:ti(m TO THE MENTARY JNCTIONS fcGLENON AND RUi c.«^*^ MI6 Gfarnell Untocrattg Slibrarg BOUGHT WITH THE INCOME ©F THE SAGE ENDOWMENT FUND THE GIFT OF HENRY W. &AGE 1891 ^ date shows when this volume was taken. To rmew this book copy the call No. and give to the hbrarian. '-■MAY-2e«0.. HOME USE RULES All Books subject to Recall All borrowers must regis- ter in the library to borrow books for home use. All books ^ust be re- turned at end of college year for inspection and repairs. ' Limited books must be re- turned within the four week limit and not renewed. Students must return all boc^s before leaving town. Officers should arrange for the return of books wanted during their absence from town. Volumes of periodicals and of pamphlets are held in the library as much as possible. For special pur- poses they are given out for a limited time. Borrowers should not use their library privileges for the benefit of other persons. Books of special value and gift books, when the giver wishes it, are not allowed to circulate. Readers are asked to re- port all cases of boc^ marked oc mutilated. ^ De sot deface books by marics and writinc. Cornell University Library QA 331.M16 Introduction to the elementary functions 3 1924 003 988 148 u ,f INTRODUCTION TO THE ELEMENTARY FUNCTIONS BY RAYMOND BENEDICT McCLENON WITH THE EDITORIAL COOPERATION OF WILLIAM JAMES EUSK GINN AND COMPANY BOSTON • NEW YORK • CHICAGO • LONDON ATLANTA • DALLAS • COLUMBUS • SAN FRANCISCO COPYRIGHT, 1918, BY GINN AND COMPANY ALL HIGHTS RESERVED 118.8 Vjt gtjitnguiii grend GINN AND COMPANY • PRO- PRIETORS ■ BOSTON • U.S.A. PREFACE This book is an attempt to solve the problem of the first-year collegiate course in mathematics. That the problem is a very real one is attested by the many discussions constantly taking place among teachers of mathematics and others interested in education. The traditional Freshman course, consisting of " college algebra," trigonometry, and solid geometry or elementary analytic geometry, is very generally regarded as unsatisfactory. There are three main objections to this traditional course : first, it is not unified, so that it sacrifices time and fails to hold the student's interest; secondly, much of the subject matter should come after a first course in calculus, when it would gain vastly in significance; thirdly, the usual plan has deprived the large majority of college students of any introduction to the calculus, which is the heart and soul of modern mathematics and natural science. Only that small number electing to go beyond the first year of collegiate mathematics have the opportunity to become acquainted with the subject, which unquestionably represents one of the most important lines of development of human thought during the past two centuries. Accordingly, we decided to construct a course with the funda- mental idea of functionality as its unifying principle, and leading up to some elementary work in calculus as its culmination. The advantages of this arrangement are that it not only meets the objections stated in the preceding paragraph, but saves time by avoiding the repetition inevitable in the triple arrangement of subjects ; and, what is more important, it leads to a deeper understanding of the significance of mathematical principles and relations than the student is likely to gain through the tra- ditional course. Thus, whether he goes farther into the study of iv THE ELEMENTARY FUNCTIONS mathematics or has but one year for the study, we believe that he will be the gainer by taking the unified course. The course presented in this book is the result of our expe- rience in Grinnell College, mimeographed copies having been used and revised during five successive years. The material included comprises the simpler and more important parts of plane trigo- nometry and analytic geometry, followed by an introduction to the differential calculus, including differentiation of the simpler algebraic functions and applications to problems of rates and maxima and minima. The conic sections are not studied as extensively as in most textbooks of analytic geometry, but enough has been given to make the student feel familiar with these important curves. The trigonometric functions are introduced early, and the gen- eral definitions are given at once, instead of those valid only for the acute angle. Thus the connection with the coordinate system is established from the first, and a clearer idea of the meaning of the trigonometric functions is obtained than if the student's atten- tion is for some time confined to the case of the acute angle. The applications of the trigonometric functions to the solution of right triangles and problems depending upon them is made without the use of logarithms, as experience shows that the early introduction of logarithms may easily lead to mechanical methods of work. The number of numerical exercises is not large, as the teacher can easily supplement those in the text by as many others as he wishes. "We feel that the purely computational work can easily be overdone in the first-year course. Four-place tables may be used in this work ; the Wentworth-Smith tables, or others of like nature, are very satisfactory. The arrangement is almost exclusively inductive, and the style direct and informal throughout. The explanations are not always as detailed as in many texts, the object being to lead the stu- dent to supply the connecting links for himseK, where they are not explicitly given in the text. Moreover, many of the important results are altogether left to the student as exercises. In such cases, as in the case of all the most important formulas PEEFACE V throughout the book, attention is called to their importance by the use of black type. The course will be found suitable for an advanced course in the secondary school, as well as for the first year in college, and in this case it might very well be made a five-hour course. For the student who has somewhat lost touch with his previous work in mathematics, a small amount of review matter has been placed in appendixes. If this work has to be taken, it will probably be unwise to attempt to cover all of the text, and certain paragraphs have accordingly been starred, to indicate that they may be omitted without interfering with the unity of the course. We have not gone so far in the way of radical changes in subject matter as our personal feelings would lead us, because we believe that progress in the teaching of mathematics, as in everything else, should be in the nature of evolution rather than revolution. For instance, no work in the integral calculus is given, although we firmly believe that this topic should eventually be included in the first-year course; but it seemed to require a greater departure from the traditional course than is as yet practicable. We hope that the present book may prove a contri- bution to the solution of the problem presented by the first year of college mathematics, and that experience will indicate what further steps may advantageously be taken. R. B. M. W. J. R. »1 Cornell University Library The original of tiiis bool< is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924003988148 CONTENTS PAGE CHAPTER I. THE GRAPHICAL REPRESENTATION 1 Measurement. Construction of segments of given length. Directed segments. Correspondence between numbers and points on a straight line. The coordinate system. Application to some problems of ele- mentary geometry. Point that divides a segment in a given ratio. CHAPTER II. FUNCTIONS AND THEIR GRAPHS 19 Variables and the equation of a locus. Graphical representation of functions Importance of functional relation. Graphical representation of statistical data. CHAPTER III. APPLICATION OF GRAPHICAL REPRESENTA- TION TO ELEMENTARY ALGEBRA 28 Graphs of linear equations. Graphical solution of simultaneous linear equations. Determinants ; applications to solution of simultaneous linear equations with two or three unknowns. The quadratic func- tion. Graphical and algebraic solutions of quadratic equation. Alge- braic solution by completing the square, by formula, and by factoring. Test for solvability of quadratic equation. Graphical interpretation of discriminant test. Construction of tangent to a parabola. Sum and product of the roots of a quadratic equation. Factor theorem for quadratic equation. Maximum and minimum values of a quadratic function. CHAPTER IV. INTRODUCTION TO THE TRIGONOMETRIC FUNCTIONS 61 Definition and measurement of angles. Definition of the trigonomet- ric functions. Simple relations among the trigonometric functions. Applications of the trigonometric functions to the solution of right triangles, to problems in velocities and forces, and to the slope of a straight line. CHAPTER V. SOME SIMPLE FRACTIONAL AND IRRATIONAL FUNCTIONS. THE LOCUS PROBLEM ... 85 Graphs of rational functions ; asymptotes. Graphs of irrational func- tions. Definition of locus. Simple examples of locus problems. vii viii THE ELEMENTAEY FUNCTIONS PAGE CHAPTER VI. THE STRAIGHT LINE AND THE CIRCLE ... 97 Equation of straight line through a given point and having a given slope. Proof that the equation of every straight line is of the first degree in x and y, and conversely. Normal form of equation of straight line. Distance from a line to a point. General equation of the circle. CHAPTER VII. THE PARABOLA, ELLIPSE, AND HYPERBOLA 114 Equation of parabola. Construction of parabola. Definition and equation of ellipse. Definition and equation of hyperbola. Equation of its asymptotes. CHAPTER VIII. SIMULTANEOUS EQUATIONS 141 Intersection of straight line and conic. Tangent to a conic. Alge- braic solution of certain types of simultaneous quadratic equations. CHAPTER IX. FURTHER STUDY OF THE TRIGONOMETRIC FUNCTIONS. POLAR COORDINATES 154 The Law of Sines. The Law of the Projections. The Law of Co- sines. Solution of the oblique triangle. The half-angle formulas. Functions of the sum and difference of two angles. Sum and differ- ence of two sines or cosines. Graphical representation of the trigo- nometric functions. Polar coordinates. CHAPTER X. THE EXPONENTIAL AND LOGARITHMIC FUNC- TIONS 188 Laws of operation with exponents. Definition of logarithms. Laws of logarithms. Use of logarithms in computation. Derivation of the half-angle formulas for the triangle. MoUweide's Formulas and the Law of Tangents. CHAPTER XI. INTRODUCTION TO THE DIFFERENTIAL CAL- CULUS 203 Average rate of change, and instantaneous rate of change, of a func- tion. Limits. Theorems on limits. The derivative. Its use in find- ing the slope of the tangent to a curve. Rules for finding derivatives. Application of the derivative to drawing graphs, and to maximum and minimum values. Differentiation of irrational and implicit func- tions. Various applications. CONTENTS ix PAGE APPENDIX A. PROOF THAT THE DIAGONAL OF A SQUARE IS INCOMMENSURABLE WITH ITS SIDE 233 APPENDIX B. LAWS OF OPERATION WITH RADICALS ... 235 APPENDIX C. to CONSTRUCT A SEGMENT HAVING A RATIONAL LENGTH 236 APPENDIX D. TO CONSTRUCT THE SQUARE ROOT OF ANY GIVEN SEGMENT 237 APPENDIX E. SIMULTANEOUS LINEAR EQUATIONS IN TWO UNKNOWN QUANTITIES 238 APPENDIX F. THE QUADRATIC EQUATION IN ONE UNKNOWN QUANTITY 240 INDEX 243 THE ELEMENTARY FUNCTIONS CHAPTER I THE GRAPHICAL REPRESENTATION 1. Measurement. The study of algebra is concerned witli nuTn- bers ; that of geometry, with space. The object of this chapter is to bring into closer relation these two portions of elementary mathematics, in order that each may be used to help the other. The simplest process in which number and space are combined is measurement. 2. The important r61e which measuring plays in the work of the world is so apparent to everyone that it needs no mention. The nature of this important process, as applied to distances, is made clear ' ' ' ' ' ' by the following considerations : If c D we wish to measure a line segment (that is, a limited portion of a straight line) AB, we begin by choosing any convenient line segment CD, which we call a unit ; we then apply CD to AB so that C coin- cides with A, when D will fall upon H, a point between A and B, in case AB is > CD. Thus AE=CD. We next apply CD to EB, C coinciding now with E, and D falling at F. Thus EF=CD, and AF= 2 • CD. By continuing this process again and again we may eventually find that D falls upon B, let us say, the mth time we applied CD successively. Then AB = n • CD, and we say the length of AB is n units. We also say the unit is contained exactly in the segment AB. Whenever this happens, the length of the segment is an integer. But in practice this does not usually happen. More often the unit will be contained exactly, let us say n times, in a segment AB', less than AB, whereas the remainder B'B of AB is less than ■1 Fig. 2 2 THE ELEMENTARY EUNCTIONS the unit CD (Fig. 2). We can then assert that the length of AB is greater than n units but less than n + 1. That is, we have measured the segment AB to the nearest integer less than its length. If we wish to get a closer approximation, we may divide the unit CD into any number of parts, ^, ^ as k, and proceed to measure the seg- ' ' ' ' ' ^~^ ment B'B, using one of these parts of ^ ? CD as a new unit. If this part of the unit is exactly contained in B'B, say m times, then B'B = m kt\\s of a unit, and the length of AB equals w + — • This number can be written ; > that is, in k k the form of the quotient of two integers. If it happens that the ^th part of CD is not contained exactly in B'B, m of these parts being less than B' B, while w + 1 of them exceed B'B, we can at any rate assert that the length of AB is yyt fyry I "1 greater than « + — units but less than n -\ ; — : that is, we k k have measured AB to the nearest kt\\ part of the unit less than its length. If we wish a stUl closer approximation, we have only to repeat the same process, taking a smaller fractional part of CD than we did before, that is, taking a larger value of k. How large a value of k we take (that is, how small a fractional part of CD we use) is a question that purely practical considerations answer. If the unit is the foot, often k = 12 would be sufficient ; that is, we should be satisfied with measurement which is carried out to the nearest inch. We might, however, wish to know the length of our segment to the nearest tenth of an inch, in which case we should of course take ^ = 120, the unit being a foot. The careful thinking through of this process will prepare the student to realize the truth of the following fact : The length of any line segment (measured by any unit whatever) is always given, either exactly or approximately, as an integer or as the quotient of two integers (that is, either as an integer or as a fraction). The actual measurement of several concrete, objects should be carried out by the student with special reference to the degree -5 la S Fig. 3 THE GEAPHICAL EEPEESENTATIOK 3 of accuracy attained in cases where the segment measured does not seem to contain exactly the unit or the part of the unit chosen. Evidently a segment might happen to have such a length that some fractional part of the unit would be contained exactly in it, even though repeated trials might not show just what part is so contained. Thus, if a segment were exactly 3^3^ units long, a measurement that gave the length to the nearest third of a unit would seem nearly exact, as 3| is only ^L too small ; and if we used tenths of a unit, we should get 3J^ as the length, — a result that differs from the actual length by only -^^^. For most pur- poses this would be an entirely sufficient degree of accuracy, but we should never find an exact measure of the segment ' unless we divided the unit into precisely 13 parts (or some mul- tiple of 13). Considerations of such examples of measurement as this one (and each student may easUy work out several for himself) may very well suggest the conclusion that any segment could be measured in terms of a given unit and fractional parts of it, if we could only dis- cover the correct fractional part to try. This conclusion, plausible though it seems, is, however, not correct; there exist segments which can never he measured hy a particular unit nor ly any fractional part of it. One example of such a segment is the diagonal of a square whose side is the unit.^ Such segments are said to be incommensurable with the unit. The student will no doubt recall numerous examples. Of course these incom- mensurable segments can be measured to any desired degree of approximation, exactly as is done in the case of such a seg- ment as the one mentioned just above, that was assumed Sj-^^ units long. 3. Summarizing the results thus far obtained, we see that in measuring a line segment, after choosing a unit length, one of three things can happen : (1) the unit is contained exactly in the 1 See Appendix A for a proof that the above statement is true of the diagonal of a square. 4 THE ELEMENTARY FUNCTIONS segment to be measured; in this case the length of the segment is an integer; or (2) the unit itself is not contained exactly in the segment, but some fractional part of the unit is ; in this case the result of the measurement is a fraction, that is, the quotient of two integers ; or, (3), neither the unit nor any fractional part of it is contained exactly in the segment ; in this case the segment is incommensurable, and the length can be only approximately given as an integer or fraction. This length is, however, still spoken of as a number, but is called an irrational number, while the lengths of commensurable segments (that is, either integers or fractions) are called rational numbers. The laws of elementary algebra include rules for working with irrational numbers, and with these rules the student is assumed to be familiar.! 4. Construction of segments of given length. The converse problem to measurement is constriction of segments having a given ratio to the unit segment, that is, having a given length. The constructed segment will be said to correspond to the number given as its length, so that a segment of length 4 units will be said to correspond to the number 4. (a) Rational numbers. Any segment whose length is a rational number can be constructed at once, because elementary geometry provides us with a method of constructing any fractional part of a unit segment. If the student has forgotten how to do this, let him review the method carefully. A brief statement of it is found in Appendix C. (b) Irrational numbers. One would naturally take it for granted that, inasmuch as the segments whose lengths are irrational num- bers are incommensurable, they would also be inconstructible ; and in general this is true. But many such segments can be very easily constructed, namely, all those depending on square roots alone. The Pythagorean Theorem ^ gives us the means of doing 1 See Appendix B for statement of these rules, with exercises. 2 " The square on the hypotenuse of a right triangle is equal to the sum of the squares on the two legs." This very important theorem was discovered by the famous Greek philosopher and mathematician Pythagoras, in the sixth century b.c. THE GEAPHICAL REPRESENTATION 5 this for numbers like ^2, Vd, y/}, Vf , etc. For iostance, V5 is the length of the hypotenuse of a right triangle whose legs are 1 and 2 in length ; Vf is the length of one leg of a right triangle in which the hypotenuse is f and the other leg f ; Vs is the hypotenuse of a right triangle in which the legs are 1 and V2. To construct a segment whose length is the square root of the length of any given segment, see Appendix D, where the method is explained and illustrated. Finally, any segment whose length involves only rational combiuations of square roots, that is, only additions, sub- tractions, multiplications, or divisions of square roots, can be con- structed by combinations of the above-mentioned constructions, — for instance, such lengths as 1-1-^2,^ — ^^, \^(=\/V9) -■ 3_V5 ^ ^'2-V3 EXERCISES Construct accurately segments of the following lengths (using the same unit for all): ^- i- 6 ^~^ . 8- "^^ 11- 3-9- 2- V3 2-ff ' ^ 9.^. 12. 3-V3. '^■2 + V3' V7. 6.1.7. 3^ ^ 15.4^. 10. — ^- 13. 3 4. 1-1- V2. 7. Vs. ' VE' ' 5-V6 5. This is as far as we can go in the construction of seg- ments of given length, with the means of elementary geometry, — namely, straightedge and compass ; for no cube root, fifth root, or other irrational num- ber not reducible to square 4 S. £ f roots, can be so constructed.^ -pia. i But we are none the less instinctively certain that there exists a definite segment corre- sponding to any length, even to these inconstructible lengths. For example, if AB = BC=1, we are convinced that there exists 1 This statement cannot be proved here, as it involves more advanced considerations than the student is yet prepared for. 6 THE ELEMENTAEY FUNCTIONS a point X between B and C such that AX='^; and similarly for other such numbers. We can thus say that to any number, rational or irrational, corresponds a definite length, when once a unit has been chosen. Negative Numbees 6. Directed segments. If we start from any point on a straight Une to lay off a distance, it often makes a great difference in which direction we proceed to take the distance in question. It is accordingly useful to have some way of distinguishing, in such cases, which direction is to be taken. This is done by prefixing to each number used a label, or sign, + if the distance is to be taken in the one direction, and — if in the opposite direction. For instance, segments directed toward the right on a horizontal line may be given the sign + and called positive segments, while those directed toward the left will then be given the sign — > 1 1 and called negative segments. Fig. 5 In Fig. 5, AB, AC, and BC are positive segments, while BA, CB, and CA are negative. The numbers expressing the lengths of such directed segments are then given the same signs as the segments themselves, and are spoken of as positive or negative numbers, as the case may be. Thus, if the measure of the segment AB is 5 units (Fig. 5), and if that of BC is Z units, AB = + 5, while BA = - 5 and CB = - 3. A line, such as ABC in this illustration, on which every segment has a direction as well as a length, is called a directed line. 7. Theorems on directed segments. If A, B, and C are any points on a directed line, then AB = -BA (1) and AC + CB = AB. (2) The first theorem results from the definition of a directed seg- ment. The second is self-evident when C is between A and B, and it is true whatever be the relative positions of the three points. Thus, in Fig. 5, ^C = -f- 8, C5 = - 3, and ^5 = -(- 5 ; and it is true that -|- 8 + (— 3) = + 5. This theorem (2) is of great THE GRAPHICAL EEPEESENTATION 7 importance in our further work, and hence the student should verify it by numerous examples showing the various possible relative positions of the points A, B, and C Indeed, the truth of (2) is obvious geometrically, since to pass from A to C, and then from C to B, will give the same displacement to the right (or left) as the single amount AB. 8. The student must be careful to notice that whenever we speak of negative numbers as the lengths of segments, the word "negative" refers merely to the direction in which the corre- sponding distance is measured ; it is a qualitative, not a quanti- tative, adjective. It wiU be remembered that this is the case with all the illustrations of negative numbers that are given in text- books of elementary algebra ; for example, if we call temperature above zero +, then temperature below zero is called — ; if an amount of capital (asset) is considered +, then a liability is called — ; and so on. In fact, whenever two sets of numbers can be associated with each other in such a way that one set gives one kind of number and the other set the opposite hind, we may call the numbers of the one set + and those of the other set — . Whatever we agree shall be the meaning of the sign +, the sign — means the exact opposite. 9. Correspondence between numbers and points on a line. Let us take any point as starting-point, and (with any con- venient unit) construct on the same straight line the segments 0A= + 1, 0B=+2, 0C= + 3, 0A' = -1,0C'=-S, etc. As B is the end-point of the seg- .. w ■ J v. i-u c' B' A' o A B c , ment determmed by the _j 1 ■ ■ > 1 > > > — length + 2, we may say no. 6 that the number + 2 de- termines the point B, and, similarly, that the number + 3 deter- mines the point C, the number - 1 the point A', the number - 3 the point C", and so on. We see that in the same way any number determines some one definite point on the straight line (and, con- versely, any point on the straight line determines one definite number), provided only that a starting-point, as 0, and a unit segment, as OA, are assumed. This gives us a definite one-to-one 8 THE ELEMENTARY FUNCTIONS correspondence between numbers and points on a (directed) straight line. Note that the number ^ corresponds to the point 0. 10. The phrases " greater than " and " less than." Since, as we have seen, the word "negative" appKed to numbers or distances is a qualitative, not a quantitative, adjective, it is evi- dently impossible to use the words " greater than " or " less than " in their ordinary quantitative sense when referring to negative numbers. A distance — 3, for instance, is neither less than nor greater than a distance + 3 in the ordinary sense, but, rather, an equal distance in the opposite direction. A meaning of these phrases that will be useful when applied to negative numbers is developed by the f oUowing considerations : When all numbers are represented as points on a horizontal straight line, in the way that we have just illustrated, we find that, of two positive numbers, the point corresponding to the lesser is always to the left of that corresponding to the greater. Thus, the statement " 3 < 4 " is equivalent to the statement " the point 3 lies to the left of the point 4 " ; and so for all positive numbers. Now it is equally true that — 1 lies to the left of 0, — 2 to the left of — 1, and so on ; and so it is natural to express these facts also ly the same words that are used in the case of the positive numbers: " — 1 is less than 0," "—.2 is less than —1," etc. In every case it should be remembered that the expression has no longer a quantitative, but a directional, meaning. EXERCISES 1. Restate the last paragraph for the case where the numbers are represented as points on a vertical line ; take the upward direction as positive. 2. Arrange in order of magnitude : 2,-3, Vs, 4, — 1.1, — VlO, V3, - 2, - 2.8, 3.9, Vi5, - VH, -v^, ^. 3. State the laws of addition, subtraction, multiplication, and division with negative numbers. 1 Many students are in the habit of thinking of "zero" as if it were not a number at all — sometimes even reading it " nothing." This is a mistake ; zero is a number, in some respects indeed the most Important number there is. THE GEAPHICAL REPRESENTATION C' 11. The system of coordinates. We have now seen that every number determines a certain point on a directed straight line, and that, conversely, to every point on a directed straight line corre- sponds a number, a starting-point (or zero-point) and a unit segment having been assumed. We now extend this correspondence be- tween points and numbers to include all the points of a plane. Let XX' and YY' be two per- pendicular lines intersecting at 0, and suppose for convenience that B r \ A XX' is horizontal. These lines will be referred to as the X-axis and the y-axis respectively. Their intersec- xA 1 — ^i — ^^^1 1^^ — i hX tion is called the origin. Then any point in the plane can be located exactly by giAring its distances from the X-axis and from the F-axis. Thus, the point ^ is -I- 1 from the r-axis and -|- 2 from the X-axis, while the point 2f is — 1 from the r-axis and -|- 2 from the X-axis. Similarly for any point in the plane. To locate the point it is sufficient to mention these two distances, imt forgetting the proper sign. It wiU also save words if we agree once for aU that the horizontal distance, or distance from the T-axis, shall always be given first. Thus, merely mentioning the pair of numbers ^^^^ g (1, 2) is sufficient to locate the point A, (- 1, 2) locates the point B (Fig. 7), while (2, 1) locates the point E (Fig. 8), and (2, - 1) the point F (Fig. 8). The student will be able to decide for himself what pair of numbers locates the point C, or the point D, in Fig. 7, or the point E' in Fig. 8. It is thus clear that a pair of numbers can be considered as determining a definite point, and, conversely, that a given point r Fig. 7 Y E E H \-X »P 10 THE ELEMENTARY FUNCTIONS determines a definite pair of numbers, namely, its distances from the X- and Y-axes. The horizontal distance, or (directed) distance from the Y-axis to the point, is called the abscissa of the point ; and the vertical distance, or (directed) distance from the X-axis to the point, is called the ordinate of the point. Thus, the abscissa of the point B (Fig. 7) is — 1 (the directed segment BB or OS), and its ordinate is 2 (the directed segment SB or OR). Give the abscissa and the ordinate of the points E and F (Fig. 8). The abscissa and ordinate of a point are called its coordinates. This system of connecting points with pairs of numbers was devised by Descartes, a great French mathematician and philoso- pher (1596-1650), and published by him in 1637 in a work called " La Geometrie,'' which is justly regarded as a milestone in the history of human thought. In honor of him the system of coor- dinates which has just been outlined is called the Cartesian coordinate system. Locating points in a drawing by means of their coordinates' is called plotting the points. For the sake of convenience in plot- ting points, squared paper is used, as it enables us to lay off any required distance both rapidly and accurately. In making all drawings a hard lead pencil with a fine point should be used, ink being reserved for certain lines which are meant to stand out prominently. In this first set of drawing exercises, draw the X- and y-axes in ink. It is scarcely necessary to say that the greatest neatness and accuracy are absolutely indispensable in all problems involving drawing, as indeed in all mathematical work. EXERCISES 1. Plot the points (2, 1), (2, - 1), (- 2, 1), and (- 2, - 1). What kind of figure do they form ? 2. Plot the points (2, 0), (-1, 0), (V2, O), (5, 0), (- 4^, 0), (- V2, O), (3f, 0). On what line are they all found? 3. On what line are all the following points located? (1, 1), (-2, -2), (3, 3), (-i, -i), (5, 5), (2.1, 2.1), (-V2, -V2), (H-V2, l-i-V2),(-^,-J). THE GEAPHICAL EEPEESENTATION 11 4. Plot the points (- 4, - 3), (3, - 4), (4, - 3), (3, 4), (4, 3), (— 3, 4). Prove that they all lie on a circle, and give the coordinates of other points on the same circle. 5. Plot the points (0, 0), (y2, V2), (-V2, V2). What kind of figure do they form ? 6. An equilateral triangle whose side is 4 units is placed with one vertex at the origin and with the opposite side perpendicular to the A"-axis. Find the coordinates of the other two vertices. (Two solutions.) 7. A square whose side is 2 units is placed with one vertex at the origin and with a diagonal lying along the A' -axis. Find the coordinates of the other three vertices and of the center. (Two solutions.) 8. A regular hexagon whose side is the unit is placed with one vertex at the origin and with its center on the A'-axis. Find the coordinates of the other 5 vertices, and the area of the hexagon. (Two solutions for the vertices.) 9. Plot the points (3, 1), (0, 4), and (3, 4). What kind of figure do they form ? Find the lengths of its sides, its area, and the coordinates of the center of the circumcircle.^ 10. How far is the point (3, 4) from the origin in a straight line ? the point (1, 5)? the point (-1, - f)? the point (a, b)? 11. Plot the points (3|, 3f), (h 3f), and (3*, -1%), and find the center and radius of the circle through the three. 12. What are the coordinates of the points P, R, and S in Fig. 7 ? What are the coordinates of the origin ? 13. What are the coordinates of the point halfway between the origin and the point (4, 4)? of the point halfway between the origin and the point (3, 6) ? 14. Plot the points (2, 1) and (4, 5). What are the coordinates of the point midway between them ? 15. Answer the same question for the points (1, 3) and (— 3, 1); for th,e points (1, 0) and (0, 1); for the points (3, - 2) and (- 5, 4). 1 Circumcirole : a circle passing through the vertices of a triangle or of any polygon. 12 THE ELEMENTARY FUNCTIONS B C \ A Y X D E O 16. Let (cCj, ^j)^ and (x^, y^ be any two points. Find the coordi- nates of the mid-point of the segment joining them. 2 2 17. Find the distance between the points (—1, 2) and (— 4, 6). Solution. If A (Fig. 9) represents the point (— 1, 2) and B the point (— 4, 6), and if we draw BD and AE perpendicular to the X-axis, and AC perpendicular to BD, we have formed a right triangle ABC whose hypotenuse is the required distance AB. But the lengths of .i4C and CB can easily be found, as follows : 0E = -1 and Oi» = - 4. Therefore ED=AC=-Z. Likewise DB = 6 Fig. 9 and DC = 2. Therefore CB = 4. By the Pythagorean Theorem, AB^ = AC^ + CB\ Therefore ABfi = 9 -(- 16 = 25. Hence AB = 5. 18. Find the distance between the points (3, —1) and (4, — \); between the points (6, 1) and (— 6, 6). 19. Find the distance from (0, 3) to (4, 2); from (1, 1) to (- 2, - 3); from (3, 5) to (- 4, 1). 20. Find the distance between the points (x^, y^ and (x^ yX Am. ^{x^ — x^y + (y^ — yj'. This result is easily established when the given points (x^, y^ and (x^ y^ both have positive coordi- nates, but the student should show that it is true for all positions of the two points. The theorems in § 7, p. 6, will be found useful for this purpose. 12. Application of coordinates to some problems of elementary geometry. This method of representing points by pairs of num- bers is very useful because it gives us a means of handling many ' Read "z-one, j/-one," meaning "the first s," and so on. Such subscripts are often very convenient, and should not be confused with exponents. THE GEAPHICAL EEPKESENTATION 13 geometric problems in an algebraic way. The significance of this statement will become much more apparent as we proceed farther in our work, but the simple examples which follow will provide at least a basis for appreciation of it. Example 1. Let us take the well-known problem of elementary geometry, Prove that the diagonals of a rectangle are equal. In such problems, where the object is to prove geometric theorems by the aid of coordinates, the secret of a simple solution lies in a wise choice of the coordinate axes. We are at liberty to choose them in any position we please with reference to any figure already given. In this case we choose as X- and F-axes two adjacent sides of the given rectangle, and suppose the length of OP (Fig. 10) is a and that of OR is b. Then the coordinates of O are (0, 0), those of P are (a, 0), those of R are (0, 5), and those of Q are (a, b). We now apply Ex. 20, above, which gives us Fig. 10 RP = V(a - 0)2 + (0 - by = Va^ + i and Therefore OQ = V(a - Of + (6 - 0)2 = Va2 + ( RP = OQ. Q.E.I>. Example 2. Another well-known theorem of elementary geometry which can easily be proved by the use of coordinates is. The diagonals of a paral- lelogram bisect each other. Proof. Let OPQR (Fig. 11) be the parallelogram. Choose one side OP as the X-axis, and as the origin. The coordinates of P, the other vertex on the X-axis, may then be repre- sented by (a, 0), those of R by (b, c), and those of Q bj (a + b, c). (For MR = b and RQ= 0P = a. .: MQ = a + b, the abscissa of Q.) Then the coordinates of the mid-points of the diagonals can be found by Ex. 16, above. For the diagonal RP the mid- point is thus I 1 -|i and for the diagonal OQ it is (^-^^ |)' that is, the same point. points coincide, the diagonals bisect each other. q.e.d. Y M R Q r \ ^ -J U ^-^ v/ 1 X O 1 Fig. 11 p Since their mid- 14 THE ELEMENTARY FUNCTIONS EXERCISES Prove the following theorems by means of coordinates : 1. The line joining the vertex of any right triangle to the mid- point of the hypotenuse is equal to half the hypotenuse. 2. The line joining the middle points of two sides of a triangle is equal to half the third side. 3. The distance between the middle points of the nonparallel sides of a trapezoid is equal to half the sum of the parallel sides. 4. In any quadrilateral the lines joining the middle points of the opposite sides and the line joining the middle points of the diagonals meet in a point and bisect each other. 5. If the lines joining two vertices of a triangle to the middle points of the opposite sides are equal, the triangle is isosceles. 13. Point that divides a segment in a given ratio. Ex. 16, p. 12, enables us to find the coordinates of the mid-point of any Y I^(X2 y,> p^-.y/^^^ "^ ^ B X o c 7 1 3 J i Pi^^'Vi) Fig. 12 segment; we can do more, and get the coordinates of the point that divides a given segment in a»y given ratio. Let ^ = («!, 2/j) ^ and ^ = {x^, y^ be the end-points of the given segment, and let P be the required point dividing the segment ij-^ in the ratio m : n, that is, so that P^P : PI^ = m:n. Let {x, y) be the coordinates of P. Then, if P^G, PD, and I^E are per- pendiculars from Jl, P, and J^ respectively, upon the X-axis, 00 equals the abscissa of ij (that is, x-^, OE=x^, and OD = x. Form the right triangle I^AI^ by drawing through ij the line 1 Read "Pj, which is the point (x^ y^" or "P^, whose coordinates are THE GRAPHICAL REPRESENTATION 15 I^A parallel to the X-axis, meeting FD in B and J^U in. A. Then IIA=C£J= x^~ x^ (notice that this is true even if ^ is to the left of i^, as in Fig. 12 6, since x^- Xj^ is then negative), ^j5= CD= X — ajj, and BA = DE= x^ - x. By hypothesis ^P:PP^ = m:n, and since the triangles I^BP and P^A^ are similar (why?), we have P^P:PP^=I[B:BA. But this last ratio is equal to Solviag (1) for x, X- ) — - ■''! hence «2 — X X- - X^ _ m X^ — X n x = TTUTj + nx^ (1) (1') m + n In exactly the same way, drawing PF perpendicular to ^A (not shown ia Fig. 12), we find AF=BP= y— y^, FP^ =y<i—y, and thus y - yi _ m Solving (2) for y, y-i-y n (2) (2') m + n Notice that these results (!') and (2') reduce to the results of Ex. 16, p. 12, when m = n. Observe also that m and n are not necessarily the lengths of the segments P^P and Pi^, but are merely numbers in the same ratio as those lengths, m correspond- ing to the segment JJP (that is, to the segment nearest i^), while n corresponds to the segment F^ (that is, to the segment nearest ^). Example 1. Find the coordinates of the point which divides the segment from (1, 3) to (6, 2) in the ratio 2 : 3. Solution. Here m = 2, n = 3, ^1 = 1> yi = 3, ^2 = 6, 2^2 = 2. Substituting these values in (1'), 2-6 + 3-1 2 + 3 = 3, Pi(i5r Pi."!^ fi(6,2) Fig. 13 and from (2'), Therefore _ 2 ■ 2 + 3 ■ 3 ^~ 2 + 3 P = (S,^^). 13 5 ■ 16 THE ELEMENTARY EUNCTIONS Example 2. Find the coordinates of the point ^ of the way from (-4, 2) to (2, -1). Solution. The point P being ^ of the way from P^ to P^, the segment PjP = JPiPj, and hence PP^ = § PjPj, so that PjP : PP^ = 1:2. (A rather natural although careless mistake that is often made in problems of this kind is to say m = 1, re = 3, since the point P is to be ^ of the way from Pi to Pj.) The student can now finish the work for himself. The result isPs(-2, 1). Example 3. Prove that the medians of any triangle meet in a point f of the way from a vertex to the mid-point of the opposite side. Proof. Let A = (Xj, y^), B = (x^, y^), and C= (xg, j/g) be the vertices of the triangle. Then the coordinates of M, the mid-point of AB, are (Ei±Ii, yi + yi \ ; those of A', the mid-point of BC, are (^s±^, ^^^) ; and those of Q, the mid-point of CA, are (^^, 2^)- Next, the coordinates of the point f of the way from C to M are as follows (m = 2, n = 1) : 2 • ^L±^ + 1 • Xg 3 ' 2' = 2-H 2 . a±l2 + 1.3, '.+ y> Fig. 14 2-1-1 Working out the coordinates of the point § of the way from A to iV, also of the point § of the way from B to Q, we get in each case the same pair of coordinates. This proves the theorem. EXERCISES 1. Find the coordinates of the point which divides the segment from (1, 3) to (— 2, 6) in the ratio 1:2; in the ratio 2:1. 2. Find the coordinates of the point f of the way from (— 2, — 3) to (7, 3) ; § of the way from (^, 3§) to (- 1, - ^). 3. Prove that in any parallelogram A BCD, if M is the middle point of the side AB, the line MD and the diagonal AC trisect each other. THE GRAPHICAL EEPEESENTATION 17 4. rind the coordinates of the point that divides the segment from (0, 3) to (2, 5) externally in the ratio 3 : 2. Hint. Here the point P is on the line P^P^ produced, and since F-^P : PP^ equals numerically |, P^P must be greater than PP^ ; that is, P must be nearer to Pj than to Pj, or, in other words, it is beyond the point P^. P^P and PP^ now being measured in opposite directions, we may think of the ratio m : n as being negative. With this understanding it will be found that the reasoning by which the formulas (1) and (2) were obtained is still valid, and they may be used in these cases also. Here m = 3, n = — 2 (or else m = — 3, m = 2), so that we have 3-2-2.0 „ 3-5-2-3 . x = = 6, y = = 9. 3-2 ' " 3-2 Hence the required point is (6, 9). 5. Find the coordinates of the point that divides the segment from (— 2, — 3) to (3, — 5) in the ratio —2:3; in the ratio —1:2; in the ratio —2:1. 6. A line AB is produced to C so that AC = SBC. Find the coordinates of C if ^ s (2, 0) and B=(— 3, 1). 7. Find the coordinates of the point that divides in the ratio \ the segment from (1, 2) to (— 1, 3) ; from (aj^, y^) to (x^ y^. 8. Find the point of intersection of the medians of the triangle whose vertices are (2, 3), (4, - 5), and (3, - 6). (Do not use the result of Ex. 3, p. 16, but follow the same method, using the special numbers of this problem.) 9. In what ratio does the point (2, 3) divide the segment from (-2, -3) to (4, 6)? Hint. Use equations (1) and (2), p. 15. 10. In what ratio is the segment from (—2, 1) to (3, —9) divided by the point (1, — 5) ? \\. li A = (2, —1) and B = (5, \), in what ratio is AB divided by the point C = (4, 0)? 12. Using the same segment AB as in the preceding problem, show that, for the point Z)=(-5, -4), |^=-io' ^^""^^ y ~y\ = _ ? , a different ratio. What conclusion can you draw y^-y ^ from this peculiarity? 13. Test as in the preceding problem the three points A = (3, 2), B = {-1, -10), and C= (0, - 8); the points (5, -1), (2, 2), and (-1, 5). 18 THE ELEMENTARY FUNCTIONS 14. Formulate the results of Exs. 12 and 13 in the form of a theorem that gives a test as to whether three points A, B, and C are in the same straight line or not. 15. Given the three points A = (2,1), B = (-1,3), and C = (1,-2), find the coordinates of D, the fourth vertex of the parallelogram determined by ^, B, and C. (Three solutions.) 16. Proceed as in Ex. 16 for the three points (0, 5), (7, 3), and (- 2, - 3). 