, taking x = 0.2, 0.4, 0.6, . . . , 1.4. Use as large a scale as possible and plot the parts of the graphs in the first quadrant only. 116 ELEMENTARY FUNCTIONS 10. Using Huntington's Tables, construct a table of values of x^ for x = 0.05, 0.10, 0.15, 0.20, 0.25, obtaining the values of the function to two decimal places. Construct the graph from a; = to a; = 0.25 on as large a scale as possible. 11. Plot the graph of x^ - Ax. On the same axes sketch the graph of the inverse function, and state several of its properties by interpreting its graph. Can you find the inverse function? Note. In finding the inverse of a function, it is necessary to solve an equation. At any stage of mathematical development, the solution of an equation may be impossible by means of functions already studied. Such an equation defines a new function, whose fundamental properties are de- termined by the equation. Thus it is impossible for a student beginning algebra to solve the equa- tion y^ = x for y. It is first necessary that he should become acquainted with the, to him, new function x^ = =t\/x. A person unacquainted with the solution of cubic equations cannot find the inverse of the cubic function in Exercise 11. But the theorem on the graphs of inverse functions enables us to get the graph and some of its properties, even though we do not know the function. This point of view will be useful later in studying certain transcendental functions. The inverse of a cubic or biquadratic function (page 39) is an algebraic function, but the inverse of a polynomial of higher than the fourth degree is usually transcendental. 12. What is the inverse of the function l/rc? What therefore can be said of the symmetry of the graph of the function? 13. Show that the graph of an equation is symmetrical with respect to the line y = xii the equation is unchanged when x and y are interchanged. Plot the graph of xy -2x -ly = 0. 14. What is the form of an equation in x and y if it defines a function of X which is its own inverse? 15. Plot the graphs of x^ and x^ on the same axes. Show how the fol- lowing facts are illustrated, (a) Wiy = x. (b) V^ = x. 16. Plot the graphs of x^ and xl on the same axes^ Show how the fol- lowing facts are illustrated, (a) (\^xy = x. (b) \/x^ = x. 17. For what value of x does x^ increase at the same rate as a:? For what value does x^ increase less rapidly? more rapidly? interpret the re- sults graphically. 18. Find the value of x for which the tangent lines to the graphs of x* and x^ are parallel. For what values of x is the graph of x^ "flatter " than that of x2? for what values is it steeper? 19. The horse power of a gas engine is sometimes determined by the equation H.P. - ihi/2.5,. where d is the diameter of a cylinder and n is the number of cylinders. I ALGEBRAIC FUNCTIONS 117 "~" "t" X ^ / * / J -: -■ - s r-J _ ~— ^ f f ^ i / / ' 1 _ Fig. 62. (a) Plot the graph for four-cyhnder engines, taking d = 2.5, 3, 3.5, 4. On the same axes, plot the graphs for six-, eight- and twelve-cylinder engines. (b) Plot the graph for d = 3, taking n = 4, 6, 8, 12. On the same axes, plot the graph if d = 2.5, or d = 4. 40. Graph of x", n<0. Graphs of Reciprocal Functions. Consider first n = - 1, the function x~^ = 1/x. Let y = l/x. Replacing a; by - x and yhy - y, and changing the signs of both sides of the equation, we obtain the given equation. Hence the graph is symmetrical with respect to the origin, and the table of values need include only positive values of X, Since the function becomes infinite as X approaches zero, the 2/-axis, x = 0, is an asymptote. Solving y = l/x for X we get X = \/y, from which it fol- lows that the x-axis, 2/ = 0, is also an asymptote. The figure shows the graph, which appears to be symmet- rical with respect to the fine y = X, the bisector of the first and third quadrants. That this is the case is seen as follows: The equation y = l/x, or xy = I, is unchanged if x and y are interchanged. Hence if (x, y) is on the graph, so also is (y, x). But these points are symmet- rical with respect to the fine y = x (Section 39), and therefore ^ihe graph is also. The graphs of x and l/x illustrate the following properties of any two reciprocal variables : j I. Reciprocal variables have the same sign. For both graphs Eire above the x-axis, or both are below, for any value of x. II. If a variable increases, the reciprocal decreases, and vice ')ersa. For as x increases, that is, as the graph of x rises, the ?raph of l/x falls; and as x decreases, the graph of l/x rises. III. If the numerical value of a variable approaches and be- comes unity, so also does that of the reciprocal. For the graphs ntersect at the points (1, 1) and (- 1, - 1). X 0, i i i, 1, 2, 3, 4, l/x », 4, 2, 1, 1, h I i 118 ELEMENTARY FUNCTIONS IV. // a variable approaches zero, its reciprocal becomes in- finite, and vice versa. For both axes are asymptotes. n = - 2, the function x~^ = 1/x^. The function 1/x^ IS called the reciprocal of the function x^ in accordance with the Definition. Two functions are said to be reciprocal if their product is unity. The following properties of the graphs of reciprocal func- tions are proved by the hke numbered facts above: la. Corresponding parts of the graphs of reciprocal functions lie on the same side of the x-axis. Ila. // the graph of a function rises, the graph of the reciprocal function falls, and vice versa. Ilia. If the graph of a function approaches a point on either of the lines y = d= 1, the graph of the reciprocal function ap- proaches the same point from the opposite side of the line. The points of intersection of the graphs lie on these lines. IVa. If the graph of a function crosses, or is tangent to, the X-axis at the point (a, 0), the line x = a is an asymptote of the graph of the reciprocal function, and vice versa. If the graph of a function has been plotted, these considera- tions enable us to sketch, roughly, the graph of the reciprocal function, without building a table of values. The general form of the graph o: 1/x^ may be obtained from that of as follows: At the extreme left of the figure, thi graph of x^ Hes above the rr-axis, and i falHng. Then the graph of l/x"^ \i above the x-axis, and is rising (by la and Ila). As x increases up to a; = — 1, the graph of x"^ falls until it reaches the line y = I, and hence the graph of l/x^ rises until it reaches this line (by Ilia). From these facts we can sketch the graph of 1/a^ from the extreme left of the figure up to a; = - 1. As x increases from - 1 to 0, the graph of x^ falls and be- V 8 4 zwi ----vM---- •'1-^r Mir Fig. 63. ALGEBRAIC FUNCTIONS 119 comes tangent to the x-axis. Hence the graph of Xjx^ rises indefinitely, and approaches the 2/-axis as an asymptote (by Ila and IVa). As X increases from to 1, the graph of x^ rises to the Une y = 1, and hence that of 1/x^ falls to this line. As x increases from 1 on, the graph of x^ rises indefinitely, and that of 1/x^ continues to fall, but remains above the x-axis. The right hand half of the figure might also be obtained from the sym- metry with respect to the y-axis. Notice that the graph of l/x^ hes between the graph of 1/x and the x-axis to the right of x = 1, while between x = and x = I, the graph of 1/x lies between it and the 2/-axis. 41. Summary of Graph of x\ The figure shows the typical form of the graph in the first quadrant for each of the cases n>l, 0l. The graph is tangent to the x-axis at the origin, and rises as it runs to the right. The larger the value of n, the flatter the graph is near the origin, and the steeper it is elsewhere. Fig. 64. 120 ELEMENTARY FUNCTIONS 0 - hh-^/x^, where r is the radius of the orifice in the outlet pipe, and h is the head, or depth of water. The head changes from instant to instant, so that the curve of whirl con- stantly changes. Assuming r = 1, plot the graph for /i = |, 1, 2. 11. Plot the graphs of the functions following, and then sketch the graphs of the reciprocal functions. Find the reciprocal function in each (a) x-3. (b) 2-x. (c) X^-1. (d) x2 - 4. (e) x/(x^ + l). (f) x2-7x + 10. (g) X' - 4x. (h) x^ - 4x. 12. What is a simple way of sketching the graph of l/(2a; + 3)? Of 3/(2x + 3)? 42. Interpolation. The process of finding, for example, the cube of such a number as 2.647, which lies between two suc- cessive nmnbers 2.64 and 2.65 whose cubes are given in the table, is called interpolation. According to Huntington^s Tables, 2.64^ = 18.40 and 2.65^ = 18.61. The difference be- tween these two values oi y = x^ is Ay = 0.21, and is called the tabular difference. In general, the tabular difference is the difference between two successive niunbers in the body of a table. For the process of interpolation it is essential that the 122 ELEMENTARY FUNCTIONS successive values of the tabular difference should be very nearly equal. The table gives one row of Huntington's table of cubes and the successive tabular differences. Since Ay changes only shghtly while Ax is always equal to 0.01, the average rate of change, Ay /Ax, is nearly constant, and the part of the graph of x^ constructed from this table would be very nearly a straight Une. Some of the successive values of Ay are equal, due to the fact that the values of x^ are approxi- mate to four figures and not exact, and hence some of the points plotted from the table would actually He on a straight line. For the process of interpolation we dssume that the part of the graph lying between two successive points is straight. To find 2.647^, consider that part of the graph lying between x = 2.64 and 2.65, as given in Figure 65. We seek the value of the ordinate EF at the point E for which X = 2.647. It may be obtained by adding GF as a correction to AB = 2M\ To find GF, we have the slope of BF is the same as that of BD, since the graph is assumed straight. Hence X 2/=a:3 Ay 2.60 17.58 .20 2.61 17.78 .20 2.62 17.98 .21 2.63 18.19 .21 2.64 18.40 .21 2.65 18.61 .21 2.66 18.82 .21 2.67 19.03 .22 2.68 19.25 .22 2.69 19.47 .21 2.70 19.68 whence (?F = GF[ BG 0.007 HD BH or GF 0.007 0.21 o.or XO.21 =0.7x0.21 =0.147 = 0.15. I 0.01 Then 2.647^ = EF = AB+GF = 18.40 + .15 = 18.55. EF may also be obtained by subtracting ID as a correction from CD = 2.653. To find ID, we have the slope of FD is equal to that of BD, since the graph is assumed straight, so that ID ^HD ID 0.21 BH' 0.003 and hence FI ID or 0.01 0.003 xO.21 0.01 0.3x0.21 =0.06. ALGEBRAIC FUNCTIONS 123 2.64 2.647 2.65 FlQ. 66. Then 2.647^ = EF = CD - ID = 18.61 - 0.06 = 18.55, as be- fore. The latter procedure is usually preferable if the figure for which we are interpolating, 7 in the illustration, is greater than 5, while the former is -^ used if it is less than, or equal to, 5. No- tice that the correc- tion GF is found by taking seven-tenths of the tabular differ- ence, HD = 0.21, and the correction ID by taking three-tenths of the tabular differ- ence. The arithmetical operations involved in interpolation may be described as follows : To find 2.64a^, where a stands for the digit for which we are interpolating, apply a-tenths of the tabular difference as a cor- rection to 2.64^ if a<5, while if a>5, apply (10 — a)-tenths of the tabular difference as a correction to 2.65^. In either case, the correction must be applied (added or subtracted) in such a way that the result lies between 2.64^ and 2.65^. In practice the decimal point is usually omitted in x, x^, and the tabular difference, and inserted in the proper place at the end of the operation. Example 1. Find 26.02'. The figure for which we must interpolate is 2, and the tabular difference is 20. Two-tenths of the tabular f difference is 4, the correction to be applied to 1758, the cube of 260. "^ i [n order to obtain a result between 1758 and 1778, the cube of 261, i: phe correction must be added, giving 1762 as the first four digits in ijl ,:he required cube. From the rule at the top of the table, we see that J6.023 = 17,620. li Example 2. Find 2.687'. Instead of applying 7-tenths of the tabular jfJifference as a correction to the cube of 2.68, we apply 3-tenths of it to fhe cube of 2.69. Three-tenths of the tabular difference is 6.6 = 7, and ' -lence 2.687' = 19.40. 124 ELEMENTARY FUNCTIONS Except for purposes of explanation, there is no need of writing anything but the desired result. All the tables in Huntington's Tables are arranged so that the numbers in the body of any table increase as we read from left to right, for the reason that a uniform procedure is obtained for applying the correction. In interpolating, the desired re- sult lies between two numbers in the same row; a correction applied to the left-hand number is always added, while a cor- rection to be applied to the right-hand ninnber is always subtracted. EXERCISES 1. Plot the graph of x^ from x = 4.50 to x = 4.60, using the values of x^ given in Huntington's Tables. 2. If we plot the graph of Vx for values of x from 2.30 to 3.40, using the square roots given in the Tables, why will the graph be nearly straight? 3. Find the values of the numbers given below, illustrating the inter- polation graphically. (a) 3.1722. (b) VOTS. (c) 6.49^ (d) ^^^0.02814. (e) 1/926. 4. Find the numbers following without illustrating the interpolation graphically. The tabular difference and the necessary number of tenths of that difference should be obtaified mentally. „ (a) 1.5342, V0.468, 8.433, ^^^:46, 1/647. (b) Squares of 2.784, 3762, 0.01388, 3846000, 0.00003728. (c) Square roots of 0.634, 3.248, 42.7, 384.3, 279,000, 0.001876. (d) Cubes of 3.143, 0.774, 1683, 0.00004592, 4889000. (e) Cube roots of 0.02258, 0.226, 19.34, 0.00176, 328000. (f) Reciprocals of 31.76, 0.00647, 35990, 0.0004325, 647. 6. Find the square of 4732, and check the result by finding its squ root. Find the cube root of 3479, and check the result by cubing it, Find the reciprocal of 25.63, and check the result. 6. FindVSTTS^, ^0.02478^, 1/73-262, 1/V2438, 1/48-363. 7. Solve the following equations, using the Tables to simplify the com putations: (a) x^ + 32x + 19-0. (b) 2x^ - A7x - 27 = 0. (c) 3x2 ^ 29x - 40 = 0. (d) 5x2 + 54x - 31 = 0. 8. How long an umbrella will go into a trunk measuring 31.5 x 18. X 22.5 inches, inside measure, (1) if the umbrella is laid on the bottom! (2) if it is placed diagonally between opposite comers of the top anl if the graph is tangent to the a:-axis at the origin, 0l, a simple value to choose for n is 2. By the above property, if the quotients obtained by dividing the values of y by the squares of the corresponding values of x differ by very little, the relation connecting x and y resembles y = kx^^ and an approximate statement of the law may be given in this form. Each quotient, y/x^, is the value of k for which the graph of y = kx^ passes through the point (x, y). A very good value to use for k is obtained by finding the average of all the quo- tients. If, however, the quotients differ widely, it is well to try n = 3, or n = I, or some other value of n, in order to see if an approximate statement of the law can be found which is better than y = kx^. If 0 height of a column of mercury in the two sides of a [/-tube. In the table, x denotes the difference in height and Q the discharge. Determine the law. 7. From the following data determine the resistance R, in ohms per d I 0.083, 0.120, 0.148, 0.220, mile, of a telegraph wire of diameter r\ 51,24.42, 16.1, 7.26, d, in inches. 8. Find the relation between the pressure p and the volume F of a gas T/ I K AQ o Kft 1 oi A on A Ao ^^ ^^^ pressure on the walls of a con- — n Ki A n^ o A^ Q AQ Q aA taming vessel for different volumes V I 0.51, 0.97, 2.04, 3.03, 3.94, - j . u • ii. . 1 1 was found to be as m the table. 9. In the following table, x denotes half the distance between two tele- X I 20, 30, 40. 50, 60, ^^"^^ P"'"'' f^^ ^ *';'= ^°'^"''' ^^ irl o.98, 2.21, 3.95, 6.19, 8.97, T'^ ^' ^* ^^^ •'"°'"'- ^""^ *'^« ' law. 10. The table gives the number of revolutions per minute, R.P.M., at h I 11.63, 62.8, 138.8, 174.5, "!;'* ^ ''<^'*1° ^°™ f 7f ' wheel runs without any load, for R.P.M. I 440, 1020, 1525, 1680, ,.«: ^ , r ^u u j i, different values of the head, h (the height of the surface of the mill pond above the bottom of the wheel). Find the relation. 11. The horse power developed by the water wheel in Exercise 10 is Ji I 69.8, 104.8,, 139.8, 174.5, given in the table. Find the law H.P. I 0.64, 1.18, 1.81, 2.56, in the form i/ = fcxi. 12. (a) Economists define supply as the amount of anything which the r> • I A ^^ 1 ^n -I ^r. i on producers will offer for sale at a given Price 0. 60, 1 . 00, 1 . 40, 1 . 80, *^ . rru .^ ui • j i. o p- — :r-^ — ^ — 5—^ — J— price. The table gives an assumed set ' ' * * of values of the supply of wheat, in mil- 130 ELEMENTARY FUNCTIONS lion bushels, at various prices in dollars. Plot the graph and determine the law. What price per bushel will stimulate the producers to furnish 5 miUion bushels? (b) The term demand is used to denote the quantity of anything which T> • 1 n ^r^ A on 1 f^f> o AA ^^^ consumers will purchase at a Price 0.40, 0.80, 1.00, 2.00, . . ^r j ^i j j ' given pnce. How does the demand fluctuate as the price rises? As the price falls? The table gives an assumed set of values of the demand for wheat, in million bushels, at various prices, in dollars. Plot the graph and determine the law. What price will stimulate the consumers to de- mand 7 miUion bushels? (c) Assuming the conditions in the preceding parts of this exercise, plot both graphs on the same axes, and find the price and the quantity of wheat transferred from producer to consumer. 13. If the apple crop amounts to 3 million bushels, then, at the harvest Price I 0.5 0, 1, 1.25,2.50, ''^°°' *" ™PP'>; '** ^7, P"'=<' ^'f K5S5^ 5, 2.5, 2, IT r"f K? ^"^ T T" "/• ihe table gives assumed values for the demand in million bushels. Plot both graphs on the same axes, and determine the equations giving the laws. At what price will apples sell under these conditions? 14. Some industries develop approximately to a point where any amount, within certain limits, can be supplied at a constant price. Sup- T, . I o .. tr ^ o 1A pose that any number of shoes will be Price 3, 4, 5, 6, 8, 10, ^ ,. , , ,^ . , ^_ . , ^Pi j-\ ' - . „ ' — z-^ supphed at the price of $5 per pair, and Demand 6, 5, 4, 3, 2, 1, ,, ^f^ xi, j 5 • -n- • ' that the demand, m milhon pairs, at a price, in dollars, is as given in the table. Plot both graphs on the same axes, and find the price and the number of shoes sold. Why cannot the price be less than $5? Assuming competition, why will it not exceed $5? (b) If a monopoly arose, with the object of making as much money as possible instead of serving the public to the best of its ability with reason- able profits, the output would be restricted to the point of maximum profits. Assuming the conditions of the preceding part of the problem, suppose that the cost of manufacturing a pair of shoes is $4. Construct a table giving the total profits at various prices, plot the graph, and determine the price and profits. 16. Plot the graphs and determine the laws for the data given in the tables. Determine the freight rate and the amount of freight handled. Price per ton mile in cents I 0.5, Q.6, 0.7, 0.8, Supply of freight service in thousand tons 14, 3, 2, 1, Price per ton mile in cents I 0.5, 0.6. 0.7, 1, Demand for freight service in thousand tons | 3, 2.6, 2.1, 1.5, ALGEBRAIC FUNCTIONS 131 45. The Linear Fractional Fxmction ax + b In order that V 1 l\ l^ \V yfe 1 1 -/ -3 -i -1 / N. ?i-| «. j;;; -^ / 6 i ^1 ' li X •^ ' / ^ -J _J Fig. 66. cx + d this function should really be fractional, it is essential that c 7^ 0, for if c = 0, it reduces to the linear fiuiction "j ^ + j The simplest Unear fractional function is \/x, obtained by- setting 6 = c, and a = rf = 0, whose graph has been considered in Section 40. If 6 ?^ c, while a = c? = 0, we have the function h/cx, or fc/x, where k = h/c. Its graph may be ob- tained by multiplying by k the ordi- nates of points on the graph of \/x (Theorem, page 89). If we set y = k/x^ whence xy = k, it is seen that the graph is symmetrical with respect to the line y = x, since the equation is unchanged if x and y are interchanged. The graph is also synametrical with respect to the origin. The geometric significance of the constant A; is obtained by solving the equations xy = k and y = x; the solutions are (V/c, Vk) and (- Vk, - Vk), which are the coordinates of the points of intersection of the graphs of the equations. The distance from the origin to the point of intersection A is found from the right triangle OAB to be V2k. Hence, for a small value of k, the graph lies close to the origin and axes, while if k is large, it lies at some distance from them. If A; is negative, the graph will lie in the second and fourth quadrants. The graph oi xy = k is called a rectangular, or equilateral, hyperbola. The former term is derived from the fact that the asjrmptotes are perpendicular. In order to determine the form of the graph of the general Hnear fractional function, set ax + h ex + d' (1) 132 ELEMENTARY FUNCTIONS Multiplying both sides by ex + b, and subtracting ax + b from both sides, cxy - ax + dy - b = 0. (2) Let us now see if we can simpUfy this equation by translat- ing the axes (Section 31, page 89). Setting X == x' + h and y = y' + k, (3) we get cix' + h) {y' + k) - a(x' + h)+ d{y' + k) - b = 0, (4) or, removing the parentheses and collecting like terms, cxY + {ck - a)x' + {ch + d)y' + chk - ah + dk - b = 0. (5) Equating to zero the coefficients of x' and y\ we get ck - a = and ch + d = 0. (6) Solving for h and k, which is always possible since c ?^ 0, h = - d/c, k = a/c. (7) Substituting these values in (5) we get ex v-(-9©-«(-^)-^'^^^=o. Simplifying, subtracting the constant term from both ' sides and dividing by c, , , be — ad ,„. ^y = — ^-- (8) This equation, referred to the new axes, has the same graph as (1). It is of the form x'y' = k, where fc = (6c - ad)/e^, or y' = k/x'. The graph is therefore a rectangular hyperbola. Hence we have the Theorem. The graph of a linear fractional function (1), or of an equation of the form (2), is a rectangular hyperbola whose asymptotes are parallel to the axes of coordinates, EXERCISES 1. Plot the graph of ?/ = {2x - 4)/(a; + 3), (a) by translating the axes; (b) by finding the asymptotes (by the method on page 27), plotting one branch of the curve from a table of values, and the other by means of the symmetry with respect to the point of intersection of the asymptotes. ALGEBRAIC FUNCTIONS 133 2. Plot the graphs of the following equations: a; + 4 ., ^ 3x + 6 , . x - 4 (a) 2/ = ^35- (b) 2/ = ^-3^- (0 y = ^^^g- (d) x!/ - 2x + 42/ - 8 = 0. (e) 2a;y - X + 3?/ + 8 = 0. 3. The linear fractional function may be written in the form a{x-{-'b/a)/c{x-\- d/c). Prove that the linear factors cancel, and the function is a constant, if and only if be - ad = 0. For this reason, it is always assumed that he - ad 5^ 0. 4. The relation between the radius of curvature of a concave mirror, R, the distance from the center of the mirror to the object, x, and the dis- tance from the center to the reflected image, y, is 211 ft R i^y If JR = 2 feet, construct the graph of the equation. 46. Integral Rational Functions. The general form of an integral rational function or polynomial is (see page 39) P^ Six) = oox" + aia;"-i + . . . -H an-ix + a„, where n is an integer, and ao, ai, . . . a„ are constants, posi- tive, negative or zero, except that ao 7^ 0. It is said to be of • degree n. We have already studied polynomials of the first and second degree in considering linear and quadratic functions. The calculation of a table of values for a polynomial of higher degree than the second by direct substitution presents no new features, but because of the greater number of terms that may be present there is more labor involved. An alterna- tive method which is simpler, in general, than that of direct substitution is developed in the next sections. 47. The Remainder Theorem. Example. Let us divide the polynomial 2x^ -6x^ + 11 by x - 2, ar- ranging the work as usual, 2rr3 - 6x2 ^ H \x -2 2a^-4a;2 |2x2 - 2x - 4 ' -2x2 . - 2x2 ^ 4a ; — 4x -4x + 8 + 3 134 ELEMENTARY FUNCTIONS The result may be put in the form ^-^^^±11 = 2.-2.-4.^, (X) where 2a:2 - 2a: - 4 is the quotient, denoted by q{x), and the remainder is Equation (1) is an identity in accordance with the Definition. An identity is an equation which is satisfied by all values of the variable or variables. Substituting 2 for a; in the polynomial in the example, /(2) = 2.23 _ 6 22 + 11 = 16 - 24 + 11 = 3. Hence the remainder, i2 = 3, obtained by dividing the poly- nomial by x — 2 is equal to the value of the polynomial when 2 is substituted for x. This result is a verification of a theorem which is proved for any polynomial as follows : Remainder Theorem. If a polynomial f(x) is divided hy X — a, the remainder is f{a) . If f{x) be divided by x — a, the quotient q{x) is a poly- nomial of degree one less than that oi f(x), and the remainder 22 is a constant. The result of the division may be expressed in the form X - a X - a ^ Multiplying both members by a: - a, we obtain the identity fix) = (x-a) q(x) + R. (2) Substituting a for x, we have f{a) = (a - a)q{a) + R. (3) Since a - a = 0, we have (a - a)q{a) = 0. Hence /(a) = R. (4) 48. Synthetic Division. The form of the division in the example of the preceding section is too cumbersome to be of practical use in calculating the table of values of a polynomial. The division may be simplified as follows: ALGEBRAIC FUNCTIONS 135 We note that the operations involved in the division were performed on the coefficients of the dividend, divisor and quotient, the powers of x serving merely to determine the proper positions of these coefficients in carrying out the details of the division. If missing terms are suppHed by zero coefficients, the division can be performed by the so-called method of de- tached coefficients as follows: 2-6 + 0+11 2-4 -2 -2 + 4 1 2-2-4 -4 -4+ 8 + 3 The first term in each partial product is canceled in the ibtraction and hence may be omitted. The second term of Lch partial product may be written immediately under the [term in the dividend from which it is subtracted. The first jrm in the divisor is not now needed, so that the process is iher condensed as follows: 2* - 6 +0 +11 I -2 -4+4-8 2-2-4 _ 2* - 4* + 3 Noting that the coefficients of the quotient are identical ith the numbers marked with an asterisk (*), it is unnecessary write them under the divisor. As we shall use this process to find the value of /(2), it is convenient to replace — 2 in the divisor by + 2 and add throughout instead of subtracting. Also bringing down the first coefficient of the dividend, the work is now arranged in the form 2-6 + + 11 I 2 +4-4- 8 2-2-4+ 3 The numbers in the last row are called partial remainders, le last being the remainder 3 (the value of the polynomial 136 ELEMENTARY FUNCTIONS when 2 is substituted for x). The preceding partial remainders 2, - 2, - 4 are the coefficients of the quotient, the first being the same as the first coefficient of the dividend. The process of synthetic division may be described as follows: To divide a polynomial by x — a, Write the coefficients in the order of descending powers of x, supplying missing terms with zero coefficients. To the right write a. Bring down the first coefficient of the dividend. Multiply the first coefficient by a, and add the product to the second coefficient. Multiply the sum so obtained by a, and add the result to the third coefficient, etc. The last term will be the remainder, R, and the preceding partial remainders will be the coefficients of the quotient, q{x), which will be of one degree less than the degree of the polynomial. The remainder theorem, with the aid of synthetic division, affords a simple method of calculating the value of a poly- nomial for a given value of x. For example, if fix) = 3x* - 9x^ - 4^2 - 17a; - 37, find /(4). By the remainder theorem, /(4) 3_ 9_ 4_i7_37i4 is the value of the remainder, R, when + 12 + 12 + 32 + 60 fix) is divided by x - 4. By synthetic 3+3+ 8 + 15 +"23 division we find that /(4) = 23. Verify this result by substitut- ing 4 for a; in the given polynomial. If the multiplications and additions are performed mentally, the partial products in the second row of numbers omitted, and the partial remainders alone put down, the computation may be arranged still more compactly in the following form: The second arrangement is more convenient in calculating a table of values of a polynomial for integral values of x, while we shall find that the first is better adapted to finding zeros of the function if the zeros are irrational. 49. Graph of a Pol3momial. Example. Construct the graph of fix) = x3 - 3x2 _ 10a; + 24. 3_9_4-17-37 3+3+8 + 15+23 ALGEBRAIC FUNCTIONS 137 The table of values is constructed by means of synthetic division and the remainder theorem, the details of the work and the graph being given below. 1 _ 3 _ 10 + 24 + 1 - 2-12 1-2-12 + 12 1 1 - 3 - 10 4- 24 f 2 ■1.2- 2-24 1 _ 1 _ 12 + Hence we find that /(I) = 12, and/(2) = 0. The calculations may be arranged compactly by writing the coefficients of the polynomial once for all as in the top row of the following table and by operating on this top row with each value of X, performing the multiplications and additions men- tally and entering the partial remainders on the same row with the value of x, under the proper coefficient of the dividend. X fix) 1-3-10 24 1 -2-12 12 2 - 1 - 12 3 0-10 - 6 4 + 1-6 5 + 2 24 6 + 3+ 8 72 -1 -4- 6 30 -2 -5 24 -3 -6+ 8 -4 -7 + 18 -48 Fig. 67. 60. Extent of the Tables. In the example of the preceding section an examination of the synthetic division by x — 6 shows that the successive remainders 1, +3, +8, +72 are aU positive. If a larger value of x be used, by the nature of syn- thetic division, it is evident that the successive remainders will be positive and larger. Hence as x increases beyond 6 f{x) will also increase and the curve will run up to the right indefinitely. The synthetic division by a; + 4 gave remainders 1, - 7, + 18, - 48 which have alternating signs. As x de- 138 ELEMENTARY FUNCTIONS creases below - 4, f{x) also decreases and hence the curve runs down indefinitely to the left. The extreme values of x that should be included in the table are determined by the following rule: The largest value included in the table is such that in the syn- thetic division the signs of the partial remainders are all the same. The smallest value {algebraically) included is such that these signs alternate. EXERCISES 1. Plot the graphs of each of the polynomials below. From the graph find approximate values of the roots of the equation obtained by equating the polynomial to zero. (a) x2-2x + 7. (b) X* - 3x3 + 2x2. (c) x3^3a.2 _ jq^. _ 24. (d) 2x3 - 8x2 - 12a; + 17. (e) x^ + 3x2 + 3^; + 1. (f) x^ -Sx*-x + 3. 2. The graphs of the functions x^ - x2, 2x^ - 3x2 ^ ^^ 2x^ - x"^ - x pass, through the origin and the point (1, 0). Find the slope of the tangent line to each graph at these points, and plot the graphs. How does the slope of the tangent line assist in drawing the graphs? 3. Plot the graph of x^ - 4x3 4- 4^2^ ^n^j ghow that it is tangent to the X-axis at two points. 4. A body moves so that its position, s, with reference to a fixed point at any time t, is given by one of the following equations. Find the velocity, V, and the acceleration, a, at any time t. Construct the graphs of s, v, and a on the same set of axes. When and where is the body at rest? What is the value of the acceleration when the velocity is a minimum? Where is the body at this moment and what is its velocity? (a) s = f3 - Zt. (b) 5 = <3 4- 3^2. (c) s = <3 _ 6^2 + 9f. (d) s = <3 + 3<2 _ ^ _3^ Suggestion. To find the acceleration, a, at any time t, find the limit of the average rate of change of v with respect to t. Note. Rate of change of a polynomial. It was seen on page 111 that the rate of change of x" is nx""-^, and it may be shown that the rate of change of ax'' is anx""'^. Thus the rate of change of ^ is 12.^2, and that of - 2x* is - Sx^. It will be show in a later chapter that the rate of change of a pol5aiomial, or the slope of the line tangent to the graph, may be obtained by adding the rates of change of the successive terms, noting that the rate of change of the constant term is zero. ALGEBRAIC FUNCTIONS 139 For example, the rate of change m of the polynomial 2/ = ^ - 3a;^ + 4a;2 - 5a; + 7 is 7W = 4x3 _ 9^2 _|. 83. _ 5^ 6. Find the slope m of a line tangent at any point to the graph of each of the following equations. Find the rate of change of the slope m with respect to z. Symbolize, the rate of change of m with respect to a; by F and plot the graphs of y, m and F on the same axes. Find the ordinate of the point on the graph of y, and also of the point on the graph of F, which has the same abscissa as the minimum point of m. Where is the line tangent to the graph of y horizontal? (a) 2/ = x3 - 6x2 + 9x - 2. (b) 2/ = x' - 3x2 - 9x + 4. — (c) 2/ = 4x3 - 15x2 ^ i2x. (d) 2/ = x' - 12x + 5. m Note. The point on a graph where m ceases to decrease and begins to increase (or vice-versa) is called a 'point of in- flection. The graph is concave downward on one side of this point and concave upward on the other side. For this value of X the value of F is zero. Hence, to find the abscissas of the points of inflection on the graph of a function^ equate F to zero and solve for x. Substitute these values of x in the equation y = f(x) to find the ordinates, 6. Find the coordinates of the point of inflection of the graph of each of the following equations. Translate the axes to this point as a new origin. Show that the graph is symmetric with respect to the point of inflection, and plot the graph of the equation on the new axes. (a) 2/ = a;' - 3x2 - 6x + 6, (b) 2/ = x3 - 6x2 + i2x - 11. R (c) 2/ = a:^ + 6x2 + i2x + 8. (d) 2/ = - a:^ + 3x2 + X - 6. ' 7. Find m and F for the function ax^ + 6x2 + ex + d. jf ^j^g g^xes are ' translated to the point of inflection of the graph of this function, show that the equation of the graph referred to the new origin is y' = ox'^ + m'x', where m' is the slope of the line tangent to the graph of y at the point of inflection, and that the graph is symmetrical with respect to the point of inflection. 8. Given s = t^ - it, find (a) the average velocity from ^ = 2 to < = 4, (b) the average of the velocities at « = 2, and t = 4, (c) the velocity at t = 3, (d) the value of t at which the velocity is equal to the average velocity for the interval t = 2 to t = 4. 9. Find the points of inflection of the following functions and plot their graphs (a) x4 - 8x» + 3x2. (b) 3.6 _ ioa.2 _ g, (c) x^ - 4x». 140 ELEMENTARY FUNCTIONS 10. For what positive integral values of n does x" have a point of in- flection? 61. Solution of Equations. Rational Roots. Any algebraic equation in one variable may be reduced to the form f{x) = 0, ^ (1) where f{x) is a polynomial, by the elementary rules for sim- plifying an equation involving fractions and radicals. The real roots of the equation (or zeros of the function) are repre- sented graphically by the intercepts on the a;-axis of the graph of /(a:). (See page 23.) If any integers are roots of (1), they will appear when the table of values for the graph is being constructed. Thus the integral roots of the equation ^3 _ 3^2 _ lOa; + 24 = obtained by equating to zero the polynomial in the example of Section 49 are 2, 4, and — 3, since the table of values shows that /(2)=0, /(4)=0, and /(- 3) = 0. That these are all the roots follows from Theorem 1. An equation of degree n has n roots, of which some may he imaginary and some equal to each other. This theorem is assumed without proof. Returning to equation (1), any real roots other than integers are either fractions or irrational numbers. In this section we shall see how to find the fractional roots of an equation, which with the integral roots, constitute the rational roots. Theorem 2. If x = a is a root of an equation of degree n, f(x) = 0, thenf{x) = (x - a)q{x), where q{x) is of degree n - 1. By the definition of a root of an equation f(a) =■ 0. Then by the remainder theorem, if f(x) is divided by x - a, the re- mainder is zero, so that f(x) is exactly divisible by x - a. Since the degree of the divisor is 1, the quotient q(x) will be a polynomial of degree n - 1. Since the dividend is equal to the product of the divisor and quotient, we have the identity fix) - (X - a)q(x), (2) ALGEBRAIC FUNCTIONS 141 Corollary 1. Any root of the given eqvMion f{x) = 0, except X = a, is also a root of the equation q{x) = 0. Let X = h be. any root of f(x) = different from x = a; then fib) = 0. Substituting a; = 6 in (2) = (6 - a)qih). But if the product of two numbers is zero, one of the numbers is zero, and since 6 - a is not zero, we have q(b) = 0, and hence a; = 6 is a root of the equation q{x) = 0. Corollary 2. Any root of the equation q{x) = is a root of ike equation f{x) = 0. If a; = c is a root of q{x) = 0, then q{c) = 0. Setting x = c in (2) we have /(c) = (c - a)q(c) = (c - a)-0 = 0. Hence a; = c is a root of the equation fix) =0. This last theorem and its corollaries simplify the solution of many equations hj enabUng us, as soon as a rational root is known, to replace the given iequation by one of lower degree. Example 1. Solve the equation f(x) = 12x3 - 11x2 _ 13a. + 10 = 0. The graph shows that the three roots are all real, and the table of values shows that one root is - 1. Divid- ing fix) by X - (- 1) = X + 1, we have from the table the quotient q{x) = 12x2 _ 23^ + 10. u y .0 ,» / . . / \. J/ i 1 ^ X , / / / / 30 Fig. 68. 1 2 1 S.2 12 - 11 - 13 + 1-12 + 13 + 13 -23 + 10 -35 + 57 Kx) + 10 - 2 + 36 -104 By Corollary 2, the roots of the equation q{x) = 12x2 - 23x + 10 = are also roots of /(x) = 0. Factoring q{x), we have (3x - 2) (4x - 5) = 0. Hence 3x - 2 = and 4x - 5 = 0, and the roots are therefore X - f and X = |. The roots of the given equation are therefore x - ~ 1, f , and |. 142 ELEMENTARY FUNCTIONS We note that the denominators of the fractional roots in Example 1 are both factors of the coefficient of the term of highest degree, and that the nmnerators are factors of the con- stant term. This conclusion is a special instance of the fol- lowing theorem which we assmne without proof. Theorem 3. If a/h is a root of an equation with integral coefficients, then h is a factor of the coefficient of the highest power of X and a is a factor of the constant term (or of the coefficient of the lowest power of x if the constant term is zero). Corollary. If the coeffixnent of the highest powers of x is unity and the other coefficients are integers j then the only rational roots are integers. Fractional roots may be determined by the use of this theorem, synthetic division and the remainder theorem as in Example 2. Solve the equation 3a;* - 5x^ - 2^^ + 16x + 40 = 0. The table of values and the graph are given below. X fix) 3 _ 5-24 + 16 + 40 1 - 2-26-10 + 30 2 + 1-22-28 - 16 3 + 4-12-20 - 20 4 + 7+ 4 + 32 + 168 -1 - 8-16 + 32 + 8 -2 - 11 - 2 + 20 -3 -14 + 18-38 + 154 u m 150 1 100 — [^ ^ V /. -- 1 \ [ y X ^ Fig. 69. One root is 2. Divid- The graph shows that there are four real roots. ing fix) by a; + 2, we have from the table the quotient qix) =3x3 -11x2 -2a; + 20. By Corollary 2, the roots of the equation qix) =0 are the other roots of fix) = 0. One root Hes between 1 and 2. A fraction between 1 and 2 whose numerator is a factor of 20 and whose denominator is a factor of 3 is |. Dividing the polynomial by x - f, it is I " 3 - 11 - 2 + 20 seen that x = f is a root and that the quotient is f i - 6 - 12 3x2 _ 6x - 12. Equating this function to zero, dividing by 3, and solving by the quadratic formula we have _ 2 V4 + 16 = 1 =fcV5 = 1 =*= 2.236 Hence the roots of the given equation are x = 3.236 or - 2, 1, 3.236, 1.236. - 1.236. ALGEBRAIC FUNCTIONS 143 Example 3. Solve the equation Qx^ - 16x4 + ii^z ^ 37a;2 _ 116a; + 60 = 0. (3) X m 1 2 3 - 1 -2 6 - 16 + 11 + 37 - 116 -10+ 1 + 38- 78 _ 4 + 3 + 43 - 30 + 2 + 17 + 88 + 148 -22 + 33+ 4-120 - 28 + 67 - 97 + 78 + 60 - 16 + 504 + 180 - 96 f 6 - 4 + 3 + 43 + + 3 + 45 -30 _ 3 2 + + 1 + 15 - 3 + li y h\ 160 1 m j UP \ i ' \ / 1 \ ^ -• ? -2 - w X M Fig. 70. An examination of the table of values of the function j{x) shows that a; = 2 is a root, and that the quotient obtained by dividing /(x) by a; — 2 is q{x) = 6x* - 4d;3 + 3x2 + 433. _ 30. By Corollary 2 all the roots of q{x) =0 are identical with the remaining )ts of the given equation. The graph shows that one of these is between and 1. If this root is rational, it must be one of the fractions \, \, f i (Theorem 3). From the graph it appears that the most probable of these values is f and we therefore test it first. As shown in the table, if q{x) is divided by x - f the remainder is zero, and therefore x = f is a root of f(x) = and hence of (3). Applying Corollary 1 to the equation q{x) = 0, it is seen that the re- laining roots of this equation, and hence of the original equation, are Identical with the roots of 6x3 4- 3a; + 45 = 0, )r the simpler form 2x' + x + 15 = 0. (4) The graph shows that there must be a real root of (3), and hence of (4), between — 1 and - 2. Theorem 1 shows that the only possibility of a rational root between — 1 and — 2 is — f. Dividing 2x' + x + 15 by x + f :Kleads to a remainder that is not zero and hence x = - | is not a root. The division of the left member of (4) by x + | in the table stopped with e appearance of a fraction among the partial remainders. This is suf- ficient to show that the final partial remainder cannot be zero. For if a partial remainder is prime to the denominator of a fractional divisor in lowest terms, it will be prime to the powers of that denominator which ppear in the further multiplication by the divisor. Hence the remaining I 144 ELEMENTARY FUNCTIONS partial products must be fractions and the last integral coefl&cient cannot be canceled by a fraction. The rational roots of (3) are therefore x - 2 and |, and the irrational root is, from the graph, nearly - 1.8. The other two roots (Theorem 1) are imaginary. EXERCISES 1. Find all the roots of the equations below, in each case constructing the graph of the polynomial on the left. (a) :j? - a;2 - 7x + 3 = 0. (b) a;' + 5x2 + a; _ 12 = o. (c) x^ - 4a:3 + &c - 3 = 0. (d) 7^ - 2x^ ■{- 2x^ - x - 6 = 0. (e) 6x3 _ Yj^i + 7a; _ 5^0. (f) Zx^ + 4x2 - 21x + 10 = 0. (g) 6x' + 20x2 + X - 20 = 0. (h) 4x4 + 4x3 - 5x2 - 9x - 9 = 0. (i) 6x4 _ 8x3 - 5x2 - 4x - 4 = 0. (j) q^ + 23x3 - 7x2 _ n^; ^ 4 ^ q 2. Find all the rational roots of the equations below, in each case con- structing the graph of the polynomial on the left, and estimating the ir- rational roots, if any, from the graph. (a) 2x* - x3 + 4x2 + 24x - 13 = 0. (b) 3x4 + 2x3 + 3a;2 _ iQx - 14 = 0. | (c) 5x5 _ 7a;4 _ a;3 _ 12x2 + ^j + 6 = 0. J (d) 2x5 + 5x4 - 6a;3 _ n^ja _ 3x + 9 = 0. (e) 6x3 _ i7a;2 + 7x - 5 = 0. (f) 3x4 ^ 10x3 + 7x2 - 3x - 7 = 0. Using synthetic division, find the irrational root to the nearest tenth of a unit. 3. Show that the graph of a polynomial of degree n cannot have more than n - 1 maximum and minimum points, and that it cannot have more than n — 2 points of inflection. 52. Translation of the y-Axis. For the purpose of determin- ing approximately an irrational root of an equation f{x) = 0, where f{x) is a polynomial, a method is used which reduces the roots of an equation by a constant amount c. A translation of the 2/-axis along the x-axis in the positive direction diminishes the intercepts on the a;-axis of the graph of f{x), and hence such a translation represents graphically a diminution of the roots of an equation. We shall consider the method first from a graphical point of view. ALGEBRAIC FUNCTIONS 145 Example. Construct the graph of 2/ = a;3 - 9x2 + 23a; _ 15 (1) and find the function whose graph is the same curve referred to a new axis 2 units to the right of the old one. The graph is constructed by the method of Section 49. The equations for translating the axes are (Sec- tion 31, page 89) a: = x' + 2, y = y'- Substituting in (1) we have y' = (x' + 2)3 - 9(x' + 2)2 4- 23(x' + 2) - 15 (2) - x'3 + 6x'2 + 12x' + 8 -9x'2-36x' -36 + 23x' + 46 or -15 y' = x'3 - 3x'2 - x' + 3, (3) which is the required polynomial. If /(x) represents the function in (1), then f{x' + 2) will represent the function in (2). The intercepts on the x-axis of the graph referred to the new origin are 2 units less than the intercepts referred to the old origin. Hence the roots of the equation f{x' + 2) = 0, or x'3 - 3x'2 - X + 3 = are 2 units less than the roots of the equation /(x) = or x3 - 9x2 + 23x- 15 = 0. We shall now derive a method of obtaining the coefficients of (3) which is simpler in general than the preceding method. V u' 16 \ i 5 / 1 ( ■1 V/ S 6 xl . \ 1/ ■S -IS Fig. 71. Let fix)* = OoX** + oix^-i + + an_2X2 + aw_iX + On. (4) Substituting x = x' + c we have /(x' + c) = ao(x' + c)** + ai(x' + c)"-* + . . . + a„_2(x' + c)2 + a„_i(x' + c)-\-an (6) Expanding the terms on the right by the binomial theorem, collecting like powers of x', and representing the coefficients of the powers of x' by &o, hi, hi, . . . hn, \we have f{x' + c) = feox'" + hx'""-^ + • • • + 6,^2x'2 + hn-ix' + &«. (6) For any value of x and x' connected by the relation x = x' + c, or X - c = x', (7) 146 ELEMENTARY FUNCTIONS we have f(x) =fix' + c). (8) Dividing (8) by (7) -^ = ^-^^^^' (9) X — C X If fix' + c) = box'" + bia;'"-! + . • . + 6n-2x'2 + 6n_ix' + 6„ be divided by x', as indicated on the right of (9), the remainder is 6„ and the quotient is qi{x') = 6oa:''»-i + bix'''-^ + . . . + bn-2x' + bn-i. If the quotient qi{x') is divided by x', the remainder is ?)r_i and the quotient is q2{x') = box'"*-^ + bix'""-^ + • • • + fc«-2. Continuing this process it is seen that the successive remainders are the coefl&cients of (6) in the order bn, 6n-i, 6n-2, . . . 5o. But if the indicated divisions be performed on both sides of (9), if the respective quotients thus obtained be divided by x - c and x', and so on, the successive remainders obtained on the left side of (9) will equal those obtained on the right, namely, bn, 6n-i, bn-2, . . . bo. Hence we have the Theorem. The coefficients of f{x' + c) may he obtained by dividing J{x) by x — c, the quotient by x — c, etc. 1-9 +23-15 |_2 The computation of the coefficients + 2—14+18 in the example may be effected by means of this theorem and synthetic division in compact form as indicated. The successive remainders in the divi- sions, marked with an asterisk, are 1* - 3* the coefficients of the function in (3). EXERCISES 1. Construct the graph of each of the functions below, and find the function having the same graph referred to a new 2/-axis c units to the right. Verify the result by direct substitution of a;' + c for x. (a) x' - 3x2 + a: - 2, c = 3. (c) x* + Sx^ - 7x^ + ISx - 5, c = 2. (b) 2x3 _ a;2 + 3, c = 1. (d) 3x^ - Sa^ - 9x - 12, c = 1. 2. Show that the second term of the function y = x^ - Qx^ + 7x + 4i will be removed if the y-axis is translated 2 units to the right. Deduce a rule for removing the second term of ax^ + bx^ + cx + d. 3. Determine a translation of the 2/-axis so that the graph ofx^ + Sx^-4: in the old system will be the graph of x' - 3x + 2 in the new system. Plot the graphs of x' and 3x - 2 on the same axes and from them determine approximately the roots of the equation x' - 3x + 2 = 0, and hence of the equation x' + Sx^ - 4 «• 0. 1-7 + 2 + 9 10 + 3* 1 -5 + 2 — 1* ALGEBRAIC FUNCTIONS 147 4. Plot the graph of x^ - 2x3 - Sx^ + 4x + 2 ^^j^^j determine the co- ordinates of the maximum and minimum points and of the points of in- flection. Find the function which has the same graph referred to a ?/-axis 1 unit to the left. 53. Homer's Method of Solution of Equations. This method enables us to compute irrational roots as acciu'ately as may be desired. j Example. Find, correct to two decimal places, the real root of the equation fix) = x' + X - 47 = 0. (1) Plotting the graph of /(x) we get the curve in the figure, which shows that there is a real root between 3 and 4. As the coeflScients of /(x) are in- tegers, and that of x^ is unity, this root is not frac- tional, and it therefore must be irrational. Now move the y-sods 3 units to the right. 1 + 0+1-47 |3 3+ 9+30 3 + 10-17 3 + 18 6 + 28 3 The new equation, omitting the primes on the x's, is fix +3) = x' + 9x2 + 28x - 17 = 0. (2) No confusion should arise from omitting the primes Fig. 72. if it be remembered that the graph of the poljniomial in (2) is the curve in the figm-e referred to the axes O'X and O'Y'. Equation (2) has a root between and 1, which from the graph appears to be about 0.5. Dividing (2) by x - 0.5, we have 1 + 9 +28 -17 [0.5 0.5+ 4.75 + 16.375 9.5 + 32.75 - 0.625 As the remainder is negative, the graph lies below the x-axis at x = 0.5 and hence, from the figm-e, the root is larger than 0.5. Try x = 0.6. 1 + 9 +28 -17 10.6 + 0.6+ 5.76 + 20.256 + 9.6 + 33.76+ 3.256 The remainder is now positive and hence the graph lies above the x-axis at X = 0.6. Therefore the root of (2) lies between 0.5 and 0.6. " y V f 1 -. o -e '^'1 a' X 1 ( i - -30 •36 / / A > V " ■^ 148 ELEMENTARY FUNCTIONS Move the ^-axis 0.5 to the right. 1 + 9+28 -17 10.5 0.5+ 4.75+ 16.375 9.5 + 32.75 - 0.625 0.5+ 5 10 +37.75 0.5 10.5 The new equation is a^ + 10. 5x2 ^ 37 75^. _ 625 = 0. (3) The graph of the polynomial on the left is the same curve referred to the axes 0"X and 0"Y", and it shows that equation (3) has a root between and 0.1. As the square and cube of a number less than 0.1 are very small, an approximate value of the root may be obtained by neglecting x' and x^ in (3) and solving the resulting linear equation 37.75X- 0.625 = 0. This gives x = „_' = 0.02 approximately. Substitute 0.02 in the left member of (3) by dividing by a; - 0.02. 1 + 10.5 +37.75 -0.625 [0.02 + 0.02+ .2104 + 0.759208 + 10. 52 + 37. 9604 + 0. 134208 As the remainder is positive, the graph shows that the root is less than 0.02. Try 0.01. 1 + 10.5 +37.75 -0.625 | 0.01 + 0.01+ 0.1051 + 0.378551 + 10. 51 + 37. 8551 - 0. 246449 This remainder is negative, and hence the graph shows that the root is greater than 0.01. Hence the root of (3) hes between 0.01 and 0.02. Then the root of equation (2) lies between 0.51 and 0.52, and that of (1) between 3.51 and 3.52. Hence the real root of (1), correct to two deci- mal places, is a; = 3.51. Which is the closer approximation to the root, 3.51 or 3.52? Why? EXERCISES Find all the real roots of the equations following, obtaining irrational roots to two decimal places. 1. a;' + 2x - 17 = 0. 6. x' - 4^2 + 3x + 3 = 0. 2. x' + 3x - 31 = 0. 6. X* + x« - 15x + 2 = 0. 3. 2x' + X - 37 = 0. 7. x« - 5x« + 2x2 + 1 = 0. 4. x» - 6x2 + 8x + 1 - 0. ALGEBRAIC FUNCTIONS 149 Note. To find negative roots by Horner's method, replace X in the equation by - x. The graph of /( - x) is symmetrical to that oi f{x) with respect to the y-sods, and the roots of the equation f{—x) =0 will be equal numerically to those of f(x) = 0, but have opposite signs. Hence the negative roots of an equation f(x) = may be found by finding the positive roots of f(-x)= 0, and changing their signs. 8. a;3 ^. 2x -f 23 = 0. 9. a;' + x2 + 7 = 0. 10. x3 - x2 - 6x + 1 = 0. 11. x4 _ 32.3 _ 4^.2 + i2x - 10 = 0. 12. x^ - 2a:' - 7 = 0. Note. If an equation has both rational and irrational roots, it is advisable to find the rational roots first. Suppose they are a, fi, 7, etc. Then divide the equation by x — a, the re- sulting equation by x — j3, the new equation by x — 7, etc., thus obtaining a simpler equation whose irrational roots are the same as those of the given equation, and then solve this simpler equation by Horner's method. 13. a:* - 3a;3 + a:2 + 1 = 0. 14. x^ + 2x3 + 3a;2 _ 43a. _ 93 = 0. 15. 3x5 - x* + 4a;3 _ i6a.2 _ 333. + 13 = Q. 16. 6x* - 31x3 ^ 40x2 + 2x - 5 = 0. 17. x^ - 3x3 - 4x2 + 14x - 6 = q. 18. 6x5 ^ 17^ + 4x' - 39x2 - 297x + 210 = 0. 19. Find the fifth root of 279. 20. A cast iron rectangular girder (breadth = \ depth) rests upon supports 12 feet apart and carries a weight of 2000 pounds at the center. In order that the intensity of the stress may nowhere exceed 4,000 pounds per square inch, it is determined that the depth d of the girder in inches must satisfy the equation SOd' — 81^2 — 17,280 = 0. Find d and the cross- sectional area. 21. The depth of flotation of a buoy in the form of a sphere is given by the equation x^ - 3rx2 + 4r3s = 0, where r is the radius and s is the specific gravity of the material. What is the depth for such a buoy whose radius is 1 foot and specific gravity is 0.786? 22. The cross section of the retaining wall of a reservoir is designed as indicated in Fig. 73. The allowable height x of the upper portion is given by the equation x' 4- 32x2 _ qq^ - 88 = 0, where x is expressed in terms of a unit of 10 feet. Find the allowable height to three significant ires. 150 ELEMENTARY FUNCTIONS 23. The allowable height of the lower portion of the wall in Exercise 22 is given by the equation y^ - 2.7 y^ + 18.4i/2 _ I23i/ - 1.46 = 0, where y is in terms of a unit of 10 feet. Find the height of the lower portion to three significant figures. 24. In Exercise 22 the slope of the water front BC to the vertical is jV, of J^E is -^^, of AE is xV^. The width of the top is 6 feet. What must be the width of the lower portion? 25. The load P, concentrated at the center, which a homogeneous elliptical plate can support is given by the formula 3 + 2w2 + 3m4 Fig. 73. 8m 8 + 4m2 + 3m* h'^R, where h is the thickness of the plate = ^q \tl., R is the maximum safe unit stress for the material = 16,000 pounds per square inch, P = 6007r, and m is the ratio of the breadth to the length of the elhpse. Find the value of m. 26. The maximum stresses on a parabolic arch of a bridge are given by the roots of the equation 2r^ - 5r* + Or'^ + 8r + 2 = 0, where r is the ratio of the length of the arch occupied by a moving load such as a train. Find r. 54. Graph of the Function f{ax). Before proceeding to the summary in the next section, we shall see how the graph of f{ax) may be found from that of fix). This will complete the study we shall make of pairs of re- lated functions and their graphs. Con- sider the Example. If f{x) =x^ - 6x, then j{2x) = (2x)2 - 6(2x). Show that the graph of /(2a;) may be obtained by bisecting the ab- scissas of points on the graph off(x). If we substitute 4 for x in the first func- tion, and half of 4, namely 2, in the second, the results are both equal to - 8. Hence the point (4, -8) lies on the first graph and (2, - 8) on the second, and the abscissa of the latter point is half that of the former. If we substitute any value for x in the first function, and half that value in the second, the results will be the same, namely x^ - 6x. Hence if (xi, yi) is a point on the first graph the point (a;i/2, t/i) will be on the second. Fig. 74. T 1 f f ' - / / \s / WT / ' 1 oi 3 « \ ' 'X u *1\ / 1 / I n / / \ / / J[ \ \ / / ' \ / \) \/ / ALGEBRAIC FUNCTIONS 151 As the latter point may be obtained by bisecting the abscissa of the former, it follows that the graph of /(2x) may be obtained by bisecting the abscissas of several points on the graph of /(x) and drawing a smooth curve through them. The reasoning employed in this example may be used to prove the Theorem. The graph of f(ax) may be obtained by dividing by a the abscissas of points on the graph of f{x). Corresponding paints on the two graphs lie on the same or opposite sides of the ., y-Qjxis according as a is positive or negative. This theorem may also be established as follows: Let y = f(x) and suppose that the result of solving this equation for X is X = (j)(y). The graphs of these two equations are identical. If we solve the equation y = f{ax) for ax the re- sult must be ax = {y), from which it follows that the abscissas of points on this curve are one-ath of the abscissas of the points on the graph of (l>{y). This theorem will find appUcation in the study of some of the transcendental functions. It is also applied in an alter- native method of finding the rational zeros of a polynomial in Exercises 2 to 6 below. If a has such a value as f , the division of the abscissas by J amounts to multiplication by 3. 55. Related Functions and their Graphs. We have studied several functions which may be obtained by transforming, or changing, a given function, and whose graphs may be obtained from that of the given function by simple geometric construc- tions or transformations. These are given in the tables below. In order that the statements may be concise and accurate, it is assumed that the constants involved in the table are positive. The changes necessary if the constants are negative should cause no difficulty. 152 ELEMENTARY FUNCTIONS The graph of may be obtained from the graph of f{x) by (1) fix) + k moving it up k units (page 19, Exercise 3) (2) fix + h) moving it to the left h units (Theorem 2, page 92) (3) am multiplying ordinates by a (Theorem, page 89) (4) Kox) dividing abscissas by a (Theorem, page 151) (5) 1 by the principles for the graphs of reciprocal functions Six) in Section 40 (page 117). The graph of (3a) -Kx) (4a) fi-x) (4b) -f(-x) (6) inverse of fix) is symmetrical to the graph of fix) with respect to the X-axis the i/-axis the origin the bisector of the first and third quadrants (Theorem, page 114). It should be noticed that (3a) and (4a) are the special cases of (3) and (4) respectively, obtained when a = - 1, and that (46) may be regarded as a combination of (3a) and (4a). Properties (1) and (2) are closely related to the translation of the axes. If we set (Theorem 1, page 89) X = x' -hh, y = y' + k in y=f(x), we get y^+k = f{x' + h). The graphs of these two functions are identical if x' and y' are plotted on axes h units to the right and k units above the old axes. But the second equation may be written y' =f{x' + h)-k', and hence, if x' and y' are plotted on the old axes the graph of y' may be obtained by moving that of y to the left h units and k units down (properties (1) and (2)). Properties (3) or (4) may also be interpreted as giving es- sentially the same distortion to a curve as is obtained if un- equal units are chosen on the coordinate axes. With these properties might be associated the method of ob- ALGEBRAIC FUNCTIONS 153 taming the graph of y = f{x) =*= g{x) by first plotting the graphs off(x) and g(x), and then adding or subtracting the correspond- ing ordinates (see Exercise 4 of the Miscellaneous Exercises following Chapter I). 56. Some Operations of Algebra regarded as Properties of Functions. The identity (a + by = a^ + 2ab + ¥ is usually thought of in elementary algebra as a rule for ob- taining the square of a binomial. It may also be regarded as a property of the function x^, that is, as the means of expressing the value of the function when x is the sum of two given num- , bers in terms of the squares and first powers of the separate nimibers. It will be well to group together a few such relations, whether we regard them as rules of operation or as properties of func- tions. As typical of such relations we select the following properties of the function a;": If fix) = x"", where n is a positive integer, then (1) /(a + 6) = (a + 6)- - a» + »a»-ft + " ' ' + ^ Binomial theorem. (2) /(a - 6) = (a - 6)" = a" - na" 16 + • • • ± fe" j (3) f(ah) = (ab)** = a«5". (4) f(a/b) = (a/5)'» = a»/b\ (5) /(aP)_ = (aP)« = a«P. (6) /( ^ B C . ^ _, or BC-3jp, AC. The width would then be obtained by subtracting from BC the distance CD from C to the bank of the river. 158 TRIGONOMETRIC FUNCTIONS 159 The solution of the example depends upon finding the ratio of two sides of a right triangle which has the same acute angle BAC as the given triangle. It would be tedious, and some- times diflScult or impossible, if every time we wished to deter- mine an inaccessible side of a triangle we had to construct a similar right triangle whose sides could be measured and their ratio determined. To avoid this difficulty, tables have been constructed which give the ratios of the sides of a right triangle with a given acute angle. Let us consider the construction of such a table. Suppose that one acute angle of a right triangle is 30°, and that the triangle be placed in the position OMP, with the vertex of this angle at the origin of a system of coordinates, the adjacent leg lying along the positive part of the x-axis, and the hypotenuse falling in the first quadrant. The triangle is placed in this position so that the pro- cedure will be in accord with fundamental definitions to be given in Section 59. Let the coordinates of P be X = OM and y = MP, and let OP = r. We seek the values of the rations y/r, x/r, and y/x. Fig. 76. Since the angle at is 30°, that at P is 60°. Hence AOMP is half an equilateral triangle OPQy and therefore ^ ^ 2y and x' + y' = 7^, (1) The ratios required may be determined by solving these luations for any two of the sides in terms of the third. We ilready have r in terms of y. Substituting in the second luation, a;2 + 2/2 = 4?/2, whence x ^ Vs y. We may now find the values of the ratios: y ^y_^l X ^ Vsy ^ Vs y^ y ^ _1^ ^ V3 (2) r 2y 2' r 2y 2 ' ^ VSy Vd 3 ^ ? r/^ y ^^ M X ^ ^^^^ ^^> Q 160 ELEMENTARY FUNCTIONS As all right triangles whose acute angles are 30° and 60° are similar, these results are true for all such triangles. Problems in geometry which can be solved by means of equa- tions (1) may be solved more expeditiously by means of the ratios (2), as in Example 2. Find the height of a flag pole MP if the pole subtends an angle of 30" at a point 100 feet from the foot M. y ' .f*" Ty^ y >^^ • X-^IOO M X Fig. 77. By the method of plane geometry we obtain the equations r = 2y, 1002 + y^ = r\ The solution is completed by solving these equations for y. Using the third of the ratios (2), which contains the known and required sides, we have 2/ V3 , 100V3 ^- = -5-, whence y 5 — 100x1.732 100 = 57.7 feet. The values of and y/x for two readily obtained plane geometry. If Z MOP = 45^ y = X and These equations and r in terms of P the ratios y/r, x/r, y^ other angles may be by the methods of we have rr2 + 2/2 = r2. (3) may be solved for y X, and the values of the ratios determined. If ZMOP = 60°, we have r = 2x and y^ + x^ =r2, (4) which may be solved for y and r in terms of a;, and the ratios determined. The details of the work are left as exercises. TRIGONOMETRIC FUNCTIONS 161 e= Z atO y r X r g X 30** U.50 ^ = O.ST f-..8 45** f = 0.71 f-0.71 \-uoo 60° f.0.87 i-0.50 V3 = 1.73 The values of the ratios for the three angles already con- sidered are collected in the table. In computing the decimal fractions, we have, for example, V|^ 1.732 2 = 0.866. The first two decimal places are 0.86, but 0.87 is a better approximation to two figures. 162 ELEMENTARY FUNCTIONS Approximate values of the ratios for any acute angle 6 may be found as follows: By means of a protractor construct the angle 6 so that the vertex is at the origin, one side lying along the X-axis, and the other in the first quadrant. From any point P on the latter side drop the Une PM perpendicular to the X-axis. Measure the lengths of the sides of the triangle OMP to find X, y, and r, and then divide y hy r, x by r, and y by X. Fig. 80 simpHfies and systematizes the process. It con- sists of a portion of a circle, whose center is the origin and whose radius is 10, and radii making angles of 10°, 20°, 30°, etc., with the x-axis. To find the values of the ratios for 6 = 10°, for example, we read off the coordinates of the extremity of the radius making an angle of 10° with the x-axis. They are, approximately, X = 9.8, and y = 1.7, while r = 10. We then have, approxi- mately hi 10 = 0.17. ? = 9_^ = 0.98. r 10 hi 9.8 = 0.17. e 17 The filling in of the remainder of the table, which is to be used in the exercises below, is left as an exercise. A method of constructing extensive - tables of values of these ratios is beyond — the scope of this course. But from the ^^ preceding considerations, it should be clear that, if a value of an angle 6 is given, then values of the ratios y/r, x/r, and y/x are determined. Hence these ratios are func- tions of 6 (Definition, page 5). In this chapter we shall study these functions, their reciprocals and inverses, using the graphs of the functions to tie their properties together. We shall also consider some of the important applications^ of the functions. 10** 20° SO** 40° 60** 60* 70* 80* TRIGONOMETRIC FUNCTIONS 163 EXERCISES 1. Find the sides of a right triangle if one acute angle is 30° and the hypotenuse is 10. 2. Find the hypotenuse of an isosceles right triangle if one side is 12. 3. One acute angle of a right triangle is 60** and the leg adjacent to this angle is 25. Find the other leg and the hypotenuse. 4. Find the side of an equilateral triangle whose altitude is 8. 5. Knd the side of a square if a diagonal is 30. 6. A path runs up the side of a mountain at an angle of 40° to the horizon. If a man chmbs along the path for 500 yards, how high will he be above the starting point? 7. A canal makes an angle of 20° with an east and west line. If a barge moves at the rate of 4 miles an hour, how far will it move east in 5 hours? How far north? 8. Two roads cross the canal in the preceding exercise at points 200 rods apart, one running east and west, the other north and south. How large is the triangular field bounded by the canal and the roads? 9. To find the height of a tree, a line 75 feet long is paced off from the foot of the tree. At the end of the line the angle subtended by the tree is 40°. How high is the tree? 10. How high is the sim (i.e., how many degrees above the horizon) if a pole 54 feet high casts a shadow 20 feet long? 11. A balloon is anchored by a rope 1000 feet long which makes an angle of 70° with the ground. How high is the balloon? If a wrench happened to drop from the balloon, how far from the point of anchorage would it hit the ground? 12. Find the area of a rhombus whose side is 15 inches if one angle is 60°. 13. What is the length of the edge of a hex- agonal nut that can be cut from a piece of circular stock one inch in diameter? What is the distance across the flats (i.e., between parallel edges)? 14. Holes are to be drilled through a piece of metal at the vertices of a regular hexagon, 3 inches on a side. The metal is fitted in place with a side AB along the bed of a milHng machine. What displacements of the bed of the machine, in the direction of and perpendicu- lar to AB, will bring the metal into the proper positions for drilling the holes in rotation? 15. An approximate geometric method of determining x is the following: Draw the diameter AB of a circle and the tangent CD at B. Construct I 164 ELEMENTARY FUNCTIONS Fig. 82. LBOC equal to 30", and make CD = 3r, where r is the radius. Then AD is approximately equal to the semicircumference ttt. Find to three decimal places the approximate value of tt given by this construction. 58. Angles of any Magnitude. Let OX and OF be two lines drawn from an initial point 0. Let a line start from co- incidence with OX, the initial line, and rotate about 0, coming to rest finally in co- incidence with OP, the termi- nal line. The line may rotate in either direction, and it may make any number of revolutions before coming to rest. The Hne is said to generate an angle whose mag- nitude is determined by the amount and direction of the rota- tion. The numerical value of the magnitude may be given in degrees, right angles, or revolutions. The sign is positive or negative according as the direction of rotation is counter-clockwise or clockwise, i.e., in the opposite or in the same direction as the hands of a clock rotate. If the terminal line of a first angle is the initial line of a second the sum of the angles is defined to be the angle whose initial line is that of the first, and whose terminal line is that of the second angle. This is analogous to the sum of two lines (page 13). Two lines determine a count- less nmnber of angles. If ^ is any one of them, the others differ from B by an integral multiple of 360°. They may all be 6 + n360°, where w = =*= 1, =^ Fio. 83. represented by 3, TRIGONOMETRIC FUNCTIONS 165 f The arcs in Fig. 82 indicate the three angles: e = 225°, 6' = 6 - 360° = - 135°, 6" = 6 -{- 360° = 585°. If the initial Hne of an angle coincides with the positive part of the x-axis, the angle is said to lie in the quadrant in which the terminal Hne Ues. Thus in Fig. 83, the angle 300^ lies in the fourth quadrant, — 210° in the second. The positive direetion on the terminal line in such a figure is defined to be away from the origin. For example, the positive direction on the terminal line of an angle of 180° is to the left. We shall make an important use of the angles whose terminal lines bound, bisect, or trisect the four quadrants. 59. Trigonometric Fimctions of any Angle. Let 6 be the number of degrees in any angle whose initial line coincides with the positive part of the x-axis, let P{x, y) be any point on the termi- nal Hne, and let OP = r. Consider the ratio y/x, which is a negative number for the case indi- cated in the figure, since x = OM is negative and y = MP is positive. The numerical value of y/x may be found approximately by measuring the lengths of MP and OM and dividing the former by the latter. If P'(x', 2/') is any other point on the terminal line, the ratios 2/Vx' and y/x have the same sign, and also the same numerical value, since the triangles OMP and O'M'P' are similar. Hence if ^ is given a definite value, the value of y/x is determined, and therefore the ratio y/x is a function of 6. Similarly, the ratio of any one of the num- bers X, y, r, to any other is a function of the angle 6. These functions are called trigonometric functions. They are named in accordance with the definitions below, which hold for Fig. 86 A, B, C, D. Fig. 84. 166 ELEMENTARY FUNCTIONS Definitions. If ^ is any angle whose initial line co- incides with the positive part of the x-axis, if P{x, y) is any point on the terminal line, and if OP = r, then - = sine of = sin ^: - = cosecant d = esc ^ ; r y - = cosine of ^ = cos d\ - = secant of 6 = sec 6; - = tangent of ^ = tan ^; - = cotangent of ^ = cot 6. To these are sometimes added X 1 — = l-cos^ = versine of ^ = vers 0, r l_:2 = l_sin^ = coversine of ^ = covers 6, r M X B in quadrant I. yjL 6 in quadrant III. Fig. 86. in quadrant IV. TRIGONOMETRIC FUNCTIONS 167 Since the three ratios on the right are the reciprocals of those on the left, we have the reciprocal relations: *="*^ = ter9' ^^''^^^' *="'^ = sT5^- (^^ The first table in Section 57 gives the sine, cosine, and tangent of the angles 30°, 45°, and 60°. The values of the sines may be easily remembered by noticing that they are respectively ^Vl, iV2, §V3, while the cosines are the same numbers in the reverse order. The tangent of any one of the angles may be ob- tained by dividing the sine by the cosine. For we always have X x/r COS 6 If d is in quadrant I, x, y, and r are positive, and the values of the six ratios, or functions, are positive. But if ^ is in one of the other quadrants, either x or y, or both, are negative, al- though r is always positive, and hence these ratios may be nega- tive. For example, if 6 is in the second quadrant, tan 6 = y/x is a negative number, since x is negative and y is positive. The reciprocal relations show that the signs of cot 6, sec 0y CSC 6, for a given value of 6, agree respectively with the signs of tan 6, cos 6, sin 6. Hence it is necessary to fix in mind the signs of the latter functions only. This is readily done in connection with the graphs (see Section 61). If the terminal Hne of 6 bounds, bisects, or trisects one of the quadrants, the functions of 6 may be found directly from the definitions, as in the examples following, by methods of elementary geometry. Example 1. Find the functions of 0**. Let P{x, y) be any point on the terminal line, which coincides with the positive part of the x-axis, since B = 0. Then 2/ = 0, and r = x. Hence, by the definitions, V* ' sin 0° = y/r = 0/r = 0. cos 0° = x/r = x/x = 1. tanO** = 2//x = 0/a; = 0. ^ — o >■ M X sec 0° = r/x = x/x = 1. But the definition of cot 6 involves divi- sion by y, and as !/ = if ^ = 0, co< 0° does ^^°' ^'^• not exist. However, cot 6 is the reciprocal of tan 6, by (1). Aa B ap- 168 ELEMENTARY FUNCTIONS Fig. 88. preaches zero, tan 6 approaches zero, and hence cot 6 becomes infinite as 6 approaches zero (IV, page 118). This is sometimes indicated by the symbols cot 0° = <». In like manner, as d approaches zero, esc 6 becomes infinite. Example 2. Find the functions of 210°. The acute angles of the triangles 0PM are 30° and 60°. Hence the numerical value of OP is twice that oi MP. If we let the numerical value of MP be 1, then that of OP is 2, and hence that of OP is Vs. But the coordinates of P are negative. Hence we may take X = -V3, 2/ = - 1, r = 2. Then by the definitions sm 210° = y/r = -1/2. CSC 210° = r/y = 2/(- 1) = - 2. cos 210° = x/r = -\/3/2^ sec 210° = r/x = 2/(_-V3) = - 2V3/3. tan 210° = y/x = - l/(- V3) = V3/3. cot 210° = x/y = - V3/(- 1) = V3. Since the functions of two angles with the same terminal line may be defined by means of the same triangle, and since all angles with the same terminal line may be expressed in the form ^ + n360° (Section 58), we have the Theorem. The trigonometric func- tions of an angle are unchanged if the angle is increased or decreased hy an integral multiple of 360°. Syrrir bolically, sin (6 + n360°) = sin 6, cos (d + n360°) = cos dy where n = ± 1, =*= 2, =fc 3, etc. Definition. A function is said to be periodic if its value is unchanged when the value of the variable is increased by a constant, that is, if fix + c) =f{x). If c is the smallest con- stant of this sort, it is called the period of the function. For example, the height of the tide at the seashore is a periodic function of the time. For if approximately 12 hours and 25 minutes are added to the time, the height of the tide will be the same. TRIGONOMETRIC FUNCTIONS 169 The theorem shows that the trigonometric Junctions are periodic, for if is increased by 360° the values of the func- tions are unchanged. It will appear later that 360° is the period of the sine and cosine, and their reciprocals, while 180° is the period of the tangent and cotangent. This is a characteristic property of the trigonometric functions, which distinguishes them from all algebraic functions. EXERCISES 1. Find all the functions of each of the so-called quadrantal angles: (a) 0°; (b) 90°; (c) 180°; (d) 270°; (e) 360°. 2. Find all the functions of the angles: (a) 135°. (b) 330°. (c) - 45°. (d) 240°. (e) - 210°. (f) 480°. (g) 315^ (h) - 150°. (i) 300°. (j) 990°. 3. What positive angles less than 360° have the same functions as (a) 540°, (b) - 60°, (c) 1320°, (d) - 675°, (e) 653°? 4. Determine the sign of: (a) cos B, 6 in quadrant II. (c) cos 6, 6 in quadrant IV. (e) sin 6, 6 in quadrant IV. (g) cot 6, 6 in quadrant IV. (b) tan 9, 9 in quadrant III. (d) tan 9, 9 in quadrant II. (f) sec 9, 9 in quadrant III. (h) CSC 9, 9 in quadrant II. 120' p 6. Determine the sign of each of the functions in each of the quadrants. Tabulate the results. 6. Build a table of values of sin 9 for 9 = 0°, 30°, 45°, 60°, 90= 135°, 150°, 180°, 210°, 225°, 240°, 270°, 315°, 300°, 330°, 360°. 7. Build a table of values, for the angles given in Exer- '^ ^^ cise 6, for (a) cos 9; (b) tan 9] (c) cot 9) (d) sec 9] (e) esc 9. 8. Construct all the posi- tive angles less than 360° for which sin = - f . Since sin 9 = y/r, the problem is equivalent to constructing right triangles with hypotenuse equal to 5, and vertical side equal to 3 and lying below the x-axis. Hence, describe the circle with center and radius 5, and draw the line parallel to the x-axis and 3 units below it. Let their intersections be P and P'. Then the lines Fig. 90. 170 ELEMENTARY FUNCTIONS OP and OP' are the terminal lines of the required angles. How many such angles are there? How many positive and less than 360°? 9. By the method of the preceding exercise, construct all the positive angles less than 360° for which (a) cos 6 = \. (b) tan = - |. (c) esc = 3. (d) sin = 4. (e) cos = - 0.3 (f) cot = 2. 10. Find all the functions of 6 given : (a) cos = - i^j, and in quadrant II. (b) tan = 1, and in quadrant III. (c) sin = - H> and in quadrant IV. (d) cot = 2, and in quadrant I. (e) sin = t, and in quadrant II. (f) sec = ^-, and in the fourth quadrant. 11. The intensity of light, /, varies inversely as the square of the dis- tance, d, from the source. If a street light is at the top of a concrete post 10 feet high, AB, express / at a point C on the pavement as a function of = AACB. 12. An aeroplane rises along a straight line, which makes an angle with a straight road directly below the path of the aeroplane, at the rate of 50 miles an hour. Express as a function of the speed at which an automobile must move along the road to keep under the aeroplane. 13. If the hypotenuse of a right triangle is 10, express the area as a function of one of the acute angles. 14. Construct the path of a point on the rim of a wheel rolling on a level road, by rolling a coin along a ruler, without slipping, and marking a number of positions of a point on the edge of the coin. This curve de- fines y, the height of the moving point above the road, as a function of x, the distance measured along the road. Show that this function is periodic, and find its period. What can be said of the graph of a periodic function? 60. Radians. To find the number of degrees in the angle u subtended at the center of a circle of radius r by an arc whose length is r, we com- pare the angle with the complete angle about the center. Since angles at the center are proportional to the subtending arcs, and since an angle of 360° is subtended by the entire circumference, we have TRIGONOMETRIC FUNCTIONS 171 J£ ^ /.x 360 27rr ^ ^ and hence u = 180/7r = 180/3.1416 = 57°.295. (2) Hence this angle does not depend on r, but is the same for all circles, and it may therefore be used as a unit angle. Definition. A radian is the angle subtended at the center of a circle by an arc whose length is equal to the radius. Equation (2) enables us to reduce radians to degrees. The most convenient form of this equation is the important relation w radians = 180 degrees. (3) From this we have, for example, that 90° = ^72 radians, and it is customary to speak of 7r/2 radians rather than 3.1416/2 = 1.5708 radians. Similarly, It is customary to express in terms of ir the number of radians in any angle which is a simple multiple or submultiple of 180°. When the degree or the right angle is used as the unit that fact is usually indicated. Thus we write 6 = 180°, or 6 = 2 rt. A . But if the unit is the radian we merely write 6 = w with- out indicating the unit. The number of radians in an angle is called its circular measure. It is customary to use the radian as the unit angle in draw- ing the graphs of the trigonometric functions (see the follow- ing section). Another elementary use is given by the Theorem. The length of an arc of a circle is equal to the radius multi- plied by the number of radians in the angle subtended at the center. Symbolically, Arc il5 = r X /.AOB {m radians). For if A B is any arc, if ^ is the p^^ 92 circular measure of the subtended angle AOB, and if arc AC = r, so that /.AOC = 1, we have = 7' whence AB = rd, r 1 172 ELEMENTARY FUNCTIONS The radian is the unit used in all theoretical work in the calculus and higher mathematics. Tables for converting radians into degrees, and degrees into radians, are to be found on page 32 of Huntington's Tables. EXERCISES 1. Reduce the following angles to radians: (a) The quadrantal angles, namely, 0°, 90"*, 180*, 270*, 360*. (b) The angles whose terminal lines bisect the quadrants, namely, 45*, 135*, 225°, 315°. (c) The angles whose terminal lines trisect the quadrants, namely, 30°, 60°, 120°, 150°, 210°, 240°, 300°, 330*. 2. Reduce the following angles to degrees: (a) 7r/6, 37r, 77r/2, 27r/3. (b) 27r/3, 7r/4, Stt/S, 57r/4, ll7r/3. 3. Using Huntington's Tables, page 32, express the following angles in radians, as decimal fractions: (a) 10°, 75°, 110°, 340°, 15°.34, 3°.77. (b) The angles in Exercise 1. 4. Using Huntington's Tables, express the following angles in degrees: 0.25, 1.13, 1.465, 0.8327, 3.2476. 5. If the radius of a circle is 10 inches, find the angle subtended at the center by an arc 15 inches long. 6. If the radius of a circle is 4 inches, find the length of an arc which subtends an angle of 173* at the center. 7. A strip of tin 8 inches wide is bent into the form of a trough whose cross section is an arc of a circle. Express the angle subtended at the center as a function of the radius. 8. Find the angle at the center of the earth subtended by an arc of the equator one mile long. In doing this, which one of the radii of the earth given on the inside of the back cover of Huntington's Tables should be used? 9. If the radius of a circle is 12 inches, find the length of an arc subtend- ing a central angle of 27r/3, and the area of the sector bounded by the arc and the radii drawn to its extremities. 10. Show that the area of a sector of a circle of radius r is § r^O, where 6 is the number of radians in the central angle of the sector. Hint. The area of a sector is one-half the product of the radius and the arc of the sector. 61. Graphs of the Trigonometric Functions. In construct- TRIGONOMETRIC FUNCTIONS 173 ing these graphs it is customary to use the same unit on both axes, and to measure 6 in radians. Graph of sin 6, A sufficiently extensive table of values is obtained by taking the values of 6 for which the terminal Une bounds, bisects, or trisects one of the quadrants. The values of sin d for the angles 30° = 7r/6, 45° = 7r/4, 60° = 7r/3 are given in the first table in Section 57, and for the angles 0° and 210° = 77r/6 in the examples in Section 59. The table of values below is identical with that asked for in Exercise 6, follow- ing Section 59, except that the angles are now expressed in radians. To plot the pairs of values in the table, choose a convenient unit on the vertical axis, and on the ^-axis lay off tt = 3-f units and 27r = 6f units. The points B = 7r/2 and B = 37r/2 are obtained by bisecting these segments, giving four segments on the ^-axis corresponding to the four quadrants. The remain- ing abscissas may be constructed by bisecting and trisecting these segments. Having determined the points on the ^-axis corresponding to the values of B in the table, ordinates are erected equal to the respective values of sin B, and the curve is then drawn. Fig. 93. sin 6 0, 7r/6, 7r/4, 7r/3, 7r/2, 27r/3, 37r/4, 5t/6, x, 77r/6, 57r/4 0, 0.5, 0.7, 0.9, 1, 0.9, 0.7, e ! 47r/3, 37r/2, 57r/3, 77r/4, 0.5, 0,-0.5,-0.7 ll7r/6, 2x ^1 -0.9, 1, -0.9, 0.7, 0.5, Periodicity of sin B. If the terminal hne of B starts in co- incidence with the positive part of the a;-axis (Section 58), the angle B may have any one of the values: . . . - 47r, - 27r, 0, 27r, 47r, . . . 174 ELEMENTARY FUNCTIONS As the terminal line makes a complete revolution, 6 may in- crease through any one of the intervals: . . . - 47r to - 27r, - 2w to 0, to 27r, 2t to 47r, . . . But no matter through which of these intervals 6 increases, sin 6 will vary through the same set of values (Theorem, Section 59). Hence the part of the graph in each interval is congruent to that in the interval from to 27r, which was plotted above. Starting at the origin, the graph does not begin to repeat until = 2w. Hence 27r, or 360°, is the smallest constant such that sin (6 -h 27r) = sin 6, and hence 27r, or 360° is the period (Definition, Section 59) of sin 6. The part of the graph from to 2t should be fixed in mind carefully, and the portion corresponding to each quadrant noted, as the graph affords a simple means of remembering the following properties of sin 6 (see page 42). Zeros of sin 6. For angles less than 27r, sin ^ = if ^ = or TT. From the periodicity of the function, the other zeros are therefore + 2n7r = =*= 27r, ± 47r, =»= Qir, . . . and TT + 2n7r = =t Stt, =t 5t, ± Ttt, . . . Sign of sin 6. Sin 6 is positive if 6 is in the first or second quadrant, negative if 6 is in the third or fourth quadrant. The periodicity shows that this holds whether or not 6 is posi- tive and less than 27r. Maximum and minimum values of sin 6. The maximum value between and 27r is sin (7r/2) = 1, and the minimum value sin (37r/2) = - 1. The other values of 6 for which sin ^ = =b 1 may be found by means of the periodicity. Changes of sin 6. As 6 increases from to 7r/2, sin 6 increases from to 1. As 6 increases from ir /2 to tt, sin 6 decreases from 1 to 0. As 6 increases from t to 37r/2, sin 6 decreases from to - 1. TRIGONOMETRIC FUNCTIONS 175 As 6 increases from Stt /2 to 27r, sin 6 increases from — 1 to 0. Notice that the numerical value of sin 6 cannot exceed unity, and, in particular, that it does not become infinite. Symmetry of the graph. The graph appears to be synmietrical with respect to the origin. Hence, probably, sin {- 6) = - sin 6. This may be proved as follows : Construct any angle 6, and then the angle - 6. On their termi- nal Hnes take OP = O'P', so that r = r' . Then since P' is sym- metrical to P with respect to the a;-axis (why?), we have x = x' and y = - y'. Then sin (■ 0) = ^ = r — sm 6, Fig. 94. The graphs of cos 6 (Fig. 95) and tan 6 (Fig. 96) are con- structed in hke manner, and should be fixed in mind. The various properties of reciprocal functions (page 118) enable us to get certain properties of cot d, sec d, and esc 6 directly from the graphs of tan 6, cos 6, and sin d respectively. (See also Exercise 2 below.) cos d I Fig. 95. 6 1 0, 7r/6, 7r/4, x/3, 7r/2, 27r/3, 37r/4, 57r/6, tt, 77r/6, 57r/4 1, 0.9, 0.7, 0.5, d I 47r/3, cos d \ -0.5, 0, -0.5, -0.7, -0.9, - 1, -0.9, 37r/2, 57r/3, 77r/4, ll7r/6, 27r 0.7 0, 0.5, 0.7, 0.9, 176 ELEMENTARY FUNCTIONS e tan d Fig. 96. 0, 7r/6, 7r/4, 7r/3, 7r/2, 27r/3, 37r/4, Stt/G, tt, Ttt/G, 5t/4 0, 0.6, 1, 1.7, 00, -1.7, -1, -0.6, 0, 0.6, e I 47r/3, 37r/2, 57r/3, 77r/4, IItt/G, 27r tan ^ I 1.7, 00, -1.7, -1, -0.6, EXERCISES 1. Discuss the periodicity, zeros, values of 6 for which the function becomes infinite, sign, maxima and minima, changes and symmetry of cos 6 and tan d. 2. Sketch on the same axes the graphs of the pairs of functions follow- ing, and discuss the second function with respect to the properties listed in Exercise 1. (a) sin 6 and esc B. (b) cos B and sec B. (c) tan B and cot B. 3. Construct the graphs of the six trigonometric functions on the same axes. 4. What properties of the functions can be inferred from graphs or the same axes of (a) sin B and cos 0? (b) tan B and cot ^? (c) sec B and esc Bl 6. Describe the motion of a particle on a straight line if its distance s from a fixed point on the line at any time t is given by s = sin t. Does such a motion approximate any motion occurring in nature? 6. On the same axes sketch the graphs of the fimctions: TRIGONOMETRIC FUNCTIONS 177 (a) sin 6 and 2 sin 6. (b) cos 6 and 3 cos d. (c) tan $ and i tan ^. (d) cot 6 and 0.2 cot 6. (e) sec ^ and ^ sec ^. (f) esc 6 and | esc 0. 7. Construct and discuss the graph of (a) vers ^ = 1 - cos d. (b) covers 6 = 1 - Bin 6. 8. By the addition of ordinates (see Exercise 4, page 44) construct the graph of (a) sin X + cos x. (b) 2 sin x - cos x. (e) sin x + 3 cos x. 62. Functions of Complementary Angles. Construct an acute angle 6 and its complement 90° — 6 with their initial lines coinciding with the positive part of the x-axis. On their terminal lines take OP = 0T\ so that r = r'. Then P and P' are symmetrical with respect to OA, the bisector of the first quadrant (why?), and hence x = y' and y = x\ Then sin (90° - 0) = and cos (90° - 6) = y -, = cos r (1) sin 6. (2) Using (2), page 167, with (1) and (2), we then have sin (90° - 6) tan (90°- d) cos (90° - 0) 1 Also, cot (90° - 6) = — .^o n^ - . n ' ^ ^ tan (90 - 6) cot 6 and in like manner, sec (90° - ^) = CSC 6, esc (90° - 6) = sec d. (5) The cosine, cotangent and cosecant of an angle are so named because they are respectively the sine, tangent, and secant of the complementary angle, as is shown by these relations. The former functions are called the cofunctions of the latter, re- spectively, and vice versa. With this terminology, the six relations (1) to (5) may be stated as the 178 ELEMENTARY FUNCTIONS Theorem. The functions of any acute angle are equal re- spectively to the cofunctions of the complementary angle. A method of extending the proofs of these relations for any value of Of not necessarily acute, will be given in Section 68. 63. Tables of Trigonometric Functions. As a consequence of the theorem in the preceding section, tables of values of the trigonometric functions may be printed in very compact form. Since cos (90° — 6) = sin 6, a table of sines of any set of angles is also a table of cosines of the complementary angles. The complements of 0°,1°, 2°, . . ., 88°, 89°, 90° are respectively 90°, 89°, 88°, . . ., 2°, 1°, 0°. Hence: 1. The table of sines on pages 8 and 9 of Huntington's Tables is also a table of cosines if read backward. 2. The sines and cosines of angles from 0° to 45° in the Condensed Table on the inside of the back cover of the Tables are, if read upward, the cosines and sines respectively of the angles from 45° to 90°. In like manner, tan 6 and cot B may be given in one table, and so also may sec 6 and esc 6. Thus the theorem on functions of complementary angles makes it practicable to reduce by one-half the space devoted to a table of trigonometric functions. Fractional parts of a degree, in Huntington's Tables, are given in tenths and hundredths instead of in minutes and seconds. One of the merits of this decimal method of sub- division is that the process of interpolation, in finding a func- tion of 6, is identical with that used earher in the other tables (page 121). Notice that, in the body of the tables, the decimal point, and any figures preceding it, are usually printed only in the first column. For example, tan 53°.32 = 1.3426. ^'°- ^^- The process of finding 6, if a function of 6 is given, is illustrated in the examples following. J^ F^^ --^ D 7? -""^ I i 1 H 1 d A E C ^ TRIGONOMETRIC FUNCTIONS 179 Example 1. Find d if sin0 = 0.4321. Searching through the body of the table of sines, we find that 0.4321 is given in the table, and reference to the margin shows that 6 = 25°.6. Example 2. Find ^ if sin ^ = 0.4332. A search in the body of the table of sines shows that 0.4332 lies between 0.4321 and 0.4337, which are the sines of 25°.6 and 25''.7. Fig. 98 shows the graph of sin 6 between these angles, on the assumption that it is straight (compare the assumption on page 122). To find 6, graphically, is to find the value of ^ at J^J if EF = 0.4332. The slope of the graph is the difference of the ordinates of B and D, HD = 0.0016, divided by the difference of the abscissas, ^C = 0.1 (Defini- tion, page 50). It is also equal to the difference of the ordinates of F and D, ID = 0.0005, divided by the difference of the abscissas, EC. Hence 0.0016 _ 0.0005 AC ~ EC ' whence ^^ ^ 0^ ^^ = 4 AC = 0.3AC = 0.3 x 0.1 = 0.03. 0.0016 16 Hence 0,1 E, d = 25^.67. The arithmetical processes used in interpolating to find an angle of which a function is given are: Find the two successive numbers in the proper table between which the given number lies. Find the difference between the given number and that one of these two numbers to which it is nearer, and divide this difference by the tabular difference. Apply the result as a correction to the last digit of the angle whose function was used in getting the difference above, in such a way that the result lies between the two angles corresponding to the two numbers in the table. Example 3. Find 6 if cos 6 = 0.4815. Searching through the body of the table, we find that 0.4815 lies between 0.4818 and 0.4802, the cosines of 61*^.2 and 61°.3. It is nearer the former, the difference being 3, neglecting the decimal point, while the tabular difference is 16. The correction to be applied to 61°.2 is therefore x% of one-tenth of a degree, so that d = QV^j^ = 61'*.22. In using the table of tenths of the tabular difference, in the margin, find the difference 3 as above. Then look for 3 in the margin. It lies in the column headed 2. Hence 3 is 2-tenths of the tabular difference, and the digit 2 is annexed to 61*'.2, giving 61°.22. The use of the tables for other than acute angles will be considered in Section 69. 180 ELEMENTARY FUNCTIONS EXERCISES 1. Find the cosine, cotangent, and cosecant of 72° A3. Find the sine, tangent, and secant of 17°.57. Compare the two sets of results. 2. Find the six functions of 73°.26, and verify the fact that the last three functions are the reciprocals of the first three in the reverse order. 3. Find the acute value of 6, illustrating the interpolation graphically, if (a) sin e = 0.9235. (d) cot 6 = 0.3603. (b) cos 6 = 0.3129. (e) sec 6 = 1.5165. (c) tan ^ = 1. 1603. (f) esc d = 5.515. 4. The arithmetical operations in this exercise should be performed mentally. Find the acute value of 6 such that (a) tan 6 = 0.8134; tan 6 = 1.8134; cos 6 = 0.3019. (b) sin e = 0.8436; sec d = 4.057; cot 6 = 0.6142. (c) cos d = 0.0589; esc 6 = 1.6397; sin 6 = 0.16543. (d) cos d = 0.0630; cos 6 = 0.06300; tan d = 1.4628. 5. Construct the line through the origin whose slope is 2, and find the angle between the line and the a;-axis. 6. Construct a table of values of and sin 6 for values of $ taken every 15° from 0° to 90**, expressing 6 in radians decimally instead of in tenns of TT (see Tables, page 32), and giving the values of 6 and sin 6 to two decimal places. Construct the graph as accurately as possible from this table, using a large scale. On the same axes draw the graph of 6. What ap- proximate value of sin for very small angles is suggested by these graphs? Using the Condensed Tables on the inside of the back cover of the Tables, determine for how large an angle this approximation is correct to three decimal places; to four decimal places. What is the limit of sin 0/0 as 6 approaches zero? 7. Solve Exercise 6 replacing sin by tan 0. 8. Using the properties of the functions sin 0, cos and tan suggested by the symmetry of their graphs, find the sine, cosine, and tangent of the negative angles, - 24**.32, - 48**.27, - 68**.46. 64. Solution of Right Triangles. It is customary to denote the magnitude of the angles of a triangle ABC hy A, B, C and the lengths of the sides opposite by a, 6, c respectively. In a right triangle, the right angle is usually denoted by C, and the hypotenuse by c. A right tri- Fio. 99. ^^gjg j^^Q jjjg^y bg placed with reference to coordinate axes so that A is at the origin, AC lies along y ^ C/^ 1 3 a A b C / X TRIGONOMETRIC FUNCTIONS 181 the positive a;-axis, and the hypotenuse AB Ues in the first quadrant. Then the definitions on page 166 show that sin A = a /c, cos A =6 /c, tan A = a/b) ^ . cot A = b/ttf sec A = c/b, CSC A = c/aj ^ ^ It is not always convenient to place the triangle on coordi- nate axes, and sometimes other letters must be used, and hence it is desirable to remember these formulas in words, as follows : The sine of an acute angle of a right triangle is the ratio of the side opposite the angle to the hypotenuse. The cosine of an acute angle of a right triangle is the ratio of the side adjacent to the angle to the hypotenuse. The tangent of an acute angle of a right triangle is the ratio of the side opposite the angle to the adjacent side. Analogous statements for the last three functions are readily obtained by the reciprocal relations (page 166). The sides and angles of a triangle are called its parts. In order to construct a right triangle, we must be given two parts, in addition to the right angle, of which at least one must be a side. To solve a triangle is to find the unknown parts from the known. Every right triangle may be solved by means of formulas (1) and the fact that A -\- B = 90°. But the Pythagorean Theorem is sometimes convenient. In solving right triangles but two essentially different cases arise: f I. Given a side and an angle, to find the other two sides use those two of equations (1) which contain the unknown sides in the numerators and the given side in the denominator. II. Given two sides, to find an angle, use that one of the two equations (1) containing the given sides which leads to the simpler division, and then to find the third side use either of the two equations containing the third side in the numerator. In either case the second angle is found from A + B = 90°. The use of the equations indicated in these rules will lead to multiplication, and division by a value of sin 6, for ex- ample, is avoided. 182 ELEMENTARY FUNCTIONS It is desirable that figures be constructed accurately, as the figure may show the absurdity of an incorrect result. A good check on the accuracy of the computation may be ob- tained by constructing the triangle with scale and protractor, and measuring the unknown parts. Example 1. Solve the right triangle A = 37°.24, b = 9. Solution. 5 = 90** - A = 52°.76. tan A , whence a h tan A . 9 X 0.7601 6.8409. 6=9 Fig. 100. = sec A, whence c = 6 sec A = 9 X 1.2561 = 11.3049. Check. The accuracy of the computation may be checked by finding b from a and c. By the Pythagorean Theorem, and tables of squares and square roots, VSLO" 6 - Vc2 - a2 = \/127.7 - 46.7 9, which agrees with the given value of b. Example 2. Solve the right triangle a = 7, & = 9. The given sides occur in the third of formulas (1). Hence ^ = 1 = 0.7778, whence A = 37^87, and hence B = 90° - 37^87 9 52°.13. To find c we have tan A a=7 CSC A = -, whence c = a esc A ^ =7x1.6290 = 11.403, Check. Find b from B and c. Since sin B = b/c, we have 6 = c sin B = 11.403 x 0.7894 = 9.0015, which agrees reasonably well with the given value of b. It is to be noted that the angles and fines in these examples and in the exercises following have not been measured with the same degree of precision. The angles are given to four figures in order to afford practice in interpolation. Problems involving isosceles triangles and regular polygons may be solved by means of right triangles, for such figures may be divided into congruent right triangles. TRIGONOMETRIC FUNCTIONS 183 Fig. 102. 65. Applications. An instrument known as a transit enables surveyors to measure angles in vertical and horizontal planes. If A and B are points not in the same horizontal plane, and if AC and BD are horizontal lines in the vertical plane through A and B, then Z BAG is called the angle of depression of B at A, and ZABD is called the angle of elevation of A at B. If the eyes are at A, looking horizontally over B, the angle of depression of B is the angle through which the eyes must be lowered to see B. While if the eyes are at B, looking hori- zontally below A , the angle of eleva- tion of A is the angle through which the eyes must be raised to see A. If A A' and BB' are the vertical lines through two points A and B, meeting the horizontal plane through a third point C at A' and B' respectively, then /.A'CB' is called the horizontal angle between A and B at C. If we think of A A ' and BB' as two trees, of different heights usually, for an observer at C, the horizontal angle between the tree tops A and B is the angle through which one must turn, if one faces first toward the tree A A' and then turns to face the tree BB'. The hearing of a line is the angle the line makes with some fundamental line of the figure which is called the base line. For example, if the base fine is north and south, the bearing of a fine running northeast is 45° east of north. Fig. 103. EXERCISES 1. In any right triangle, if c and A are given, show that a = c sin A and 6 = c cos A. Express these equations in words. If a and A are given, find b and c. If h and A are given, find a and c. 184 ELEMENTARY FUNCTIONS 2. Solve and check the following right triangles: (a) A = 22°.13, 6 = 5. (h) B = 40°.28, c = 4. (c) 6 = 17, c = 30. (d) a = 10, 5 = 19. (e) A = 66^47, = 20. (f) a = 30, c = 37. 