17. Prove that the sum of the squares of the medians of any triangle equals three fourths the sum of the squares of the sides. CHAPTER II FUNCTIONS AND THEIR GRAPHS 14. In the preceding chapter we have seen how to associate with every segment a number, — its length; and by means of the system of coordinates we have associated with every point in the plane a pair of numbers. We apphed this to the proof of a few theorems of elementary geometry. In this chapter we shall consider not merely fixed points, as heretofore, but variable points, that is, points that are free to occupy any number of positions, subject to certain conditions. ^ p, For example, let us say that a ' *~ point is free to move so as to be x always at the distance +2 from ^'(^ ^ the X-axis. There are evidently Pm ^5 an unlimited number of such points, but they will all be found on one straight line, the line ABC, parallel to the X-axis and 2 units above it. We use the expression. The locus of points at the distance -|- 2 from the X-axis is the line parallel to the X-axis and 2 units above it. Furthermore, to specify that the point shall be at the distance 2 from the X-axis is the same thing as saying that its ordinate shaU equal 2 ; or, writing y for the ordinate of any one of the points in question, the original condition is equivalent to the statement that y = 2 (or y — 2 = 0). We may say, then, that the locus of the equation y=2(ovy—2 = 0) is the straight line parallel to the X-axis and 2 units above it. We call the line also the graphical representation, or simply the graph, of the equation. 15. As another example, let us make the following condition: A point moves so as to be always twice as far from the X-axis as from the Y-axis. Here again we know from elementary geometry 19 20 THE ELEMENTARY EUJSTCTIONS P(x,y) Q Fig. 16 that the locus of these points is a straight line P'OP, for it P is twice as far from the X-axis as from the Y-axis (that is, if QP= 2 • BP= 2 • OQ), then the same thing is true for any point on the line OP, and for no other points.^ Since the given condition is equivalent to the condition that the ordinate of the point shall equal twice its abscissa, we can express this condition in the form of an equation, thus : y=2x, where x represents the abscissa of the moving poiat, and y its ordinate. 16. This example illustrates the use of the general numbers («, y) to represent the coordinates of a variable, or moving, point, — a notation which is very convenient and which will be frequently employed throughout this book. The quantities that we have to deal with in physics, engineering, astronomy, and indeed in aU applications of mathematics, are largely variable, and hence it is necessary to learn to use and understand symbols that represent such variable quantities. To represent constant quantities we shall use the earlier letters of the alphabet or letters with subscripts, as x-y, y-^, etc. 17. As a third example let us make the following condition : A point moves so that its ordinate is always equal to the square of its abscissa. If the coordinates of this variable point are represented by {x, y), then evidently the equation of the locus is y = o?; but elementary geometry gives us no information as to just wliat this locus is, that is, just how the point can be located if it satisfies this condition. We can, however, get a good idea of the locus by starting with the equation y = o^ ' For the present this assertion may be accepted without proof. The proof is given on page 92, where this example is taken up again more in detail. Fig. 17 PUNCTIONS AND THEIR GRAPHS 21 itself and finding a number of pairs of values of x and y which satisfy the equation. If we then plot the points having such values {x, y) as their coordinates, we shall get a series of points satisfying the given condition; that is, we shaH get points on the locus. Thus, when x = l,y = l- when x=2,y = 4:; when a; = 1 y = i ; when a; = 0, y = ; etc. Hence (1, 1), (2, 4), (i, i), and (0, 0) are points on the locus. We may conveniently arrange these pairs of values in the form of a table, thus : X h 1 1 2 -1 -2 -h _ 3 y \ 1 1 4 1 4 i 9 etc. We now plot the points carefully, and when enough points have been located, it will be possible to see that they suggest a curved line, concave upward, symmetrical with respect to the r-axis, and with its lowest point at the origin. This curve should now be carefully drawn on a large scale, and we can then infer that we have a fairly accurate drawing of the locus of the equa- tion y = a?. The curve is called a parabola. (It must not be over- looked, however, that we cannot prove that all points on the curve satisfy the condition y = a^ ; but we asswme this to be the case, and if the drawing has been made accurately, this assump- tion will be in fact approximately correct.) 18. As a fourth example, let us suppose that a point (x, y) moves so that the following relation between x and y always holds : y = a?—29i? — x + 4=. As before, we proceed to compute a table of values of x and y that satisfy this equation. The result may be exhibited thus : X 1 2 3 -1 - 2 - 3 i 1 -i y 4 2 2 10 2 -10 -38 H If H Plotting the points given by this table, and connecting them by a smooth curve, as in Fig. 18, we get approximately the graphical representation of the equation given. 22 THE ELEMENTARY FUNCTIONS la the following exercises a point is to be understood as varying subject to the condition given each time in the form of an equation between x and y. Draw the graph of the equation carefully, determining the exact locus in Exs. 1-10, and a very close approximation in the others, as was done in the last two ex- amples above. Also, in the first 12 exercises state in words the condition which the equation gives in symbols : for example, in No. 1, " If a point moves so that its ab- scissa equals 2, what is its locus ? " Fig. 18 1. a; = 2. 2. y=-3. 3. a; = 0. 4. y — 5 = 0. 5 . 2/ = 3 a;. 6. x=— 2. 7. y = 0. 8. y=—2x. 9. 2/ = — a;, and y = — x + 2, in the same figure. 10. y = i X, and y = ^ x — 3, in the same figure. 11. 2/ = 2a; —1. 12. y = 3x + i. 13- y=-ix+l. 14. y = 2x'^-2. 15. y=-Sx^-5. EXERCISES 16. y = x^ — X. 17. y = x''—2x. 18. y =— x^ — x. = — a;^ — a; + 5. = -2x' + x +1. 19. y^=—x + 3x. 20. y = 2 x^ — 3 a;. 21. y = 2cc^ — a;-|-3. 22. y =—x 23. y 24. y = x^. 25. y = a;'— 1. 26. y = 7? — X. 27. y = x^ -^x. 28. y =zx^ — x^. 29. y = - x' + 3 xl 30. y=:x''-3x2-|-l. 31. y =— 2x''— 6a;. FUNCTIONS AND THEIR GRAPHS 23 32. y = x^ + x\+x+l. 35. y=-x^ + 2x' + 3. 33. y = x''-x^' + x-l. 36. y==-2x' + x'-x + 2. Si. y = x'' — x^—x—l. 37. 2/ = - a;= - a;2 - a; - 3. 19. Definition of function. In all of these later exercises, pairs of values (x, y) that would satisfy the equation given were deter- mined by taking a set of values of x and then using the equation to compute the corresponding values of y. Whenever a pair of (variable) numbers x and y are related in this way, that is, in. such a way that to a value of the one corresponds a definite value of the other, the second is said to be a function of the first, or the relation is called a functional relation. The symbol for this relation between x and y is y—f{x), read "3/ is a function of X " or « y =f of x." For example, if y = 2 «, y is a function of X, because to a given value of x corresponds a definite value of y ; or, again, ily — a?—2x'^ — x + 4:,y\sa. function of x. In these examples x is called the independent variable, and y, or the function of x, is called the dependent variable, because its value depends upon that of x. 20. Importance of functional relation. The idea of this functional relation between two variable numbers is of very far- reaching importance, because it can be applied to any pair of quantities that are representable by numbers if one of these numbers is determined by the value of the other. For instance, the temperature at any place is a function of the time, be- cause at any definite time there is a definite temperature at that place ; or, again, if a man walks at the rate of 3 miles per hour, the distance he has gone is a function of the time he has walked ; or, again, the population of a city is a function of the time (date) when the estimate or count is made. Considera- tion of the wide variety of such possible illustrations wiU make clear the very great importance of a study of various kinds of functional relations. 21 . For such a study the graphical representation just illustrated is a most valuable aid, because it gives us a vivid ("graphic") picture of the functional relation, thus assisting materially in forming a clear 24 THE ELEMENTARY FUNCTIONS idea of the nature of that relation. This may be seen in the ex- ample mentioned above of the variation of temperature at a certain place ; suppose the weather observer had made the following record : Time Tempebatuke 7 a.m. 51° 8 a.m. 55° 9 a.m. 60° 10 a.m. 62° 11 A.M. 70° 12 m. 72° 1 P.M. 72° 2 p.m. 73° 3 p.m. 71° 4 p.m. 66° 5 p.m. 61° 6 p.m. 58° 80 1 70° 60" SO'-f 40'- 30°' ■ 20" 10° 7A.1 -+- -+- ::It -t- -4- H 1- 10 1112M.1F.M. 2 3 4 6 Fig. 19 Since the first observation was taken at 7 a.m., we represent that time as the origin and construct the times as abscissas and the temperatures as ordinates, thus getting the above figure. A single glance at the figure evidently gives us a clearer and more comprehensive idea of the variation of the temperature on that particular day than the table of values does. 22. To take a more elaborate illustration of this use of the graphical representation, we find in the United States Statistical Abstract for 1915 the following table showing the progress of shipbuilding in the United States from 1900 to 1915: Year Tonnage Built Yeak Tonnage Built 1900 393,790 1908 614,216 1901 483,489 1909 238,090 1902 468,831 1910 842,068 1903 436,152 1911 291,162 1904 378,542 1912 232,669 1905 330,316 1913 346,155 1906 418,745 1914 316,250 1907 471,332 1915 225,122 FUNCTIONS AND THEIE GRAPHS 25 Here the tonnage built is a function of the time. Plotting the number of the year as abscissa (starting of course with 1900 at the origin) and the corresponding tonnage as ordinate, we get a figure like the one adjoining. Let the student complete the figure, using as large a scale as possible. 400,000 Now a glance at the broken hne joinmg these points 100,000 600,000- ■ 600,000 gives a much clearer idea iLix izoiui, is i, isi, io ii L iz I .X Fig. 20 of the variation in the ship- building during these years than does the mere table from which the diagram was constructed. 23. In drawing these graphical representations of statistical tables it is not necessary to try to draw a smooth curve joining the points located, since we have no possible way of telling how the curve should look, between the located points ; hence we draw only a broken line, that is, join each point to the next following by a straight line. In the graphical representation of equations, however, as in the exercises on page 22, it would not be right to do this, because we can locate points just as close together as we need, thus determining the shape of the curve to any desired degree of accuracy. For instance, in the example of § 18 we had (among others) the points. (0, 4) and (—1, 2), which are so close together that it would be natural to connect them by a straight line (just as we should do if there were no way of tell- ing how the curve reaUy lies between these points) ; but by taking x = — ^ we found y = 3|, which shows that the straight line joining (0, 4) and (—1, 2) would be decidedly incorrect. Between x = l and a; = 2 it would be natural to make the same mistake again, and join (1, 2) and (2, 2) by a horizontal Hne; but the point (|, 1|) corrects this. The detailed study of the graphical representation of functions in this way is made much simpler and more practical by the use of a chapter in mathematical analysis which is called the Differential Calculus. Until the use of that very powerful method has been learned, however, the student 26 THE ELEMENTARY FUNCTIONS must be content to plot accurately a sufficient number of points so that the form of the curve is evident. Summarizing, the graphical representation of functions given by mathematical equations can be carried out to any required degree of accuracy, whUe in the case of functions given by tables of observations or statistics we cannot, of course, locate more than the number of points given in the tables. EXERCISES 1. The Statistical Abstract for 1915 gives the following figures for the values of exports and imports of merchandise for the years 1900-1915 : Year Exports Imports Year Exports Imports 1900 1,394,483,082 849,941,184 1908 1,860,773,346 1,194,341,792 1901 1,487,764,991 823,172,165 1909 1,663,011,104 1,311,920,224 1902 1,381,719,401 903,320,948 1910 1,744,984,720 1,556,947,430 1903 1,420,141,679 1,025,719,237 1911 2,049,320,199 1,527,226,105 1904 1,460,827,271 991,087,371 1912 2,204,322,409 1,653,264,934 1905 1,518,561,666 1,117,513,071 1913 2,465,884,149 1,813,008,234 1906 1,743,864,500 1,226,562,446 1914 2,364,579,148 1,893,925,657 1907 1,880,851,078 1,434,421,425 1915 2,768,589,340 1,674,169,740 Make a graphical representation of these statistics. 2. We find in the Statistical Abstract that the number of immi- grants admitted to the United States during each of the years from 1892 to 1915 was as follows : Year Number Year Number Number Admitted Admitted Admitted 1892 623,084 1900 448,572 1908 782,870 1893 502,917 1901 487,918 1909 751,786 1894 314,467 1902 648,743 1910 1,041,570 1895 279,948 1903 857,046 1911 878,587 1896 343,267 1904 812,870 1912 838,172 1897 230,832 1905 1,026,499 1913 1,197,892 1898 229,299 1906 1,100,735 1914 1,218,480 1899 311,715 1907 1,285,349 1915 326,700 Draw a graphical representation of these facts. rtJNCTIONS AND THEIR GRAPHS 27 3. The number of persons killed in railway accidents in the United States during each of the years from 1892 to 1910 was as follows : Yeak Number of Persons Year Number of Persons 1802 2554 1902 2969 1893 2727 1903 3606 1894 1823 1904 3632 1895 1811 1905 3361 1896 1861 1906 3929 1897 1693 1907 4534 1898 1958 1908 3405 1899 2210 1909 2610 1900 2550 1910 3382 1901 2675 Make a graphical representation of these statistics. 4. Find statistics for the growth of the population of the United States since 1790, and draw the graphical representation of these figures. 5. Answer the same question for three or four of the states, from the time they were admitted to the Union to the present. 6. Make a graphical representation of the amount of flOOO at simple interest for one year, as a function of the rate per cent (from 1% to 10%). 7. Answer the same question for the amount of flOOO at 6% as a function of the time (from 1 yr. to 10 yr.). Note. Ample material for further work in statistical graphs may be found in the government census reports, crop reports, and in the financial journals, etc. CHAPTER III APPLICATION OF GRAPHICAL REPRESENTATION TO ELEMENTARY ALGEBRA 24. In the preceding chapter we saw how to use the coordinate system to draw the graphs of many simple functional relations be- tween X and y ; and we observed that such graphical representation of equations in the form y=f{x) is of very wide application iu various fields. In this chapter we shall use this geometrical work of drawing graphs for the purpose of throwing new light upon certain problems of elementary algebra. As a preliminary step it will be foimd useful to make a simple classification of functions according to their degree. 25, Degree of a function. The expressions 2a;— 3, —?>x+2, 1 X Z X, -x — 1, — + 7 are all functions of x, according to the defi- Ji nition of the word " function," for the value of any one of these quantities is determined by the value of x. In each of them the variable number x occurs to no higher power than the first. Functions in which this is the case are said to be "of the first degree " or " linear " in x. A general form of such functions of x would be ax+h, where a and h are general numbers, that is, may have any value we please. A function which contains x^ but no higher power of x is said to be " of the second degree " or " quadratic " in x. A general form for it would be aa^+hx+c. A function which contains a? but no higher power of x is said to be " of the third degree " or " cubic " in 03, a general form being aoc? + hx^ + cx + d. In the list of prob- lems on page 22 we had examples of all these three kinds of functions, — linear, quadratic, and ciibic. In the same way it is possible to write down functions of the fourth, fifth, • • • degree, but we shall seldom need to go beyond the third degree in this book. 28 If this equa- c a y = b- T" ELEMENTARY APPLICATIONS 29 26. Linear equations. An equation of the form ax + hy = c (1) is called a linear equation in the variables x and y. tion be solved for y, a linear function is obtained : For example, if we had 2x + 3y= — 1, then y= — ^ — f a^, which is a linear function of x. Let the student write down, at random, three or four equations of the form (1), choosing any numbers whatever for a, h, and c, and then make a careful graph of the function in each case. Each graph will turn out to be a straight Hne. Later on we shall be able to prove that this is necessarily the case, but for the present we merely assume it to be a fact, — that is, we make, as yet without prx)of , the following assumption : The graph of every equation of the first degree in x and y is a straight line. This is the reason why such an equation or functional relation is called "linear." Accordingly, in making the graph of such an equation it is sufficient to plot two points whose coordinates satisfy the equation, and then the straight line joining these two points wiU be the graph of the equation. (In practice a third point shoiild also be plotted, as a check; if it is not exactly on the graph, a mistake has been made.) For the two points it is often most convenient to choose the points where the hne crosses the X-axis and the T-axis. For the point of intersection with the X-axis the ordinate equals (since the ordinate of any point on the X-axis equals 0) ; hence let y = in the equation of the hne, and solve for the value of x. Similarly, to get the point of inter- section with the F-axis, let a; = in the equation. The (directed) distances from the origin to these points of intersection with the X- and F-axes are called the X- and F-intercepts of the line. In general, if in the equation of any locus we let a; = 0, we obtam the value of the ordinate of the point where the locus meets the F-axis ; and if we let y = 0, we get the abscissa of the point where the locus meets the X-axis. These two points are especially useful in the practical work of drawing graphs. 30 THE ELEMENTARY FUNCTIONS 27. Simultaneous linear equations. It is a famUiar and simple problem of elementary algebra to find a pair of values {x, y) that will satisfy each of two linear equations in the variables x and y. We can now solve this problem graphically, that is, geometrically, in a very simple manner. Let the equations be ax + hy = c (1) and a'x + h'y = c'. (2) We draw in the same figure the graphs of the two equations; then the graph of equation (1) contains all points whose coordi- nates {x, y) satisfy that equation (for this is the definition of the graph of an equation), and the graph of equation (2) contains all points whose coordinates (x, y) satisfy that equation. Hence any point whose coordinates {x, y) satisfy both equations at once must be found on both graphs, that is, must be their point of intersection. The solution of the problem can accordingly be obtained, at least approximately, by a glance at the figure; and since the graphs of linear equations are straight lines, there will be one and only one such point of intersection, unless the lines are parallel, when there will be none. The equations (1) and (2) have, therefore, in general one and only one solution, or, in case the graphs are parallel lines, they have no solution. Example 1. Solve graphically the simultaneous equations \2x-%y = T, (1) L3 a; + !/ = 5. (2) Solution. To make the graphs, we may determine the X- and Y- intercepts of each of the lines (1) and (2) 1; in (1), when ^ = 0, a; = J, and when a; = 0, y = — J ; in (2), when y = Q, X = ^, and when x = 0, ^ = 7. As a check point take any value of X at random and compute the corresponding value of y\ the point thus determined should lie upon the graph. If correctly drawn, 1 The expression "line (1)" is used for brevity, instead of the complete expression "line whose equation is (1)." G Fig. 21 ELEMENTARY APPLICATIONS 31 the lines (1) and (2) will result as in Fig. 21. The coordinates of the point of intersection P are evidently about (2, — 1), which fact gives (approximately) the solution of the problem, namely, x = 2, y = —X. In this case the graphical method has given us the exact solution, as we can easily verify by substituting a; = 2, t/ = - 1 in the equations (1) and (2). Example 2. Solve graphically the simultaneous equations p + 4)/ = 10, (1) 1.5 a; -8 2^ = 1. (2) Solution. Making tables of values for x and y in order to determine points on the graph, we have in tabulated form : Check. Y (1) X 10 2 y n 2 Check. (2) X i 2 y -4 1 Fig. 22 The coordinates of the point of intersection P are about (3, 2), which is in fact very nearly the correct solution, although not exactly so (as was the case in Ex. 1). Algebraic solution ^ shows that x = 3, y = 1^ is the exact solution. Example 8. Solve graphically the simultaneous equations ■4y = 5, c (1) x-Sy = 7. (2) The graphs result as in Fig. 23, and apparently the lines are parallel. That they actually are so is shown by attempting an algebraic solution. If we mviltiply equation (1) by 2, we have 6 a; — 8 y = 10, whereas (2) requires that 6 a; - 8 2^ = 7. Fig. 23 Both cannot be true at once; hence there is no pair of values (x, y) that will satisfy both equations; that is, the lines (1) and (2) do not intersect, as was indicated by the figure. 1 See Appendix E for the method, in case it has been forgotten. 32 THE ELEMENTARY FUNCTIONS EXERCISES Solve the following pairs of simultaneous equations both graphi- cally and algebraically : f3x + t ,y=n, , f4x -7^=11 , ^ rSx-l- 9^ = 93, ■U + y = 3. "■ l3x + 2y=l. " l7x-5y = n. fx-4y = l, f3x-z/ = 3, r|x-|y=7, ■\2x + y=-7. \6x-3y = 2. ' Ux-5y = 20. (x + 2y = l, fl5x+7y=ll, r7.2 x + 1.5 y= 0.42, ■ l2x + 8y = 5. 'l^x = 2y. * U.8x- 2.5 y= 0.98. 10. r3x 4y_ fx-l 2/4-2 11 5 6 30' 2x + 3 3y-l , [7 ' 5 -^• 12. ■ 3^4 4 4x 2?/ 3a; 19y 3 6 - 4 ' 40 ■x + 32/ 2x + y x + y 3x — 5y 8 5 2 7 ' 13. 3x + 2,v 1-32/3 5 [7 ' 4 "*"4 x+7 5X + 82/ 6 ' 2 ' ^'" 14. Prove algebraically that the simultaneous equations ax + by + c = and a'x + b'y + c' = will have one and only one solution unless ab' = ba'. Determinants 28. Before passing on to another application of the graphical representation to elementary algebra, it will be useful to take up a new method of solving simultaneous linear equations. This new method is indeed a very valuable aid in many more advanced mathematical studies. It is the method of determinants, which are defined as follows : A determinant is a symbolic expression in the form a b c d ' which is understood to represent the quantity a • d — b • c. Thus, ELEMENTARY APPLICATIONS 33 -4. 2 = 15-8=7; is a determinant and equals 3 ■ 5 is a determinant and represents x^ — y^. Before reading farther the student should write down some other determinants and give their values, so that this symboKo expression may become somewhat familiar. 29. Solution of simultaneous linear equations by use of deter- minants. If we solve by the ordinai'y elementary algebraic method the pair of equations rax + ly = c, (1) \a'x + Vy = c', (2) we obtain the values cV - he' aV — ha' y- ac' ca ah' — ha' Now both numerator and denominator of each of these fractions are differences of products, and hence can be written in the determinant notation, as follows : x = c I a c c' h' y = a' c' a h a h a' V a' h' These fractions can be written down by inspection of the equa- tions (1) and (2) if we observe the followiag directions : The denominator is the same for x and y, and is formed by copy- ing the coefficients of x and y ia the equations, in the exact order in which they occur. V The numerator for x is formed by writing, instead of the coefficients of x, the numbers on the right- hand side of the equations (c and c'), thus : , , , > the second colunm of the determinant being the coefficients of y {h and h'). Finally, the numerator for the value of y is formed by writing in the first column the coefficients of « in the equations (a and a'), and in the second column, instead of the coefficients of y, the numbers on the right-hand side of the equations (c and c'), thus : 34 THE ELEMENTARY FUNCTIONS Example 1. Solve in this way the equatidns {: 4 a; + 5?/ = 17. (X) (2) By the preceding rule the values of x and y are written down directly, thus : 10-(-119)_129_g 2 -7 17 5 ,15 -(-28) 43 3 2 4 17 3 -7 4 5 51-8 43 = 1. The solution is accordingly (3, 1), which pair, in fact, satisfies both equations. Example 2. Solve by determinants : if- 1, I- 4 • a: + 2 = 0. (1) (2) Here the second equation is not in the form a'x + Vy = c', so we first write it in that form : — x -\- y =—2. The solution can now be written down : 1 -i _ 2 1 i -i -1 1 i- f = 6, i 1 -1 -2 4 -i -1 1 =s=*- Therefore (6, 4) is the solution, and this pair of values satisfies both equations. EXERCISES Solve by the method of determinants the problems in the exercises on page 32. *30.i Determinants of the third order. This method of deter- minants can be used to solve a system of threfe linear equations in three unknown quantities. The determinants thus far used 1 Starred paragraphs may be omitted if desired, without interfering with the unity of the course. ELEMENTARY APPLICATIONS 35 are called determinants of the second order, and we now define determinants of the third order as being symbolic expressions of the form «2 \ "2 ' (1) a, hr, c. *3 which is imderstood to represent the quantity (2) This rather long expression can be written down without bur- dening the memory at all, by observing that it contains three products with + sign and three products with — sign, and that the first three products can be read off by following the directions of the arrowheads lq the following scheme : (I) (in (III) 2 2 ^2 (3) jug^s ; from the From the line marked (I) we get the product a line marked (II), the product liC^a^ ; and from the line marked (III), the product c-fi^a^. These are the three products that have the + sign in the value of the whole determinant. The other three products are read off in a similar way by following the direction of the arrowheads in this figure: (4) (IV) The lines marked (IV), (V), and (VI) give us the products a^\cy, l^c^a^, and c^^"-^, which, except that the negative sign must be prefixed to each, are the last three terms in the expression (2) for the whole determinant. 36 THE ELEMENTAEY FUNCTIONS *31. Two illustrative examples will make this procedure entirely clear, and will show that it is in reality very simple. Example 1. Write the value of the determinaiit 3 4 1 2 3 4 6 6 8 It is better not to draw the guide lines marked (I), (II), (III), etc. in (3) and (4) above ; we merely think them into the figure. Thus, the value of the determinant is 3 -3 -3 + 4-4-6 +1-6-2-6-3-1-6-4-3-3-4-2 = 27 + 96 + 12 - 18 - 72 - 24 = 21. Example 2. Write the value of the determinant ■1-6 2 -3 1 -2 Here some of the numbers a, b, etc. are negative, but of course this fact introduces no new difficulty, except that care must be taken about the sign of each product. The result is (- 1) (- 3) (- 1) + (- 6) - 5 - 1 + 4(- 2) ■ 2 - 1 (- 3) . 4 - (- 2) - 5 (- 1) - (- 1) (- 6) - 2 = - 3 - 30 - 16 + 12 - 10 - 12 = - 59. EXERCISES Write the value of each of the following seven determinants : 3 6 7 2 13 4 3 7 12 3 2 3 4 3 4 6 4. 7. 5 -3 -2 3 -1 7 -6 6 -3 X y 1 3 2 -1 2 -1 5 -6 6 1 4 -2 1 z 4 1 1 a a^ 1 b P a h 9 h b f 9 f ELEMENTARY APPLICATIONS 37 8. Prove that interchanging two adjacent rows (or two adjacent columns) of the terms of a determinant changes the sign of the determinant; that is, h «i «1 h "i h «2 = — % h "8 = + K «8 «2 h "2 «3 "1 etc. (Note that the same statement can be made about a determinant of the second order as well.) 9. Prove that if any two rows (or two columns) in a determinant are identical, the determinant equals 0. 10. Prove that ttjW b^n CjU «2 K ^ % h "s that is, if a number is a factor of every term in a single row (or column) of a determinant, it is a factor of the determinant. 11. Prove that a^ b„ G„ 12. A minor determinant is a determinant obtained from a given one by suppressing all the terms in any one single row and also all those in any one single column. Thus, obtained from is a minor determinant by suppressing all the terms in3<the first row and all those in the first column. This minor determinant will be symbolized by A^ and is said to " correspond " to the term a^ at the intersection of the row and the column struck out. Similarly, to 5j corresponds Bj= ^ ^ and soon. Prove that aj^, — 6j5j + c,Cj=a.j^j — a242+a3^3= and also that a^B^ — a^^ + aji^ 0. o «8 "1 \ «2 h 38 THE ELEMENTARY FUNCTIONS y- *32. Solution of simultaneous linear equations in three vari- ables. If we solve, by the method of elementary algebra, the system of equations '' a^x + \y + c^z = (ij, (1) a^x + \y + c^z = d^, (2) ^a^x + l^y + c^z = d^, (3) we obtain the values (after dividing out a common factor in numerator and denominator) ^ ^ d^h^c^ + \c^dg + Cjbgd^ - dgb^c^ - b^c^dj^ — c^\d^ ^ a-fi^c^ + \c^a^ + HhH — HK^i — hV\ — ^3*1*2 (Al-tQ/nCn ~\~ C(/-iOnUin —\~ l/-< C&q wa ttottnCl CX/oCnt^-t CnLt-tiA/n and a similar fraction for z. Each numerator and denominator can be written in the form of a determinant, as follows: x = where the denominator is in each case the determinant formed by copying the coefficients of x, y, and z in the equations (1), (2), and (3); and the numerators are formed in an exactly analo- gous way to that used in solving two equations in two variables. Example. Solve the set of equations '2x-Zy + 5z = -\0, (1) x-^2y-z = 9, (2) lbx-y + iz=7. (3) Writing down the values of x, y, and z according to the method just explained, d. h Cl d. h h d. h h «1 h h «2 b. h «3 bz h «i d. «1 «2 d. ^2 «3 ds «8 «1 \ «1 «2 h C2 «3 bs «3 «1 *1 d, ^2 \ d. «3 bs ds «1 b^ «i «2 b. «2 «3 bs «3 -10 -3 5 9 2 -1 7 -1 3 2 -3 5 1 2 -1 5 -1 3 60 + 21 - 45 - 70 + 10 + 81 _ - 63 12 + 15-5-50-2 + 9 ~ - 21 ELEMENTARY APPLICATIONS 39 -21 2 -10 5 1 9 -1 5 7 3 -21 2 - 3 -10 1 2 9 5 - 1 7 54 + 50 + 35 - 225 + 14 + 30 _ - 42 - 21 ■ ~ - 21 ' ^2, 28 - 135 + 10 + 100 + 18 + 21 _ 42 - 21 ~ - 21 :-2. Therefore x = d, y = 2, z = — 2 is the solution of the problem, and in fact this set of values satisfies all three equations. EXERCISES Solve by the method of determinants : 3a; -2/ -2s =-7, [x + y + 3z = i. 3. [ix + iy + ip = 38. . + 1 + 1 = 20, - + __«_8. ( ax + by — cz = 2 ah, 4. ■i.by + cz — ax = 2 be, \^cz + ax — by = 2 ac. Ua + b)x+(a — 6)s = 2Sc, 5. \(b + c)y+(b — c)x = 2ac, [ (c + a) » + (c — a) y = 2 a&. *33. Linear equations in three variables do not lend themselves to graphical representation in the same way as linear equations in two variables, because a point is determined by a pair of values, — its abscissa and ordinate. In three-dimensional geome- try, however, a set of three coordinates may be introduced, corre- sponding to the three variables x, y, and z, and then any equation containing these variables may be given a graphical interpretation. This is found very useful in all higher study of geometry, but we shall not make any use of it in this book. The Quadratic Function 34, The next application of the graphical representation to ele- mentary algebra arises when we consider the quadratic function. We have drawn the graphs of a number of such functions (see 40 THE ELEMENTARY FUNCTIONS Exs. 14-23, p. 22), and in every case the locus was a curve of a particular form, which we called a parabola. We assume that this is always the case, just as we assumed that the graph of every hnear function is a straight hne, without being able as yet to prove the correctness of the assumption.^ Stated in the form of a theorem the assumption is as follows: The graph of every function of the second degree, that is, of the form aa? + 'bx + c (or, what is the same thing, the graph of every equation of the form y = ay? + 6a; + c), is a parabola. Knowledge of this fact will be useful in checking the accuracy with which the coordinates of points on the curve have been com- puted ; for if the points, when plotted, do not lie on a curve of the general shape that we have learned to recognize as a parabola, a mistake has been made, either in the computing or in the plotting. 35. Intersection of parabola and straight line. Let us draw in the same figure the graphs of the following equations : 'y = a;^ — a; — 4, (1) Vy = 1x. (2) The tables of values for determining points on the graphs are (1) (2) I' X y = x^ — X -4 -4 1 -4 2 -2 3 2 4 8 -1 -2 -2 2 \ -4i -i -31 Check. Fig. 24 Now the graph of (1) contains all points whose coordinates {x, y) satisfy that equation, and the graph of (2) contains aU 1 It is to be hoped that the student has not failed to notice that the word " parabola" has as yet no precise meaning at all ; it is, so far, only a word used in order to give a name to a more or less vaguely defined curved line. For this reason, if for no other, it would be impossible at this point to prove the assumption above. ELEMENTARY APPLICATIONS 41 points whose coordinates satisfy the second equation. Hence any point whose coordinates satisfy both, equations must be found on both graphs, that is, must be one of their points of intersec- tion. By inspection of the figure we see that these points are approximately (—1, — 2) and (4, 8) ; and in fact these are the exact points of intersection, for these pairs of values wUl satisfy both equations. EXERCISES Find in this way the points of intersection of the following : f 2/ = x^ + X - 3, ^ry = 2.x^ + 2x-l, '\y= — x. ■ 1?/ = — 3x +1. f2/ = 2x^-5x-|-l, ^ r2/=6cc2-8x-12, ^' ly = X - 3. ■ ty = - X + 8. 36. Algebraic solution. Let us return to the example above and consider how it could be solved algebraically. To this end. we may substitute the value of y, obtained from one equation, in the other equation ; thus, putting 2 x, the value of y obtained from the second equation, in place of 2/ in the first equation, we have 2x = 3?— X— 4:; that is, x^ — Zx = 4. This equation is an example of what is called in elementary algebra a " quadratic equation in one unknown quantity," and the student should be able to solve such an equation.^ The result is a; = 4, or a; = - 1. Since y=2x,y=B, when a; = 4, and y = - 2 when a; = — 1. Hence the solutions are (4, 8) and (— 1, — 2), exactly as we had already discovered by the graphic method. EXERCISES Solve the above four problems algebraically. 37. The general quadratic equation. Any quadratic equation in the one unknown quantity x can be reduced to the form ax^ + hx + e = 0, (1) 1 If not, sufacient time should be taken to review this process thoroughly ; for this purpose Appendix F gives a full explanation, with examples. 42 THE ELEMENTARY FUNCTIONS which is accordingly a general quadratic equation. In this equa- tion a, b, and c may be any numbers whatever, either rational or irrational (except that a cannot equal zero). Usually, however, these numbers will be rational, in the problems with which we shall have to deal. When we have solved the equation (1), we shall have a solution for every possible quadratic equation in one unknown quantity, since (1) represents any such equation. To solve (1), we use the method of elementary algebra known as " completing the square " (see Appendix F). Dividing (1) by a and transposing, we have a;2 + -« = -£. (2) a a To " complete the square," that is, to make the left side of (2) a perfect square, we must add \-z—] : a \2a) \2a) a ! a) \2 a) that is, {a; + -— )=-—--- = J2 — 4 ac 4a2 (3) m, ( , ^ ^^"^ — Aac Therefore x -\ = ± . 2a 2a -6±V6'-4ac or x = (4) rn, . 1 4 4-1, — h + y/H^ — 4: ac , —h — y/V^—Aac Ihese two values of x, then, and , 2a 2a wUl satisfy the equation (1). The values of the unknown quan- tity which will satisfy a given equation are known as the roots of that equation. So the results (4) are the roots of the quadratic equation ax^ + hx + c^ 0. Since these are the roots of a general quadratic equation, the form (4) may be used as a symbolic representation, or formula, for the roots of any particular quadratic equation whatsoever, as the next paragraph illustrates. ELEMENTAEY APPLICATIONS 43 38. Solution of quadratic equation by formula. If we take the equation that we have solved before (§ 36), x^ —3x = i, we can write it in the form aa^ + bx + c = by mere transposi- tion, thus: x'-Sx-A=0, so that a = 1, 6 = — 3, and c = — 4. Now the roots of the equation ax^ + bx + c= are, as was just found, — b ±V62 _ 4 ac 2a hence, using the values of a, b, and c that apply to our special equation. -(-3)±V(-3)^-4-l(-4 ) 3±V9 + 16 . X = -^ '- "^-^ ^ '- = = 4, or - 1, which are of course the same results that we obtained before (§ 36). Again, take the equation x + -=3. X Here the equation is easily reduced to the form ax^ + bx + c = by multiplying both sides by x and transposing, thus : a^-3a: + l=0. By the formula the roots are 3±V9-4 3 ±V5 " = 2 = ^— (These results should be checked.) EXERCISES Solve by the formula the exercises on page 41, and also the following : 1. 3a;2-7x + 2 = 0. 5. x^ - 6x - 775 = 0. 2.7.^ + 6.-1 = 0. e.^ + ^ = ^^. cc — 1 X — 2 X + 1 3. 4a;^-a;-3 = 0. ^ ^_ 2 2-x 9 7. — = -• 4. 10x'-13a; + | = 0. .-1 x + 1 4 44 THE ELEMENTAEY FUNCTIONS 39. Solution by factoring. Suppose the quadratic function aoP' + bx + c can be resolved into two rational factors, thus : ax^ + bx + c = a{x — a)(x — ^). (1) Now a product can equal zero only if one of its factors equals zero ; hence, if aa? + bx + c=^0, that is, if a{x — a) (x — 0)= 0, then either „ X — a = or a; — /3 = ; that is, a; = a and x = ^ are the only values of x that can satisfy the equation ax^ -\-lx + c = 0. This means that a and /3 are the roots of the equation ax^ + 6a; + c = 0. Example 1. Let us take once more the equation a;2 - 3 a; - 4 = 0, which we have already used several times. The factors of a:'' — 3 a; — 4 are (X - 4) (a: + 1). Hence a; — 4 = and a; + 1 = are the only possible ways in which our equation can be satisfied. Thus the roots are 4 and — 1, agreeing of course with the results already found. Example 2. 5 a:^ - 2 a; - 3 = 0. The factors of 5 x^-2 a;-3 are (5 a: + 3) (x-1); hence, if 5x^-2 a; -3 = 0, then either 5a; + 3 = 0ora: — 1 = 0. Hence a; = — | or x = 1. This method is evidently the simplest way that we have met of finding the roots of a quadratic equation. The only difiiculty is that it is often impracticable to find the factors of the given quadratic function. In that case the previous method, that of the formula, should be used. EXERCISES Solve the following quadratic equations by factoring : 1. x'' + 3a; + 2 = 0. 5. 2x2 - 5x + 2 = 0. 2. x^ + i a; - I = 0. 6. 3 a:^ - a; - 2 = 0. 3. a:''-6r + 6 = 0. 7. 3 «'' + 2a; - 1 = 0. 4. a;' -5a-- 6=1 0. 8. 4x2 - 203- + 9 = 0. ELEMENTAKY APPLICATIONS 45 40. Summarizing the results thus far obtained, we have seen that a quadratic equation iu one imknown quantity can be solved (1) by " completing the square "; (2) by the formula ; (3) by factor- ing. Of these only the third or the second is suited to practical use. Of course the imknown quantity need not be x, as it was in the examples we have considered ; on the contrary, the student shoiild practice solving quadratic equations in various letters, until the process becomes thoroughly familiar. A number of prob- lems are added, as an aid in this work and for review purposes. EXERCISES Solve and check (algebraically) : 1. 2t' + t-2 = 0. 6. s^-10s-M6 = 0. 2. 5y^-3y-2 = 0. 7. 3 _ 2 ^^ r — <o r — o 3. 7)5^ -8!! + 8 = 0. y 21 1 4. 92/^-ll2/-12 = 0. ^' im'^2E^j^~^' 4 111, 5. r-|-3 = T- 9. q i + z. + q-— = 4. r — 1 1 — s^l — sl + s Solve the following pairs of simultaneous equations both alge- braically and graphically : 10 ly = 6. ■ I: ^-3a;, ry = x'-3x + 4:, nj = x'-3x, ^'■{2/ =2. ^^•\2x + y = 2. ry=.x% ^g ry = 2x^-6x + l, 12- •[y_a;-2=0. \x-h2y = 3. 41. Has every quadratic equation a solution? Let us try to solve this quadratic equation : a;^ - 2 a; + 2 = 0. (1) Applying the formula, we get 2 ± V4 - 8 2 ± V-4 x = = t: ■• \'^) 46 THE ELEMENTARY FUNCTIONS Now (2) gives an expression for the roots of the equation (1) ; but what is the actual value of that expression? V4 is equal to 2 or to - 2, since 2^ = (- 2)^* = 4 ; but what is the value of V— 4 ? Not — V4, as is sometimes hastily concluded, for the simple reason that — V4 = — 2, and (— 2)* is + 4, not — 4. As a matter of fact there is no positive and no negative number whose square is — 4. Hence the " solution '' (2) for x is merely a formal ex- pression, without any value among aU the numbers that we are acquainted with. The same thing is evidently true of any indi- cated square root of a negative number. We shall call such expressions " complex expressions." They are not, for us, subject to laws of operation, like numbers ; but whenever they enter into a problem, we shall consider that the problem has no solution. It is indeed found useful in higher mathematics to introduce new numbers such that their squares are negative ; these new numbers are called " complex numbers," and simple, practical rules are adopted for performing arithmetical operations upon them. We shall not need to use them, however, and so we shall, as stated, consider as unsolvable any problem which leads to complex expressions. We accordingly conclude that our quadratic equation a;2-2« + 2 = is unsolvable, because its roots are complex expressions. EXERCISES Show which of the following equations are unsolvable, that is, have complex roots : 1. x'' - a; + 1 = 0. 4. a;' + 3 a; + 2 = 0. 2. 2a;^-3a;-3 = 0. 5. 5a;^ + 3a; - 1 = 0. 3. 2x''-5x + 6 = 0. 6. 10a;2-7a; + 2 = 0. 42. Test for solvability of a quadratic equation. The situa- tion described in the last paragraph can only happen when in — b± V&2 — 4 ac the formula , for the roots of the equation, a ELEMENTARY APPLICATIONS 47 negative number appears under the square-root sign. But the quantity under the square-root sign is b^ — 4:ac; hence the roots will be complex when h^ — iac is negative. On the other hand, if 6^ — 4 ac is positive, the formula gives two different values for the roots of the equation, these values being rational if 6^ — 4 ac equals a perfect square, and irrational if 6^ — 4 ac is not a per- fect square (on the assumption that a, h, and c are all rational numbers). One .)ther possibility must not be overlooked, — that is, J^ — 4 ac may equal zero ; in this event the formula takes the very simple form > and thus there is only one root, ^ ^ 2a " 2a We describe this situation by saying, " The roots of the equation ax^ + hx + c = Q are equal when 6^ — 4 ac equals zero." Summarizing, we have the following facts: ' >0, the equation has two unequal roots. If 6"— 4ac J =0, the equation has equal roots. < 0, the equation is unsolvable (has complex roots). The quantity 6" — 4 ac is called the discriminant of the quadratic equation ax^ + 'bx + c = (i, since it enables us to determine the nature of the roots of the equation. EXERCISES Determine the nature of the roots of the following equations (without actually finding the roots) : 1. a;2_3a;+l=0. 6. 4x=' -12a; -f- 9 = 0. 2. a;^ — 4x4-4 = 0. T-Sas^-I-Sx — 4 = 0. 3. 2x^-x-l=0. 8. ^x'' -I- 5a; -f 3 = 0. 4. a;'*-f-x=l. 9. 8a;''- 8x -I- 2 = 0. 5. 3x^- 5a; -I- 6 = 0. 10. §x' - |a; -f 1= 0. 43. Graphical interpretation of discriminant test. Let us return to the problems concerning the intersection of a parabola and a straight line, from which we were led to the study of quadratic equations. Take again the example on page 40, § 35. 