3. To find the width of a river, two points, A and C, are taken on one bank 100 feet apart. If 5 is the point on the other bank directly opposite 1 C, and if ZCAB equals 72°. 16, how wide is the river? 4. From the top of a lighthouse 35 feet high, the angle of depression i of a ship is 11°.38. How far is the ship from the lighthouse? 5. What is the angle of elevation of the sun if a pole 49 feet high casts a shadow of 11 feet? 6. An iron wedge for splitting rails is to have a base two inches wide and a vertex angle of 15°. How long will each side be? 7. A rustic summer house, or shelter, is to be built with an octagonal floor, 6 feet on a side. Determine the amount of flooring necessary, mak- ing an allowance of 25 % on account of the " tongue and groove " and for waste. 8. Solve Exercise 7 if the floor is to be pentagonal. ^Ive Exercise 7 if the floor is to be a regular seven-sided polygon. Can this Exercise be solved by plane geometry? 9. If the radius of a regular polygon of n sides is r, find (a) the perimeter in terms of n and r; (b) the area. 10. If a side of a regular polygon is a, express the area as a function of a, assuming n to be constant. 11. What is the angle of inclination of an ordinary gable roof, if its pitch (the ratio of its height to its entire width) is f ? ^? 12. A point P moves with uniform speed around a circle of radius one foot, making a complete revolution every 36 seconds. The projection M of P on a diameter AOB, moves along the diameter. Find OM for the values 50°, 60°, 70°, if ^ = ZAOP. Find the average velocity of M as ^ increases from 50° to 60° and from 60'' to 70°. Find the average velocity of M as 6 increases from 59° to 60° and from 60° to 61°. From 59°.9 to 60° and from 60° to 60°.l. Approximately what is the velocity of M at the instant when 6 = 60°? 13. Two life saving stations are 10 miles apart on a beach running 22°.5, east of north. A lightship anchored off the beach lies 33°.75 north of east from one station, and 11°.25 east of south from the other. How long would it take a boat to run from one station to the other if its speed is 10 miles per hour, and if it passes outside the lightship? How long would it take the boat to go from the ship to the nearest point on the beach? 14. (a) Two sides of a triangle, not a right triangle, are 12 and 20» TRIGONOMETRIC FUNCTIONS 185 and the included angle is 28°.48. Find the altitude on the latter side and the area. (b) Find the area of any triangle ABC in terms of the sides b and c and the included angle A. 15. Find the area of a parallelogram in terms of two adjacent sides and the included angle. 16. Two ships are on a line w^ith a lighthouse, which is 40 feet high. At the top of the lighthouse, the angles of depression of the ships are respectively 4°.26 and 6°.31. How far apart are the ships? 17. A regular octagonal tower is 4 feet on each side. The roof is 10 feet high. How long should the rafters be? 18. A bridge is to be built across a ravine between two points A and B on the same level. Two stations C and D are chosen in the ravine which lie in the vertical plane through A and B. At A, the angle of depression of C is 42°.37, and AC = 50 feet. At C the angle of depression of D is 4**.52, and CD = 40 feet. At D the angle of elevation of B is 37°.89, and DB = 60 feet. Find the length of the bridge. 19. Check the accuracy of the measurements in Exercise 18, by finding the difference in altitude of A and D, and also of B and D. 20. To find the width of a river, a tree is selected on one shore. At a point 50 feet from the tree, the angle of elevation of the top is 48°.72. At the point on the other shore directly opposite the tree, the angle of elevation of the top is 17**.39. How wide is the river? 21. From a station A the horizontal angle between a mountain top C and a second station B, 1000 feet from A and at the same altitude, is 64''.37. At B the horizontal angle between the mountain top and A is 90°, and the angle of elevation of the mountain top is 37*^.24. How high is the mountain? 66. Parallelogram Law — Velocities, Accelerations, Forces. The law to be considered in this section is illustrated in Example 1. The current in a river flows at the rate of 2 miles an hour. A man rows across at the rate of 4 miles an hour, keeping his boat at right angles to the shore. Show that the boat moves in. a straight line. Find how fast it moves, and in what direction. Take the starting point for the origin of a system of coordinates, and let the x-axis lie along the bank of a river. At the time t, the boat will be at a point P, whose coordinates give the distance the boat has been carried by the current, x = 2t, and *the distance the man has rowed from shore, y = 4^. Eliminating t, we JllOOOft.S Fig. 105. AM Fig. 106. 186 ELEMENTARY FUNCTIONS obtain y = 2x, which is true for all values of t. Hence the boat moves in a straight line (Corollary 1, page 57). Let C denote the position of the boat one hour after starting, and let OA and OB be the coordinates of C. The line OA represents the velocity of the water with reference to the earth, the direction of the line being that in which the water flows, and the length of the hne being the niunber of miles per hour the water moves. Similarly, OB represents, in direction and magnitude, the velocity of the boat vrith reference to the water. (If the water were at rest, OB would repre- sent the actual motion of the boat, in one hour, with reference to the earth.) With reference to the earth, the boat moves along the line OC, and in one hour moves from to C Hence OC represents, in direction and magnitude, the actual, or resultant, velocity of the boat. The resultant velocity is readily computed. Its magnitude is OC = VOA^ + AC^ = V22 + 42 = 4.47 miles per hour. Its direction may be given by ZAOC. Since tan AAOC = AC/OA = 4/2 = 2, we have ZAOC = 63°.43. This illustration exemplifies the law known as the Parallelogram of velocities. If a body is subjected to two differ- ent velocities, represented in direction and magnitude by two lines OA and OB, the actual or resultant velocity i^s represented by the diagonal OC of the parallelogram determined by OA and OB. If a body is moving through the air in any way, it has an accel- eration of 32 feet per second per second directed toward the center of the earth (see Section 22, page 63). This means that its vertical velocity is increased or decreased each second by 32 feet per second, according as it is falling or rising. Its horizontal ve- locity is uniform, and is not affected by the action of gravity. No account is taken here of the resistance of the air, which would make the discussion very much more complicated. The acceleration due to gravity may be represented by a vertical line running downward whose length is 32. In hke manner, any acceleration may be represented, in direction and magnitude, by a line. A force may also be represented, in direction and magnitude, by a line OA, representing the point of application of the force. Accelerations and farces may also be combined according to the parallelogram law. TRIGONOMETRIC FUNCTIONS 187 Finding the resultant of two velocities is known as the compo- sition of velocities, the two given velocities being called com- ponents of the resultant. The converse problem of determining two component velocities which have a given resultant is called the resolution of velocities. The same terms are also used with reference to accelerations and forces. If the components are at right angles, the parallelogram is a rectangle, and the composition or resolution may be effected by solving a right triangle. If the parallelogram is not a rectangle, it is necessary to use the methods to be developed in Sections 71-73. Problems of this sort will be found in the exercises following Section 73. Example 2. A ball rolls down a plane whose inclination is 30°. Re- solve the acceleration due to gravity into two components parallel and perpendicular to the plane. Let OC = 32 represent the acceleration due to gravity. Through and C draw lines parallel and perpendicular to the plane DE, forming the rectangle OABC. Then OA and OB represent the required components, for the resultant of OA and OB is OC (parallelogram law). In the triangle OAC, OC = 32, and ZACO = ZEDF = 30° (why?). A Hence OA/32 = sin 30* and AC/32 = cos 30^ Then OA = 32 x | = 16 and 05 = AC = 32 X W3 = 16\/3. As there is no motion in a direction perpendicular to the plane, the component OB is neutralized by the plane. The component parallel to the plane, OA = 16 feet per second per second, gives approximately the effective acceleration with which the ball rolls down the plane. If the ball starts to roll from rest, how fast will it be moving at the end of one second? At the end of two seconds? At the end of four seconds? If a ball is rolled up the plane, how will its velocity be affected during any second? If it is started up with a velocity of 50 feet per second, how long will it roll up the plane? 67. Conditions of Equilibrium of a Particle. In measuring , a force we shall use the pound as the unit. Fig. 107. I 188 ELEMENTARY FUNCTIONS If a number of forces act on a particle, which we take as the origin of a system of coordinates, each of the forces may be resolved into two components, one acting along each axis. If the particle is in equihbrium, there is no motion in the direc- tion of the X-axis, and hence the algebraic sum of the com- ponents along that axis must be zero. Similarly, the sum of the components along the y-Sixis must be zero. And conversely, if each of these sums is zero the particle must be in equilibrium. Hence we have the Theorem. A particle is in equilibrium under the action of any number of forces if and only if the sum of the components in each of two perpendicular directions is zero. In applying this theorem, first determine all the forces acting on the particle, and then choose the perpendicular direc- tions. If two of the forces are at right angles to each other, choose these directions. Example 1. A particle weighing 4 ounces is supported on a smooth plane whose inclination is 30° by a cord parallel to the plane. Find the tension of the cord and the pressure on the plane. The forces acting on the particle are (1) Its weight, TF = 4, acting vertically. (2) The tension of the cord, jT, acting parallel to the plane. ^(3) The resistance of the plane, R, acting perpendicular to the plane. Since R and T act at right angles, we proceed to resolve all the forces into components paral- lel and perpendicular to the plane. Choosing these directions facilitates the work, because W is then the only force whose components must be determined. By the parallelogram law and the solution of a right triangle, the component of W parallel to the plane is found to be 4 sin 30**, and that perpendicular to the plane is 4 cos 30**, both acting downward. Then by the theorem, taking the directions of R and T a» positive, we have r - 4 sin 30** = 0, and R -4 cos 30** = 0, whence T = 2 ounces and R = 2\/3 = 3.4 ounces. Fig. 108. TRIGONOMETRIC FUNCTIONS 189 In pointing out the part played by mathematics in the illustration of the scientific method in Example 1, page 79 (see bottom page 80), it was stated that a more satisfactorj^ verification of the law obtained would be indicated later, by deducing from the law some fact that may be verified by an experiment of a different nature. • This deduction will be macle in Example 2 below. Friction between an object and a plane acts parallel to the plane, and in the direction opposite to that in which the object moves or tends to move. When motion is about to take place, the force of friction is equal to the coefficient of friction for the surfaces in contact (see page 81) multiplied by the pressure of the object on the plane in the direction perpendicular to the plane. It is this law that we wish to verify. Example 2. A block of wood weighing 20 grams rests on a horizontal board. If the coefficient of friction is 0.29, the result obtained in Example 1, page 79, at what angle may the board be tipped before the block will be on the point of sliding? The forces acting on the bloc are: (1) The resistance of the plane, fi, (2) The friction, F, and (3) The weight W = 20. It is known that Fig. 109. F = 0.29 R, (1) since the pressure on the plane is numerically equal to the resistance of the plane. Resolving W into components parallel and perpendicular to the plane, as in Example 1 above, we get F = 20 sin e, R = 20 cos d, (2) where 6 is the angle at which the block is on the point of sliding. Substituting in (1) the values of F and R given by (2), we get, 20 sin d = 0.29 X 20 cos d. Dividing both sides by 20 cos 6, we obtain ?i5-^ = o.29. cos u 190 ELEMENTARY FUNCTIONS By equation (2) on page 167, n = tan 6. Hence tan 6 = 0.29, whence 6 = 16*'.2. This result may be readily tested by experiment. If it is found that when one end of the board is gradually raised the block begins to slide when the incUnation is a httle over 16°, then we have a verification of the correctness of the deduction by which 6 was found, and also a verification of the law (1) on which the deduction was based. That is, we have a verifica- tion of the law obtained in Example 1, page 79, which was re-stated on page 81. EXERCISES 1. A boy kicks a football so that it would roll across a street at the rate of 15 feet per second, and simultaneously a second boy kicks it so that it would roll along the street with a velocity of 12 feet per second. Find the actual velocity of the ball (magnitude and direction). 2. A man walks across a canal boat. If his velocity with reference to the earth is 7 feet per second in a direction inclined at 30** to the bank of the canal, find the velocity of the boat and that at which he walks. 3. A ball is thrown into the air at an angle of 40* with a velocity of 30 feet per second. Find the horizontal and vertical components of the velocity. 4. If a ball is placed on a plane inclined at 20°, find the acceleration with which it rolls down the plane, and how fast it will be moving at the end of 3 seconds. 5. A ball is rolled up a plane inclined at 10** with an initial velocity of 20 feet per second. How long will it roll up the plane? 6. What is the inclination of a sidewalk if a ball rolls down it with an acceleration of 8 feet per second per second? 7. A cake of ice weighing 200 pounds is held on an ice slide at an ice house by a rope parallel to the slide. If the inclination of the shde is 45", find the pull on the rope and the pressure on the slide. 8. An automobile weighing 2500 pounds stands on a pavement in- clined at 12**. What force do the brakes exert? 9. Will a box slide down a board inclined at 15** if the friction is 0.3 of the pressure of the box on the plane? 10. A bar of iron weighing 5 pounds rests on a rough board. As one end of the ^ .ard is gradually raised, it is found that the bar is just ready to slide when the inclination is 20**. Find the force of friction and the coefficient of friction (Definition, page 81). TRIGONOMETRIC FUNCTIONS 191 11. A boy and his sled weigh 50 pounds. What force is necessary to hold them on an icy sidewalk whose inclination is 10° by means of a rope inclined at 30** to the walk? 12. (a) A rifle is fired at an angle of 20° to the horizon. If the muzzle velocity of the ball is 2000 feet per second, find the horizontal and vertical components of the velocity at the muzzle. (b) How long will a ball rise if it is thrown vertically upward with a velocity of 32 feet per second? of 80 feet per second? How long will the rifle ball in (a) rise? How far will it move horizontally in this time? 13. A rifle with muzzle velocity of 1600 feet per second is fired at an inclination of 30°. Find the horizontal and vertical components of the velocity at the muzzle. How long will it rise? When will it hit the ground (see Exercise 13, page 104)? How far from the point where it is fired will it hit? 14. A rope is tied to a heavy weight lying on the ground. A boy pulls on the rope with a force of 50 pounds in such a way that the rope is in- clined at 25° to the vertical. Find the horizontal and vertical compo- nents of his pull. What is the force with which he tends to lift the weight? To drag it? If the weight weighs 100 pounds, what is the pressure of the weight on the ground when the boy pulls? 16. A body weighing 25 pounds rests on a rough plane inclined at 5°. Find the components of the weight parallel and perpendicular to the plane. What is the pressure of the body on the plane? What force is tending to make the body move down the plane? Why does it not move? (The weight of a body is the force with which the earth attracts it.) 16. Three bales of cotton, a total weight of 1400 pounds, are raised from the hold of a ship by a derrick until they are just above the deck, when they are pulled one side by a horizontal; rope. Find the pull of the rope and of the cable on the derrick when the cable is inclined at 12° to the vertical. 17. A stone weighing 300 pounds is raised to the top of a building by means of a derrick on top of the building, while a man on the ground keeps it away from the side of the building by means of a rope tied to the stone. Find the pull on the cable of the derrick and on the rope if the cable is inclined to the vertical at 5° and the rope at 60°. 18. Two cords are tied to a weight of 10 pounds. The weight is held by two boys who pull on the cords. If one cord is inclined at 20° to the vertical and the other at 14°, find the force with which each boy pulls. 19. Two narrow boards are 2 feet long. A boy places their upper ends together and their lower ends on the ground 1 foot apart, and balances a weight of 5 pounds on top of them. Find the pressure along each board. 20. A body weighing 5(X) pounds is suspended from the center of a horizontal beam. The beam rests on two V-shaped supports, inverted, which are made of pieces of "2x4" each 4 feet long. Find the thrust 192 ELEMENTARY FUNCTIONS or pressure along each 2x4, assuming that the effect is the same as if half of the weight were placed directly over each support. 21. Just above the door on the second story of a barn, a beam projects horizontally for 3 feet. Objects are raised to the second story by a rope which passes over a pulley at the end of the beam and enters the bam over a pulley in the wall at a point 4 feet directly above the beam. Find the thrust along the beam when an object weighing 60 pounds is suspended by the rope. 68. Functions of n90° =t 6, In Section 62 we saw how to express the functions of the complement of 6, 90° — 6, in terms of functions of 6. Let us now consider the functions of 180° - 6, the supplement of 6, employing the same method. Construct the angles 6 and 180° - 6 with their initial Hnes coinciding with the x-axis, and on their terminal lines take P(x, y) and P'(x\ y') so that r = r'. Then P and P' are sym- metrical with respect to the 2/-axis (why?), and hence x' = - Xy and y' = y. Then >s. P' y ' P^ y' \ X V 180 y 3 1' x' X iV I X Fig. 110. sin (180° -6) =% = sin 6. cos(180°-^)=|-i = = - cos 6, y r (1) X r (2) Dividing (1) by (2), and applying (2), page 167, we obtain tan (180° - ^) = - tan ^. (3) Similar formulas may be obtained for the other functions from these formulas by means of the reciprocal relations. Inspection of these re- sults shows that The numerical values of the trigonometric functions of an obtuse angle are equal respectively to the functions of the supplementary angU. In the figure above, 6 Fio. ill. M X 'A ^. x' M' A -e X y /, r/ SV N y J p p k TRIGONOMETRIC FUNCTIONS 193 ft was taken acute. But the proof applies without change to the second figure, in which d is in the third quadrant. It would also apply to properly constructed figures in which 6 lies in the second or fourth quadrants, so that these formulas hold for all positive values of 6 less than 360°. The proof may be extended to include all values of d by using the periodicity of the functions. Any function of one of the angles - d, 90° =t e, 180° ^ d, 270° =t 6, 360° - 6 may be expressed in terms of a function of 6. Any one of these forty-eight formulas may be written down by the fol- lowing rule: If any one of the trigonometric functions he denoted by f(6), and its co-function (page 177) by co-f{6), then =*= /(^) if ri is zero or even, or , ± co-fid) if n is odd. /(n90° ^ 6) ? y p 2/1 X \<1 ry^ y M^ «1 ^!w X M x^ >\^ ^3 X 2/2 ^ ^Jy^ ^x:3 2/8 i 2 1 3 Fig. 112. The sign on the right is to agree with that of /(n90° =t d) when d is acide. A comprehensive survey of these rela- tions is afforded by the figures following. To construct the first figure, construct the angles 6, 180° - 6, 180° + 6, and 360° - 6, OX being the initial line of each. On their terminal Unes take, respectively, f*(^, 2/), P\ixi, 2/i), P2{x2, 2/2), and Pz{xz, ys), so that r = n = r2 = rz. Then Pi, P2, and P3 are symmetrical to P with respect to the y-axis, the origin, and the x-axis respectively (why?). Hence the numerical values of x, Xi, X2, and X3 are equal, and 80 also are those of ?/, 2/1, 2/2, and ys. Hence the numerical value of any function of one of the angles 180° - 6, 180° + d, or 360° - 6 194 ELEMENTARY FUNCTIONS is equal to that of the same function of 6. Whether the signs agree or not will depend on the angle and the function in question. For example: Since 2/2 = 2/3 = -?/, we have ^ = ^^ = _ ^ whence sin (180° + ^) = sin (360° - 6) = - sin B. (4) And since 2/2 = - y, and X2 = - a;, so that 2/2/^2 = y jx, we have tan (180° + 6) = tan 6. (5) To construct the second figure, lay off the angles 6, 90° - 6, 90° + B, 270° - B, and 270° + B, with OX for the initial line. On their terminal lines take P{x, y), Pi(xi, 2/1), ^2(^:2, 2/2), Psixs, 2/3) and P4{x4, 2/4), so that r = n = r2= rs = r4. Since Pi is symmetrical to P with respect to the bisector of the first quadrant (why?), Xi = y and 2/1 = x. P2, P3, and P4 are symmetrical to Pi with respect to the 2/-axis, the origin, and the a>-axis respec- tively. Hence the numerical values of Xi, x^, X3, and Xi are equal to y, and those of 2/1, 2/2, 2/3, and 2/4 are equal to x. In- spection of the definitions on page 166 shows that if x and y are interchanged, each function is replaced by its co-function. Hence the numerical values of any function of one of the angles 90° =i= B or 270° =*= B is equal to that of the co-function of B, For example, since 0:2 = 0:3 = - xi =• - 2/, 2/2 \ //Xr P y M2 *2 \ m-^ Ml Xi y \ «4 xj^x 2/3 ^3/ \ 1 Fig. 113. we have y Ti r$ r whence cos (90° + ^) = cos (270° - ^) = - sin ^. (6) In this survey we have taken B acute, so that all the func- tions of B are positive. Hence fin90° ^ B) will equal ^ fiB), or =*= co-fiB), according as /(n90° =*= B) is positive or negative, which can be determined readily in a given case. TRIGONOMETRIC FUNCTIONS 195 To illustrate the use of the rule, express cos (270° — 6) in terms of a function of 6, Since 270° = 3-90°, and 3 is odd, we will obtain the co-function, ± sin 6. If 6 is acute, 270° - 6 is in the third quadrant, in which the cosine is negative (see graph). Hence cos (270° - 6) = - sin 6. (7) Writing down a formula by the rule does not constitute a proof. The proof of (7), for example, which is identical with part of (6), is given above. The graphical significance of any one of the formulas is easily found. Consider, for example, formulas (1) and (4), and construct the graph of sin 6. Let OA = 6, and construct OB = 180° - d,OC = 180° + e,OD = 360° - 6. Then by (1), the ordinates at A and B are equal, and by (4), the ordinates at C and D are equal nu- merically to that at Af but are opposite in sign. If 6 increases from 0° to 90°, it follows that the graph consists of four congruent parts. Periodicity of tan 6. From (5) it follows at once that the period of tan 6 is 180°, or tt. The following formulas are stated for purposes of reference: sin (180° + ^) = - sin d. (8) cos (180° + <9) = -cos^. (9) tan (180° + ^) = tan d. (10) sin (360° - d) = - sin 6. (11) cos (360° - ^) = cos d. (12) tan (360° - 6) = - tan d. (13) If 360° are subtracted from the angle on the left in (11), (12), and (13) we obtain sin (- ^) = - sin d, ^ cos (-6) = cos 6, [ (14) tan (- 0) = - tan d, J the relations proving the synometry of the graphs. Fig. 114. 196 ELEMENTARY FUNCTIONS 69. Application to the Use of Tables. A positive angle less than 360° may be put in the form 180° - e, 180° + 0, or 360° - 6, (1) where 6 is acute, according as the angle Hes in the second, third or fourth quadrant. Hence the functions of such an angle may be expressed in terms of the functions of an acute angle 6, and the latter may be found from the tables. For example: sin 150° = sin (180° - 30°) = sin 30° = i. cos 213° = cos (180° + 33°) = - cos 33° = - 0.8387. tan 312° = tan (360° - 48°) = - tan 48° = - 1.1106. The functions of a positive angle greater than 360° may be found by using the periodicity of the function, and then pro- ceeding as above; for example, sin 985° = sm (2-360° + 265°) = sin 265° = sin (180° + 85°) = - sin 85° = - 0.9962. The functions of a negative angle are found by first using either the relations sin {- B) = - sin 6, cos (- 6) = cos 6, tan (- 0) = - tan 0, etc., or the periodicity. Thus tan (- 225°) = - tan 225° = - tan (180° + 45^) = - tan 45° = - 1, or tan (- 225°) = tan (360° - 225°) = tan 135^ = tan (180° - 45°) = - tan 45° = - 1. The fimctions of a positive acute angle less than 360° may also be found by putting the angle in one of the forms 90° + d, 270° -6, or 270° + 6, (2) where 6 is acute. The form (1) is somewhat less confusing because the application of the rule in Section 68 does not in- volve a change to the co-function. The examples following show how to find all the angles for which a given function has a given value. The solutions TRIGONOMETRIC FUNCTIONS 197 depend upon the fact that the numerical value of a function is the same for the three angles (1) as for the acute angle 6. Example 1. Find all values of B for which sin B = 0.4332. First find the positive values less than 360°. From the tables, one value is ^ = 25°.67. The graph of sin B shows that a line parallel to the 0-axis and 0.4332 unit above it cuts the graph in two points, one in the first quadrant, corresponding to the value B = 25°.67 found from the tables, and one in the second quadrant. Formula (1) Section 68, shows that the second value is = 180° - 25°.67 = 154°.33. All values of B for which sin ^ = 0.4332 are given by B = 25°.67 + n360° and B - 154°.33 + n360°. Example 2. Find ^ if cos ^ = - 0.5. If we neglect the negative sign and seek an acute angle whose cosine is 0.5, we know it to be 60° (table, page 161). The graph of cos B shows that a line parallel to the B-axis and 0.5 unit below it cuts the graph in two points, one in the second and one in the third quadrant. Then formulas (2) page 192, and (9), page l95, show that the values of B correspond- ing to these points are B = 180° - 60° = 120°, and B = 180° + 60° = 240°. All values of B are then, by the periodicity of cos B, B = 120° + n360° and B = 240° + n360°. If we notice that one of the angles of the second set is, for n = - 1, - 120°, which may also be obtained from B = 120° by the relation cos {- B) = cos B, all the angles may be expressed by the single equation B= ^ 120° + n360°. Example 3. Find all the values of B if tan B = - V3. Neglecting the negative sign, we recognize that one value of B is 60° (table, page 161). A line below the rc-axis cuts the graph of tan B in two points, one in the second quadrant, and one in the fourth. Hence, by formulas (3) page 192, and (13), page 195, the required values of B less than 360° are 180° - 60° = 120° and 360° - 60° = 300°. Then all the re- quired values are B = 120° + n360° and B = 300° + n360°. If we use the fact that the period of tan B is 180°, noticing that 300° = 120° + 180°, all these angles may be expressed by the single equation B = 120° + nl80°. EXERCISES 1. Prove formulas (8) - (13), Section 68, for B acute. 2. State, prove, and give the graphical significance of the formula for 198 ELEMENTARY FUNCTIONS (a) cos (90° + 0). (b) cot (- 0). (c) sec (180" - 0), (d) sin (270° - 0). (e) cot (360° - 0). (f) cec (90° - 0). (g) cot (180° + 0). (h) sec (- 0). (i) cos (270° + 0). 3. Prove the six formulas for the functions of the angle: (a) - 0. (b) 180° - 0. (c) 180° + 0. (d) 360° - 0. (e) 90° - 0. (f ) 90° + 0. (g) 270° - 0. (h) 270° + 0. 4. By means of the proper formulas and the table on page 161 find (a) sin (- 60°), cos 300°, cos 240°, tan 315°. (b) sin 330°, cos (- 120°), cot 210°, sec 150°. (c) cos (- 135°), tan 120°, tan (- 150°), sin 225°. 5. By means of the formulas for the sine, cosine, and tangent, and the reciprocal relations, derive the formulas for the cotangent, secant and cosecant of (a) - 0; (b) 90° - 0; (c) 180° - 0; (d) 180° + 0; (e) 360° - 0. 6. Find all the functions of 142°.30; of 118°.17. 7. Find all positive values of less than 360° for which (a) cos = 1; tan ^ = 1; sin ^ = - ^. (b) sec 2; cot ^ = - 1; esc = \/2. (c) sm = 0.3486; cos = - 0.8111; tan = 0.4770; tan ^ = - 1.4770. 8. Find all the values of for which (a) cos ^ = - \/3/2; cos = 0.4761; tan = 2. . (b) sin = - 0.6460; cos = 0.5348; tan = 2.638. (c) sec = 1.4788; esc = 4.865; cot = 33.96. 9. Express the following as functions of 0. (a) sin (0 - 90°). SoliUion. Smce ^ - 90° -= - (90° - 0), we have sin {0 - 90°) = sm [- (90° - ^)] = - sin (90° - 0) = - cos 0. (b) cos {0 - 180°). (c) tan {0 - 270°). (d) sin (0 - 180°). 10. Construct a table of values of and sin for values of taken every 10° from 0° to 360°, expressing in radians decimally instead of in terms of TT (see Tables, page 32), and giving the values of and sin to two deci- mal places. Construct the graph as accurately as possible from this table of values. 11. As in the preceding exercise, construct a table of values and draw the graph of (a) cos 0f (b) tan 0, (c) cot 0, (d) sec. 0, (e) esc 0. TRIGONOMETRIC FUNCTIONS 199 70. Inclination and Slope of a Straight Line. The function tan 6 enables us to express precisely the relation between the slope and the direction of a line (see page 52). The upper, right-hand angle (the northeasterly angle) which a line makes with the a:-axis is called the inclination of the line. Parallel lines have the same in- clination, and conversely (why?). Consider any line through the origin. Its slope m is the ratio of the difference of the ordinates of any two of its points to the difference of the abscissas. For these points take the origin and any point P{x, y) on the line, above the x-axis. We then have, in either figure. Fig. 116. t m = y ^ = ^ = tan ^. X Hence the slope of the line through the origin is the tangent of the inclination. As parallel lines have the same slope and the same inclination, the result holds for any hne. We thus have Fig. 116. Theorem 1. The slope of a line is the tangent of the incUnor Hon, i.e., m = tan 6. In most of the applications of the linear equation y = mx + b, the slope is of greater importance than the incUnation, be- cause the slope is the rate of change of y with respect to x. 200 ELEMENTARY FUNCTIONS The chief importance of the theorem above is in geometry. For from the slope the direction of the line may be found, provided the same unit has been used on the x and y-axes. Let a Une start in a horizontal position and rotate about one of its points through half a revolution. Its inclination d will increase from 0° to 180°. The variation of tan 6 (see graph, page 176) shows that the slope of the Hne, m = tan dj will increase from zero through all positive values, and become infinite as the inclination approaches 90°, when the hne be- comes vertical. As the incHnation increases from 90° to 180°, the slope is negative, and its numerical value decreases to zero. Example 1. Find the angle formed by the lines I and V whose equa- tions are 3x - 4y + 12 = and 2x + Z/ - 8 = 0. Solving the equations of the lines for y we get °i 45 1 ^ -b^ 5^-^ Z ^"^ i^^::^: > » \ ^ '. 5-« ^S: 3::^- /"' ^ ■ - 1 . ^ ^ - y = to; + 3, and y = - 2x + 8 and hence the slopes of the Hnes are m = tan ^ = r = .75 and m' = tan 6' = - 2. 4 From the tables and the rule in Section 69, the inclinations of the lines are e = 36^87; d' = 180^ - 63^43 = 116°.57. Fig. 117. If a is the angle between the lines, then a = 6' - d (why?). Hence a - 116**.57 - 36*'.87 = 79°.70. If two lines are perpendicular, then, using the notation in Example 1, a = 90°, so that d' = 90° + d. Hence m' = tan 6' = tan (90° + 0) = - cot ^ = - -^ = - 1. ^ tan u m That is, the slope of one is the negative reciprocal of the slope of the other. The converse may be proved by retracing the steps in the reverse order. Hence we have Theorem 2. Two lines are perpendicular if and only if the slope of one is the negative reciprocal of the slope of the other. ii TRIGONOMETRIC FUNCTIONS 201 EXERCISES 1. Find the equation of the line passing through the point (2, 3) whose inclination is (a) 30°; (b) 135*'. 2. Construct each of the lines below, and find its inclination. (a) 2z-2y + 7 = 0. (b) 4a; - 3^ - 12 = 0. (c) 4x + 3y -12 = 0. 3. Find the angle between each of the pairs of lines: (a) VSx - 32/ + 12 = and VSx - 2/ - 3 = 0. (b) 5x -2y + 10 = and x + Sy -Q = 0. 4. Find the angles of the triangles formed by the lines below. How can the results be checked? (a) X + y - 4 = 0, a; - Vdy -3 = 0, VSx -y -3 = 0. ih)2x-y -Q = 0, x + 2y -3 = 0, 2x + 3y + 9 = 0. Definition. The angle between a line and a curve at a point of inter- section is the angle between the line and the tangent to the curve at that point. The angle between two curves at a point of intersection is the angle between the tangents to the curves at that point. 5. At what point on the graph oi y = x - x^ will the tangent line make an angle of 30** with the x-axis? Find the angles at which the curve cuts the X-axis. 6. Find the angle made with the x-axis by the tangent to the graph of y = x^ -XB,t the point for which x = 1. At the point of inflection (Defini- tion, page 139). 7. Find the angle between the graphs of x^ and x^ at the point (1, 1); between the graphs of a^ and x^ at the same point. 8. Find the points on the graph of y = x^ - x'^ at which the inclination of the tangent line is 45**. 9. Find the angles at which the straight line y = 4x cuts the graph of = a:' at the three points of intersection. 10. Find the equation of the line through the origin which is perpen- dicular to the tangent to the parobola y = x^ at the point (1, 1). 71. Law of Sines. A theorem of geometry states that if two sides of a triangle are unequal, the angles opposite them are unequal in the same order, and conversely. The exact relation between the sides and angles is given by the Law of sines. The sides of a triangle are proportional to the sines of the opposite angles. Symbolically ^ a b c sin 4 ~ sin 5 ~ sin C 202 ELEIVLENTARY FUNCTIONS Let ABC be any triangle, and draw the altitude CD Then in the right triangles ADC and BDC we have C h. Fig. 118. fe = 6 sin A and h = a sin B, whence 6 sin A = a sin B, and therefore a _ b sin A ~ sin 5 If A is obtuse (Fig. 1185), then ZCAD = 180'* - A, so that /i =. 6 sin (180° - A) = 6 sin A ((1), page 192) as before. In like maimer, by drawing the altitude from A or B, we get h c a c ,^. "= 5 = - — n 01" ~ 7 = -= — 7y- (1) Sin B sm C sin A sin C 72. Law of Cosines. This law replaces the two theorems in geometry concerning the square of a side of a triangle op- posite an acute angle and the square of a side of a triangle opposite an obtuse angle. Law of cosines. The sqitare of a side of a triangle equals the sum of the squares of the other two sides less twice their product times the cosine of the included angle. Symbolically fl2 = 62 + c2 - 26c cos A. 62 = c2 + a2 _ 2ca cos B. c2 = a2 + 62 _ 2a6 cos C. If A is acute (Fig. 118A), we have a^^h^-\- DB^ = ¥ + (c- ADY (since DB = c - AD) ^h^^c"- 2c'AD + AD^ s= 52 _|_ g2 _ 2bc cos A (since h^ + AD^ = ¥, and AD = 6 cos A). TRIGONOMETRIC FUNCTIONS 203 If A is obtuse (Fig. USB), we have =^h^ + (c + Any (since DB = c + AD) ^h^ + c' + 2c'AD + AD^ = 62 + c2 _ 26c cos A (since h^ + AD^ -= ¥ and AD ^ h cos (180° - A) = - 6 cos A, (2), page 192.) EXERCISES • 1. Show that the law of sines reduces to the first two of formulas (1), page 181, if C = 90°. What does the third form of the law of cosines be- come? 2. Prove the first of equations (1), Section 71, if B and C are both acute. 3. Prove the second of equations (1), Section 71, if A is acute and C is obtuse. 4. Prove the second form of the law of cosines if 5 is acute. 6. Prove the third form of the law of cosines if C is obtuse. 6. Find the ratio of two sides of a triangle, a/6, if (a) A = 40** and B = 20°. Is one side double the other? (b) A = 60° and B = 20°. Is sin 30 = 3 sin 67 Can this question be readily answered from the graph of sin 6? 7. If a particle is in equilibrium imder three forces, OA, OB, OC, prove that -: — 5777; = -; — TTTTf = -; — TrTfT (The Tcsultant, OD, of OA and OB sm BOC sm ADC sm AOB. ' must be equal and opposite to OC, and hence the sides of the triangle OAD will represent the forces numerically. Use law of sines.) 73. Solution of Oblique Triangles. An oblique triangle is one none of whose angles is a right angle. The constructions of plane geometry show that an oblique triangle may be con- structed if three of its six parts (sides and angles) are given, provided that at least one of them is a side. The laws derived in the last two sections enable us to solve the triangle, that is to compute the three unknown parts from those given. It is necessary to distinguish four cases, according as there are given: LA side and two angles. II. Two sides and the angle opposite one of them. III. Two sides and the included angle. D IV. Three sides. 204 ELEMENTARY FUNCTIONS The first two cases may he solved by the law of sines, and the last two by the law of cosines. The last two may also be solved by using first the law of cosines and then the law of sines. Case I. Given a side and two angles, the third angle is found from A + B + C = 1S0°, and the other two sides by the law of sines. Example 1. Two forts by the sea, A and B, are 12 miles apart. At A the angle between B and a target C anchored off the coast is 37°.24, and at B the angle between A and C is 42°.87. Find the distance from each fort to the target. We are given a triangle determined by A = 37°.24, B = 42°.87, c = 12. We have C = 180° - (A + B) = 99°.69. From the law of sines, c sin A J - c sin B a = — : — 77-j and = — : — 7^> sm C sm C , 12 X 0.6052 „ „^- ., , , 12 X 0.6803 ^ oot 1 whence a n qq^j " ^-^^^ miles, and b = — ^ - = 8.281 miles. Check. Find c from a, 6, C. By the law of cosines, c2 = a2 + 62 - 2ab cos C - 7.372 ^ 8.282 - 2 X 7.37 x 8.28 x (- 0.168) = 54.4 + 68.6 + 20.5 = 143.5, and therefore c = 11.98. This value agrees reasonably well with the given value of c, especially when we take into account the fact that the table of squares does not en- able us to use all four figures in a and b. Case II. Given two sides and an angle opposite one of them, the angle opposite the other given side is found by the law of sines, then the third angle is found from the relation A -\- B -\- C = 180°, and finally the third side is found by the law of sines. Example 2. Two straight roads diverge at an angle of 35°. An automobile starts from the fork in the road and runs along one road until a cross road is reached, the odometer showing the distance to be 2.1 miles. It turns into the cross road and after running 1.4 miles comes to the other road. How far is the automobile from the starting point, and at what angles does the cross road meet the other two? TRIGONOMETRIC FUNCTIONS 205 If A denotes the fork in the road, B the point 2.1 miles from A, and C the third point reached, a triangle is determined by the parts A = 35^ c = 2.1, a = 1.4. To construct the triangle, construct angle A = 35° with one side hori- zontal, and on the other side lay off AB = c = 2.1. Then with J5 as a center, describe the circle with radius o = 1.4, which, in this problem, cuts the other line in two points C and C. Either BC or BC may represent the cross road, so that the problem has iwo solutions, AC and AC. These may be found by solving the triangles ABC and ABC To find C we have, by the law of sines, csinA 2.1x0.5736 sin C = 0.8610, a 1.4 whence C = 59°.43 or 120°.57 (compare Example 1, page 197). Smce BCC is isosceles, C = ZAC'B' = 180° - C, and hence C = 59°.43 and C = 120°.57. Whence B = 180° - (A + C) and B' = 180° - (A' + C) = 85°.57 = 24°.43 To find b = AC and h' = AC, we have, by the law of sines, in AABC 6 = ^^^ and in AAB'C h' = ^-^ sm A sin A 1.4x0.9969 1.4 X 0.4136 ~ 0.5736 " 0.5736 = 2.433 = 1.009 Check. By the law of cosines we have: In triangle ABC c^ ^a^ + b^ - 2ab cos C = 1.42 + 2.4332 - 2 X 1.4 X 2.433 X 0.5085 = 1.960 + 5.91 - 3.464 = 4.406, and hence c = 2.099. In triangle AB'C c"^ = a^ + 6^ - 2ab' cos C = 1.42 + 1.0092 _ 1 X 1.4 X 1.009 X (- 0.5085) = 1.960 + 1.018 + 1.437 = 4.415, and hence c = 2.101. Both of these values agree very well with the given value of c. In order to illustrate the method of solution, the computations have been 206 ELEMENTARY FUNCTIONS carried out with all the accuracy permitted by the tables. Do the given data warrant us in saying that the automobile is either 2.433 miles or 1.009 miles from the fork in the road? // two sides of a triangle and the angle opposite one of them are given, there can he two solutions only if the angle is acuie and the Me opposite it is less than the other given side. Under these conditions, there will be two solutions unless in solving for the second angle (C, in Example 2), it is found that its sine is equal to or greater than unity (sin C^ 1). In the first case there is but one solution, a right triangle; and in the second case there is no solution, since the sine of an angle cannot exceed unity. A complete statement of the possible solutions is given in Exercise 3 below. Case III. Given two sides and the included angle, the third side is found by the law of cosines, and then the remaining angles hy either the law of cosines or the law of sines. Example 3. To find the distance across a bay, AB, a point C is taken j whose distances from A and B can be measured. It is found that AC is 70 yards, BC is 120 yards, and Z AC Bis 50**. Find AB and also the angles at A and B. A triangle is determined whose given parts are a = 120, 6 = 70, C = 50**. By the law of cosines c2 = a2 + 6^ - 2ab cos C = 120"* + 702 - 2 X 120 X 70 X 0.6428 = 14,400 + 4900 - 10,800 = 8500, whence c = 92.2. To find A and B, we use the law of cosines: Substituting the values of the sides in a^ = ¥ + 0^ - 26c cos A, we get 14,400 = 4900 + 8500 - 2 X 70 X 92.2x cos A, whence cos A = - 0.0774 and hence A - 180° - 85^55 = 94^45. Substituting the values of the sides in fe2 =, c2 ^ ci* _ 2ca cos B, we get 8500 + 14,400 - 2 X 92.2 x 120 x cos B, 0.8136 35°.55. Fig. 121 4900 whence cos^ and hence B TRIGONOMETRIC FUNCTIONS 207 Check. The angles having been found without using the fact that the sum of the three angles is two right angles, we have A-{-B + C = 94°.45 + 35^55 + 50** = 180°. Having found A as above, B might have been obtained from A + B + C = 180". But then it would have been necessary to use the law of sines as a check, and little would have been gained. Case IV. Given the three sides^ the three angles are found by the law of cosines, or one may he found by the law of cosines and then the others by the law of sines. Example 4. The lengths of a triangular lot are found by pacing to be 80 feet, 50 feet, and 100 feet. Find the angles at the corners and the area. A triangle is determined by a = 80, 6 = 50, c = 100. Substituting these values in the three forms of the law of cosines we get 802 = 502 + 1002 - 2 X 50 X 100 cos A, 502 = 1002 + 802 - 2 X 100 X 80 cos B, 100* = 802 + 502 - 2 X 80 X 50 cos C, and hence cos A = 0.6100, whence A = 52*'.41 (COS B = 0.8687, B = 29°.69 cos C = - 0.1375, C = 97°.90. Check : A+B + C = 180". 00. To find the area, draw the altitude from C and denote it by h. Then In the right triangle ACD, h = b ein A = 50 X 0.7924 = 39.62. Then the area is i ch = 50 x 39.62 = 1981 square feet. The lengths of the lines in the examples preceding and in the fol- lowing exercises have been chosen ^ cioo with but one or two significant p^^ 122. figures, in order to simplify the computations, and the angles to four significant figures for the purpose of obtaining drill in interpolation. In the next chapter, a labor saving device to assist in the computations will be considered, and further exercises given. 208 ELEMENTARY FUNCTIONS EXERCISES 1. In Example 4, find B and C by the law of sines. What point in the solution might be overlooked, which one is more likely to notice in using the law of cosines? 2. Is a triangle always determined if values are given for a side and two angles? For two sides and the included angle? For the three sides? 3. If two sides, say a and 6, and the angle opposite one of them, say A, are given, show that (a) If A is obtuse, there is one solution if a>5, and there is no solution if a^b. (b) If A is right, there is one solution if a > 6, and there is no solution if a^6. (c) If A is acute, there is one solution if a ^ 6, there are two solutions if 6 sin A 42°. Find the length of the tunnel. 7. A man starts from camp and walks N.E. for 5 miles, and then 22°.5 east of south until he reaches a point from which the camp is visible in a direction due west. How far has he walked, and how far is he from camp? 8. Two batteries of artillery, A and B, are four miles apart. An enemies' battery is located at a point C such that ZBAC = 64°.22 and ZABC = 43°.17. Find the range for each battery. 9. A field is bounded by two roads intersecting at right angles, and by two other straight lines. Find its area if the lengths of the sides, be- ginning at the corner at the crossing of the roads and measured in order around the field, are 30, 60, 70 and 40 rods. TRIGONOMETRIC FUNCTIONS 209 10. A schooner sails 10° west of north at the rate of 7 knots an hour across the Gulf Stream at a place where it flows N.E. with a velocity of 4 knots an hour. Find the actual velocity of the schooner in direction and magnitude. 11. The current in a river flows at the rate of 2 miles an hour, and a man rows at the rate of 4 miles an hour. If he desires to cross the river at an angle of 70° with the bank, in the direction of the current, in what direction should he row? 12. Solve the preceding exercise if the man desires to cross at the same angle with the bank but in the upstream direction. 13. Resolve a velocity of 50 feet per second into two components in- clined at 10° and 40° respectively to the direction of the given velocity. 14. A boy in an automobile moving 40 feet per second throws a ball in a horizontal direction inclined at 50° to the road with a speed of 30 feet per second. At what angle to the road will the ball move? 16. A road runs up a hill at an angle of 20°. At a point on it, 500 feet from the foot, the angle of depression of a horseman on the road leading to the hill is 5°. How far is he from the foot of the hill? 16. To find the width of a river, two points A and B are taken on one bank 100 feet apart. If C is a point on the opposite bank such that IBAC = 62°.34 and /.ABC = 49°.82, find the width of the river. 17. At a certain point the angle of elevation of the top of a mountain is 45°, and at a point 1000 feet nearer the mountain and at the same level as the first, the angle of elevation is 54°. 13. How high is the mountain? 18. A ship steams due east at the rate of 25 miles an hour, and the smoke from its funnel is blown in a direction 20° south of west. The wind gauge shows an apparent velocity of 35 miles an hour for the wind. Find the actual velocity of the wind in direction and magnitude. 