48 THE ELEMENTARY FUNCTIONS Example 1. X — 4:, (1) (2) We saw that by eliminating y between the two equations there results the quadratic equation in x, a;2-3x-4 = 0, (3) which must be satisfied by the abscissas of the points of intersection of the graphs of (1) and (2). Now the discriminant of this equation is equal to 9 — 4(— 4) = 25, a positive number, and accordingly the equa- tion has two unequal roots. Therefore there exist two abscissas (the roots of (3)) which determine points on both of the graphs ; that is, the graphs intersect in two different points, as of course we proved by actually finding the points (4, 8) and (— 1, — 2). But the method of this section enables us to prove that the graphs really intersect, without actually find- ing the coordinates of their point of intersection, by merely noting the value of the discriminant 6"— 4 ac. { Example 2. Let us consider the graphs of the two equations fj/ = 2 x2 _ 3 a: _ 4, (l) x-y-Q = 0. (2) Substituting the value of y from (2) in (1), we have 2a;2_42; + 2 = 0, that is, z^ - 2 X -1- 1 = 0, (3) which must be satisfied by the abscissas of the points of inter- section of (1) and (2). In (3), ft^ — 4 ac = ; hence there is only one root of (3), which means that j-jq 25 there is only one point common to the parabola (1) and the straight line (2). In other words, the line is tangent to the parabdla. The figure, of course, verifies this conclusion. Example 3. Let us find the points of intersection of 2/ = 2a;2-3x-4 and y = 2 a; — 8. Substituting the value of y from (2) in (1), we obtain 2x2-5x-l-4 = 0, (1) (2) (3) which must be satisfied by the abscissas of the points of intersection of (1) and (2). But in (3), i^ _ 4 „£ = 25 - 82 = -7; hence there is no ELEMENTAEY APPLICATIONS 49 solution to (3), which means that there is no point common to the parabola (1) and the straight line (2). Moreover, a drawing of the two graphs verifies this conclusion, as of course it must. This verification is left to the student. SuMMAEY. Given two equations, representing a parabola and a straight line, after eliminatiag one of the variables from the two given equations we obtain a quadratic equation in one unknown quantity (the equation (3) in each of the above examples), whose roots are the abscissas (or ordinates) of the points common to the two graphs. If the discriminant of this equation is positive, the graphs iatersect in two distinct points ; if the discriminant equals zero, they have only one poiut in common, which means that the liue is tangent to the parabola; and if the discriminant is negative, the line does not meet the parabola at all. EXERCISES Determine by the discriminant test, before drawing the graphs, whether each of the following pairs of lines will intersect, be tangent, or fail to meet. Verify by drawing the graphs. ry = 2a^-x-3, ^ Cy = o^-2x+l, '\y=Zx. ■l2/ = 0. ry = a? + x+l, ry = 3x' + Bx + 3, ^•\y = 3. '"•lcc + y = 0. , ry = x'-5x-3, ry = x' + x+l, ^■\y=-3i. ■Uy-3 = 0. ry = x^-ix + 2, , ^2 ry = ar' + 5aj+l, \2x + y+l=0. '1^ + 3 = 0. ry = of + 3x+l, ry = x^ + x, ^■iy = ix-3. '^•ly-2 = 0. ^- [y = 0. 12/ + 2 = 0. ry=:x^-2x-3, jg ry=3x^-4:X + 2, ■^^ |y + 4 = 0. ■ \x + y + 3=0. ry=2x'-3x + 8, ig ry=ix'-12x + 9, ^•\y = 5x. ■l2/ = 0. 50 THE ELEMENTAEY FUNCTIONS 17. 18. 19. fy = 4:x' -12 X + 9, Cy = ix'-12x + 9, 12/=- 2. \y = x^ ■{ ■{ 2a; -3. y = x', X + y = 0. y = 3?-l, X + y + 3 = 0. 23 24 25 { 2/ = 2x^-3x + 3, a; + 2/=l. = 2a:'=-3x + 3, -y+l=0. 2/ = 6a;^- 3a; -2, 2/ + X = 6. 3/ = 4x2-3x-2, 2/ = X - 3. y = x'' + X + 1, 1 44. Construction of a tangent to a parabola. This discriminant test is especially useful for discovering tangents to a given 'parab- ola (and also to certain other kinds of curves, as we shall see later). Thus, let us draw the parabola y = a;2 + a; + 2 (1) and the straight lines y = 3 X + c (2) for several values of c. Fig. 26 shows the lines for c = 0, c = 1, c = 2, and c = — 1. We observe that the line for c = 1 seems to be tangent to the curve ; to prove that it actually is so, we take equations (1) and (2) simultaneously, and, substituting «/ = 3 x + c in the equation y = x^ + x + 2, we get x2 - 2 a; + 2 - c = 0, (3) which has for its roots the abscissas of the points of intersection of (1) and (2). If (2) is tangent to (1), the roots of (3) will be equal, that is, 62-4acwill equal zero. But &2_4«c = (- 2)^— 4(2 — c) = 4c-4, which will equal zero when c = l. Hence the value c = l does in fact make the line (2) tangent to the parabola (1), just as was indicated by the graphical solution. We have thus found a line, namely, the line y = 3 X + 1, which is tangent to the parabola y = 3?--\-x-\-1. Fi6. 26 ELEMENTARY APPLICATIONS 51 Next, let us try the same parabola (1) with the lines y = 2x + c. (2') On drawing the lines (2') for several values of c it will be seen that we have, as before, a set of parallel lines ; but these make a smaller acute angle with the X-axis. Fig. 27 shows the lines for c = 0, c = 1, c = 2, and c = 3. The line for c = 2 seems to be tangent to the parabola, but we apply the discriminant test as above, to verify this conclusion. Eliminating y by substituting its value from (2') in (1), we get ic2_ic + 2-c=0. (3') The discriminant of (3') is 4 c — 7, which will equal zero when and only when c = |. This shows that the Mne y = 2x + ^ is tan- gent to the parabola, while the -^la. 27 line y = 2 a; -I- 2 is not actually tangent. This illustrates again the fact that the graphical method cannot always be relied upon to give accurate results. Of course this is an unavoidable defect of graphic methods in general, whereas the algebraic method is precise. EXERCISES 1. Eind for what value of c the line y = — a; -|- c is tangent to the parabola y = x^ + x + 2. 2. Answer the same question for the parabola y = ar' — 4a; — 3 and the line y = ^x + c; for the line y =— ^x + o; for the line y = c. 3. The same question for the parabola y = 2x^—Zx + 2 and the line y = mx. ^ns- m = 1 or - 7. 4. For what value of k will the line y = mx + k(m being a fixed number) be tangent to the curve y = x^ ? 62 THE ELEMENTARY FUNCTIONS 5. Find the coordinates of the point where the tangent line of Ex. 4 meets the parabola. Ans. 2'TJ- 6. Find the coordinates of the point where the tangent line of Ex. 4 meets the F-axis. , /. rrv'\ Ans. [0, - — j- Draw a conclusion from the results to Exs. 5 and 6. *45. Sets of curves. In the equation y = 3 a; + c of the example in § 44 the number c was understood to be capable of assuming any value we chose, thus giving rise to a set of straight Hues. Such a general number as c in this equation, which can be given any particular value desired, is called a parameter. The presence of a parameter in an equation, then, indicates that the graphical representation of the equation wiU consist of an unlimited set of lines, straight or curved according to the nature of the equation. In § 44 we considered the prob- lem of discovering which one of a set of straight lines is tangent to a given parabola, but the dis- criminant test can equally weU be used to discover which one of a set of parabolas is tangent to a given straight line. Example 1. Cy = x^ ■ ■ ix + c, (1) (2) Fig. 28 Fig. 28 shows the parabolas for the values 2, 3, 4, and 5 of the parameter c. The curve for c = 4 seems to be tangent to the line (2), that is, to the X-axis ; and the discriminant test verifies the result indicated by the figure. For, substituting in (1) the value y = from (2), we get • 4 X + c = 0. (3) The discriminant of this is 16 — 4 c, which evidently equals for c = 4, and only for that value. ry = 2x^ + kx + 8, Example 2. < Putting y = in (1), we have 2 x^ + ifcx + 8 = 0. (1) (2) (3) ELEMENTARY APPLICATIONS 53 If (1) is tangent to (2), the discriminant of (3), which is k^ - 64, must equal 0. This will happen when ^ = ± 8, giving two curves of the set (1) that are tangent to (2), that is, to the X-axis. (If k is any number > 8, the discriminant of (3) will be positive, and the cor- responding parabola will meet the X-axis in two dis- tinct points ; while if k is any number between — 8 and 4- 8, the discriminant will be negative, and hence the corresponding parabola will not meet the X-axis at all ; if k<- 8, there will again be two intersections.) Draw the figure on a large scale. Fig. 29 EXERCISES Find, by means of the discriminant test, for what values of the parameter k the curve and straight line will be tangent in each of the following exercises. Draw the graphs for these and also for several other values of k. "^y = a? + 5x + k, fy = x' — .lx-\-k, 0. ' \y = 2. 2. 3. 5. 6. 5, ' y = a^ — kx + i, i = o. y = lj(^ — kx + 12, Ly = 3. "'\y^2x-3. *46. Sum and product of roots of a quadratic equation. The roots of the quadratic equation am? + l)x + c = iy = Cy = kx^ — 3 a; — \y = x + l. {y = kx^ — a; -f I, 2/ = are — h -f V62 _ 4 «c , X-, = and 2a The sum of the two is and their product is h - V&a -4tac 2a 26 -b a X^Xo — 62. 2a a ■ (h^ — A ac) _4: ac _ c 4a2 ""4a2~a' (1) (2) 54 THE ELEMENTARY FUNCTIONS The results (1) and (2) enable us to write down hy inspection the sum and the product of the roots of a quadratic equation. Example 1. 2 a;" _ 5 a; + 3 = 0. Here = - , which is accordingly the sum of the roots ; and - = - , which is the product of the roots. We can verify this by actually finding the roots 2 x" — 5 a; + 3 = (2 a; — 3) (a: — 1) = 0. Hence the roots are x = 1 and a; = 5. Their sum is in fact J, and their product ^, as just found. Example 2. Use the results (1) and (2) to find the values of x^ + x^ and Solution. in terms of a, We have h, and 1 + ^2 -b a and X, x„ _ c (1) (^> To get Xj2 + x|, begin by squaring (1) : Xi^ + 2x1X2 + x| = -. Multiply (2) by 2, 2 x,x„ = — . a Subtracting, x ^ + x^ = ^ . (•3) To get — + — > which equals ^ ^ ^^ , we need only to divide (3) by Xj Xg Xj Xg x,^xi, that is. by '-^ . This gives ^^tfl = Ijzl^ ^f.^ '^-^ac EXERCISES 1. Write down the sum and the product of the roots of each of the following equations, and verify the results by actually finding the roots : (1) x''- 335-4 = 0. (5) 3cc^-lla: + 6 = 0. (2) a;^ + 10a; + 9 = 0. (6) 2a;^ - 3a; = 0. (3) 2a5»-3x-2 = 0. (7) x'' + jsx + ? = 0. (4) 5a;2 - 6a; + 1= 0. (8) aiV + 2ma; + 1 = 0. 2. Write down the sum and the product of the roots of each of the following equations : (1) mV + 2 (m - 2 a)a; + 1 = 0. (2) {h^ + aV)x^ + 2 a?mx + a\l - b") = 0. (3) (a + b + c)a^+(a + b- c)x + a' + 6^ - c^ + 2 aj = 0. ELEMENTAEY APPLICATIONS 55 3. Find two numbers whose sum is 22 and whose product is 117. 4. Find two numbers whose sum is 47 and whose product is 420. 5. Find the value of each of the following expressions in terms of a, b, and c (it being understood that x^ and x^ are the roots of the equation ax^ + ftx + c = 0) : (1) xl + xl (4) xixl + xlxl. (x, , X, (2)xt + xl .5. J_,^ ^H^/^ (3) x',x?, + xlxl ^"^ x-^x^ "^ x^xl ■ *47. Factor theorem for quadratic equatibn. On page 44 we learned that any factorable quadratic equation can be solved by- inspection. We used the principle that " a product of two factors wiU equal zero when and only when one of the factors equals zero." The result of that investigation may be stated as follows : If x — a is a factor of the function aoP- + 6a; + c, then a is a root of the equation ax^ + 6a; + c = 0. (I) The object of this paragraph is to show that the converse^ of this theorem is also true, that is, to prove : If a is a root of the equation ax^ + 6a; + c = 0, then x — a is a factor of the function ax^ + bx + c. (II) The statements (I) and (II) are called the factor theorem for the quadratic equation. Proof ot (11). ax^ + bx + c = a(x^ + -x + -\ \ a a/ If a and P are the two roots" of the equation ax'^ + hx + c = 0, then, by the preceding paragraph, a and aR = -• a Therefore ax^ + bx + c = a^x'^ - (a + /3)x + ar/3] = a (a; - a) (x - /3). This proves the theorem. 1 The converse of a theorem is a new theorem in which the hypothesis of the original theorem becomes the conclusion of the new one, and vice versa. Thus, the theorem "If the sides of a triangle are equal, its angles are equal" has for its converse the theorem " If the angles of a triangle are equal, its sides are equal." Here both the original theorem and its converse are true, but that is not always thfe case. The student should think of examples in which a theorem is true but its converse false. " p will equal a in case 6^^ — 4 oc = 0, but the proof is valid in this case also. The case b^ — iac<0 cannot happen under the hypothesis we have made. 56 THE ELEMENTARY FUNCTIONS A second proof. Another method of proof of this theorem can be worked out by the student as follows : Divide ax^ + bx + chj x- a, using the ordi- nary process of long division ; the quotient will be found to be a:c + aar + b, with the remainder aa^ + ba + c. Now, by the hypothesis, a is a root of the equation ax^ + 6x + c = ; hence aa^ + ba + c = 0; that is, the remainder when ax^ + bx + cis divided by x — a is 0. This proves the theorem. A third proof. This theorem can also be proved in the following manner, which is especially instructive, and which, moreover, leads us to another result of importance. Begin with ax^ + bx + c = a(x^ + -x + -j a,a before. This time we will " complete the square " of the first two terms in the parenthesis, that is, add ( — ) , so as to make a perfect square. We then have \ a a/ L a i a" i a^ aA _ r _ -b-W-iac l r _ -b + Vb^-iac l = a (x — a) (x — /3), - 6 - VJ2 -iac , o -b + Vi2 - 4 ac , , 4. - ^i. where a = and p = are the roots of the 2a 2a equation ax^ + bx — c = 0. This completes the proof of the theorem. But it does more than that. The form (A) shows that the function aso^ + bx + c can be written as the difference of two squares whenever 6^— 4ac is positive ; while if S^ — 4 ac is negative, the form (A) will be the sum of two squares; and if 6^— 4ac equals zero, the function aoa^ + bx-\-c is shown to be itself a perfect square. It follows from this that in each of the last two cases (6^ — 4 ac ^ 0) the sign of the function aa^ + bx + c wUl be the same as the sign of a for all values of x (for neither the sum of two squares nor a single perfect square can be negative). This is often a useful fact to know concerning such a function^. For example, the function a? — x + 1 will be positive for all values of x, since J^ — 4 ac< and a is positive; and the function — 2 a? + 10 x — 13 will be ELEMENTARY APPLICATIONS 5T negative for all values of x, since 6^ — 4 ac < and a is negative. In the case where 6^—4 ac>0 the result is almost equally simple, for then (A) gives us asi? + bx + c = a(x -- a) (x — /3), so that the sign of the function will be the same as that of a if both the factors x — a and » — /3 are positive or if both are negative, that is, if x>^ or <a; but the sign of the function will be opposite to that of a if a; — /3 is negative, while as — a is positive, that is, if X is between a and /3 (a < a; < /8). Example. — x"^- 5x - 6 = -l(x + B)(x + 2). Here a = — 3, /3 = — 2, and evidently the sign of the function will be negative (the same as that of a) except when x is between — 3 and — 2 (— S<x<— 2), in which case the last factor a; + 2 is negative; the factor X + S, however, is positive, so that their product is negative, and hence the function — 1 (a; + 3) (a; + 2) is positive. This result is well illustrated by the graphic representation of the function. If we take as usual x for the abscissa and y =f(x') = — x^ — 5x ^ 6 for the ordinate, the graphical representation will give the parabola shown in Fig. 30. Of course ?/ = when a; = — 3 or — 2, that is, at the points A and B (because a; + 3 and a; + 2 are factors of /(a;)) ; and we see that the function is positive (the curve is above the J^T-axis) for all values of x between — 3 and — 2, but that it is negative (the curve is below the .X-axis) for all other values of x. Y Table of values X y -6 1 -12 -1 -2 -2 -3 -4 -2 -H + i -^ + feto. Fig. 30 EXERCISES 1. Test both algebraically and graphically the signs of the following functions : (1) ar'-Saj + S. (3) a;^ - 10 a; - 7. (5)3x^-x+l. (2) x^-3a; + 2. (4) 2a;^ - 3x - 4. (6) 6a;' - 6x +1. 58 THE ELEMENTARY FUNCTIONS 2. Test algebraically in a similar manner the product and apply the information gained to make a rough sketch of the graph of the function, without computing any pairs of values exactly. 3. Test in the same way the product 2(x—iy(x — i); the product -3(x +l)\x - Sy-, the product -x{x+l)(x + 3f. *i8. Maximum and minimum values of a quadratic function. The expression (A) on page 56 enables us to draw still another conclusion. Since we have ax^ + bx + c-- 7 bV h^-4: acl \"+2^j — 4^r and since the term I x + ^r— 1 is always positive, or at least zero, the least possible value of the whole expression in the square / i V brackets will be obtained when ( x + — — ) is equal to zero : that \ 2 a/ is, x = — —- will give the least possible value of the expression which multiplies a, and therefore also the least possible value of ax^ + bx + c it a is positive. On the other hand, this value of x will obviously give the greatest possible value of aa^ -\-hx + G if a is negative, because the term I x + -— 1 is then a subtractive term, so that its least value will give the greatest value of the whole function. Example. -2x^ + i,x -1 = -2{x'^ - ^x + \) = -2[(x-i)2-ff+H Since the least value that (x — J)^ can have is zero, x = f will give the least value of that term, which, being subtractive, gives then the greatest value of the whole function. Thus, in this example the greatest value of the function is -V-, the value obtained when x = f . This agrees with the result just proved in general; namely, that x = — —- will give the greatest _ J _ 5 5 2 a value of the function, for here = = -. 2a -4 4 ELEMENTAEY APPLICATIONS 59 It is always better to go through with the complete process of reducing the given function to the form (A) for each special case, as was done in the above example, rather than to use the value x = ^— as a formula. 2 a EXERCISES Find the maximum or minimum values of each of the following functions, and verify your result by making a graph of the function : (1) x'-Sx + G. (3) -x^-6a;+7. (5) Sx'' - 6x + 5. (2) 2a;2-3x-4. (4) - 3 a;'' + a; - 1. (6) - a;^ - 10 a- + 1. Note. An easier method of finding maximum and minimum values of functions is developed in the Differential Calculus. MISCELLANEOUS PROBLEMS INVOLVING QUADRATIC EQUATIONS A few concrete problems leading to quadratic equations are here given. In solving such problems the most important thing is to be sure of understanding the problem itself thoroughly, then to trans- late the concrete language of the problem into the more abstract language of algebra. 1. The length of a rectangular field exceeds its breadth by 2 rd. If the length and the breadth were each increased by 4 rd., the area would be 80 sq. rd. Find the dimensions of the field. 2. The area of a certain square may be doubled by increasing its length by 10 ft. and its breadth by 3 ft. Find the length of its side. 3. A rectangular grass plot 12 yd. long and 9 yd. wide has a path of uniform width around it. The area of the path is f of the area of the plot. Find the width of the path. 4. A farmer sold a number of sheep for |120. If he had sold 5 less for the same money, he would have received $2 more a head. How much did he receive per head ? 5. A man agrees to do a piece of work for $48. It takes him 4 days longer than he planned, and he finds that he has earned $1 less per day than he expected. In how long a time did he plan to do it ? 6. A circular grass plot is surrounded by a path of a uniform width of 3 ft. The area of the path is J the area of the plot. Find the radius of the plot. 60 . THE ELEMENTAEY FUNCTIONS 7. A boat steams down a river 12 mi. and back in 2 hr. 8 min., and its rate in still water is 4 times the rate of the current. Find the rate of the steamer and that of the current. 8. A straight line AB, of length I, is divided at a point X in such a way that AX is the mean proportional between AB and XB (that is, AB:AX=AX: XB). Find the lengths V5-1 ^ 5 ^ AX and XB. Am. AX = — l. 2 Fig. 31 AB is said to be divided in extreme and mean ratio at X. Prove that if BA is produced beyond A by a length AX' = AX, then X'B is divided in extreme and mean ratio at A. 9. In a right triangle ABC whose hypotenuse AB equals 20, a perpendicular CD is drawn from C to AB. If BC = f AD, find the lengths of BC, AD, and AC. 10. Divide 100 into two such parts that their product shall be a maximum. 11. Find the dimensions of the largest rectangle that can be inclosed by a perimeter of 60 ft. Hint. If x represents the number of feet In the length, then 30 — x will represent the number of feet in the breadth, and x (30 — x) is to be made a maximum. 12. An isosceles triangle has the dimensions 10, 13, 13. Find the dimensions of the largest rectangle that can be inscribed in this triangle, one side of the rectangle lying in the base of the triangle. 13. A window is to be constructed in the shape of a rectangle surmounted by a semicircle. Find the dimensions that will admit the maximum amount of light, if its perimeter is to be 48 ft. CHAPTER IV INTRODUCTION TO THE TRIGONOMETRIC FUNCTIONS 49. Definition and measurement of angles. We have considered in some detail the question of measurement as applied to distances ; we have now to consider its application to angles, — a question which is of equally fundamental importance. We shall find it convenient to think of an angle as resulting from a rotation of a straight line from one position to another, the vertex of the angle being the center of rota- tion. Thus, the angle BAG (Pig. 32) may be considered as generated by the rotation of a line about A, from the initial position AB to the ter- fig, 32 minal position ^C; and the numerical measure of the angle BAC gives the amount of this rotation in terms of a conveniently chosen unit. The most common unit of measure is the degree, which is g^^ of a complete rotation about the vertex of the angle. Thus, an angle of 90°, or a right angle, is produced by one fourth of a complete revolution about the vertex ; an angle of 180°, or a straight angle, is produced by one half of a complete revolution about the vertex; an angle of 60° is produced by one sixth of a complete revolution ; and so on. Any angle can thus be measured by the amount of rota- tion from its initial line to its terminal line; and it is evident that an angle can have a numerical measure greater than 180° or even greater than 360°, for we can easily have an amount of rotation that is greater than 360°, that is, greater than one com- plete revolution. Thus, an angle of 450° would mean one complete revolution (360°) and 90° more. 50. Moreover, as rotation can take place in either of two opposite directions, we are led to make a distinction between positive and 61 62 THE ELEMENTARY FUNCTIONS negative angles. It is usual to regard angles that are produced by a counterclockwise direction of rotation as positive, and those that are produced by a clockwise direction of rotation as negative. Thus, in Fig. 33, if the rotation is from AB to AC, the angle is positive ; if the rotation is frovi AC to AB, the angle is negative. Fig. 33 EXERCISES Construct (with a protractor) angles equal to 250°, — 30°, 390°, - 430°, - 400°, 1000°, - 180°, - 350°, 490°. 51. Relation to coordinate system. For convenience of reference the four parts into vs^hich the plane is divided by the X- and Y-axes are called quadrants and are numbered in the counterclockwise order, as shown in Fig. 34. If we now consider an angle to be formed by a rotation about the origin, the initial position being the positive halt of the X-axis (OX), then the terminal Unes of all angles between ~ 0° and 90° will lie in the first quadrant; those of all angles be- tween 90° and 180° will lie in the second quadrant ; and so on. Thus, the terminal line of the angle 225° wUl be in the third quadrant, bisecting the angle between the X- and r-axes (Fig. 35). We shall say, " The angle 225° is in the third quadrant," and similarly for any angle, meaning that the terminal line of the angle is in that quad- rant when the initial line coincides with the positive half of the X-axis. Likewise, an angle of — 225° is in the second quadrant, and so on. Fig. 35 III IV Fig. 34 TEIGONOMETKIC FUNCTIONS 63 EXERCISES 1. In what quadrant is each of the following angles : 150°, 300°, 550°, - 20°, - 150°, - 200°, 1040°, - 500°, - 36,280° ? Using a protractor, construct each terminal line, and mention the number of complete revolutions in case it is more than one. 2. If a wheel makes 80 revolutions per minute, through how large an angle does it turn in 15 sec. ? in ^ sec. ? in 5 min. ? in 1 min. 60 sec. ? 3. Through what angle does each hand of a clock rotate in 1 hr. ? in24hr.? 52. Definition of the trigonometric functions. Having become familiar with the idea of an angle of any magnitude, including A Ke X Fig. 36 the distinction between positive and negative angles, we are now prepared for the study of some important numerical values that depend upon angles, or are determined by angles. 64 THE ELEMENTARY FUNCTIONS Let ^ ^ be any angle, and suppose it has been formed by a rotation from the initial position OX to the terminal position OA (Fig. 36). Let P be any point on the terminal line of the angle Q, and con- struct line segments representing the ordinate and the abscissa of P(Fig. 37). These segments are QP and OQ in each of the ^v fp^ Q'Q \P' QQ'O Fig. 37 four parts of Fig. 37. Now the ratio of QP to OQ will be the same for any one angle 0, whatever point P is taken on the ter- minal line OA. For if P' is any other point on OA, the triangle OQ'P' is similar to the triangle OQP (why ?), and hence Q'P' OQ' QP OQ 1 See Greek alphabet at end of book. TRIGONOMETRIC FUNCTIONS 65 In other words, the ratio — • that is, the ratio — ; — : > OQ abscissa of P depends for its value only upon the magnitude of the angle 6, and not upon the particular point F that we take. This ratio is accord- ingly a function of the angle 0. The same thing can evidently be said of the ratio ^— > which is the ratio — ;; > OP distance from O to P and also of the ratio -— > which is the ratio — ;; — - — - ■ OP distance from to P (The distance from to P is called the radius vector of P.) All these ratios are thus functions of the angle 0. They are called the trigonometric functions and are named as follows (see A (a) Y Q X A-^P 6 (b) y c ^ X Q k ^y~^ •^A (<=) (d) Fig. 38 Fig. 38, where x, y, and r symbolize the values of the abscissa, ordmate, and radius vector of the point P) : The ratio ordinate of P ^ gf ^ y i^ ^^^^^ tl^e tangent of the abscissa of P 0(^ X angle and is written tan 6. 66 THE ELEMENTARY FUNCTIONS _, ^. ordinate of P QP y . n i ii . c t.\. The ratio = ■^— = - is called the sine or the radius vector of P OP r angle 6 and is written sin 6. mi i- abscissa of P OQ x . ,, j ,, . « , i The ratio = — — = - is called the cosine or the radius vector of P OP r angle 6 and is written cos 6. The reciprocals of these ratios are named as follows : 1 abscissa oi P X . ■,■,■,.■, ^ ., in = = - is called the cotangent of the angle tan ordinate of P y and is written cot 6. 1 radius vector of P r . ,i j ,, j, ,, , /, = = - IS called the cosecant or the angle sin ordinate of P y and is written esc 0. 1 radius vector of P r . n , ,, » i, , /, = = - IS called the secant of the angle cos abscissa of P X and is written sec 0. These names must be thoroughly memorized and the definitions made famUiar both in terms of the words used and in terms of the corresponding lines in the figure. Angles should be constructed in various positions, and the trigonometric functions of each one obtained. In doing this it must never be forgotten that both the ordinate and the abscissa of any point are directed distances, so that particular attention has always to be paid to the question of sign. It will be noticed that some of the ratios are negative on this account. The radius vector is always to be taken as positive in determining the signs of the ratios. For example, in Fig. 38, (c), oc cos = -> and x is negative ; hence, r being positive, the ratio is r negative, which means that the cosine of an angle in the third quadrant will be negative. EXERCISES 1. Which of the ratios are negative when 6 is in the second quadrant? when $ is in the third quadrant? in the fourth? 2. With the aid of a protractor, construct the following angles and determine as accurately as possible, by careful measurement, the values of the six trigonometric functions of each : 20°, 60°, 100°, 230°, 280°, 350°, - 30°, results in decimal form. TEIGONOMETEIC FUNCTIONS 67 300°, - 1300°, 2000°. State the 3. Show that sin (-20°) = -sin 20°; sin (- 70°) = - sin 70° • sin (-100°) = -sin 100°; sin (-210°) = -sin 210°. Generalize these results to apply to any angle a ; that is, show that sin (-«) = - sin a. 4. Show that cos (- 20°) = cos 20° ; cos (- 70°) = cos 70° • cos (-100°)= cos 100°; cos (- 210°) = cos 210° ; and generalize these results to apply to any angle a; that is, show that cos (— a) = cos a. 5 . Prove that sin^ o + cos'' o = 1 . (We usually write sin'' a, cos^ a, etc. for (sin ay, (cos ay, etc.) XI MP Solution, sin a = - = , r OP X OM cos a: = - = r OP Therefore V^ x^ sin^ a = ^ and cos- a = —\ 1/2 . hence sin'' a + cos^ a = Fig. But, by the Pythagorean Theorem, Therefore y'^ + x^ = r^. sin^o: + cos" a = 1. The student should draw figures with a in different quadrants, and show that the proof holds in these cases also. 6. Prove that tan^a + 1 = sec" a. 7. Prove that cot" a + 1 = esc" a. „ T. . , , sin a 8. Prove that = tano. Note. The results of Exs. 5, 6, 7, and 8 are such important relations among the trigonometric functions that they should be memorized. 9. Construct an angle whose tangent is 2 ; find its value in degrees, with the aid of a protractor. What are the values of the other five trigonometric functions of this angle ? 10. Construct an angle whose sine is J ; find its value in degrees. Give the values of the other trigonometric functions of this angle. 11. Proceed as in Ex. 10, for an angle whose sine is — f ; for an angle whose cosine is — ^ ; for an angle whose tangent is — |. THE ELEMENTARY FUNCTIONS 53. Functions of 45°, 30°, and 60°. (I) If we construct an angle of 45°, we observe that the ordinate and the abscissa of any point on the terminal line are equal {0Q= QP, since the triangle OQP is isosceles); that is, y = a;. Therefore tan 45° = 1 and also cot 45° = 1. Now r^z=a? + y^=2y'^; hence i' = y • ^• Therefore sin 45° = -^ = and hence Since Therefore Hence yV2 V2 CSC 45° = V2. ^H Y » / / A r/ / y Xb' X O X Q Fig. 40 x = y, cos 45° = sin 45°. J__V2 V2~ 2 = V2. cos45° = — = = — -■ sec 45° (II) If we construct an angle of 30°, we find that r = 2 y, as is easily seen by producing PQ its own length beyond OX (giving the point P') and joining OP', thus completing an equi- lateral triangle OP'P; whence 0P=P'P=2y. But x^ — r^ — y\ Therefore oi? = 4:y^ — y^=3y^, and hence x = y • V3. Now the trigonometric func- tions of the angle 30° can be written down as follows : Pig. 41 tan30°-^- y.- L = » yv3 V3 4^. sin30° = ^ = ,2/ ^1, r 2y 2 cos30°-:^-^^-^l r 2y 2 TRIGONOMETRIC FUNCTIONS 69 The other three ratios can then be written down as reciprocals of these values respectively. The values of the trigonometric functions of 60° can be obtained in a similar way. The work is left as a problem for the student. 54. Functions of 180° -ff etc. (I) A very simple relation exists between the trigonometric functions of an angle 6 and those of its supplement 180° — ^. Thus, let d be any angle XOA', and let ZXO^ = 180°-^. Denoting BA by y, OB by X, and OA by r, as usual, we have, from the defini- tions of the trigonometric functions, tan(180°-6l) = ^- (1) Now take A' on the ter- minal line of 6 so that 0A!= OA, and draw A'B' perpendicular to OX. Then the triangles OAB and OA'B' are equal (being right triangles with the hypote- nuse and an acute angle of one equal respectively to the hypotenuse and an acute angle of the other), so that B'A' = BA, that is, y' = y. Further, B0 = OB'; and, since x=OB (not BO), therefore x = —x'; and r = r', since every radius vector is positive. Hence Fig. 42 y_. X y - — tan 0. — X' Therefore, from (1), tan(180°-6l) = - tan 0. Again, that is, and that is, sin (180° -6')=^ = ^ r r sin (180° - cos,(180°- cos(180°- •^)=sin(9 r sin ^ : • = — cos ; 0) = - cos 0. In the figure, was an angle in the first quadrant; but the proof holds, word for word, if ^ is in any other quadrant. (If 70 THE ELEMENTARY FUNCTIONS is in the third quadrant, for example, 180° — ^ is constructed by rotating first through 180°, then through an amount equal to —6, which will yield a terminal line in the fourth quadrant, as Fig. 43 shows. The student should draw figures illustrat- iag the various possible values of e and 180° -0.) (II) An equally simple rela- tion exists between the trigono- metric functions of an angle 6 j-ig 43 and those of the angle 90° -t- 6. Let e be any angle, and let Z XOA= 90° -|- 6. Then (Fig. 44) tan(90°-f-^) = ^ = — • ^ ' X OB (1) Now take A' on the terminal line of 6 so that OA' = OA, and draw A'B' perpendicular to OX. Then the triangles OAB and OA'B' are equal (why?), and hence BA = OB', that is, y = x'; also BO=B'A', that is, x = -y' {iov BO = —x and B'A'=y). Hence ^ = - cot 61. (2) y. X y Comparing (1) and (2), tan (90° -h 61) = -cot 6'. Similarly, Fig. 44 sia(90°-h6l) = ^ = ^ = cos6l and cos(90° + 6')=' = — sin ^. Fig. 44 shows in this case also an angle in the first quad- rant, but, as in (I), the student should construct figures showing that the proof stiU holds in case 6 is in the second or any other quadrant. TRIGONOMETRIC FUNCTIONS 71 EXERCISES 1. Prove that sin (180° + 6) = - sin ^, cos (180° + «) = - cos 6, tan(18O° + e)=tan0. 2 . Prove that tan (270° - «) = cot $, sin (270° - ^) = - cos 6 eos(270°-e) = -sin^. 3. Give the exact values of the following : sin 120°, tan 150°, sin (-120°),. tan (-150°), tan 225°, cos 240°, sin 300°, tan 300°, cos 330°, cos (-300°). 4. Show that sin 150° + cos 240° = 0. 5. Show that tan 60° + sin 240° + cos 150° = 0. 6. Show that sin 150° • cos 60° + sin 60° ■ cos 150° = sin 210°. 7. Show that cos 330° ■ cos 210° + sin 330° • sin 210° = cos 120° 8. Show that -, f"^'''^!^" = tan 60°. 1 + cos 120 9. Show that 2 • sin 120° ■ cos 120° - sin 240° = 0. 10. Show that sin 0° = 0, cos 0° = 1, tan 0° = 0. 11. Show that sin 90° = 1, cos 90° = 0, cot 90° = 0. Applications of the Trigonometric Functions 55. Having learned the meaning of the trigonometric functions, we now take up the question of their applications. These are very numerous, large parts of physics, surveying, and astronomy being indeed based entirely on the use of these functions. We shall here consider only a few of the simplest applications. 56. Solution of the right triangle. To "solve" a triangle means to find the unknown parts (angles, sides, etc.) from sufficient data. But what are "sufficient data"? Elementary geometry answers this question for us by showing what combinations of sides and angles are sufficient to determine a triangle. In the case of the right triangle it is proved that any of the following combinations is sufficient: 1. A leg and an acute angle. 2. A leg and the hypotenuse. 3. The hypotenuse and an acute angle. 4. The two legs. 72 THE ELEMENTAEY EUNCTIONS EXERCISES 1. Construct accurately, witli rule and compass, a right triangle corresponding to each of the four problems indicated in § 57. 2. State what would be "sufficient data" for constructing an isosceles triangle ; a scalene triangle not right-angled. Make con- structions for each case. 57. Notation. The notation of Fig. 45 will be found con- venient and will be used throughout this section. The angles a, y8, and 7 are opposite the sides a, h, and c respectively, and 7 is the right angle. 58. The way in which right tri- angles may be solved by the help of the trigonometric functions may be best explained by giving some examples. Example 1. In the right triangle ABC, given b = 150, a = 75°, to find the unknown parts. Solution. Since the value of the angle a is given, all of its trigonometric functions are determined, and can be computed after the manner of Ex. 2, p. 66 ; or their values may be read out of a printed " table of trigono- metric functions." From now on we shall use the latter method.^ We find tan 75° = 3.7321. But tan a = 7 > and hence we have tan 75° = 3.7321 a 150 Therefore u jhe value of = 150 ■ 3.7321 = c : = 559.8. Next, to find 1 cos a h = - = cos c 75° = 0.2588. Therefore 150 c = 0.2588, c = 150 .2588 579.6 (1) (2) Finally, as a check on the accuracy of these results, a 559.8 ...„ sin a = - = = .9658. c 579.6 1 Any one of the printed tables on the market may be used. Four decimal places give ample accuracy for this work. TRIGONOMETRIC FUNCTIONS 73 Referring to the table, we find sin 75° = .9659. Thus the results check. (A discrepancy of 1 or 2 in the fourth place may be expected when four-figure tables are used.) Another method of checking would be to use the Pythagorean Theorem, c^ = a^+b^; this also verifies the correctness of the results obtained above. 59. We observe that the method of solving such problems consists in writing down the value of one of the trigonometric functions of the given acute angle, being careful to choose a ratio that contains the given side. Thus, in the above example we started with tan a = — rather than with sin a = - > because neither c a nor c was given. In case two sides are given, the modification of the method to be used is nearly self-evident and is illustrated by the following example: Example 2. In a right triangle ABC, given a = 15, c = 20, to solve the triangle. Since the two sides a and c are known, we write down a trigonometric function which contains both a and c : - = sin a. (1) c Therefore sin a = ^^ = .75. Using the tables, we find that the angle whose sine is .75 is 48° 35' ; hence a = 48° 35' . Therefore /8 = 41° 25' . To find 6, - = cos a = cos 48° 35' = .6615. c Therefore 6 = 20 • .6615 = 13.23 . Check. tana = J = ^ = 1.134. And as 1.134 is in fact the value of tan 48° 35', the above results are checked. 60. In dealing with right triangles in practice, it will not always be convenient to turn the figure about so that the vertex of the acute angle we are using shall fall on the origin, and one side 74 THE ELEMENTAEY FUNCTIONS along the X-axis. Thus, in Fig. 46, to write down the ratios tliat form the trigonometric functions of the angle a we should consider A to be the origin and ^C to be the X-axis, when of course AC would be the abscissa, and CB the ordinate, of the point B. Then CB a . CB a tan a = — = - > sm a = = - 1 etc. AC h AB c This is not altogether an easy process, however, especially for such an angle as /3 in Fig. 46, and we shall find it more practical to think of the definitions of the trigonometric functions in terms of the sides of the triangle themselves. We shall then regard a as " the side opposite the acute angle a," b as " the side adjacent to the angle a," and c as " the hypotenuse of the right triangle." Using these terms, we can ^^^ ^g easily restate the definitions of the trigonometric functions for an acute angle in a right triangle in such a way that we shall not need to think of an X- or I'-axis at all. Moreover, since the angles concerned are necessarily acute angles, we shall have positive values for all the trigonometric functions, and so may think of each side as undirected; that is, we may take its length as positive. Careful consideration wiU show that, in whatever position the angle may appear, the values of the trigonometric functions may be stated as follows: side opposite a _a tana = : — > side adjacent a b side opposite a a sm a = £i = - , hypotenuse c side adjacent a: b cos a = = - • hypotenuse c And the values of cot a, esc a, and sec a are the reciprocals of these three respectively. This way of thinking of the trigonometric functions wUl be found useful in problems that involve right triangles, but it must not be forgotten that it applies only to right triangles. TRIGONOMETRIC FUNCTIONS 75 EXERCISES 1. Construct a right triangle with the sides 3, 4, and 5, and give the values of the six trigonometric functions of each acute angle. 2. Solve the same problem, the given sides being 5, 12, and 13 ; 8, 15, and 17. 3. In Fig. 46, read oE the values of the functions of a and of jS. 4. Compare the values of the functions of a with those of ^, and thus show that for any acute angle a the following relations hold : sin (90° — a) = cos a, cos (90° — a) = sin a, tan (90° — a) = cot a. 5. Prove that the results of Ex. 4 are true for any angle a. 6. In Fig. 47, ZACB = 90° and ZADC = 90°; show that the three right triangles formed are similar, and hence write down the trigonometric functions of the angles a and yS, each in three ways. „,, . h a a Thus, sin a = 7 = -^ = — ; ' b a p + q In each of the following prob- lems, through Ex. 22, construct the triangle determined by the given data, and make an estimate of the values of the unknown parts; then compute these values and check the results arithmetically. 7. a = 6 in., a = 30°. 8.b = 75 ft., /3 = 15°. 9. c = 1.3 ft., a = .9 ft. 10. a = 1 ft., J = 2 ft. 11. i = 2fin., c = 5in. 12. a = 67°, a = 356 ft. 13.^=88°, c = 110ft. 14. a = 15.8 mi., b = 6.3 mi,. 15. a = 72 ft., /3 = 25°. 16. c = 10 in., a = 70°. 17. c = 40 in., a = 6 in. 18. a = 1 ft., 19. a = 40°, 20. p = 16°, 21. a = 73°, 22. « = 42°, c = 2 ft. a = 1 mi. b = 23.4 ft. b = 17.3 ft. = 3950 mi. 76 THE ELEMENTAEY FUNCTIONS Each of the following problems depends for its solution upon the solution of a right triangle. In most cases merely drawing the figure will give the clue to the method to be employed. One technical term needs explanation; the "angle of elevation" of an object means the angle formed by the line of sight to the object, and the hori- zontal line, — the angle CBA in Eig. 48. Fig. 48 23. Determine the height of a tree if its angle of elevation, from a point 200 ft. away, is 20°- 24. A standpipe 100 ft. high stands on the bank of a river. Erom a point directly opposite, on the other bank, the angle of elevation of the standpipe is 29°. How wide is the river there ? 25. A rectangle is 40 in. x 17 in. What angle does the diagonal make with the longer side ? 26. In a circle of radius 5 in., how long is a chord that subtends an angle of 20° at the center ? 27. How long is the chord that subtends an angle of 1° at the center of a circle of radius 100 ft. ? 28. Eind the side and the area of a regular nonagon (9-sided polygon) inscribed in a circle of radius 16 ft. ; circumscribed about the same circle. 29. Solve the same problem for the regular dodecagon (12-sided polygon). 30. Solve the same problem for the regular 15-sided polygon. 31. Eind the radius of the fortieth parallel of latitude; of the eighty-fifth. (Assume the earth a sphere of radius 3960 mi.) 32. Eind the length of the perimeter of a regular inscribed poly- gon of 24 sides when the diameter of the circle equals 1. Solve the same problem for the regular circumscribed polygon of 24 sides. 33. Solve the problem of Ex. 32 for the regular inscribed and circumscribed polygons of 48 sides. 34. Solve the problems of Exs. 32, 33 for the regular inscribed and circumscribed polygons of 96 sides (given sin 1^° = .032719 and tan 1|° = .032737). 35. Use the results of Ex. 34 to determine an approximate value of TT. Ans. IT is between 3.1410 and 3.1428. TRIGONOMETRIC FUNCTIONS 77 61. Velocities and forces. A second useful application of the trigonometric functions is to physical problems involving velocities or forces. These are indeed, as we shall see, merely special cases of the solution of right triangles, but as they involve certain difficul- ties of their own, it is better to consider them in a separate section. 62. Suppose a body is moving with known velocity v in the direction AB, making an angle 6 with the northerly direction, as in Fig. 49. Then the problem arises, How fast is the body moving toward the east ? If AB represents the distance traveled in a unit of time, then AB = v. If A]^ repre- sents the direction north from A, the angle NAB will equal the given angle d. Drawing BC perpendicular to AN, we have a right triangle ABC in which CB = x represents the dis- tance the body has traveled toward the east in unit time ; that is, x is the '**' numerical value of the required velocity toward the east. Hence the solution of the problem is the same thing as the solution of the right triangle ABC. The side AC = y will give the numerical value of the velocity toward the north, x and y are called the eastern and northern components of the given velocity v; v ia called the resultant of the velocities x and y. Note that 3? + y^ = v^. Directions are usually given with reference to north or south as standards ; thus, the direction 20° east of north is written N. 20° E. (read " North, 20° east "), and the direction 20° south of east, S. 70° E. (not E. 20° S.). EXERCISES 1. A train is running at a speed of 40 mi. per hour in a direction 10° north of east (N. 80° E.). How fast is it moving eastward, and how fast northward ? 2. A man walks, at the rate of 3^ mi. per hour in the direction S. 36° E. How far south has he gone in 3 hr.? 3. A boat is steaming in a direction N. 70° E. at the rate of 20 knots per hour. What are the eastern and northern components of this velocity ? 78 THE ELEMENTAEY FUNCTIONS 4. A point moves in a vertical plane at the rate of 20 ft. per second in a direction inclined 53° with the horizontal. Find the horizontal and vertical components of this velocity. 5. The horizontal and vertical components of a velocity are re- spectively 30 ft. per second and 40 ft. per second. Find the resultant velocity and its direction. 6. A balloon is rising vertically at the rate of 660 ft. per minute and encounters a wind blowing horizontally at the rate of 15 mi. per hour. In what direction will the balloon continue to rise and with what velocity ? 7. From the platform of a train going at the rate of 40 mi. per hour a boy throws a stone in the direction at right angles to the train's motion with a velocity of 50 ft. per second. In what direction will the stone go, and how fast ? 8. A projectile from an 8-in. gun on a warship has a velocity of 2000 ft. per second, and the ship is moving 22 knots per hour (1 knot = 6080 ft.). If the gun is fired in a direction perpendic- ular to the ship's motion, in what direction will the projectile actually go ? 9. A man rows at the rate of 4 mi. per hour, and the rate of the current in a river is 3 mi. per hour. If he starts to row straight across at a point where the river is 350 ft. wide, how far down will he reach the other bank ? 10. If in Ex. 9 the man wishes to land directly opposite his starting-point, in what direction must he row ? 11. A hunter is traveling straight north in an auto at the rate of 20 mi. per hour, when he notices a rabbit in a field about 100 ft. away. He fires when he is due east of the rabbit. If the velocity of the shot is 1000 ft. per second, in what direction must he aim if he is not to miss ? 63. Problems involving component forces are identical mathe- matically with these problems in velocities. For example, if two forces, one of 40 lb. and the other of 30 lb., act at right angles to each other upon a point P, the effect is equivalent to that of a single force F acting upon the point P in the direction PQ, the diagonal of the rectangle PAQB, and with an intensity TRIGONOMETEIC FUNCTIONS 79 numericaUy equal to the length of PQ. F is called the resultant of the component forces PA and PB. Here, evidently, F=50 lb., and by solving the right triangle PAQ we find e = 36° 52'. This relation between the component and resultant forces is familiar to students of physics under the name " parallelogram of forces." It may be stated as follows : If a point P is acted upon by two forces, represented ia magnitude and direction by PA and PB, then the diagonal PQ of the parallelogram PAQB represents, in magnitude and direction, the resultant force. Fig. 50 EXERCISES 1. Two forces, of 45 lb. and 75 lb., act at right angles to each other upon a point. Find the direction and intensity of the resultant force. 2. The resultant of two forces at right angles to each other is 100 lb., and it makes an angle of 30° with the horizontal force. Find the horizontal and vertical components. 3. A weight of 250 lb. lies on an inclined plane whose angle is 20°. With what force does it press against the side of the plane, and with what force does it tend to slide down the plane ? (In Fig. 61, ii WW rep- resents 250 lb., then WR and WS represent the required forces.) 4. Two forces, one of 100 lb. and the other of 75 lb., act on a body, one force pulling N. 20°E., the other N. 40° E. Find the residtant force. Hint. Find the eastern and nortjiern components of each force. 64. Slope of a straight line. Another important application of the trigonometric functions is in the study of the linear equa- tion in ,« and y. We saw on page 29 that the graph of such an equation is a straight line, although we have not yet proved this to Fig. 51 80 THE ELEMENTARY FUNCTIONS Fig. 52 be true. The question we shall now consider is the determination of the angle Q which a straight line makes with the X-axis. (By "the angle which one line (1) makes with another line (2)" we mean the positive angle through which (2) must rotate to come into coincidence with (1). Avoid the expression, "the angle he- tween (1) and (2)," because that is ambiguous, there being always two angles between any two intersecting lines, unless the lines are perpendicular.) 65. Now, to find the angle which a given straight line^ makes with the X-axis, take any two points F and Q on that Hne, and suppose their coordinates to be (x^, 2/i) and (x^, y^. Draw PM and QN parallel to the F-axis, and PB parallel to the X-axis, thus obtaining the right tri- angle PBQ, in which the angle BPQ = 6. Using this triangle, tan0 = ^ PB and BQ=NQ-NB-. Fig. 53 ■■ y% - Vv while that is. PB = MN= ON- 0M= x^-x^; tan^ = ^?-^l- (A) This simple result is of great importance, as we shall see. The truth of the equation (A) is not dependent upon which particular points P and Q we choose, on the line ABPQ. Thus, if we choose P and Q as in Fig. 54, the reasoning is as follows : In the triangle PBQ, tan = ^ ^ PB and BQ=:NQ-NB = y^-y^, while PB = MO+ON = -x^ + x^ = :r^-x^, 1 The given line is assumed not to be parallel to either coordinate axis. TRIGONOMETRIC FUNCTIONS 81 the only difference being that PR is the sum of the absolute lengths of MO and ON, but x^, the abscissa of P, is not MO, but OM'^ ; that is. MO = - Therefore BQ Vi- Vi PR x^- ^1 Fig. 54 SO that formula (A) is true in this case also. The student should ex- periment further with points P and Q in various other positions on the line, and satisfy himself that in every case formula (A) is true. When the angle 9 is obtuse, the same result will again be found to hold. Thus, in Tig. 55, let P and Q be any two points on the given straight line, and let their coordi- nates be (a;^, y-^ and {x^, 2/2) respectively. Draw perpendiculars to the JT-axis from P and Q, and the perpen- dicular to the F-axis from Q, forming the triangle PRQ. Eegarding Q as the vertex of the obtuse angle 6, the definition of tan 6 is R P ^ B Q R I J / vl---^..^^^ Pig. 55 Now and RP I RP' tan 6 = ( observe that it is not — — ■ QB\ RQ RP=MP--MR = y^- 2/2, QR = NO + 0M= —x^ + x^ (since NO = — x^). Therefore tan. 6 = ^ — ^ . which equals ^ — ^ ■ x^ — ^2 *2 ^ *i Thus, formula (A) is correct in this case also. If the line is parallel to the X-axis, we shall say 0=0, and formula (A) stUl holds true. 1 A very common mistake in work of this kind is to write OM = - x^, on account of the abscissa of P (or of M) being negative. By de/imtion the abscissa is OM, and hence MO = — Xj. 82 THE ELEMENTARY FUNCTIONS In work of this kind it will be noticed that the all-important thing is to keep in mind the fact that every segment parallel to the X- or the Y-axis is a directed segment, and that abscissas are always measured from ilie Y-axis, ordinates from the X-axis. 66. The number tan 6 is called the slope of the line PQ. In words, the slope of a straight line is the tangent of the angle which it makes with the X-axis. We shall hereafter usually designate the slope by m. Formula (A) is then 1/ — y tan ff = m = slope of PQ — _ , or, in words, the slope of the straight line through two points is the difference of their ordinates divided by the difference of their abscissas (both differences being taken in the same order). EXERCISES Draw careful figures in all cases. 1. Find the slope of the line through the points (2, 3) and (5, 6); through the points (3, 1) and (— 3, — 1). 2. Find the slope of the line through the origin and the point (4, 3); through the points (- 2, - 3) and (1, 0). 3. Find the slope of the line through (2, 0) and (0, — 3); through (a, 0) and (0, b). 4. Write down the values of sin 6 and cos in each of the Exs. 1-3 above. 5. What is the slope of the line x — y = 2? (In drawing the graph you necessarily get two points on the line; hence its slope can be found.) 6. What is the slope of the line 2x — 3y = 5? of the line a! + 22/ = 3? 7. What is the slope of the line x — ni/ = 5? of the line 2/ = mx + k? ot the line ax + by + c — 0? 8. Find the angle which the line ix-i-Gy = 7 makes with the X-axis. TRIGONOMETRIC FUNCTIONS 83 9. Prove that if two straight lines are perpendicular to each other, and if the slope of one is J, the slope of the other is — |. 10. Prove that if two straight lines are perpendicular to each other, and if the slope of one is m, that of the other is REVIEW PROBLEMS ON THE TRIGONOMETRIC FUNCTIONS 1 . Prove that tan + cot 6 = sec $ • esc 0, being any angle. 1 Solution. By definition, cot = Therefore tan 5 + cot 6 = tan 6 + tand 1 tan^e + l tan tan ■, sec^'fl „ sec0 . cos 5 = - — ^ = sec e • - — ^ = sec • -r-^ tan 6 tan sin tr I cos 5 = sec 6 • y: = sec ^ • CSC 0. Q.E.D. sin 6 In problems 2-14, and a represent any angle. Prove : 2. sec fl — cos = sin ^ • tan 0. 1 — sin cos 3 • = • cos fl 1 + sin 4. sin^e — cos''^ = sin''^ — cos^fl. 5. (sin^a — cos'^a)^ =1 — 4 sin'^a eos'^a. 6 ^ + ^ =1. 1 + tan^a 1 + cot^a 7. (sin a + cos a)^ + (sin a — cos af = 2. sec e — esc 5 tan ^ — 1 Q ^ • ■ sec ^ + CSC tan e + 1 9. (cot e + 2) (2 cot ^ + 1) = 2 csc''^ + 6 cot 0. tan ^ — sin g _ secg ^®" sin»e ~l + cos(9' tan a — cot a _ 2 _ . ' tan a + cot a csc^a 12. sec'^e csc^'e - 2 = tan^e + cotf^^. 13. cot e - sec esc 6(1 - 2 sin'^^) = tan 6. 14. (3 sin a - 4 sin'a)'' + (4 cos'a - 3 cos af = 1. 84 THE ELEMENTARY FUNCTIONS In each of the following problems, a and (3 are the acute angles of the right triangle ABC, and a, b, and c the sides. 15. Prove that tan - = ^ Z a Solution. Bisect ^a and draw the perpen- dicular from B upon this bisector, producing it to meet AC (produced) at F. Then ABCF is similar to AAEF, and ta,n^ = ta,nZFBC = — = - — - ■ 2 EC a Prove : 16. sin 2 a; = cos (/8— a). 17. cos 2a = -^^ 'hr -• Q.E.D. {P>a) 18. tan 2 a = 0' 2 ah s 2 a6 19. cos(j8 — a)= — ;^- (b + a)(b — a) 20. tan- « _ io — b 2~yc + b + b 4y=12 makes with 21. Find the angle which the line 3x the A'-axis. 22. Find the angle which the line x + i/ = 3 makes with the line 3 X — 1/ = 5. Find also the coordinates of the point of intersection of the two lines. 23. The equations of the sides of a triangle are 2x — 3y = 5, X + 5y = 9, and 3 a; + 2 // = 1 ; find (a) the coordinates of the vertices, and (b) the angles of the triangle. CHAPTER V SOME SIMPLE FRACTIONAL AND IRRATIONAL FUNCTIONS; THE LOCUS PROBLEM 67. Review questions. What is the meaniag of the statement " y is a function of a; " ? What is meant by the " graph of a function " ? Classify functions according to degree. What is the nature of the graph of a function of the first degree ? of the second degree ? How can a quadratic equation be solved graphi- cally ? How can the nature of the roots of a quadratic equation be determined without solving it? Define the six trigonometric functions of an angle. State the most important relations among these functions. Define "slope of a straight line." When wiU a line have a negative slope ? 68. Rational and irrational functions. The functions whose graphs we have constructed (Exercises, p. 22) involved only the operations of addition, subtraction, and multiplication, so far as the independent variable x was concerned. Such functions are caUed integral rational functions. The most general form of such a func- tion of X is /(«)= aga;"-f aia;"~i-Fa2*"~^+ • • • +*«-i* + *»- As in Exs. 10 and 13, p. 22, the coefficients of certain terms may contain fractions, but the function is still called integral. Thus, the quadratic function y = ax^ + bx + c is an integral rational function, whatever may be the values of a, b, and c. If the independent variable occurs in the denominator of a frac- tion in its lowest terms (that is, if the operation of division has to be performed with the independent variable in the divisor), then the function is called a fractional rational function, — for example, 2 B + X X^ r> ii, 4-1, I, ^ * J ^^ + ^ y = -, y = — ■ — , y = s • t)n the other hand, - and — - — are integral rational functions of x. 85 86 THE ELEMENTARY FUNCTIONS Either of these two kinds of function is called a rational function, so that a rational function may be defined as one that involves any or aU of the four fundamental operations, — addition, subtraction, multiplication, and division. If the functional relation is such that a root must be extracted in order to arrive at the value of the function, it is said to be an irrational function, — for example, y = v a;, y = v 2 x^ + 1, y 69 ■ for example, y =^ ■ ' = -vl , y = Vuj^ — 5 X, y = \x + 'Vx. Graphs of rational functions. Example 1. Draw the graph of the function - 1 ^ sponding values of x and y, when »/ = - > is as follows : The table of corre- X 1 2 3 -^ i i -1 - 2 -3 -i y=f(x) 1 i i -i 2 3 -1 _ 1 ■J -i -2 etc. The value x = 0, it should be observed, gives no value of y at all, since division by zero is impossible. This means that the graph does not meet the I'-axis, because, for points on the y-axia, ^ = 0. But for very small positive values of x, y is very large, so that the graph is very far above the ^-axis when near the y-axis in the first quadrant. For negative values of x near to the value of y is very large numerically, but negative, so that the graph is very far helow the A'-axis when near the F-axis in the third quad- rant. Thus the curve is separated into two parts, or " branches," each of which approaches the F-axis more and more closely as the numerical value of y increases.^ Fig. 57 When a curve continually approaches a straight line in such a way that as either x or y increases without limit the distance between the curve and 1 What was just stated was'of course the converse of this, namely, that when X decreases, y increases; but the converse statement is equally obvious from the equation y =-• X FRACTIONAL AND IRRATIOKAL FUNCTIONS 87 the line eventually becomes and remains less than any assignable value, the line is called an asymptote to the curve. Thus, the I'-axis is an asymptote to the graph of the function - ■ It can easily be seen by writing the equa- 1 ^ tion in the form a; = - that the A'-axis is also an asymptote to this curve, since, as x increases without limit, y decreases and becomes less than any assignable value. The curve is called a hyperbola. a: + 1 Example 2. Draw the graph of the function X —% X + 1 The table of corresponding values of x and y, when y = , is as follows : X — 2 X 1 2 3 4 5 6 -1 -2 -3 -4 1 3 V -\ -2 No value 4 S 2 I i f i -5 7 Here the value x = gives a definite value for y (jj = ^), but x = 2 makes the denominator zero and hence gives no value of y. As in the case of the I'-axis in the preceding example, so here the line x = 2 is easily seen to be an asymptote to the graph, because, for very large values of y, X is very near the value 2. There is also another asymptote, which can be found by letting x in- crease without limit. To do this it will be found convenient to change the form of the fraction as fol- lows : divide both numerator and denominator by x, thus obtaining 1 1 -I- X -I- 1 H-- Now as X in- --2 i_2 X 12 •^'''- ^^ creases without limit, both - and - 1 X diminish and become eventually less than any assignable value. Thus, 1 + i ^ + 1 i approaches 1, that is, y approaches 1. This means that y = \ x-2 i_2 X is an asymptote. This curve is also a hyperbola. (The fact that ?/ = 1 is an asymptote could also have been discovered by solving the equation y = ^^^ for x as a function of y, which gives x = -j--^ , and the form of this fraction shows that !/ = 1 is an asymptote to the curve.) 88 THE ELEMENTARY FUNCTIONS EXERCISES Draw the graphs of the following, showing the asymptotes if there are any : 1. xy = 4. 2. xij = — 10. 1 3. y = 4. 2/ = 5. y = 6. y = '•y-^x-1 X -4 1 X + 4 X + 3 X + 5 X -2 X -1 2 x-1 8. y = 9. y = 10. y = 11. y = 12. 2/ = 13. y = 2x + S 3 X — 6 — X — 1 2^- + 5 a; 1-a; 3j-4 5-2x' 7 « — 1 a;-l 20. Prove that the graph of the function y of , a asymptotes x = — - and. y = -• C G ax + b -d x' + X - will have as 70. Irrational functions. We shall consider only such irra- tional functions as involve square roots. Several examples of irrational functions will be given, in order to bring out all the details that must be attended to in the study of this class of functions. Example 1. /(x)=±Vl + : Write, as usual, y =f(x), that is,y=± Vl + x. For any value of x that is < — 1, y will be complex ; hence no point to the left of x = — 1 can be found upon the graph. For x =—1, y = 0; and for any value of x which is > — 1, y will have two values that are equal numerically but of opposite sign, as in the following table : etc. Plotting these points and joining them by a smooth curve, we have the graph of the function (Fig. 59). The curve is symmetrical with respect to X -1 1 2 3 4 5 6 7 8 y ±1 ±V2 ±V3 ±2 ±V5 ±V6 ±v^ ±V8 ±3 FRACTIONAL AND IRRATIONAL FUNCTIONS 89 the X-axis because of the fact just noted, that for every vahie of a; (> — 1) y has two values which are numerically equal but of opposite sign, thus giving two points symmetrically located with regard to the X-axis. The curve is a parabola. The functional relation of this example can be written in the form y^ = \-\- X, or y^ — x = 1, or y^ — x — \ = (i. In any of these forms y is said to be an implicit function of x, because the func- tional relation is definitely im- plied by the equation ; when the equation is solved for y, y is said to be an explicit function of x. Example 2. f{x) ■ Fig. 59 ±vr Writing y =f(x), we see that y is complex if x^ > 1, that is, if x > -1- 1 or < — 1 . Hence the only values of x that will give points on the graph are those between — 1 and -I- 1 (inclusive). The table of values is as follows : X -1 i 1 i 1 (any negative value > — 1) y ±1 ±Vi| ± VI iVxV (same as for corresponding -|- value) Plotting these points and joining them by a smooth curve, we find that the graph seems to be a circle (Fig. 60). This is in fact the case, as wiU be shown later, y is given as an implicit function of x by the equa- tion y^ = 1 — x^, or x"^ + y^ = 1. Example 3. f{x) = ± V4 - 4 x\ If, as usual, f(x) is represented by y, then the equation can be written in the implicit form 4 a;2 -I- / = 4. Within what limits must x lie in order that y may have real values? Having answered this question, the table of corresponding values of x and y should be drawn up, as follows : Fig. 60 X ±v ±i ±1 ±1 y ±2 ± Vi ±V^ ±vi 90 THE ELEMENTARY FUNCTIONS Joining the points by a smooth curve, we have the graph of this function as in Fig. 61. It is a closed curve, symmetrical with respect to each of the coordinate axes. The curve is called an ellipse. Fig. 61 Fig. 62 Example 4. f{x) = ± Vx^ — 1, or x^ — y^ = 1, when written in the im- plicit form. What are the limitations on the values of x here ? The table of values is as follows : X ±1 ±2 ±3 ±4 ±9 V ±V3 ±V8 ±VT5 ± VI The graph of this function consists of two parts, symmetrically located with respect to the F-axis, each part being symmetrical with respect to the X-axis. The curve is a hyperbola ; its similarity in form to the graph of y = - is evident, the only difference being in the position of the curve. A characteristic feature of the hyperbola is the presence of two asymp- totes. In the case of ^ = - they were the A^-axis and the F-axis, while here X they are the lines through the origin with slopes -f 1 and — 1 respectively. The distances from points on the curve to either line become and remain less than any positive distance that can be mentioned, as the values of x or y increase without limit. This assertion will be proved later (see p. 134). Fig. 62 shows the asymptotes as well as the curve. FEACTIONAL AND IREATIONAL FUNCTIONS 91 EXERCISES Make a careful graph of each of the functions of x given by the following equations, and state in every case for what values of x, if any, the function fails to have a value. If the graph is a hyper- bola, draw the asymptotes, as near as you can get them. l.x' + 3y' = 12. S. 2x''+5y^=5. 15. 23^- 3y'+5 = 0. 2. ar" - 2 2/2 = 4. 9. f + x = i. 16. 4x2- 9 j^'' = 36. 3.x^ + 4:f = i. 10. 4y2_a; = 8. IT. ix" + 9t/ = 36. i.i/-ix = 4:. 11. 7a:2 + y2=28. 18. 4a! - 9/ = 36. 5. x^ + y^ = 9. 12. 4y_a;2 = 4. 19. 4a; + 9?/* = 36. &.3x^+f=6. 13. 4 a;" -2/2 = 4. 20. 4 x" - 9 y'' ^ - 36. 7. f -0^ = 1. 14. x^ + 3y''=5. 21. 2x' + 2y^ = 11. The Locus Peoblem 71. One of the most important uses of the graphical repre- sentation of functions is in the study of geometric loci. The word "locus" has already been used (p. 19), but before going farther it will be well to review carefully a few examples which will aid in making the precise meaning of the word clear. 1. The locus of points equidistant from two fixed points is the perpendicular bisector of the line segment joining the two points. 2. The locus of points equidistant from two fixed intersecting straight lines is the bisectors of the angles formed by the lines. 3. The locus of points at a constant distance from a fixed point is a circle about the fixed point as center, and with the constant distance as radius. 72. Definition of locus. As these illustrations remind us, the locus of points that satisfy a certain condition means the totality of points satisfying that condition. The locus must, first, contain aU the points that satisfy the condition, and, secondly, must not contain any point that fails to satisfy the condition. No statement about a " locus of points '' can be justified unless it can be shown that both these things are true of the alleged locus. Each of the three examples given above should be carefully tested to see if 92 THE ELEMENTARY FUNCTIONS it fulfills both requirements of a locus. Notice that 2 is not true if the italicized word bisectors be changed to bisector. Notice also that 1, 2, and 3 are not true if we consider points outside of the plane. What is the corresponding locus in space in each case ? 73. It was stated above that the graphic representation can be used in solving problems concerning loci. This fact has been partly brought out in the work at the beginning of Chap. II, p. 19, and those examples wiU now be considered again. Example 1. The locus of points at the distance 2 from the A'-axis is evidently the straight line parallel to the A'-axis and 2 units distant from it, because, first, all points on the line ABC (Fig. 15, p. 19) are at the distance 2 from the A'-axis, and, secondly, every point that is at the dis- tance 2 from the A'-axis lies on this line, since a point above the line ABC will be at a distance greater than 2 from the A'-axis and a, point below the line ABCynW be at a distance less than 2 from the A'-axis. (Of course, if we were not dealing with directed distances, the locus would consist of two straight lines, one above and one below the A'-axis ; but the latter is the locus of points whose distance from the A'-axis is — 2.) Example 2. The locus of points that are twice as far from the A'-axis as from the F-axis is a straight line through the origin, with slope 2. The result of this problem was stated on page 20, and indeed it is nearly self-evident that this locus is a straight line ; but it is important even in these simple cases to make sure that the essential nature of the locus problem is not lost sight of. To prove that the line P'OP (Fig. 16, p. 20) is the locus of points that are twice as far from the A'-axis as from the F-axis, it is necessary to show, first, that every point on this line satisfies the condition, and, secondly, that every point that satisfies the condition is on the line. First, if P is any point on the line, and if OM is the abscissa and MP the MP OM hypothesis ; that is, MP = 2 OM, which means that P is in fact twice as far from the A'-axis as from the I'^axis. Secondly, let P' (Fig. 63) be any point that is twice as far from the A-axis as from the F-axis. Then M'P' = 2 OM' ; hence triangle OM'P' is similar to triangle OMP ; and ordinate of P, then : tan 9 = 2, by M' M Fig. 63 THE LOCUS PROBLEM 93 therefore /.M'OP' = ZMOP. Therefore I" is on the line OP. This completes the proof of the theorem as stated. It was seen on page 20 that the equation of this locus is !/ = 2 z ; hence we have now proved that the graph of the equation y = 2 a; is the straight line through the origin with slope 2. Example 3. What is the locus of points at a constant distance of 3 units from a fixed point? Let the fixed point be chosen as origin, and let (x, y) be the coordi- nates of any point P on the locus. Then the geometric condition of the problem is Qp _ g ^-i -, Now, by the Pythagorean Theorem, OP=V^^Tf. (2) Hence Vx^ + i/ = 3. (3) This algebraic condition is thus p,„ qa equivalent to the geometric condi- tion of the problem, so that the graph of equation (8) is the locus required. (In this case the locus is a circle, from the very definition of that curve. The same reasoning proves that the graph of Example 2, p. 89 (Fig. 60), was a circle, as was there asserted. The student should state the locus problem corresponding to that example.) 74. These examples illustrate the process of obtaining equations corresponding to geometric locus problems. This process may be summarized thus : First, represent by the variables (x, y) the coor- dinates of any point on the required locus (the X- and F-axes being chosen in any convenient position). Secondly, translate the geometric condition of the problem into an algebraic condition or equation involving x and y, or either one of them alone. The graph of this equation will then be the locus required. The equation is called " the equation of the locus." To be sure, no advantage of this process is evident in cases like those just given, where the locus was easily obtained by the apphcation of well-known theorems of elementary geometry. But in many cases it is difficult, perhaps even impossible, to gain definite knowledge of the locus directly ; whereas the equation of the locus can be obtained, and then the graph of this equation. 94 THE ELEMENTARY FUNCTIONS found by plotting points whose coordinates satisfy the equation, will be the desired locus. In later chapters we shall see that many important facts about a locus can often be established by drawing conclusions from the equation of the locus. For the present, however, the derivation of the equation is the main thing. EXERCISES Draw a figure for each. 1. Find the equation of the locus of points at a distance 2 from the F-axis ; from the .Y-axis. 2. Find the equation of the locus of points at a distance — 5 from the -Y-axis. 3. What is the equation of the -Y-axis ? of the F-axis ? 4. Find the equation of the locus of points that are 3 times as far from the F-axis as from the A'-axis. What is the loous ? Prove the fact. 5. Find the equation of the locus of points that are — 2 times as far from the -Y-axis as from the F-axis. What is the locus ? Give proof. 6. Find the equation of the locus of points at a distance c from the F-axis. 7. Find the equation of- the locus of points that are at a constant distance equal to 6 from the origin. 8. Find the equation of the locus of a point that moves so as to be always equidistant from the points (2, 1) and (3, 5). Solution. Let A = (2, 1) and B = (3, 5) ; also let P= (x, y) be any point on the required locus. Then the condition of the problem is PA = PB. Each of these distances can be expressed algebraically by means of the result of Ex. 20, p. 12, which proved that the distance between two points (xj, j/j) and (xj, 2/2) is Therefore PA =^(x-2y-\-{y-\y and PB=^(x-Zf+{y-by. Since PA = PB, the correspond- ing algebraic equation is y \ ~~-~-\p(^,y) A X Fig. 65 V(x-2)2+(y-l)2 = V(x-3)■^-t-(2/-5)^ or, simplifying, 2 x -t- 8 y = 29. THE LOCUS PROBLEM 95 This is accordingly the equation of the locus. The graph is found to be a |traight line, which, as we know, is the perpendicular bisector of the line segment AB. (Of course the mid-point of the segment AB must lie on the locus ; its coordinates are (—- — > —~\, that is, (5 ' 3 ) , and these coordi- nates must therefore satisfy the equation of the locus, 2x + Sy = 29; in fact, 2 • § -I- 8 • 3 = 5 -I- 24 = 29, checking partially the accuracy of the result obtained.) 9. Find the equation of the locus of a point that moves so as to be equally distant from the points (2, 3) and (6, 1). 10. Find the equation of the locus of points equidistant (a) from the points (1, 3) and (— 1, 4); (b) from the points (3, 2) and (— 5, — 1) ; (c) from the points (— 1, — 2) and (3, — 5) ; (d) from the points (2, 2) and (0, 0) ; (e) from the points (2, 0) and (0, 2); (f) from the points (0, — 4) and (— 6, 0); (g) from the points (a, 0) and (0, b). 11. A point moves so that its distance from the origin is con- stantly equal to 10. Find the equation of its locus. 12. A point moves so that its distance from the point (2, 4) is always equal to 6. Find the equation of its locus. 13. Find the equation of the path of a point that moves so as to be equally distant from the points (— 2, — 3) and (2, — 5). 14. A point moves so thatits distance from the point (2, 0) is always equal to its distance from the F-axis. Find the equation of its locus. Solution. Let P = (x, ?/) be any position of the moving point (that is, any point on the required locus). Then the condition of the problem is AP=BP (Fig. 66). The algebraic equivalents of these geometrical quantities are AP=V(x-2y+f and BP = x. Hence the required equation is y/(x - 2)2 + y'^ = x. That is, x'^-i:X + i:-\-y'^= x\ or ^2 = 4 1 — 4, „ _ y' + 4 Y P(x.v) / X A (2,0) Fig. 66 96 THE ELEMENTARY FUNCTIONS This is therefore the equation of the required locus. Here we have an illustration of a problem whose solution could not be obtained by using any of the theorems of. elementary geometry merely, but by drawing a careful graph of the equation of the locus we can gain a very good idea of the appearance of the locus itself. It will be found to be a parabola. 15. A point moves so that its distance from the X-axis is always equal to its distance from the point (0, 4). Find the equation of its locus. 16. Find the equation of the locus of a point which moves so that its distance from the point ( 6, 1) is always equal to its distance from the line x = 2. In each of the following problems, find the equation of the locus of a point that moves according to the condition given, and draw the graph of the equation : 17. Its distance from the point (— 2, 5) is always equal to its distance from the line y = — 3. 18. Its distance from the point (4, 0) is always twice its distance from the line x — 2. 19. Its distance from the point (5, 3) is always one half its distance from the line y = 1. 20. Its distance from the point (—1, 4) is always one half its distance from the F-axis. 21. Its distance from the point (3, 0) is always two thirds its distance from the line a; = 4. 22. The sum of its distances from the points (2, 0) and (— 2, 0) equals 6. 23. The difference of its distances from the points (2, 0) and (- 2, 0) equals 3. 24. Its distance from (3, 0) is constantly equal to twice its distance from (— 3, 0). 25. The sum of the squares of its distances from the points (3, — 2) and (— 3, — 4) is always equal to 70. 26. The difference of the squares of its distances from the points (2, 1) and (— 3, — 6) is always equal to 19. ' 27. The square of its distance from the origin is always equal to the sum of its distances from the X-axis and the F-axis. CHAPTER VI THE STRAIGHT LINE AND THE CIRCLE 75. Having become somewhat familiar with the locus ^oblem in a general way, we now take up the study of the simpler types of loci in detail, in order to learn the great power of the methods of algebra as applied to geometric problems. We shall see that to certain types of equations between x and y correspond certain loci, the connection being so simple that we can tell at a glance what the locus of a given equation will be. Example 2, p. 92, proved that the locus of points twice as far from the X-axis as from the Y-axis corresponds to the equation y=1x, and that the graph is the straight line through the origin with slope 2. In the same way we are led to the general theorem : The equation of the straight line through the origin with slope m is y = mx. The proof wiU be given, although it is practically a repetition of the work of the problem just referred to. We must show, first, that if P = (x, y) is any point on the straight line AOB (the straight line through the origin with slope m), then the coordinates (x, y) of P satisfy the equation y = mx ; and, secondly, that if the coordinates of any point P'= (of, y') satisfy the equa- tion y = mx, then the point P' lies on the line AOB. Proof. First, if P is any point on the line AOB (Fig. 67), =^ = tan /.XOA = m; X that is, y = mx. 97 Fig. 67 then 98 THE ELEMENTAEY FUNCTIONS Secondly, if the coordinates of P' (x' and y') satisfy the equation y = mx, then ,/ that is, X tan Z XOP' =tanAXOA. Thereioie ZXOP' = ZXOA or else 180° + ZXOA, andin either case the point P' is on the straight line A OB. This completes the proof that the equation y = mx is the equation of the straight line through the origin with sjcpe m. EXERCISES Write down the equations of the straight lines through the origin with each of the following slopes : 1,-2, §, ^, — 5, 7, — 1, — J, V3 3 V2, Draw a figure for each. 76. As our next problem, let us find the equation of the straight line through the point (1, 3) with the slope 1. If P s (aj, y) is any point on the straight line (Fig. 68), then, by theorem (A) on page 80, the slope P(x,y) of the line is „ y-3 x-l Since the slope is equal to ^, y-3. x-l 1 "2" That is, 2 y - 6 = = X- -1, or as — 22/ + 5 = -0, which is accordingly the equation of the line AP. Fig. I EXERCISES 1. Find the equation of the line (a) through the point (2, - 3) with slope 1; (b) through the point (0, 3) with slope 2 ; (c) through the point (-1, - 2) with slope |; (d) through the point (7, 1) with slope - 3 ; (e) through the point (- 2, - 3) with slope - f THE STRAIGHT LINE AND THE CIRCLE 99 2. Find the equation of the straight line through the point (0, k) with slope m. Am. y = mx+k. This equation is called the slope-intercept form of equation of the straight line, because it shows at a glance the slope (m) and the F-intercept (/c) of the line (cf. p. 29). Any equa- tion of a straight line can be reduced to this form (if y is present in the equation) by solving for y. Thus, 3 a; — y = 2 is equivalent to 2/ = 3 a; — 2, in which form m = 3 and k = — 2. 3. By reducing each of the following equations to the slope- intercept form, read off the value of the slope and the F-intercept. Verify by the graph, as usual: x-\-y = 2, 3a5-|-4y = 5, 7a; — 3y = l, ^x + ly = Z, Ix + y = b, X — ny = q. 4. Find the equation of the line through the point (— 3, — 5) with slope m. 5. Find the equation of the line through the point (x^, y^ with slope m. ^^g y_ZLJL = ^ or y-y, = m(x-xX x — x^ " "'■ ^ " This is called the slope-point form of equation of a line. 6. Write down the equation of the line through the point (5, 3) with slope 2. 7. Write down the equation of the line through the point (— 1, 2) with slope — 1. 8. Write down the equation of the line through the point (2, — 3), parallel to the line y=^x-\-&. 9. Find the equation of the line through the origin, parallel to the line x -|- 2 y = 3. 10. Find the equation of the line through the point (1, —1), parallel to the line ^ x — 3 y = 6. 1 1 . Find the equation of the line through the point (- 5, 1), perpen- dicular to the line 2 x — 3 y = 4. Hint. The slope of the required line can be foujid from that of the given line by means of Ex. 10, p. 83. 13. Find the equation of the line through the point (1, 4), perpen- dicular to the line x — 3 y = 10. 13. Find the equation of the line through the origin, perpendicular to the line x -f- y = 4. 100 THE ELEMENTARY FUNCTIONS 14. Find the equation of the straight line through the points (3, 4) and (4, 1). Hint. Find the slope of the line by the theorem on page 80, then follow the method of the preceding exercises. 15. Find the equation of the line through the points (a) (4, 1) and (- 1, - 6) ; (c) (4, 3) and (4, - 1) ; (b) (4, - 4) and (3, 2); (d) (0, 0) and (- 6,-3); (e) (3, 4) and (-2, 4); (f) i'^v yd ^"^^ (^2' yd- Ans. '^^^ = y^^^y^, or x„ y„ 1 = 0. x^ 2/i 1 ^1 y. 1 X y 1 16. Find the equation of the line whose X- and F-intercepts are a and b respectively. j^^^g ^ ^y. — i ah 17. Find the equation of the line through the point (— 2, 3) and making an angle of 136° with the A'-axis. 18. Find the equation of the line through the point (1, 2) and making an angle of 60° with the X-axis. 77. These exercises illustrate the fact that the equation of a straight line can be found if its slope and a point on it are given. Ex. 5 brought this out definitely, and it is of fundamental impor- tance in the study of straight lines. We can now state as a theorem what we have hitherto tacitly assumed : The equation of any straight line is of the first degree in x and y. For every straight line has a definite slope (unless it is perpen- dicular to the X-axis, in which case its equation is x = k, which is of the first degree), and hence its equation is y — y^ — m (x — x-^, where (x^, y{) is any fixed point on the line. But this equation is of the first degree ; hence the theorem is proved. 78. Conversely, any equation of the first degree in x and y has for its graph a straight line. Proof. The general equation of the first degree in x and y can be written ax + by + c = 0. If J ;!t 0, we can divide by b, getting that is, y = ^ ~ 7' b b THE STRAIGHT LINE AND THE CIECLE 101 This is now in the form y = mx -V k, where m = and k= But b b Ex. 2 above proved that this is the equation of the straight line whose CL c slope is m, that is, — - 1 and whose F-intercept is k, that is, Hence • ft the equation ax + by + c = represents a straight line whose slope is c * and whose i'-intercept is , provided ft ^ 0. But if 6 = 0, the equation is simply ax + c = 0; that is. (Naturally a cannot equal 0, else the equation would not have been of the first degree.) This is the equation of a line parallel to the F-axis, so that the theorem is proved also for the case ft = 0. Note that if o = 0, the line is parallel to the .X-axis. Moreover, a A since m = — - , m = U ft in this case, so that lines parallel to the X-axis have the slope zero. This of course results also from the definition of " slope,'' as we observed on page 81. 79. Normal form of the equation of a straight line. Let AB be any straight line not passing through the ori- gin, and OiV the perpendic- ular from the origin upon the line. Let the length of OiV be denoted by p, and the angle XON by a. We shall, moreover, consider ON as being a directed line, the positive direction being from 102 THE ELEMENTAEY FUNCTIONS to N, in whatever position AB may lie. The angle a may have any value from 0° to 360° (0 ^ (u < 360). The line ON is called the normal to the line AB. In case we have to do with a line AB through the ori- gin, the perpendicular to AB through {OC in Fig. 70) is still called the normal to AB, and it is directed so that the angle m is between 0° and 180°(0Sa)<180). Evidently, under these stipu- lations, any line AB determines a pair of values o and jp ; and, conversely, any pair of values of m and j> determines a straight line. Fig. 70 EXERCISES Construct the lines for which (1) <u = 30°, _?J = 3 ; (4) o) = 200°, ^ = f (2) 0, = 165°, i? = 1 ; (5) a. = 90°, p = 5 (3) 0. = 330°, p = 2; (6) 0. = 100°, ^ = (7) o. = 180°, J? = 4 ; (8) «, = 0°, j9 = 0. (9) «) = 90°, ^ = 0. Since <» and p determine the straight line AB completely, it must be possible to express the equation of AB in terms of these values. This is easily done by considering that the line AB passes through the point iV (Fig. 71), and that its slope is the negative recipro- cal of the slope of ON; that is, it equals — : > which tan a) equals cos ft) Hence the sm o) equation of AB is COSO) y-y\ = -~ — sm<u {x-x^). (1) Fig. 71 THE STRAIGHT LINE AND THE CIRCLE 103 But 2/1 = MN (Fig. ll) = psmm, and ajj = OM=p cos tu. (1) is accordingly the same as cos ft) , y — « sin o) = -, (x — p cos eo) : sino) ' ' that is, X cos a> -\-y sin ft) — j? (sin^ft) + cos^ft)) = 0. Since sin2ft) + cos2&) = 1, the equation takes the simpler forin Of cosw + j/sinw— />=0, (2) which is known as the "normal form of the equation of a straight line." Write in this form the equation of each of the nine lines in the above exercises. 80. Reduction of general linear equation to normal form. Any equation of a straight line can be reduced to the form (2). For example, suppose we have the equation 3 x — 4 ?/ = 5, and that we wish to reduce it to the form (2). The only change that we can make in an equation that will not change its graph is to add the same quantity to both sides of the equation or to multiply both sides by the same constant quantity. Since the right-hand side of (2) is 0, we first write our equation so that its right-hand side becomes : 3a;-4y-5 = 0. (3) The only other change that can be made is to multiply both sides by the same number, say k ; 1 that is, to write (3) in the form 3kx-4:ky-5k=0. (4) 1 A precise statement of the fact used here is, If a^x + 6j2/ + Cj = (1) represents the same line as a^x + b^y + Cj = 0, (2) then O2 = tei, \ = k\, and c^ = kc^ ; or (2) can be obtained from (1) by mul- tiplying it by a constant k. This theorem may be proved as follows: The slope of (1) is — ^, and that of (2) is ^ . Hence, if the lines are the same, ^ = ^ ; that is, ^ = ^ . Also, the F-intercept of (1) is - ^ , while that of (2) is _ ^. Hence, if the Mnes are the same, — = -^; that is, -^ = -?• Therefore ^ = 2? _ ^ . Call the value of this common ratio fe, and we see that a^ = fca„ flj 6j Cj 62 = A;6i, and c^ = kCj . Q'E-d. 104 THE ELEMENTARY FUNCTIONS If (2) is the same line as (4), h must have such a value as to make each term of (4) equal to the correspondiug term of (2) ; that is, cos ft) = 3 ^, sia o) = — 4 ^, p = Zlc. Therefore cos^w + sin^w = 9 ^ + 16 Z:;^ = 25 A;^. But cos^a) + sw?m = 1. Therefore Ic^ = ^, and p = 5k = + l, the sign being determined by the fact that p is always positive. Hence only the + sign may be chosen for k. Therefore cos <» = f , sin (u = — l^, p = l, and the normal form of the equation Sa? — 4^ = 5 is accordingly |x-|2/-l=0. (5) By inspection of equation (5) we see that the length of the normal is 1, and that it makes with the X-axis an angle whose sine is — |^ and whose cosine is |-; that is, an angle in the fourth quadrant. General case. The general case is treated in a similar way. Let the equation of a straight line be ax + by + c= 0, and let it be required to reduce this equation to the form (2) of § 79. If (2) and ax + by + c=0 represent the same line, we must be able to get the one equation from the other by multiplication by a constant k. The equation akx + bky + ck = must then be exactly the same as X cos Q) + ysma>—p = 0. Therefore cos to = ak, sin a) = bk, p=— ck. Therefore cos^ea + sin^co = (a^ 4. 