74. Inverse Trigonometric Functions. To find the inverse of sin X (pages 40 and 114) we set y = sin x, and interchange X and y, obtaining x = sin y. The solution of this equation for y in terms of x requires the introduction of a new function which is called the angle whose sine is x, and which is denoted by arc sin x. Hence, if jc = sin y, then y = arc sin x. A table of sines may be regarded as a table of angles whose sines are given (see " finding 6 if sin 6 is given " page 178). Thus Example 2, page 179, might have been stated: find B = arc sin 0.4332, the result being arc sin 0.4332 = 25°.76 + n360° or 154°.33 + n360°. k 210 ELEMENTARY FUNCTIONS arc sin x This example illustrates the fact that for a given value of x arc sin x has not only one but a boundless number of values. This is apparent from the graph of arc sin x, which is sym- metrical to that of sin x with respect to the bi- sector of the first and third quadrants (see page 114). The graph also shows that the function is defined only for values of X from - 1 to + 1 in- clusively. The inverse of cos x is denoted by arc cos x (read "the angle whose cosine is x"); of tan x, by arc tan X, etc. Definition. The principal valve of any ono of the inverse trigonometric functions for a given value of x is that one of the boundless number of values of the function which is smallest numerically. If two values of the function are equal numerically, but opposite in sign, the positive value is the principal value. Unless the contrary is indicated, the symbols arasin x, arc cos x, etc., will he used in this work to denote the principal values only. Thus arc sin J = tt /6, Fig. 123. and arc sin (— 1) = — 7r/2. The part of the graph which represents the principal values of arc sin x is given in the figure. Inverse trigonometric functions are of much importance, although we shall use them but little in this course, venient in stating a general result, as in this FiQ. 124. They are con- TRIGONOMETRIC FUNCTIONS 211 Example. What angle is subtended at the center of a circle of radius r by a cliord c units long? Choose the center of the circle as the origin of a system of coordinates, and let the ic-axis be perpendicular to the chord. Then the a>-axi8 bisects the 'chord and the angle formed by the radii drawn to its extremities. From the figure sin d/2 = (c/2)/r, whence 6/2 = arc sin (c/2r) and hence 6 = 2 arc sin (c/2r). If c = 3 and r = 5, we would have ^ = 2 arc sm 0.3 = 2 x 17*'.46 = 34°.92. The notation sin"^a;, cos^^o;, etc., is used sometimes for arc sin x, arc cos X. etc. Fig. 125. EXERCISES 1. Construct the graph of arc sin x, and indicate on it the part which represents the principal values of the function for all possible values of x. State as many properties of the function as can be readily obtained from the graph. 2. Proceed as in Exercise 1 for the function (a) arc cos x. (b) arc tan x, (c) arc cot c. (d) arc sec x. (e) arc esc x. 3. Find all the values of arc sin 0; arc cos ^; arc tan (-1). 4. Find the value (noting the convention with regard to principal values) of (a) arc cos (- 5); arc tan 2, arc sin (- 0.3215). I' (b) arc sin f ; arc sec 2; 3 x arc sin (- VJ). 6. A rope I feet long is stretched from the top of a building to the ground, the lower end being d feet from the building. Find a general expression for the angle which the rope makes with the ground. What is the angle if I is 50 feet and c? is 17 feet? 6. A mountain h feet high is viewed from a point d miles away (hori- zontally). What angle does the line from the point of observation to the peak make with the ground? 7. What is the value of sin (arc sin a)? Of arc sin (sin a)? 8. Recall the method of solution of Exercises 9 and 10, page 170. Find the value of (a) sin (arc cos ^). (b) tan (arc sin ■^^). (c) cos (arc tan - 2). (d) tan [arc cos (- i)]. 9. If the maximum distance from a point on an arc of a circle to the chord of the arc is d, show that the central angle subtended by the arc is 2 X arc cos (r - d)/r, where r is the radius of the circle. 212 ELEMENTARY FUNCTIONS MISCELLANEOUS EXERCISES 1. If a creek is 20 feet wide, and if from a point 4 feet above the water's edge on one side the angle of elevation of the top of the bank on the other side, directly opposite, is 7°.27, how high is the bank? 2. An object weighing 60 pounds is supported on a smooth plane whose inclination is 60° by a man who pushes against it horizontally. Find the force exerted by the man and the pressure on the plane. 3. A board is just strong enough to bear an object weighing 100 pounds at its middle point when the board is supported horizontally at its ends. How heavy an object will it bear at its middle point if it is supported at its ends in a position inclined at 30", the object being held in position by a rope parallel to the plane? 4. Solve Exercise 3 if the object is held in position by a man pushing against it horizontally. 5. The hatchway into the hold of a ship is 16 feet wide. To raise an object weighing 200 pounds from the hold, two men on opposite sides of the hatchway pull on a rope which passes through a smooth ring fastened to the object. Find the tension of the rope when the ring is 6 feet below the deck. What force does each man exert? Hint: See the note following. Note. The tension of a cord or rope which passes over a smooth peg, or over a pulley, or through a smooth ring is assumed to be the same on both sides of the peg, pulley, or ring. It is usually not the same if the rope is tied to the ring. 6. A cord is tied at a point A, passes through a smooth ring B weighing 3 pounds, over a pulley C at the same height as A, and to the end is tied a weight of 2 pounds. Find ZBAC when the ring and weight are in equiUbrium. 7. A man drags a trunk across a room by pulling on the handle in a direction at 35** to the horizontal. If the trunk weighs 150 pounds, and if the friction is 0.1 of the pressure on the floor, find the force he exerts at the instant the trunk is about to move. Note that the pressure on the floor will be less than the weight of the trunk, because a part of the man's effort tends to lift the trunk. 8. Find the force exerted in pushing the trunk in Exercise 7, when the trunk Is just on the point of moving, If the man pushes down on the trunk in a direction inclined at 25° to the horizontal. 9. To find the width of a river, a point A is taken on one bank directly opposite a tree on the other bank, and a point B is taken 100 feet from A In the line of the tree and A. At B the angle of elevation of the top of the tree is 32°. 19, and at A it is 4r.33. Find the width of the river. 10. If a body on a rough inclined plane is just on the point of moving down the plane, show that the coefficient of friction is equal to the tangent of the angle of inclination of the plane. TRIGONOMETRIC FUNCTIONS 213 11. Rain drops are falling straight down with a velocity of 20 feet per second. At what angle would they appear to fall to a man walking at the rate of 3 miles per horn*? To a man in an automobile moving at the rate of 20 miles per hour? To a man in an express train moving at the rate of 60 miles an hour? 12. Prove that the area of any quadrilateral is equal to one-half the product of the diagonals and the sine of the angle between them. 13. The wind is blowing down a lake 5 miles wide at a rate which would blow a row boat a mile an hour. A man who rows at the rate of 3 miles an hour desires to go straight across. In what direction should he row, and how long will it take him to cross? 14. Two girls hold a traveling bag weighing 40 pounds. One puUs on the handle at an angle of 10* to the vertical, the other at 15°. What force does each exert? 15. If the distance an ivory ball rolls down a smooth plane inclined at < I 1, 2, 3, 4 5** in various times are as given in the table, s I 1.4, 5.5, 12.5, 22.2 the units being feet and seconds, find the ac- celeration due to gravity. Analysis of the problem: Find s as a function of t from the table, from this find the velocity at any time, then the acceleration with which the ball rolls down the plane, and finally the acceleration due to gravity. 16. Charles' law states that the rate of increase of the volume of a gas under constant pressure per degree (Centigrade) rise of temperature is sJj of the volume at 0", vq. Hence the volume v at any temperature d is V = ^r;^ 6 + vo. Boyle's law states that the product of the pressure and volume of a gas at constant temperature is constant, pv = k. If a quantity of oxygen occupies 200 cubic centimeters at a temperature of 17" Centi- grade under a pressm-e of 742 millimeters (barometric height), find its volume at 0** temperature under a pressure of one atmosphere at sea level, or 760 millimeters. Illustrate graphically, plotting the graphs of both laws on the same axes, taking v on the vertical axis for each law. First find the volume Vo under a pressure of 742 millimeters; the graph of Charles' law being determined by the given data and the fact that v = when ^ = - 273**, the " absolute zero." The value of k in Boyle's law is determined by the point whose coordinates are 200 and this value of vo. 17. Construct a square 8 inches on a side. Its area is 64 square inches. Cut the square into two rectangles of widths 3 inches and 5 inches. Cut the first rectangle along a diagonal into two triangles. Mark points on the 8 inch sides of the second 5 inches from opposite vertices, and cut along the line joining them. The four parts of the original square may be arranged in the form of a rectangle whose dimensions are 5 inches and 13 inches, and whose area Is 65 square inches. Explain the fallacy. CHAPTER V EXPONENTUL AND LOGARITHMIC FUNCTIONS 75. Introduction. The formula for the nth term, I, of a geometrical progression whose first term is a, and whose ratio is r, is I = ar^-\ (1) If we are given a = 3, r = 2, Z = 96, then n may be found. Substituting the given values, we have 96 = 3 X 2«-S whence 2"-^ = 32 = 2^, (2) and hence n - 1 = 5, from which n = 6. Equation (2) differs from most of the equations arising in algebra, in that the unknown n appears in an exponent. The possibiHty of finding n by elementary methods is due to the fact that 32 is recognized as a power of 2. If we set a = 3 and r = 2 in (1), we get l = 3x 2--K (3) Regarding n as variable, this equation defines I as a func- tion of n. For integral values of n, I is the nth term of a geometrical progression, but in studying I as a function of n we need not restrict n to integral values. Thus if n = |, Z = 3 X 23-1 = 3 X 2* = 3 X 1.414 = 4.242. Thus the value of I may be found readily by algebraic methods for many fractional values of n. But if n is irrational, for ex- | ample, if n = \/2 or w = tt, we cannot compute I by algebraic j methods. Hence Z is a transcendental function of n (definition, ! page 39). i 214 i EXPONENTIAL AND LOGARITHMIC FUNCTIONS 215 Definition. An exponential function is one in which the variable occurs in an exponent. For example, equation (3) defines I as an exponential function of n. The simplest form of an exponential function is 6*, where 6 is a constant, other than unity, called the base. It has the property, in common with more general forms of exponential functions, that if x increases in arithmetical progression, the function increases in geometrical progression. For example, values of a; in the table are in arithmetical ^progression, since successive values of Ax are equal, while a; I 1 2 3 4 ^ » •^-\ ' ' ' — j^ the values of 2^ are in geometrical pro- gression, since the ratio of any value to the preceding is 2. Exponential functions are important because in many fields, especially in physics, one variable changes (increases or de- creases) in geometrical progression as another, frequently time, changes in arithmetical progression. The former is always an exponential function of the latter. Examples of such changes are Newton's law of cooling of a heated body; the variation of atmospheric pressure with the altitude; the law of chemical reaction; and the law of organic growth. In these examples, and generally in nature, the change proceeds continuously. The amount of a sum of money at compound interest is an exponential function of the time, but the changes come at stated intervals instead of continuously. However, the b\!lsiness of some very large firms, which make many loans a day, approximate a condition of compounding interest every instant. And this property of an exponential function is often called, following Lord Kelvin, the compound interest law. Another name which has been suggested is the snow-hall law. The inverse of the function 6* is called the logarithmic func- tion, and a table of values of this function is spoken of as a table of logarithms. Such a table is an invaluable labor-saving device. It enables us to replace the laborious processes of multipHcation and division by the simpler operations of addic- tion and subtraction. It also makes it possible to reduce the very tedious operations of computing powers and roots to 216 ELEMENTARY FUNCTIONS multiplication and division, and through these to addition and subtraction. For this labor-saving device we are indebted to John Napier (1550-1617), who arrived at it by considering a function which increased in arithmetical progression as the variable decreased in geometrical progression. Following Napier, this tool was put in a more serviceable form by Henry Briggs (1561-1631). We shall use a table of logarithms for simplifying many computations, especially in connection with the solution of triangles. The introduction of the exponential and logarithmic functions completes the list of functions to be studied in this course (see Classification, page 38). 76. Graph of the Exponential Function b'^j b>l. In the following table of values of the exponential function 2", the value of the function for a; = is obtained by means of the definition 6^ = 1, and the value for any negative value of x is obtained by means of the definition ft"** = j-- By the first defi- nition, we have 2P 1, and by the second 2'^ ^ 02 '^ 4* 2x 2, ~ 1, 0, 1, 2, 3 h h, 1, 2, 4, 8 The table of values is readily com- puted and the graph plotted as usual. The process is so simple that it is hardly worth while to discuss the graph in advance. The following properties of the graph are apparent. 1. The intercept on the y-axis is 1. 2. The graph Hes entirely above the x-axis. 3. The X-axis is an asymptote. 4. Any fine parallel to the y-axis cuts the graph once and once only. 5. The graph rises to the right more and more rapidly. The form of the graph of the exponential function 6^, 6>1, is very much like that of 2'. As indicated in Fig. 127, the graph always cuts the y-axis at 2/ = 1, Hes entirely above the m g -y- Fig. 126. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 217 • y T — g t& 3' i' 1 n_j 1 / JT/ it/" fjl i uy J 9 1 A .. ., 0| 1 2 1 J,. X FiQ. 127. X-axis, which is an asymptote, and runs up to the right. If b>2, the curve rises more rapidly to the right of the y-axis than in the case of 2*, and the larger the value of b the more rapid is the rise. The values of b of most importance in mathematics are 10 and a number denoted by " e" whose value to four figures is 6 = 2.718. It was proved in 1844 that there are numbers which are not the roots of any algebraic equation no matter how high the degree, and the first number definitely proved to be of this sort was e, in 1873. In 1882 this same fact was proved about TT, and by means of this it was proved that it is impossible to " square the circle '* with ruler and compass. A table of values of e^ is given in Huntington's Tables, page 30. 77. Properties of the Exponential Function 6', 6>1. The properties of the function which correspond to the above properties of the graph are as follows: 1. For all exponential functions 6^ = 1. 2. The exponential function 6* is positive for all real values of X. 3. As X decreases indefinitely through negative values 6* approaches as a limit. 4. For each value of x there is one and only one value of 6=^. 5. The function b^ increases as x increases and the rate of change of b' also increases. Of two exponential functions the one with the larger base has the greater rate of change for x>0. Other important properties of the function which are not so apparent from the graph are the following: 6. If the values of x be chosen in arithmetical progression, the corresponding values of the function are in geometrical rrogression. Let the values of x be a, a + d, a i- 2d, a -\- Sd . . , 218 ELEMENTARY FUNCTIONS Then the values of the function are 6«, 6°+^, ¥+^, ¥+^, . . . which form a geometrical progression, for any one of these values may be obtained from the preceeding by multiplying byt*^. 7. 6*" • fe'^ = 6"*+". 8. 9. (b*")" = fe*"". 10. \/b^ = h^/n Properties 7, 8, 9, proved in elementary algebra for integral values m and n of x, are true for all values of x. The defini- tion 10 holds also for all values of x. If fix) = 6*, the last four properties may be written, in the reverse order: 11. f(m + n) = 6»«+» = 6"»6«. 12. f(m -n) = &♦«-" = 6«/6«. 13. f{mn) = 6"*" = (fc"*)". "• 1? if 6 = 1? if 6 < 1? 6. Plot the graphs of e*, e~*, i(e* + e~*) and i(e* - e"*) on the same axes for the range x = -2toa; = +2at intervals of 0.5. (See Tables, p. 30.) The functions i (e* + e~*) and \ (e*- e~* ), have somewhat the same rela- tion to the equilateral hyperbola that the trigonometric functions have to the circle. They are called respectively the hyperbolic cosine and hyper- bolic sine and are symbolized by cosh x and sinh x, 6. If x', y' and x", y" are two sets of values satisfying the equation y = lf, show that the value of the function corresponding to the arithmetic mean of x' and x" is the geometric mean of y' and y". 7. Find the value of 2* for x = 1.50, using Exercise 6 and the values for a; = 1 and x^2. Then find the value of 2' for x = 1.25; for x = 1.75. Arrange the results in tabular form. 8. Referring to the table in Exercise 7, between what two values of x does X = \/2 lie? The value of 2^-*^ is an approximate value of 2^. Find it by applying Exercise 6 several times, using the table in Exercise 7, and choosing the successive values of x' and x" so that their arithmetic mean approaches 1.41. 9. By means of the table in Section 78 find the value of each of the following: (a) (524288) (0.015125). (b) 4096/0.0078125. (c) 16*. (d) V16384. (e) (V'(32768) (0.0625) /5 12)'. (f) VW^- 10. A glass marble falls from a height of 4 feet and rebounds one-half the distance fallen, falls again and rebounds one-half the preceding distance fallen, and so on. Express the distance fallen each time as a function of the number of times it has fallen and draw the graph. How far does it fall the eighth time? 11. If the planets are numbered in the order of their distance from the sun the distance from the sun to the nth planet is approximately 4 + 3(2)'*~2^ the distance of the earth being represented by 10. Compare the results of substituting in this formula with the following table. Planet Mercury Venus Earth Mars Asteroids Jupiter True distance 3.9 7.3 10 15.2 27.4 52 Saturn Uranus Neptune 95.4 192 300 If there is a planet external to Neptune, at what approximate distance may we expect it to be found? EXPONENTIAL AND LOGARITHMIC FUNCTIONS 221 12. Show for the exponential function e* that the average rate of change ^w /gAx— \\ is -^ = e* I —7 1° Find the value of the second factor if Ax = 0.1, 0.01. Ax \ Ax I Assuming that the limit of the second part is 1 as Ax approaches 0, find m for the exponential e* and determine the angle that the graph makes with the 2/-axis. 13. Plot the graph oi y = ke"*^, first letting k = 1 and choosing m suc- cessively equal to - 2, - 1, 0, + |, + 1, + 2, and then letting m = I and choosing k successively equal to - 1, -h 5, + 2. How do changes in m and k affect the graph? 14. The pull, P, needed to check a weight, W, being lowered by means of a rope wrapped around an iron drum, is given by the equation P = We""^, where the coefficient of friction is m = 0.3, and d, the angle of contact of the rope with the drum, is measured in radians. If W = 500 pounds plot the graph of P as a function of d. What is the value of P if the rope is wrapped 4 times around the drum? 79. The Logarithmic Function, the Inverse of the Exponential Function. To find the inverse of 2^, let 2/ = 2^ and interchange X and y, obtaining x = 2^. The solution of this equation for y in terms of a; is a new function called the logarithm of x to the base 2, and it is denoted by log2 x. Thus X = 2y and y = log2 x are forms of the same equation, in the one case solved for x and in the other for y. In general, we have the Definition. If 6"* = n, then m is said to be the logarithm of n to the base b. Hence the logarithm of a number to the base b is the exponent of the power to which b must be raised to equal the given number. It is frequently convenient to change from the exponential to the logarithmic form of this relation, and vice-versa. For example, from 3^ = 243, we have logs 243 = 5. Again, to find logs 625, we let logs 625 = m, whence 5"* = 625. Since 5^ = 625, we have m = 4 and hence logs 625 = 4. 80. Graph of the Logarithmic Function. The graph of y = log2 X may be readily obtained from that of 2* by the rule in Section 39, page 113, and from it certain properties of the graph are obvious. 222 ELEMENTARY FUNCTIONS "' -/- .-l.l\l -l^ ^ ^ - / / : // ^-T;r, ^ «, ^<^ '■' ^ .| ,^ -4 . J ^ W. ± / 1. The intercept on the a;-axis is a: = 1. 2. The graph Ues entirely to the right of the y-axis. 3. The 2/-axis is an asymptote. 4. The graph is above the x-axis to the right of a: = 1 and below the ic-axis to the left of a; = 1. 5. The graph rises to the right but at a decreasing rate. 6. A Une parallel to the a;-axis with an intercept on the y-axis at y = 1 Fia. 128. cuts the graph at a point whose abscissa is x = 2. As indicated in Fig. 129, which shows the graph of log5 x for 6 = 2, 3, 10; the graph of any logarithmic function, y = logftX, 6>1, cuts the a;-axis at x = 1 , lies entirely to the right of the y-axis, which is an asymptote, lies above the a;-axis to the right of the line a; = 1, and below to the left of a: = 1, rises at a decreasing rate as it moves to the right, and is cut by the line ?/ = 1 at a point P whose abscissa is b. 81. Properties of the Log- -piQ. 129. arithmic Function, log6 x, b>l. Corresponding to the properties of the graph are the following properties of the function: 1. For any base, logb 1=0. For 6^ = 1. 2. Logb a; is a real number for positive values of x only. 3. As x approaches 0, logb x approaches - 00, that is, log6 = - 00, 6>1. 4. The logarithm of a positive number greater than unity is positive. The logarithm of a positive number less than unity is negative. 5. The function log6 x increases and the rate of change of log6 X decreases as x increases, and of two logarithmic func- y ■ 1 ^-^ "^ lolfsX I ^=d ■ 1 '> ~~ ""■ '4.- 1 "/ • 4 i 1 ' f 1 " 7 f r r EXPONENTIAL AND LOGARITHMIC FUNCTIONS 223 tions, the one with the larger base has a smaller rate of change fora;>l. 6. The logarithm of the base itself is unity. For since h^ = h, we have logb 6 = 1. Other properties of the function are easily deduced from the corresponding properties of the exponential function as follows: 7. Theorem. The logarithm of the product of two numbers is equal to the sum of the logarithms of the numbers. Let p = h"", whence log6 p = my and q = 6", whence log6 q = n. Then pq = h"'b'' = 6'"+". Therefore logb pq = m +n why? = log6 p + logb q. " 8. Theorem. The logarithm of the quotient of two numbers is equal to the logarithm of the dividend minus the logarithm of the divisor. Let p = b'^ and q = 6", whence Iog6 p = m and log6 q = n. Then p/q = &""/&" = &"*""• Therefore logb pjq = m - n why? » = logft p - log6 q. ^ 9. Theorem. The logarithm of the nth power of a number equals n times the logarithm of the number. Let p = 6"*, whence log^ p = m. Then p" = (6"*)'* = 6"*". Hence logb p" = mn why? = n logb p. 10. Theorem. The logarithm of the nth root of a number equals the logarithm of the number divided by n. Let p = 6"*, whence logb p = m. Then v^p = v^6^ = fc"'/". Hence logb\^p = m/n why? = - logb p. 224 ELEMENTARY FUNCTIONS If }{x) = logftX, properties 7, 8, 9, 10, may be written Jijpq) = logft vq = logb J) + log6 q, (11) Jijplq) = logft p/g = logo p - logb g. (12) }{jp^) = logb ^"=71 logb p. (13) /(\^p) = log5 <^V=]; log6 p. (14) These properties are respectively analogous to those of x" given by (3), (4), (5), (6), page 153. If the first two be written in the reverse order, we have f{v) +fiQ) = logfe p + logb q = logb pq, (15) and /(p) -fiq) = log?, p - logb q = log^ p/g, (16) which are analogous to (7) and (8), page 153. By means of the theorems in 7 and 9 the solutions of Ex- amples 1 and 3, page 219, may be written as follows: Example 1. log2 0.0078125 = - 7 Example 3. log2 8=3 log2 1048576 = 20 loga 8« = 6 logoS loga (product) =13 =18 Therefore product = 8192 Therefore 8« = 262144 EXERCISES 1. Express the following exponential equations in logarithmic form. 23 = 8, 3^ = 81, 4-2 = 3^6, 10-3 = 0.001, 16i = \/2. 2. What are the logarithms of 1, 2, 16, 1024, 0.125 with respect to the base 2? Express the answers in exponential and logarithmic form. 3. Find logz 2048, logs 81, loge 2.718, log^ 9, log^ 16. Express the answers in exponential form. 4. Find lo&, 6^, b^^^', log2 (log4 16), loge (1/e), b^^'f^''^^^''. 5. (a) Change the equation log« 6= -kt + log« ^o to the form ^ = ^o e~^, (b) Show that - ^lo©, (l/x^) = log, x. (c) Find/(a:) if \ogbf{x) = log^ (l - x) - 21og6 x - ^logj. (1 + x). 6. Plot the graphs of log2 x, loge x, logs x, on the same set of axes. (See page 31 of the Tables for values of log* x.) Calculate the value of Ay /Ax for each function for the interval 1 to 2. 7. Plot the graph of logi/aX. In what two ways can this graph be obtained from that of log« x? 8. Illustrate each of the theorems in 7, 8, 9, 10, Section 81, on the graph of log2 z. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 225 9. Write the solutions of Examples 2 and 4, Section 78, in logarithmic forms. 10. Show by Theorems 7 and 8 that logt ( — ) = loe> V + log& 3 - logt **• 11. Using the table m Section 78 find (0.03125) (262144)/ (32768). 12. Prove the theorem, if the numbers x, x', x", etc., are in geometric progression, their logarithms are in arithmetic progression. 13. Plot the graphs of logs x, logs x^, logaVx on the same axes. How can the graphs of the last two be obtained from that of the first function? 14. Plot the graph oi y = log2 {x + k) for the values fc = 0. 1, - 1 on the same axes, and discuss the change effected in the graph by a change in A;. 16. The equation for the economic law of diminishing utiUty is y = k logb (x/c), where x denotes income; y happiness; c the income suf- ficient for necessities; and A; is a constant depending on individuality. Plot a graph of the law, determine the graphical significance of c and A;, and discuss the law for x^c. 82. Common Logarithms. Logarithms to the base 10 are called common logarithms. The first table of logarithms to the base 10 was published in 1617 by Henry Briggs. It con- tained logarithms of nmnbers from 1 to 1000. In writing logarithms to the base 10, the base is usually omitted. Thus y = logio x is written y = log Xy which means that X = 10^. That is The common logarithm of a number is the power to which 10 mitst he raised to obtain the number. A four-place table of common logarithms of numbers from 1 to 10 is given on pages 16 and 17 in Huntington's TableSc The graph of this table of logarithms is shown in the figure. The method of inter- polation is the same as t/ in the tables already used. For example, from the tables we find log 2.342 = 0.3696 which means ' Fjq^ 13q^ that 100-3698 = 2.342. The number 2.342, and its logarithm, the number 0.3696, are the coordinates of a point P on the graph. 226 ELEMENTARY FUNCTIONS Either from the graph or from the table we see that The logarithm of any nmnber between 1 and 10 is a positive decimal fraction less than 1. We proceed to develop, by illustration, the rule for finding from the table the logarithm of any positive number <1 or > 10. If we translate the exponential equations 10-^ = o.Ol, 10-1 = 0.1, W = 1, W = 10, 102 = 100, W = 1000, into the equivalent logarithmic equations log 0.01 = - 2, log 0.1 = - 1, log 1 = 0, log 10 = 1, log 100 = 2, log 1000 = 3, we see that The Ipgarithm of an integral power of 10 is an integer. Since 23.42 = (2.342) (100, we have, using the theorem in 7, Section 81, log 23.42 = log (2.342) (10^ = log 2.342 + log 101 = 0.3696 + 1 = 1.3696. Similarly 234.2 = 2.342-102, whence log 234.2 = log 2.342 flog W^ 2.3696, 2342. = 2.342. lO^, whence log 2342 = log 2.342 + log 10^ = 3.3696, 0.2342 = 2.342- lO'S whence log 0.2342 = log 2.342 + log 10-^ = 0.3696 - 1, 0.02342 = 2.342-10-2, whence log 0.02342 = log 2.342 + log lO-^ = 0.3696-2. The values of x considered above have the same sequence of, digits 2, 3, 4, 5 and the corresponding values of log x have th same decimal part 0.3696. Multiplication of 2.342 by an integral power of 10 mere! moved the decimal point without altering the sequence of digits, and corresponding to this operation, the logarithm 0.3696 was increased or diminished by an integer without altering the decimal part. Any positive number x10 can be expressed in the form X = x'-lO** where re' is a number in the range l Use the vah of 1" in radians found at tl bottom of the front cover the Tables.) If light trave at the rate of 186,320 mil€ per second, find the length of time necessary for light to traverse tl distance. 7. The annual parallax of the North Star is 0.073". Find the distance to the North Star in light-years. * Inspection of the Condensed Table on the inside of the back cover of the Tables shows that values of Q, in radians, and of tan Q agree to three decimal places if ^ = 6°; to four places if = 3°. Fig. 138. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 237 8. Find the time that it takes light to reach the earth from the sun; to reach Neptune, the most remote planet, distance 2,788,000,000 miles. B? 86. Exponential Equations. An equation in which the un- known enters as an exponent is called an exponential equation. Exponential equations of the type a^^^^ = 6^^*^, where f{x) and F{x) are hnear or quadratic algebraic functions may be solved by equating the logarithms of the two sides of the equa- tion and solving the resulting algebraic equation, as in the fol- lowing examples. Example 1. Find the function of the type y = x" whose graph passes through the point (3, 5). Substituting the coordinates of the point in the equation, we have S** = 5. Equating the logarithms of both sides of this equation, we have log 3" = log 5 J or n log 3 = log 5. Hence 0.4771n = 0.6990 _ 0.6990 log 0. 6990 = 0. 8445 - I ^ ~ 0.4771 log 0.4771 =0.6786 -1 = 1.4653 log n = 0.1659 Hence the required function is y = xi-<653. The value of n is here calculated from the quotient by logarithms rather than by long division. Notice that if 0.4771 were subtracted from 0.6990, the result would be log (5/3) and not log 5/log 3. Example 2. Find the logarithm to the base e of any positive number X assuming a table of common logarithms to be given. Let y = log« X, hence e" = x. Equating the common logarithms of both sides of the second equation we have logio e" = logio X, whence y logio e = logw x. Therefore y = , logw x, logio e Since logw e = 0.4343, we have log, x = ^^^ logio x. (I) That is, the logarithm to the base e of any number can be found by multi- plying the common logarithm of the number by 1/0.4343. 238 ELEMENTARY FUNCTIONS For instance we have log. 6.75 = 1 logio6.75 0.8293 1.91. y —3 ^ lo 1i flfeX F L _ ^ ^ Qio xsE 2.3( m ^ ■^ fl' x~ O y T i 0.4 e 343. ^ '■ I ) s f 0.4343 ^'"" 0.4343 The constant ^ ^^.^ is called the modulus of the base e with respect to the base 10. The figure shows the graphs of logio x and log« x. If x = OP J then logio x = PQ, and log, x = PR. If OA = e, then AB = logio e. Hence equation (1) may be written in the form PR = -~ = — - — PQ AH \j . 4:o4o and the graph of log* x can be obtained from that of logio x by the theorem on page 89. Example 3. The variation of at- mospheric pressure p (in millimeters of mercury in a barometer) with respect to the altitude h above sea level (in Fig. 139. meters) is given by the equation p = jhe'^, where po and m are con- stants and e is the base of the natural system of logarithms. If p = 719 when h = 450, and p = 594 when h = 2000, find p U h = 8000. Before starting the solution, notice that po is the atmospheric pressure at sea level; for ii h = 0, p = poe~^ = poe° = po. A more convenient form of the given equation, for purposes of compu- tation, is obtained by equating the common logarithms of both sides of the equation, which gives: log p = log Po - mh log e. (1) Substituting the given data we have the two equations log 719 = log Po - 450 X 0.4343 m, (2) log 594 = log Po - 2000 X 0.4343 m. (3) These equations may be solved simultaneously for the constants m and poi as follows: Subtracting log 719 - log 594 = 1550 x 0. 4343w 2.8567-2.7738 0.0829 Hence m log 1550 = 3. 1903 log 0.4343 = 0.6378-1 1550x0.4343 wi = 0.0001232 1550x0.4343 log 0.0829 log m 2.8281 0.9186 0.0905-4 Substituting this value of m in (2) we have log Po = 2. 8567 4- 450 x 0. 4343 X . 0001232 = 2. 8567 4- 0. 02407, log Po = 2. 8808 log 450 x 0. 4343 m = 0. 3815 - 2 Po = 760. 450 X 0. 4343 m = 0. 02407 log 450 = 2.6532 log 0.4343 = 0.6378-1 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 239 Hence p = 7Q0e-^-^^^'^'^. Uh = 8000, p = 760e-o-985«, log 0. 9856 = 0. 9937 - 1 i log p = log 760-0. 9856 log e. log 0. 4343 = 0.6379 -I r =2.8808-0.4282 0.6316-1 = 2. 4526. 0. 9856 log e = 0. 4282 Hence p = 283.5 if /i = 8000. EXERCISES 1. Solve for x, (a) 7* = 4. (b) 4*-i = 9. (c) 4^ = 7. (d) S^'-^ = 5*"'. 2. Each of the persons receiving one of the letters mentioned in Exercise 18, page 232 was requested to forward 10 cents. Find the number on the letters at which the chain should have been completed if the object had been to raise $1,000,000. 3. The population of the United States in 1898 was approximately 75,000,000. Find the number on the letter at which the chain in Exercise 18, page 232, should have been completed if it was expected that every man, woman, and child in the country would receive one. 4. Would the method of Section 85 enable one to solve the equation 2* + 3* = 7? 5. Find a function of the type y = z^ which passes through the point (3, 2). Find the average rate of change of this function from x = 3 to a; = 4. 6. Find the common logarithm of any positive number x, assuming that a table of logarithms to the base e is given. 7. Find the numerical value of M = logio e (e = 2.718) from the table of common logarithms. By the method df Example 2, Section 85, find the value of m = loge 10. Find the reciprocal of M, and compare it with m. 8. Assuming that a table of logarithms to the base h is given, show that the logarithm of any positive number x to the base a is , logt X l0gaX = ,-^ logi a Hence show that log* 10 = , , or log« 10 logio e = 1. logio e 9. By means of the preceding exercise show that loge X = m logio X, where m -2. 3026, and logio x = JVf loge X, where M = 0.4343. i Illustrate these two equations on the graphs of loge x and logw x. Note: The system of common logarithms, in which the base is 10, and that of natural logarithms, in which the base is e = 2.718, are the most important systems of logarithms. The former is used for numerical 240 ELEMENTARY FUNCTIONS computations, the latter for most theoretical work. The relations be^ tween these two systems obtained in Exercises 6 to 9 may be summed up as follows: If the common logarithm of a number is given, the natural logarithm is found by multiplying it by m = ,— i- =logJ0 = 2.3026. logix) e The constant m is called the modulus for changing from the common to the natural system. If the natural logarithm of a number is given, the common logarithm is found by multiplying it by which is called the modulus for changing from the natural to the common system. 10. Solve Example 3, Section 85, using the table of natural logarithms (page 31 of the Tables). 11. Find log2 3, log2 5, log2 7, and then by means of some of the funda- mental properties of logarithms (Section 81, properties 7, 8, 9) construct a table of logarithms to the base 2 for the integers 1 to 10 inclusive. 12. A hawser of a ship is subjected to a strain of 6 tons. How many turns must be taken around a post in order that a man who can not pull more than 200 pounds may keep the hawser from slipping, if the coeflBcient of friction m = 0. 175 and S = Pe"^, where S =• strain in the hawser in pounds, P = pull of the man in pounds, 9 = the angle of contact of the hawser and post in radians. 13. A steamer approaching a dock has a velocity of 20 feet per second at the instant the power is shut off, and 10 feet per second at the end of 1 minute. Find the velocity at the end of 2 minutes if the law of motion is y = Voe~*^K 14. In a chemical experiment it was found that at the end of 1 hour there were 35.6 c.c. of a substance in solution and at the end of 3 hours, 18.5 c.c. If the amount A of the substance at any time t is given by the equation A = /ce"^', determine the time that elapsed before the amount was reduced to 5 c.c. 16. The intensity of light / after passing through a medium of thick- ness T is 7 = 7oe~^r. If light loses 3 % of its intensity in passing through a lens, what per cent of intensity will remain after passing through four lenses? 16. Radium decomposes so that the amount A remaining after a time t is A = Aoe~^. If I % disappears in 20 years, how long before one-half of the original amount will be gone? EXPONENTIAL AND LOGARITHMIC FUNCTIONS 241 17. As a body cools, at any time t the difference in temperature $ be- tween the body and the surrounding medium is given by the equation 6 = doe~"^. If a fireless cooker oven is 430*' above room temperature when t = 0, and 216** after one hour, how long before it will be 10° above room temperature? 86. Compound Interest. Interest is the money paid for the use of borrowed capital which is called the principal. Interest that is paid only on the principal is called simple interest. When interest is paid on unpaid interest, which is added to the principal periodically, it is called compound interest. If un- paid interest is added to the capital at yearly intervals it is said to be compounded annually. If the interest is added every six months or three months it is said to be compounded semi- annually or quarterly. In any problem in compound interest where the interest is compounded annually, four quantities are involved: P, the principal ; i, the rate of interest, which is the sum of money in dollars paid for the use of one dollar for one year. Thus, if the rate is4%,i = $0.04; n, the number of years; S, the amount, which is the sum of the principal and interest for n years. Theorem 1. The amount, S, of a principal of P dollars j in- terest compounded annually for n years at the rate ^, is S = P(l + i)\ The amount of one dollar for one year at the rate t is 1 + i. Hence the amount of P dollars for one year is P(l + i). The amount at the end of the second year is obtained by multiplying the new principal, P(l + i)> by the amount of one dollar for one year 1 + 1. Hence the amount at the end of the second year is p(i + i)(i + i) =p(i + iy. Similarly, the amount at the end of the third year is P(l + iy{l + i) = P(l + iy. Comparing these results, we see that the exponent of (1 + i) is the same as the number of years, and hence 242 ELEMENTARY FUNCTIONS The amount S at the end of n years is >S = P(l + i)". This equation involves four variables. If any three are given, the fourth may be determined by substituting the given values and solving for the unknown. The value of S, P, or i may be determined by algebraic processes while the determina- tion of n involves the solution of an exponential equation. Example 1. At what rate will $50 amount to $75 in 7 years? Here P = 50, *S = 75, and n - 7. Hence 75 = 50 (1 + iy, log 1 . 5 = 0. 1761. Then (1 + i)7 = ^ 5_ ^15^ ^ log 1 . 5 = 0. 0252. So that (1 + i) = v^O v^Es = 1 . 0598. = 1.0598, Hence i = 0. 0598, so that the rate is 5. 98 %. If the interest is compounded semi-annually (as in some savings banks), in n years there are 2n periods and the interest on one dollar for each period is i/2, so that the amount in n years is If the interest is compounded quarterly, the amount in n years is In general, we have Theorem 2. If interest is compounded m times a year, at the annual rate i, the amount after n years is the same as if the interest were compounded annually at the rate i /m, for mn years. That is S = P(1 +i7m)'"\ Example 2. Find the amount of $50 for 10 years with interest con- vertible into principal semi-annually at 4 %. Here P = 50, i = . 04, n = 10, m = 2, i/m = . 02 and mn = 20. I Hence ^1 = 50 (1 + . 02)2o log 50 = 1 . 5690 = 50 (1.02)20 20 log 1.02 = 0. 1720 = $74.30. log 5 = 1.8710 Definition. The present value, P, of a sum of money, S, due after n years, is the principal which must be placed at EXPONENTIAL AND LOGARITHMIC FUNCTIONS 243 compound interest at a given rate i, in order to amount to the sum S at the end of n years. Thus in the formula in Theorem 1, P is the present value of S. EXERCISES 1. A principal of $100, deposited in a trust company, bears interest at the rate of 4% compounded semi-annually. What will the balance be at the end of ten years if no withdrawals are made? 2. A mother promises her twelve year old boy that she will present him with $100 on his twenty-first birthday provided he abstains from smok- ing until that time. How much should she deposit in the trust company of the previous example in order to have an amount of $100 when the boy is twenty-one? ^ ■' ^s^ 3. At what rate would a siun of money double itself in 25 years if inter- est is compounded annually? 4. In how many years will a sum of money double itself if placed at com- pound interest if (a) the rate of interest is 5%, compounded annually? (b) the rate is 4 % compounded semi-annually? 5. If $100 be deposited in a trust company at 4 %, compounded semi- annually, how long before it will amount to at least $150? 6. A building and loan association offers an opportunity for an invest- ment to yield 8 %, compounded quarterly. (a) If $100 is invested, to what will it amount in 5 years? (b) What sum should be invested to amount to $120 in 7 years? (c) How long must $50 be invested to amount to $175? »(d) How long will it take any principal to double itself? (e) How long will it take for a sum to treble itself? 7. A man borrows $50 from a friend, and three years later returns $65. What is the equivalent rate of interest, if interest is compounded quarterly? 8. A man bought a diamond for $150 in 1900, and sold it for $400 in 1915. What rate of interest did he realize, assuming it to be compounded annually? 9. If a building lot is bought for $500, and its value increases by 10 % annually, what will it be worth in 5 years? 10. The number of students in a college increased from 275 to 460 in 10 years. At what rate did the number increase, assuming that the per- centage of increase each year was constant? 11. What sum should be deposited in a bank paying 3 % compounded semi-annually in order to pay off a debt of $500 due three years later? 12. Construct the gra,ph of the amount of one dollar, interest com- pounded annually at 6 %, as a function of the number of years n. Build the table of values for n = 10, 20, 30, 40. 244 ELEMENTARY FUNCTIONS 13. Verity the fact that the values of S obtained in Exercise 12 are In geometrical progression. 87. Annuities. Definition. An annuity consists of a series of equal payments made at equal intervals of time. The first payment of an annuity is made at the end, not at the beginning, of the first period of time, unless otherwise specified. Annuities are common in commercial life. The rent of a house or store, the premium of an insurance policy, the dividend on a bond, wages and salaries, the payments of interest on a mortgage, are examples of annuities. Theorem 1. The amount of an annuity of R dollars, payable annually, accumulated for n years at the rate i, interest compounded annually is, 1 The compound interest on each payment is obtained by the method of the preceding section. Since the first payment is made at the end of the first year, the first payment is at interest for n — 1 years and amounts to R{1 + i)^-K The second payment is at interest f or w — 2 years and amounts to R{1 4- ^>-^ etc. The next payment to the last is at interest 1 year and amounts to R{1 + i). The last payment bears no interest and amounts to R. Adding these amounts in the reverse order we obtain K = R + R{l+i)+ . . . -\-R{l+ ^>-2 + R(l + i^-K The terms on the right form a geometric progression whose sum is obtained by the formula S = — — r, where the first term 18 a = R, the last term is I = R(l + t)"~S and the ratio is r = 1 + i. Substituting, we have „ (l4^^)ig(l+^)"-^-/^ p (1 +iy-l EXPONENTIAL AND LOGARITHMIC FUNCTIONS 245 Theorem 2. The amount of an annuity, if R dollars are paid m times a year, accumulated for n years at the rate i, interest compounded m times a year, is equivalent to the amount of an annuity of R dollars, payable annually, accumulated for mn years at the rate i /m, interest compounded annually. Definition. The present value of an annuity of R dollars a year for n years is the amount that must be deposited in a bank, interest compounded annually at the rate i, so that, if R dollars be withdrawn annually, there will be no balance left after the nth withdrawal. Theorem 3. The present valu£ of an annuity of R cbllars payable annually for n years is I The present value of the annuity is the sum of the present values of R dollars due 1 year hence, R dollars due 2 years hence, etc., for n years. We have from the formula for com- pound interest that the present value P of a sum S due in n years is P = S{1+ t)~". Hence if A denotes the present value of the annuity, we have A = P(l + i)-i + R{1 + i)-^ + • • • -f P(l + i)-\ The sum on the right is a geometric progression for which a = P(l + i)-\ ? = /2(1+ ^)-^ and r = (1 + i)-\ Substi- tuting these values in the formula for the sum of a geometric progression, we have , _ (1 + i)-'R{l + ^)-" - P(l + i)-' (1 + i)-i - 1 'Multipl3dng the numerator and denominator of the right-hand side by (1 + i)+i and simpUfying, we have 1 - (1 + 1) I Theorem 4. The present value of an annuity, if R dollars fare paid m times a year for n years, at the rate i compounded m times a year, is equivalent to that of an annual annuity of R dollars for mn years at the rate i/m compounded annvully. 246 ELEMENTARY FUNCTIONS EXERCISES 1. If $50 be deposited annually in a building and loan association pay- ing 6%, compounded annually, what will the savings amount to in 10 years? 2. How much must a man save annually, and deposit in a savings and loan company paying 5%, compounded annually, in order to pay off a mortgage for $2000 after 5 years? 3. How long will it take to accumulate $2000 if $75 are deposited an- nually in a savings and loan company paying 6 %, compounded annually? 4. A father buys a bond for $1000, due after 18 years, which bears interest at 5 % payable semi-annually, for the college expenses of his in- fant boy. The interest payments are deposited in a bank paying 4% compounded semi-annually. How much money will be available IJ years later? 5. A man buys a house and lot, paying $1000 down, and agreeing pay $1000 annually for 4 years. What is the equivalent cash price money is worth 6 % per annum? Note that the first payment is not p of the annuity, since the first payment of an annuity is due at the end the first period. 6. A piano is sold for $100.'cash and $25 quarterly for 2 years. What is the equivalent cash price, if money is worth 4% compounded quarterly? 7. A man buys a house, giving back a mortgage for $5000 with interest payable annually at 6%. If the mortgage is to be paid off by five equal annual payments, covering principal and interest, what should be the amount of the annual payment? Hint: The annual payments consti- tute an annuity whose present value is $5000. 8. Which is the better offer for a house from the seller's standpoint, $5000 down and $1000 annually for 4 years, or $5000 down and $2000 annually for 2 years, if money is worth 4 %? 9. Six months before a boy enters college, his father wishes to deposit a sum in a savings bank paying 4 %, interest compounded semi-annually, which will enable the boy to draw $300 every 6 months during his college com'se. How much should he deposit? 10. It is estimated that a certain mine will be exhausted in 10 years. If the mine yields a net annual income of $2000, what would be a fail purchase price, money being worth 5 %? 11. A man holds a mortgage for $6000, due after 3 years, interest 6 % payable quarterly. If he disposed of it how much might he expect to rei- ceive, if money is worth 4 % compounded quarterly . Hint: The present value of the mortgage is the sum of the present value of $6000 due after 3 years and of a quarterly annuity of $90 (one fourth of a year's interest) . 12. A bond for $100 is to be redeemed at the end of 10 years, and bears interest at 5 per cent, payable semi-annually. At what price for the bondj i ^ EXPONENTIAL AND LOGARITHMIC FUNCTIONS 247 will the purchaser realize 4 per cent on his investment? Hint: The present value of the bond consists of the present value of $100 due 10 years hence and the present value of the annuity consisting of the semi-annual in- terest payments. 13. What is the purchase price of the bond in the preceding exercise if the buyer is to reahze 6 per cent? 14. An insurance company desires to offer, at the time a policy matures, an option between a 20-year bond for $1000, bearing interest at 5 per cent, payable semi-annually, and an equivalent cash payment. If the computa- tions of the company are on a 3 per cent basis, what should the cash pay- ment be? 15. At the maturity of an insurance policy, the company offers a cash payment of $1000 or the option of five equal annual payments, the first to be made one year after the policy matures. If the company assumes that money is worth 3 per cent, what should be the amount of the annual payment? 16. Three months before a boy enters college, his father deposits on the boy's account $1200 in a bank paying 4 per cent interest, compounded quarterly. The boy wishes to withdraw the money quarterly, in equal amounts, during the four years of his course. How much should he draw at a time? 17. A city issues 20-year bonds to the amount of $100,000 in order to raise money for the improvement of its water supply. A sinking fund to provide for the extinction of the debt can be accumulated at 4 per cent, interest compounded annually. How much must be deposited in the sinking fund at the end of each year? 18. A man buys an automobile for $1000, and estimates that he will be allowed $400 for it in purchasing a new car three years later. How much should he save every three months for the purchase of the new car, if he deposits his savings in a bank paying 4 per cent interest, compounded • quarterly? 