52^ p. Therefore ky^ = and k = a2 + 62 1 ±Va2 + j2 Since p is positive and equals — ck, the sign of k must be opposite to that of c. If c = 0, the angle <o is between 0° and THE STRAIGHT LINE AND THE CIECLE 105 180° (§ 79); hence sin to is positive, so that the sign of k must be taken the same as that of b. 81. This result may be summarized as follows: To reduce the equation of a straight line ax + hy + c = to the normal form, divide each term by ± v a^ _j_ 52^ choosing the sign of the radical op- posite to that of c, or, if c= 0, choosing the same sign as that of b. EXERCISES 1. Reduce each of the following equations to the normal form : 5x + 12y = l, x + j2/ + 5 = 0, x + y + l = 0, 2x — 3y — 8 = 0, 3 X + y = 4. Draw a figure for each. 2. Find the distance between the parallel lines x + 3y — 3 = and 2a; + 62/ +1=0. 3. Find the equation of a line parallel to the line 3x — Ay = 5 and at a distance 3 from it. (Two solutions.) 82. Distance from a line to a point. One of the most impor- tant uses of the normal form of the equation of a straight line Fig. 72 is to determine the distance from a given line to a given point. Let the equation of the given line AB be a; cos a) + y sin Q) — J? = 0. (1) Let P= («!, y■^) be the given point, and d its distance from AB; draw through P the line PP' parallel to AB; the distance from 106 THE ELEMENTARY FUNCTIONS the origin to PP' will then equal p + d. The equation of PP' is accordingly xcos<o + y Bm(i>-{'p + d)= 0. (2) Since P is on tliis line, its coordinates {x.^, y^) must satisfy the equation (2); that is, ajj cos a) + 2/i sin oj — p — d={). Therefore d = Jf^ cos <o + j/i sin « — />. (3) The result (3) may be stated in words thus : In the normal form of the equation of the given line, substitute for x and y the coordinates x-^ and y-^ of the given point; the result will he the distance from the given line to the given point. If the equation of the given line is in the form aa; + &?/ + c = 0, it must be reduced to the norfiial form ax hy : + - : + :=0. Then, substituting the coordinates of the given point (ajj, y^) for {X, y), we have ax^+by^ + c ... a= , » W iVa' + b" the sign of the radical being opposite to that of c. 83. In Fig. 72 the point P was on the opposite side of the line from the origin, so that the direction from the line AB to the point P was the same as the positive direction of the normal ON. Hence d, that is, QP, may also be considered as positive, while in case P were on the same side of AB as the origin (as in Fig. 73), the direction from the line AB to the point P would be opposite to the positive direction of the normal ON. Hence d may in this case be considered negative. Thus, for all points P on the opposite side of AB from the origin, d would be positive, and for all poiuts P on the same side Fio. 73 THE STRAIGHT LINE AND THE CIRCLE 107 of AB as the origin, d would be negative. The student should now verify the fact that the formula (3) above gives the sign of d correctly accordiug to this agreement. Every straight line thus divides the plane into two parts, which may be called the positive and negative sides of the line. The origin is, then, always on the negative side of any line (unless the line passes through the origin, in which case the upper side of the line is the positive side) (see Fig. 70). EXERCISES 1 . How far is the point (1, 4) from the line x + y + l = 0? Solution. Using the formula (4), ± Va2 + /,2 1 + 4 + 1 6 r- we have in this case d = == = — = — .3 V2 = — 4.24 + . - Vl + 1 V2 Draw the figure and note that the result thus obtained by the formula is correct in sign as well as numerically. 2. Find the distance from the line 3x + iy + 1 = to each of the points (- 1, - 3), (2, 5), and (0, - 4). 3. Find the distance from the line y — 2x = to each of the points (3, - 2), (1, 4), and (- 2, 2). 4. Find the equation of the locus of a point that moves so as to be equally distant from the lines 3a; -42/ + 5 = (1) and 5a; + 122/-6 = 0. (2) Solution. Let P = (x, y) be any point on the locus, and d^ and d^ the distances from the given lines to the point P (Fig. 74). „, , 3 1 - 4 ?/ + 5 Then d^ = —^ , 5 X + 12 ?/ — G and «2 = z^ 108 THE ELEMENTAEY FUNCTIONS Since the point P is equally distant from the lines (1) and (2), we must have either or that is, either d,-- --d. d,-- -— d^\ 3a; -42, -5 + 5 5 a; + 12 2/ - 13 6 3a; -42/ -5 + 5 5a; + 12y 13 — 6 (A) (B) (A) contains all points equally distant from (1) and (2) for which the distances are liotli positive or both negative ; (B) contains all points equally distant from (1) and (2) for which one of the distances is positive and Fig. 74 the other negative. These loci are of course the bisectors of the angles formed by the lines (1) and (2), and (A) is the one passing through the angle containing the origin. The simplified forms of the equations are and 64a; + 82/ + 35 = 14 a; - 112 y + 95 = 0. (A) (B) 5. Eind the equations of the bisectors of the angles formed by the lines 3 a; + 4 2/ = 6 and 3 a; — 4 3/ = 10 ; by the lines x — 2y + Z = Q and 2a;— 2/— 7 = 0. 6. Find the equations of the bisectors of the angles formed by the lines 3 a; — y = and a; + 3y — 4 = 0. THE STRAIGHT LINE AND THE CIRCLE 109 7. Find the equations of the interior bisectors of the angles of the triangle formed by the lines 6 x — 12 y = 0, 6 a: + 12 y + 20 = 0, and 12 a; — 5 y — 30 = 0. Show that they meet in a point. 8. Find the equations of the interior bisectors of the angles of the triangle formed by the lines 4x — 3y=5, 5a; — 12^ = 10, and 4 a; + 3 2/ = 12. Show that they meet in a point. 9. Answer the same question for the triangle formed by the lines y = 4, y = 2x, and y =—2x. MISCELLANEOUS PROBLEMS ON THE STRAIGHT LINE 1. Find the equations of the medians of the triangle whose vertices are the points (2, 3), (4, — 5), and (— 2, — 1). Show that they meet in a point. 2. Find the equations of the perpendicular bisectors of the sides of the same triangle, and show that they meet in a point. 3. Find the equations of the altitudes of the above triangle, and show that they meet in a point. 4. Show that the three points of intersection obtained in Problems 1-3 lie on a straight line. 5. Find the center and radius of the circumcircle of the triangle whose vertices are the points (0, 6), (7, — 3), and (— 2, 2). 6. Find the area of the triangle whose vertices are the points (0,0), (-2,3), and (-4,-1). 7. Prove that the area of the triangle whose vertices are the points (0, 0), (Xj, y^, and (a-^, y^) is h (x^i/^ - x^^- 8. Find the center and radius of the inscribed circle of the tri- angle whose sides are the lines 3x + 4y — 8 = 0, 4a; + 32/ + 6 = 0, and 5 X - 12 y - 13 = 0. 9. Find the center and radius of each of the escribed circles ^ of the triangle in Problem 8. 10. Prove that the medians of any triangle meet in a point. (Take the origin at one vertex and let the A'-axis coincide with one side ; that is, let the vertices have the coordinates (0, 0), (a, 0), and {b, c).) 1 An escribed circle of a triangle is a circle which is tangent to one side of the triangle and to the other two sides produced. 110 THE ELEMENTARY FUNCTIONS 11. Prove that the perpendicular bisectors of the sides of any triangle meet in a point. 12. Prove that the altitudes of any triangle meet in a point. 13. Prove that the three points of intersection found in Problems 10, 11, and 12 lie on a straight line. 14. Show that the locus of the equation (ax + bi/ + c) (a'x + b'lj + e') = consists of the two lines ax + Sy + c = and a'x + b'y + c' = 0. 15. Prove that the locus of the equation ax^ + bxij + cy'' = is a pail- of intersecting lines if i^ — 4 ac > 0, that it is one straight line if ^2 _ 4 oc = 0, and that it contains no real point except the origin if 6^ - 4 ac < 0. ie. Prove that the two straight lines bx^ — cxy + ay" = are respectively perpendicular to the lines as? + cxy + by'^ = 0. 17. Taking the vertices of a triangle as (cCj, y^), (x^, y^), and (x^, y^), find the equations of the perpendicular bisectors of the sides. Show that they meet in a point. 18. Prove that the locus of the equation ax ^- by -\- c ■\- k {a'x + b'y + c') = is a straight line through the points of intersection of the lines ax + by + c = and a'x + b'y + c' = 0. 19. Prove that the area of the triangle whose vertices are the points (»i> yi), (a:^, y^, and {x^, y^) is \ (x^y^ - x^^ + x^^ - x^^ + x^y^ - x^y^, that is, 1 ^x Vx 1 2 % 2/2 1 "=s Vz 1 The Circle 84. Several of the locus problems in the preceding chapter (for example, Exs. 11, 12, p. 95) led to circles as their result. Indeed, it is easy to determine the equation of any circle in terms of its radius and the coordinates of the center. THE STRAIGHT LINE AND THE CIRCLE 111 Let C={cc,^} be the center and r the radius (Fig. 75). If P = {x, y) is any point on the circle, then CP. P(x,y) But CP=yf{x- af + (y - ^f. Therefore {x-ay + {y-pf = r\ (1) which is accordingly the equa- tion of the circle. CoEOLLAEY. The equation of the circle whose center is the origin and whose radius is r is x-^-\-y^ = r\ (2) Fig. 75 EXERCISES 1. Draw the graphs of the following equations : le' -\- i^ = 1, x^ + f = 2, x' + f=l x' + i/' = 3.24. 2. Write the equation of the circle whose center is (a) (4, 6) and whose radius is 5 ; (b) (— 1, — 2) and whose radius is 3 ; (c) (2, 0) and whose radius is 2 ; (d) (a, 0) and whose radius is a. 3. Show, by reducing to the form (1), that the locus of each of the following equations is a circle, and construct it. (a) x'' + f-ix + 2j/+l=0. Solution. To get the standard form (1), complete the square of the terms in x and also of those in y, thus : (i2 - 4 a; + 4) + (^2 + 2 ,y + 1) = 4. That is, (x - 2)2 + (y + 1)^ = 4. This is in the form (1), with a = 2, ^ = — 1, r^ = i ; hence the graph is a circle whose center is (2, — 1) and whose radius is 2. The student may now draw the figure. (h) x^ + 2f-x-iy=2. (f) x'' + f=3(x + 3). (c) a;'^ + y^-|-16x-122/ + 84 = 0. (g) 3x^+32/'-5a;+ 24?/= 0. (d) x^ + y^ + Ax + Sy-il = 0. <h) x^-|-y^-|-16y + 36 = 0. (e) x' + f — x — 'i/ = 0. 112 THE ELEMENTARY FUNCTIONS 4. Find the equation of the circle whose center is (3, 0) and which passes through the origin. 5. Find the equation of the circle whose center is (2, — 1) and which passes through the point (— 1, — 5). 6. Find the equation of the circle whose center is (3, — 2) and which is tangent to the F-axis. 7. Find the equation of the circle which has the line joining the points (3, 4) and (— 1, — 2) as diameter. 8. Find the equation of the circle whose center is on the A'-axis and which is tangent to the lines y = 3 and x = l. (Two solutions.) 9. Find the equation of the circle whose center is on the F-axis and which is tangent to the lines // = 2 a; — 1 and y =— 2x; of the circle whose center is on the A'-axis and which is tangent to the same two lines. 10. Find the equation of the cii-cle whose center is on the line y = 1 and which is tangent to each of the lines 3x — y = 6 and X — 3y = 3. 11. Find the equation of the circle that is tangent to the three lines X = 0, y = 0, and x = 3. 12. Find the equation of the circle that is tangent to the three lines X = 5, y = X, and y = — x. (Four solutions.) 13. Find the equation of the circle that is tangent to the three lines 3x-4y — 5 = 0, ix + 3y-8 = 0, and 4a; - 32/ -|- 12 = 0. 14. Find the equation of the circle whose center is on the F-axis and which passes through the points (3, 4) and (2, — 1). 15. Find the equation of the circle which passes through the three points (4, - 1), (- 2, - 3), and (- 1, 3). 16. Find the equation of the circle which passes through the three points (0, 6), (3, 0), and (- 2, - 4). 17. Find the equation of the circle which passes through the points (— 1, 9) and (6, 8) and is tangent to the A'-axis. Am. x" + y^ — ix — 10 1/ + i = or x'^ + y^- 244 ,r - 1690 y -|- 14,884 = 0. THE STRAIGHT LINE AND THE CIRCLE 113 18. Find the equation of the circle which passes through the points (8, 8) and (1, 7) and is tangent to the line 3 as + 4 y = 6. Am. aj^ + j/'- lOz- 8y + 16 = OTx^ + y^-2x-64:i/ + 400 = 0. 19. Find the equation of the circle which passes through the point (2, 4) and is tangent to the X- and F-axes. (Two solutions.) Prove that each of the following loci (Exs. 20-22) is a circle, by showing that its equation assumes the form (1) (p. 111). To this end, choose the X- and F-axes in a convenient position with refer- ence to the given points or lines. 20. A point moves so that the sum of the squares of its distances from two iixed points is constant. 21. A point moves so that the ratio of its distances from two iixed points is a constant (not equal to 1). 22. A point moves so that the square of its distance from a fixed point, divided by its distance from a fixed line, is constant. 23. Find the equation of the circle passing through the mid-points of the sides of the triangle whose vertices are (4, 0), (— 2, 0), and (0, 6). 24. Prove that the circle of Ex. 23 passes through the feet of the altitudes of the triangle, and also through the points halfway between the vertices and the orthocenter of the triangle. This circle is called the Nine Points Circle of the triangle. 25. Do as in Exs. 23, 24 for the triangle whose vertices are (a, 0), (b, 0), and (0, c). ^^^ ^j^^ equation of the Nine Points Circle is CHAPTER VII THE PARABOLA, ELLIPSE, AND HYPERBOLA 85. The parabola. A particularly important locus is that of a point which moves so as to be equally distant from a fixed point and from a fixed line. The student should construct carefully such a locus, taking the fixed point and the fixed line quite at random, and locating a sufficient number of points on the locus to make the shape of the curve clear (cf. Ex. 14, p. 95). This locus is called a parabola, and thus we have for the first time a definition of that word, which we have often used heretofore merely for the sake of giving a name to a curve of a certain shape. We have now the definition : A parabola is the locus of a point wJiose distance from a fixed point is always equal to its distance from a fixed line. The fixed point is called the focus, and the fixed line the directrix, of the parabola. 86. Equation of the parabola. The loci of Exs. 14-17, pp. 95, 96, were parabolas, and accordingly the equations obtained were equations of parabolas. In order to get the simplest possible form of equation, proceed as follows : Let F be the focus and DD' the directrix. Draw the per- pendicular from F to DD', and choose this line as the X-axis, with the direction from DD' to F as the positive direction. Choose as origin the point mid- way between F and DD', and represent the distance OF by a (a, 0), and the equation of DD' is x 114 D Q Y \ , N D' F Fig. 76 Then the coordinates of F are a. Now if P s (x, y) THE PARABOLA, ELLIPSE, AND HYPERBOLA 115 is any point on the locus, the two distances MP and FP must be equal, by the definition of the parabola. But MP=x-\'a and FP = y/(x-af+y\ Therefore V(a; -af+if=x + a (1) Simplifying, we get y'^ = ^ax. (2) This equation must then be satisfied by the coordinates of all points on the parabola; and, conversely, all points whose coordi- nates satisfy (2) must lie on the parabola, because we can reason back from (2) to (1). The possible double sign in (1) does not make any difference here, as a; + a cannot be negative. Equation (2) is therefore the equation of the parabola, and it may be considered as a standard form for the equation of the curve. By analyzing the equation several important properties of the curve may easily be obtained. 87. (1) Solving (2) for y, we have y =±V4aa;. Hence, for any positive value of x, y has two corresponding values, which are equal numerically but of opposite sign. This means that the curve is symmetrical with respect to the X-axis. The line through the focus, perpendicular to the directrix, of a parab- ola is accordingly called the axis of symmetry, or simply the axis, of the curve. The point where it intersects the parabola is called the vertex of the curve. Here the vertex is the origin. (2) The form of the equation y^=Aax shows that x and a must have the same sign, since y"^ cannot be negative; that is, X must be positive, since we have taken a as positive. Hence the curve lies entirely on the positive side of the F-axis. (3) As x increases, the positive value of y increases also, but less rapidly (since y is proportional to the square root of x), so that for very large values of a; a small change in x wUl produce scarcely any change at all in y. This means that the curve becomes more and more nearly horizontal as it recedes from the vertex. 116 THE ELEMENTARY FUNCTIONS (4) The chord of the parabola drawn through the focus perpen- dicular to the axis is called the lahis rectum. The student may show that the coordinates of B (Fig. 77) are {a, 2 a) and that those of C are (a, — 2 a), and hence that the length of the latus rectum is 4 a. Summarizing, the equation y^ = 4 ax has for its locus the parabola with focus at (a, 0), directrix x = — a, axis the X-axis, vertex the origin, and latus rectum 4 a. If we take a as a negative quantity, the parabola will be on the negative side of the F-axis, and the other inferences that have been drawn will require cor- responding modifications. Also, if we should choose the line through the focus, perpendicular to the directrix (that is, the axis of the parabola), as Z-axis instead of as X-axis, the effect would be merely to interchange x and y in the equation (2), giving jc=' = 4 ay. (3) The focus is now the point (0, a), the vertex is the origin, and the curve is concave upward if a is positive, concave downward if a is negative. Draw the figure for this choice of coordinate axes, taking the F-axis vertical as usual. What is the equation of the directrix ? Fig. 77 EXERCISES Draw the loci of the following equations, locate vertex, focus, directrix, and axis of symmetry, and find the length of the latus rectum : 1. / = 8x. 5. 2/2^1 K. 9. x'^-by. 2. 2/' = 10 a;. &.y'^ = -i,x. \0.y = A,x^. %.f=x. I.f = -Vax. 11. x^ + 32/ = 0. ^.f = \x. 8. a;' = 4y. 12. y^-|-ix = 0. THE PAEABOLA, ELLIPSE, AND HYPERBOLA 117 *88. Generalized standard equation of the parabola. We often have to deal with problems in which the X- and Y-axes cannot be chosen at pleasure but are already prescribed in a less favor- able position than that which we chose in deriving the standard equation (2), § 86. Exs. 14-17, pp. 95, 96, are examples of such problems. In problems of this kind the equation of the curve wUl not take so simple a form as (2) or (3), §§ 86 and 87. The simplest generalization, and the only one which we shall here consider, is the case where the axis of the parabola is parallel to either the X- or the F-axis and the vertex and the latus rectum are given. In this case the equa- tion of the curve can be found by using a certain f imdamental prop- erty of the parabola, as follows : If P s (x, y) is any point on the parabola indicated in Fig. 78, we have seen that the coordinates of P must satisfy the equation y'^ = 4l ax. But y = MP and x = VM, so that the equation y'^ = 4^ax is the same as MP^ = 4 a • VM. Fig. 78 This relation must then hold true for all points P on the parabola whose vertex is V and whose latus rectum is 4 a. It accordingly states an intrinsic property of the curve, which, expressed in words, is the following Theoeem. If from any point on a parabola a perpendicular is drawn to the axis of the curve, then the square of this perpen- dicular (MP^) is equal to the product of the latus rectum {4 a) by the distance from the vertex to the foot of the perpendicular (VM). This theorem is now stated in such a way as to be inde- pendent of the position of the coordinate axes. We can use it to write down at once the equation of a parabola whose axis is parallel to either the X- or the F-axis. Let the vertex be y E f f V F M 118 THE ELEMENTARY FUNCTIONS Y 1 B A ( f R k 1 ^ F M Q S Fig. 79 V={a, /8) (Fig. 79), and let the latus rectum be 4 a, the axis being parallel to the X-axis. Then ii P = {x, y) is any point on the curve, we have, by the above theorem, Jfp2 = 4 a . VM. But MP = SP-S3f = 5P - QF = 2/-/3, and VM=BM-Rr = OS-OQ = X — a. Therefore (y-^)^ = ^a{x-a). (4) Since this equation is true for all points {x, y) on the parabola, and for no other points, it is the equation of the curve. If the axis of the curve had been parallel to the F-axis instead of to the X-axis, the equation would have been (a;-a)2 = 4a(j/-/8). (5) The student should draw the figure and give the proof for this case also. EXERCISE Give the coordinates of F, B, and C, and the equation of the directrix, both for Eig. 79 and for the case where the axis of the parabola is parallel to the F-axis. *89. Summary. The equation (y — /3)2 = 4 a (a; — a) represents a parabola with vertex at the point (a, /3), but otherwise the same as the locus of the equation 2/^ = 4 ax. Its locus can be thought of, indeed, as the same curve as y^ = 4 ax, merely moved along so that its vertex takes the position (a, /8), the axis of the curve re- maining horizontal. Likewise, the equation (x — af — ^ a(jj — /8) represents the same parabola as a;^ = 4 ay, moved so that its vertex takes the position (a, /8), the axis of the curve remaining vertical. A movement of this sort is called a translation. THE PAEABOLA, ELLIPSE, AND HYPERBOLA 119 *90. By using these results we can also recognize and draw the graph of any equation of a parabola whose axis is parallel to the X-axis or to the J-axis. Example. Draw the graph oi 1/ + 2 x ~ i y = 6. This equation can be reduced to the form (4), (y-.pf = ia(x- a), by completing the square of the y-teims. Thus, y^~4:y = -2x + 6. Adding 4 to complete the square, y^-iy + i = -2x + 10; that is, (!/ - 2)^ = - 2 (2: - 5). This is now in the form (4) with o: = 5, ;8 = 2, 4 a = — 2 (that is, a = — I). Hence the graph is the parabola whose vertex is the point (5, 2), whose axis is parallel to the X-axis, and whose latus rectum is 2. The focus is to the left of the vertex, because a is negative. Hence the coordinates of the focus are (4i, 2), and the equation of the directrix is i = 5 J. With this information it is easy to draw the graph without computing the coordinates of any points on the curve. As a check, however, it is always well to determine the intercepts on the A'- and F-axes. Thus, when y = 0, 2 x = Q, x = & ; hence the point (3, 0) is on the curve. When x = 0, 3/''— 4 i/ — 6 = 0, 2/=2±VlO = 5.16+or-1.16+. Hence the points (0,5.16+) and (0,-1.16+) are on the curve. These facts agree with Fig. 80. EXERCISES Draw the graphs of the following equations and locate vertex, focus, and directrix. Check by finding the intercepts of the curve on the X- and F-axes. 1. y''-2x-2y = l. e. 2 1/ + ^x - tj + ^ = 0. 2. 2/=-5a;-82/ + l = 0. 7. x^ - Sx + Sy = 0. i. f - X - y = 0. 8. ai2 - a; + 2 2/ = 3. 4. 42/^ — 6x— -2/ = 5. 9. 4a;^— 9a; +-2/ — 1 = 0. 5. 32/^ + 4a;-52/ + 2 = 0. 10. lOx^ - 15a; + 8 // = 20. Fig. 80 120 THE ELEMENTAEY FUNCTIONS D P X A F Q D . P' s Fig. 81 91. Construction of parabola. (1) By points. Since any point on the parabola is equally distant from the focus and from the directrix, we can locate any number of points on the curve as follows : Let F (Fig. 81) be the focus and Djy the directrix. Draw the axis AFX, and through Q, any point on the axis, draw BS perpendicular to AX. From F as center, with radius AQ, describe an arc meeting RS in the points P and P'- These will both be points on the parabola whose focus is F and whose directrix is DD'. (Why ?) (2) By continuous motion. Place a right triangle JKH (Fig. 82) with KH along the axis AX and with KJ along DD'. One end of a string of length KH is fastened at H, the other end at F. If now a pencil point P be pressed against the string, keeping it taut as the tri- angle JKH is slid along the directrix, then P will trace an arc of a parabola. Prove this. J K D \ \ A F D' Fig. 82 PROBLEMS ON THE PARABOLA 1. The expression y^ — 4: ax, which equals zero for all points (x, y) on the parabola whose equation is 2/* = 4 ax, is positive for all points outside the curve and negative for all points inside. 2. Every point outside the parabola is nearer to the directrix than to the focus ; every point inside the parabola is nearer to the focus than to the directrix. 3. If PP' is any chord of a parabola passing through the focus, then the circle on PP' as diameter is tangent to the directrix. THE PAEABOLA, ELLIPSE, AND HYPERBOLA 121 4. If FP is the line segment joining the focus to any point on the parabola y^ = 4 ax, then the circle on FP as diameter is tangent to the F-axis. 5. Given the directrix of a parabola, and two points on the curve, to find the focus. 6. Given the focus of a parabola and two points on the curve, to find the directrix. 7. The straight line passing through the mid-points of any two parallel chords of a parabola is parallel to the axis of the curve and bisects all other chords that are parallel to the original ones. 8. Given a parabola, to find its axis, focus, and directrix. 9. The vertex of a parabola is 0, and P is any other point on the curve. Through P two lines are drawn, one perpendicular to the axis and the other perpendicular to OP. These lines meet the axis in Q and R respectively. Prove that QR is equal to the latus rectum. 10. Two perpendicular chords are drawn through the vertex of a parabola. Show that; the line joining their other intersections with the parabola meets the axis in a fixed point. 11. Prove that the following is a correct construction for a parab- ola : A being the vertex and F the focus, produce AF to B, making FB=AF, and draw the circle with B as center and BA as radius. At Q, any point on the axis, erect the perpendicular QR to AB (ij being on the circle), and lay off QP (and QP') equal to AR. Then P and P' are points on the parabola. In this way any number of points can be found. 12. Given A and F, the vertex and focus respectively, draw FA and produce it to C, so that ^C = iFA. Let Q be any point on the axis FA, and draw the circle having CQ as diameter. Draw the chord RAR'± CQ, and the tangent QP. The line RP W AQ meets QP in P, which is a point of the parabola. Prove this.^ The Ellipse 92. Definition of the ellipse. Another important locus is that of a point which moves so as to be always half as far from a fixed point as from a fixed line. The student should construct this locus carefully, taking a point and a line entirely at random 1 The constructions of Exs. 11 and 12 are of Arabian origin. They are found in a work written by an Arabian mathematician named Abti'I WSfS, who lived in the tenth century. 122 THE ELEMENTARY FUNCTIONS and locating enough points on the locus to make the shape of the curve evident (of. Exs. 20, 21, p. 96). This locus is called an ellipse. The locus would also be an ellipse if we changed the word "half" in the first sentence of this paragraph to "one third," "two thirds," or any other positive number less than one. The complete definition of an ellipse is accordingly: An ellipse is the locus of a point whose distance from a fixed point is to its distance from a fixed line in a constant ratio less than one. The fixed point is called the focus, the fixed line the directrix, and the constant ratio the eccentricity. The student should now construct carefully several ellipses, using different values of the eccentricity, — • for instance, ^, |, |, ^, |, ^^. Notice how the shape of the curve is affected by increasing or diminishing the eccentricity. 93. Equation of the ellipse. The simplest form of the equation of an ellipse can be obtained as follows: Let F be the focus, D-D' the directrix, and e the eccentricity, a positive number less than one. Choose as A-axis the line through the focus perpen- dicular to the directrix, with the direction from F to DD' as the positive direction. Two points on the curve can be located at once, namely, the points A and A' where the curve meets the X-axis ; for FA = e and (1) (2) r D X A' F A. B D' Fig. 83 AB A'F A'B' so that A and A' can be found by a simple construction of ele- mentary geometry. Take now as origin the mid-point of the segment AA, and represent the distance OA by a. Then A'O also equals a. In order to get the equation of the ellipse in the simplest possible form, we shall need to obtain the lengths OF and OB in terms of a and e. To accomplish this, rewrite the equations (1) and (2) above : FA = e- AB, (1) A'F==e-A'B. (2) THE PARABOLA, ELLIPSE, AND HYPERBOLA 123 (3) Adding, we get But and since Therefore (3) gives that is, A'F + FA = e{A'B+AB). A'F + FA=A'A = 2a, A'B + AB={A'0 + OB) + {0B- OA) = 2 OB, A'0 = OA. 2a = e.2 0B; a 0B=-- e In order to get the length of OF, subtract (1) from (2): A'F -FA = e {A'B - AB) = e ■ A' A = 2 ae. But A'F - FA = {A' + OF)- {OA - OF) = 2 OF, since A'0 = OA. Therefore (5) gives 20F=2ae; that is, OF=ae. (4) (5) (6) The important results of equations (4) and (6) may be stated thus : the coordinates of F are {ae, 0), and the equation of DB' is a; = - • Now, to get the equation of the ellipse, let P = {x, y) be any point on the curve; then, by the*definition of the ellipse, PF 'FQ or PF=e-PQ. (7) But : = «' Y R P(x,y) D Q \ A' F A B D' Fig. 84 and or PF = y/{x — ae)^-iry^ PQ = RQ-RP = --x. e Therefore V(a; - ae)2 -\- y^ = e(^ - x\= a - ex, {x — aef + y'^ = {a — ex)\ (8) 124 THE ELEMENTARY FUNCTIONS Multiplying out and rearranging terms, x^{l-e2) + f = a^{l-e% (9) Equation (9) is the equation of the ellipse, because it is the equa- tion which must be satisfied by the coordinates (x, y) of aU points on the curve; and it is not satisfied by the coordinates of any other points, because, if (9) is true, we can reason backwards through (8) to (7), the possible change of sign in (8) affecting only the direction of PQ, which here makes no difference, because e is the ratio of the numerical lengths of PF and PQ. 94. It is usual to make, in equation (9), the abbreviation Ifi^d^^V — ^, which is allowable because e< 1 and hence a2(l— e^) is necessarily positive. Then (9) becomes — a^ + y2 = J2. a'' that is. ^"+§=1- (10) This is the simplest form of the equation of an ellipse, and it may be considered as a standard form for the equation of the curve. By analyzing the equation, several important properties of the curve may be obtained. 95. (1) Solving (10) for y as an explicit function of x, we ob- tain « = ±-Va2_a32_ rpjjjg siiows that a? cannot be>a2; for if a it were, y would be complex, thus giving no point on the curve. This means that no point of the curve lies to the right of a; = a (the point A) or to the left of x=— a (the point A'). For any value of X between — a and + a, y has two values, numerically equal but opposite in sign, so that the curve is symmetrical with respect to the JT-axis. (2) Solving (10) for x as an explicit function of y, and reason- ing as in the preceding paragraph, we find that the ellipse is symmetrical to the T-axis, and that the points B s (0, b) and P' = (0, — h) are respectively the highest and the lowest point on the curve. The segment B'P is called the minor axis of the eUipse; THE PARABOLA, ELLIPSE, AND HYPERBOLA 125 A' A is called the major axis. The end-points of the axes, A, A', B, and B', are called the vertices of the ellipse. The intersection of the axes, 0, is called the center and is the mid-point of any chord drawn through it.^ (3) The relation among the three quantities a, b, and e (namely, ■ = a^{l-e^)-. aV) may be easily remembered by noting that in a right triangle with b and ae as the two legs, a will be the hypotenuse. Thus, in Pig. 85, the triangle BOF has OF =ae,OB = b; hence BF = a. This fact enables us to find the focus of an ellipse when the major and minor axes are given. (4) As in the parabola, the chord through the focus, perpendicular to the major axis, is called the latus rectum. The student may show that the coordinates of its end-points are Fig. 85 (a.,^')and(ae,-^') 62\ 2 &2 > so that its length is — •• a/ a r 96. Summarizing, the equation — -|- f- = 1 has for its locus the ellipse with major axis 2 a, minor axis 2 h, center at the origin, vertices (± a, 0) and (0, ± b), focus {ae, 0), and directrix x = — > where the value of e may be obtained from the fact that e jg2 a2g2 ^a2_ffl_ Thus, the equation — 4- y^ = 1 has for its locus the ellipse with center at the origin, major axis 2>/5, minor axis 2, vertices (±V5, O) and (0, ±1), focus (2, 0), and directrix aj = f (Fig. 86). 1 Because if (ij, y-^ is a poiirt satisfying the equation of tlie ellipse, E_ 4- IL = 1, then (— x,, — y^) will also satisfy the equation ; and the origin is the mid-point of the segment joining the points (x^, y^ and (- Xj, - y.^). 126 THE ELEMBNTAEY FUNCTIONS In deriving the equation of the ellipse, if we had chosen the line through the focus, perpendicular to the directrix, as F-axis instead of as X-axis, the effect would have been merely to in- terchange X and y in the equation (10) (§ 94), giving 1+^-1- (11) Thus, the equation — H =1 has for its 3 2 Fig. 86 graph the ellipse with center at the origin, major axis sVs, minor axis 2V2, vertices (O, ±V3) and (±V2, O), focus (0, 1), and directrix y = 3. It is the same curve as the ellipse — -f- ^ = 1, rotated about the origin through an angle of 90°. Draw the figure. EXERCISES Draw the graphs of the following equations, locating focus and directrix and finding the eccentricity : '1-, 8. ■lx^ + Zf= 21. 9. a? + ^>f = :^. 10. 2x' + Zif=l. H. 4x^-1-52/'= 6. 12. Prove that the point (— ae, 0) and the line x= are also X V a focus and directrix of the ellipse -; -|- t;; = 1. '•9+-i=^- 2.j + f=l. 3 ^+l!-l ^- 25 + 16"^- i ^ + r 4^2 5.| + y^=l. 6. x^ + -^=l. 13. Find the equation of the ellipse whose center is the origin and whose foci are on the A'-axis, if a =10 and e = i. THE PAEABOLA, ELLIPSE, AND HYPERBOLA 127 *97. Generalized standard equation of the ellipse. As in the case of the parabola, the simple standard equation of the ellipse gives a geometric property common to all points of the curve, and this property will enable us to obtain a more gen- eral form for the equation of the curve. Thus, ^ + § = 1 is equivalent (see Fig. 87) to 1 = 1, this relation a^ J2 being expressed in words by the following Theorem. If from any •point Fig. 87 on an ellipse perpendiculars are drawn to the rnajor and minor axes, the square of the perpen- dicular upon the minor axis (ATP^), divided by the square of the semi-major axis (a^, plus the square of the perpendicular upon the major axis (MP^), divided by the square of the semir-minor axis (JP), equals unity. Using this theorem, we can write down the equation of an ellipse whose axes are parallel to the X- and Z-axes, and whose center is any point (a, /S). Let the major axis be parallel to the X-axis and of length 2 a, y the minor axis having the length 2b. Then, by the theorem just stated, NF^ MP^_ But NP=TP-TX = OS-OQ = X— a, and MP=SP-SM Fig. 88 128 THE ELEMENTARY FUNCTIONS Therefore the equation of the ellipse is If the major axis had been parallel to the F-axis, we should have had , ., , .^^ (fi^+(LLa.i. (13) The figure should be drawn and the proof given for this case also. EXERCISE Give the coordinates of F, F', A, A', B, B', and the equation of each directrix (both for equation (12) and for equation (13)). *98. Summary. The equation ^^ — 7r-^+ ,„ =1 represents an ellipse with center at {a, /8), but otherwise the same curve as the ellipse -5 + 75 = 1- It can be thought of, indeed, as the same curve, merely translated from the position ia which the origin is the center to the position in which the point («, /S) is the center. By using these results we can recognize and hence draw the graph of any equation that is reducible to either of the forms (12) or (13). Example. Draw the graph of the equation x^ + 4 v^ — 6 a; + 4 j/ — 6 = 0. We see that by completing the square of the x-terms and of the 2/-terms it will be possible to reduce this to the form (12) or (13). Hence we write it first in the form (a:2-6x) + 4(),2 + y) = 6. To complete the square we must add 9 inside the first parenthesis and \ inside the second (which amounts to adding 9 + 1 to the left member of the equation): (x2 _ 6 X + 9) + 4 (y2 + 2, + i) = 6 + 9 + 1 = 16 I that is, (x - 3)'' + 4 (j, + \y = 16, or (^ - 3)' , iy+hy _. 16 4 which is in the form (12) with a: = 3, |8 = - ^, a = 4, ft = 2. Hence ths graph is the ellipse whose center is the point (3, — J), whose major axis THE PARABOLA, ELLIPSE, AND HYPEEBOLA 129 is 8, and whose minor axis is 4. The vertices are accordingly the points (7, - i), (- 1, - i), (3, li), and (3, - 2^). To find the focus, aV = a2 - «2 = 16 - 4 = 12. Therefore ae = Vl2 = 2 Vs = 3.46+. TV. * 2V3 V3 -_ Therefore e = = = .87-. 4 2 The foci are therefore at the distance 2v3 from the center, and the directrices are at the distance - = — — = from the center. With the help of these data the graph is easily drawn. As a check the X- and . Y- intercepts should be found. When a; = 0, 4y^ + 42/ — 6 = 0, 23^^ + 2)/ — 3 = 0, — 1±V7 y = = .8 or — 1.8, approximately. When y = 0, x^ — 6 x — 6 = 0, X = S ± Vl5 = 6.87 or — .87, approximately. These intercepts should agree with the figure. EXERCISES Draw the graphs of the following equations, and locate vertices, foci, and directrices. Check in each ease by finding the intercepts of the curve on the X- and F-axes. 1. 4a;^ 4- 9 2/" -16a; +18 3/ -11 = 0. 2. 3x' + 9f-6x-27y + 2 = 0. 3. Ax' + y^-Sx + 2y + l = 0. 4. a;'' + 152/' + 4a; + 602/ + 15 = 0. 5. x'^ + 2f + 3x + y=0. 6. 2x^ + iy^ + x-8y = 0. PROBLEMS ON THE ELLIPSE 1. Eind the equation of an ellipse, given (a) foci at (3, 0) and (— 3, 0), one directrix a; = 4 ; (b) foci at (1, 1) and (— 1, 1), one directrix x = 2; (c) major axis = 8, foci (4, 3) and (— 2, 3); (d) major axis = 2, foci (0, ^) and (0, — J). 2. Find the equation of the locus of a point which moves so that the sum of its distances from the points (3, 0) and (— 3, 0) is con- stantly equal to 10. What kind of curve is this locus ? Draw it. 130 THE ELEMENTAEY FUNCTIONS 3. Find the equation of the locus of a point which moves so that the sum of its distances from the points (c, 0) and (— c, 0) is constantly equal to 2 a. (a > c) . x'^ y^ _ . A.71S, — r -| ^ T — 1. This result, being the equation of an ellipse, establishes the following Theorem. The locus of a point which moves so that the sum of its distances from two fixed points is a constant greater than the dis- tance between the points is an ellipse having the fixed points for foci and the constant distance for its major axis (of. Ex. 22, p. 96). 4. Prove the theorem of Ex. 3 by showing that the distances from any point (x^, y^ on the ellipse ^ + ji = 1 to the foci are a + eXj and a — ex^, so that their sum is the constant 2 a. 5. Use the theorem of Ex. 3 to construct an ellipse by continuous motion, with the help of a piece of string and two thumbtacks. x^ If^ 6. Given an ellipse "i + f^ = 1- (a>b-) Let b be gradually in- creased until it finally equals a. How will the ellipse change ? What will happen to the foci ? to the eccentricity ? to the directrices ? 7. The lines joining a point on an ellipse with the ends of the minor axis meet the major axis in 5 and T. Prove that the semi- major axis is the mean proportional between OS and OT, being the center of the ellipse. 8. A and A' are the ends of the major axis of an ellipse; P is any point on the curve, and PM and PN are perpendiculars to PA and PA' respectively, M and N being on the major axis. Prove that MN is equal to the latus rectum. 9. The same construction as in Ex. 8, except that perpendiculars are drawn to PA and PA ' at A and A ' respectively. Let these per- pendiculars intersect in Q. Prove that the locus of Q is an ellipse whose semi-axes are a and f 10. If a point P moves around an ellipse, starting from one end of the major axis, its distance from the center will decrease until it reaches the end of the minor axis. 11. The circle on any focal radius as diameter is tangent to the circle on the major axis as diameter. THE PARABOLA, ELLIPSE, AND HYPERBOLA 131 The Hyperbola 99. A third important locus is that of a poii>t which moves so as to be always twice as far from a fixed point as from a fixed line (cf. Ex. 18, p. 96). This locus should now be constructed carefully, enough points being located so that the shape of the curve is clear. The locus is called a hyperbola. It would also be a hyperbola if we changed the word " twice," in the first sentence of this paragraph, to " three times," or " 1 J times," or any other number greater than 1. The complete definition of a hyperbola is accordingly : A hyperhola is the locus of a point whose distance from a fixed point is to its distance from a fixed line in a constant ratio greater than one. The fixed point is called the focus, tlie fixed line the directrix, and the constant ratio the eccentricity. The student should now construct carefully several hyperbolas, using different values of the eccentricity, — for instance, e= 3, e = 4, e = 11, e = l^,e = 5. Notice how the shape of the curve is affected by increasing or diminishing the eccentricity'. (In drawing these curves be sure to take account of the whole of the locus, and not only of part of it.) 100. Equation of the hyperbola. Since the definition of the hyperbola differs from that of the ellipse only in the fact that the eccentricity is greater than one instead of being less than one, the equation of the hyperbola can be obtained in exactly the same way. The details are left to the student to carry out, and care should be taken that the figure drawn corresponds to the facts. For instance, the points A and A' will now be on opposite sides of the directrix; but we obtain, exactly as for the ellipse, the facts that 0B = -, OF=ae e (using the same letterings as for the case of the ellipse. Fig. 83). That is, the coordinates of the focus are (ae, 0), and the equation of the directrix is a; = - (do not overlook the bearing of the fact that e>l). ^ 132 THE ELEMENTARY FUNCTIONS The equation of the hyperbola takes accordingly the same form as that of the ellipse, in (9), p. 124, namely, a,'2(l-e2) + 2^2 = a2(l-«2). If we wish to abbreviate this, however, as we did in the case of the ellipse, we cannot do as we did before and put V^ = a^{l— ^), because 1— e^ is negative in the. case of the hyperbola. We may, however, let 6^ = a^ (g2 _ \\^^ \^ which case our equation takes the form _ w _^a^ + y2^_J2. a'' that is. ^"fe^"^" ^^^ This is the simplest form of the equation of a hyperbola, and it may be considered as a standard form for the equation of the curve. By analyzing the equation we can obtain several impor- tant properties of the hyperbola. 