19. A furnace costing $450 must be installed in a house every 10 years. * How much should the landlord save each year for this purpose, if the sav- ings can be accumulated at 5 per cent, interest compounded annually? 20. A man owes $2000 on which he pays 5 % interest. If he pays the debt, principal and interest, in 6 equal annual installments, what is , the amount of each payment? How much will he owe at the beginning of the second year? At the beginning of the third year? I 21. Find the Hmit of the present value of an annuity as the number of f payments increases indefinitely. Verify the result by finding the sum ? of the infinite geometric progression whose terms are the present values of the several payments. 22. A college graduate wishes to provide for a scholarship of $90 a year. Find the amount which he must present to the college, if collegiate 248 ELEMENTARY FUNCTIONS --f/f "vir wM jpi^ ' ^ J-S^L --- — \- ' r funds can be invested at 4^ %, using the result obtained in the preceding exercise. By what other method can the amount be found? 23. A raih-oad spends S900 annually to provide a watchman for a grade crossing. If money is worth 5 per cent, how much can they reasonably be expected to spend for the elimination of the crossing? 88. Graph of the Exponential Function kh""" . The form of the graph of If was determined on page 216. From it the graph of Mf may be found by multiplying ordi- nates by k (Theorem, page 89). Fig. 140 gives the graph of /c2^ for several values of fc. Notice that i/ = A; is the intercept on the 2/-axis. The form of the graph of the function 6"* may be found by dividing by n the abscissas of several points on the graph of ¥ (Theorem, page 151). The graph of 2"^ for several values of n is given in Fig. 141. The combination of the two theorems quoted shows that the graph of kb^ may he obtained as follows: Plot the graph of If, divide the abscissas of points on it by n, which gives the graph of 6"^, and multiply the ordir rmtes of points on the new graph by k. Example. Sketch the graph of 2 {2'"^). Here k = 2 and n = \. Dividing the ab- scissas by \ amounts to multiplying them by 3. Hence we first construct the graph of 2* and multiply the abscissas of points on it by 3. The ordinates of points on the curve so obtained are then multi- plied by fc = 2. The form of the graph of A;6"* depends on the value of 6, but for suitable values of k' and n' the same curve is the graph of k'c^'*. We shall use FiQ. 140. J/> _[_ " 7- ?»/ 7- .Am.1 '■I'.L -ujtt Mi 3^ I^JII /^ 7 Mtt y 7 -'4/ 2^7 -IM^-^ ^r -f-jo 2 ; 4 .'■■ |_ ^ Fig. 141. JL _j / L ■ ■fLj$g__ih' ty / -J7 ^7 7^ y ^ ^ ":^ T ■ 1 2 3 i i 6 7 r i FiQ. 142. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 249 b =- 10 as a standard value, and we shall now show that we can determine a constant m such that the graphs of kb"^ and A; 10"^ are identical. The graphs will be identical if and only if or b'^ = 10'^. This is essentially an exponential equation to be solved for m. Equat- ing the logarithms of both sides of the equation, nx log b =' mx log 10 = mx, whence m = n log 6. •» Then kb"^ = fclO" * '"«' \ Hence it is always possible to use the base 10, and we shall do so in the future unless the contrary is indicated. The most characteristic properties of the graph of y = klO""" are that it does not cross the x-axis, that it does cross the^-axis, and that it always rises or always falls. EXERCISES 1. Plot the graphs of the functions: (a) T^(22*). (b) K2^). (c) 3(20-6'). (d) |(2o-»*). (e) 2(3'^*). (f) 3(2-0-25*). (g) 2(3-0-^). (h) 0.25(2^). (i) 0.5(10^). 2. Plot the graph oi y = e~^. How can the graph ol y = ke~ ^ be obtained from it? This graph is called the probability curve. 3. Find the inverse of the function A:10^. What are the most im- portant characteristics of the graph of i/ = c log nx? 4. (a) Construct the graph of k log2 x for several values of k. (b) Construct the graph of log2 nx for several values of n. (c) Construct the graph of ^ log2 3x. 89. The Logarithmic Scale. The ordinates of points on the raph of logio x corresponding to integral values of x from 1 to 10 2/ A - .. _^_,.i^.^ ij ~ "" 1 ^^^ T 12 3 4 » ) I 1. .. . ^ X Fig. 143. inclusive, are a set of segments on the 2/-axis whose lengths from the origin are equal to the logarithms of the corresponding abscissas. 250 ELEMENTARY FUNCTIONS The line O'A' is an enlargement of the segment OA on the y-axis, the numbers on the right being the logarithms of those on the left. 10^1.00 A scale is called a uniform scale if the distance of a number from the point marked is equal to the ^^ number, the distance from to 1 being the unit segment. A scale is called a logarithmic scale if the distance of a number from the point marked 1 is equal to the logarithm of the number, the distance from 1 to 10 being taken as unity. The logarithms are spaced uniformly along the Hne O'A'j while the integers are spaced non-uni- formly. The utility of a logarithmic scale lies in the fact that the addition and subtraction of the logarithms of nimibers, and hence the multiplication and division of numbers, may be effected mechanically by the addition and subtraction of line segments on the Fig 144 scale. For instance, to find the product 3x2, we add the segment 1 - 2 to the segment 1-3. For this purpose, place a pair of dividers so that the points are on the extremities of the segment 1-2, and then with this opening place one end of the dividers on the point 3 and the other will touch the scale exterior to the segment 1 - 3 in the point 6, the required product. To find the quotient |, subtract the segment 1-2 from the segment 1—3 in a similar manner by means of the dividers. The logarithmic scale is sometimes called Gunter's scale after Edmund Gunter, who first made use of it for purposes of calculation in 1620. The dividers may be dispensed with if two identical logarith- mic scales are arranged to slide along one another as shown in Fig. 145, which shows the method of finding the product 3x2 and the quotient 3/2. For the product 4x5, the above method gives a point out- o' EXPONENTIAL AND LOGARITHMIC FUNCTIONS 251 side the scale. To avoid this we divide one of the factors by 10, then multiply by the other factor and move the decimal point one place to the right in the result. 7 8 910 910 4 5 7 8 910 i 5 6 7 8 910 3x2 = 6. 3 -^ 2 = 1.5. Fig. 145. To find the quotient %, divide 10 by 5, multiply by 4 and move the decimal point one place to the left in the result. The following figures illustrate the methods: 4 S S 7 8 9 1 4 S 6 7 8 9 10 4 S 6 7 8 9 1 5 6 7 8 9 10 4 X 5 = (A X 4)10 = 20. I = (V- X 4)tV = 0.8. FiQ. 146. The logarithmic scale finds practical application in the slide rule, an instrument much used by engineers in calcu- lations. The slide rule consists of a ruler with a central portion which slides back and forth in grooves on the rule, the slide and rule 252 ELEMENTARY FUNCTIONS being graduated with equal logarithmic scales which are labeled with the corresponding numbers. The ordinary sHde rule has four scales A, B, C, D, as shown in the figure. Each of the scales C and Z) is a single logarithmic scale. Each of the scales A and B consists of two logarithmic scales, which are equal. If we give to the left index on all four scales the value 1, then the scales A and B extend from 1 to 100 while scales ^ ^MmMnifi.Ui i ..ii! . ui..fi./.|^,r , |1lV ^ 1 I* Is %. 5 6 7 S 9 'l £"""" ' 1^' i^fmmimA^ ^ f\ [^ i 5 6 7 8 91 TTT FiQ. 147. C and D extend from 1 to 10. Hence the scales A and B, numerically considered, are twice as long as scales C and D, and a number on the upper scales is the square of the number below it on the lower scales. Conversely, a number on the lower scales is the square root of the number directly above it on the upper scales. The methods employed to keep the result on the scales in calculations of the type 4x5 and 4 /5 are not necessary if the upper scales are used. The lower scales give more accurate results, since the graduations are not so fine as those of the upper scales. The instrument is provided with a runner, R, which enables coinciding points to be found on any of the scales, and alsc permits of extensive calculations being worked out without the necessity of recording the intermediate results. The slide rule makes use of matissas only, the characteristic^ being determined by inspection. j Logarithmic scales are also used on the axes of cross-sectioij paper. If the scales on both axes are logarithmic the cross| section paper is called logarithmic paper. If the scale on on^ of the axes is logarithmic and on the other is uniform it is call«j semi-logarithmic paper. The methods of using these cross section papers are shown in the following examples. I EXPONENTIAL AND LOGARITHMIC FUNCTIONS 253 Example 1. Plot the graph of the equation obtained by taking loga- rithms of both sides of the equation y = 3x^. The required equation is log ?/ = log 3 + 2 log x. Letting log y = Y and log x = X, we have Y = 2X + log 3, which is a linear equation, and hence its graph is the straight line whose slope is 2 and whose intercept on the F-axis is log 3. The tables give corre- sponding pairs of values of x, y and of X, Y, and the figures show cor- responding parts of the graphs. Only the part of the graph of y = Sx^ in the first quadrant corre- sponds to the graph in the X, Y system, as the logarithm of a nega- tive number is not a real number. y 0.1 .03 1 3 2 12 300 X Y — 00 — 00 -1 0.4771 - -2 0.4771 0.3010 1.0792 1 2.4771 If logarithmic scales are con- tructed on the X and F-axes the graph may be plotted from the table of values of x and y since the logarithmic paper serves the pur- pose of finding the logarithms as is shown in the following example. Example 2. Plot the graph of the equation y = Sx^ on logarithmic paper. Let X and y represent the numbers on the logarithmic scales on the axes, the unit lengths 1 to 10 being the same, and X and Y the num- bers on the uniform scales attached to the figure, so that X = log x and Y = log y. We follow the same procedure as in plotting the graph of the equation on ordinary cross-section paper, except that negative values of x and y are not used, and the lines through the point (1, 1) are usually chosen as the axes since log 1 = 0. 254 ELEMENTARY FUNCTIONS Y y 12 •10 9 8 7 6 5 4 3 i 1 1- / / / / 1 1 / / 2 3 456789 10 OJ ] L jr 3 12 The point (1, 3) will be at the inter- section of the lines a: = 1 and ?/ = 3. The points cor- responding to other pairs of numbers are similarly- plotted. These points lie on the straight line whose equation Y = 2X + log 3 was obtained in the preceding example. Note that the uniform scales and not the logarith- mic scales determine the slope. Thus m ■= log 12 Fig. 149. log 2 log22 log 2 - log 3 _ log 4 - log 1 ~ log 2 2 log 2 log 2 2. Theorem 1. The graph of the power function, y= fcx", plotted on logarithmic paper is the straight line whose slope is n and whose in- tercept on the y-axis is k. Let X and Y represent numbers on uniform scales attached to the axes so that X = log x, and Y = log y. Taking logarithms of both sides of the equar tion y = kx'^ we have log y = \ogk + n log x, or Y = nX -\- Ky the graph of which referred to the uniform scales is the straight Une with slope n and intercept K. Hence, 1 i 4 :\ "1 \ 1 . 1 \ \ .^ 5^ \ A B \ SI A^ 7 \ yj ^^ x/ 7 ~ ^ ^-^ X ^ A% , / c^ c\/ \ y 2^-- \ » Cy/^ / ) \ / / \ \ s O 2 3 4 5 6 7 8 9 10 a> " Fiq. . L50. i X EXPONENTIAL AND LOGARITHMIC FUNCTIONS 255 using the notation of the logarithmic scales in place of X, Y and K, we have the theorem. Example 3. Find the equation of the line marked c in Fig. 150. The intercept on the t/-axis is 2. Since the line passes through the points (1, 2) and (10, 10) the slope is log 10 - log 2 _ l og 5 _ 0.6990 log 10 - log 1 " log 10 ~ 1 = 0.6990. Hence the equation is y = 2{lQP-^^'>). Example 4. Plot the graph oi y = SCIO"-***) on semi-logarithmic paper with the scale on the x-axis uniform and on the y-axis logarithmic. Let Y represent numbers on a uniform scale attached to the y-axis, so that Y = log y. The unit hne on the x-axis is equal to the imit segment 1 to 10 on the i/-axis. Constructing the table of values and plotting the points as usual, we have the graph in the Y figure. X y 3.00 0.5 5.33 1.0 9.49 Taking the logarithms of both sides of the equation y = 3-10°-5*, we obtain the equation log y = \ogS + 0.5a;, or F = 0.5x + log3, whose graph is a straight line referred to the uniform scales on the axes. Hence the graph oi y = 3* 10°*^* plotted on the semi-logarithmic paper is the ^ straight line whose slope is 0.5 and whose intercept on the 2/-axis is log 3. Note that the uniform scales determine the slope. Theorem 2. The graph of the exponential function y = fclO*"* b plotted on semi-logarithmic paper is a straight line whose slope is m and whose intercept on the y~axis is k. The proof of this theorem is left as an exercise. 256 ELEMENTARY FUNCTIONS EXERCISES 1. Make a paper slide rule by copying the logarithmic scale given in Fig. 144 on two paper rules of an adequate degree of stiffness. With this instrument calculate the following and describe the process. 4 X 6, I, f, 6 X 9, (5 X 7)/(8 x 9). 2. Write the equations of the lines a, 6, c, . . . g, in Fig. 150. 3. Write the equations of the lines in Fig. 152. 4. Plot the graphs of the following equations on logarithmic paper. (a) 2/ = ar», (b) y - 3x2, (c) y-2x\ (d) y - hx^ (e) y = 3x, (f) 4 (g) pv = 4, (h) pv vr A .5 .6 .7 .8 Fig. 152. 6. Plot the graphs of the following equations on semi-logarithmic paper. (a) 2/ -2(102-), (b) 2/ = 3(10-^) (c) y = K10-), (d) y = e^. 6. Measure the length of the loga- rithmic scale in Fig. 144 and calculate the radius of a circle whose circumfer- ence is equal to this length. Fasten a circular disk of this radius to a larger disk by means of a pin through their centers so that they may rotate freely. On the circumference of the smaller disk and on the circle of equal radius of the larger copy the logarithmic scale of Fig. 144. Make the calculations of Exercise 1 with this instrument. What advantage has the circular slide rule over the slide rule, of the first exercise? 7. Find the equations of the lines in Fig. 150 if the given ranges of the scales are as follows: (a) X-axis, 10 to 100, (b) 0. 1 to 1, (c) 0.01 to 0. 1, y-axis, 1 to 10. 10.0 to 100. 0.1 to 1.0. 8. Find the equations of the lines in Exercise 3 if the ranges of the scales are as follows: (a) 10 to 20, (b) -5toO, (c) 4 to 8, 0. 01 to 0. 1. 10 to 100. 100 to 1000. \ X-BXIS, j/-axis, 90. Empirical Data Problems. If the points representing a! given table of empirical data appear to lie on the graph of I EXPONENTIAL AND LOGARITHMIC FUNCTIONS 257 an equation of the form y = klO""'^ the constants k and m may- be determined as in the Example. Determine the law representing the table, and find the value of y ii X = 2.5. Plotting the points whose coordinates are x\ 1, 2, 3, 4 the pairs of values in the table we get a curve y |2.37, 3.75, 5.97, 9.45 which appears to be of the form 2/ = fclO^. (1) To test this assumption, we plot the table on semi-logarithmic paper. As the points obtained lie very nearly on a straight line we conclude that the law may be represented by an equation of the form (1) with a fair degree of accuracy (Theorem 2, Section 89). y —9 p 1 —7 -6 1 / / —3 —a —1 / f j / ' ■t Y r y 10 6 1 ^__^..^'^ '*'* ^ "jrl^ ' i > ,< ? ^ I X Fig. 163. Fig. 154. ?o determine fc and m, equate the logarithms of both sides of (1), which (2) (3) log y = tnx 4- log fc or Y = mx-\-K, where F = log y and K = log k. Equation (3) is linear in the variables x and Y, and the constants m and K can be determined by the method in Section 27, page 78, from a table of values of x and Y. We therefore look up the logarithms of the given values of y, which are the values of Y. From this table we obtain, by the method referred to, the values = 0.175. to the Since K = \ogk '= — - 1, 2, 3, m = 0.200 and iC _ .. #0.175, reference to the table shows that ^ ' "^'^' '^^^' ' ' '^' '^'^ k = 1.49. Substituting the values of k and m in (1), we have the required law, !/ = l. 49(10°- 20to) (4) If X = 2.5, 2/ - 1.49(100-200 X 2.5) = 4.73. If the points representing a given table appear to he on the graph of an equation of the form y = A;x" (page 119), a value of n may be chosen by means of the form of the graph, and the 258 ELEMENTARY FUNCTIONS value of k determined, as in Section 44, page 127. The method is unsatisfactory in that the only way to tell which of two values of n is the better, for example, whether n = 2 or n = | is the better, is to try them both. The following procedure is preferable. If the points representing the table appear to lie on the graph of an equation of the form y = kx^'y (5) plot the points on logarithmic paper. If the graph is now ap- proximately straight (Theorem 1, Section 89), we conclude that the law may be well represented by (5). As the logarithms of the two sides of (5) must be equal, we have logy = n log X + log kj (6) or Y = nX + K, (7) where Y = \ogy, X = log x, and K = log k. With the table of logarithms build the table of values of X and Y corresponding to the given table of values of x and y. From it the values of n and K may be found by the method in Section 27, page 78. The value of k is then found from that of K. It should be noticed that if we suspect that the law has the form (1) or (5) it is not essential that the table should ba plotted on semi-logarithmic or logarithmic paper, respectively. But this plotting strengthens our feeling of having a suitable law, and approximate values of the constants may be obtained easily directly from the gra ph (see Example 3, Section 89, and Exercises 2 and 3 of the preceding set). Of course, if the new graph is not approximately straight the choice of one of these two laws does not give a very good representation of the given , data. If the points representing the table, plotted on ordinary cross- section paper, appear to he on the graph of y = clog nx, (8) we interchange x and i/, and proceed as in the example above. The graph of (5) may be distinguished from that of (1) or EXPONENTIAL AND LOGARITHMIC FUNCTIONS 259 (8) by the fact that it passes through the origin, if n is positive, and that it does not cut either axis, if n is negative. EXERCISES 1. Find the law of growth of the population of the United States from the following data: Year 1 1880, 1890, 1900^ 1910, Population in millions | 50, 62, 76, 92, Let X = in 1880. Assuming the law of growth does not change find the population in 1920. 2. Water flows out of a sharp-edged circular opening in the vertical side of a tank. The table gives pairs of values of the head h, the height in feet of the surface of the water above the center of the opening, and the discharge Q, the number of cubic feet of water flowing through the opening in a second. Determine Q as a function of h. h I 0.390, 0.709, 1.013, 1.540, Q I 0.0376, 0.0505, 0.0602, 0.0740, 3. A rectangular weir is a rectangular notch in the wall of a reservoir or channel. The table gives pairs of values of the head h, the height in feet of the surface of the water above the bottom of the notch, and the dis- charge Q in cubic feet per second. Find the law. Find Q if ^ = 0.265. h I 0.053, 0.103, 0.160, 0.217, q\ 0.0422, 0.1095, 0.2065, 0.3200, 4. A triangular weir is a triangular notch in the side of a tank or channel. 'The table gives pairs of values of the head, h, and the discharge, Q. The angle of the notch in this case was 90°. Find the law. h \ 0.095, 0.169, 0.249, 0.330, 0.392, q\ 0.00615, 0.0302, 0.0780, 0.1590, 0.2445, 5. In a chemical experiment it was found that the concentration x of a solution was connected with the amount of precipitation y of a, metal as in the table. Determine the law and find y ii x = 5. X \ 1, 2, 4, 8, • 2/ I 1, 3, 9, 27, Find the velocity, v, of a certain chemical reaction as a function of the temperature, T, from the data given in the table. Find v ii T = 50°. T I 0°, 10, 20, 30, 40, V \ 1.00, 2.08, 4.32, 8.38, 16.19, 260 ELEMENTARY FUNCTIONS 7. Find the law connecting the maximum speed, v, of an electric vehicle and the total weight, W, in thousand pounds from the following data. What weight will reduce the maximum to 6 miles per hour? W \ 2.0, 3.0, 4.2, 6.5, 8.5 , V I 20, 14, 12, 10, .8, 8. Find the law connecting the cost, C, in cents per miles for tires, repairs, battery, electricity, of electric vehicles of weight, W, in thousand pounds. What will be the cost per mile of a vehicle weighing 10,000? W \ 1, 2, 4, 7, C\ 7.9, 9.1, 11.5, 16.0, 9. From the table find the law connecting the resistance, R (in pounds per ton), which a passenger train encounters at a speed, v (miles per hour). Find i2 if y = 60. V I 5, 10, 15, 20, 25, 35, r\ 5.9, 5.5, 5.4, 5.5, 5.6, 6.2, 10. Airships are provided with air bags within their hulls for regulat- ing the height of ascent. The total cubic contents of these air bags when filled is V = mV where V is the cubic contents of the airship. Values of m for various heights, H, are given in the table. Find the law and the value of m when H = 8. H in thousand ft. 1 1, 2, 4, 10^ m=V'/V I 0.04, 0.075, 0.141, 0.318, 11. Find the law connecting the load, L (in pounds per square foot of wing) of an aeroplane with the area, A, of wings (in square feet). Find the safe load ii A = 200. A I 150, 300, 400, 700, L I 9.4, 6.8, 6.2, 5.1, MISCELLANEOUS EXERCISES 1. The temperature of a body cooling according to Newton's law, 6 = 6(fi~^\ fell from 125° to 94** in 8 minutes. Find the equation con- necting the temperature and the time of cooling. 2. Construct a scale for the function x^ following the method used in constructing the logarithmic scale. Find by means of the scale (2.3)' and VTT. 3. Construct a table of values for the function y = 2x'' + 3 and plot the graph. Let u = x^ and hence y = 2u -\- Z. Construct on the hori- zontal axis a uniform scale of u and a corresponding non-uniform scale of X. Plot the table of values of x and y using the non-uniform scale of x and the uniform scale of y. What is the character of the graph? EXPONENTIAL AND LOGARITHMIC FUNCTIONS 261 4. Determine scales on the axes so that the graph oi y = S{1QP'°^ ) is a straight line. 5. Determine the constants my = /clO'"*^ by means of the pairs of values (2, 0.5) and (3, 0.8). Choose scales on the axes so that the graph of the equation is a straight line. From the graph read the value of t/ if x =2.5. 6. Find the radius of the moon in miles, given that the diameter of the moon subtends an angle of 32' as seen from the earth, the distance of the moon from the earth being 240,000. Find the ratio of the mass of the moon to that of the earth if the density of the moon is 0.6 that of the earth. 7. If the distance of the sun from the earth is 92,800,000 miles and the diameter of the sun subtends an angle of 32.4' as seen from the earth, find the diameter of the sun, and the ratio of the volume of the sun to that of the earth. 8. If the length, I, of a range finder is 20 feet and the distances, d, are calculated by the formula d = I tan 6, find the values of 6 corresponding to the extreme values 400 ft. and 20,000 ft. on the dial of the instrument. (Use the " log rad " table on page 28 of the Tables.) 9. The following table gives the collegiate running records, the distance d being in yards and the lime t in seconds. Find the law and the value of t\id = 600. Find f if d = 3520. d I 100. 440, 880, 1760. t I 9.75, 47.4, 113, 254.4, 10. An observer on a destroyer moving at the rate of 35 miles an hour notes that the line of sight to a ship makes an angle of 41°.56 with the for- ward path of the destroyer and that one minute later the angle is 74°.83. He estimates that the ship is moving on a parallel path at the rate of 14 miles an hour. Find the distance between the paths. 11. The economic law of diminishing utility is stated as follows: The total utility of a thing to any one (that is, the total pleasure or other benefit the thing yields) increases with any increase in one's stock of it but not so fast as the stock increases. If one's stock increases at a uniform rate, the benefit derived from it increases at a diminishing rate. Another way of stating the law is: The increase in the utility of a thing, or marginal utility, diminishes with every increase in the amount of it any one already has. Plot the graphs of these two statements of the law. 12. Find approximate values of the real roots of the following equa- tions, which can not be solved by the ordinary methods. Hint. Plot the graph of the function on the left and from it locate an intersection with the a;-axis as accurately as possible. Enlarge the table of values until the coordinates of two points are obtained such that the intersection lies between them and such that the part of the graph 262 ELEMENTARY FUNCTIONS 2-. Ix m between them may be assumed straight. Then determine an approximate value of the root by interpolation. (a) fix) = 2~* - ix = 0. Solution. The graph of f{x) shows that there is a root between 1 and 2, near 1.2. Enlarging the table of values, it is seen that the root is between 1.2 and 1.3, since the corresponding values of f{x) have opposite signs. In order to find a value of x by interpolation, construct the graph between x = 1.2 and x = 1.3 on the assumption that it is a straight line BD. To find AE we have, by similar triangles, AE^AB 6 BF~ DF' '--- Substituting BF = AC = 0.1, 035 AB = 0.035, -0.027 and DF = 0.027 + 0.035 = 0.062, -1 1 2 1.2 1.3 2 1 _i 0.435 0.406 0.400 0.433 2i 1 ""- ^ _,s ^v ^^^ 2 \S x\ eSv ^^^ ^ 1^^ L 5^^ iki- ^5^"^ - ^^ ^-'•' S^ ^ ~ ^ ' V 2 or : :_ :_± we get whence AE 0.1 0.035 0.062 AE = 0.056. = 0.56, Then the abscissa of ^ is x = 1.2 + 0.0c6 = 1.256, which is an approxi- mate value of the root of the given equation. .035 "\' -— 1 1 1 1 A ^*"**^,^ C X 1 £^^N ■->^-.027 Fig. 156. (d) 2-» - x/2 -1=0. = 0. (g) 2 sin X - 3x/2 = 0. ). (j) sin X - x"^ = = 0. Fig. 1-55. (b) 2-* - x = 0. (c) 3-*/2 -x = 0. (e) 4-*-a;/5 = 0. (f) sin x - 3a;/4 (h) cos X - X = 0. (i) tan a;- 2x = (k) 2~* - sin X = (smallest positive root). (1) cos X - 2~* = (smallest positive root), (m) e-* - a; = 0. (n) e"* - x/3 = 0. 13. Some properties of many functions can be expressed entirely in terms of the notation f(x) . If analogous properties of two functions can be so ex- pressed an abstract point of view is obtained which gives a deeper insight into the differences between the functions. Establish the following relations: » (o) e-*/2 - x/3 = 0. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 263 (a) If J{x) = x^ or cos x, If J{x) = a:' or sin x, (b) If/(x)=rc«, If fix) = log X, (c) If/(x)=x«, fi-a)-m. /(-a) = - /(a). fiab)=f{a)xm. f(p)_f(A m ft). (d) If /(a:) = Vx, kj{a) ^f{k%), Jifix)=logx, fc/(a)=/(a*). 14. The speed of an aeroplane is 80 miles an hour and the wind is blow- ing from the north with a velocity of 20 miles per hour. The pilot desires to move S. E. Find the direction in which he should head the machine and how fast he will move. 15. Solve the preceding exercise supposing that the pilot desires to sail N. W. If the aeroplane can stay in the air 4 hours, find the greatest dis- tance the pilot can sail S. E. and be able to return to the starting point. 16. The speed of an aeroplane is 100 miles per hour and the wind is blowing from the west. The plane can stay in the air 5 hours. How far can the pilot sail in the direction 10° S. of W. and return to the starting point? CHAPTER VI DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 91. Introduction. If ?/ is a function of x, the average rate of change of y with respect to x in any interval Ax is Ay /Ax (page 35). If the average rate of change is constant, its value is the rate of change of y with respect to x (Definition, page 48), and the graph of 2/ is a straight line (Theorem, page 50) whose slope is the rate of change. Numerous appHcations of uniform rate of change were given in Chapter IT. If the average rate of change, Ay /Ax, is not constant, then the rate of change of y with respect to x is defined to be the limit of Ay /Ax as Ax approaches zero (page 94). It is repre- sented graphically by the slope m of a fine tangent to the graph of the function y. In this chapter we shall consider this limit more formally than heretofore, and derive rules for finding it expeditiously if y is an algebraic function of x. The appHcations are based either on the geometric interpretation of the limit of Ay /Ax as the slope of the tangent line or on the physical interpretation of the limit as the rate of change of y with respect to x. The ideas to be considered in this chapter, and the one following, are among the most fundamental and far-reaching concepts in mathematics. They were developed by the famous Englishman Sir Isaac Newton (1642-1727) and the noted German Gottfried Wilhelm von Liebnitz (1646-1716), and a bitter controversy lasting for many years was waged over the question as to which one of these men should be accorded the honor of the discovery. Leibnitz was the first to publish some of his results, in 1684, but Newton had written a paper on the subject and submitted it to some of his friends in 1669. The 264 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 265 I followers of each claimed that the other had been guilty of claiming ideas not his own, but most historians of mathe- matics are agreed that the work of Newton and Leibnitz was independent. On the basis of the work of these men, there followed a period of rapid and extensive mathematical develop- ment. »! Let us first consider the underlying concept of the limit of a variable. 92. Limits. By definition (footnote, page 93) the limit of a variable x is sl constant such that the numerical value of the difference between x and a becomes and remains as small as we please. This explains what is meant by saying that '' x ap- proaches a as a limit," or in more compact form, '^ x approaches a." It is immaterial whether or not x becomes equal to a. A function of a: is a second variable y (Definition, page 5). The functions we have studied are such that if x approaches a limit, so also does y, provided that y does not become infinite as X approaches a. The notation Hr hmy = b (!) is used to indicate that " the limit of y, bs x approaches a, is 6." This means that the numerical value of the difference be- tween y and b can be made as small as we please by taking x sufficiently near to a. Graphically, the difference between the ordinates y and h can be made as small as we please by making the ; difference between the abscissas x and a sufficiently small. In other words, Ay = b - y approaches zero ii Ax = a — X approaches zero. For example, the average rate of change of the function ax^ in an interval Ax beginning at a definite point x is (see (5), page 95) Ay FiQ. 157. Ax 2ax + a Ax, 266 ELEMENTARY FUNCTIONS The rate of change of y with respect to x for a given value of X is therefore m = Um -T^ = Hm (2ax + aAx). Ax=0 i^X Aa;=0 We are thus led to find the limit of 2ax + a Ao;, a function of Ax, as Ax approaches zero. In computing this limit, x has a given value and is regarded as constant. In order to prove the assumption made earher that the limit is 2ax, it must be shown that a Ax, the difference between the variable 2ax + a Ax and the constant 2ax, can be made as small as we please by making Ax sufficiently small. This follows readily. For if we wish to make a Ax as small as 0.001, it is sufficient to take Ax = 0.001 /a, which is possible since Ax approaches zero and can therefore be made as small as we please. Ay The limits encountered in computing lim -r- , where y is a, Ax=0 ^X polynomial or a rational function of x, can be computed by means of the following theorems, which we assume without proof. Theorem 1. The limit of the sum of several variables is the sum of their limits. Theorem 2. The limit of the product of several variables is the product of their limits. Theorem 3. The limit of the quotient of two variables is the quotient of their limits, provided the limit of the divisor is not zero. The limit of the difference of two variables may be found by Theorem 1, since u - v = u + (- v). In applying these theorems, it is frequently convenient to regard ?/ = c, a constant, as a function of x, whose limit, as x approaches a, is c. Example. If y >= x^, then Ax Give the details of the computation of the limit of Ay/ Ax as Ax approach* sere. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 267 Denoting the limit by w, we have m = lim ^ = lim (3x2 + 3x Ax + ^x'^) = hm 3x2 ^ lim 3a; ^ ^ ]^j^ ^ ^ (Theorem 1) Ax=0 Ax=0 Ax=0 = 3a;2 + lim Zx hm Ax + hm Ax hm Ax Ax=0 Ax=0 Ax=0 Ax=0 (Theorem 2) = 3x2 (Since lim Ax = 0) Ax=0 93. Derivative of a Function. Definition. If 2/ is a func- tion of X, and if At/ is the change in y corresponding to a change of Aa; in x, then the limit of Ay /Ax, as Ax approaches zero, is called the derivative of y with respect to x. Denoting it by Dxy (read " the derivative of y with respect to a: '0; the definition may be expressed by the equation le results obtained in Section 33, page 94, may now be ited as follows: \The derivative of y with respect to x, Dxy, measures the rate of mge of y with respect to x. The derivative of y with re- spect to X is represented graphi- cally hy the slope of a line tangent to the graph of y. That is Dx2/ = m = tan 0. (2) jj The process of finding the If derivative of a function is called differentiation, and the succes- sive steps in the process have been given in the section cited above. Example. Differentiate y = Substituting x + Ax f or x and y + Ay for y, 1 y+ Ay (x + Ax)2 268 ELEMENTARY FUNCTIONS Subtracting the value of y, _ 1 _ L _ a^^ - (a^ + Ax)^ -2x Ax- Axy ^ ~ (a; + ^xY x^ ~ x\x + Aa;)2 " x^a: + Ax)^ Dividing by Ax, Ay _ — 2x - Ax Ax "" x2(xTax)2* Passing to the limit as Ax approaches zero, lim (- 2x - Ax) D = hm ^ = lim " ^^ " ^^ ^ Ax==o *^ Ax=0 Ax Ax=0 x^ix + Ax)2 " lim x2(x + Ax)2 Ax=0 by Theorem 3 of the preceding section. Applying Theorem 1 in tl numerator and Theorems 1 and 2 in the denominator, we get n _ 2±- 2 ^^y - " x2x2 ~ " x3* EXERCISES 1. Evaluate the following limits, indicating in detail the use of tl theorems in Section 92. (a) lim (2-3x + x2). (b) lim a;' - 3x x=0 x=2 x+l (c) lim (2x + Ax - 4) . (d) Hm (3x2 + 3a; ax + Ax^ - 6x - 3 Ax)il AX = AX:S:0 (e) lim x2 + X Ax + 1 (f) lim 2x+ Ax-x2-xAx Ax=0 x{x + Ax) * Ax=0 x2(x + Ax)2 2. Differentiate the functions: (a) y = x3 - 3x. (b) y = 1/x. (c) y = (x + l)/x. (d) y = c. (e) 2/ = x. (f) mx + 6. (g) y = y/x. Hint. Rationalize the numerator of Ay /Ax before passi to the limit as Ax approaches zero. 3. If /(x) = (x2 + 3x)/(x2- 1), find lim /(x). Note that the value obtained is f(a), provided o 5^ ± 1. What happens if o = :*= 1? 4. If fix) = x2, prove the relations lim fix) - fid) and lim Ay -= x^a Az^O and interpret them graphicall}- . 6. Prove that the relations in Exercise 4 are true if fix) is any qi ratic function. 5. . I DlFFEREMTiATlON OF Al.GEBRAlC FUNCTIONS 269 6. If f{x) is any function, and Ax = a - x, show that the first relation in Exercise 4 is true if and only if the second is true. What is the graphi- cal significance of each relation? Note. A function is said to be continuous at x = a ii the first relation in Exercise 4 is true for the given function. It can be proved that the algebraic and transcendental functions studied in this course are con- tinuous for all values of x for which they do not become infinite. If a function becomes infinite as x approaches a, then f{a) has no meaning. Hence x = a is a point of discontinuity. There are other types of points of discontinuity. 7. Prove that (a) any polynomial, (b) any rational function, is con- tinuous for all values of a, except, in (b), for the values for which the func- tion becomes infinite. /\u 8. If w is a function of x, what is the value of lim Aw? of lim — — ? of AxaO Ax=o Ax ,. Au^ ,. . Aw„ lim — — = lim Au -T— ? Ax=0 Ax Ax=0 Ax b 94. Fundamental Formulas for Differentiation. The rules in this section are useful in differentiating a function without the labor involved in computing the limit considered in Section 93. I Theorem 1. The derivative of a constant is zero; that is, D:,C = 0. (1) For the graph of i/ = c is a straight Hne parallel to the x-axis, whose slope, m = Dxy = DxC, is zero. Theorem 2. The derivative of the in- dependent variable is unity, that is, D^x = 1. (2) For the graph oi y = x is the straight line bisecting the first and third quad- rants, whose slope, m = Dxy = DxX, is unity. Theorem 3. The derivative of a constant times a function is equal to the constant times the derivative of the function. Symbolically, if u is any function of x, Fig. 159. Let DxCU = cDxU. y = cu. (3) 270 ELEMENTARY FUNCTIONS Then y + Ay = c{u + Au), and hence, subtracting, Ay == c Au. Dividing by Ax, S = ^S* XT T^ T Aw T Au ,. Au T^ Hence Dxy = hm -r^ = hm c -r- = c um -r— =cDxU. Ax=o Ax Ax=o Ax Ax=o Ax What is the graphical interpretation of this theorem, in the Ught of the Theorem on page 89? Theorem 4. The derivative of the sum of two functions is equal to the sum of their derivatives. Symholically, if u and v are any two functions of x Dx{u + y) = D^u + DxV. (4) Let y = u + V. Then y -\- Ay = u + Au + v -{■ Av, and hence Ay = Au + Av. •r^- • T 1 A Av Au Av Dividing by Ax Sc^Ai+M' Therefore D.y = ^n^ g = jim (^ + ^) ,. Aw , ,. At; y. , y^ = hm -7 — h hm -r- = D^u + D^v, Ax=iO ^^ Ax=0 ^a; Corollary. The derivative of the sum of several functions is the sum of their derivatives. Theorem 5. The derivative of the nth power of a function of X is n times the (n - \)st power of the function times the deriva- tive of the function with respect to x. Symholically, if u is any function of x Z)xi/" = nW"-^ DxU. (5) Let y = w". Then y + Ay = {u + Au)"" . = w" + nw"-i Au + ^^^ " ^^ w"-2Aw2 + • • • + Au", by the binomial theorem. Subtracting, Ay = nu^-^'Au + ^ ^V ^^ w'^-^Aw^ + • • • + Au\ DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 271 Hence, A?/ , Au n (n — 1) „ Au . Au . , Ax Aa;2 Ax Ax and therefore, passing to the Limit as Ax approaches zero, Dxy = nu''~WxU, Corollary. The derivative of the nth power of x is n times the in - l)st power of x; that is, DxX'' = nx«-i. (5a) For, by (5), DxX" = nx^'-^DxX = nx^-^ since DxX = 1, by (2). Example 1. Differentiate y = 3^ + Sx^ + 5. We have I>x2/ = -DxX^ + D^Sx^ + D^5 by (4) = 3a:2 + 3D.X2 + by (5a), (3), (1) = 3x2 + 3-2x by (5a) = 3x2 ^ 6^.^ Example 2. Differentiate y = (x^ + 4x - 5)'. If we think of x^ + 4x - 5 as a function u, the given equation has the form y = y?. Applying (5). D^y = DM + 4x - 5)5 = 3(x2 + 4x - 5)2Dx(x2 + 4x - 5) = 3(x2 + 4x - 5)2[Dxx2 + DAx + D^{- 5)] ^^ = 3(x2 + 4x - 5)2(2x + 4). ™ 95. Derivative of a Poljmomial. Any polynomial has the form (page 133) y = oox" + aix"-i + • • • + an-ix + a„. Then D^l/ = I>x(aoa:" + aix""^ + • • • + an-ix + an) = D:c(aoX-) + Z)x(aia;"-i) + • • • +Dx(an-ia:)+Dxan by the corollary to Theorem 4, Section 94, = OoDxX'' + aiOxX""-^ + • • • + an-iDxX + »by Theorems 3 and 1, Section 94. = aonx""-^ + ai{n - l)a;"-2 _|_ . . . 4. an-i by (5a), Section 94. Hence, The successive terms of the derivative of a polynomial may he found from the corresponding terms of the polynomial by multiply- ing the coefficient by the exponent and decreasing the exponent by 272 ELEMENTARY FUNCTIONS one. In applying this rule, the constant term, whose derivative is zero, may be regarded as the coefficient of a^ = 1. For example, if y = 3x^ - 4x^ + 3a; - 7, then D^y = 12x3 _ i2a;2 + 3. 96. Corresponding Properties of a Function, its Graph, and its Derivative. The derivative is useful in expressing some of the properties given on page 42, and some other properties also. Rate of Change. The steepness of the graph at any point is measured by the slope of the tangent line, and the rate of change of the function is measured by the value of the derivative. Changes of the Function. The graph rises (or falls) to the right if and only if the slope of the tangent line is positive (or negative). Hence A function increases (or decreases) as x increases if and only if its derivative is positive (or negative). Maxima and Minima. A line tangent to the graph is hori-_ zontal if and only if its slope is zero. Hence, To find the abscissas of the points at which the tangent line horizontal, set the derivative equal to zero, and solve the resulth equation, Dxy = 0, for x. The roots of this equation may be values of x for which the function has a maximum or minimum value. At a maximum point, the function ceases to increase and begins to decrease, and hence (see Changes of the Function above), as x in- creases, the derivative must change sign from positive to negative. Similarly, at a minimum point, as x increases, the de- rivative must change sign, from negative to positive. If, as X increases in the vicinity of a horizontal tangent, the derivative does not change sign, the point of contact is neither a maximum nor a minimum point. y ^ 1 /B\ J / ^\ J^ J " \J « D Fig. 160. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 273 For example, the graph oi y = x^ has a horizontal tangent at the origin (page 112), since Dxy = Sx^ = ii x = 0. But as X increases through x = 0, the derivative Sx^ does not change sign, and the origin is therefore not a maximum or minimimi point. Concavity. The curve in the figure is concave downward from A to C. If the tangent line at A rolls along the curve until the moving point of contact reaches C, the slope m de- creases, and hence Dxtn is negative (see Changes of the Function above). Similarly, from C to E the curve is concave upward. The slope of a tangent hne roUing from C to E increases, and Dxm is positive. Hence // the graph is concave upward {or downward) ^ Dxm is positive {or negative). It is also true that the curve is concave upward (or downward) only if Dxm is positive (or negative). It follows that a point at which the tangent line is horizontal will he a minimum point if Dxm is positive, or a maximum point if Dxm is negative. Since m = Dxy is the rate of change of y, and since m in- creases or decreases according as Dxm is positive or negative, it follows that The rate of change of a function increases or decreases, as x increases, according as the graph of the function is concave up- ward or downward. Points of Inflection. At a point of inflection (Definition, page 139) m ceases to increase and begins to decrease, or vice versa. Hence m has a maximum or minimum value, and there- fore Dxm = (see Maxima and Minima above). The curve is concave upward on one side of a point of inflection and concave downward on the other (see Concavity). Not every root of Dxm = is a point of inflection. It is necessary that as x increases through the root in question Dxm shall change sign (see Maxima and Minima). For ex- ample, if 2/ = x^, then m = Dxy = 4a;^ and Dxm = Vlx"^; Dxm = if a; = 0, but it does not change sign as x increases through zero. Hence the origin is not a point of inflection. It is, in fact, a minimum point, as is apparent from the graph (page 112), 274 ELEMENTARY FUNCTIONS or' from the fact that DxV = 4x^ changes sign, from negative to positive, as x increases through zero. Since m is the rate of change of y, At a point of inflection on the graph of a function the rate of change of the function has a maximum or minimum valu£. The corresponding properties of the graph of a function and the deriva- tive considered in this section may be indicated as follows: Property of Graph Slope of tangent line. Maximum or minimum point. Graph rises or falls to the right. Point of inflection. Graph concave upward or down- ward. Property of Derivative Value of Dxy. D,y = 0. Dxy is positive or negative. Derivative of m = Dxy is zero. Derivative of m is positive or nega- tive. This table supplements the table of corresponding properties of a func- tion and its graph given in Section 15. Example. If find the points of maxima, minima, and inflection. Plot the graphs of y, m = Dxy and Dxm, and discuss their relations to each other. Differentiating (1) m = Dxy and differentiating (2) Dxm Setting Dxy = 0, ^x' and solving for x, x Since Dxm = 6(2a;2 + x-l) the roots of Dxm = are x = 4rc3 + 3x2 _ 6^^ = 12x2 + 6x - 6. + 3x2 _ 6x = 0, = 0, 0.9, - 1.7. = 6(2x-l)(x + l), = I = 0.5, and - 1. (2) (3) (4) (5) (6) (7) y Dxy Dxm 4 -6 2.9 9.1 -1.2 18.5 3.4 -1.8 1 5 31 -69 84 -8 30 3 1 12 16 32 54 The values of y, Dxy, and Dxm cor- responding to the values of x in (5) and (7), which are found by substitution in (1), (2), and (3), are given in the table. The point A (0, 4) is a maximum point since the tangent line is horizontal iPxy = 0) and the curve is concave downward (Dxi/<0). While the points B(0.9, 2.9) and C(- 1.7, - 1.2) are mini- mum points, because the tangent line is horizontal {Dxy = 0) and the curve is concave upward (Z>xm>0). As X increases through 0.5 or - 1, in- DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 275 spection of the changes in sign of the factors of (6) shows that D^m changes sign in each case. Hence the points D(0.5, 3.4) and E{- 1, 1) are both points of inflection. rhe table of values is then extended to include the necessary integral values (Section 50, page 137), the computation of the entire table being made by means of synthetic division and the remainder theorem. The graphs are then drawn. The maximum and minimum points oi y, A, B, C, are directly above or below the points O, B\ C, at which DxP cuts the x-axis. The vertical v>^ -V D. m -ft -■^* " iv^ \ ^ t t \ \ tit \ \ ^ I t-i \ \ t ti \ z i 4^1 \ \ e' 6 h 1 / \ _J_ \ J.^ 4 17 \ f y K \ ^ 4 \ 1 / \ . V T" \ 7_ % t- ^r \ _d\ ^ -T- ii ■1 \A~y ' '\ ^^^' TiZ J i ' ^c ~ _ j[ 1^ T \ i / r \ *~t \~ \ r \ -^ 1 i>.v 1 Fig. 161. le through the maximum point A cuts the graph of Dxin below the x-axis and the vertical hnes through the minimum points B and C cut it above the X-axis. The points of inflection of y, D and E, the minimum and maximum points of Dxy, D' and E\ and the points at which Dxtn cuts the x-axis, D" and E", are respectively on the same vertical hnes. The graph of y rises (from C to ^, and from B to the right) or falls (from the left to C and from A to B) according as the graph of Dxy lies above or below the a;-axis. And the graph of Dxy = m rises or falls ac- cording as that of D^m hes above or below the x-axis. 276 ELEMENTARY FUNCTIONS The graph of y is concave upward (from the left to E and from D t6 the right) or downward (from E to D) according as the graph of Dxm lies above or below the x-axis. What further relations exist between the graphs of Dxy and D^m? 97. Velocity and Acceleration. The velocity of a body moving along a line is the rate of change with respect to the time of its distance s measured along the line from a certain station (Section 32, page 93). Hence, V = Dts, (1) The acceleration a of a moving body is defined to be the rate of change of its velocity v, and therefore a = Dtv. (2) Example. The position of a body moving on a line at the tune t is given by s = <3 - 5f2 + 2« + 8. (3) Find the velocity and acceleration at any time. Determine the posi- tion, velocity, and acceleration when i = - 1. When is the body at rest? Differentiating, we get V = Dts = 3^2 _ lof + 2, (4) and a = DtV = U - 10. (5) When « = - 1, we find that s = Q, v = lb, and a = - 16. Hence at that time the body passed through the station, moving in the positive direction at the rate of 15 feet per second, which was decreasing at the rate of 16 feet per second per second. The body will be at rest at any instant at which » = 0; that is when 3^2 - lOi + 2 = 0. (6) Solving for t, t = ?_±v^ = 0. 2 or 3. 2 seconds. EXERCISES ' 1. Differentiate the functions: (a) 3x^-20^- 4x2 + 5^ - 3. (b) ^ -Sx^ + 2x - 5. (c) 3^ix^ + 5x + 6). (d) {x -I) (x^ + x-h 1). / N x3 - 7x2 + 4 a; - 2 X + 1 (e) ^ (f) ^-X-^- 2. Differentiate 2/ = (x2 - 3)2; (a) by applying (5), Section 94, regarding tt = x2 - 3; (6) by first removing the parentheses. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 277 3. Differentiate the functions : (a) 2/ = (a;2 + 2x)\ (6) y = {2x^ -Sx + 7)^ (C) ?/ = (X2 - 4)5, {d) 2/ = (3X3 - 5X - 3)3. 4. Find the slope of the line tangent to the graph oi y = x^ - 4:X at any point P{x, y); at the point for which x = 1. Plot the graph, and draw the tangent at the latter point. 5. The distance fallen in t seconds by a ball thrown downward with a velocity of 48 feet per second is s = 48< + 16^2 Yind the velocity at any time. How fast will it be moving when t = S? 6. A billiard ball rolhng down a smooth plane inclined at a httle less thaa half a degree moves according to the law Find the velocity at any time. Plot the graphs of s and v on the same axes, using a large scale, from t = to t = 5. From the graphs answer the following : (a) What is the position and velocity of the ball after 4.5 seconds? (b) With what velocity will the ball be moving after it has rolled 3 feet? (c) How far must it roll to acquire a velocity of 2 feet per second? (d) When is the distance equal to the velocity? How many solutions? (e) Draw a tangent to the graph of s and measure its slope. Measure the ordinate on the point on the graph of v which has the same abscissa as the point of tangency. How do the two compare? 7. Find the points at which the tangents to the graph of ^ = a:^ - 27a; are horizontal. Construct the figure. 8. The height after t seconds of a body thrown vertically upward with a velocity of 96 feet per second is s = 96^ - 16i-. Find the velocity at any time. When will the velocity be zero? How high will the body rise? 9. Find the angles which the graph of y = — x^ + 5x^ — Qx makes with the X-axis. 10. At what angles does the line y = 4x cut the graph oi y = x^? 11. Oil dropped on the floor spreads out in a circle. Find the rate at which the circumference increases with respect to the radius, and the rate at which the area increases with respect to the radius. 12. The kinetic energy of a body of mass m moving with a velocity v is given by K = |m»2. Find the rate at which the kinetic energy changes with respect to the velocity. 13. If P is the pressure of a body on a surface and F the friction between them, what does the derivative of F with respect to P represent? 14. If s is a quadratic function of t, show that the acceleration is uni- form (constant). 278 ELEMENTARY FUNCTIONS 15. Find the points of maxima, minima, and inflection on the graphj of each of the equations below. On the same axes, plot the graphs of m = Dxy, and DxW, and discuss the relations between them. (a) y =^ x^ - 4x + 5. (h) y = x^ + 2x^ - 4x - 3. (c) y^x^- Qx^ + \2x + 3. {d) y = a^ ■\- Zx^ + 5z - 2. (e) 2/ = 3a:* - 4a;3 _ 63^2 + 12a; - 2. (f) y =^ x^ - x^ -2x + Zx - 1. (g) y = x^ -bx^ + 2x + 8. 16. The distance from the starting point, after t seconds, to a ball rollec up a plane inchned at a Uttle less than half a degree with an initial velocitj of 8 feet per second is How far will it roll up the plane, and how fast will it be moving when i\ returns to the starting point? 17. The position of a body moving on a line is given by one of the equations following. Find the velocity and acceleration at any time. Oi the same axes plot the graphs of s, », and a, and discuss the motion witl reference to each of them. (a) s = f-U-\-4.. (b) s = <3 - 6^2 + i2t _ 8. (c) s = <3 + ^2 _ 6f, (d) s = 2^ - 8^3 - 9^2 + 54i. 18. Find the largest possible number of horizontal tangents to the grapl of a polynomial of degree n; the largest possible number of points of ia flection. 19. If the graph of a function y is concave toward the a;-axis, show tha y and D^m have opposite signs. 20. li y = ax^ + dx"^ + ex + d, show that the abscissas of the points oi the graph at which the tangent line is horizontal satisfy a quadratic equa tion. Find the condition that the number of horizontal tangents is two one, or zero. Apply this condition to determine the number of horizonta tangents in 15, 6, c, d. 21. li y = uv, show that Dxy = uDxV + vDxU. 98. Derivative of a Rational Function. In order to dif ferentiate any rational function we need but one more rule. Theorem. The derivative of a fraction is a fraction whoSi numerator is the denominator times the derivative of the numerata less the numerator tim^s the derivative of the denominator, am whose denominator is the square of the denominator. Symbolically p. u _ vDxU - uDzV v v^ T X ^ Let 2/ = 7 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 279 len u -\- Au whence Ay = Hence u + Au V + Av ' ' ^'- v + Av' u uv + vAu - uv - uAv vAu — uAv V v{v + Av) v{v + Av) Au Ay Ay '^Ax'^'Ax Ax v(v + Av) ' Finding the limit as Ax approaches zero, we get vDxU — uDxV Example 1. -(fel) D.y = I/- _ (x^ + 5) D,(Sx - 4) - (3a; - 4) D:,(x^ + 5) {x^ + 5)2 (x^ + 5) (3) - (3a; - 4) (2x) (x2 + 5)2 - 3x2 + ga; + 15 (a;2 + 5)2 After one fixes the rule in mind, it becomes easy to write down the second ^ fraction on the right without taking the time to write out the first. Example 2. Find the derivative of y = 1/w", where it is a function of x. 1 w" DA - 1 Dxw« re have D. w* - nu'^~^DxU — nDxU by (1) and (5), Section 94. Since the result obtained in Example 2 may be written in the Dxiu-"") = - nir^-^DxU, see that Theorem 5, Section 94, holds for negative as well as positive, integral values of the exponent n. 99. Derivative of an Irrational Function. In order to dif- ferentiate an irrational function it is sufficient to show that Theorem 5, Section 94, holds also for a fractional exponent, n = p/q. 280 ELEMENTARY FUNCTIONS Let y = wP/« (1) Raising both sides to the qth power, 2/« = w^. (2) As 2/« and u^ are merely different notations for the same function of x, their derivatives are the same. Hence L D.r = D,u^' (33 Applying Theorem 5, Section 94, qy^-^DxV = puP-^DxU, (4 Dividing (4) by(2), qy-^D^y = pu-^DxU. (6| Multiplying (5) by (1), and dividing by q, V - —1 Dxy = -UQ DxU = nU^~^DxU, where n = -• Q Hence we have the Theorem. If y = W^, where in is a fraction^ then Dxy = nW'^DxU, Example. Differentiate y = Vx^ - a?. Since y = {x^ - a2)l, and since we may regard x^ - a^ as a function u, we have, by the theorem, J>*y = Ka:^ - a')-i2x. X Vx^ - a2 EXERCISES 1. Find the derivatives of the following fimctions: (^) ^^- (^^ 3^34- (^) ^• ^^^iiTs' ^"^^m:-!- (^)^^32i- (g) Va:2 _ 2a;. (h) Vl - xK (i) Vx. (j) a;V2x - 3. Hint: xV2x - 3 = V2x» - 3x2. (k) aj^-j-z ._ 5^;. (1) x^{2x + 4)|. (m) -=4==- (^) (^' + 3x - 5)1 ^"^ ^^%^- "*-(r-» 286 ELEMENTARY FUNCTIONS box deep. In either case the volume will be small. Somewhere between these two extremes will be a box whose volume is greater than that of any other, that is, a box of maximum volume. Let X be the side of the square cut out. Then the depth of the box X, and the side of the base is 4 - 2x. Hence the volume, expressed as function of a; is F = x(4 - 2x)2 = 16x - 16x2 + ^3?. The graph of F is readily plotted. From it, the value of x which makej F a maximum appears to be somewhat less than unity. By computing F for a large number of values of x, we could approximate the best value of x. By means of th( derivative we can avoid this labor, and obtain thei exact value. At the maximimi point the tangent line horizontal, and hence its slope is zero, so that DxF = 0. We therefore compute the derivative, set it equal to zero, and find the value of x which pro^ duces this result. We then have -.1 T 4- t ^^p- t l-^ J~ S .^1 ^ ii.txl? Fig. 166. or, dividing by 4, Factoring and hence DxF = 16 - Z2x + 12x2 ^ 3x2 - 8x + 4 = 0. (3x - 2) (x - 2) = 0, X = 2/3 or 2. From the figure, we see that x = f makes F a maximum, and x = gives a minimum. That x = 2 gives a minimum follows also from the fact that if X = 2, then F = 0, that is, the box will not hold anjd^hing. The preliminary discussion showed the existence of a maximum, and as x = ^ is the only other possibility it must be the value of x for which F is a maxi- mum. Either of the criteria in Section 96 may also be appUed to show thati F is a maximum if x = f . In order then, to have a box of maximum capacity, we must cut oul squares from the corners f of an inch on a side. The depth of the box will be I of an inch, the side of the base will be 4 - 2 x f = 2| inches,^ and the capacity will be F = (2f)2 x f = ^^^ cu. in. Example 2. As large a rectangular stick of timber as possible is to be sawed from a log 10 inches in diameter at the smaller end, the length of the stick to be the same as that of the log. Find its other dimensions. Let F denote the volume of the stick of timber, I its length, and A the area of an end. Then F = Z^l, and since I is constant, the volume F will be a maximum if and only if A is a maximum. The end of the stick is a rectangle inscribed in a circle whose diameter is 10, and the problem re- duces to a determination of the dimensions of the maximum rectangle which can be inscribed in this circle. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 287 The area of the rectangle is A = xy, (1) rhich involves the two variable dimensions x and y. These are connected )y the relation x"^ + y^ = 100. Solving for y, y = V 100 - x^ . and substituting in (1) A = xVlOO - x\ In order to find the derivative of A we write it in the form A = Vl00a;2 - x* = (lOOx^- x*)K lifferentiating, DxA = K100a;2 - a:4)-j(200x - 4^^) lOOx -27^ 100 - 2x2 (2) (3) Vl00x2 - X* VlOO - x^ luating the derivative to 100 - 2x^ 0. viOO -x^ [ultiplyin g both sides by ioo - x\ 100 - 2x2 = 0, id solving for x, x = ^ 5 V2. The negative value of x has 'ho meaning in this problem. That the value of A is a maxi- mum if X = 5 V2 may be seen as follows: If x is very small, or very near 10, A is very Pj^ j^gy small, while larger rectangles lie between these extremes. Hence as x increases from to 10, A first increases and then decreases, and as A does not become infinite it therefore has a maximum in this interval. But x = 5 V2 is the only point in this interval at which the tangent to the graph of A is horizontal, and it must therefore give the maximum value. _ The corresponding value of y, from (2), is ?/ = 5 V2. Hencethe end of the largest stick of timber will be a square whose side is 5 V2 inches, or very nearly 7 inches. 288 ELEMENTARY FUNCTIONS EXERCISES 1. A box is to be made by cutting squares from the comers of a piece of cardboard 6 by 8 inches, and folding up the sides. Find the dimensions if the capacity is to be a maximum. 2. A chicken yard is to be made from 36 feet of poultry fencing, the side of a bam being used for one side of the yard. Find the dimensions in order that the yard may be as large as possible. 3. Find the dimensions and capacity of the largest box which can b< made with a square base and no top if the total amount of cardboard ii the box is 48 square inches. 4. What should be the dimensions of a rectangular garden plot with perimeter of 12 rods, in order to have the greatest area possible? 5. A two acre pasture in the form of a rectangle is to be fenced off alonj the bank of a straight river, no fence being needed along the river. Find the dimensions, in rods, in order that the fence may cost as little as possible 6. The legs of an isosceles triangle are 6 inches long. How long must the base be in order that the area may be a maximum? 7. The height of a rifle ball fired vertically upward with an initial velocity of 1200 feet per second is s = 1200« - 16^^^ How high will it rise? 8. By the Parcel Post regulations, the combined length and girth of 9 package must not exceed 6 feet. Find the dimensions and volume of the largest parcel which can be sent in the shape of a box with square ends. 9. A farmer has 150 rods of fencing. Find the dimensions and area ol the largest rectangular field he can enclose and divide into two equal pai by a fence parallel to two of the sides. 10. If the total area of the field in the preceding Exercise is to be 1^ square rods, find the dimensions and the amount of fencing needed if the latter is to be a minimimi. 11. A rectangular cistern is to be built with a square base and open top./ Find the proportions if the amount of material used is to be a minimum. 12. A rectangular piece of ground is to be fenced off and divided into three equal parts by fences parallel to one of the sides. What should the dimensions be in order that as much ground as possible may be enclosed with 16 rods of fence? 13. If the total area enclosed in Exercise 12 is an acre, find the dimensions in order that the total length of the fence should be a minimum. 14. The number of tons of coal consumed per hour by a certain ship is 0.3 + 0.001 V^, where V is the speed in knots. For a voyage of 1000 knots at V knots per hour, find the total consumption of coal. For what speed is the consumption of coal least? 15. Divide a string 16 inches long into two parts, so that the combined area of the square and circle with perimeters equal to the parts shall be a minimum. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 289 16. Find the coordinates of the maximum or minimum point of ?/ -= I ax^ + bx + c. 17. At what height should a light be placed above a writing table in order that a small portion of the table, at a given horizontal distance d from the point directly below the light, may receive the greatest illumina- ' ' tion possible? (It is known that the intensity of illumination varies in- i versely as the square of the distance and directly as the sine of the angle f between the line from the light to the point in question and the table. i Express the illumination as a function of the height of the light, and find I the value of the height which gives the maximum illumination.) 18. A rectangular box is to be made by cutting out the comers of a I rectangular piece of cardboard and folding up the sides. If the depth of ; the box is to be 2 inches, find the dimensions of the smallest piece of card- I board which will make a box to contain 72 cubic inches. I 102. Related Rates. If two variables x and y are functions I of the time t, then their derivatives with respect to t, DtX and I Dty, measure the rates at which x and y are changing. These I rates will be related, that is, connected by a relation, if y is I a function of x. ! For example, if x and y are functions of t such that y = x^ ! then the rates at which x and y change satisfy the equation j Dty = SxWtX, which is obtained by differentiating the given i equation with respect to t. The rate of change of y depends on the value of x as well as on that of the rate of change of x. , In solving problems involving the rates at which two varia- |i bles change, it is first necessary to determine the relation y = f{x) I connecting the variables. Then the relation between the rates of change of x and y is found by differentiating y = f{x) with respect to t. Example 1. Oil dropped on a smooth floor spreads out in the form of a circle. If the radius is increasing at the rate of | an inch per second > when it is 6 inches long, how fast is the area increasing? If r denotes the radius and A the area, the question asked may be ex- [> pressed symbolically as follows: If Dtr = | when r = 6, what is the value ) of DtA'f In order to answer this question we must have a relation between DiA and Dtr. And to obtain this relation we must first express .4 as a function of r. From plane geometry we have A = Trr^. 290 ELEMENTARY FUNCTIONS Differentiating both sides with respect to t, we get DA = 2TrDtr. Substituting the values of tt, r, and Dtr, the oil covered area is increasing at the rate of DtA = 2 x 22/7 x 6 x | = 18.85 square inches per second. Example 2. A man walks along a sidewalk at the rate of three miles an hour (4.4 feet per second), approaching a house which stands back 7 feet from the walk. When he is 24 feet from the walk leading to the house, how rapidly is he' approaching the house? The rate at which he approaches the house is the derivative of his distance from the house with respect to time. If y denotes his distance from the house, we seek Dty. The rate at which the man walks is the rate of change of his distance from some point on the walk with respect to the time. This point is conveniently chosen as the point on the sidewalk directly in front of the house, because we know this distance at the time, we wish to determine D 1, 0) cuts the bisector of the first and third quadrants at B, then OPi = BPi. 6. The line y = mx + \a passes through the focus (0, la) of the parabola y = ax^. Find the abscissas of the points at which it intersects the parab- ola. Show that the tangents at these points are perpendicular, and that they intersect on the directrix. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 299 7. If Vq is the volume of a quantity of water at 0° Centigrade then the volume at a temperature 0° Centigrade is given by V = Fo(l - 0.00005758 d + 0.000007560^ - 0.0000000351 0^). Show that the volume is least, and hence the density greatest, when 6 = 3°.92C. (Let the decimals be denoted by a, b, c respectively until the end of the computation, and write 0.00005758 in the form 5.758 X 10-5, etc.) 8. A bandit is walking along a street at the rate of 3 miles an hour toward the intersection with a second street making an angle of 45° with the first. A tree stands 100 feet from the intersection and 25 feet toward the second street from the first. A timid citizen walks along the second street en- deavoring to keep the tree between himself and the bandit. How fast does he walk? 9. There are two sources of fight at the points A and B. At a distance X from the first the illumination is /i = S/x^, and at a distance x from the second the illumination is /2 = 27/a;2, Find the point on the segment AB where the sum of the illuminations is a minimum. 10. Suppose that Ui, a-i, as, . . . a„, are the values of n measurements of a magnitude whose true value is x. Then the errors in the measure- ments are respectively x - ai, x - a2, x — az, . . . x - an, some of which are positive, and some of which are negative. The theory of least squares asserts that the most probable value of a: is such that the sum of the squares of the errors is a minimum. Show that the most probable value of x is the arithmetic mean (or average). 11. Find the dimensions of the strongest beam which can be sawed from a log 12 inches in diameter, assuming that the strength of the beam varies as the breadth and the square of the depth. 12. Find the dimensions of the stiflfest beam which can be cut from a log 10 inches in diameter, if the stiffness is proportional to the breadth and the cube of the depth. 13. Plnd the dimensions of the rectangle with maximum perimeter which can be inscribed in a circle of radius r. 14. A point moves along the straight fine y = 2x + 10. If its abscissa is increasing at the rate of 3 inches per second when x = 1, find the rate at which its ordinate changes. 15. Two railroad tracks intersect at right angles. A train on one track is 24 miles from the intersection and is approaching it at the rate of 30 miles an hour. A train on the other track is 7 miles from the intersection, and is receding from it at 45 miles an hour. Is the distance between the trains increasing or decreasing? How rapidly? 16. A 16 foot ladder resting against the side of a barn begins to sfip. When the foot of the ladder is 10 feet from the bam it is moving 2 feet per second. How fast is the top of the ladder moving? 300 ELEMENTARY FUNCTIONS 17. The side of a square is 39.51 inches, with ap error of not more than 0.01. Find the error in the area of the square. Is the value of 39.51^ given in Huntington's Tables suflSciently accurate for the area? 18. The intensity of heat at a distance of x feet from a source of heat is / = lOO/a:^. If a body moves directly away from the source at the uni- form rate of 3 feet per second, how rapidly is the intensity of the heat changing when x = h feet? When a; = 20 feet? 19. The side of an equilateral triangle is 5.4 ±0.1 inches. What is the percentage of error in the area of the triangle? 20. If a body moves so that v^ = k8^ show that the acceleration is con- stant. An automobile moving v miles per hour on a slippery pavement should be able to stop in s feet, where v* = 17s. Find the acceleration in feet per second per second. 21. A rectangular grain bin, without a cover, is to be divided into two equal parts by a partition parallel to the ends. The bin is to be 3 feet deep and to hold 98 cubic feet. Find its dimensions if the amount of lumber required is a minimum. 22. A rectangular pan is to have its width two-thirds of its length and its capacity is to be {^^ of a cubic foot. Find its dimensions if the amount of tin used is a minimum. 23. The thrust of an aeroplane is given by the equation (see Exer- cise 11, page 157) * = TF(0 + ig). Find the value of B for which ^ is a minimum. Substitute this value of Q in the given equation, and determine how t varies as the fineness / changes. 24. The power required for oblique flight of an aeroplane up a slope m is T =m (^ +yl^) + ^^^ = ^^(^ + j4 + '^)' ^ For gliding flight T = 0; what relation must eonnect x/(x) = F(x), then }{x) is an integral of F{x). Example 1. Given Dxy = 2a:, find y. Since DxX" = wic""^, we have DxX^ = 2x. Hence x^ is an integral of 2x. But the derivative of each of the functions x"^ + 5, x^ - 7, x^ + f , is 2x. Hence these functions are also integrals of 2x. In fact, any function of the form y = x^ + C, where C is any constant, has the derivative 2a:, and is therefore an integral of 2x. Thus we see that the process of differentiating the function x"^ gives rise to the single derivative 2x, but that the inverse process of integrating 2x gives rise to an indefinite number of integrals x"^ + C which differ by a constant. Theorem. If two functions f(x) and F{x) have the same derivor tive^ their difference is a constant. Let y = fix) - F{x). Then D,y = DJ{x) - DJ'ix), By hypothesis, DJ{x) = D^Fix). Hence Dxy = 0. But a function whose derivative is equal to zero for all values of the variable is a constant. For the tangent line at e very- point on the graph is horizontal, so that the graph of the func- tion is a straight line parallel to the a;-axis. 301 302 ELEMENTARY FUNCTIONS Hence y = f{x) - F{x) = C, or fix) = Fix) + C. That is, every integral can be obtained from a given integral by adding the proper constant. The constant C which is added to a known integral of a func- tion is called the constant of integration and it should not be omitted in performing the operation of integration. If y is a function of x whose derivative with respect to a; is given, say Fix), then D.y = Fix). The notation employed to indicate that 2/ is an integral of Fix) is y = fFix)dx, the equation being read ''y equals the integral of Fix)dxy In the illustration above, since Dzy = 2x, then y = f2x dx = x'^ + C, In a particular problem sufficient data are usually given to determine the constant of integration and to enable us to find a particular integral which satisfies the given conditions. The graphical signi- ficance of the constant of integration and the method of determining its nu- merical value so as to satisfy given conditions is shown in Example 2. Find y if DxV = z'^ - 4 and interpret the result graphically. Find the equation of that one of the resulting curves which passes through the point (1, 2). By dififerentiating a^ we obtain 3x^, and hence we must differentiate Ix^ in order to get x^. Differentiating 4x gives 4, and hence if D,y = x^-4, y='ix^-4x + C, where C is the constant of integration. This represents a set of curves which may be obtained from one of them I by moving it up or down (Theorem, page 19) . Several curves of this set] are shown in the figure. 1 y 1/ -i % 7^ / /> ^ // 4 ^ ^/X t-^^-'\ LL 'A-n \«../, 3 ^ ^2n* i/f?^ °\ ' FT A~\\A~ i 'A t ^ J- ::----!£ Fia. 173. INTEGRATION 303 These curves have parallel tangent lines at points with the same ab- scissas, since the derivative of y with respect to x is the same for each fimc- tion, namely, Dxy = x^ — 4. The intercepts of these curves on the 2/-axis are the respective values of C. If (1, 2) lies on one of these curves then 2 = i-l'-4-l + C, whence C = 5| and the required equation is i/ = i^^ - 4a: + 5|. Example 3. If Dxy = ax**, show that y = .. x"'''^ + C. The result is correct because if we differentiate it we get D^y = ^j^T^i (n + l)x'' = ox". Hence, to integrate a term of the form ax^'j increase the exponent by one and divide the coefficient by the new exponent. In like manner, if Dxy = aw" D^u, then y = 1^"+^ + C. n -f- JL If the integral notation is used, these results may be written Tax" dx = — ^ a:"+i + C, faW" D^u dx = — ^ w"+i + C. Example 4. (a) Find y if Dxy = 4:3^. 4 From the preceding example we have y = ^ — z x* + C <= x* + C. (b) Integrate f(Sx^ + Sx)dx. We have f{Sx^ + Sx)dx = fSx^ dx + fSx dx = x^ -\- 4x^ -{- C. (c) Find y, if Dxy = i{x^ + Sx)i (2x + 3). If we think of x^ + 3a: as a function u, then 2a: + 3 is DxU, and therefore y = l^-^i^-^c={x^ + 3x)i + c. Differentiation is a direct process while integration is an in- verse process. The former can be carried out according to a general method of procedure biit there is no general method of procedure in the case of the latter. Integration in the last analysis is a matter of trial in which attempts are made to re- duce the given function to a form the integral of which is known. A thorough knowledge of differentiation is essential for the process of integration. EXERCISES In Exercises 1-12, find y if Dxy is the given function, and illustrate the result graphically. Determine the equation of that member of the set of curves which passes through the indicated point. 304 ELEMENTARY FUNCTIONS 1. 4. x + 1, (1, 2). x3 + a:, (- 1, 11). 2. x2 - 3, (- 1, 3). 5. A (3, 4). 3. x^-a:, (0,2). 6. ^7^, (1, 3). 7. Va^ + 1, (4, -2). 8. ^„ (1, 1). 9. ^-, (2, 1). 10 ^ , (0, ^). 11. — ^, (2, 4). y/x^ - 4 12. 2x-3 V2a: + 3' ' ' V(a:2 _ 3^)3' U> "^ 13. 14. For what value of n does the result in Example 3 above fail? Given D:,y = (x^ + 3x- 2Y {2x + 3), find y. I . 15. Given D^y = {Sx^ + 4Wix^ + ^)\ find y. 16. Find the equation of the curve whose slope at any point is 2a; + 3 and which passes through the point (2, 3). Find the equation of the tan- gent hne at this point. Plot the curve. 17. The slope of a curve at any point is 6x^ + \x and it passes through the origin. Find the equation of the curve and the equation of the nor- mal line at the point of inflection. Plot the graphs of the two equations. 18. Find the cubic function whose graph has a maximum point at (1, 4) and a minimum point at (3, - 5). Find the equation of the tangent line at the point of inflection. Suggestion. D^y = a{x - l)(x - 3). 19. The slope of a curve at any point is 4a: - 3 and it passes through the point (1, 2). Find the equation of the curve and the coordinates of the minimum point. Plot the graph. 20. Integrate the following: (a) y(6x2 + 8x - h)dx. (b) fxl dx. (c) f^y/x - 2 dx. (d) r| dx. V I 1 fi Q / Ay s J /^ A AA M N R "d < — a > r ri 1 ' (e) f2V4:-3xdx. 106. Area under a Curve. Consider the an A, bounded by any curvl y = fix) 7 the X-axis, the fixed ordinate MB at X = a, and the movinji ordinate at NP. Since A changes as x changes and is determined when x h fixed, A is a function of x. As N moves to R, x increases by an amount Ax, A by ai amount PNRQ = AA, and y by an amount SQ = Ay. Fig. 174. INTEGRATION 305 Since PNRQ is less than the rectangle TNRQ and greater than the rectangle PNRS, we have PNRSxA = x^. (2) . x' Integrating, + C. (3) Since A = when x = 1, we have = ^ + C or C = - i . ^ 1 ^ X u Si -U 18 M M 15 t^ii u 12 ^'- 9- t J ^ J- z 3 A ^s -W^ - ^ ^'^ 2 i ~V X ■""^i- : Hence (4) Fig. 175. Equation (4) gives the area under the curve starting at the ordinate MP, at X = 1, and continuing to any second ordinate. In this case the second ordinate is fixed at x = 4. Hence substituting x = 4 in (4) we have .43 1 "*- 3~3 55 = 21 3 ^^' (5) 306 ELEMENTARY FUNCTIONS M N Fig. 176. The graph of (4) is the curve LMD which crosses the a:-axis at x - 1 where the area MPBN begins. The number of square units in MBPN is the same as the number of linear units in the ordinate ND, which is 21 units in length. Equation (1) may be interpreted thus: Suppose that we had y = c, whose graph is a straight hne parallel to the a;-axis. Then the rate of change of A with respect to x, DxA, would be uniform, since DJi = y = c. It would be B D measured by the change in A due to a unit change in x (page 48), that is, by a rectangle with base unity and altitude c. "?J' Returning to the figure used in proving the theorem, let NR = 1. Then the area of the rectangle PNRS = y X I = y is the amount by which A would increase when X increases by unity, provided that A increased uniformly for x>ON. The method of proving the theorem above may be used to show that an area A bounded by a curve, the 2/-axis, and two abscissas (one fixed, the other not), is such that DyA = X. (6) EXERCISES 1. Find the area under each of the following curves from x = 1 to a; = 3. (a) y = 3x2, (b) y = 2/x\ (c) y ^ x%, (d) ?/ = x^ + x + 7. 2. Find the area bounded by the curve y = x^, the x-axis and the line X = 2, and the area bounded by the curve, the 2/-axis and the line y = 4. Check by finding the area of the rectangle formed by the axes and the lines X = 2 and y = 4. 3. Find the area above the x-axis and below the parabola y = - x^ + 4. 4. If the positive intercepts of the parabola y =- - x^ + 2x + 3 on the coordinate axes are denoted by A and B, and if a tangent parallel to the chord AB cuts the ordinates at A and B in C and D, and touches the parabola in E, show that the area of the parabolic segment ABE is two- thirds that of the parallelogram ABCD. 6. If the tangent to the parabola ?/ = 3x* - 6x + 8, which is parallel to the chord joining the minimum point A to the point B on the curve INTEGRATION 307 whose abscissa is 2, cuts the ordinates of A and B m C and D, find the relation between the area of the parabolic segment from A \jo B and the area of the parallelogram ABCD. 6. Find the area bounded by the following pairs of curves: 7 15 1 (a) y = x\ y" = x, (b) y = xi, y ^ x\, (c) y = - ^x^ + — x, y = -• 7. Find the area bounded by the curve y = a? - %z^ + 9a; + 4, the tangent to the curve at the point of inflection, and the ordinates at the maximum and minimum points. 8. Show that the area under the parabola y = ax^ + 6a; + c between x~h and x = 1c, ik>h), is A = — = — (?/i + 4^2 + 2/3), where yi, y^, ys h + k are the values of 2/ at a; = A, a; = — 2~> ^ = ^ respectively. Suggestion: Sub- stitute the values of y in the above expression and show that the result is the same as the area under the curve. 9. Find the area under the following parabolas and check the results by means of the formula for the area given in Exercise 8. (a) y = 2x^ + 4:X + 1, from a; = 1 to a; = 3. (b) y = - a;2 + 9, from a; = to a; = 3. (c) 2/ = 3a:2 - 4a; + 3^ from x = to a; - 2. 10. Show that the area formula in Exercise 8 is true for any cubic function y = ax^ + bx^ + cx + d. 11. Find the area under the curve y = a^ + 12a; + 4 from a; = 1 to a; = 3. Check the result by means of the formula in Exercise 8. Definition. The average ordinate of a curve y = f(x) from X = a to X = h is the height of the rectangle with base b - a, whose area is equal to the area under the curve from x = a to X = b. If y represents the average ordinate and A the area under the curve from x = ato x = b, then A y b-a 12. Find the average ordinate of the following curves for the ranges indicated. (a) 2/ =» 2a; + 3, from a; = 1 to a; = 4. (b) y = -x^ + 16, from a; = to a; = 4. (c) y = 1/x^, from a; = 1 to a; = 3. (d) y = xl, from a; = to a; = 4. 308 ELEMENTARY FUNCTIONS 13. Given 6 = t^ - 9t^ + 15t + 30, where 6 is the temperature at any time t, find the average temperature for the range from maximum to mini- mum temperature. 14. The pressure p and the volume v oi & gas are connected by the equa- tion pv^'^ = k. and p = 15 pounds per square inch when v = 0.5 of a cubic foot. Find the average pressure of the gas in expanding from 1 to 3 cubic feet. 15. The tension T oi a spring is connected with the amount of stretch- ing s by the equation T = |s + 3. Find the average tension as s changes from to 3. 16. Find the percentage of error in the area under the curve y = x^ -2x + 3 from x = to a: = 2, due to an error of 1 per cent in the range. 107. Motion in a Straight Line. The fundamental notions in- volved in the motion of a particle in a straight line are: The distance, s, measured from a convenient station on the line. The time, t, which has elapsed from a fixed time. The velocity, v^ given by the equation v = DtS. i The acceleration, a, given by the equation a = DtV. The motion may be described by an equation involving two or more of the magnitudes t, s, v, a. We have already considered motions described by an equa- tion of the form s =f(t), and obtained the velocity and ac- celeration by differentiation. We shall now consider two other types in which integration is involved. If the velocity is given as a function of the time by an equa- tion of the form v = f{t), the acceleration, a, is found by dif- ferentiating while the space, s, is found by integrating this equation. The constant involved in the integration is de- termined as in the following example. Example 1. A body moves from rest in a straight line so that its velocity after t seconds is given by the law v = 4<. Assuming that a ball placed on a plane inchned at arc sin I would roll according to this law, determine: (a) How far it will roll in 2 seconds. (b) Its distance from the upper end of the plane after 1.5 seconds, if it is placed 4 feet from that end. (c) Its distance from the lower end after 1 second, if it is placed 12 feet from the lower end. INTEGRATION 309 We know that if s represents distance from a certain point at the time t, then Dts = V, and hence, in this example, Dis = At. The required function, s, is therefore one whose derivative with respect to t is 4:t. One such function is 2t^, and adding the constant of integration (page 302), we have s = 2f + C. (1) The Value of the constant of integration may be determined if we know a pair of values of t and s, and these depend on when and where the ball starts to roll. We shall assume that t = when the ball begins to roll. The distance s is measured from some chosen station 0. (a) The distance the ball will roll in 2 seconds is the same as its distance from the starting point after 2 seconds. Hence for this part of the problem we choose the station at the starting point, so that s = when ^ = 0. Substituting these values in (1) we find that C = 0. Hence, substituting this value of C in (1), the distance from the starting point after t seconds is s = 2P. (2) Then at the end of 2 seconds, s = 2 x 2^ = 8 feet. (b) Take at the upper end of the plane. Since the baU starts 4 feet from 0, we have s = 4 when t = 0. Substituting these values in (1), we find that C = 4. Setting this value of C in (1), the distance from the upper end of the plane to the ball after t seconds is s = 2t^ + 4. (3) After 1.5 seconds, s = 2(1.5)2 + 4 = 8.5 feet. That is, the ball is 8.5 feet from the upper end of the plane after 1.5 seconds. (c) If is taken at the lower end of the plane and the ball placed 12 feet above it, then s = - 12 when 1^=0(8 being negative since the positive direction for s is the same as that of v, namely, down the plane). Substituting in (1) we find C = - 12, so that the distance from the lower end of the plane at any time is s = 2«2 _ 12. (4) If t = 1, s - - 10, that is, after one second the ball is 10 feet above the lower end of the plane. Graphically, the given relation v = At is repre- sented by the straight Une through the origin whose slope is 4. The graph of (1) is given for C = 0, 4, - 12, which are the values of C in (2), (3), (4), respectively. The intercepts of these curves on the s-axis are the respective values of 'C, which give the ■ distances from the station to the ball when t = 0- V , i \_ c- r l\ 10 L \\ '/ \v 1 / \\ /4 -T-, -19 \ \ S' '// _J \ ^- -'\-f (y : >', Y — g i r~ K / V . -. Fig. 177. 310 ELEMENTARY FUNCTIONS If the acceleration is given as a function of the time, t, by an equation of the form a = f{t), the velocity is obtained by integrating this equation and the distance by integrating the result. Two constants of integration are introduced in the process which are determined by given conditions as in Example 2. A balloon is ascending with a uniform velocity of 28 feet per second and at the height of 720 feet a ball is dropped. When will the ball strike the ground and with what velocity? Let s be the height of the ball above the ground, and let ^ = when the ball is dropped. The direction of the acceleration of gravity, being toward the center of the earth, is opposite to the positive direction of s, and hence the accelera- tion is negative. The initial conditions and required data are collected in the table. As soon as the ball leaves the balloon it is subject to the law a = - 32. (1) Integrating (1), » = - S2t + d. (2) When t = 0,v = 2S, and therefore Ci = 28. Hence v= - S2t + 28. (3) Integrating (3), s = - 16^2 ^ 2St + d. (4) When t = 0,s = 720, hence'^Ca = 720. Therefore s = - 16^2 + 2St + 720. (5) When the ball strikes the ground s = 0. Substituting this value of s in (5) = - 16^2 ^ 2St + 720, or 4<2 -7t- 180 = 0. Solving for t we have ^ 7 =*= V49 + 2880 54.2 __ , t = 5 = — ^— = 6.8 seconds, o o the negative value of t having no meaning in this problem. Substituting this value of t in (3) « = - 32(6. 8) + 28 - - 189. 6 feet per second. EXERCISES 1. A ball is rolled up a plane inclined at an angle of 10° with an initial velocity of 15 feet per second. Find the distance it rolls up the plane and the time that elapses before it returns to the starting point. (Sug- gestion: Find the 'component of the acceleration of gravity acting along the plane.) t s V a 720 28 -32 ? ? -32 I INTEGRATION 311 2. Solve Example 2 of the preceding section if the balloon is descending. 3. A high jumper raises his center of gravity 3 feet. How long is he off the ground, and with what velocity does he hght? 4. A baseball dropped from one of the windows of the Washington monument, 500 feet from the ground, has been caught. Compare the velocity with which it struck the catcher's hand with the velocity of 120 feet per second which is said to be the maximum velocity that a pitcher has imparted to a ball. 5. A man descending in an elevator whose velocity is 10 feet per second drops a ball from a height of 6 feet above the floor. How far will the elevator descend before the ball strikes the floor of the elevator? 6. In the preceding problem, suppose the elevator is ascending instead of descending. 7. A balloon is ascending with a velocity of 24 feet per second when a ball is dropped from it. The ball reaches the ground in 5 seconds. Find the height of the balloon when the ball was dropped. Determine the highest point that the ball reached. 8. An automobile reduces its speed from 35 miles an hour to 20 miles an hour in 8 seconds. If the retardation is uniform how much longer will it be before it will come to rest, and how far will it travel in this length of time? 9. How high will a ball rise if thrown vertically upward with an initial velocity of 60 feet per second? 10. A street car in going from one stop to another 400 feet distant is uniformly accelerated at the rate of 2 feet per second per second for a dis- tance of 320 feet and then brought to rest with a uniform retardation. Find the time it took to go the 400 feet. 108. Motion in a Plane. If a particle moves along a curve in a plane, two rectangular axes are chosen and the position of the particle at any time determined by means of the coordi- nates of the particle. If the coordinates of the particle on the axes are given by the equations X = m and y = F{t) (1) the position of the particle in the plane is determined. The discussion of the motion of a particle in a plane is thus resolved into the discussion of two motions along straight Unes. The components of the velocity, obtained by differentiating equations (1) are Vx = DtX and Vy = Dty. (2) 312 ELEMENTARY FUNCTIONS They are represented in Fig. 178 by directed lines parallel to the axes. Since the components are perpendicular the magnitude of the resultant velocity, v, is given by the equation V = Vy^x + v\ (3) rit '1l p ^ / ax ' X Fig. 178. Fig. 179. and the direction of the velocity can be found from the equation tan 6 = Vyjvx. (4) The components of the acceleration, obtained by differentiat- ing equations (2) are ttx = DtVx atid ay = DtVy. (5) They are represented in Fig. 179 by directed lines parallel to the axes. The magnitude of the resultant acceleration, a, is given by the equation a = Va^ + a'y (6) and its direction can be found by means of the equation tan = Qy /a^. (8)| Example 1. If a particle moves in accordance with the law x = <', y = t^, find the equation of the path, the position of the particle when] t = 2, and the magnitude and direction of v and of a at this point. From the first equation t = a;i and hence y = (xi)^ = xi, which is the] equation of the path shown in Fig. 180. Differentiating the given eqiu^l tions with respect to t we obtain 3t\ and INTEGRATION 2t. (1) 313 Hence » = V9J* + 4^2 = < V9i2 + 4 (2) and tan = 2t/3t^ = 2/3<. (3) Differentiating equations (1) we obtain, ax - Qt, and Cy = 2 Hence a = VSQt^ + 4, and tan = 2/6t = l/3«. V n F ( p.4) ^ r — -3 ^ '^^ p] a ^^ ^ N ^^ > ^/ - ^ _ u ^ u ' 1 t VH Fig. 180. (4) '(5) (6) \ Substituting < = 2 in equations (2), (3), (5), (6), we find v = 12.64, e = 18°.43, a = 12.17, = 9^46, at the point where a; = 8 and y = 4. The inverse problem of determining the path of a particle, given the equations of the component accelerations, is illus- trated in Example 2. Find the equation of the path of a projectile fired at an angle of 30'' to the horizontal with a muzzle velocity of 1200 feet per second. Find the range of the pro- y\ jectile and the maximum height attained. Let the origin be taken at the initial point of the path, the a;-and y-axes being hori- zontal and vertical. The horizontal and vertical components of the initial velocity are 1200 cos 30°, and 1200 sin 30°, respectively. If the resistance of the air be disregarded there will be no horizontal acceleration. From the instant the projectile leaves the gun gravity is acting on it vertically in the direction opposite to the positive direction on the 2/-axis. Hence the acceleration equations of the motion are a* = and ay = - 32. (1) Integrating these equations we have Vx = Ci and Vy = - 32t + d. (2) When < = 0, tJx = 1200 cos 30°, and Vy = 1200 sin 30°. Therefore Ci = 1200 cos 30°, and Cz = 1200 sin 30°. Hence Vx = 1200 cos 30°, and r„ = - 32« + 1200 sin 30*. (3) -range — Fig. 181. 314 ELEMENTARY FUNCTIONS Integrating (3), X = 1200 cos SO''-^ + Cs and y=- \U^ + 1200 sin 30°-« + C4. To determine the constants we note that when i = 0, a; = 0, ^ = 0, and therefore Cz = 0, and C4 = 0. Hence X = 1200 cos SO°t and y=- 16^ + 1200 sin aO*'-^. (4) These equations give the position of the particle at any time. Solving the first of these equations for t and substituting in the second we find the equation of the path to be ^- - 1200^^30° ^' + ^^°^"°-^' ® which is the equation of a parabola. To find the range, let y = in (5) and solve for x. ^ 1200* C0S2 30° tan 30** 00 n^rn r i. ^00 -1 Hence x = —; 38,970 feet - 7.38 miles. 16 At the highest point Vy = 0, hence from (3), = - 32f + 1200 sin 30°. Therefore the highest point is reached when t = ^ = 18| seconds. Substituting this value of t in (4), we obtain for the maximum height attained 2/ = - 16(-V-)' + 1200 sin 30°-(^''-) = 5625 feet. EXERCISES 1. In the following exercises find the equation of the path in terms of x and .y by eliminating t from the given equations, plot the path, find the magnitudes and directions of v and a for the given value of t, and at the point on the path corresponding to this value draw lines representing v and a in magnitude and direction. (a) a; = 2f, 2/ = f2, f = 1. (b) ^ = f, t/ = 1/t, t = 1. (c) x~St^ + 2, y 2t\ « - 2. (d) a; = 2t, y = 1/(^2 + i), ^ . 0. (e) x = 3,y-6'. (1) I INTEGRATION But by Theorem 1, b = |or6'. -1- Hence, substituting in (1), D.V -i.-- Integrating V - ^ X3 + c. As 7 = when x = = 0, we have C = 0, and hence x3. 323 When X = h, the value of V is the volume of the given pyramid. Hence the required volume is V =— h^ = -bh ^ 3/i2 ^ S^''- 112. Volume of a Solid of Revolution. A solid of revolution is a solid generated by revolving an area about an axis. Such a solid may be turned out on a lathe. A section of the solid by a plane perpendicular to the axis is always a circle. Theorem. The volume V generated by revolving about the X-axis the area bounded by the curve y = f{x), the x-axis, and the ordinates to the curve at the points whose abscissas are a and X is such that D^V = 7ry2. Let the soUd V be bounded on the left by the plane per- pendicular to the a;-axis at A (a, o). The radius of the circle bounding V on the right is y = f{x). Denote the area of this section by 6' (Fig. 189). If X increases by Aa;, V increases by an amount AT which is the volume of that part of the solid lying between V and a section b" at a distance Lx from b\ If y increases as x in- creases, then b'DS? In biology, we learn that a cell subdivides after a certain time. Why? 4. The diameter of a sphere is found by measurement to be 8.3 inches with a probable error of 0.1 inch. What is the error in the computed value of the area of the sphere? In the computed value of the volume? Is the percentage error in the area greater or less than the percentage error in the volume? 5. What is the relation between the rates of increase of the radius and volume of a soap bubble? When the radius is 3 inches and is increasing at the rate of 0.1 inch per second, how fast is the volume increasing? The surface? 6. A washbowl is in the form of a hemisphere of radius 8 inches. How much water is there in it when the water is 5 inches deep? 7. When the depth of the water in the bowl of Exercise 6 is 4 inches the surface is f alHng at the rate of | of an inch per second. How rapidly does the water run off through the drain pipe? 8. Find the area and volume of the earth, assuming that it is a sphere whose radius is 3957 miles. Find the error and relative error in each case if the error in the radius is not more than 7 miles. How does the relative error in the area and in the volume compare with the relative error in the radius? 9. How many lead shot i of an inch in diameter can be made from a piece of lead pipe 2 feet long whose outside and inside diameters are re- spectively 1.25 inches and 1 inch? MISCELLANEOUS EXERCISES 1. Find the volume generated by revolving the area bounded by the curves y = x^ and x = y^ about the rc-axis. 2. A ball rolls down a smooth plane inchned at an angle of 18°. Find how far it will roll in t seconds. 3. The diameter of a cone is found to be 5 inches, and the altitude 8 inches. If the error in the diameter is 0.06 of an inch and the altitude is exact, find the error in the computed value of the lateral area. Definition. At the center of an equilateral triangle ABC erect a line perpendicular to the plane of the triangle. Take a point D on it such that AD = AB, and join D to A, B, and C. The figure so obtained is called a regular tetrahedron [Fig. 193 (a)]. 330 ELEMENTARY FUNCTIONS 4, Find the altitude DO and the volume of a regular tetrahedron (a) whose edge AB is 6 inches; (b) whose edge is e. Definition. At the center of a square ABCD erect a hne perpen- dicular to the plane of the square, and extend it on both sides of the plane to points E and F such that EA =- FA = AB. Join E and F to the vertices of the square. The resulting figure is called a regular octahedron. Fio. 193. 6. Find the volume of a regular octahedron (a) whose edge ^iS is 6 inches; (b) whose edge is e. (Note that EF is a diagonal of the square AECF.) 6. Express the total area of' the octahedron in terms of the edge. Do the same for the cube and the regular tetrahedron, and plot the graphs of these three functions of e on the same axes. For the same value of e, which soUd has the greatest area? Which area increases the most rapidly as e increases? 7. The graph of xi + yi " ai is a parabola tangent to the coordinate axes whose axis of symmetry bisects the first quadrant. Find the area bounded by the curve and the axes. 8. Find the volume of the solid generated by revolving about the y-axis the area bounded by the parabola y = ax^, x = 0, and y = h. Show that this volume is one-half the volume of a cylinder whose altitude is h and whose bases are equal to the circle forming the upper surface of the solid. 9. A point moves so that its coordinates at any time t are given by 2t 1 + fi' 1 -t \ 1 + 1^' Find the components parallel to the axes of the velocity and acceleration. Show that the point moves in a circle (to eliminate t, square both equations INTEGRATION 331 and add the results) and determine the part of the circle it describes from t = to t = 1. Find the velocity and acceleration (direction and magni- tude) at these times 10. A water resevoir is in the form of the figure generated by revolving the curve 12y = x* about the y-axis. How much does it contain if the water is 10 feet deep at the center? If water is entering at the rate of 10 cubic feet per second when the depth at the center is 12 feet, how rapidly is the surface rising? CHAPTER VIII PROPERTIES OF TRIGONOMETRIC FUNCTIONS LOGARITHMIC SOLUTION OF TRIANGLES, CASES m AND IV 116. Introduction. In this Chapter we shall derive formulas which express properties of the trigonometric functions anal- ogous to certain properties of algebraic, exponential, and logarithmic functions with which we are already familiar. These properties enable us to change the form of expressions involving trigonometric functions, a process of great importance in those parts of more advanced mathematics where these func- tions appear. These properties, together with the formulas for the functions of n90° =*= d (page 192), are used more, perhaps, than the solution of triangles, except in such fields as surveying. As an application of these formulas we shall obtain formulas for the solution of triangles, which are adapted to the use of logarithmic tables, for those cases in which we have hitherto used the law of cosines. In the sections immediately following we shall consider the interdependence of the trigonometric functions of an angle 6, with applications to the solution of equations and the proof of identities. 