101. (1) Solving (1) for y as an explicit function of x, we obtain y = ±- V'a;2 _ (i\ This shows that x^ cannot be < a^ ; for if it a were, y would be complex, thus giving no point on the curve. This means that no point on the curve lies between the points where x = a and where x = — a (that is, between the points A and A'). The points A and A' are called the vertices of the hyperbola. For any value of x> a or <— a there are two values of y, numeri- cally equal but of opposite sign, so that the curve is symmetrical with respect to the X-axis. Since this axis crosses the curve (at A and A'), it is called the transverse axis. As x increases indefinitely, the positive value of y also increases without limit; hence the curve extends indefinitely far from both the X- and the Y-axis. (2) Solving equation (1) (§ 100) for x as an explicit function of y, we get ^ , « ,/ a . m h ^ This shows that for every value of y there are two values of x, equal numerically but of opposite sign ; hence the curve is sym- metrical with respect to the F-axis. This axis of symmetry is called the conjugate axis of the hyperbola, and it does not meet THE PARABOLA, ELLIPSE, AND HYPERBOLA 133 the curve at all, since x cannot equal 0. The length B'B = 2 6 is called the length of the conjugate axis, while A' A = 2 a is called the length of the transverse axis. The point of intersection of the transverse and conjugate axes is called the center and is the mid- point of every chord drawn through it (cf. footnote, p. 125). (3) The relation among the three quantities a, b, and e, namely, 62 = a2 ^g2 — 1) = a2g2 _ ^2^ jjjj^y 133 easily remembered by noting that in a right triangle with a and b as the two legs, ae will be the hypotenuse. Thus, in Fig. 89 the triangle OAC has OA=a, AC=b, and hence OC=ae. This fact enables us to find the focus of a hyperbola when the transverse and conjugate axes are known. (4) The chord pass- ing through the focus, perpendicular to the transverse axis, is called the latus rectum. 2 &2 The student may show that its length is (5) As we have already noticed (pp. 87, 90) in the preliminary study of functions whose graphs were hyperbolas, we may expect to find that a hyperbola has two asymptotes intersecting at the center of the curve. We have not yet seen how to locate them exactly, however. This is accomplished by the following Theorem. The equations of the two asymptotes of the hyperbola — = i are obtained by replacing the 1 by in the equation Fig. 89 rfi of the curve, thus : ^~&2 Proof. The graph of the equation ^ - tj = {- + r I ( " ~ i ) = ^ consists of the two straight lines - + ^ = and a b y = 0. (Ex. 14, p. 110) 134 THE ELEMENTARY FUNCTIONS These are the lines through the origin with slopes and - respectively. (Note that the latter line is in fact the line OC of Fig. 89, of course pro- duced indefinitely.) To prove that the line - + j = (that is, bx + ay = 0) is an asymptote to the hyperbola — — ^ = 1, it is sufficient to show that as the point (x^, y,) recedes indefinitely along the curve, the distance from the line bx + ay = to the point (ar^, j/j) approaches more and more closely. Now the distance from the line bx + ay = ix> the point (Xj, y^) is ^ ^ b^ i + aVi . (Formula (4), p. 106) By the equation of the hyperbola, a2J2 that is, ftij + ayi - a^b"^ Replacing bx-^ + ay^ in d by d = aW (6xi - ay^) Va2 + J^ Since the line Ja; + a^ = passes through the second and fourth quad- rants, we are concerned with points {x^, ^j) for which either x^ < and y^ > 0, or else Xj > and y^< 0. In either case, as Xj and y.^ increase in- definitely in numerical value, bx^ — ay^ increases indefinitely also, and hence d approaches 0. This proves that the line bx + ay = Q is an asymp- tote to the hyperbola ~; ~ f^ = 1- The proof for the line bx— ay = is similar and should be carried through by the student. 102. Summarizing the facts obtained from the standard form of the equation of the hyperbola, — — ^ = 1, the locus of this equation is the hyperbola whose center is the origin, with trans- verse axis 2 a, conjugate axis 2 I, vertices (± a, 0), focus {ae, 0), directrix a; = - (where aV = a^ + P), and asymptotes — = 0. Thus, the equation -- — ^ = 1 has as its locus the hyperbola whose center is the origin, transverse axis 4, conjugate axis 2V5, and vertices (± 2, 0). aV = ^2 ^ j2 ^ 4 ^ 5 Therefore ae = 3, THE PARABOLA, ELLIPSE, AND HYPERBOLA 135 3 a 4 e = -, — = -• Hence the focus is the point (3, 0), and the direc- 2 e 6 . 4 trix is the line « = o " '^^^ asymptotes are given by the equation — — ^ = 0, and are the lines through the origin with slopes ± -— ■ If in deriving the equation of the hyperbola we had chosen the line through the focus, perpendicular to the directrix, as F-axis instead of as X-axis, the effect would have been merely to inter- change X and y m the equation (1), giving ^-^ = L (2) Thus, the equation = 1 has for its locus the hyperbola with vertices (0, ± 2), focus (0, 3), directrix y = ^, and asymp- totes =0. It is in fact the same hyperbola as that of the example above, merely rotated about the origin through an angle of 90°. Draw the figure. EXERCISES Draw the graphs of the following equations, and locate focus, directrix, and asymptotes : 1. -- — ^ = 1. 4 9 X' X _ ■ 2. — -^ = 1. 16 9 7. ,2 6. x' x^ 144 y^ = 2. 9. 4 12 ~ • -S-- 10. y'-x'=2. 11. 32/"-4a;= = l. f-- 12. 9 4~-'- 3. ^ - 2/= = 1. * 8 ^ 4. a;=-92/= = 4. " 4 Compare the graph of Ex. 12 carefully with that of Ex. 1, and note that they have the same asymptotes, the transverse axis of the one coinciding with the conjugate axi'S of the other. Two such hyperbolas are called conjugate hyperbolas. Draw them both in the same figure. 13. — — — = 1. What is the equation of the conjugate hyperbola ? 9 7 Draw them both in the same figure. 14. 4:x^ — ^y^ = 12. What is the equation of the conjugate hyperbola? Draw both curves. 136 THE ELEMEKTARY EUKCTIONS 15. If e and e' are the eccentricities of two conjugate hyperbolas, prove that - + — = 1. 16. Prove that the point (— ae, 0) is also a focus, and the line X = is a directrix, of the hyperbola ~^ — f^ = 1- *103. Generalized standard equation of the hyperbola. As in the case of the parabola and the ellipse, so the simple standard form of the equation of a hyperbola can be generalized' to apply to any hyperbola with axes parallel to the X- and Y-axes. The student should work through the details and obtain the result that the equation of the hyperbola whose center is the point (a, /3), whose semi-transverse axis is a, and whose semi-conjugate axis is & is (^ (y-)g)' , .„. if the transverse axis is parallel to the X-axis, and if the transverse axis is parallel to the Z-axis. The graphs of equations (3) and (4) can be thought of as the same curves as (1) and (2) respectively, merely translated from the position in which the origin is the center to the position in which the point (a, /3) is the center. By using (3) and (4) as standard forms the graphs of many equations can be recognized and drawn very easily. Example. 9 a:^ - 4 j/^ - 18 a; + 24 y - 63 = 0. This equation can be reduced to one of the forms (3) or (4) by com- pleting the square of the x-terms and of the ^-terms. Hence we write the equation 9(^2 _ 2:,) _ 4(^2 - Gy) = 63 and note that to complete the square of the x-terms we must add 1 inside the first parenthesis, and that to complete the square of the ^-terms we must add 9 inside the second parenthesis. This gives 9 (x2 - 2 X + 1) - 4 (2^2 _ e y + 9-) = 63 4- 9 - 36 = 36 ; that is, 9 (s _ 1)2 _ 4 (2, _ 3)2 _ 36^ or (x-1)^ (y-3)2_, 4 9 ~ ' THE PARABOLA, ELLIPSE, AND HYPEKBOLA 137 ■which is in the form (3), with or = 1, ^ = 3, a = 2, 5 = 3. Its graph is accordingly the hyperbola whose center is the point (1, 3), whose transverse axis is parallel to the X-axis and of length 4, and whose conjugate axis is of length 6. The vertices are accordingly the points (3, 3) and (—1, 3). Moreover, c?e^ = a^ + b^ = 13. Therefore ae = v 13. Hence the distance from the center to either focus is VlS. Further, Vl3 ' = -2-' hence - = -4= = ^ ^^IS = 1.11-, e Vl3 13 the distance from the center to either directrix. The asymptotes are the lines through the center, that is, through the point (1, 3), with slopes | and — |. What are their equations ? With these data, the graph is readily drawn. As a check the intercepts should be found. When a; = 0, 4 j/^ — 24 )/ + 63 = 0, an equation which has complex roots (b^ — iac being negative) ; hence the graph does not meet the K-axis. When ^ = 0, 9 x'' — 18 a; — 63 = ; that is, x2 — 2 X - 7 = 0, I = 1 ± VS = 3.83, - 1.83, approximately. These intercepts agree with the figure if it is correctly drawn. EXERCISES Draw the graphs of the following equations, and locate vertices, foci, directrices, and asymptotes. 1. x'-2/''-2x + 8y-3 = 0. 2. x'- 2 if +10^ = 0. 3. ix^-f + 8x-2i/-l = 0. i. x' - 5if + 6x -lOy =: 0. 6. 3x^-2/^+12a; + 2y+14 = 0. 6. 2x^-3f + x + ij+10 = 0. PROBLEMS ON THE HYPERBOLA 1. Find the equation of a hyperbola, given (a) foci at (3, 0) and (— 3, 0) and directrix x=l. (b) foci at (0, 2) and (0, — 2) and directrix y = i- (c) transverse axis = 3, foci (2, 3), and (— 2, 3). (d) vertex (4, 0), asymptotes y = 3x, and y = — 3x. 2. Find the equation of the locus of a point which moves so that the difference of its distances from the points (6, 0) and (— 5, 0) is equal to 6. What kind of curve is this locus ? 138 THE ELEMENTARY FUNCTIONS 3. Find the equation of the locus of a point which moves so that the difference of its distances from the points (c, 0) and (— c, 0) is constantly equal to 2 a. (a <. c) x' 1/^ _i Arts, —z ^ 5 — 1. a^ r — a' This result, being the equation of a hyperbola, establishes the following Theorem. 27i6 locus of a point which moves so that the difference of its distances from two fixed points is a positive constant less than the distance between the points is a hyperbola having the fixed points for foci and the constant distance for its transverse axis (cf. Ex. 23, p. 96). 4. Prove the theorem of Ex. 3 by showing that the distances from any point (x^, y^ on the hyperbola -;; — ^ = 1 to the foci are eXj + a and ex^ — a, so that their difference is the constant 2 a. This theorem suggests a mechanical construction of a hyper- bola by continuous motion, thus : fasten pegs or thumb tacks at the foci, then pass around both a string whose ends are held together. If now a pencil point be fastened at P and both ends of the string be pulled down together, the point P will move along an arc of a hyper- bola, because PF'-PF will remain" constant. Fie. go 5. What is the eccentricity of a hyperbola in which a = b? Such a hyperbola is called equilateral. 6. What is the angle between the asymptotes of an equilateral hyperbola ? X^ 11^ 7. Show that the foci of the hyperbola -i; — 7; =1 and those of a^ b' its conjugate all lie on the circle whose equation is x^ + y^=a' + h^. 8. Show that the circle of Ex. 7 meets either of the two hyper- bolas on the directrix of the other. THE PAEABOLA, ELLIPSE, AND HYPEEBOLA 139 9. The latus rectum of the hyperbola -^ — ^ =1 is extended by the amount k so that it just reaches the asymptote. Prove that k is equal to the radius of the circle inscribed in the triangle formed by the asymptotes and the line x = a. 10. Prove that the foot of the perpendicular from a focus of a hyperbola upon an asymptote lies on the directrix corresponding to that focus and also upon the circle described upon the transverse axis as diameter, and that its length is equal to the semi-conjugate axis. 11. The distance of any point of an equilateral hyperbola from the center is the mean proportional to its distances from the foci. 12. Through any two points P and Q of a hyperbola, lines are drawn parallel to the asymptotes, forming the parallelogram PRQS. Prove that the diagonal RS passes through the center. 13. If a straight line cuts a hyperbola at the points P and P', and its asymptotes at R and R', prove that the mid-point of PP' will also be the mid-point of RR'. 14. If a point moves along a hyperbola, the product of its distances from the two asymptotes remains constant. 104. The curves that have been studied in this chapter and the preceding — namely, parabola, ellipse, hyperbola, and circle — are called conic sections, or simply conies, because they can all be obtained as plane sections of a circular cone. They were origi- nally studied from that point of view, and nearly all their ele- mentary properties that are known to-day were proved by the Greek geometers more than two thousand years ago.^ The conic sections are of especial interest because of the fact that the paths of all the heavenly bodies are curves of this kind. This fact was first established by the great German astronomer and mathematician Johannes Kepler, in 1609. He showed that the planet Mars moves in an ellipse; the other planets, including of course the earth, do the same, while many comets move in parabolas or hyperbolas. 1 The most complete study of the conic sections among the Greeks was made by Apollonius of Perga, about 200 b.c. 140 THE ELEMENTARY FUNCTIONS 105. From the point of view of their equations it will be noted that all these curves have one thing in common : the equation of every conic section is of the second degree in x and y ; that is, when cleared of fractions and radicals and reduced to its simplest form, each of our standard equations has been of the second degree. We cannot as yet prove that every conic section must have as its equation one of the second degree, but this is a fact, and the proof of it will be possible at a later stage of the mathematical course. The converse statement, that every equation of the second degree in X and y has for its locus a conic section, is not true, because many equations of the second degree have no locus at all.- If there is a locus, however, it must be either a conic section or something simpler — a pair of straight lines (as a^— 3/^= 0), a single straight line (as o?—1xy + y'^ = ^), or a point (as a^ + 2/^^=0). This asser- tion also will not be proved here. The most general form of equation of the second degree in x ^ ao^ + hxy + ey"^ + dx + ey +/ = 0, where at least one of the coefficients a, h, c is different from 0. The student should show how to get each one of the standard forms of equation of the conies by giving special values to a, i, c, d, e, and /. For example, the circle a^ + y^ = r^ is obtained if we put a = l, h = 0, c = l, d = e = 0,f = —7^. CHAPTER VIII SIMULTANEOUS EQUATIONS 106. Review questions. Give the standard forms of the equa- tions of straight line, circle, parabola, ellipse, and hyperbola. How are the asymptotes of a hyperbola determined? How can the coordinates of the point of intersection of two straight lines be found ? of a straight line and a parabola ? 107. This last question brings us to the problem of this chap- ter, which is, to find the points of intersection of a straight line and any conic section, or the points of intersection of two conies. Some important geometric results can then be obtained, based on the solutions of these problems. As a preliminary step the prob- lems on page 45 should be reviewed, that there may be no difficulty in the algebraic solution of a quadratic equation in one unknown quantity. 108. Intersection of straight line and conic. The same method which was used in Chapter III, to find the intersections of a straight line and a parabola, can be used in the case of a straight line and any conic section. An example will make this clear. Example. Find the points of intersection of the straight line X + 2/ = 2, (1) and the conic i x^ + y'' = i. (2) Making the graphs of (1) and (2), we see that their intersections are the points A and C (Fig. 91). A is evidently (0, 2), and C is not far from ( j, 1 J). To determine algebraically the exact values of the coordinates of A and C (A we have found exactly, because (0, 2) satisfies both equations (1) and (2), but (|, j) does not satisfy the equations) we obtain from (1) the value of y as a linear function of x, thus : „ y = '2-x. Substituting this value of y in (2), 4 x2 + (2 - rf = 4. (3) 141 142 THE ELEMENTARY FUNCTIONS Since (3) has been obtained by using hoth (1) and (2), the values of x which it determines will be the abscissas of points lying both on (1) and on (2) ; in other words, the roots of equa- tion (3) are the abscissas of A and C, the points of intersection of (1) and (2). Simplifying (3), 5 a;2 - 4 a; = 0. 3;(5z — 4) = 0. Therefore a; = or 4- (4) Hence the abscissas of A and C, the points of intersection of the straight line and the ellipse, are and I respectively. Sincex + !^ = 2, ^^^ g^ X = corresponds to y = 2, and I = I to y = f . Therefore (0, 2) and (5, f) are the points of intersection. Evidently this method can be used in any case where it is desired to find the points of intersection of a straight line and a conic. We may solve the linear equation for y as a function of x (or for a; as a function of y) and substitute in the equation of the conic, getting a quadratic equation in the one variable x (or y), whose roots will be the abscissas (or ordinates) of the required points of intersection. The other coordinate is then found by sub- stituting in the linear equation the value of the coordinate that has been found. (The equation of the conic should not be used for this last substitution, because it would usually give two values where only one is correct. Thus, in the example above, if we had substituted a; = in the equation of the ellipse, we should have found y = + 2 or — 2, whereas y = + 2 is the correct value.) EXERCISES Solve both graphically and algebraically the following pairs of simultaneous equations : ra;^ + 2/' = 25, ^ fy^^Sx-Sx' + T, 0. rx' + f = : \x — y=l. y = 2x'~3x-i, ry = ^x-~ \y-x = 2. ry — ax — ox-\- ' i3a;-2y + 5 = 4. 25, l2a- -y = 4. SIMULTANEOUS EQUATIONS 1-3x^+22/^=11, " U-3y=7. 10. 36 20 ' 1. ] 25 ^ 9 ' .2x-2/=14. 11. .a; — y = 4. rx-3y=l, a;y + 7/2 = 5. r3x''+16y2==192, \x + 2y=lQ. 12. •4y = 5x+l, l2xy = 33-a;l ■ \x-2y+l=0. 13. r7a;'^-8jci/=159, \hx-\-2y=l. rx'^-2tf = i, ' 3x-2y=10. 14. ra;2/ + 3 2/^ = 42, l2x + 2/=13. 143 8. 109. Tangent to a Conic. Just as in the case of the parabola (p. 49), so here for any conic, the sign of the discriminant of the quadratic equation corresponding to (3), p. 141, enables us to tell whether the straight line (1) and the conic (2) have two points in common, or only one, or none. In case they have only one common point (which happens when the discriminant equals zero), the line will, in general, be tangent to the conic. Example. For what values of c will the line Sx + 'i:y = c (1) be tangent to the curve x^ + y'^ = 251 (2) c — S X Solving (1) for y, y = — Substituting this value of y in (2), that is, 25x2-6cx + c2-400 = 0. . (3) The roots of (3) are the abscissas of the points where (1) meets (2). If the line is tangent, these abscissas must be equal, and this requires that the discriminant of (3) shall equal zero. The discriminant of (3) is D = (- 6 c)2 - 4 • 25 (c2 - 400) = 36 c" - 100 6-2 + 40,000 = 40,000 - 64 c\ If 2) = 0, 64 c" = 40,000, c' = 625. Therefore c = ± 25. 144 THE ELEMENTARY FUNCTIONS Hence the line Sx + iy = 25 or 3x + 4y = — 25 will be tangent to the circle (2). (Verify this by a figure.) If c^ > 625, that is, if c > 25 or < - 25, the value of the discriminant will be neyatioe, and the straight line will not meet the circle at all. If c^ < 625, that is, if — 25 < c < 25, the value of the discriminant will be positive, and the straight line will meet the circle in two distinct points. Each of these possibilities should be illustrated by a figure. EXERCISES Determine for what values of k the following lines and conies will be tangent, and discuss the values of k that will give intersection or nonintersection of the two graphs : ' [3y = ix + k. ' \x + 2i/ = k. ' \kx + y = 3. rx'-,f=9, \^,m1 = , 8 (2x'-3y'=5, l5x-42/=A-. 5. 36"^25 ' Ux + h/=5. 3. 16 "'"25 ' jy^=4x, 9. ■! 9 4~' [kx + 4:y=20. ^'\t/ = 2x + k. [6x-kif=9. 10 r2/' = 4x + 8, ^^ r4:a^ + 9f=36, ' \x + ky = 2. ' \tj=ix + k. 12/ = mx + k (th a, fixed number). ry^ = 4 ax (a a fixed number), a \y = mx + k (m a. fixed number). ' m This result means that the line ?/ = mx -\ is tangent to the m parabola y" = 4 ax for any given value of m (not 0). 14. Eind the coordinates of the point of contact of the tangent line of Ex. 13. / a 2 a\ Am. ( — , — I- \nr m I 15. Find the F-intercept of the tangent line of Ex. 13. Draw a conclusion from the results of the preceding exercise and this one (cf. Exs. 6, 6, p. 52). 16. Find the Z-intercept of the tangent line of Ex. 13. Compare this with the abscissa of the point of contact. 17. Use the results of the preceding problems to find the equation of the tangent to the parabola y'^ = 6x at the point (f , 4) ; at the point (6, 6); at the point (x^, y^. SIMULTANEOUS EQUATIONS 145 18. Prove that the tangents to the parabola y^ = 4 ax at the ends of the latus rectum are perpendicular to each other and intersect on the directrix. 19. Give a geometric construction for the tangent to a parabola at any given point. 20. Discover a construction for the two tangents to a parabola from an external point. 21. The normal to a curve means the perpendicular to the tan- gent, drawn through the point of contact. Find the equation of the normal to the parabola y" = 4: ax, at the point (x^, y^). 22. Prove that the subnormal is constant for all points on the parabola ■i^ = iax. (The subnormal is the distance from M, the foot Y Fig. 92 of the perpendicular from the point of contact to the Z-axis, to N, the intersection of the normal with the X-axis. See Fig. 92.) 23. Prove that the tangent to a parabola bisects the angle between the focal radius at the point of contact and the line parallel to the axis through the same point. 24. For what values of k will the line y = mx + k (m a fixed number) be tangent to the ellipse 4 a;^ -|- 9 y^ = 36 ? x^ y^ 25. Answer the same question for the ellipse -^ + Ti = 1 and the line y = mx + k{ma. fixed number). {Ans. k = ± Vs^ + aW.) This result shows that the \mey = mx ± VPT^Wis tangent to the ellipse 2 2 * ?- + ^ = 1 for any given value of m. The double sign shows that a^ b^ there are two such tangents, that is, that there are two tangents with any given slope m, — a fact which is geometrically self-evident. 26. Find the coordinates of the point of contact of the tangent line of Ex. 25. 146 THE ELEMENTAEY FUNCTIONS 110. Construction of a tangent to an ellipse. The results of Exs. 25 and 26 enable us to construct the tangent to an ellipse, thus : The tangent line FT (Fig. 93) has the equation y^mx+^h^ + a^m?. (1) This meets the X-axis at the point T, whose ab- scissa is the X-intercept of (1). This is found by setting y = in (1), giving = mx + y/j)^ -h ahn^, or x = — - Vz,2 + ahiv^ m Fig. 9.3 0T = - that is, But, by Ex. 26, 0M= Therefore OT V&2-I-, m mAtn V62. 0M= a^ or 0T-. a'' X, where x-^ = OM = the abscissa of the point of contact. This is a very important and remarkable result, because it shows that the distance OT is independent of b, that is, does not depend upon the minor axis of the ellipse. In other words, if we have a set of ellipses, aU having the same major axis A' A but different minor axes, the tangent to any one of the ellipses, at the point whose abscissa is x-^, wUl pass through T. If b = a, the ellipse be- comes the circle with Pig. 94 SIMULTANEOUS EQUATIONS 147 A' A as diameter, and hence its tangent (at P', the point whose abscissa is x^) is easily constructed. It is the line P'T, perpen- dicular to OP' (Fig. 94). The intersection of this tangent with the axis A' A produced is the common point T where the tangents to all the ellipses meet the axis. Hence we only need to join this point T with the point P on the ellipse to get the required tangent TP. PROBLEMS 1. In Eig. 93, prove that F'T: FT = F'P : PF, and hence that PT bisects the exterior angle of the triangle FPF' This theorem gives another construction for the tangent to an ellipse at any point. 2. Obtain the equation of the tangent to the ellipse -^ + y^ = 1 SC OC 1/7/ at the point (Xj, yj in the form -\ + ^ = 1. a Hint. It passes through (Xj, y^) and ( — , Oj 3. Find the equation of the normal to the ellipse -^ + y^ =1, at the point (x^, y^. 4. Find the ■ JT-intercept of the normal of the preceding problem. 5. Prove that OF, the distance from the center to the focus of the a.2 ,,2 ellipse -^ + y^ = 1, is the mean proportional between the -Y-intercepts a' V" of the tangent and the normal. 6. State and prove the corresponding theorem for the F-intercepts of the tangent and the normal. 7. Construct the two tangents from an external point to an ellipse. A similar set of problems can be worked out for the hyperbola, but the results do not show enough difference from those for the ellipse to justify stating them explicitly here. This whole subject is considered from a higher point of view in the chapter, Introduc- tion to the Differential Calculus. 111. Intersection of two conies. To obtain graphically the coordinates of the points of intersection of two conies is a sim- ple matter; unfortunately the algebraic method, by which alone 148 THE ELEMENTARY FUNCTIONS we could be certain of obtaining exact results, is not practicable ^ except in a comparatively few special types of problem. In each of the following exercises the graphical solution will be found simple, in some cases even leading to exact results. EXERCISES Find (graphically) the coordinates of the points of intersection of each of the following pairs of conies : la { ix + if=ll. a?-tj= 17, 112. Solvable type; both equations homogeneous. If we have two equations of conies in which every term containing the vari- ables is of the second degree, then algebraic solution is always possible. Tlie most general form of such an equation is ax^ + hxy + cy'^ = d, which is called a homogeneous equation of the second degree in X and y. When each of the given equations is of this type, we can eliminate the constant term and. get an equation of the form a' a? + h'xy + c'y^ = 0, which can be factored if the points of intersection have rational coordinates, and often even when they do not. Example. f 2 a;^ - 3 a;y + 5 / = 14, (1) [ x2 + 4 a;2, _ 2 2/2 = 19. (2) To eliminate the constant term, multiply (1) by 19, and (2) by 14, and subtract. The result is 24 x^ - 113 xy + 123 y'^ = 0. (3) 1 That is, it would involve more advanced algebraic work than we are yet prepared for. SIMULTAKEOIJS EQUATIOi^S 149 Solving (3) for a; as a function, of y, we get _ 113 y ± Vl2769 y^ - 11808 y^ 48 _ 113y ^VQGly" 48 _ 113yi:31y 48 = 3y or |l y. (Equation (3) could of course have been factored, thus : 24 x' - 113 x^ + 123 ;/•- = (x - 3 ?/) (24 a: - 41 y) = 0. Therefore a: = 3 2^ or |1 y.) (a) Using a; = 3 y, we get, from (1), 2(3 3,)2-3j,(3y) + 52/^ = 14, 14 / = 14, 2/^ = 1, y = ±l. Since a; = 3 y, when y = 1, x'=%, and when 3^ = — 1, x = — 3. Check these pairs of values by substituting them in both (1) and (2). (b) Using x = %^y, we get, from (1), 2(ti3^)''-3y(li3r)+5 2/>' = 14, V235 23o Since ^ = ti ^> x = ±^\V2m. The graphical solution is not so simple as the algebraic one in this case, because (on account of the xy term) the equations can- not be reduced to any of the standard forms of the equations of conies. To draw the graphs we shall therefore have to compute the coordinates of enough points so that the form of each curve becomes clear. Solving (1) tov y as, a, function of x, _3a;±V9a^-20(2a!2-14)_3a;±V280-31a;2 ^~ 10 ~ 10 ' 150 THE ELEMENTAEY FUNCTIONS Using this, we get the table of values as follows : X 1 2 3 ±2.6+ y ±1.7 1.8 or - 1.2 1.8+ or - 6+ 1 or .8 Negative values of x give the same results but with opposite sign. It is evident from the value of y above that x cannot be > S'*'- The graph is an ellipse. r Fig. 95 Similarly, from (2), y = 4a;±Vl6a^-8(19-a^)_2a:±V6a^-38 The table of values is as follows: X 1 2 3 4 ±4.3+ y complex 1 or 5 .2 or 7.8 The graph is a hyperbola. Fig. 95 shows the graphs of (1) and (2), their four intersections corresponding to the algebraic solution. The straight lines given by (3) (p. 148) are also shown. SIMULTANEOUS EQUATIONS 161 EXERCISES Solve for x and y, checking graphically where practicable : ^_ (x^-f=l, \j)i? — xy + y^ =1. f 2 2/^ -4x2/ + 3x^=17, Ixy + y' = A. Cx^-xy-f = 5, ■ 12x^ + 3x2/ + 2/^ =28. Cx' + 2xy-if=li, rx^-x2/ = 35, • 1x2/ + 2/^ =18. f x^ + xy + 2 y2 = 44, rx^-x2/ + 2/^ = 21, • V- 2x2/ =-15. *113. Algebraic solution of some equations of higher degree. In the case of simultaneous equations, one or both of which is of higher degree than the second, it is only in special cases that alge- braic solution by elementary methods is possible. In some cases, however, it is so very simple that it is worthy of consideration. As for the graphical solution, we shall neglect it entirely, because it involves in nearly all cases too difficult a process of computing tables of corresponding values of x and y. *114. The following illustrative examples should be carefuUy studied, until the general methods used are well understood. It will be noticed that in each case the given equations are combined in such a way as to lead to a linear and a quadratic equation, which pair can then always be solved by the methods with which we are already familiar. Example 1. x' + y^ = 133, (1) x + y = 7. (2) Here we notice that if (1) is divided by (2), we obtain a quadratic equa- tion. In fact, division gives ^^ - xy -V y"^ = 19. (3) We may now solve (3) with (2) as usual, or we may proceed thus : Squaring (2), x' + 2xy + y'^ = 49. (4) Subtracting (3) from (4), Zxy = 30. xy = 10. (5) Multiplying (5) by 4 and subtracting from (4), x^ — 2xy + y^ = 9. X - 2/ = ± 3. (6) 152 THE ELEMENTAEY FUNCTIONS Adding (6) and (2), 2 a; = 7 ± 3 = 10 or 4. Therefore a: = 5 or 2. Therefore 2/ = 2 or 5 (since x -\- y = 7). Hence the solutions are (5, 2) and (2, 5). Another method is to cube (2) and subtract (1) from the result. This gives ^xy(x + y) = 210, and hence 3 xy = 30, xy = 10, since x + y = 7. From here we can proceed as usual, substituting the value of x (or y) from (2) in the equation xy =\0\ or we can continue as from equation (5) above. Example 2. x^ + y* = 706, (1) a- + y = 8. (2) Raising (2) to the fourth power, a;* + 4 x^i/ + 6 xh/ + 4 a;y^ + j/* = 4096. Subtracting (1), ^x^y + Q xV ^^xy^ = 3390. (3) Squaring (2) and multiplying by 4 xy, 4 x^y + 8 xhf + 4 x!/S = 256 xy. (4) Subtracting (3) from (4), 2 xV = 256 xy - 3390. xy - 128 xy + 1695 = 0. {xy - 15) {xy - 113) = 0. (5) We can now use (5) and (2) together, getting (5, 3) and (3, 5) as the only real solutions. Example 3. x< + x^ + y* = 481, (1) x''-\-xy + y^ = 37. (2) Dividing (1) by (2), x'' - xy + y^ = 13. (3) Subtracting (3) from (2), 2 xy = 24. xy = 12. (4) We can now obtain (x + yy and (x — yy by using (4) with (2) and (3) ; the remainder of the work is left to the student. EXERCISES Solve for x and y, and check : ' \x + y=5. ■15x2 + 6 2/^=50. 2 [»' + / = 72, ra;^ + y2 + ^ + 3/=36, ' \x + y=&. • Xxy = 10. \a;2/(a; + 2/) = 30. ' \xy — 4.. SIMULTANEOUS EQUATIONS 153 la; + 3/ = 4. ' Ixy (x — y) = 30. g f a;s - y5 = 3093, ^^ f x^ - 2/^ = 14 - X, ■ \x-y = 3. ' \f=^2x + 6. p-a;y + / = 7, f a;^ + ^^ + 2^^ = 19, ■ Ix* + x'y' + if = 133. • la: + xy + y = 1. 13. The sum of two numbers is 28, and their product is 147. Find the numbers. (Of. also Exs. 3 and 4, p. 55.) 14. The product of two numbers is 180, and their quotient is #. A^'hat are they ? 15. The diagonal of a field is 89 rd. long, and another field which is 3 rd. less both in length and in breadth has a diagonal 85 rd. long. What are the dimensions of each field ? 16. The diagonal of a rectangle is 68 cm., and if the length were increased by 2 cm. and the breadth diminished by the same amount,' the area would be diminished by 60 sq. cm. Find the dimensions. 17. A sum of money and its interest for one year amount to fl3,520. If the sum is increased by |200 and the rate of interest by ^%, the amount will be |13,794 for one year. Find principal and interest. 18. The fore wheel of a carriage turns in a mile 132 times more than the rear wheel, but if the circumferences were each increased by 2 ft., it would turn only 88 times more. Find the circumference of each. 19. A merchant buys a certain amount of wheat for |322. The market goes up 3^, and he can then for |323 get 10 bushels less than he could before for |322. What was the price per bushel ? CHAPTER IX FURTHER STUDY OF THE TRIGONOMETRIC FUNCTIONS. POLAR COORDINATES 115. Review questions. Define the trigonometric functions. Give their signs in each of the four quadrants. State four rela- tions among the functions of an angle. What are the values of the functions of 180° - 6 and of 90° + ^ in terms of functions of 6 1 Hov? can a right triangle be solved ? How is the resultant of two forces found ? Give the slope of the straight line joining the points {x^, y^) and {x,^, y^. What is the slope of the line ^a; + my + m = ? State the relation between the slopes of two perpendicular lines ; of two parallel lines. 116. Chapter IV included a few of the simplest applications of the trigonometric functions. In this chapter we shall take up some other problems m which they are used, and also study their graphs. 117. Solution of the oblique triangle. We saw in Gliapter IV how to solve any right triangle. Inasmuch as any oblique-angled triangle can be divided into two right triangles by drawing an altitude, the unknown parts can be found without the use of any new principles. A more practical method, however, is obtained if we discover the relations that exist among the sides and angles of the oblique triangle itself, thus avoiding the necessity of solving the two right triangles separately. Moreover, the study of these relations and of many others involving the trigonometric functions has very great value for its own sake and on account of its wide application in mathematical work of a more advanced character. I. The Law of Sines 118. The following discussion applies only to Fig. 96, (a), whicli represents an acute-angled triangle. Draw the altitude CD, thus forming the right triangles BCD and ACD. 154 THE TRIGONOMETfilC FUNCTIONS 155 h In the triangle A CD, Therefore In the triangle BCD, Therefore From (1) and (2) it follows that a sin /S = & sin a, or, dividing by sin a sin/3, a h — = sm a. h = b sin a. h . a - = sm p. a h = a sin yS. (1) (2) (3) (4) sin. a sin /3 If we draw the altitude from B, we shall get, in the same way a c Therefore sm a sm 7 a b c (5) (6) sin a sin j3 sin y This extremely important relation is known as the Law of (a) Pig. (6) Sines. It may be stated in words as follows : In any triangle the sides are pro20ortional to the sines of the opposite angles. 119. We have proved only that this law is true for acute- angled triangles ; in Fig. 96,(&), where the triangle ABC is obtuse- angled (a being the obtuse angle), we see that the altitude CD falls outside the triangle. Accordingly, - = sm ZJ»^C = sin (180°- a) = sin a (p. 69). b "With this hint the student may complete the proof for himself, thus establishing the truth of the Law of Sines for any oblique triangle. 156 THE ELEMENTARY FUNCTIONS II. The Law of the Peojections 120. Let the projections ^ of the sides a and h of the triangle ABC upon the side c be ^ and q respectively ; then (Fig. 97, (a)) p = a cos /S and q = h cos a. Since p -\-q = c, c = a cos ^ + & cos o. (7) By drawing the altitude upon the side I we get, in the same way, 6 = a cos y + c cos a, (8) and by drawing the altitude upon the side a, a=bcosY + ccosfl. (9) In the obtuse-angled triangle (Fig. 97, (6)) the side c is not the sum, but the difference, of the projections p and q ; but equation (7) c c (a) Fig. 97 (6) is true in this case also, because q=h cos (180°— a) = —h cos a (p. 69). The details should be worked out by the student. III. The Law of Cosines 121. If we apply the Pythagorean Theorem to the right triangle BCD (Fig. 97, (a)), we have a^=p^ + h\ (10) Ar Eeplaciug p by its value a^ = (c- qf + W = c'-2cq + q^ + h\ (11) p,^.98 1 The projection of a line segment AB upon a line CD is the segment A'B, between the feet of the perpendiculars from A and B upon CD (Fig. 98). THE TEIGONOMETRIC FUNCTIONS 157 But <f'+'h? = &2 Therefore a? = c^ - 2 c^ + J^. (12) And, finally, g' = J cos a. Therefore tf = 6^ + c" - 2 6c cos a. (13) By starting with the triangle ACD instead of the triangle BCD we obtain, in exactly the same way, 6^ = a'' + c" - 2 ac cos ^. (14) And by drawing the altitude from ^ or £ we get, likewise,^ c" = a" + 6" - 2 a& cos y. (15) Equations (13), (14), and (15) are known as the Law of Cosines. This law may be stated in words as follows : The square of any side of a triangle is equal to the sum, of the squares of the other two sides, minus twice their product times the cosine of the included angle. As in the case of the other theorems of this section, the student should carry through the proof for Fig. .9 7, (b), and show that the same relation holds true in an obtuse-angled triangle. 122. By the aid of Theorems I-III it is possible to solve any oblique triangle without first drawing an altitude and solving the auxUiary right triangles thus formed. The student should state what Laws I, II, and III give when apphed to a right triangle. Example 1. Given a = 36°, ji = 69°, a = 35 ft., to find y, h, and c. y = 180° - (a + /3) = 75°. h a Using the Law of Sines, sinjS sin a' XI. ^ . 35 -sin 89° 35-0.9.836 ,, _. therefore b = . = ^ .„„„ = 55.59 . sindD O.ooYO ^=^= Again, c a sm y sm a a sin 7 35-0.9659 ^, , asmy do - u.aooa ^ ^. therefore c = —. 1- = — t-^t^t;^— = 57.54. sm a O.OOYO =^ Check = — - — '■ — = '- Hence the results obtained ■ sin 13 sin y sin 69° sin 75° are verified. 1 Equations (14) and (15) are the same formula as (13), only expressed in different letters ; it is logically correct to derive them from the latter by merely changing the letters suitably. 158 THE ELEMEISITARY FUNCTIONS Example 8. Given a = 15, b = 20, y = 51°, to find c, a, and ^. Using the Law of Cosines, c'^ = a^ + V^ — 2 ab cos y = 225 + 400 - 600 cos 51° = 247.42. Therefore c = 15.73. To find a, use the Law of Sines, thus : sin a _ sin y n c mv, jr 15 sin 51° „„„ Therefore sin a = — — = .7410. Hence a = 47° 50'. SUuilarly, sin^ ^ siiiy Therefore sin 3 = ^ V° '"'^° = -SSSO- 15.73 Hence ^ = 81° 10' . Check. a + l3 + y = 180°. EXERCISES Solve the triangles in Exs. 1-16, drawing an accurate figure for each and checking both by measurement and by computation : 1. a = 7 in., 13 = 68°, y = 54°. 2. a =16 ft., a = 80°, /? = 37°. 3. a = 62 ft.. b = 55 ft., y = 42°. 4. a = 8.6 ft., c = 11.3 ft.. ;8 = 70°. 5. b = 21.2 ft., y = 71°, /3 = 60°. 6. a = 5 in.. 6 = 6 in.. c = 7 in. (Use the Law of Cosines.) 7. a = 13, 6 = 15, 0=16. 8. a = .82, 6 = .59, c = .71. 9. y = 38°, 6 = 14, c = 20. 10. /3=19°, y = 83°, 6 = 35.2 ft. 11. a = 15, a = 150°, /3 = 12°. 12. 6 = 26, /3 = 16°, y = 125°. 13. a = 21, 6 = 35, y = 110°. 14. 6 = 17, r = 19, .; = 115° 15. « = 12, 6 = 14, r = 22. 16. a = 1.6, 6 = .3, c = 2. THE TEIGONOMETEIC FUNCTIONS 159 17. Prove that the area of any triangle is equal to ^ab sin y, or ^ ao sin j8, or ^ be sin a. 18. Derive the Law of Sines by circumscribing a circle about a a b c , where triangle ABC and showing that 2 R R is the radius of the circle. sin a sin fi sin y Hint. In Fig. 99, ZI) = a. (Why ?) Z CB2> = 90°. Using the right triangle GBD will lead to the required result. Or use Fig. 100, prov- ing that Z BOD = a and then using the right triangle BOD to establish the result. Fig. Fig. 100 19. Use the Law of the Projections to derive the Law of Cosines algebraically. Hint. Multiply the equations (7), (8), and (9) by a, b, and c respectively, and combine the three resulting equations by addition and subtraction. 123. In general three independent data are sufficient to deter- mine, and hence to solve, any triangle (but observe that the three angles are not three independent data). There are four possible combinations of sides and angles that are essentially different, and it may be found convenient to regard them as four " cases," or groups of data, for the solution of a triangle. The four are as follows : Case I. Given two angles and one side. Case II. Given two sides and the included angle. Case III. Given the three sides. Case IV. Given two sides and the angle opposite one of them. 160 THE ELEMENTARY FUNCTIONS It will be noted that examples of each of these possible com- binations have occurred in the exercises on page 158. Before read- ing farther the student should make constructions of triangles from given parts, according to each of the first three cases. In each case, at least one example of an acute-angled triangle and one of an obtuse-angled triangle should be taken. 124. The "ambiguous case" in solution of a triangle. In Case IV appears a slight difference from the other three cases. The construction is, to be sure, equally simple. For instance, if we are given a, h, and a, we construct first the given angle a and lay off on one side the length AC =h; then, from C as center, with radius equal to the given opposite side a, we describe an arc which may cut the side AX in two points B^ and B^. Then either of the triangles ACBj^ or ACB^ is a correct solution, for either contaius the two given sides and the angle a opposite the side a. Case IV is accordingly often called the " ambiguous case" ia the construction and computation of trian- gles, because, when the sides and angles are as in Fig. 101, there are two equally correct solutions or constructions. 125. This ambiguity may, however, not occur if the relations of the given sides and angles are different from what they were in the figure above. If, for example, the side a is equal to or greater than the side b, there will be only one intersection with the line AX, and hence only one triangle can be constructed. Or, again, the side a may be exactly equal to the perpendicular distance CD from C to the opposite side AX, and in this case the arc with radius a will be tangent to AX, thus determining but one point BD. The right triangle ABC is then the only construc- tion. Note that in case this happens, a = CD = 6 sin a. Finally, the side a may be shorter than the perpendicular distance from THE TRIGONOMETRIC FUNCTIONS 161 C to AX, and then the arc with radius a will not meet AX at all, so that no construction is possible. In this case a < 6 sin a. A careful figure should be drawn for each of these possibilities. 126. Summarizing, in Case IV, when a, h, and a are given, a being an acute angle, there wiU be two solutions when and only when a is less than h and at the same tune greater than h sin a. If a=i there will be only one solution ; if a = & sin a there wUl also be one solution, in this case a right triangle ; and, finally, if a < & sin a there wiU be no solution. EXERCISES 1. Show that in Case IV, if the given angle is obtuse, there cannot be two solutions. When will there be none ? 2. Determine the number of possible constructions (solutions) in each of the following cases : (1) a = 30°, 6 = 10, a =12. (2) a = 30°, 6 = 5, a = 3. (3) a = 30°, 6 = 100, a = 50. (4) P = 30°, 6 = 25, a = 50. (5) y = 30°, 6 = 25, c = 30. (6) ^ = 30°, 6 = 15, c = 25. (7) ^=60°, 6 = 7, a = 10. (8) y = 20°, a = 20, c = 10. (9) y = 110°, 6 = 60, c = 100. (10) y = 150°, c = 25. a = 30. 3. In Kg. 101, writing Cj for AB^, and c^ for AB^, prove that Cj + Cj = 2 6 ■ cos a. 4. Show that (in the same figure) Cj — c^ = 2 a • cos j8^. 5. Solve the following triangles, checking the results both by measurement and by computation. In the two-solution case, check by the formula of Ex. 3. (1) a = 39, 6 = 65, a = 25°. (2) 6=23, c = 4.1, fi = 31°. (3) a = 16, c = 24. y = 49°. (4) 6 = 78, c = 50. y = 62°. (5) a = 153, 6 = 136, P = 40°. 162 THE ELEMENTARY FUNCTIONS MISCELLAHEOUS PROBLEMS In the following list of miscellaneous problems, draw an accurate figure whenever possible, and in every case check the result by computation : 1. In order to find the distance between two objects A and B, sepa- rated by a swamj), a station C is chosen, and the distances CA = 355 ft., CB = 418 ft., and Z A CB = 36° are measured. Find the distance from A toB. 2. Two objects, A and B, were observed from a ship to be in a line bearing N. 15° E. The ship then sailed N. W. 5 miles, when it was found that A bore due east and B bore N. E. Find the distance from A to B. 3. In a circle with radius 3 find the area of the part included between two parallel chords (on the same side of the center) whose lengths are 4 and 5. 