117. Fundamental Trigonometric Relations. As an im- mediate consequence of the definitions of the trigonometric functions we obtained the reciprocal relations cote = ^^; (1) CSC e = -j^- (3) sin (7 332 PROPERTIES OF TRIGONOMETRIC FUNCTIONS 333 In the same section, we saw that . /, sin ^ tan 6 = (4) cos 6 To these relations we now add sin2 d = cos2 ^ = 1, (5) in which sin^^ denotes the square of sin d, or (sin dy, and cos^ 6 denotes (cos dy. To prove (5), we have, for any value of d, sin2 6 + cos2 6 2/2 X^ -2 +72 y^ + x' These five formulas are known as the fundamental formulas of trigo- nometry. They are independent, that is, no one of them is a consequence of the other four, and they may be solved for any five of the six func- tions in terms of the sixth. This is illustrated in the Fig. 194. Example. Express each of the trig onometric f unctions in terms of sin d. From (5) we obtain cos = ±\/l - sin^ 6. sin 6 Substituting in (4), Then by (1), (2), (3), tan0 = VI - sin2 e cot ty = : — ^ , sec ^ = sm 6 ' CSC 6 1 sin VI - sin2 The sign of the radical must be chosen from a knowledge of the quadrant in which lies. These results may also be obtained from a figure by means of the definitions of the functions (see Exercises 9 and 10, page 170). The given function, sin 0, may be re- garded as a fraction with unit denominator. Describe a circle about the origin with radius r = 1, and draw the line y = sin 0, regarding as constant. Let the line cut the circle in P and P'. Then the angles 334 ELEMENTARY FUNCTIONS XOP and XOP' are such that their sines are equal to ^^, and hence these angles are values of 6. From the right triangles, using the Pythagorean Theorem, we get OM = VI - sin2 e and OM' = -y/l - sin^ d. Then the results obtained above may be written down from the figure. In this figure it is assumed that sin 6 is positive. If sin 6 is negative, the angles would he in the third and fourth quadrants. This example shows that it would be entirely possible to get along with only one of the trigonometric functions, but fre- quently it would be inconvenient. The formulas (l)-(5) are identities (Definition, page 134). There can not be more than five independent equations con- necting the six functions, for if there were six, it would be possible to solve them for the six functions, which would then be constants. Hence any other identity involving functions of a single angle 6 must be a consequence of these five, and may be proved by them. • Three other identities are of sufficient importance to be listed with the fundamental formulas. They are: J. a cos ^ cot 6 = ^—A' (6) l + tan2 = sec2^. (7) 1 + cot2 d = csc2 e. (8) The first two may be proved as follows, the proof of (8) being analogous to that of (7). 1 1 cos ^ cot^ tan 6 sin 6 sin 6 cos 6 1 + tan^ ^ = 1 4- — 2-2 = 2Q = — rh = sec^ 6. cos^ cos^ a cos^ The formulas in this and the following sections should be memorized. 118. Trigonometric EquatioQS. No specific rules for the solution of equations involving trigonometric functions can be given, but the following remarks may be of service. PROPERTIES OF TRIGONOMETRIC FUNCTIONS 335 Express the equation in terms of a single function by means of the formulas in the preceding section, and solve for this function. The values of the angle may then be found from the tables. //, when all the terms are written on the left of the equality sign, the left-hand member can be factored, equate each factor to zero, and solve the resulting equations. For any solution makes the product of the factors zero, and hence it must make one of the factors zero. Conversely, any solution of one of the resulting equations makes one of the factors zero, and hence also the product. Example 1. Solve tan a; - 2 sin x = 0. By (4), Section 117, ^-^^-^ - 2 sin x = 0. •^ ^ ' ' cos X Multiplying by cos x, sin a; - 2 sin x cos x = 0. Factoring, sin x (1 - 2 cos x) = 0. Equating each factor to zero and solving, sin X = 0, or cos x = \. Hence a; = 0, tt or x = 7r/3, 57r/3. All the other values of x may be obtained by adding 2mr to each of these four solutions. To check the results, we substitute each of the values of x in the given equation. This gives: If X = 0, tan X - 2 sin X = - 2 X = 0. If X = -TT, tan X - 2 sin X = - 2 X = 0. If X = 7r/3, tan X - 2 sin X = V3 - 2 X V3/2= 0. If X = 57r/3, tan X - 2 sin X = - VS - 2(- V3/2) = 0. Hence the results obtained are correct. Example 2. Solve the equation 2 sin x - cos x = 1. By (5), Section 117, 2 sin x =fc Vl - sin^ x = 1. Transposing, 2 sin x - 1 = ±\/l - sin^x Squaring both sides, 4 sin^ x-4sinx + l = l- sin^ x, whence 5 sin^ x - 4 sin x = 0. Factoring, sin x (5 sin x - 4) = 0, so that sin X = or sin x = | = 0.8, and henc^ x = or tt, or x - 53°. 13 = 0.9272 or 126.'*87 = 2.2144. 336 ELEMENTARY FUNCTIONS The values of the last two angles in radians are obtained from page 32 of the Tables. It is essential that these results be checked, as extraneous roots are frequently introduced when an equation is squared. We have: lfa; = 0, 2sinx-cosa; = 2x0-l = -l. If X = TT, 2 sin x - cos X = 2 X - (- 1) = 1. Jfx^ 0.9272, 2 sin a; - cos X = 2 X 0.8 - 0.6 = 1. If X = 2.2144, 2 sin 0? - cos X = 2 X 0.8 - (- 0.6) = 2.2. The first and last values of x do not satisfy the given equation, and they are therefore discarded. The values x = ir and x = 0.9272 do satisfy the equation. All other solutions may be obtained from these by adding 2n7r. 119. Trigonometric Identities. Identities constitute an im- portant part of mathematics. If we encounter a complicated fraction in the solution of a problem, we proceed to simplify it. The reduction of the fraction is essentially a proof of the identity obtained by equating the original fraction to the simpler result which is foimd. The logarithmic identity, , a sin B , i • r> ^ • a log — : — J- = log a + log sm B - log sm A, which is proved by Theorems 7 and 8, page 223, is the basis of computing b, sl side of a triangle, if a, A, and B are given. Trigonometric identities are of the same importance in dealing with expressions involving the trigonometric functions as algebraic and logarithmic identities are in working with algebraic and logarithmic functions. Trigonometric identities involving functions of a single angle may be proved by means of the fundamental formulas in Section 117 and the operations of algebra, by transforming one member of the identity into the other. The proofs of equa- tions (6) and (7) in the section cited are illustrations of the method. Example 1. Prove the identity, 1 sec* X - sec a; tan z 1 +sina; PROPERTIES OF TRIGONOMETRIC FUNCTIONS 337 By the formulas in Section 117 we obtain sec*x Example 2. Prove Proof. 1 1 sec z tan z = — r~ cos2 z cos z sinx C08X 1 - sin x COS^ X 1 - sin X ~ 1 - sin * a; 1 1 + sinx sin a; 1 + cos x 1 - cos X "" sin X sin X sin x (1 + cos x) 1 - cos X " (1 - cos x) (1 + cos x) sin X (1 + cos 1 - COS^ X sin X (1 + cos x) sin^ X 1 + cos X Another method of establishing the truth of an identity is to show that both members can be reduced to the same form. Still another is to transform the given equation until it is reduced to some known identity. EXERCISES 1. Express formulas (l)-(8), Section 117, in words. 2. By means of the formulas in Section 117, find all the functions of 0, given (a) cos e = j%. (b) tan d = 1 (c) esc ^ = 3. Check the results by the method used in Exercise 10, page 170. 3. Using both methods in the Example in Section 117, express each of the functions in terms of (a) cos 0. (b) tan 0. (c) cot 0. (d) sec 0. (e) esc 0. 4. Express , V tan + cot . . e • o J a W ~T~5 — I a 111 terms of sm u and cos u. ^ cot - tan ,, V sec - tan^ . . . n (b) r-T-fl 111 terms of cos 0. sm^ t7 , V 2 sin cos ^ . . t i. a (c) — r-5 ^-r-a in terms of tan 9. cos'' u - sm^ c' 338 ELEMENTARY n^NCTIONS 5. Solve the equations: ^ (a) 2 sin2 a: + 3 cos a; = 0. (b) cot X -2 cos a: = 0. (c) sin X tan x - cos a; = 0. (d) sin X + cos x = \. (e) sin X + 3 cos a; = 1. (f) sin a; + 3 cos X = 2. 6. Prove the identities: (a) sin2 Q - cos^ ^ = 2 sin^ ^ - 1. (b) 2 sin2 ^-1 = 1-2 cos^ B, (c) tan + cot = sec Q esc ^. 1 - sin a: cos x (d) cos a; 1 - sin a; 1 (e) csc2 + cot esc ^ = - ^ 1 — cos a (f) (2 sin X cos a;)* + (cos^ x - sin^ a:)^ = 1. 120. Functions of the Sum of Two Angles. The relatioD (a + 6)2 = a2 4- 2a& + h^ expresses in a different foim the square of the sum of two num- bers. We seek now an analogous expression for the sine of the sum of two angles, sin (0 + <^), where ^ and are any two acute angles. The sum ^ + is constructed by using the terminal line of as the initial line of <^. It may be either an acute or an obtuse angle, as indicated in the figures. Let A be any point on the terminal line of ^, and draw AB perpendicular to X. Then 8in(e + <« = ^- To express this ratio in terms of and <^, draw AC per- pendicular to the initial line of , and CD perpendicular to OX. Then the functions of B and <^ may be found from the right triangles OT>C and OCA respectively. Draw CE perpendicular to AB. Then the triangles ODC and AEC are similar (why?), so that Z.EAC = 6, and the functions of 6 may also be found from the triangle AEC, We then have PROPERTIES OF TRIGONOMETRIC FUNCTIONS 339 BA sin {6 + 0) OA DC + EA OA PC sin e + CA cos B OA . nOC ^ ^CA Fig. 196. or sin (^ + ) = sin ^ cos + cos ^ sin . In like manner, 1) cos(e + 4>)-^ SO that OP- EC OA PC COS d -CA sin 6 OA nOC • nCA cos a 7^-r - sin ^ TT-j OA OA cos (^ + <^) = cos 6 cos (l> - sia. 6 sin 0. (2) 340 ELEMENTARY FUNCTIONS The proofs hold for either figure. In deriving (2) when B -\- (j) is obtuse, it should be noted that OB is negative, so that we will still have OB = OD - BD. We shall assume that these formulas hold for all values of 6 and , negative as well as positive. (See Exercises 5-8 below.) Dividing (1) by (2), and using (4), page 333, we have ffi ^^ - sJQ (^ + ) _ sin 6 cos (f) + cos 6 sin tan ( +q>) - ^^^ (^q ^ ^^ - (>os cos <^ - sin 6 sin (j)' Dividing numerator and denominator by cos 6 cos , we obtain sin 6 sin <^ //I . j.\ cos 6 cos (b tan (^ + <^) = : — s — ^^' 1 sm sm (t> cos 6 cos , , ,^ , , . tan + tan ,_ . whence tan id + )=^ l^tan^tanV ^^^ EXERCISES 1. Express formulas (1), (2), (3), in words. 2. Using the functions of 30°, 45**, 60°, as given on page 161, find all the functions of (a) 75°, (b) 105°. Hint: sin 75° = sin (45° + 30°), etc. 3. If ^ = arc sin f and <}) = arc cos /a, find the functions of 6 + (f>. 4. What force acting parallel to the plane is necessary to support a body weighing 100 pounds on a smooth plane if the inclination is ^ = arc sin f ? If the inclination is <^ = arc sin jf ? If the incUnation is ^ + ^? 6. Prove that (1) holds for all positive values of and . Solution: We will first prove that: // (1) and (2) are true for a pair of values 6 and , then (1) is true when 6 is increased by 90°. Let 6' = 90° + d. Then sin (6' + 0) - sin (90° + + ) = cos (^ + = cos {9' - 90°) cos (j) - sin {$' - 90°) sin = cos (90° - 6') cos = sin 0' cos (j) + cos 6' sin . Since (1) holds for all acute values of 6 and , by the above it holds for all obtuse values of and all acute values of 0. PROPERTIES OF TRIGONOMETRIC FUNCTIONS 341 Applying the fact proved above once more, (1) holds for all values of d less than 270° and of (f) less than 90°; etc. In Uke manner we may prove that may be increased by 90° and thence that (1) holds for all positive values of 6 and . 6. Prove that (2) holds for all positive values of 6 and (f). 7. Prove that (1) holds if (f> is negative. Solution: No matter what the value of may be, we can choose an integer n such that n 360° + ^ is positive. Then sin (0 + <^) = sin {n 360° + 6 + )= an Id + {n 360° + <^)] = sin d cos (n 360° + , positive and negative? 10. What 'property of (a) a;**, where n is a positive integer; (b) 6*, is analogous to the properties of sin x, cos x, and tan x given by formulas (1), (2), and (3)? 11. If sin 1° is known, how complete a table of values of the trigo- nometric functions can be computed? 12. Prove the identities: (a) Sin (^ + ^) - cos (^ + ^) = sii (b)Tan(|+x) = ][-±- + tana; tanx (c) Sin {x + y) cos y + coa{x + y) sin t/ =» ein (a; + 2y). 13. Solve the equations: (a) Sill f a: + M = (b)Sin(a: + |)-cos(a; + ^) = ^ V2 2 * V2 2 * 14. Prove that the force to make a sailboat move forward will be greatest when the direction of the sail bisects the angle between the keel and the apparent direction of the wind. 121. Fimctions of the Difference of Two Angles. We now seek an expression for the sine of the difference of two angles, analogous to the formula in algebra for the square of the dif- ference of two numbers. 342 ELEMENTARY FUNCTIONS Since 6 - (j) = 6 + (- (jy), and since the formulas in the pre- ceding section are true for all values of the angles, we have sin (0 - 0) = sin {d + (- 0)) = sin 6 cos (- 0) + cos 6 sin (- ). But cos (- <^) = COS0 and sin (-) = - sin 0. And hence, sin (6 - - cos 6 sin 0. (1) In like manner it may be proved that cos {6 - ) = cos cos <^ -f sin ^ sin 0, (2) 122. Functions of Twice an Angle, or the Functions of Any Angle in Terms of Half the Angle. Since 20 = ^ + 0, we have sin 26 = sin {B + 6) = sin cos + cos 6 sin 6, (1), page 339, and hence sin 20 = 2 sin cos 0. (1) In like manner, it may be proved that cos 26 = cos2 6 - sin2 0, (2) a.d tan2» = j?^. (3) Since sin^ 6 + cos^ = 1, (2) may be written in the forms cos 20 = (1 - sin2 6) - sin^ = 1 - 2 sin^ 0. (2a) cos 20 = cos2 - (1 - cos2 0) = 2 cos^ 0-1. (2b) The graph of sin 20 may be obtained by bisecting the abscissas of points on the graph of sin (Theorem, page 151). The graph suggests that the period (Definition, page 168) of sin 20 is TT = 180°. This is, indeed, the case. For if we re- place by + 180°, we get sin 2(0 + 180°) = sin (20 + 360°) = sm 20. A general expression for the sine of the product of two numbers would be analogous to the theorem giving the log- PROPERTIES OF TRIGONOMETRIC FUNCTIONS 343 arithm of the product of two numbers (page 223), or the nth power of the product of two numbers [(3), page 153]. Formula (1) is the special case of such a theorem obtained when one Fia. 197. of the numbers is 2. Another special case is given in equation (1) of the section following. 123. Functions of Half an Angle, or Functions of Any Angle in Terms of Functions of Twice the Angle. Since all that is essential in the formulas of the preceding section is that the angle on the left be double that on the right, formulas (2a) and (2b) may be written cos = 1 -2sin2 612, cos 6 = 2cos2 612- 1. Solving these equations for sin 6 /2 and cos 6 /2 we have sin 612 cos 612 cos 6 + cos 6 Dividing (1) by (2) and using the formula (4), page 333, tan 'i^-4\t cos 6 cos 6 (1) (2) (3) EXERCISES 1. Express the formulas in Sections 121, 122, and 123 in words. In Section 122, describe (a) B as any angle, (b) 20 as any angle. In Section 123, describe (a) 6 as any angle, (b) B/2 as any angle. 2. Find all the functions of 15° from those of 45° and 30°. 344 ELEMENTARY FUNCTIONS 3. Find the functions of 7r/8 from those of 7r/4; of 7r/12 from those of 7r/6. The table given by the Hindu Aryabhata (476 a.d. -) gives the values of the sines of angles at intervals of 3° 45'. How could this table be obtained? 4. If ^ = 90", show that (1), (2), (3), Section 121, reduce to formulas in Section 62, page 177. 5. If ^ = 180°, show that (1), (2), (3), Section 121, reduce to (1), (2), (3), page 192. 6. State the properties of 7^ analogous to each of the equations (1) in Sections 120, 121, 122, 123. 7. In the following, find sin B and cos B without finding B given (a) Tan 2B = f. (b) tan 2^ = - f . (c) tan 2B = -V- Hird: Find cos 2B from a figure (see Exercise 10, page 170) and then use formulas (1), and (2), Section 123. 8. Express sin 4^ in terms of functions of 20; sin B in terms of functions of B/2', sin 30 in terms of functions of 30/2. 9. Transform (3), Section 123, into the forms ra /ON 1 - cos B sin B tan (0/2) sin 1 + cos Then transform each of the fractions into tan (0/2) by expressing sin 6 and cos in terms of 0/2, using the formulas in Section 122. In what respect is the latter procedure preferable? 10. If tan 20 = 1.4123, find and cos 0/2. 11. A body is placed on a rough plane which is incUned at any angle greater than the angle of friction. (The angle of friction is an angle such that tan (f) = m, the coeflficient of friction.) If the body is supported by a force acting parallel to the plane, find the limits between which the force must he. Let be the angle of inclination of the plane, W the weight of the body and R the reaction perpendicular to the plane. (a) Let the body be on the point of moving down the plane, so that the force of friction acts up the plane and is equal to mR. Let P be the force required to keep the body at rest. Resolving W into components parallel and perpendicular to the plane, we have P + mR = W sin B, R = W cos 0, and therefore P = W (sin B - m cos 0). Since m = tan we have, P «= PT (sin - tan cos B ,^ sin (0 - ) cos cos (f> (b) Let the body be on the point of motion up the plane. Complete the solution. PROPERTIES OF TRIGONOMETRIC FUNCTIONS 345 12. What is the maximum and the minimum force which will hold a weight of 12 pounds on a plane incUned at an angle of 40°, if the coefficient of friction is 0.5, and if the force acts parallel to the plane? 13. A block W rests on a horizontal plane. If an obhqae force P acts upon W, making an angle 6 with the direction of sUding, and if the coef- ficient of friction is m = tan (f), prove that the magnitude of P that will cause the block to shde is sin (/) P = W cos {d - (t>y Find the least pull that will make the block slide. Suggestion: Show that P will be a minimum when B = and p = 6 -(t). 346 ELEMENTARY FUNCTIONS Then sin a + sin /3 = sin( 6 + (j)) + sin {d - 0) = sin 6 cos 4> + cos ^ sin <^ + sin 6 cos <^ - cos ^ sin <^ = 2 sin ^ cos (j). Solving the first two equations for 6 and , by adding and subtracting, and dividing by 2, we have e = i{a + ^), = i(a - ^), and hence, substituting, sin a + sin /3 = 2 sin i(a + 1^) cos J(a - jS). (1) In hke manner, prove sin a - sin jS = 2 cos J(a + jS) sin |(a - /3). (2) cos a + cos jS = 2 cos i(a + ^3) cos i(a - ^), (3) cos a - cos jS = - 2 sin ^{a + /3) sin i(a - ^3) . (4) Equations (1) and (2) give properties of the function sin x. What are the analogous properties of log x? To what property of x^ is (2) analogous? 125. Logarithmic Solution of Triangles, Case III. The law of sines (page 201) may be written sin A _ a sin B b , Applying division and composition to this proportion sin A — sin jB _a — h sin A + sin B ~ a + 6 Then by (2) and (1), Section 124, 2 cos \{A + B) sin \{A - B) a-h 2 sin J(A + B) cos \{A - B) a + 6 From (6), (1), and (4), page 333, we then have tan \{A - B) _ a-h tan i(A + j5) ~ a + 6 whence tan |(il --5) = ^-^ tan | (il +5). (1) PROPERTIES OF TRIGONOMETRIC FUNCTIONS 347 If two sides and the included angle of a triangle are given, the other angles may be found by means of (1), as is the Example. Solve the triangle, given 6- 96, ^- . 34°.24. We have: a + 6 =» 214, a-h= 22, Then by (1), - 72°.88. tan \{A -B)=^ tan 72^88. log 22 = 1.3424 log tan 72^88 = 0.5115 2.8539 - 1 log 214 = 2.3304 p^ log tan K^ -B) =0.5235-1. Hence h{A - B) = 18°.46. But KA+B)=72^88, 6 = FlQ, = 96 . 198. and hence A = 91^34 (adding) and B - 54^42 (subtracting) To find c, we have 6 sin e 96 X sm 34°.24 ^ sin i^ sin 54^42 log 96 = 1.9823 log sin 34''.24 = 0.7502 - 1 2.7325 - 1 log sin 54^42 = 0.9102 - 1 log c = 1.8223 whence c = 66.42. Check. Find one of the given sides from c, A or B, and C. c sin A 66.42 x sin 91*'.34 sin C sin 54°.42 log 66.42 = 1.8223 log sin 9r.34 = 0.9999 - 1 2.8222 - 1 log sin 34°.24 = 0.7502 - 1 log a = 2.0720 a = 118.0, which agrees with the given value of a. 348 ELEMENTARY FUNCTIONS 126. Logarithmic Solution of Triangles, Case IV. We seek an expression for tan ^A, A being an angle of a triangle ABC, in terms of the sides. We have /I — cos A VlT^^ (by (3), page 343), I b^ + <^ -a 26c (substituting the value of cos A en by the law of cosines) (by multiplying numerator 6M-c2--a2 given by the law of cosines) = V (6^ + 26c + o^)-a^ ^^^ denominator by 26c, and ^ groupmg the terms) -4 (a-6+c)(a + 6-c) (f^^^^rfng) (a + 6 + c) (6 + c - a) ^^ Denote half the perimeter by s, so that a + 6 + c = 2s. Subtracting 26, a - 6 + c = 2s - 26 = 2(s - 6). Subtracting 2c, a + 6 - c = 2(s - c). Subtracting 2a, 6 + c — a = 2(s — a). Substituting above, ^.A=,^^^^^^ i 2s2(s - a) -4 (s - a)(s - 6)(s - c ) s{s - ay 7^a\ (s - a)(s - 6)(s - c) Hence tan^il = f where r = y- — — (1; s - a V s ^ ^ Interchanging the values of a, 6, c, does not affect the values of 8 and r and hence tan J5 ^ , tan K = -^ — (2) PROPERTIES OF TRIGONOMETRIC FUNCTIONS 349 It is proved in most elementary geometries that the area of a triangle in terms of its sides is A = Vs{s - a){s - b){s - c), whence A = rs, (3) A simpler proof is indicated in Exercises 6 and 7 below. Example. Find the angles and area of the triangle a = 34.25, b = 42.91, c = 50.82. The angles are found by (1) and (2), the first step in the computation being to find log r from its value in (1). a= 34.25 log (s -a) = 1.4734 6= 42.91 log(s-6) =1.3239 c= 50.82 log(s-c) =1.1196 2s = 127.98 3.9169 s= 63.99 log 8 = 1.8061 s-a= 29.74 2 [2.1108 s-6= 21.08 logr = 1.0554 s-c= 13.17 logr =1.0554 logr =1.0554 logr =1.0554 log {s-a) = 1.4734 log(s-6) = 1.3239 log (s-c) = 1.1196 log tan M = 0.5820-1 log tan 1 B = 0.7315 - -1 log tan ^ C = 0.9358 - 1 |A = 20^90, iB = 28°.32 ^C = 40°.78. A = 41°.80, B = 56^64 C = 8r.56. Check. A+B + C = 180^00. By (3), the area is A = rs log r = 1.0554 = 727.0. log s = 1.8061 log A = 2.8616 EXERCISES 1. Solve the equations: (a) sin 3a; + sin a: = 0. Hint: Use (1), page 346. (b) cos 3a; + cos x = 0. 2. Prove the identities: (a) sin f ^ + a; j - sin f - - a; j = sm x. ... sin 3a; + sin a; , „ (b) 5 — ■ = tan 2a;. cos 3a; + cos x 3. Solve and check the following triangles, and find the areas. (a) a = 34.34, 6 = 18.96, C = 50°.68. (b) a = 543. 2, 6 = 496. 7, c = 568. 3, (c) b = 7634, c = 8427, A = 102°. 16. (d) a = 1243, b - 1497, - c = 2046. 350 ELEMENTARY FUNCTIONS 4. Two men start to walk from the same point in directions inclined at 45° to each other. When one has gone 792 yards, the other has gone 846 yards. What is the inclination of the line joining them to their paths, and how far apart are they? 5. The sides of a triangular field are 84.3 rods, 77.5 rods, and 102.1 rods. What are the angles of the field? How many potatoes could be raised on it, if the yield were 200 bushels per acre? 6. Find the radius of the circle inscribed in a triangle in terms of its sides. Let the points of contact with the sides be L, M, N, Then AN = AM, BL = BN, CL = CM, hence s^AN + BL + CL^-AN + a, whence AN = s - a. Then if r' is the radius and the center of the circle, we have tan 2^ A = AN But tan U' r » -a and hence ■/ s -a) (s - 6) (s - c) 7. Prove that the area of ABC is A = rs. Hint: If is the center of the inscribed circle, ABC = OAB + OBC + OCA. 8. Find the diameter of the circle circumscribed about a triangle. Hint: Draw the diameter CD through C, and join D to B. What are the values of the angles CBD and CDB1 9. A and B are two inaccessible points. If CD = 1000 yards, ZACD = 82^34, ZADC = 58^22, ZBCD = 72°.78, ZBDC = 105M3, find the direction of AB (the angle AB makes with AC) and the distance from A to B. 10. The altitude of a triangular prism is 13.48 inches, and the sides of the base are 11.23 inches, 8.43 inches, and 4.79 inches. Find the volume. 11. The sides of a field measured in succession are 22.5, 20.4, 18.7, and 21.5 rods. The angle between the first two sides is 84°.5. How many acres does the field contain? Fig. 200. I PROPERTIES OF TRIGONOMETRIC FUNCTIONS 351 127. Miscellaneous Identities and Equations. Example 1. Express sin 3^ in terms of sin 6. Since SO = 20 + 8, we have sin 30 = sin (2^ + ^) = sin 26 cos + cos 20 sin = (2 sin cos 0) cos + (cos^ - sin^ 0) sin = 3 sin cos2 - sin3 = 3 sin (1 - sin2 0) - sin' = 3 sin - 4 sin' 0. This identity, like the formula for sin 2^, is a special case of the sine of the product of two nmnbers (compare (3), page 153, and (7), page 223. It may be used in the proof of the fact that it is impossible to trisect an angle with ruler and compasses, as this phrase was understood by the ancient Greeks. ^ o T» xu -J x-x tan 2x + tan x sin Sx Example 2. Prove the identity j^ — ^ = — : • tan 2x - tan x sin x sin 2x sin x ^ . tan 2x + tan x cos 2x cos x Proof. 7 — n 1 z = "■ — n = tan 2x - tan x sin 2x sin x cos 2x cos X _ sin 2x cos x + cos 2x sin x ~ sin 2x cos x - cos 2x sin x (by multiplying numerator and denominator by cos 2x cos x) sin (2x + x) _ sin Sx '^ sin (2x - x) ~ sin X Example 3. Solve the equation cos x — cos 2x + cos 3x = 0. Applying the formula for the sum of the cosines of two angles to the first and third terms, we get ' 2 cos 2x cos X - cos 2x = 0. Factoring, cos 2x (2 cos x - 1) =0, whence cos 2a: = or cos a; = |. TT TT Then 2a; = ± - + 2mr or a: = =*= -5 + 2n7r. Hence x = ^ -: +mr or a;==*=^ + 2n7r. 4 3 EXERCISES 1. Prove the following indentities. / X ^ X sin (x + y) /un i. . x sin (a; + y) (a) tan a; + tan 2/ = ^ • (b) cot x + coty = . ^ ^ -'^^ . ^ ' 'J/ gQg ^ gQg ^ Sin a: Sin iy 352 ELEMENTARY FUNCTIONS . , sina; + sin 2x . T' sin x ^x (C) 1 ; ^ = tan X. (k) = cot p: • ^ 1 + cos a; + cos 2a; 1 - cos a; 2 ,,. sin x + sin 2a; + sin 3a; , - , . . . _ (d) ; ^ — ■ 5- = tan 2x. /,>. + / , ttX 1 - sin 2a; cos X + cos 2x + cos 3a; (1) cot I a; + - I = \ 2/ cos 2a; sin A + sin ^ _ 1 . . ^®^ cos A + cos 5 ~ *^° 2 ^"^ "^ ^''- (m) cot a; + tan a; = 2 esc 2a;. (f) 1 + tan 2a; tan x = sec 2a;. .X. X cot ^ - tan .J (g) r Z = cos a; n) sin f ^ + a; j = cos f ^ - a; y X X = ^^° •^- / ^ + /^ . 7r\ VI +sin a; ,, > sin 2a; , 3. W ; FT = tan a;. o +„„ _ ^ ^ 1 + cos 2a; ^ xan ^ , (p) = sin X. (i) tan X 2 cot I 1 + taii2 1 cot2 - - 1 ,. tan 3a; - tan X 1 (q) 2 X tan 3a; + tan x 2 cos 2a; . tan^l (j) f = cog a; . (r) cos 3a; = 4 cos' x - 3 cos x. 1 + tan2 1 (s) 2 sin f X + 2 ) sin ( 2; - 7 J = sin^ x - cos^ x. 2. Solve the following equations. (a) cos 2x - 2 cos X - 3 = 0. (h) sin 2x + sin 3x = 0. (b) cos 2x - cos X = 0. (i) sin 3x + sin 5x = 0. (c) cos 2x - sin X = 0. (j) 2 sin x - 3 cos x = 1. (d) tan 2x = s in x sec x. (k) sin x + cos x = \/2. (e) 12 sec2 x + 7 tan x - 24 = 0. (1) 4 sin x + 3 cos x = 5. (f ) sin X + sin 2x + sin 3x = 0. (m) tan x + tan 2x + tan 3x =« 0. (g) cos X 4- sin 2x - cos 3x - 0. , . , x . , x , „ , (n) cos2 2 - sin^ 2 = 1 - 2 cos^ x. 3. In the following identities transform the left member into the right. (a) sin2 X = i - ^ cos 2x. (b) cos^ x = i + | cos 2x. (c) sin* X = f - I cos 2x + I cos 4x. (d) cos* x = f 4- ^ cos 2x + 1 cos 4x. 128. Differentiation of Trigonometric Functions. In order to find the derivative of sin x we shall need to show that lim !!]L^_i ■ m 0^0 e ~ ^^" I PROPERTIES OF TRIGONOMETRIC FUNCTIONS 353 For this purpose we construct the figure, which is self-ex- planatory, and assume that P'P< arc P' NP a cos (at, y = a sin Oit, show that P moves in a circle, and find the magnitude and direction of the velocity and acceleration. Squaring and adding the given equations, we get x^ + y^ = a^ cos* cot + a^ sin* (at - a*, which is the equation of a circle (page 327), and the point P therefore moves on a circle. PROPERTIES OF TRIGONOMETRIC FUNCTIONS 355 The radius OP makes an angle oi 6 = 03t with the ar-axis. Since Dfi = co, the rate of change of 6, or the angular velocity of OP, is the constant co. The table of values gives the times at which P crosses the axes in its first revolution. Velocity of P. The components of the velocity parallel to the axes are (page 311) Vx = Dtpc = Dt (a cos oit) = - a sin (tit Dt03t = - aco sin co<, and Vy = ao) cos cat. Then the magnitude of the velocity is Fig. 202. t? = v(- aco sin coO^ + a^ca^ cos^ o)t = aco. As ao) is constant, the point P moves uniformly in its path. This re- sult may be verified as follows: Since the angular velocity is co, in one second the radius OP will sweep over CO radians, and hence the arc described by P will be aoi (Theorem, page 171); that is, P will move a distance aco in one second. The slope of the direction of the velocity is t e=u)t X y a TT 2« 2 a TT IT —a ^ 3«- — —a 27r -axis. I t :^ ^ ? A compound harmonic curve is obtained by adding the ordinates of points on two harmonic curves. It is the graph of an equation of the form 2/ = a sin (hx + c) Fig. 206 +dsm (ex+f), (5) Fig. 207 shows the compound harmonic curve 2 sin a; + sin (2x + tt). Harmonic curves find important apphcations in such theories as sound and electricity which depend on wave theory. The » L Z^^ -/ % 2 ^t \- '' V \2sBi:-s'n [2a; t-7r) / / *> \l 2sinx' '^ \ \ -1 Z 2 „^V J- ^53 Z % ~ t I 2 Si ■ ^^ ^ ^ ^ it ^ \ tJ t JX^ -<-- ^^-^ O^'' 2 3>l . / / 6 /27f .. . ^ . _^ . . . n. . ^. <- y^ ''8n2lr-V) \ / / ; .. -1 :^-^ ifc-^ 7 ^ J >: z^ 2 ^^ -4 V t ^ z^ ^ -2 ^ ^>-'^ ^ t ^^7 -.-± Fig. 207. period is frequently called the wave length. They are applied, also, in many motions, for example, in the theory of th(? pendulimi. PROPERTIES OF TRIGONOMETRIC FUNCTIONS 361 130. Empirical Data Problems. We shall content ourselves with pointing out the method of solution of some very simple problems in determining the constants in the equation of a harmonic curve. The treatment of empirical data problems in which the points representing a given table of values appear to lie on a harmonic curve furnishes sufficient material for a book involving considerable higher mathematics. Example. Determine the constants a and b if the graph of y = a sin X •{- b sin 2x is to pass through the points (7r/6, 1 + a/3/2) and (7r/4, 1 + V2). Since these points are to lie on the graph their coordinates must satisfy given equation. Hence 1+ -^ = -a + —b, l+V2 = ^a + 6. Solving these equations for a and 6, a = 2 and 6 = 1. Hence the required equation is 2/ = 2 sin x + sin 2x. I --^ - - v"-^ / \^2sinx +feir 2j; -2 -/■ ^ ^ t ^^V>^ i-^^ S " 1 // \ \ismx \/^ "^ X V A^^^ 7^ \ \ -, X jj \ ^ \ / sp^-^ l ^ '^^^SS XX £ i \ 2 sA'-v ' ^5 e <2Tr V ' X 5 \ 1 ^ Z ^ ^ \ 27 ■--1 ^-.^ ^ _5 '^^y^ ^ ^ 21 i 5 ^ A^t ^ ^ ^' ■:2 VV^^ 7 V L ^ J ^'^ Fig. 208. If several points are given which appear to lie on or near the graph of an equation of the form given in the example, we may determine a pair of values of a and b from each pair of points. If the values of a agree closely, and also those of b, by using the average value of the a's and that of the 6's we obtain a good ap- proximation of the law connecting the coordinates of the points. 362 ELEMENTARY FUNCTIONS EXERCISES 1. Show that the equation y = a sin (bx + c) may be put in the form y = A smbx + B cos bx, where A = a cos c and B = a sin c. 2. If 2/ = A sin bx + B oos bx, find the values of a and c for which the given equation has the form y = a sin {bx + c). 3. Show that y = a Bin {bx + c) may be put in the form y = o cos {bx + c'), where c' = c - x/2. 4. On the same axes sketch the graphs of (a) sin X, sin ^x, 3 sin Ix. (b) cos X, cos 2x, I cos 2x. (c) sin X, sin ttx, 4 sin irx. 6. Determine the phase, period, and amplitude of each of the following functions, and sketch the graph. (a) 3 sin irx. (b) 2 sin irx/2. (c) | sin ttx/S. (d) 2 sin {ttx + tt). (e) | sin (rc/2 + x/2). (f) 2 sin (x - 1). • (g) 3 sin {ttx/S - tt). (h) I sin (7ra:/2 + x/2). (i) 4 sin {irx - tt/S). 6. Sketch the following compound harmonic curves by the addition of ordinates. (a) sin a; + 2 sin 2x. (b) 2 sin x + sin x/2. (c) 3 sin 2x - sin x/2, (d) § sin x + sin 3x. (e) sin TTX - I sin 7rx/2. (f) 2 sin ttx/S + | sin ttx. (g) 2 sin 27ra;/3 + ^ sin 27rx. (h) 1.4 sin .7x + .3 sin 1.2x. (i) 2 sin {ttx/ 4: + 7r/2) + | sin (27ra; + tt). 7. Find the symmetry, intercepts, and the points of maxima, minima, and inflection for each of the curves below. Sketch the curve, and find the area of the arch to the right of the origin. (a) 2 sin a; + sin 2x, axis, and suppose that P starts on the positive x-axis. Then the abscissa of P, which gives the position of M at any time, is s •» a cos o)t, where a is the radius of the circle. PROPERTIES OF TRIGONOMETRIC FUNCTIONS 363 If P starts at the point for which OP makes an angle of Bq with the x-axis then s = a cos (co< + ^o). If the diameter on which M moves is taken for the 2/-axis, the distance OM is given by 5 = a sin (co< + do). 8. A point moves according to one of the laws below. Find the time of one oscillation, and when and where the distance, velocity, and accelera- tion have maximum or minimum values. Describe the motion. (a) s = 8 sin 2t. (b) s = 10 sin \t. (c) s = 2 sin irt. (d) s = 4 cos xV3. (e) s = sin (3i + 7r/2). (f ) s = 2 sin {irt/Z + 7r/4) . 9. If s = o sin {(Jit + ^o) show that (a) the acceleration is proportioned to the distance, (b) that the acceleration and distance have maximum or minimum values when the body is at rest, (c) that the body is at the cen- ter of its motion and has no acceleration when it is moving fastest. 10. The wind has left long swells moving along the surface of a lake when a slight breeze produces a ripple. The equations of cross-sections of the waves before and after the breeze are y = 2 sin a:/4 and y = 2 sin x/4 + \ sin 2x. Plot both curves on the same axes. 11. Determine the constants a and h if the graph of the given function is to pass through the given points. Sketch the graph. (a) asmx + h sin x/2, (7r/2, 3 - V2), (tt, - 2). (b) a sin xJ2 + h sin 2x, (tt/S, 1 - \/3/4), (27r/3, V3 + V'3/4). (c) a sm a; + 6 sm 2x, (.157, .309), (.332, .632), (.611, 1.04). (d) a sin X + 5 cos x, (.122, .570), (.349, 1.20), (.646, 1.47). MISCELLANEOUS EXERCISES 1. A and B are two points on the opposite sides of a hill which are to be connected by a tunnel. Both points are visible from a third point C which is 1356 feet from A and 2582 feet from B. li AACB = 47°.34 and if the angles of elevation of A and B &t C are respectively 12°.35 and 9°. 82, find the length of the tunnel and the inclination of the tunnel. 2. Construct a table of analogous properties (see Section 56) of the functions e*, log x, and sin x. 3. Contrast the formulas for 6"+* and tan (o + h) by expressing each entirely in terms of the notation /(x) . 4. Find approximate values of the real roots of the following equations (see Exercise 12, page 261). (a) sin 2rc - X = 0. (b) cos ^x - x^ = 0. (c) sin 3x - e~* => 0. (smallest positive root.) (d) cos 2x - 2x = 0. (e) sin ^x - log x = 0. (f) 2 cos 2x - e* = 0. (g) tan |x - x2 = 0. 364 ELEMENTARY FUNCTIONS 6. To find the distance between two towers on opposite sides of a river a base line 300 feet long was laid off. The following angles were measured, C and D representing the towers and AB the base line: AABC= 35°.24, ABAD = 29°.61, AABD = 108°.47, ABAC = 97°.59. ComputaCD. 6. Find the number of acres in a field in the form of a quadrilateral ABCD if AB = S15 yards, ZZ)AC = 123°.6, ZCAB = 54°.68, ZABD = 65°.23, and ZABC = 112° A. 7. If a projectile is fired at an angle d with the horizontal with an initial velocity of vo, the equation of its path is y = X tan a - cos^ d Find the range as a function of 20, and the value of 6 which gives the maximum range. 8. A corridor turns at right angles. It is 8 feet wide on one side of the .turn and 6 feet wide on the other. How long a beam can be carried hori- zontally around the corner? 9. The path of a point on the rim of a wheel rolling along a level road is given by the equations X = a(ojt — sin o)t), y = o(l - cos oot), where a is the radius of the wheel, co is the angular velocity of the wheel about the axle, and t is the time. Plot the equation of the path taking t = 7r/6co, tt/Sco, 7r/2aj, etc. Find the velocity of the point at any time, and its maximum and minimum values. Where is the point when it is moving slowest? fastest? A bit of mud is thrown from the highest point of a wheel on an automobile. Compare the velocity with which the mud leaves the wheel with that of the automobile. 10. If a is the angle between two lines whose slopes are mi and mz, show that mi — mi tan a 1 + Wi mj CHAPTER IX THEORY OF MEASUREMENT 131. Statistical Methods, The object of an experiment is to replace a complex system in which a nmnber of causes are operating by a simpler system in which causes are controlled and allowed to vary only at the will of the investigator. In physics, for instance, after the isolation of two related vari- ables, an endeavor is made to determine the relation between the variables and to express this relation in the form of a function y =/(x). In some sciences it is difficult or impossible to isolate two related variables from the complex system in which they odcur. In such cases the variation of each variable is observed independently of the others. Statistics has to deal with variables affected by a number of causes. Some of the methods of statistics which have been developed for the analysis of such variations are discussed briefly in this chapter, three of the principal ends sought being the determination of (1) An average value which will represent the values of a variable. (2) A measure of the variability of the items with respect to the average value chosen. (3) A measure of the extent of the relationship between two variables which are associated. The first few sections are given to some essential prerequi- sites. To illustrate the fundamental principle on which the following sections are based, consider the Example. There are five pitchers and three catchers on a bass ball team. In how many ways can a battery be chosen for a particular game? 365 366 ELEMENTARY FUNCTIONS The position of the pitcher can be filled in 5 ways, and with each of these there is a choice of 3 catchers. Hence the two positions together can be filled in 5 X 3 = 15 ways. Fundamental Principle, If one act can he done in A ways and a second act in B ways, the total number of ways in which the two acts may be performed in succession is A-B. The following theorem is a generalization of the fundamental principle. Theorem. If one act can be done iri'A ways, a second in B ways, a third in C ways, and so on, they can all be done together in the order stated in A-B C- . . . ways. EXERCISES 1. If two coins are tossed, in how many ways can they fall? 2. If two dice are thrown, in how many ways can they fall? 3. A committee of three is chosen from seven physicians, eight lawyers, and twelve business men, so that each group is represented. In how many ways can the committee be chosen? 4. If two dice and three coins are tossed, in how many ways can they fall? 132. Permutations. Definition. Each arrangement which can be made of all or part of a number of things is called a permutation. Thus the permutations of the letters a, b, c, taken three at a time are abc, acb, bac, bca, cab, cba, and their permutations taken two at a time are ab, ba, ac, ca, be, cb. The number of permutations of n different things taken r at a time is denoted by nPr- Theorem. The number of permutations of n things taken r at a time is „Pr = n(n - 1) (n - 2) . . . (n - r + 1). There are r places to fill and n things from which to choose. For the first place there is a choice of n things. The second place can be filled with any one of the remaining n - 1 things. The third can be filled in n - 2 ways, and so on. For the rth place there is a choice ofn- (r-l)orn-r + l things. THEORY OF MEASUREMENT 367 Hence by the theorem of Section 131, nPr = n(n - 1) (n - 2) . . . (n - r + 1), Corollary. The number of permutations of n things taken all at a time is „P„ = n{n-l) (n - 2) . . . 2 x 1 = nl The symbol n ! is read factional n, and represents the product of all the integers from 1 to n inclusive. Example, (a) In how many ways can the letters of the word triangle be arranged? (b) How many of the arrangements will begin with the letters tri in this order? (c) How many arrangements will have the letters tri together in any order? (d) How many arrangements con- sisting of three letters can be made from the word triangle? (a) Here n = 8 and r = 8, and hence the number of arrangements is gPs = 8! = 40,320. (b) The first three places are filled and there remain five letters to be permutated in five places, hence the number of arrangements is sPs = 120. (c) The letters tri can be permuted as a group with the remaining five letters, and then the three letters tri be permuted within their own group. Hence the number of arrangements is aPs X ePe = 3! 6! = 4320. (d) Here n = 8 and r = 3; hence the number of arrangements is gPa = 8x7x6 = 336. EXERCISES 1. In how many ways can five flags of different colors be arranged five in a line? Three in a line? 2. How many arrangements of all the letters of the word English will t)egin with a vowel, and end with a consonant? How many will have the vowels together? 3. How many different numbers less than 1000 can be formed from the digits 1, 2, 3, 4, 5, 6? 4. In how many ways may first and second prizes be awarded, if there are 12 competitors in a race? 133. Combinations. Definition. A group of things which is independent of the order of the elements is called a com- bination. The combinations of a, b, c, taken two at a time are ab, ac, be. The selection 6a is a different permutation from ab, but the same combination. 368 ELEMENTARY FUNCTIONS The six permutations of the letters ahc taken three at a time, namely, abc, acb, bac, bca, cab, cba are different arrangements of one combination. The number of combinations of n things taken r at a time is denoted by nCr- Theorem 1. The number of combinations of n things taken T at a time is n(n - 1) (n - r) . . . (n - r + 1) nCr- ^^— Each combination of r things chosen from the n things can be arranged in r! ways. Therefore all the combinations can be arranged in r !„(7r ways. But this is the number of ways in which n things taken r at a time can be arranged, so that r\nL>r = ni^r Pr r — I rl n^r _ ^(^ - 1) • - ♦ (n - r + 1) r\ ' Theorem 2. The number of combinations of n things taken r at a time is the same as the number of combinations of n things taken n — r at a time. Multiplying numerator and denominator of the formula for nCr by (n - r) ! we have _ n(n - 1) . . . {n - r + 1) (n - r)(n - r - 1) . . . 2-1 "^'■~ r\{n-r)\ n\ r\{n — r)\ By Theorem 1, we have ^ _ n(n - 1) . . . (n - (n - r) + 1) _ n(n - 1) . . . (r+ 1) '^^"-'" " {n-r)\ ~ {n-r)l Multiplying numerator and denominator by r! n(n - I) . . . (r + l)r(r - 1) ... 1 ?i! »C n—r — ' {n — r)\r\ {n — r)\r\ THEORY OF MEASUREMENT 369 Example 1. A committee of 5 is to be ciiosen from 7 lawyers and 6 physicians. How many committees will contain (a) just 3 lawyers, (b) at least 3 lawyers? (a) The number of ways o^ selecting three lawyers is The number of ways of selecting two physicians is Hence the number of committees which will contain exactly three lawyers is 7C3 X 6C2 = 35 X 15 = 525. (b) At least three lawyers are present in each committee of the types: 3 lawyers and 2 physicians, 4 lawyers and 1 physcian, and 5 lawyers. Hence the number of committees which contain at least three lawyers is 7C3 X 6C2 + 7C4 X eCi +7 Cfi = 756. Example 2. How many arrangements can be made consisting of two vowels and three consonants chosen from the letters of the word triangle? The vowels may be selected in 3C2 = 3 ways. The consonants may be selected in sCs = 10 ways. Hence the total number of selections consisting of two vowels and three consonants is 3C2 X 5C3 = 30. Each of these selections can be arranged in 5! ways. Hence the required number of arrangements is 5! 3C2 X sCs = 3600. EXERCISES 1. How many alloys can be made from thirty of the known metals chosen two at a time counting one alloy only for each pair of metals? Solve the problem if there are three metals in each alloy. 2. In how many ways can a basket ball team be selected from 9 candi- dates? If A plays center in every combination, in how many ways can the team be chosen? 3. How many arrangements can be made of 3 vowels and 4 consonants chosen from 5 vowels and 8 consonants? 4. How many straight lines are determined by (a) 5 points no 3 of which are in the same straight line? (b) n points, no 3 of which are in the same straight line? 5. In how many ways can a baseball nine be chosen from 13 candidates, provided A, B, C, D are the only battery candidates, and can play in no other position? 370 ELEMENTARY FUNCTIONS 6. In how many ways can a committee of 5 be chosen from 7 democrats and 7 republicans, so that there will be (a) three democrats, (b) no more than three democrats, (c) at least three democrats on the committee? 134. The Binomial Expansion. In finding the product of the binomial factors {x + ai)(x + a2){x + as), each partial product is obtained by choosing one and only one term from each factor and multiplying the three terms together. The sum of the partial products gives the desired product. There is only one term containing x^, since the three x^s can be chosen from the three factors in but one way. The terms of the product containing x^ are obtained by choos- ing X from two of the factors and an a from the third factor, which gives the partial products x^ai, x^a^, x^as. The number of such terms will be the number of ways we can choose an a from the three a's, or 3^1. To obtain the term in x, we choose an x from one binomial and two a's from the two remaining in all possible ways, which gives the partial products xaia2, xaias, xa2az. The number of such products is therefore 3C2. There is only one way of choosing the three a's. Hence, {x + ai){x + a2){x + az) = x^ + (ai + ^2 4- az)x'^ + (01^2 + aiaz + a2az)x + aia^fiz- If we let ai = a2 = ^3 = «, we have {x + ay = x» + 3ax2 + ZaH + a^. And since, from the preceding, the coefl&cient of x^ is the num- ber of ways we can choose an a from the three a's, the coeffi- cient of X is the number of ways we can choose two a's out of three a's, and so on, we can write the expansion in the form (x + af = x^ + sCiax^ + zC^a^x + zCza\ In a similar manner it can be shown that when w is a positive integer, we have The binomial expansion: (x + c)" = X" + wCiCX"-! + „ Cio'x"-^ + . . . THEORY OF MEASUREMENT n{n - 1) 371 where nCi = 1, nC2 21 n(n - 1) . . . (n - r +1) 7i ' nCr The values of the coefficients are given in the following table for several values of n. The table is called Pascal's triangle. n. 123456789 lOlr 1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1 5 1 5 10 10 5 1 6 1 6 15 20 15 6 1 7 1 7 21 35 35 21 7 1 8 1 8 28 56 70 56 28 8 1 9 1 9 36 84 126 126 84 36 9 1 10 1 10 45 120 210 252 210 120 45 10 1 The coefficients in each row in the table may be calculated from those in the preceding row by the following rule: In any row add to a coefficient the following coefficient and place the sum below the latter. As an application, consider the Theorem. The total number of combinations of n things taken one at a time, two at a time, and so on up to n at a time is 2"— 1. In the binomial expansion for {x + a^ let x = a = 1. Hence (1 + Ij" = 1 + nCi + nCa 4- . . . + nCn. Therefore „Ci + nCa + . . . + nCn = 2" - 1. EXERCISES 1. What is the middle coefficient in the expansion of (a + 6)"? Using Pascal's triangle write the coefficients of the expansion (a + 6)". 2. Plot the terms of the expansion (^ 4- 1)^^ as ordinates at equal dis- tances along the x-axis. 3. How many compounds consisting of two elements could be made from eighty-three chemical elements? How many consisting of three elements? 4. From eight men, in how many ways can a selection of four men be made (a) which includes two specified men? (b) which excludes two specified men? 372 ELEMENTARY FUNCTIONS 5. How many symbols would be available for a cipher if each symbol is an arrangement of the letters a, b, in a group of five. Thus, A = aaaaa, B = aaaab, etc. 6. Twelve competitors run a race for three prizes. In how many ways is it possible that the prizes may be given? 7. In how many ways can a baseball nine be arranged if each of the nine players is capable of playing any position? If A must pitch and B, C, D, play in the outfield? If A or B must pitch, B or C catch, and D, E, F, play on the bases? 8. How many dominoes are there in a set from double blank to double six? 9. How many melodies consisting of four notes of equal duration can be formed from the eight tones of the major scale? From the thirteen tones of the chromatic scale? 10. A Yale lock contains 5 cylinders, each capable of being placed in 10 distinct positions, and opens for a particular arrangement of the cylinders. How many locks of this kind can be made so that no two shall have the same key? 11. The combination of a safe consists of figures and letters arranged on three wheels, one bearing the numbers to 9 inclusive, another the letters A to M inclusive, and the third the letters N to Z inclusive. If the safe opens for but one of these arrangements, how many different com- binations can be used? 12. How many arrangements of all the letters of the word Columbia (a) begin with a vowel? (b) begin with a consonant and end with a vowel? (c) have the vowels together? 13. Show that the number of ways in which n things can be arranged in a circle is (n - 1)! In how many ways can six persons be arranged in a line? In a circle? 14. Show that if of n things a are alike, h others are alike, c others are alike, etc., the number of distinct permutations, taken all at a time, is n! alblcl .. 16. How many distinct arrangements can be made of all the letters of the word Mississippi? International? 16. How many signals can be made by arranging 2 white flags, 3 red, and 1 blue in a row? 17. Prove that nCr + nCr-i = n+iCr. Compare this with the rule fo: finding numbers in Pascal's triangle. 18. Prove that 2nCn+r+i = 2»Cn+r ( ^~ ^ . ), a result which will be used later. THEORY OF MEASUREMENT 373 19. How many jdiflferent sums can be formed with a penny, a nickle, a dime, a quarter, a half dollar, and a dollar? 20. A set of weights consists of 1, 2, 4, 8, and 16 ounce weights. How many different amounts can be weighed? 21. If three coins are tossed in how many ways can they fall? Solve the problem for 4 coins. 22. If two dice are thrown in how many ways can they fall? 23. In how many ways can the hands of whist be dealt? 24. How many four-figure numbers can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, which are (a) divisible by two? (b) divisible by five? 135. Probability. On one of the faces of a cube is placed the letter A, on two of the faces the letter B, and on the re- maining three faces the letter C. If the cube is thrown the total number of ways the cube can fall is six, all of which we will assume are equally likely to occur. The number of ways that the letters, A, B, C, can turn up are respectively one, two, and three. In a great number of trials the letter A would turn up approximately in ^th of the total number of trials, the letter B in fths and the letter C in fths. This does not mean that in every set of six trials A turns up once, B twice, C three times, but that in the long run, as the number of trials is increased, the frequency with which A, B, C turn up ap- proximates to i, I, I of the total number of trials. Definition. The ratio of the number of ways in which a particular form of an event may occur to the total number of w^ays in which the event can occur (all assimaed equally likely) is said to be the probability of the particular event. Theorem 1. If the probability that an event will happen is p and the probability that it will not happen is q, then q = 1 — p. Let T denote the number of ways the event can happen, F the number of ways in which the favorable form of the event can happen, and U the number of ways in which the favorable form of the event cannot happen. U 7 Since F + t/ = T, Then, P = ^, q = f 374 ELEMENTARY FUNCTIONS we have, dividing by T, F U or p + q = 1. ,\ q = 1 -p. Corollary. If the favorable form of an event is certain to happen, then q = and p = 1. Example 1. Find the probability of not throwing a sum of five with two dice. If the dice fall 1, 4, or 4, 1, or 2, 3, or 3, 2, the sums will be five. Hence, F = 4, T = 6.6 = 36, p = A = i, 9 = 1. Hence, the probability of not throwing a sum of five with two dice is |. If a sum ot money s is paid upon the happening of an event whose probability is p, the product sp is called the mathemati- cal expectation. Example 2. According to the experience of the American life insur- ance companies, of 100,000 children 10 years of age, 749 die within a year. Neglecting interest on money and the cost of administration, what would be the cost of insuring the life of a 10-year old child for $1000 for one year? 749 The probability of a child dying is p 100,000 749 The mathematical expectation is • 1000 = 7.49, and hence the re- quired premium is $7.49. That is, if each of the 100,000 children paid $7.49, then the sum $749,000 resulting would be sufficient to make 749 payments of $1000 each. Example 3. From a bag containing 8 white balls and 6 black balls, 5 balls are drawn at random. What is the probability that 3 are white and 2 are black? From 14 balls, 5 can be selected in uCg ways. From 8 white balls, 3 can be selected in gCa ways, and from 6 black balls, 2 can be selected in eCz ways. Hence the probability of drawing 3 white and 2 black balls is 8.7.6 6.5 _ sCs-eCa 1.2.3 12 60 P" ua ^ 14. 13. 12. 11. 10 1.2.3.4.5 143 THEORY OF MEASUREMENT 375 EXERCISES 1. From a bag containing 3 white balls and 7 black balls one ball is taken at random. What is the probability that it will be white? Black? What is the sum of these probabilities? 2. From a bag containing 4 white balls and 8 black balls 2 balls are drawn at random. What is the probability that they will both be black? both white? one white and one black? What is the sum of these probabilities? 3. If two dice are thrown what is the chance of throwing a sum of seven? double sixes? 4. If the probability of an event is ^i, what is the probability that the event will not happen? 5. Six persons are about to seat themselves in a row. • What is the prob- ability that two specified persons will be together? Will not be together? 6. If a prize of $10 is given for drawing a red ball from a bag containing 2 red balls and 6 white, what is the value of the expectation? 7. If of 92,637 people living at the age of 20 there are 85,441 living at the age of 30, what should be the premium for insuring the life of a person of age 20, for $1000 for 10 years, neglecting interest and administrative charges. 136. Compound Events. Sometimes it is convenient to consider an event as made up of two or more simpler events. Thus if two balls are drawn from a bag containing 4 white balls and 5 black balls, the double drawing may be regarded as a compound event made up of two single drawings per- formed in succession. The component events into which a compound event is resolved are said to be dependent or independent according as the occurrence of one does or does not affect the occurrence of the others. If two balls are drawn from the bag mentioned above, the probability of drawing a white ball the first time is f . If a white ball is drawn the first time and not replaced, the prob- ability of drawing a white ball the second time is f . But if a black ball is drawn the first time and not replaced, the prob- ability of drawing a white ball the second time is f . Hence if the ball drawn the first time is not replaced the events are dependent, as the form of occurrence of one does affect the occurrence of the other. If the first ball is drawn and then replaced the probability 376 ELEMENTARY FUNCTIONS of drawing a white ball the second trial is the same as on the first trial. The events are independent as the occurrence of one does not depend on the occurrence of the others. Several events are said to be mutually exclusive if only one of the events can happen. Thus in drawing two balls from the bag mentioned above there are three ways in which at least one white ball might be drawn. Both balls might be white, the first white and the second black, the first black and the second white. These events are mutually exclusive for the oc- currence of one necessarily excludes the occurrence of the others. Theorem 1. 'If pi, P2, - . -, Pn are the probabilities of n mutvxilly exclusive events, the probability that one of these events will occur is equal to the sum of their separate probabilities. Let T be the total number of ways in which the event can happen, Fi the number of ways favorable to the first event, F2 to the second, and so on. Then Fi F2 Fn Since each one of the mutually exclusive events is favorable, the total number of favorable ways is /^i + /^2 + . . . -f Fn. Therefore the probability that one of them will happen is _ Fl + i^2 + . . . i^n _ ^1 , ^ , ^. P ~ rp ~ rn \ rp i ' • * rp' Hence p = pi + P2 -f . . . + Pn. Example 1. If a cube has the letter A on one face, the letter B on two faces, and the numeral 5 on three faces, what is the probability that a letter will turn up when the cube is tossed? There are two mutually exclusive events favorable to this contingency. (1) The turning up of the letter A. (2) The turning up of the letter B. The probability that A will turn up is |. The probability that B will turn up is |. The probability that a letter will turn up is therefore Theorem 2. If pi and p2 are the probabilities of two inde- pendent events the probability that both events happen is the product of their separate probabilities, pipi. THEORY OF MEASUREMENT 377 Let Ti and Fi be respectively the total and favorable num- ber of ways the first event can happen. Let T2 and F2 be respectively the total and favorable num- ber of ways the second event can happen. Then the total number of ways the compound event made up of these two events can happen is T1T2 and the number of the favorable ways for the compound event is F1F2. Hence the probability of the compound event is Fi' F2 Fi F2 In general if pi, p2, . . ., Pn are the respective probabilities of n independent events, the probability that all the events will happen is the product of all the separate probabilities. Example 2. Find the probability of throwing a six in the first trial only in two throws of a single die. The probabiUty of throwing a six the first time is |. The probability of not throwing a six the second time is | . These events are independent, and the probability of throwing a six the first trial only is if = j\. Example 3. If the probability that A will solve a problem is f and the probability that B will solve it is f , find the probability that the problem will be solved. The problem will be solved if A and B both succeed, if A succeeds and B fails, or if A fails and B succeeds. The probability that they will both succeed is |4 = ^|, since A's success or failure is independent of B's success or failure. Likewise, the probability that A succeeds and B fails is |-f = ^3^, and the probability that A fails and B succeeds is | • f = /g . Since the three possible ways in which the solution of the problem may occur are mutually exclusive the probability that the problem will be solved is A second method of solving the problem is as follows: The probability that the problem will not be solved is f -f = ^^5. Hence, by Theorem 1, Section 135, the probability that the problem will be solved is 1 - j% = |f . Theorem 3. If there are any number of dependent events and if pi is the probability of the first, p2 the probability that when the first has happened the second will follow, pz the probability that 378 ELEMENTARY FUNCTIONS when the first and second have happened the third will follow, and so on, then the probability that the events will occur in succes- sion in this way is pi, P2 . . . Pn- The proof is similar to that of Theorem 2. Example 4. If a bag contains 4 white balls and 5 black balls, and if 2 balls are drawn, what is the probability of drawing a white ball and a black ball? Consider the compound event as resolved into two successive drawings. The probabiUty of drawing a white ball first is f. The probabiUty of drawing a black ball second is |. The probabiHty of drawing the balls in this order is therefore | -f = tg. But since the balls may be drawn in 2 ways, the probability of drawing a white ball and a black ball is x\ -2 = f . Theorem 4. If the probability of the happening of an event in one trial is p and the probability of its failing is q, the proba- bility of its happening exactly r times in n trials is nCr p^q''~^. The probability that the event will happen r times and fail n - r times in a given order (for example, happen the first r times and fail the last n - r times) is p'"g"-'' (Theorem 2). The number of ways this can happen is the number of ways that the r trials can be selected from the n trials, which is nCr. Hence the event can happen exactly r times in nCr mutually exclusive ways, and the required probability is therefore nCrP^q''-^ (Theorem 1). Example 5. What is the probability of throwing six twice in five throws with a single die? Here p = i, g = t, n = 5, r = 2, and hence P_ r rM2 m3_ 5^4. 1.125 ^625. r-^K.2 u; Kt) "1.2 36 216 3888 Theorem 5. // the probability that an event will happen in one trial is p, then the probability that it will happen at least r times in n trials is P = p" + np"-i^i + . . . + nCrP'q''-^. An event will happen at least r times in n trials, if it happens n times, or w - 1 times, and so on down to r times. These are mutually exclusive events. THEORY OF MEASUREMENT 379 HencG; applying Theorems 3 and 1, the probability that an event will happen at least r times in n trials is which reduces to what is required. EXERCISES 1. What is the chance of making a throw with two dice that will be greater than 9? 2. A bag contains 3 dimes and 4 quarters. Three coins are drawn. Find the value of the expectation. Suggestion: Find the value of the expectation of drawing 3 quarters, 2 quarters and 1 dime, 1 quarter and 2 dimes, 3 dimes, and add. 3. Johnny may or may not receive a birthday gift of $1.00 from each of five relatives. What is the value of the expectation? 4. In a bag are five white and five black balls. If two balls are drawn what is the chance that they will both be white? Both black? one white and one black? 5. In a bag are five white and four black balls. If three balls are drawn in succession what is the chance that they will be white, black, white, in this order, if (a) the balls are not replaced each time, (b) the balls are replaced each time. 6. A man is sent 8 keys on a ring for eight locks. What is the proba- bility that he will be able to unlock the first lock with the first key he tries? With one of the first two keys? With one of the first three? 7. Five cards are drawn from a pack of 52 cards. Find the probability that (a) there is a pair, (b) three of a kind, (c) two pairs, (d) three of a kind and a pair, (e) four of a kind, (f ) a flush, (g) a straight, (h) a straight flush. 8. Three dice are thrown. What is the most probable throw? What is the probability of throwing exactly 15? At least 15? 9. If 5 letters are chosen from a group of 4 vowels and 6 consonants, what is the probability that a set will begin with a consonant and end with a vowel? 10. If there were eight independent chances in youth of growing one inch above 5 feet, find the probabilities of the statures from 5 feet to 5 feet 8 inches. 137. Mortality Tables. An important application of the theory of probability is the appUcation to problems concerned with the duration of human life, such as life insurance, pen- sions, life annuities, and inheritance tax laws. Such prob- lems are based on tables called mortality tables which show the 380 ELEMENTARY FUNCTIONS number of deaths that may be expected to take place during a given period, among a given number of persons of a given age. These tables differ for different countries, different races, in the same country, different periods of time, and for the two sexes. Some tables are constructed from the experience of insurance companies, others from census and vital statistics reports. The American Experience Table is based on the records of the Mutual Life Insurance Company of New York. If a mortality table is based on a sufficiently large number of observations, the difference between the result furnished by the tables and actual mortality is negligible. But this must be understood to apply to large groups of people and furnishes no surety to an individual. The following table is a selection from the American Ex- perience Table of mortality, which gives the number of people Ix living at age x out of 100,000 living at the age 10. The probability that a person of age x will be alive at the end of n years is denoted by _ Ix+n nPx — 1 ' I'X The probability that a person of age x will not be alive at the end of n years is 'x+n I'x i'x+n Age Number living X I. 10 100,000 15 96,022 20 92,637 25 89,032 30 85,441 35 81,822 40 78,106 45 74,173 50 69,804 55 64,563 60 57,917 65 49,341 70 38,569 75 26,237 80 14,474 85 5,485 90 847 95 3 nQx = I - nVx = I - Ix h Example. An inheritance of $20,000 is to be paid a boy 10 years of age when he becomes 20. What is the present value of the inheri- tance, money being worth 5% compounded semi-annually? If the sum were certain to be paid to the boy in ten years the present value would be P - 20,000(l + '-^y - $12,220. Since the payment is contingent upon the probability that the boy will live to receive the THEORY OF MEASUREMENT 381 sum, the present value will be the mathematical expectation of receiving $12,220 contingent upon the probability that he will live 10 years. The probability that a person 10 years old will live at least 10 years is ^ ^20 92,637 Hence the present value of the inheritance is (12,220) (.92637) = $11,320. EXERCISES 1. Plot the graph of the mortality table given in Section 137. By means of the graph estimate your own chance of living to the age of 75. 2. A man is 45 years of age and his son is 15. What is the probability that both will be alive 10 years hence? What is the probability that at least one will be alive? 3. A man and his wife are 40 and 35 years old respectively, when their child is 10 years old. What is the probability that all will be alive until the 20th anniversary of the child's birth? What is the probability that at least one will survive? 4. What should be the minimum cost of insuring the Ufe of a person 20 years old for $1000 for five years? For insuring a, couple against the death of either or both for the same sum and period if the husband is 30 and the wife 25? (Neglect interest, etc.) 5. A man makes a will leaving $40,000 to his wife in case she survives him. A son is to inherit the money if he survives both parents. If the ages of husband, wife, and son are 60, 50, 25 respectively, and if money is worth 5% compounded semi-annually, what is the present value of the expectation (a) that the wife will inherit the money in 10 years? (b) that the son will inherit the money in 15 years? 6. In each of the following exercises plot a graph with the probabilities as ordinates at arbitrary equal intervals along the x-axis. (a) A coin is tossed six times. Find the probabilities of the various ways in which it can turn up heads. (b) Find the probabilities for the various ways in which two dice can fall in one throw. (c) In the long run A wins 3 games out of 4 from B at chess. Find the probabilities of the various numbers of games which A might win in 8 successive games. (d) If a die is tossed six times, find the probabilities of the various ways in which an ace can turn up. (e) If the quantity of a trait in an individual of a group is the result of a chance combination of seven causes, determine the probabilities of the ways in which the trait may occur. 382 ELEMENTARY FUNCTIONS Number of Number of seeds apples 4 9 5 4 6 14 7 21 8 24 9 25 10 13 138. Frequency Distributions. At an agricultural experi- mental station 110 apples were classified with respect to the number of seeds each contained, and the number of apples in each class was determined. The results are given in the table. Thus 9 apples had 4 seeds each (the minimum number found in this investigation), 13 apples had 10 seeds each (the maximum number), while 25 apples had 9 seeds each (the most frequent number occurring). Such an arrangement of the indi- viduals of a group, classified with re- spect to some characteristic which gives the number of individuals in each of the classes is called a frequency distribution and the table in which the classes and frequencies are given a frequency table. A graphical representation of this analysis, called a fre- quency polygon, is obtained by plotting the magnitudes of the classes as abscissas, the frequencies as ordinates, and connecting the points by straight lines as in the figure. The magnitude of each class in this case is an integer and the class in- i 2 s 4 5 e ? 8 9 lo n tervals are said to vary discretely. In case the characteristic measured varies continuously, as for instance the stature of a group of men, the size of the class intervals and their mid-points are chosen arbitrarily. The magnitude measured may vary discretely but by such small amounts that the number of classes is so great that the variation of the group with respect to the characteristic cannot be easily determined. In such a case the class interval is en- larged by grouping the frequencies in two or more adjacent ,„20 -s |10 e / \ / \ / \ / \ / \ / \ Number of Seeds Fig. 209. THEORY OF MEASUREMENT 383 classes and associating the resulting frequency with the mid- value of the resulting larger class. Example. The grades in geometry of 30 students were 30, 42, 48, 55, 60, 64, 68, 71, 72, 74, 75, 76, 77, 77, 78, 78, 78, 79, 80, 82, 82, 83, 84, 85, 86, 87, 88, 88, 91, 95. Collect the data in frequency tables with class intervals of 5% and 10%. Table 1 Table 2 Table 3 Grades FreqiLenqj Grades Frequenqj Grades Frequency 28-32 1 30-34 1 30-39 1 33-37 35-39 40-49 2 38-42 1 40-44 1 50-59 1 43-47 45-49 1 60-69 3 48-52 1 50-54 70-79 11 53-57 1 55-59 1 80-89 10 58-62 1 60-64 2 90-100 2 63-67 1 65-69 1 68-72 3 70-74 4 73-77 5 75-79 7 78-82 7 80-84 5 83^87 5 85-89 5 88-92 3 90-94 1 93-100 1 95-100 1 Tables 1 and 2 show the data collected in class intervala^of 5 % with different mid-points. In table 3 the class interval is 10 %. The class interval of 1 % is too small for an adequate presentation of the data. Even class intervals of 5 % leave some classes empty. The class interval should be chosen so as to avoid empty classes. The smaller the number of measurements the larger the class interval should be, and vice versa. The starting points of the intervals are not so material, but it is convenient to take them so that the mid-points of the intervals are integers. In age distributions the returns usually cluster about the multiples of 5, which are taken as the mid-points of the intervals. A second method of representing frequency tables graphi- cally is indicated in the Figs. 210-212 representing the tables above. Rectangles are constructed on the class intervals as bases with altitudes equal to the frequencies. Such diagrams are called histograms. The area of a histogram is the sum of the frequencies times the class interval. 384 ELEMENTARY FUNCTIONS If the number of observations be increased and the class interval decreased the frequency polygon or histogram will 1 — 1 "^ Ll 1 1 1 1 1 1 z: 1 1 30 35 40 45 55 60 65 70 75 6 % Class Interval Fia. 210. 80 85 95 100 30 35 40 45 50 55 60 65 70 75 6^ Class Interval Fia. 211. 95 100 10 f 1 3U 40 50 60 70 10% Class Interval Fig. 212. 80 90 100 approach more and more closely to a smooth curve. Such a curve is called a frequency curve. A frequency histogram is sometimes smoothed by drawing a curve through the mid-point of the upper base of each rec- tangle in such a way that the area under the curve is the same as the area of the histogram. A frequency table is smoothed by replacing the middle frequency of three adjacent frequencies by the average of the three frequencies. The two end values are counted twice and averaged with the adjoining value. THEORY OF MEASUREMENT Thusii A, B,C,D,E, . the smoothed values are are the values of the frequencies 2A-\-B A+B + C B + C + D etc Magrnitudes Fig. 213. If the polygons or histograms corresponding to successive smoothings of the table are plotted we can approximate closely to the frequency curve which best represents the data. The types of frequency distribu- tions which are most common are: (a) The symmetrical distribution in which the frequencies decrease to zero symmetrically on either side of a central magnitude. It is found in the distribution of errors in chemical and physical measurements and in biological measurements, par- ticularly the measurements of anthro- pology. (b) The moderately asymmetrical distribution in which the frequencies decrease more rapidly on one side of the value of maximum frequency than on the other. ^'°-2^*- It is the most I conamon of all the distributions occurring s in all forms of statistics. (c) The J-shaped _ distribution in which the frequency con- stantly increases or decreases. It is found in economic statistics and is characteristic of the distribution of wealth. (d) The U-shaped distribution in which the frequency decreases to a minimum value and then increases. It is rare. It occurs in meteorological statistics and in sta- tistics pertaining to heredity. Magnitudes Magnitades Fig. 215. Magnitudes Fig. 216. 386 ELEMENTARY FUNCTIONS EXERCISES 1. The following grades were obtained in a spelling test given to third- grade pupils; 65, 85, 55, 60, 100, 23, 92, 74, 73, 75, 76, 94, 82, 17, 10. 63, 97, 77, 96, 75, 90, 85, 90, 75, 86, 82, 94, 100, 90, 100, 74, 100, 100. Construct the frequency polygon, giving frequencies for class intervals of 5% arranged along the a;-axis with multiples of 5 (a) at the first points of the intervals, (b) at the middle points of the intervals. Construct the smoothed curves. Were the words used a good test of the ability of the pupils in spelling? 2. Construct the histogram for the data in Exercise 1, with class in- tervals of 10%, (a) with mid-points at 55, 65, etc.; (b) with mid-points at 60, 70, etc. Calculate the smoothed values in the first table, and draw a curve through the ordinates resulting. 3. The number of seeds per apple in normal and aphis-injured Rome apples, as determined by an investigation, are given in the table. Number of seeds per apple.. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12 Frequencies of normal apples 0, 0, 0, 4, 31, 30, 38, 63, 64, 37, 31, 1, 1 Frequencies of aphis-injured apples 5, 14, 20, 29, 44, 37, 28, 44, 38, 25, 13, 3, Plot the histograms based on these samples. What is the most probable number of seeds in the normal and in the aphis-injured apple? State two dififerences between the two variations. 4. The following table gives the degrees of cloudiness of the sky observed at Breslau during the years 1876-1885, on a scale of 10. Degrees I 0, 1, 2, 3, 4, 5, 6 , 7, 8, 9, 10 Number of days, I 751, 179, 107, 69, 46, 9, 21, 71, 194, 117, 2089 Plot the histogram and the smoothed curve. What is the type of the distribution? Would the arithmetic average of the measurements repre- sent the variation well? What is the probability that a day at Breslau will be clear? Very cloudy? 5. Individuals A and B were tested by hearing a series of 10 letters read at the rate of 1 per second and being required to write as many as they could remember in the proper order of the letters, as soon as the reading was finished. Their scores were: Letters correct 4, 5, 6, 7, 8, 9, 10, A frequencies 1, 4, 5, 8, 11, 4, 3, B frequencies 0, 2, 4, 10, 16, 3, 1, Draw the histograms and the smoothed curves. Which is the better performance? What would be the probability that A and B could each get 8 correct in a particular trial? 6. The following table gives the age distributions for deaths from typhoid, measles, scarlet fever, influenza, tuberculosb, chrculatory and respiratory diseases in New York state in 1916. THEORY OF MEASUREMENT 387 Age Typhoid Measles Scarlet fever Influenza Tubercu- losis Circula- tory Respira- tory diseases diseases 0-1 2 293 9 143 371 165 4161 1-2 3 404 26 58 301 37 1726 2-3 3 136 31 39 194 35 638 3-4 4 64 31 11 129 32 274 ^5 5 25 30 9 78 39 149 5-9 22 54 52 33 285 285 327 10-19 112 17 14 43 1267 603 299 20-29 170 8 9 62 3634 788 837 30-39 127 4 5 83 3754 1252 1379 40-49 76 12 123 2936 2222 1761 50-59 56 3 193 1748 3748 2084 60-69 23 291 889 5564 2257 70-79 8 448 302 6464 2137 80-89 365 45 3536 1203 90-99 1 73 2 497 212 100 3 11 10 Unknown Age ... ... 2 4 25,282 3 Total 611 1,021 208 1,977 15,937 19,457 Plot the curves given by these tables, identify the types of curves, and state characteristics of the diseases. 7. The errors of observation in 471 astronomical measurements made by Bradley were distributed as in the table, in which the mid-values of the magnitudes of the errors are given in decimals of a second of arc. Mid-values of errors I 0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5 , 10.5 Frequencies 1 94. 88, 79, 58, 51, 36, 26, 14, 10, 7, * 8 Assuming that positive and negative errors occur with equal frequency, plot the histogram and the frequency curve. Discuss their forms. 8. Discuss the form of the frequency curve obtained by plotting the probabilities given by the terms of the expansion (p + g)'*, for p = h 9 = l> ^ = 6, at equal intervals along the x-axis and drawing a smooth curve. What is the most probable number of times the event will occur in 6 trials? If the distances along the x-axis are multiples of unity, what ■will the area under the curve from the maximum ordinate to the right hand extent of the curve represent? 388 ELEMENTARY FUNCTIONS 9. Plot the graph of the frequency curve y = e~*^. Determine approxi- mately by counting squares the abcissa of the point whose ordinate bisects the area of the right branch of the curve. 10. Plot the graph of the frequency curve 2/ = lo(l+^} M--j from X = - 3 to X = 4. What is the type? 11. Plot the graph of the frequency curve y = 10x°-^ e'^-'^ from x = to X = 10, and discuss the type. 139. Averages. The frequency gives the variation of the magnitude measured but it is convenient in most cases to have a single value to represent the table. The representative values or averages which are commonly used for various kinds of frequency distribution are (a) the arithmetic mean, (b) the median, (c) the mode, (d) the geometric mean, (e) the harmonic mean. The arithmetic mean. The word mean or average alone is generally understood to denote the arithmetic mean. It is defined by the equation . _ sum of the measurements number of measurements If there are n measurements mi, m^j . . . m„, then _ m i-\-m2+ . . . + mn _ 2m n ~ n * where the symbol Sm is used to denote the sum of the m's. If a measurement mi is obtained /i times, then the sum of these /i measurements is /iWi. If the measurements Wi, m2, . , ., mn occur with the frequencies /i, /2, . . ., /», re- spectively, then the sum of all the measurements is /iWi -1- f^m^ + . . . + fnrrin = 2/m, and the total nimiber of measurements is THEORY OF MEASUREMENT 389 Hence the arithmetic mean of the measurements is This equation defines the weighted arithmetic mean of the numbers mi, m^, . . ., mn with the weights fi, ji, . . ., /„. It is used whenever the numbers to be averaged are not of equal importance. Example. If the averages of the weekly wages of three factories are 14, 17, 18 dollars and the number of employees are 450, 360, 670, re- spectively, then the average weekly wage for all three factories is ^ 450 X 14 + 360 X 17 + 670 x 18 ^ 450 + 360 + 670 The calculation of the arithmetic mean of a frequency dis- tribution is simplified by considering the measurements of a class as concentrated at the center of the class interval, as- suming the class interval as a unit and applying an extension of the rule on page 83, given by the Theorem. If mi, m2, . . ., m„ are measurements with fre- quencies fi, /2, . . ., fn, if A is their true mean, E an estimated value of the mean, and di, d^, . . ., dn deviations of the measure- ments from E, then We have mi = E + di, so that firrii = fiE + fidi, and similarly /2m2 = f2E + f2d2, etc. Adding Ml +/2W2 + . . . +fnmn = E(fi +/2 + . . . +/n) H-(M+/2^2+ . . . +fndn). Using the symbol 2 to represent the summations. Xfm = E^f+Xfd. Dividing by 2/, A =^ = E + ^. 390 ELEMENTARY FUNCTIONS Corollary. The sum of the deviation from the arithmetic mean is zero. ^fd . For if E = A, then the correction c l^d is zero. Hence S/i, the sum of the deviation from A = E, is zero. The computation of the correction c =^ is illustrated in I Example 1. Find the arithmetic mean of the incomes of the group of famihes in the first two columns of the table. Number of Deviation from Incomes families, mean estimated, 'd (class intervals) / d $400-$499 8 -4 - 32 500-599 17 -3 - 51 600-699 72 -2 -144 700-799 79 -1 - 79 800-899 73 -306 900-999 63 + 1 63 1000-1099 31 + 2 62 1100-1199 18 + 3 54 1200-1299 8 + 4 32 1300-1399 8 + 5 40 1400-1499 1 + 6 6 1500-1599 6 + 7 42 + 299 2/ = 384 2/d = - 7 Estimated average, E = 850. = - 0.0182 class intervals. 384 .-.A = 850 -^-100 = 850 -1.8 = $848.20 In the computation it is assumed that the income of the families in each class is the income at the middle of the class THEORY OF MEASUREMENT 391 interval. With this assumption an income of $450 is re- ceived by 8 famihes, of $550 by 17 famihes, etc. Estimating the mean as $850, the deviations of the incomes of the successive classes from $850 are - 400, - 300, etc. To simplify the computation we take the class interval $100 as a unit, so that the deviations d become -4,-3, etc. To find the correction, c, which must be appUed to the estimated mean, multiply the pairs of values of / and d, ob- taining the column headed fd, find the sum of the numbers in the columns headed / and fd, and divide the latter by the former. As $100 was taken as a unit, this quotient must be multiplied by 100. The correction to be applied to the esti- mated mean of $850 is thus found to be - $1.8. The median. If all the measurements of a series are ar- ranged in order of magnitude, the magnitude of the measure- ment halfway up the series is called the median. The magni- tudes one-quarter and three-quarters up the series are called the first and third quartiles. The number of the measurement whose magnitude is the 71 + 1 median is — —, where n is the number of measurements. The numbers of the measurements whose magnitudes are the quartiles are |(n + 1) and f (n + 1). Graphically, the first quartile is the abscissa of the point whose ordinate cuts off one-fourth of the area of the frequency curve of a distribution to the left of it. The median has one- half the area on either side, the third quartile has one-fourth the area to the right of it. In a frequency table where the items are given by classes it is assumed for the computation of the median that the measure- ments are distributed uniformly through the class interval, so that after the class in which the median lies has been de- termined, its position in that class interval is found by ordinary interpolation. Example 2. Find the median and quartiles of the incomes of the group of families given in Example 1. 392 ELEMENTARY FUNCTIONS The table is conveniently arranged in the form given. Dividing by two the number of terms plus Incomes less than Number of families / $500 8 600 25 700 97 800 176 900 249 1000 312 1100 343 1200 361 1300 369 1400 377 1500 378 1600 384 one, we have 385 192.5, so that the median is the average of the 192nd and 193rd terms. As appears in the table, there are 176 measure- ments in the first four classes and 249 in the first five classes. Hence the median lies in the 5th or 800-899 class, 192.5 - 176 = 16.5 measurements from the lower end. Hence, the median is 16.5 Af.-800 + 73 100 = 822.6. This result means that half of the 384 families have incomes less than $822.6 and half have larger incomes. The quartiles are lound as follows: Since I (385) = 96.25, the first quartile, Qi, lies between the 96th and 97th terms, in the third or 600-699 class, 96i - 25 = 71 J measurements from the low^r end. •. Qi 600 + ^.100 698.9. And since f (385) = 288.75, Qs is in the 900-999 class, 288f - 249 - 391 measurements from the lower end. .-. C)3 = 900 + ^-100 963.1. The mode. The mode is the magnitude of the most fre- quent item of a distribution. It is the abscissa of the maxi- mum ordinate in a smoothed distribution. The average in- come, the average man of the newspapers usually means the modal income, the modal man. In Example 1, assuming that the incomes are all at ihe mid- points of the class intervals, the modal income would be $750. Without this assumption, the modal income would be some- where between $700 and $799. The mode or modes of a frequency distribution may not be well defined for the class intervals given in the table. In such cases the mode is roughly located by grouping in larger class intervals as shown in THEORY OF MEASUREMENT 393 Example 3. The distribution of grades in a class in mathematics are given in the following table. Locate the mode roughly. Frequency 11 1 oL \ or 1 oj 1 ■ 1 2 4 oJ ^9 11 1 1 5 1 1 10 >10 The table gives the frequencies for class intervals of 1 %. There is no well-defined maximum fre- quency, and hence no value of the mode is apparent. The numbers by the left-hand column of brackets give the fre- quencies for the class intervals 95-90, 89-85, etc., but the class containing the mode is not yet apparent so that a class interval larger than 5% must be chosen. The numbers by the middle column of brackets give the fre- quencies for class intervals from 99-90, 89-80, etc. The maximum frequency, 10, shows that the mode lies in the third class, between 79 and 70. The numbers by the right-hand brackets give another grouping of the frequencies by 10% class in- tervals, from 94-85, 84-75, etc. Here the mode appears in the class interval from 84r-75. As the mode lies between 70% and 79% and also between 75% and 84%, it must therefore lie be- tween 75% and 79%. In a distribution of the symmetrical type the arithmetic mean, the median and the mode fall together. In a moder- 394 ELEMENTARY FUNCTIONS Fig. 217. ately asymmetric group the mode, median and mean lie in the order named with the mean toward the longer branch of the curve, as in the figure. A good approximation to the value of the mode is obtained by the use of the formula Mode = Mean - 3 (Mean - Median). This formula is based on the assumption, which has been ob- served to hold approximately in a large number of cases, that the median lies one third of the distance from the mean toward the mode. The application of this formula shows that the modal income in Examples 1 and 2 is 848.2 - 3(848.2 - 822.6) = 771.4. The best method of determining the mode is to find the equation of the smooth frequency curve that best fits the data, and then find the ab- scissa of the maximum point by equating the derivative to zero. The method is tedious and difficult. A graphic method of determining the median and the mode Example 4. Find graphically the median, the quartiles, and the mode of the distribution in Example 1. Plot the data in the table in Example 2, using 500, 600, etc., as abscissas, the ordinates being respectively the number of families with incomes less than 500, 600, etc. For convenience in plotting $100 is taken as a unit so that the abscissas are 5, 6, etc. _.jw 1 -S^- ± =«- y'" ^ i - '^^ A- J t 7 T .200 ---^ :^:::::::5t: ::: :: :: :: zt t - 7 - - % U- - - ';) 1 I : . it z A ^ M^-t^ ©1 Jlfo Me ^ o\ 6 7 8 1 9 ,5; ) u 1:2 :3 11 1& X 1 1 L I 1 1' -L Fia. 218. THEORY OF MEASUREMENT 395 Let M and N be the projections on the y-axis of A and J5, the end points of the curve. Divide MN into four equal parts by the points D', E\ F'. Let the perpendiculars to the 2/-axis at these points cut the curve at D, E, F, respectively. Then the abscissa of D is the median, Me = 820, approximately, and the abscissas of E and F are the quartiles, Qi = 700 and Qa = 960. The mode is represented graphically by the abscissa of the point of inflection I. This point may be determined roughly by inspection, or by placing a ruler tangent to the curve and rolling it along the curve so that it remains tangent. The direction of rotation of the ruler changes when the point of contact coincides with the point of inflection. In this figure the mode. Mo, appears to be about 800. The geometric mean is defined by the equation, G = Vmim2 . . . Wn. . It is most easily calculated by using logarithms, since J g ^ log mi + log m2 + . . . + log m„ The geometric mean is used in averaging rates of increase such as arise in the study of the growth of population, growth of skill in an individual, and relative changes in the prices of commodities. The trend of prices in a series of years is gauged by finding the ratio of the average price of a commodity in any year to that in a particular year which is chosen as a base and given the arbitrary value 100. Numbers which are determined for the purpose of showing the trend in prices are called index numbers. If the index numbers of three commodities for one year are a, h, c, and for a second year are ar, 6s, d, then the ratios for the three commodities are r, s, t, and the geometric mean of the ratios is \^rst. The^eometric mean of the inde x num bers for the first year is \^ahc and for the second is '\^ahcrst. The ratio of these two numbers is VrsL Hence the geometric mean of the ratios of the index numbers of several commodities for two years is equal to the ratio of the 396 ELEMENTARY FUNCTIONS geometric mean of the index numbers for the second year to that of the first. This property of the geometric mean is the reason for its use in averaging index numbers. The harmonic mean is defined by the equation 1 iJ = n\mi m2 * * * mj It is used in finding the average amount of work performed in a given time, and the average amount of a commodity -pur- chased for a given price. In the equation v = -, if s is constant and t varies, then the average time for a given distance would be found by the har- monic mean. For since vi='^,V2 = ^, . . . Vn = r> h I2 In the average rate = ^ +^^ + ---+^ -- = ^(1 + 1 + . . . + LV ^ n n\ti ^ ^2 tj Hence the average time average rate n\ti^t2^ ^tj The mode determined as in Example 3 can be found roughly more easily than the median or mean can be calculated. The calculation of the precise true mode is more difficult than that of any of the averages. In the case of discrete variation, as for instance the number of seeds in an apple, the mode may be the only average that will mean anything, as the values of the other averages are quite likely not to occur in the series. The mode is the most probable value of a distribution of the asymmetrical type, since it is the value that occurs most fre- quently in the distribution, and hence it is the best repre- sentative value of central tendency. It is the typical case. It is not useful if it is desirable to give weight to extreme variation, since all the items of the group do not enter into its determination. THEORY OF MEASUREMENT 397 The median ranks after tho mode in ease of determination, and usually may be located more precisely. If the unit of measure is difficult or impossible to determine and a ranking of the items is all that can be attained, as in the measurement of scholarship, the median is the best repre- sentative value. If the values of some of the items are given vaguely, as for instance if items of an upper class are given as greater than some value, the median can be determined more precisely than the mean. All the items enter into the determination of the median but extreme cases affect its value but slightly. Changes in the values of extreme cases would affect the value of the mean without disturbing the value of the median. The arithmetic mean is the average most generally employed. It is the most familiar of the averages, is the most precise when all the items are given, gives weight to extreme deviations, which is desirable in certain cases, and is affected by every item in the distribution. It may be determined if the ag- gregate and the number of items are known though knowledge of the values of the items is lacking, and conversely the ag- gregate may be determined if the mean and the number of items is known. This property is not possessed by the other averages. The harmonic mean and geometric mean are used less fre- quently than the other averages as they are unfamiliar and more difficult to calculate. The arithmetic mean is sometimes incorrectly employed in place of the harmonic mean in averag- ing time rates, and sometimes incorrectly in place of the geo- metric mean in averaging rates of increase. If variations are measured by their ratio to, rather than by their difference from, the average, then the geometric mean is the best average to employ. EXERCISES 1. Which average is meant in the following: average student, average wage in an industry, average daily temperature, average stature, average number of potatoes of a given species in a hill, average annual rainfall, average gain per year in height of a child, average ability in arithmetic. 396 ELEMENTARY FUNCTIONS average rate of increase of population^ average time in doing a piece of work? 2. Find the mean, median, and mode of the distribution in the example in Section 138. 3. Ten men in a department can complete a piece of work in the fol- lowing number of minutes respectively, 45, 50, 60, 60, 60, 65, 65, 70, 75, and 85. What is the average time for the work? 4. The population of a city increased in a decade from 185,000 to 260,000. What was the average annual rate of increase? 5. By the probable duration of life of a man m years of age is meant the number of years which he has an even chance of adding to his life. By the expectancy of life for a man of m years is meant the arithmetic mean of the number of additional years of life enjoyed by all men m years of age. Find the probable duration and the expectancy of life of a man 20 years old (use the table in Section 137). 6. The following table gives the distribution of wages per week of a group of laborers. Wages (mid-values) I 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23 Frequency I 5, 23, 50, 80, 105, 130, 160, 165, 148, 120, 36, 3 Find the mean, median and mode of the distribution. Which average represents the table best? 7. What is the average age at death for each disease in the tables in Exercise 6, page 387. What average should be used? Determine by in- spection of the tables which are diseases of children and which of adults. Which warrants being called a vacation disease? Why? 8. The frequencies of distribution of budgets of a group of college students with a class interval of $50 starting at $350 and running to $1800 inclusive were 4, 15, 21, 26, 39, 46, 52, 32, 34, 24, 17, 17, 14, 11, 8, 10, 6, 6, 4, 4, 3, 3, 1, 1, 0, 2, 1, 0, 0, 3. Calculate the median and quartiles al- gebraically and graphically. Which average would best represent the distribution? 9. In 1913 the index numbers for steak, bacon, chickens, eggs, butter, milk, flour, potatoes, sugar were 94, 95, 97, 91, 108, 100, 100, 90, 100. In 1918 the numbers in the same commodities were respectively 131, 179, 170, 177, 151, 151, 200, 188, 193. Calculate the average index numbers in each year and the relative increase in prices. 140. Measures of Variability. Next in importance to selecting an average to represent a group of measurements is the determination of a measure of the extent to w^hich the other values cluster about or are dispersed from the average chosen, that is, a measure of the variability of the group with respect to the average. THEORY OF MEASUREMENT 399 The average obviates the necessity of stating all the measure- ments from which it is derived, and a measure of variability- is a single number characterizing the deviation from the aver- age. A series of measurements can thus be summarized by two numbers which are usually written in the form a =*= rf, where the first number gives the average value chosen and the second gives a measure of the deviations of the items of the series from the average. The principal measures of dispersion or variability of a distribution from an average are (a) the quartile deviation, (b) the mean deviation, (c) the median deviation or probable error, (d) the standard deviation. The quartile deviation is defined by the equation Q = --^-^ — -* where Qi and Q3 are the first and third quartiles respectively. It is the simplest measure of deviation to calculate. The quartile deviation for the distribution of incomes in Example 2, Section 139, is ^^ 963.1-698.9 ^^3^^ Hence the group is summarized by the median value of the group and the quartile deviation in the form 822.6 =*= 132.1. This means that approximately 50 % of the incomes lie between $822.6 - $132.1 = $690.5 and $822.6 + $132.1 = $954.7. The mean deviation is the arithmetic mean of the numerical values of the deviations from an average. The method of calculation for a frequency distribution is shown in Example 1. Find the mean deviation from the arithmetic mean of the distribution of incomes in Example 1, Section 139. In this example the estimated mean is E = 850, the numerical value of the correction is c = 1.82 units = 0.0182 class intervals and the true mean is A = 848.2. The sum of the numerical values of the deviations from E is 306 + 299 «= 605 class intervals. The deviation from A of each of the 176 items less than A is 0.0182 class intervals less than the deviation from E. The deviation from A of each of the 73 + 135 = 208 items greater than 400 ELEMENTARY FUNCTIONS A is 0.0182 class intervals greater than the deviation from E. Hence the sum of the numerical values of the deviations from A is 605 + 208 X .0182 - 176 x .0182 = 606. Dividing by n = 384, we have The mean deviation from A = ^^ = 1.6 class intervals =160 units. The median deviation or probable error. If all the devia- tions from some one of the averages are arranged in order of magnitude without regard to sign, the median deviation is calculated in the same way as the median of the distribution. The middle 50 % of the items come within the range of the median deviation if the median is the average used to repre- sent central tendency. Approximately 50 % of the items come within the range of the quartile deviation, but these cases are not necessarily the middle 50 % since the median does not lie exactly halfway between the quartiles except in a symmetrical distribution. The median deviation is usually called the probable error and will be discussed further in Section 142. The standard deviation is defined by the equation di' + ^2^ + . . . dr? 2(i2 or 2 _ n where di, d2, . . ., dn are the deviations of the measures from the average chosen. If the deviations occur with the frequencies /i, /2, . . ., fn, then the standard deviation is This measure of dispersion gives more weight to extreme cases than the other measures of variability, and while more tedious to calculate is more generally used. In all deviation measures retain at most two significant- figures. The reason for this rule is as follows. Consider a length I = 324.57 cm. with a mean deviation of .14 cm. The mean deviation indicates that the figure 5 in Z in the first position after the decimal point is uncertain by 1 unit and that THEORY OF MEASUREMENT 401 the next figure 7 is uncertain by 14 units. The next figure would be uncertain by at least 140 units and is discarded. If the first significant figure of the deviation is 8 or 9, the figure in the corresponding position of the average and the figure following are retained but only one significant figure of the measure of deviation is retained. The calculation of the standard deviation is simplified by means of the Theorem. If xi, X2, . . ., Xn are deviations from the arith- metic mean A of a series of measurements with frequencies /i, /2, . . ., /«, if di, c?2, . . -, dn — I 1 ^-* are the deviations from an estimated mean E, c *- ^ and if A = E + c, then *^ ^ Fig. 219. = o"! = (T> ^vj ^ ^ ' :^ ^ rJ > 1 1 " vj ^ ^ 1 1 =5i , b. ^ " 1 1 _ -20- -cr .lxm = -^xe 2(rW ^ ) + e 2 "^ •^ y \ ^ 1 _ \ / 1 t Y 27r 1 \ V y 1 'V s ^ 1 ■) ■\ 1 V *s - — 1 Til so that Zrfx^ = naj^. Hence the slope of li is mi = ^, where o-^^ = • (1) Similarly, the slope of h referred to the y-Sixis is m. = ^^ where