4. The angle of elevation of a tower is at one point 63° 30'; at a point 600 ft. farther from the tower, in a straight line, it is 32° 16'. Find the height of the tower. 5. A tower makes an angle of 113° 12' with the hillside on which it stands ; and at a distance of 89 ft. from the base, measured down the hill, the angle subtended by the tower is 23° 27'. Find the height of the tower. 6. To determine the distance between two points A and B, a base- line CD is measured, 500 ft. long, and the following angles are ob- served: ACB = 58''20', ACD = 95° 20', ADB = 53° SO', BDC = 98° iS'. Find the length AB. 7. Two inaccessible points A and B are visible from JO, but no other point can be found from which both are visible. A point C is taken, from which A and D can be seen, and CD is found to be 200 ft., while Z.ADC = 89° and Z 4 CD = 50° 30'. Then a point E is taken, from which D and B are visible, and DE is found to be 200 ft., ZBDE = 64° 30', ZBED= 88° 30'. At D, Z.ADB is observed to be 72° 30'. Compute the distance AB. 8. A pole 10 ft. high stands vertically, and from its foot the angle of elevation of the top of a tree is 32° 27'. The angle of elevation of the top of the pole from the foot of the tree is 14° 48'. Find the distance between the tree and the pole, and the height of the tree. THE TRIGONOMETRIC FUNCTIONS 163 Fig. 102 127. The half-angle formulas. We now turn from the appli- cations of the trigonometric functions to the study of further theorems concerning the functions themselves. Problems 5-8 on page 67 gave four important relations among the trigonometric functions of an angle. We shall now discover the rela- tions that exist between the functions of an angle and those of an angle twice as large. 128. Let ABC (Fig. 102) be an isosceles triangle, and let ZBAC = ZBCA = a. Let each of the equal sides be represented by a, and the base by h. If AB be produced, the exterior angle CBE = 2 a. (Why ?) If we now apply the Law of Sines to the triangle ABC, we get a _ 5 6 h sin a ~ sia /.ABC ~ sin (180°- 2a) ~ sin 2 a ' ^ ^ But & = 2DC=2acosa;, (2) since DC = a cos a. Putting this value of b into equation (1), a 2 a cos a sin a sin 2 a that is, sin 2 a = 2 sin a cos a. (3) Next, let us apply the Law of Projections (p. 156) to the triangle ABC: a = b cos a + a cos (180° — 2a)=2a cos^a — a cos 2 a, since 6 = 2 a cos a, by (2) ; that is, cos 2 a = 2 cos^ a — 1. (4) 129. This relation can be written in other forms by making use of the fact that (Ex. 5, p. 67) sin^a;-f-cos^a; = 1. Therefore cos 2a = 2(1— sin^a) — 1 ; that is, cos 2 a = 1 — 2 sin'* a. (5) 164 THE ELEMENTAEY FUNCTIONS Replacing the term 1 in equation (5) by cos^a + sin^a;, cos 2 a = cos' a — sin" a. (6) It will be noted that these formulas have been proved for any- acute angle a ; they are true for any angle whatever, but that fact must for the present be left without proof. EXERCISES 1. Given sin 10° = .1736, cos 10° = .9848, find the values of sin 20°, cos 20°, and tan 20° to four decimal places. -r^ , 1 . . r> 2 tan a 2. Prove that tan 2a = : — 5— • 1 — tan-'a 3. In formulas (4) and (6), change 2 a: to a, which requires that a be changed to - > thus getting cos a =: 2 cos^ 5 ~ 1 ^nd cos a = 1 — 2 sin" - > and thence derive the formulas a 11 + cos a , .a jl . — cos o cos- 4. Find the values of sin 16°, cos 15°, and tan 15° from the fact that cos 30° = — ^ (p. 68). _ _ ' ^n.. sinl5° = iV^3^=^^^^, Ii 4: cosl5° = |V^Wl= -^ + ^ , tan 15° = \ r ~ \^ = 2 - Vs- >2+V3 5. Find the values of sin 22 i°, cos 22^°, and tan 22^° from the known values of the functions of 45°. Ans. sin 22^° = ^ V2 — V2, cos22i° = i\/2 + V2, tan 22^° = -y ;^ = V2-I. 2 + V2 «T> i.i_i..^ — I-' — COS a sin a 1 — cos a 6. Prove that tan - = x| = = :. . + cos o 1 + cos a sin a 'f=^ THE TRIGONOMETRIC FUNCTIONS 165 7. Derive the values of the functions of 15° geometrically by constructing a regular dodecagon and computing the exact values of the ratios of the apothem^ and side to the radius. Compare with the results of Ex. 4. 8. As in Ex. 7, construct a regular octagon and thereby find the exact values of the functions of 22^°. 9. As in Ex. 7, use the regular decagon to find the values of the functions of 18° (cf. Ex. 8, p. 60). V6— 1 Ans. sin 18° = ; ; 4 cos 18° = \ VlO + 2 V5, tanl8°=Jl^l4 = JVi^^^I^. >10 + 2V5 5 10. As in Ex. 7, use the regular pentagon to find the values of the functions of 36°. , Ans. sin 36° = i V 10 - 2 VS, cos 3d = - — ) tan 36° = V5 - 2 V5. 11. Prove that tan 7^° = V6 - Vs + V2 - 2. 130. The problems which follow are designed to aid the stu- dent in fixing in mind the relations among the trigonometric functions which we have thus far considered, and also to develop the power of discovering new relations. The illustrative examples should be studied, as indicating the general method to be followed, the details, however, varying from problem to problem. Example 1. Prove that cos*0 — sin^O = cos 2 0. Proof. cos«e - sin^e = (cos^S - sin^^) (cos^g + ain^e) = cos^e - sin^e, since cos^S + sin^^ = 1. But eos^^ - si-D^O = cos 2 ; hence the theorem is proved. 1 Apothem : the distance from the center to a side of a regular polygon. 166 THE ELEMENTARY FUNCTIONS Example 2. Prove that 2 tan - 9 l + tan^" Proof. l + tan2? = sec2^. 2 tan - 2 tan - Therefore = 1 = 2 tan - ■ cos^ - • 1 + tan- - sec^ - . X sm- 1 2 Now tan - = 2 X cos- 2 ar 3r XX Therefore 2 tan- • cos^— = 2 sin — cos— = sinx. q.e.d. Another method. This formula can also be proved as follows : and Therefore Therefore tan- sma; 1 + cos X tan^ - 1 — cos X 1 + cos X 1 + tan^l _ ^ 1 1 — cosx 2 1 + cos X 1 + cos X 2tan| 2 sin a; 1 + cos a; 1 J- (-.5.T,2*' 2 -«'^^- 2 1 + cos X Q.E.D. EXERCISES Prove each of the following formulas : 1. tan X + cot a; = 2 CSC 2 x. 4. cot a; — tan a; = 2 cot 2 x: . 6 e I r 2. sm- + cos-= Vl+sin0. 1— tan^^ 5. = cosa;. 3. tan(45° + |)=seca; + tanx. l+tan^^ Hint, tan (45° + ^) = tan ^J^l±l. 6. ?' '^"^"'^ = 2 sin »■+ sin 2 x. \ 2/ 2 1 — cos a; THE TRIGONOMETRIC FUNCTIONS 167 7. sin - — cos ^ = Vl — sin x. 8. sin 2 a; sin x = (1 — cos 2 x) cos x. ^ . 1+sinfl— cose 9. tan - = : — . 2 l+sme+eose 10. cos^a; + sin^a; = 1 — ^ sin^2 x. 11. CSC X — 2 cos X cot 2 x = 2 sin x. lo 6 -6 o /h sin''2a;\ 12. cos°a! — sin^a; = cos 2 a; ( 1 — 1 • 13. cos^x + sin^a; =1—3 sin^x cos^'a;. 14. sin 4x = 4 sin a; cos a;— 8 sin'a; cosx=8 cos'a; sin a;— 4 cos x sin x. 15. cos 4 X = 1 — 8 cos^x + 8 cos*x =1—8 sin^x + 8 sin*x. , „ . „ . _ cos X + sin X 16. tan 2 X + sec 2 x = ■ — -. cos X — sin X sec^x 17. sec2x = jr 5--. 2 — sec^x 18. tanfl- tan 2^= sec 2^ — 1. ^ g _ 2 sin e — sin 2 e 19. tan2-2si^^^sij^2e' 20. l+sec2e+tan2e sin a + sin 2 a a 21. ^ = cot-- cos a — cos 2 a 2 l-tan^ 22. (1+ cot'' |) sin a- tan I = 2. 2 cot a 23. tan 2a = 25. eot^a —1 sin X. cos X — sin X 1 — sin 2 x cos 2 x 24. sin'' I f cot I — ij = 1 — I cos X + sin x cos 2 x 1 + sin 2 x tan a tan a 1 1 26. tan 2a = 3 1 ^- rTT = ^i i^ m: ^' 1 — tan a 1 + tan a 1 — tan a 1 + tan a 11 1 I . o._ (l+ tan g)^ ^ (l + cot 0)^ ^ (1 + tang) (1 + cote) "^ l + tan^e l + cot^0 tane + cote 168 THE ELEMENTAEY FUNCTIONS sin 2x 2 tan x 2 cot x 28. l+sin2x (1 + tanic)'' (1+cotx)^ (l + tanx) (l+cotcc) 29. 2cos^=\2 + V 2 + y/2-\ + V2 (w + 1 radical signs). Of\0 I . — 30. 2 COS -— = y 2 + V 2 H + VS (w + 1 radical signs). 131. Functions of the sum or difference of two angles. For the further study of the trigonometric functions no theorems are more important than those now to be proved, which enable us to find any function of the sum or difference of two angles from the func- tions of each angle. Two proofs will be given, one algebraic and the other geometric. 132. First proof. This proof uses the Law of Sines, and also the formula (7) on page 156 : c = a cos /3 + J cos a. By the Law of Sines (in the form given in Ex. 18, p. 159), c = 2iJsin7, J = 2 -B sin /3, and a = 2 i? sin a. Therefore 2 ^ sin 7 = 2 iJ sin a cos /3 + 2 -B sin /8 cos a. Now, since a + /S + 7 = 180°, 7 = 180°-(a + /3); therefore sin 7 = sin [180° — (a + /8)] = sin (a + /8). Therefore sin (o + ^) = sin a cos )ff + cos a sin p. (I) To find the value of cos (a + /S) : Begin with equations (8) and (9) on page 156, ' « = j oos7 + ccos/3, 6 = a cos 7 + c cos a. Multiplying these together, we get ab = ab cos^7 + c cos 7 (6 cos a -\- a cos yS) + c^ cos a cos /3. But, by (7) on page 156, h cos a + a cos/8 = c. Therefore a6 (1 — cos^7) = c^ cos 7 + c^ cos a cos ;8 = c^ (cos 7 + cos a cos /8). By the Law of Sines, as above, a = 2 i? sin a, J = 2 .B sin (S, and c = 2 ^ sin 7 ; also 1 — COS27 = w?'^. Therefore 4 if 2 sin a; sin ^ . sin27 = 4 iJ^ gin2^ ^^Qg ,y ^ cos a cos /3). THE TRIGONOMETRIC FUNCTIONS 169 Dividing both sides of this equation by ^B^sin^y, sin a sin /3 = cos 7 + cos a cos /8. And, since a+/S + 7 = 180°, cos 7 = cos [180° - (a 4- /3)] = - cos (a + /8). Therefore co6 (o + )ff ) = cos a cos ^ — sin a sin ^. (II) 133. Second proof. Let a and /8 be any two acute angles. Place them with a common vertex and a common side OiV^ as in Fig. 103, so that the angle MOP = a +y8. From any point A in OM draw AB perpendicular to ON at B, and produce it to meet OB at C Notice that this assumes that a + yS is an acute angle. Then we can express the areas of the triangles formed, as follows (Ex. 17, p. 159) : Area. AAOC = ^OA • OC sin(a:+/3). Area AAOB = ^OA- OB . sin a. Area A^OC =10-8. OC • sin/8. Since AAOC = AAOB +ABOC, therefore ^OA. DC- sin(a + /3) = J0^. 0.B ■ sina + 105 • OC- sinA Therefore ^^ OB . OB . „ sm(a; + /3) = --sma + — smA c/ Fig. 103 But and Therefore OB OC' qB_ OA ■ coS/S • = COS a. Fig. 104 sin (a + jff) = sin a cos jff + cos a sin p. (I) This same method can be used to find the value of sin (a - /8), as follows: Let ZAOC=a (Fig. 104) and ZBOC=^, AC being perpen- dicular to OC. Then ZAOB=a-0, and, just as before, lOJ. OC-sma = ^OA-OB-sm{a-^) + ^OB- OC- sin^ 170 THE ELEMENTARY FUNCTIONS Therefore sin (a — /3 ) = — sin a sin B, ^ ' OB OA or sin (a — ^) = sin a cos ^ — cos a sin p. (HI) From (I) and (III) the values of cos {a + /3) and cos (a — /S) can be found. For ^ . „.o a\ cos 6 = sm(90 — &). Therefore cos (a -f /3) = sin [90° - (a + y8)] = sin[(90°-a)-;S]. This last expression is in the form of the sine of the difference of two angles, both of which are acute, since 90°— a is acute if a is. Therefore we can apply (III), gettrug sin[(90°-a)-;S] = sm(90°-a)cos;8-cos(90°-a:)sin/S = cos a cos /8 — sin a sin /3 ; that is, cos (a+/3)= cos a cos ^ — sin a sin /S. (II) To obtain the value of cos (a — /S) : cos (a - /3) = sin [90° -{a- 0)] = sin [(90° -a) + 13]. Since the angles (90° — a) and /S are both acute, we can apply (I) : sin [(90° - a) + ^] = sin (90° - a) cos /3 + cos (90° - a) sin /3 = cos a cos /3 + sin a sin /8. Therefore cos (a — ^) = cos acos ^ + sin a sin p. (IV) 134. The first method of proof (§ 132) assumes that a and /3 can be taken as two of the angles of a triangle ; that is, that their sum is less than 180°. The second proof (§ 133) assumes that a, /8, and a +13 are acute. We can, however, remove these restrictions and establish the fact that formulas I-IV hold for any angles whatever. To accomplish this, take a =90°+ a', where a' is an acute angle, and suppose /8 to be acute, as before. Then sin(a;+^)=sm[90°+a' + /8]=cos(a;' + iS). Since a' and /3 are acute angles, (II) can be applied, giving cos {a' + /3) = cos a' cos jS — sin a' sin /3. THE TRIGONOMETEIC FUNCTIONS 171 But cos a' = cos (a — 90°)= sin a, and sin a' = sin (a — 90°) = — cos a. Therefore cos (a' + /3) = sin a cos y8 + cos a sin /S. Therefore sin (a + /3) = sin a; cos /3 + cos a sin /8. Hence formula (I) is true if one angle is obtuse and the other acute, and by exactly the same process we can establish its truth if we add 90° to any angle a' for which the formula has already been proved. Thus, we can add 90° to /3, proving the formula true when hoth angles are obtuse; then we can add 90° to a, proving its truth when one angle is between 180° and 270°, the other being either acute or obtuse ; and so on. The same reason- ing holds for successive subtractions of 90°. Hence (I) is true for angles of any magnitude whatever. The theorem is, as was said before, of the very greatest importance. It is called the Addition Theorem for the trigonometric functions. Similar reasoning can be used to establish the truth of II, III, and IV for angles of any magnitude. EXERCISES 1. Work through the same reasoning as above to prove (II) in general. 2. Prove (III) by changing ;8 to — /3 in (I). (This is allowable, since (I) has been proved true for all angles.) 3. Prove (IV) in- a similar way. 4. Eind the values of sin 76° and cos 76° from the known values of the functions of 45° and 30°. 5. Find the values of sin 16° and cos 15° from the values of the functions of 45° and 30°. Compare with the results of Ex. 4 and also of Ex. 4, p. 164. 6. Derive formulas for tan (a +/3) and tan (a — ;8) by using I-IV. tan a + tan S , . .. tan « — tan/? Ans. tan(« + ^)= j3^^^^^; *^°(^-^)= 1 + tana tan^" _ , . , „, cot a cot fl — 1 7 . Prove that cot (a+B) = — , ,, , , > ^ "^^ cot ;8 + cot a cot a cot /8 + 1 and that cot(a - ^) = ^^^J^^^ ' 172 THE ELEMENTARY FUNCTIONS 8. Prove formulas (I) and (II) by using Fig. 105. Show that „, DE +FB sm (^a + P) = OB DE OE OE OB + FB EB EB OB = sin a cos ^ + cos a sin fi ; and, similarly, derive (II) by starting with „, OC OD—FE cos(a+^)=— = — ^^— 9. In (I) and (II) let ^ = a, thus getting formulas for sin 2 a and cos 2 a. Compare with the results of §§ 127, 128. Notice that we are now for the first time able to assert that these results are true for all angles whatsoever. The " half-angle formulas " of Ex. 3, p. 164, are thereby also assured universal validity, because they were derived by purely algebraic processes from the values for cos 2 a. 10. Find the values of the functions of 75° from those of 150°, and compare with the results of Ex. 4 above. 11. Prove that sin (a — 13) — sin a cos ^ — cos a sin /3 by using Fig. 106. ABC is any triangle, CD is the altitude from C, DE=AD; show that Z.ECB = a —fi, and then apply the Law of Sines to the triangle CBE. 12. Prove the Addition Theorem from the Law of Sines, using a b' the fact that if - = - > then ; b d b -\-d Prove each of the following formulas (13-30): sin (x + y) 13. ^^ — -^^ = tana^ + tany. sin 2 a- + sin 2 y cos X cos y 14. tan(x -f- 2/) = cos 2 x + cos 2 y 15. cos (x + 30°) + cos (x - 30°) = V3 • cos x. 16. sin (a; + 60°) + sin (a; - 60°) = sin x. 17 . sin e + sin (6 - 120°) + sin (60° - 0) = 0. THE TRIGONOMETRIC FUNCTIONS 173 18. sin 3 aj = 3 sin a; — 4 sin'a;. 19. cos 3 a; = 4 cos'a; — 3 cos x. 20. sec (a + 46°) sec (a - 46°) = 2 sec 2 a. 21. tan (45° + a) - tan(45° - a) = 2 tan 2 a. 22. tan (5 + 46°) + tan (45° - ^) = 2 sec 2 0. 23. sin {a + P) + sin (a — |8) = 2 sin a cos ;8. 24. sin (or 4- /3) — sin (a — ^) = 2 cos a sin /8. 25. cos (a + y8) + cos (a — /?) = 2 cos a cos /?. 26. cos(a: + y8) — cos (a — y8) = — 2 sin « sin /3. 27. sin (a + ^8) sin (a- p)= sin'a - sin'jS. 28. cos (a + ;8) cos (a — j8) = cos^a + cos^/3 — 1 = cos^a — sin'^/? = cos'^yS — sin^a. 29. sin (a -\- p) cos (a — p)= sin a cos a + sin /3 cos p. 30. sin (a — P) cos («+/?) = sin a cos a — sin p cos /?. 31. Obtain the values of the functions of 18° by the follow- ing method : sin 36° = 2 sin 18° cos 18°, and also sin 36° = cos 64°. But, by Ex. 19, above, cos 54° = 4 cos' 18° - 3 cos 18°. Therefore 2 sin 18° cos 18° = 4 cos» 18° - 3 cos 18°. Solve this for sin 18° and compare the result with that of Ex. 9, p. 165. 32. Prove that sin (a + /?) cos p — cos (a + p) sin p = sin a. Solution. This can be proved by writing out the values of sin (o: + /3) and cos (a +/8), but it is simpler to observe carefully the combination of terms that we have here, namely, sin x cosp — cosa: sin /?, x representing a + p. This is the value of sin (x — /3), that is, of sin (or + ^ — ^), or sin a. 33. Prove: sin(a — ;8)cos/3 + cos(a: — /3)sin/3 = sina. 34. Prove : sin (a + P) sin /3 + cos (a + P) cos /3 = cos a. „^ .„ tan(a: — y8)+ tan;8 , 35. Prove: q 1 — / o\^ '"- = tana:. 1 — tan (a — P) tan p _ cos a —• cos 6 cos (Q + a) . . 36. Prove: : — 77; — ^^ ^ = sinSseca. cos a sm (6* + a) 37. Prove that the angle Q, made by a line whose slope is m^ with m — TO. a line whose slope is m^, is given by tan B = -r-j^ 38. Find the angle that the line 3x — 2 ij + 1 = makes with the line x + y — 3 = 0. 174 THE ELEMENTARY FUNCTIONS 39. Find the angles of the triangle whose vertices are the points (3, 6), (7, - 1), and (- 2, 3). 40. The theorem known as the Ptolemaic Theorem ^ is as follows : In any inscribed quadrilateral the sum of the products of the opposite sides is equal to the product of the diagonals. (a) Prove this theorem by elementary geometry. (b) Assuming its truth, prove the Addition Theorem. Hints. For (a) draw from B a line BE (not shown in Fig. 107), so that /.ABE = p, E being the point where this line meets AG; then the triangles ABE and BDC are similar, as are the triangles BEG and ABD. From these facts the theorem can be proved. Fio. 107 Fig. 108 For (b) construct the given angles a and /3 on opposite sides of the diameter through A (Fig. 108), and complete the inscribed quadrilateral ABGJD. Then apply the fact that a = 2fisina etc. to the right triangles ABC and AGD; also the Law of Sines to the triangle ABD. 135. Sum and difference of two sines or two cosines. The equations that were proved in Exs. 23-26, p. 173, were as follows : sin (a + y8) + sin (a — /3) = 2 sin a cos /3, (1) sin (a + /8) — sin (a — /S) = 2 cos a sin y8, (2) cos (a + /3) + cos (a — /3) = 2 cos a cos /8, (3) cos (a + /3) — cos (a — /3) = — 2 sin a sin /3. (4) These equations give the sum or difference of two sines, or of two cosines, in the form of a product, a form which we shall often 1 From Ptolemy, the famous Greek astronomer of the second century. This theorem is included in his great work called the "Almagest," which explained the astronomic system which, under the name of the "Ptolemaic system" was universally accepted for 1400 years. It was only overthrown after a long struggle between its adherents and those of the system of Copernicus. See the articles "Ptolemy" and "Copernicus" in the Encyclopedia Britannica. THE TRIGONOMETRIC FUNCTIONS 175 find useful. These equations are written in a more practical form by setting a + ^ = x and a — ^ = y. Therefore 2a = x + y and 2^ = x— y. x + y , ^ X — y a = " and j8 = —-^■ 2 • 2 Thus the equations (1) — (4) are equivalent to x+y x—y sin Jr + sin y = 2 sin cos » (5) sin Jf — sin tf = 2 cos sin » (b) 2 2 x + y x-y C0SX+C0SU=2 cos COB- » (/) 2 2 ^ . x + y . jr-y COS j: — cos w = — 2 sin sin • (») 2 2 EXERCISES ^^ ^ sin 33° + sin 3° , , _„ 1. Prove that 5^5- 53 = tan 18° cos 33 + cos 3 Proof. By (5) and (7), . . 36° 30° 2 sin — — cos — — sin 33° + sin 3° _ 2 2 _ sin 18° cos 33°+ cos 3° „ 36° 30° cos 18° 2 cos -^ cos -— - „ sin 3 a + sin 5 a , . 2. Prove: 5 ; ^ = tan 4 a. cos Sa + cos o a sin 75°+ sin 16° , „„» 3. Prove : . „.,o r-^^ = tan 60 . sin 75 — sm 16 ^ tan X + tan y , , 4. Prove : —r , r^ = tan x tan y. cot X + cot y x + y tan— r— „ sin r + sm y ^ 5. Prove: -. : — - = = tan 18°. Sin x - sin y . x-y 176 THE ELEMENTARY FUNCTIONS 6. Prove the formula (3), p. 174, by means of Fig. 109. This method of proof is found in an astronomical work of the tenth century, by an Arabian scholar, Ibn Junos. The formulas (1), (2), and (4) can also be proved in a similar way. Hints. Letting and it follows that and Therefore ZNCP = /.Z'CE'= a, ZE'CD" = ZECD = p, JD" = cos (a + /3) (if r = 1) D'D = cos(a-/3). J'D = cos (a: + (3) + cos (a — /3) . Next, show LD = ^ J'D = cos a ■ cos /3 hy showing that LD = BD ■ cos a. 7. cos X + cos 3 a; + cos 5x + cos 7x = 4 cos x • cos 2 x ■ cos 4 x. Proof. cos X + cos 3 x = 2 cos 2 a; cos x, cos5x + cos7:r = 2cos6a;cosx. Therefore cos x + cos 3 x + cos 5 x + cos 7x = 2 cos x (cos 6 x + cos 2 x) = 2 cos X (2 cos 4 X cos 2 x) = 4 cos X (cos 2 X cos 4 x). q.e.d. 8. sin X + sin 3 a! + sin 5x + sin 7a; = 4 cos x • cos 2 x ■ sin 4 x. tan a; + tan /8 _ sin (a + ;8) tan a — tan fi sin (a — j3) THE TRIGONOMETEIC FUNCTIONS 177 sin a + sin 2 a + sin 3 a 10. ~ — ; = tan 2 a. cos a + cos 2 a + cos 3 a , , sin a -}- sin 2 a + sin 3 a + sin 4 a , 5 a 11. ^ -. — = tan-;7-- cos a + cos 2 a + cos 3 a + cos 4 a 2 If a, j3, and y are the angles of a triangle, prove that each of the following relations is true : 12. sin a + sin yg + sin y = 4 cos ^ • cos ^ • cos ^ ■ Proof. sin a + sin /3 = 2 sin — ;— ^ cos — ^-^ • siny = sin [180°- (a + )3)] = sin (a + ;8) = 2 sin ^4^ cos 2±^. ((3), p. 163) Therefore sin a + sin/3 + siny = 2 sin " ^ cos " + cos — —^ - . 180° -yr„ a /yi = 2 sm ;r — t 2 COS "5 cos c: a B y = 4C0S- COS^COS^- Q.K.D. , . a . B . y 13. cos a + cos j8 + cos y = 1 + 4 sm - sin o s^"' o ' 14. tan a + tan j8 + tan y = tan a tan j8 tan y. 15. sina + sin/3 — siny = 4 sin - sin - cos -• . . «■ P ■ y 16. cos a + cos /3 — cosy = — 1+4 cos -cos 2 sin ^• 17. cot I + cot I + cot I = cot I cot I cot |- a B By y "^ t 18. tan - tan I + tan | tan ^ + tan ^ tan - =1. 19. cos^a + cos^yS + cos^y = 1 — 2 cos cr cos /3 cos y. 20. sin^a + sin^/S + sin'y = 2 (1 + cos a cos /S cos y). 21. 1 - sin^l _ sin^l - sin^| = 2 sin ^ sin | sin |- 22. sin2« + sin 2^8 + sin 2y = 4 sin a sin ^8 siny. 23. sin 2 a + sin 2 yS — sin 2 y = 4 cos a cos j8 sin y. 178 THE ELEMENTARY FUNCTIONS sin 2 a + sin 2 fi + sin 2 y „ . a . B . y 24. -. ■ -. — r^ = 8 sin - sin ^ sin ^ • sin a + sin yS + sin y ^ 2 z 25. cos 2 a + cos 2 j8 + cos 2 y = — 1 — 4 cos a cos ^ cos y. 26. cos 2a + cos 2/3 — cos 2y =1— 4 sin a sin /3 cos y. 27 . cot a + cot 13 + cot y = cot a cot ^ cot y + esc a esc y3 esc y. 28. sin-"^ = LI a •J/ _ (sin j8 + sin y — sin a) (sin y + sin a — sin j3) ,/8 29. cos ;- + cos ^ + COS Z u z 4 sin a sin j8 4 -t = 4 COS . ' cos T^ COS — - 4 3n; 3;3 3y 4 cos -^r- COS -h- COS -r^ ■ L Ji Z .2 1 _ (sin tt + sin /3 — sin y) (sin a + sin ^8 + sin y) 31. COS „ — A • • n 2 4 sm a sin /3 30. sin 3 a + sin 3 j8 + sin 3 y : Changes in the Trigonometric Functions as the Angle changes 136. Our experience with the use of the trigonometric func- tions has given us considerable information as to the way in which they change as the angle changes. Let us now follow out these variations more carefully. 137. The sine. Since the sine of an angle equals the quotient of the ordi- nate by the radius vector, we shall find it easy to follow the changes in the sine function if we choose points on the terminal line of the angle such that the radius vector is constant. This will be accomplished if the points are taken on the circumference of a circle, as in Fig. 110. As the angle a increases, the point P will move aroimd the circumference of the circle, from a position on the X-axis, when Pig. 110 THE TRIGONOMETRIC FUNCTIONS 179 the angle a = 0°, through the four quadrants, till it reaches the same position again, when a = 360°. The question is. What hap- pens to sin a as a; passes through this series of values ? When a = 0°, the ordinate of P is 0, and hence the ratio — = ; radius vector that is, sin 0° = 0. Now as a increases, the ordinate increases also ; and since the radius vector is constant, sin a must increase. This increase of the ordinate, and therefore of the sine function, continues through the first quadrant, until, when the angle is 90°, the ordinate equals the radius vector, and therefore sin 90°= 1. As the angle increases from 90° to 180°, the sine of the angle evidently decreases from 1 to ; as the angle passes through the third quadrant, sin a decreases from to — 1 ; and, finally, as a increases from 270° to 360°, sin a increases from — 1 to 0. 138. This review of the variation in the sine function should be completed by drawing up a table of the values of the function for all angles at 15° inter- vals, from 0° to 360°. Do not use the printed tables for this purpose, but take the re- sults obtained from our pre- vious study (cf. pp. 68, 164). When this table of values has Fto. Ill been written down, it will be found easy to construct the graph of the function y = smx from a;=0toa;=360by taking the values of the angle {x) as abscissas and the corresponding values of sin x as ordinates. Plotting the points located by the table in this way, and joming them by a smooth curve, we shall have a part of the graph of the function. Fig. Ill gives a smaU-scale representation of the curve obtained. The student should make a careful drawing on a larger scale. (The scale on the T-axis will need to be much larger than that on the X-axis, for obvious practical reasons.) 139. Since the increase of an angle beyond 360° means merely increasing the amount of rotation beyond one complete revolution, the sine function will change in precisely the same way while the 180 THE ELEMENTARY FUNCTIONS angle changes from 360° to 450° as it does while the angle changes from 0° to 90°; in other words, beyond 360° the same cycle of changes is repeated, because the terminal Une of the angle merely occupies the same positions again, in the same order. Thus the sine function repeats itself periodically at intervals of 360°. It is accordingly called a periodic function, with the period 360°. Anticipating the next paragraphs, we can see that all the trigo- nometric functions are periodic, with a period 360°. For negative values of the angle we have the fact that sin (—«)=— sin x, which enables us to extend the drawing of the sine graph to the left of the F-axis. This should be done in the final construction of the curve. 140. The cosine. By similar reasoning to that for the sine, and by referring to Fig. 110 again, the student may show that the cosine varies as follows : Angle 0° to 90° 90° to 180° 180° to 270° 270° to 360° Cosine 1 toO to -1 -1 to to 1 And, as in the case of the sine, a table of values of the cosine func- tion should be drawn up for angles at 15° intervals, from 0° to 360°. From this table points can be plotted and the graph of \q y = cos X drawn. y 141. The tangent. Since the tangent of an angle , ^, ^. ordinate equals the ratio abscissa we see at once that tan 0°=0 and that, as the angle in- creases, the tangent also increases through the first quadrant. For it we keep the abscissa constant, as in Fig. 112, the ordinate will evidently increase as the angle increases. As the angle a approaches 90°, the ordinate of the point Q, where the terminal line of the angle meets the line A^^, Fig. 112 THE TRIGONOMETEIC FUNCTIONS 181 grows larger and larger. This ordinate will eventually exceed any value that can be named if the angle a be taken sufficiently near to 90°; and thus tana will exceed any assignable value if a is near enough to 90°. If a= 90°, however, tan a does not exist, since the ratio — — -. — would then take the form (the abscissa abscissa of any point on the F-axis being 0), and a fraction whose denominator equals zero has no value whatever.^ The fact that the tangent function will exceed any assignable value, for angles in the vicinity of 90°, is expressed by saying, "tana becomes infinite as a approaches 90°," or, in symbols, tan 90° = co. (This may be read "equals infinity," but it is not to be understood as giving a valii,e to tan 90°, which, as we have just seen, is impossible ; on the contrary, it only expresses in brief symbolic form the fact that the tangent of an angle will exceed any assignable value if the angle is near enough to 90°.) As soon as the angle a passes into the second quadrant, the tangent is negative and numerically very large. Thus there is an enormous jump in the value of the function as the angle passes from a value a little less than 90° to one a little more than 90°. Contrast this behavior of the tangent function with that of the sine, which changes gradually or continuously as the angle changes from a value a little less than 90° to one a little greater than 90°. This is an example of the important distinction between a func- tion which is continuous at a point (as the sine at 90°) and one which is discontinuous at a point (as the tangent at 90°). Of course it is obvious that both functions are continuous for any value of the angle between 0° and 90°. Eetuming to the variation of the tangent: as the angle a increases from 90° through the second quadrant. Fig. 112 shows 4;hat the length of the ordinate decreases (the abscissa being kept constant), and hence tan a decreases numerically ; but as its value is negative, it actually increases from — co to 0. (The symbol — co 1 Notice that it is a very different tiling to say that a certain expression "has no value" and to say it "equals zero." Zero is a perfectly definite value (cf. note on page 8). 182 THE ELEMENTARY FUNCTIONS is to be interpreted in accordance with what was said above con- cerning the symbol oo.) As the angle increases from 180° to 270°, the tangent becomes positive and increases from to oo. Is it con- tinuous or discontinuous at 180 ? at 270 ? FiuaUy, as the angle increases from 270° to 360°, the tangent increases from — co to 0. 142. Remembering that tan (180°+ 6)= tan0, we see that this function has a period of 180°, unlike the functions sine and cosine, which do not repeat their cycle of values for a smaller interval than 360°. The student should now make a table of values of tan x for all values of X at intervals of 15°, from 0° to 360°, and make the graph of the function y = tana;. Fig. 113 gives a smaU-scale representation of a part of this curve, which should be drawn on a large scale, with considerable accuracy. 143. The graphical representation of these functions, as in the case of all others that we have studied, is a great help in forming definite and accurate ideas of the way in which the functional values change. To gain more detailed knowledge of these changes one must study that branch of mathematics which is known as Differential Calculus (see Chapter XI). That study enables us to answer questions about the rate of increase or decrease of func- tional values, whereas for the present we must be satisfied with the general information which the graph gives. Tig. 113 EXERCISE Study in the same way the variations in the other three trigo- nometric functions, and draw their graphs. THE TRIGONOMETRIC FUNCTIONS 183 Fig. 114 Polar Coordinates * 144. We have seen how any point can be located by means of its distances from two perpendicular straight lines; it is also possible to locate a point in various other ways, which are found useful in solving a great many prob- lems. The most important of these other ways is by means of _the radius vector of the point and the angle which the radius vector makes with a fixed line. Thus, the point P is located by the radius vector r and the angle 6, which is called the vectorial angle of P. The line OA is called the initial line, and the angle 6 may have any value, positive or negative. The values (r, d) are called the polar coordinates of the point P. Thus, in Fig. 115 the point P is completely located by the radius vector 2 and the vectorial angle 25°, which values are its polar coordi- nates. The radius vector is always mentioned first, so that the statement P = (2, 25°) locates definitely the point P. Similarly, Q = (i 200°) locates the point Q in Fig. 116, and Q={^, — 160°) locates the same point, as does also Q= (1 560°). It is thus clear that a point can have an indefinite number of pairs of polar coordinates (but a single pair of coordinates determines only one point). Moreover, it is customary to consider that a negative radius vector gives a point at the same distance from the origin as the corresponding posi- tive radius vector, but measured in exactly the opposite direction. Thus, the point P in Fig. 115 is not only (2, 25°) but also <^2, 205°), and Q in Fig. 116 is (-.^,20°) or (-1, 380°). Fig. 115 184 THE ELEMENTAEY FUNCTIONS EXERCISES 1. Locate each of the following points : (1, 45°); (5, 90°); (2, 0°); (^, 270°); (- 1, 100°); (2, 320°). Choose also other pairs of coordi- nates at random and locate the corresponding points. 2. Where can a point be if its radius vector is 2 ? if its vectorial angle is 180° ? 0° ? 3. Prove that the line joining the points (1, 45°) and (1, 135°) is parallel to the initial line. 4. Show that the points (0, 0), (2, 30°), and (2, 90°) form an equilateral triangle, and find the (polar) coordinates of the mid-points of its sides. 5 . The rectangular coordinates of a point (that is, its abscissa and or- dinate) are (2 V2, 2 V2). Find its polar coordinates. Answer the same question for the point (— 2 V2, 2 V2) ; for the point (2 V2, — 2 V2). * 145. If (r, d) represent variable coordinates, r being a function of expressed by the equation r=f{0), then the corresponding point will take various positions, the total- ity of which form the graph of the function The following ex- amples wlU illus- trate the way in which such graphs are studied. Example 1. r—\0. Making a table of corresponding values of r and 6, we have Fig. 117 etc. Evidently the graph is a spiral, as illustrated in Fig. 117. It is known as the Spiral of Archimedes. e 30° 45° 60° 75° 90° 135° 180° T n Hi 15 18| 22^ 33f 45 THE TRIGONOMETRIC FUNCTIONS Example 2. '• = sin 6. Making a table of values of r and 6, 185 r h .87 1 .87 i -h e 30° 60° 90° 120° 150° 180° 210° etc. Evidently all values of 6 in the third or fourth quadrant will give negative values of r, and hence the corresponding points will be in the first or second Fig. 118 quadrant. Furthermore, sin (90°+^) =sin (90°-^), so that the graph is sym- metrical with respect to the line ^ = 90° The curve is a circle (Fig. 118). Example 3. r = sin 2 d. Here the table of values is as follows : r .5 .87 1 ,87 .5 -.5 e 15° 30° 45° 60° ■ 76° 90° 105° r -.87 -1 -.87 -.5 .5 .87 6 120° 135° 150° 165° 180° 196° 210° etc. in the same order as for 6 in the first and second quadrants. We do not need to pay attention to the values of beyond 180°, because increasing B by 180° increases 2 « by 360°, and hence gives the same value 186 THE ELEMENTARY FUNCTIONS of sin 2 5 as if we had used $ itself. Thus, sin (2 • 195°) = sin (2 • 15"), sin (2 • 210°) = sin 60°, etc. Of course the values of r are negative for all values of 6 in the second quadrant (for 2 ^ is then in the third or fourth quadrant); hence the points on the curve are in the fourth quadrant (see Fig. 119). When 6 is in the third quadrant, however, 2 6 is in the first or second ; hence r is positive, and the points on the curve are in the third quadrant. The graph is the curve known as the four-leaved rose. EXERCISES Draw the graphs of each of the following equations in polar coordinates : 1. r = e. 2. r = cos 6. 3. r = l — coBO. i. r = cos 2 d. B. r = tan 6. 6. r'' = cos 2 e. ■ 6 7. r = sin-- 8. r = tan-' o 9. ?• = tan 5 • sin 5. 10. . = 1 11. r = sin- +1. Z 12. /• = sec ± 1. 13. r = 1 + sinf fl. 6 14. r = 1 — 2 cos e THE TRIGONOMETRIC FUNCTIONS 187 8 3 1 + 2 cos e 1 + cos e 3 8^ ,„ 8 22. r = acsc4- 16. »' = acos'-r- 19. r = - -• 2 o 3 — cos 6 a 17. ?-cos2fl = a''. 20. »- = 4(l-2cose). 23. r = asm''-- 24. ?• = atan^^secfl. . 6 d 29. r = sin- + C0S7^• 25. r=asm3e. , "^ ^ 26. r = a cos 3 6. an j.2 = _ a'sm^d + b^cos^e 27. »-2 = cos4e. 31. r = a tan 3 d. 28. r^ cos e = a' sin 3 e. CHAPTER X THE EXPONENTIAL AND LOGARITHMIC FUNCTIONS 146. In elementary algebra, the symbol a" was defined, where n represents a positive integer, as meaning a • a • a ■ a • • • to n factors. The number a is called the base, and n the exponent. The laws of operation are as follows : I. a*. a* = a" + * II. ^ = a''-*if A>;fc = ^ if k>h. III. («*)* = a**. . IV. («&)* = a* . bK EXERCISES The following easy exercises in review of these important laws should be worked over as often as is necessary in order to insure perfect mastery of them : a;y ' a + b ' (x — yy ' oio 2 . 2" . 2 = ? y = ? 2"' + 2 = ? (Do not multiply out.) 3 (x' + /)'■ _. [(a + bY-ir _^ • (x'-yy ■ (a + b-iy ■ i. of ■x^" = ? w" ■ av = ? 7. What is the value of (»y? of a''? (Note that (a"')" is wo< the same as a™".) 8. What is the difference in meaning between 2^ and (2'')^? 188 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 189 147. It is evident that our definition of a* has no meaning unless w is a positive integer, because we cannot multiply together a fractional number of factors or a negative number of them ; but it is equally evident, after a little thought, that the laws I-IV do Tiot lead to any absurdity if we suppose that the exponent can become fractional or negative. For instance. Law I gives for }i=zh = \, a* • a^ = «i J that is, a' is a number which, when mul- tiplied by itself, gives a. But " a number which, when multiplied by itself, gives a " is the square root of a. Hence Law I will be absolutely true for h = k = i if we simply regard a^ as meaning y/a. We shaU accord- ingly take this as the definition of a^. (Notice that Law III leads to the same resvdt, by taking h = ^, k=2.) By using the same process of reasoning with three factors we have a^ • a^ ■ a^ = a, and hence we must define a^ as meaning v a ; and by extending the method farther we find, in general, a« = "v^. (Or, still better, we may get this result by using Law III with h = - and k = n.) EXERCISES 1. Show (from Law III) that a^ ={-^T = ^ ; that a* = Vo" ; m that a* = -v^; and, in general, that a" =Var={VaJ'. 2. Write down the values of 9*, 8^, 4^, 16*. 148. Now let us proceed in the same way to find a meaning for a negative exponent by testing the laws I-IV. For instance, what should be the meaning of a-^ ? If Law I is to hold, a-^ • a^ = a2-i — a. . a 1 Therefore «" = ~2 ^ 37 ' a^ a Or h can be kept general : Therefore "'^""^"a" Similarly, it can be shown that «-» = ■ a" 190 THE ELEMENTARY FUNCTIONS EXERCISES 1. Show that the last statement is true whether w is an integer or a fraction. 2. Write down the values of i'', 4"*, 4"^, IQ-i, 10-", 9"^, 8~^, 8-* 8"*, 16-*. 3. Show that the first part of Law II holds even if A<A;. What does it become if h = k? What is therefore the meaning that we must give to a"? 4. Make a table of the powers of the base 4, using as exponents all values that give rational results, from 4"° to 4*, inclusive. The table should then have twenty-five entries. By its aid many problems of multiplication and division can be solved without any computation. Thus, 32 X 16 = ? Solution. From the table, 32 = ii, 16 = 42. Therefore 32 x 16 = 4^ . 42 = 4! = 512. Again, 4096 -=-128 = ? Solution. From the table, 4096 = 4^ 128 = 4^ Therefore 4096 h- 128 = i^-i = ii = 32. Work the following in a similar way : (1) 8 ^ 128 = ? (5) -v''4048 = ? | 32.64 ^ ^ (2) 256 • 8 = ? (6) VIO24 = ? ^ ^ N2.256 ■ (3) ^V • 2048 = ? (7) ■v'1024 = ? (10) (^ ■ 128)^ = ? (4) 5I5 . 64 = ? (8) V256 • -v/612 = ? > 149. Such a table as that of Ex. 4 is called a table of loga- rithms. For instance, in the equation 4^ = 64, 3 is called the logarithm of 64 to the base 4. In general, if a^=b, x is the loga- rithm of h to the base a. This is abbreviated j: = log„&. The number h in this equation is called the antilogarithm. Thus a logarithm is the exponent of the power to which the base must be raised in order to get the antilogarithm. "8~i=^" means EXPONENTIAL AND LOGARITHMIC FUNCTIONS 191 exactly the same as "loggi =-^." 8 is the base, - 1 the loga- rithm, and I the antilogarithm. The student should now practice translating equations from the exponential form to the logarithmic and back again. EXERCISES 8' = 2 is equivalent to log, 2 = ^. logj^lOO = 2 is equivalent to lO'' = 100, 1. Work the following in a similar way : (1) log,9 = ? (6) log,3 = ? (11) logj,8 = ? (2) log^oT^TT = ? (7) log,,a) = ? (12) log^81 = ? (3) log^^lO = ? (8) log^^lOO = ? (13) log^^lO . log,„100 = ? (4) log327 = ? (9) log^3 = ? (14) log,27 • log^3 = ? (5) log,16 . log,,8 =? (10) log,125 • log^S = ? 2. Translate into the logarithmic form, (1) 10' = 1000. (2) 16- J = f (3) 32-^ = ^. 3. What is the value of log^^l? of log„l? 150. Having thus fixed in mind the fundamental fact that logarithms are exponents, we see that the laws of operation with exponents must naturally give corresponding laws of operation with logarithms. Law I, »*.«*= «'' + * tells us that the exponent of a product is equal to the sum of the exponents of the factors, the base being the same. Hence, using the word "logarithm" for its equivalent, "exponent," we have, as the First Law of Loga- rithms, The logarithm of a product equals the sum of the logarithms of thg factors, all to the same base. a* 151. Similarly, Law II, — = a''-'', tells us that, the base beiug a* the same, the exponent in a quotient equals the exponent in the dividend minus the exponent in the divisor. Hence, using the word "logarithm" for "exponent," we have, as the Second Law of Logarithms, The logarithm of a quotient equals the logarithm of the dividend minus the logarithm of the divisor. 192 THE ELEMENTARY FUNCTIONS 152. Similarly, Law III, («'')* = «''*, gives as the Third Law of Logarithms, The logarithm of a power of a number equals the logarithm of the number, multiplied by the exponent of the power. For, calling ^ a^ = N, (1) TV* = a**. (Law III) (2) Translating (1) into logarithmic form, h = log^N. Translating (2) into logarithmic form, M = log„(iV*). Therefore log„(iV*) = h • log^N. Example, logg (4') = 3 log8 4. We can verify this by writing down the values of the logarithms on both sides of this equation, thus : 4" = 64, log,64 = 2; and log8 4 = |, 31og34 = 3-§=2 = log,64, as was to be shown. 153. Since the nth root is the same as the — th power, this law n also includes as a corollary the following : The logarithm of a root of a number equals the logarithm of the number, divided by the index of the root. 154. If we now consider the application of these laws of loga- rithms (taking for base some number > 1), we see at once that the tables we can make are very restricted in their scope. Thus, with the base 4, or the base 8, we can easUy give the values of log 2, log 4, log ^, etc., but not of log 3 or of log 5 ; and wiflh the base 9 we can give the value of log 3, but not that of log 2. We do not even know that there is any power of 4 that will give 3, or of 9 that will give 2. Thus, no matter what base we take, there will be large gaps in our table. How can this incompleteness be overcome ? Only by intro- ducing an assumption, which, although it cannot now be proved, EXPONENTIAL AND LOGAEITHMIC FUNCTIONS 193 yet leads to consistent results and is of the very highest utiUty. This assumption is, With any base greater than 0, except 1, every positive number has a logarithm ; that is, there exists a number x such that a^ = N, where iVand a are any positive numbers (a ¥= 1). It turns out that this logarithm x is in most cases an irrational number ; thus, log^3 and log9 2 are irrational. To find their approxi- mate value as rational numbers is a problem of no very great difficulty ; in fact, it is not much harder than extracting the cube root of numbers by the arithmetical method. This computation of logarithms, however, is best taken up at a later point in the mathe- matical course ; for the present we shall use the results obtained by other computers and pubhshed in tables of logarithms. (For further information see the articles "Tables'' and "Logarithms" in the Encyclopedia Britannica.) For practical purposes the base 10 is the most convenient. We shall therefore use that base altogether, so that, unless otherwise stated, logiVis hereafter to be understood as meaning logjgiV: EXERCISE Make a table of the positive and negative integral powers of 10, from 10" to 10- ^ and translate each item into its equivalent loga^ rithmic equation, thus : 10^ = 10, log 10 = 1, etc. 155. Using this elementary table, it will be seen that the loga- rithm of any (positive) number can be determined to the nearest integer at a glance. Thus, log 11, log 12, and the logarithm of any number between 10 and 100 will be between 1 and 2, that is, wiU equal 1 plus a decimal ; the logarithm of any number between 100 ^d 1000 will be between 2 and 3, that is, wiU equal 2 plus a dechnal ; and so on. The decimal part of the logarithm is called the mantissa ; the integral part, the characteristic. Using these words, we have at once the First Rule : TJie characteristic of the logarithm of a number betvjeen 1 and 10 is 0, of a number between 10 and 100 is 1, of a number between 100 and 1000 is 2, and, in gen- eral, is one less than the number of digits to the left of the decimal point. But if there are no significant figures to the left of the decimal 194 THE ELEMENTARY FUNCTIONS point (that is, if the antilogarithm is less than 1), another rule must be formulated. Noting that the characteristic of the logarithm, of a number between .1 and 1 is — 1 (since the logarithm is then — 1 plus a decimal), that the characteristic of the logarithm of a number between .01 and .1 is — 2, and so on, we have the Second Eule : When a number is less than 1, the characteristic of its loga- rithm is negative and numerically one more than the number of zeros after the decimal point before the first significant figure. EXERCISE Determine the characteristic of the logarithm of each of the followLQg numbers : 531, 24, .62, 7.1, .006, 4.06, .05001, 56.3, .403, .9, 45,000,000. 156. For the mantissa we use printed tables of logarithms, as has been said ; but their use is facilitated by the follow- ing principle : The mantissa of a logarithm is not changed hy moving the decimal point in the corresponding number ; that is, log 2, log .2, log .0002, log 200, log 20,000,000, all have the same mantissa. This important fact is very easily proved, thus : Given logiV, then we have, from §§ 150 and 151, log (10 i\^) = log ^ + log 10 = (log N) + l; and • log(^^ = logiV^-loglO=(logiV^)-l; in general, log (iV . 10*) = (log N) + k, k being an integer (positive or negative). Example. Log 2 = 0.3010, log 20 = 1.3010, log 2000 = 3.3010, log .2 = - 1 + .3010 (written 1.3010), log .00002 = - 5 + .3010 = 5.3010. 157. "We have now the ability to find, with the help of the table, the (approximate) values of the logarithms of aU positive numbers. The student should practice doing this untU the use of the printed tables becomes very easy. Then it is time to apply the laws of logarithms to problems. A few examples wUl first be worked out. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 195 Example 1. Find the product of 36 by 124. Solution. By the First Law, log (36 • 124) = log 36 + log 124. From the table, log 36 =i 1.5563 that is, IQi-^^ea = gg Also log 124 = 2.0934 that is, loa-0884 = 124 log (36 • 124) = 3.6497 ioi-6668 + 2.0934 - gg . 124 Therefore 36 ■ 124 = 4464 By the table, lO'**" = 4464 Example 2. Find the quotient of .0031 by .0925. iir -4- -0031 Solution. Write x = : .0925 then logx = log .0031 - log .0925. (Second Law) From the table, log .0031 = 3.4914 log .0925 = 2.9661 '' log X = 2.5253 Therefore x = .03352 Translate each equation also into the exponential form, as was done in Example 1. Example 3. Find the value of v3 to three decimal places. Solution. Let x = "v3 ; then log X = J log 3, by the Corollary to Law III (§ 153). From the table, log 3 = 0.4771. Therefore log x = 0.2385, and X = 1.732 , correct to four figures, that is, to three decimal places. , 213 X 3.23 Example 4. Find the value of • /594 Solution. If X is the required value, log x = log (dividend) - log (divisor) : log 213 + log 3.23 - i log 594. We find log 213 = 2.3284 log 3.23 = 0-5092 2.8376 Log 594 = 2.7738, ^ log 594 = 0-9246 log X = 1.9130 Therefore x = 81.84 + 196 THE ELEMENTARY FUNCTIONS Find (by expressions : 1. 3.14 X using .216. logs 3. 4. 5. EX] irithms) V21.5. ERCISES the value of each of the 9.003 X .3874 ■ 25 X .0067 following „ 293 </l.331. (l.Ol)i^- "• .4772 7. Vl4x 21 X 35 X 112 X 3.42. 8. Find the area of a circle of radius 2^ in. (Take log 7r = 0.4971.) 9. Find the volume of a sphere of radius 3 ft. 3 in. ; find also its weight if its specific gravity is 7.2 and a cubic foot of water weighs 62.5 lb. 10. Find the amount of $1 at compound interest for 10 yr. at 6% ; for 20 p. ; for 100 yr. Hint. The amount at the end of 1 yr. is $1.06; at the end of 2 yr., |1.06 . 1.06 = f 1.062 ; and so on. 11. Use logarithms to find the unknown parts and the area of the A ABC, given a =15.63, a = 26° 10', y = 90°. b Solution. a : cot a. Therefore b = -■ a cot a. Therefore log 6 = = log a + log cot a. Also c = a sin a Therefore logc = : log a - log sin a. log loga cot a log 6 = 1.1940 = 0.3086 = 1.5026 logo log sin a logc = 1.1940 = 9.6444 - = 1.5496 -10 b = 31.81 c = 35.45 Check. c2- -62 = a^ = {c + b){c- c + 6 = c-b = log(c + 6) = log(c-6) = log a^ = -6). 67.26. 3.64. 1.8278 0.5611 : 2.3889 loga = : 1.1944, which checks the above results. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 197 For the area, S =—■ 2 log a = 1.1940 log b = 1.5026 log I = 9.6990-10 log S = 2.3956 Therefore the area is 248.6+. Note. The student will observe that the logarithms of the trigonometric functions are given in a separate table. Negative characteristics are iisually written " 9 - 10" for " - 1," and so on. 12. Use logarithms to solve, check, and find the area, given (1) a = 421, ;8=54°35', y = 90°. (2) a = 37° 15', |8 = 7r54', a = 4.263. (3) j8 = 115°, y = 30°30', a = 1.00L (4) a =10° 40', y=150°10', c = .124. (5) a = 41° 34', a = 46.70, b = 60.03. (6) a = .698, c = .615, y = 38° 15'. (7) a = 3.16, S = 5.09, /8=147°59'. 158. These last problems make it clear that the use of loga- rithms is practical in solving triangles by the use of the Law of Sines; but it is evidently impossible to use them to so good advantage in cases where the Law of Cosines applies, because the Law of Cosines contains additions as well as multiplications, and these cannot be performed by logarithms. So it is necessary, in order to gain the advantage of the use of logarithms in the computations, to obtain modified formulas that shall include no additions or subtractions. We begin with three sides given. By the Law of Cbsines, cos a = 2 6c (Z (X but we have also cos a = 1 - 2 sin^ - = 2 cos^ - - 1. (Ex. 3, p. 1 1 9) ^ 7i2 _L rt2 /v2 Therefore 2 sin^ - =1— cosa = l r-7 2 2 oc 2lc-W--c^+a^ _ d'--(b- ef " 2 be ~ 2bc {a + b — c)(a — b + c) ~ 2 he 198 THE ELEMENTAEY FUNCTIONS The numbers a + b — c and a — b + c are very easily found, and so we have a practical formula for logarithmic computation : . 1 l(a+b -c){a-b + c) '^r=\- ibc ■ (') It is usual to abbreviate by writing 2s = a + b + c, or a + b + c s= ~ Therefore 2s — 2a = 6 + c — a, 2s—2b = a — b + c, 2s — 2c = a + b — c. Therefore (1) becomes sin - a = -\ ^ 'z^ -■ (1 ) 2 y be Similarly, by starting with 2 cos^ 1 a = 1 + cos a, let the student show that 1 rz r 1 \s(s — a) „, Dividing (1') by (2), tan^a=^<i-^J^^. (3) Using the other angles of the triangle instead of a, tani;8=-x|(^H«HiEi). 2^ \ s(s-6) tani,=JiH«)(£3. 2' \ s{s-c) The formulas for the tangent are generally the best to use, be- cause all three angles can be found with very little more work than would otherwise be needed to find one angle. One further abbreviation is very useful: Since and tanj«= 1 is-a)is-bns-c) 2 s —a y s tan^^- ^ {s-a)(s-b)is-c)^ 2 s — 6 \ s and tan^- 1 (s-a)is-b)(s-c) 2 s — cy s ' EXPONENTIAL AND LOGARITHMIC FUNCTIONS 199 therefore, if we write ^_ j(^:r^(s-b){s-c) ^ we have tan-a^ 2 s— a 2^^ s-b taii-y = ; 2 s — c or, using logarithms, log tan 1^ a = log r — log (s — a), and log r- = 1 [log {s-a)+ log {s-h)+ log (s-c)- log s]. Example. Given a = 46.34, 6 = 31.27, c = 55.03, to find the angles. a = 46.34 s-a = 19.98 6= 31.27 s-b =35.05 c = 55.03 s-c= 11.29 2s = 132.64 66.32 = s s = 66.32 (Notice that (s - a) + (s - i) + (s — c) = s, necessarily. Why so?) To find log r: logr = i[log(s - a) + log(s - b) + log(s - c) - logs]. log (s- a) = 1.3006 log (s- 6) = 1.5447 log (s-c) = 1^0527 3.8980 logs = 1.8216 log r2 = 2.0764 logr =1.0382 Therefore log tan ^ a: = 9.7376 - 10 log tan i/3 = 9.4935 -10 log tan iy =9.9855-10 Therefore ia = 28=39' i;3=17n8' ^y = 44°3' 90°00' Check Therefore a = 57° 18', j8 = 34°36', 7 = 88^- 200 THE ELEMENTARY FUNCTIONS 159. If the area is to be found, we may use the formula already proved (p. 159): Sz Now sma : 1 be sm a. 2sm-cos-; or, using (1') and (2) (p. 198), sin a = —y/s(s — a){s — b){s — c). DC Therefore S = ^s{s - a){s - h){s - c). This important result is called Hero's Formula} Notice that S = r • s also. EXERCISES 1. Solve and check, and find the area: (1) a = 13, b = li, c = 15. (2) a = .00365, b = .00846, c = .00697. (3) a = 100.1, b = 102.1, c = 104.1. (4) a = 3.194, b = 5.235, c = 6.118. J (s-a)(.s-b)(s- B 2. Prove that r inscribed circle. is the radius of the 160. The only remaining case of the solu- tion of a triangle that requires new formulas ^ for logarithmic work is that in which two sides and the included angle are given. Sup- pose the given parts are a, b, y, and a >b. Construct the triangle ABC (Fig. 120), and locate the point D on CB, so that CD = b ; then BD= a~b. Produce BC to JS, making (^ CE= b ; then 5^= a + b. The student may show that ZAEC = ZCAE = i, and that Therefore Also ABAD = a- ZDAE=W -^ a+^ a—^ Fio. 120 1 From Hero of Alexandria, who was the first to prove it (about the first century a.d.). His method was a purely geometric one. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 201 Therefore /.BAE= 90° + ^^ • Now apply the Law of Sines, first to ABDA and then to ABU A. . a — B . a — B sm — - — sm — - — We get = ^ = , (1) sm — - — cos ~ 2 2 sml90 H -— ) cos '^ a+6 \ 2 / 2 and — !— = = (2) c -7 .7 ^ ' sin 4- sm - 2 2 Formulas (1) and (2) (leaving out the second fraction in each line) are called MoUweide's Formulas.^ By dividing (1) by (2) we get a-B a-B tan — tan C ^Z^ = ?_^ ?_. (3) a+b a+fi V tan — cot — 2 2 This important result is known as the Law of Tangents ; it pro- vides an easy way of solving by logarithms in case two sides and the included angle are given. Example. Given a = 26.71, b = 20.89, y = 61° 32', to find a, p, and c. a + 6 = 47.60, a-h = 5.82, 9i±A = 90° - 3^ = 90° - 30° 46' = 59° 14'. 2 2 From the Law of Tangents, log tan ^^^ = log (a - ft) + log tan ^^^ - log (a + 6). log (a- h) = 0.7649 , ^ a + S 0.2252 logtan^ = ^;^g^ log(a + 6) = 1.6776 Therefore log tan ^^^ = 9.3125 - 10 and a - j8. : 11° 36' 1 After C. B. MoUweide (1774-1825), who, however, was not the first to dis- cover them. Formula (2) was first given by Newton in 1707, and Formula (1) by Friedrich Wilhelm von Oppel in 1746. 202 THE ELEMENTARY FUNCTIONS Therefore a = ^^ + ^^ = 70° 50^ ^ = 4^-^ = 4^38; Check. a + /3 + y = 180°. To find c, use either of MoUweide's Formulas ; thus (1) gives log c = log(a -b) + log cos ^ - log sin ^y^- log(a-i)= 0.7649 (a + J)sin5: log cos I = 9.9341-10 Check. c = ^ 10.6990 - 10 cos "-P .'^~P— a 5n94 _ 1 n 2 log sin ^^-2 = 9.3034 - 10 rc= 1.3956 "= ?M5 """ ^ 11.3865 - 10 log (a + 6)= 1.6776 logc= 1.3956 log sin ^= 9.7089-10 log cos ^^ = 9.9910-10 logc= 1.3955 EXERCISES Solve, check, and find the area : (1) a = 75, b = 70, y= 56° 18'. (2) a = 463, b = 499, y = 28° 6'. (3) b = 2.68, c = 1.69, a = 80° 54'. (4) a = 41.3, b = 28.7, y = 136° 28'. Note. Inasmuch as many of the formulas used have not been proved for the case a > 90°, the student should supply this omission here. 161. These examples illustrate but' a few of the very large number of applications of logarithmic computations. The subjects of surveying, navigation, and astronomy make constant use of th js labor-saving device. It may, indeed, be said to have revolutionized numerical work in every field. We now return to the study of the logarithmic function itself. The best way to get a general idea of this function (as of any other) is by drawing a careful graph of it. Accordingly, writing y = logi()a;, we lay off a> values as abscissas and y-values as ordinates (choosing a suitable scale, as usual). As we have had no definition of a loga- rithm of a negative number, no points can be located to the left of the y-axis. The details of forming the graph are left to the student. CHAPTER XI INTRODUCTION TO THE DIFFERENTIAL CALCULUS 162. The preceding chapters have been devoted to a study of various types of functions, — their fundamental properties, their graphical representation, and some algebraic and geometric results derived therefrom. The linear function, the quadratic function, the fractional function, the irrational function, the trigonometric functions, the logarithmic and exponential functions, all have their individual characteristics, which are in each case intimately connected with their graphs, and which must be well understood as a preliminary to the solving of many important problems. A complete vmderstanding of a function would involve a complete knowledge of the way in which the function changes as the inde- pendent variable changes. The object of this chapter is to lay the foundation for such an understanding. The branch of mathematics which investigates the rate of change of a function is called the differential calculus. 163. Although this branch of mathematics is of vast complexity and difficulty in its advanced aspects (indeed, no one can even faintly imagine the extent which future discoveries may give to its applications), yet in its simpler aspects it is exceedingly simple. This win be seen by examining one of the simple functional rela- tions, which has already been mentioned (p. 23), — the case of a man walking at the constant rate of 3 mi. per hour. If t represents the number of hours he has walked, and s the number of miles he has gone, then s is a function of t[s=f{t)]. Evidently the func- tional relation is the linear one s = 3 t. If we consider two different values of t, say t = 2 and t = 2|, the corresponding values of s are 6 and 6| ; that is, the additional distance he has gone in the extra 1 hr. is | mi. Again, if we take t = 3 and t = 3^, we find that the additional distance he has gone in the extra ^ hr. is 203 204 THE ELEMENTARY FUNCTIONS again | mi. In fact, it is evident that any ^-hr. increase in the time spent will produce an increase of | mi. in the distance cov- ered ; and that is exactly what is involved in the statement that his rate is constant. 164. Contrast with this the functional relation between dis- tance and time that exists in the case of a body falling freely near the surface of the earth. The Law of Falling Bodies ^ gives this functional relation to be (neglecting the resistance of the air) s = 16 <2, where t is the number of seconds the body has been falUng and s the number of feet it has fallen. If we give t two different values, say t=2. and t = 2\, we find the corre- sponding values of s to be 64 and 81 (that is, the additional fall during the extra \ sec. is 17 ft.) ; but if we take t=2> and t = 3|, we find the corresponding values of s to be 144 and 169, the additional fall during the extra \ sec. being thus 25 ft. In fact, we shall find that an increase of ^ in i will produce a different amount of change in s, according to the value of t chosen to begin with. This is what is involved in the statement that the rate of fall is not constant. 165. These two simple examples contain the essential material out of which the structure of the differential calculus is built; namely, the ideas of functional relation and rate of change of a function. A number of examples of a similar kind will now be considered, and although the computations involved are, as was said above, extremely simple, still they bring out the essential thing, which is, the effect upon a function of changing the inde- pendent variable by a certain amount. This " amount of change " is called an incre^nent, and is to be considered positive if the change is an increase, negative if the change is a decrease. Incre- ments are usually symbolized by the Greek letter A prefixed to the letter standing for the variable in question. Thus, " the increment of t " means " the amount of change in t " and is symbolized by At. In the examples above, A< = ^ (that is, ^ hr. in the first example and ^ sec. in the second example). As a further abbreviation the symbol 1 Discovered and proved by Galileo (1564-1642) ; published in his " Dialogues on Two New Sciences," in 10o8. INTRODUCTION TO DIFFERENTIAL CALCULUS 205 /(2) is used for « the value of f{t) when t = 2" or « the value of /(«) when x = 2." Thus, itf{t) = 16 <2, then /(2) = 16 • 2^= 64, /(O) = 16 . 02 = 0, /(a) = 16 . a2, /(J + 1) = 16 (J + 1^2^ g^^ Example 1. Take the function s = 5 < + 6, and find the effect upon the value of s of an increase of 1 in the value of i} (a) when t = 2, (b) when t = 4, (c) when t = J, (d) when t = t^. Solution, s =f(t) = 5^ + 6. (a) When < = 2, s =/(2) = 5 • 2 + 6 = 16. When < = 2 +1, s=/(8) = 5-3 + 6 = 21, Therefore the change in s, As, is equal to/(3) — /(2) = 21 — 16 = 5. (b) When < = 4, s =/(4) = 5 • 4 + 6 = 26. When / = 4 +1, s=/(5) = 5-5 + 6 =31. Therefore the change in s, As, is equal to/(5) — /(4) = 31 — 26 = 5. (c) When When Therefore (d) When When Therefore t=h> =/(*) = 5 ■ i + 6 = 8|. ( = J + 1, s =/(f) = 5 • I + 6 = 13^ ^s=/(i)-/(i) = 13i-8i = .5. ; = «i + 1, s =/(«! + 1) = 5 ft + 1) + 6 = 5 <i + 11. As=/ft + l)-/ft) = 5. This result shows, since <j is any value of t, that As = 5 for aH values of when A< = 1 ; that is, a change of 1 y in « produces a change of 5 in s, no matter what value of t we start with. In the same way it can be shown that a change of any amount A; in t pro- duces a change of 5 A< in s, whatever value of t be chosen to begin with. This means that the rate of change of s, compared with that of t, is con- stant, s changing five times as much as t does (As = 5 A(). The graphical representation of the function s will throw more light upon these results. If we use x and y to stand for the variables, instead of J and t, the functional relation will be y = 5 a: -I- 6. The graph of this is the straight line with slope 5 and F-interoept 6. The work above proved that for any value of x, Ai/ = 5 ^ In symbols this would be stated, Find the value of As when Ai = 1. 206 THE ELEMENTARY FUNCTIONS so that the rate when Aa; = 1. This means that, whatever point on the straight line we start with, whether A,B, C, D, or any other, an increase of 1 in a; produces an increase of 5 in ;/. Geometrically this is evident, because of the fact that the graph is a straight line. (The student may prove by elementary geometry that this is true.) Show further that the function y = rnx + k gives always Ay = m- Ax, whatever value of x be chosen to begin with. Example 2. Let us now consider a quadratic function, choosing y = x^ as the simplest one. Starting with any value of x we please, say x = x^, let us find out what change in the value of y will be produced by a change Ax in the value of x. When x = x^, y =f(x^ = x^. When X = Xj^+ \x, y =f{Xi + Ai) = (x^ + Aa;)^. Therefore £i.y =f{x^ + Ax) -/(a^i) = {x^ + Aa:)^ - x ^ = 2 ijAa; + (Aa;)'''. This is evidently not the same for all values of of change of this function is \ y not constant. For instance, if Xj = 1, A?/ = 2 Aa; + (Ax)^; while if Xj = J, A^ = Ax + (Ax)^, and if Xj = 3, A?^ = 6 Ax + (Ax)''. Refer- ence to the graph will make this fact still clearer. Thus,inFig.l22, if Ps(xi, ^j) and if PR = Ax, then RQ = Ay, SP=r y^ =A^i), TQ=f(ix^ + Ax) = y^ + Ay. It _ is geometrically evident that for the same value of PR the length of -RQ wiU be different accord- ing to where P is located upon the curve. This of course agrees with the result of the algebraic work above, which showed that RQ = 2x^Ax + (Ax)^, a length which is different according to the value of Xj chosen. Fig. 122 EXERCISES Work through the same study as in Examples 1 and 2 for each of the following functions, illustrating by the graph : 1. s=At + 2. 4. s = i!!-|. 7.y = 2x'-l. Z. s=-2t+l. 5. y=-6a; -f- 3. 8. 2/=l-a;». 3. y = 10x-f 25. 6.y = x' + 3. 9. y = 2-3x\ In particular, find the value of Ay (or As) when x (or ^) = 1 and Aa; (or A^) = l, .1, and .01 respectively. nsTTRODUCTIOlSr TO DIFFERENTIAL CALCULUS 207 166. Instantaneous rate of change of a function at a point. Keturning to the function y = x^oi Example 2, § 165, we had established the fact that for any given value of x, such as a; = a;,, Ay = 2 ajj . Aa; + (Aa;)^. This is the total' change in y produced by increasing x by the amount Aa;. Hence the average change in y per unit of Aa; will be found by dividing Ay by Aa;,i the average Av rate of change — ^ . or 2 a;j + Aa;. If now we take Aa; smaller and smaller, we get the average rate of change in y throughout an ever- diminishing interval, and this average rate will in this case (and generally) approach a deiinite value, here 2 a;j, as Aa; approaches 0. Av This value, which — ^ approaches as Aa; diminishes, is called the instantaneous rate of change of y relative to x at the point x — x^. This instantaneous rate of change is what we mean when we speak of the rate of change of any variable at a definite point or at a definite time. Thus, to state that a train is at a certain moment moving at the rate of 30 mi. per hour does not necessarily mean that it continues to move at that rate for an hour, or even for a single second ; it means, rather, that the instantaneous rate (namely, the limiting value of the average rate throughout an in- terval of time or of space, as the interval is conceived to diminish toward zero) is 30 mi. per hour. Of course it would be impossible by any process of observation or experiment to determiue what this limiting value is ; it is determined by a process of reasoning, which may be very simple or very difficult, according to the nature of the functional relation involved. Only in case the variable is changing at a constant rate, as in Ex. 1, p. 205, can we avoid this consideration of a limitiug value, because only then can we take any fixed interval we please and be sure that the rate of change for this interval is the same as that for any other interval. In that example (p. 205) we saw that As = 5 A<; that is, — = 5, a constant, so that the rate of change of s is always 5 times A^ the rate of change of t. But if the rate is not constant, we cannot 1 If the total change in y were 10 for Ax = 3, the average change in y per unit of Ai would be J^. 208 THE ELEMENTARY FUNCTIONS As / AwN do this, but must consider what happens to the ratio — I or —I as the interval At (or Aa;) becomes smaller and smaller. It is thus evident that a study of the rate of change of a variable will nec- essarily involve a study of limiting values, or, more briefly, limits. 167. Limits. The student has already met with a few examples of limits ; for instance, in elementary geometry it is stated that the limit of the perimeter of a regular polygon inscribed in a circle, as the number of sides is indefinitely increased, is the circumference of the circle, and that the limit of the apothem of the polygon is the radius of the circle. The following are other examples of limits. 1. If X represents a variable which assumes the series of values hhhh ' ' '• ^^^^ ^ approaches the limit ; in symbols, lim x=0, or x = Q. 2. If X takes on the series of values .9, .99, .999, • • •, then lima; = 1, or x = l. 3. If y==- and a; = 10, then y = -— ; this statement is abbrevi- ated thus : limy = 777' or lim (-) = -—• 4. If 2/ = 2 +h and h = 0, then y = 2; in symbols, lim (2 + h) = 2. 5. If y = — and x is positive and diminishes toward as limit, y ^ ■ ■ ■ /1\ will increase without limit ; in symbols, lim y = oo, or lim ( — 1 = oo. 6. If y = - and x is negative and approaches as limit, y wUl decrease without limit (since it is negative and its numerical value becomes larger and larger); in symbols, limy = — oo, or ii„(l)=. x=0-\x/ 00. '^- I^ y = oS ' "^liere m = 1, 2, 3, 4, • • •, y approaches as limit ; in symbols, lim ( — 1=0. n = Qo \^ / 8. If y = ^' where without limit ; in symbols, lim (— ) ^- ^^ 2^ =^' where ?!, = — 1, —2, —3, —4, ••-, y increases INTRODUCTION TO DIFFEEENTIAL CALCULUS 209 9. If a; = l + (-i)» where 71 = 1, 2,3,4, • . ., lima; = L 10. If a; = and z increases without limit, x approaches — 1 as limit, that is, lim a; = — 1 ; if a = 0, lim a; = |. 11. If y = &m.x and a; = 90°, y = l; that is, limy = l. 1 a: = 90 .. 12. If y = sin - and x^O, y has tw limit, because - increases iC . X without limit, and hence sin- will take on all values between X — 1 and + 1 over and over again ; that is, the values of y will oscillate as x diminishes. 13. li y = log X, lim y = 0. x = l 14. If P^ is the straight line joining any two points F and Q on a curve, and if Q approaches P along the curve, the straight line PQ will usually approach a limiting position PT, which is called the tangent to the curve at P. Also, ZBCQ=ZBDT (Fig. 123). Fig. 123 15. If two circles C and C, of equal radii, intersect, and if C approaches C, its center 0' remaining on the straight line O'O, then the points of intersection ^ and -^ will approach Q and S respectively, where Q and B are the ends of the diameter through O, perpendicular to O'O (Fig. 124). 168. Definition of limit. These examples make the idea of limit sufficiently clear so that we are prepared for a definition of the word: A variable v approaches a constant I as a limit if 210 THE ELEMENTARY FUNCTIONS \v — l\ becomes^ and remains less than any assignable positive number e (epsilon). The above thirteen algebraic examples should be thought through carefuUy to verify that this definition is satis- fied in each case. In the geometric examples (Exs. 14 and 15) it is either an angle or a length that is the variable, so that here also the definition will be seen to be satisfied. EXERCISES Show that the following limit equations are true : 1. If y =1+ -) where /i =1, 2, 3, 4, • • •, then v =1. n „ 1 -f-a; 1 2 . i± V = ) hm V = -■ 3. lim-^=l. x=oi- — x 4. lim = 0. X = cO J- -^ 5. lim-^=l. x = at J- ~r iC 6. lim [tan^(a; —■ ff)— cos a;] = sec^^. x = 180- 3-^ — 1 r'' — 1 7. lim f = 3 ; lim ~ = 2. x = 2X—i- x^lX—1 ^8 -j^ ^n '\_ 8. lim — = 3 ; lim —■ = n,n being a positive integer. x=\ ^ 1 x=i ^ — 1 9 lim ^(^+^ ) 1 ■ „^oo(»i + 2)(n + 3). -"• 169. Theorems on limits. The following theorems are easily seen to result from the definition of a limit : I. If lim v = l, then lim (v + c)=l + c, c being any constant. II. If lim V = 1, then lim (cv) = cl, c being any constant. III. If lim v■^ = Zj, and lim v^ = l^, then lim (v■^ ± v^) =l^± l^ IV. If lim'yi=Zj, and \imv^ = l^, then \\vsx.{v^v^=l.^l^. V. If lim v^ = \, and lim v^ = l^, then lim/^ j= -1 {provided l^ + 0). 1 |b — i| means " the numerical value of u — T'; that Is, i) — i itself if »>/, but I — n ii v <l. INTRODUCTION TO DIFFERENTIAL CALCULUS 211 170. The derivative. The most important case of a limiting value with which we have to do is that of the instantaneous rate of change of a function, described in § 166. It is the limit of the quotient -^ as Ax approaches 0. In the case of the function y = x^ we found that -^ = 2 ajj + Ax. Hence for this function we have the fact that Hm--^ = 2x.. This limiting value of the quotient -^ is called the derivative of y with respect to x at the point x = x^. In words, the derivative is the limit of the ratio increment of the function increment of the independent variable as the increment of the independent variable approaches 0. The symbol used is I>a:y]x^xi> ^^^^ "*''^® derivative of y with respect to X at the point x = Xj." Thus DM.-. = lim^ = ^■^/(^. + A^)-/(^,) If the function is s =/(t), the corresponding statement will be /,,.],_, =lim^*=lim>^^l±^?z£(i). Other symbols often used for the derivative of f(x) with respect to x are Dxf{x) and /'(«). The last form is especially convenient if we wish to show clearly for what value of x the derivative is to be understood ; for example, f'(Xj), /' (0), etc. mean " the derivative at the point x = x-^, x= 0," etc. 171. The derivative as the slope of the tangent to a curve. We have seen that the derivative of a function gives the value of the instantaneous rate of change of the function in relation to that of the independent variable. It can also be interpreted in another way, as can be seen by referring to Fig. 122, p. 206, which is the graph 212 THE ELEMENTARY FUNCTIONS of the function y = x^. In this figure EQ —Ay =/(x^ + Ax) — /(xj) = 2 x, Aa; + (Ax)\ and PH = Ax. The quotient — ^ = — is the ^ ' ^ " ^ Ax PB value of tan Z.BPQ ; that is, it is the slope of the line PQ. If we now let Ax diminish toward zero, the point Q will approach the point P along the curve, and the chord PQ wUl approach as limiting position the line PP', which is the tangent to the curve at P. Also, the angle EPQ will approach as limit the angle RPP' = 8, and therefore tan ZBPQ = tand; that is, Ay lim _^ = tan fl. Ax=o Ax Ay In words, the limiting value of — ^ as Ax approaches is the slope of the tangent to the curve at the point x = x-^; or, the derivative of y with respect to x gives the slope of the tangent to the curve at the point for which the derivative was formed. Thus, the derivative of the function y = ^ heing 2 iCj at any point a; = a;j (§ 170) we can say that the slope of the tangent to the curve y = a^ at the point x = x.^ is 2x^. 172. We shall now consider by means of some examples the method of finding the derivatives of a few functions. Example 1. Find the derivative of the function y = 2x''' — x at any point X = Xy Solution. Let y^ be the value of y when x = Xj. Then y^ =fixj) — 2x? — x^, and 2/i + Ay =f(x^ + Aj:) = 2 (Xj + \xf - (ij + Ax). Hence, by subtraction, A)/ = 4 XiAz + 2 (Ax)2 - Ax and ^ = 4x,-l+2Ax. Ax ^ Therefore lim ^ = 4x^-1; that is, D.y]x=..= 4xi-L Thus the slope of the tangent to the curve y = 2x^ — x at the point (Xj, y^) is 4 Xj — 1. Draw a figure. INTRODUCTION TO DIFFERENTIAL CALCULUS 213 Example 2. Find the derivative of the function y = -■ X Solution. From now on we shall drop the subscripts and write, for the point at which the derivative is to be found, (x, y) instead of (x^, y^. Then 2,=/(z) = i. X j^ y ■¥ /^y=f{xJr iix)-- Therefore Ay = X + Aa; 1 1 -Ax x + Ax X x(x+Ax) Ay^ -1 Ax X (x + Ax) and Hence lim ^ = - i (Theorem V, § 169), provided x 5^ ; a3;=oAx ^ x2 that is, Dj, (- ) = ^ > or the slope of the tangent to the curve y = - at the point (x, y) is -■ Draw a figure. X — 1 Example 3. Find the derivative of at any point (x, y). '^ 2x + o X — 1 Solution. y = 2x + 3 , . (x + Ax)-l ^ + ^^ = 2(x + Ax) + 3- Therefore X + Ax — 1 X — 1 ^2' = 2(x + Ax) + 3~2x + 3 (2x + 3)(x-l) + (2x + 3)Ax-(2x + 3)(x-l)-2Ax(x-l) (2 X + 3) (2 X + 3 + 2 Ax) 5 Ax '^(2x + 3)(2x + 3 + 2Ai;)' Ay 5_ aI~(2x + 3)(2x + 3 + 2Ax)' 1- ^y - 5 Therefore, ^lim — - ^^^_^_^y - /x-l\_ 5 that is, ^xy^ ^ ^ 3; (2 X + 3)2 The student should verify this result in a figure, using several special values for x, such as x = 1, x = 0, x = — 1. 214 THE ELEMENTARY FUNCTIONS EXERCISES Find the derivatives of each of the following functions, and draw figures : l.y = x' + 5x. ^ 2x + Z _ X '■y = ^r^r- ^^•^ = 5^^-6- 2. y = x". X ,_ 3 3.y = 3x^-4.x. ^-y^T^' ^^■y^\-x' 4. y = x^-ix+l. ^ ^ 2x 5. y = 2x^ + 3x-2. ^- y = y:^' "• y ~ T+T^ 1 cc — 2 X 6. y = -— r- 10. 2/ = — ^- 14.2/ x+1 " x-3 •' (1 + x^) 15. Find the derivatives of x^ -\-\, x' —1, x' — 3, and x^ + k. Compare with the value of Dja?). 173. Rules for finding derivatives. The derivatives of all the algebraic functions that we have studied in this book can be found ^ by methods like those illustrated by the above examples. In prac- tice these computations can be greatly shortened by breaking up more complicated functions into simple parts, according to the following rules : I. The derivative of a constant is zero. D^c = 0. For a constant does not change its value at all ; hence if y = c, Ay = for all values of A«. Therefore lim -^ = 0, which proves the theorem. II. The derivative of a variable with respect to itself is unity. For if y = x,Ay = Ax. Therefore — ^ = 1 for all values of Ax, , Av ^* and hence lim — ^ = 1. Axio Aa; III. If u and V are two functions of x, each having derivatives D^u and D^v, then the function u + v has as its derivative D^u + D,v. D, {u + v) = D^u + D^v. ^ That is, they can be found at every point for which they exist. A function inay have a value for some values of the independent variable for which never- theless no derivative exists, — for example, V* for a; = 0. INTRODUCTION TO DIFFERENTIAL CALCULUS 215 For if y = u-\-v, and if the increments of y, u, and v, caused by giving x the increment Aa;, are represented by Ay Aw, and Av respectively, then y+Ay = (u+Au) + (v JtAv) ; therefore Ay = Au+ Av, Ay Am Av Ax Ax Ax By hypothesis. 1- ^w ^ , ,. Av ^ iim — - = B^u and hm — - = B^v. Aj;ioAa; AiioAx Hence, by Theorem III, § 169, Iim -^ = B^u + B^v. Q.E.D. Ax = o Ax Evidently this result can be extended to the sum of any number of functions. Stated in words it is as follows : The derivative of the sum of any number of functions whose derivatives exist equals the sum of their derivatives. Example 1. We have found that D^ (x'^) = 2x, and that D^ (x^) = 3 x^ (Exercise 2, § 172). Therefore D^.{x^ + x^) = 3x^ + 2x, so that it is not necessary to work out the computation for the function x^ + x^ anew. Example 2. D^ (x") = 3 x% (Exercise 2, § 172) and D^(x''_3H-l) = 2x>-3. (Exercise 4, § 172) Therefore i)^(x3 + x^ - 3 x + 1) = 3 x''' + 2 x - 3. IV. If u and v are two functions of x, having derivatives B^u and B^v, then the function u • v has as its derivative uB^v + vB^u ; that is, B^{uv) = uB^v + vB^u. Proof. liy = ««, and Ay, Au, and Au have the usual meaning, then y + Ay = (« + Au) (v + Ay) = «u + uAv + v\u + A«At>. Therefore Ay = «Ak + «A« + AuAv, Ay Aw , A« j^ A« . ^-^ -i = ''-Kx^''^x^^x-^"- 216 THE ELEMENTARY FUNCTIONS AjT A?/ By hypothesis, lim —~=Dj.v, lim -— =Z)a?">^i"i) by Theorem rv,§ 169, lim — • At! = 0, since the limit of the first factor is D^u and the limit of the second factor is 0. A?/ * Therefore lim — ^ = uD^v + vD^u. q.e.d. Ax=0 Aa; In words, TAe derivative of the product of two functions is equal to the first times the derivative of the second, plus the second times the derivative of the first. Evidently this rule can be extended to a product of any number of functions ; thus, ii y = uvw, By = u ■ D (vw) + vw ■ Bu = uv • Bw + uw - Bv + vw • Bu, and so on for any number of functions. Special cases. (1) If v = c, any constant, the rule becomes B^ (c ' u) = cB^u + iiB^c. But, by Eule I, B^c = 0. Therefore I^xi'^'") = cB^u. Example. Since D^ (x^) = 2x, D^ (5 x") = 10 x, Da;(50 x^) = 100 x, etc. (2) Another special case is y = m", n being any positive integer. Then, using the extended form of the rule (for n factors), we have Z)^y = M"-ii)^M + M''-iZ)^M+M«-iX)^MH {n such terms all precisely the same). Therefore B^y = n . u'"-'^B^u. Example 1. Find the derivative of x^". D^ (xi») = 10 • a;» • D^x. By Rule H, D^ = 1. Therefore D^ (x^") = 10 x*. In general, we can write down the result for x" just as easily : Dx (x") = n • X" - ^D^x = n • x" - 1 (n any positive integer). Example 2. Find the derivative of (2 x^ + 3 x - 2)^. Let 3^ = (2 x" + 3 X - 2)= = u^, where « = 2 x^^ + 3 x - 2. Then R^y = 5 ■ u*D^u. INTRODUCTION TO DIFFERENTIAL CALCULUS 217 By Ex. 5, § 172, D^u = JD^ (2 x^ + 3 ^ _ 2) = 4 j; + 3. (This can also be obtained by direct application of Rule IV (1), Rule III, and Rule I.) Therefore Dx{2x'' + 3a; ^ 2)^ = 5(2a;2 + 3a; - 2)4(4:X + 3). -«(:)= vDm — uD^v 7 7 • „ . — ^ ^^) u and V being any two functions having derivatives D^u and D^v. In words. The derivative of a fractional function is equal to the denominator times the derivati've of the numerator, minus the numerator times the derivative of the denominator, divided by the square of the denominator (this excludes of course points at which the denominator equals zero). Proof. Let y = -• Then y + ^y ' Therefore Ay = « + Au M + A« M _ uA« — uAi) V + Ay V v(v + Aw) and Au All Ay Ax Ax Ax v(v + Ay) Therefore lim ^ = f^^^L^f^V. (Theorems III and V, § 169.) i,x=o Ax y2 Q.E.D. Special case. If u = c, any constant, this rule takes the form •ifi v^ since D^c = 0. The case d = c is not to be taken as a special case of this rule, because - is not really a fractional function at all (cf. § 68, p. 85). " 1 . . Eather it is to be treated as a product - • u, givmg c for example, D^\^j = ^ D^ = -g- 218 THE ELEMENTARY FUNCTIONS 174. By the help of the foregoing rules we can work out, with the minimum expenditure of time and effort, the derivative of any rational function. (The process of finding derivatives is called differentiation.) Example 1. Dififerentiate the function 3 :t:* — 6 a;' + 5 1'^ — 1. Solution. By Rule III, i)^(3 x*-%x^ + 0x^-1) = D^(Z X*) - D,(6 x') + D^(o x^) - D^ (1) = 3 D^ (x*) - 6 jDj; (xS) + 5 Z>a; (x^) - (Rules IV (1) and I) = 3-4x'-6-3x2 + 5.2x (Rule IV (2)) = 12 x' - is x2 + 10 X. Of course it is possible to take the intervening steps mentally, and the student should observe that we can vrrite down at once £>j:(3 X* - 6 x« + 5 x2 - 1) = 12 x3 - 18 x' + 10 x.. Example 2. Differentiate the function (4 x'' — 3 x)^ Solution. By Rule IV (2), Z>x(4: x2 - 3 x)' = 3 (4 x" - 3 x)^ Z»^(4 x" - 8 x) = 3(4x2-3x)2(8x-3). Here also the intervening step can be taken mentally, enabling us to write at once ^^ (4 x^^ - 3 x)' = 3 (8 x - 3) (4 x= - 3 x)K X" Example 3. Differentiate the function : l-x2 c, ■ 7^ / ^M (l-x2)2x-x2(-2x) 2x Solufon. D^ (^ = i L__J ) = __ EXERCISES Differentiate each of the following functions (doing as much of the work as possible mentally): ax + h 2x'-3a:'+l l+a;' 2. ©' 3. 1 4. 10 x'' 5. ex -\- d X ■ (^ • \l+xl 6. (.^V. 10.1+^ + ^' + xl 1+ X — x^ 1+x 11. (.T* - 5 x^ 4- 1)'. '''• l+x" 12. 8. x'-5x« + x'. i^ + c^x^ 13. Verify the results of Exs. 1-14, § 172, using rules I-V. Fig. 125 Fig. 126 INTRODUCTION TO DIFFERENTIAL CALCULUS 219 175. Use of the derivative in drawing graphs. Since the value of the derivative of y with respect to x (D^y) gives the slope of the tangent to the curve y=f{x) at any point, the derivative shows the direction of the curve at that point, the direction of the curve being that of the tangent line. In particular, if the derivative is positive, the tan- gent line makes an acute angle with the JPaxis, and the curve is rising as we go along it toward the right (Fig. 125). If the derivative is negative, the tangent line makes an ob- tuse angle with the X-axis, so that the curve is falling toward the right (Fig. 126). Finally, if the derivative is equal to zero, the tan- gent line is parallel to the X-axis (or coincides with it). Fig. 127 shows various forms of curve where this happens. Tlius, in drawing the graph of a given function it will often be foimd useful to compute the value of the derivative, and especially to note at what points it is positive, negative, or zero. (positive, the curve is rising, negative, the curve is falling, zero, the curve has a horizontal tangent. Example. Draw the graph of y = 2 a;' + 2^ — 8. Making a table of Gorresponding values of x and y, we find Fig. 127 X 1 2 - 1 -2 y -8 -5 12 -9 -20 220 THE ELEMENTARY FUNCTIONS These points give a general idea of the shape of the curve, but we can learn more about it by making use of the derivative of y with respect to x. This expression has the value for a; = and for x = — \, and is positive for all values of x except those in the interval from — ^ to inclusive, where it is negative. Thus the curve is rising except for values of x in the interval from a;=— Jtoa; = 0. Atar=0 (where y= — 8) the tangent is horizontal, also at a: = — ^ (where y — — 7f f ). Only between these points (A and B in Fig. 128) is the curve falling. This short interval would in all probability have been overlooked if we had not had the help of the deriva- tive in locating it. At all events, we could not have been certain what were the highest and lowest points on the little wave in the curve. AVe now know that A is the highest point in its imme- diate vicinity (because the curve is rising on the left of A and falling on the right of A), and that B is the lowest point in its imme- diate vicinity (because the curve is falling on the left of B and rising on the right of B). Such a point as A, that is, a point which is the highest point in its immediate vicinity, is called a maximum point ; while a point (such as B) which is the lowest point in its immediate vicinity is called a minimum point. 176. Maximum and minimum values. Summary. Thus we see that, in general, the value x = a will give a maximum point on the curve y =f{x) when D^fix) is zero at the point x = a, and is positive just to the left, and negative just to the right, of that point ; that is, f'{a -h)>0, f{a) = 0, /'(a + A) < 0, h being a small positive number. For in that case f{x) is increasing as x approaches a from the left, and is decreasing as x passes beyond a toward the right. Hence at x = a, f[x) must have the greatest Fig. 128 INTRODUCTION TO DIFFERENTIAL CALCULUS 221 value that it can have for any point in that immediate vicinity. For similar reasons the value x — a will give a minimum on the curve y =f{x) when f'{a -h)<0, f'{a) = 0, and f'{a + h)>0. These results are summarized in the following table: f(a - h) /'(«) /'(a + h) Maximum . . . >0 <0 Minimum . . . <0 >0 Example. Draw the graph oi y = 1- As we saw on page 87, the graph of any fractional function will have a vertical asymptote corresponding to any value of x that gives the denominator the value zero. Hence we have here the asymp- totes X = 1 and X = —1. The value of the derivative will give the direction of the curve at any point : l- x^-x(-2x) Dxy = (1 - x^y (1 -x^y Since this is always positive, the curve is always rising; hence there can be no maximum or minimum point. Fig. 129 EXERCISES Draw the graphs of the following functions, making use of all the information which the derivative gives, especially with regard to maximum and minimum points : 1. y = x'— 4a; 4- 4. 2. 2/ = 3a;'-a;^+l. 3. y = x\ 4. y = x'-a;'+l. h. y = 6. y x+_2 l-x' x + 2' 8. y- 1+x \-irX-\-x'- 9. Discuss the function ax^ + te + c for maximum and mmimum values. Compare the results with those obtained by more elementary methods in § 48 (p. 58). 222 THE ELEMENTARY FUNCTIONS 177. Differentiation of irrational functions. The only irrational functions that we have considered have been those involving square roots. Any such functions can be differentiated as follows : If y = Vm, where u is any function of x that has a derivative, then y + Ay = V w + Am. Therefore Ay = y/u+Au — ^ni Ay _ Vm + Am — Vm and Aa; Aa; Am In this form it is not easy to see what limit the quotient — ^ will approach as A« — ; but by multiplying both terms of the fraction by Vm + Am + Vm we get ^y _ (^^ + Am — Vm)(Vm + Am + 'v^m) A* Aa; ( Vm + Am + Vm ) _ (m + Am) — w _ 1 Am Aa;(VM + Am + Vm) Vm + Am + Vm ^* Now, as Aa; = 0, — = Bji and Vm + Am = Vm, since Am = 0. Aa; Therefore lim —^ = — ;= Bjw : Ax=oAa; 2V M that is, B^y = — -= • B^u. (1) 2VM In words this result is, The derivative of the square root of a function is equal to 1 divided hy twice the square root of the function, times the derivative of the function. Notice that (1) can be written i)^(M*) = lM-*D,M, which is exactly the form we should get by using Rule IV (2) (p. 216) with n = \; that is, the rule B^{u^)=n • u^-^ ■ B^u holds when n has this fractional value, as well as for the posi- tive integral values for which the rule was proved. As a matter of fact, the rule is still true if n has any fractional value, but INTRODUCTION TO DIFFERENTIAL CALCULUS 223 we do not need to use that general fact in our work. For our purposes it wHl usuaUy be found simpler to use the form of equation (1) than to change to the form of a fractional exponent. Example 1. Find the derivative of V2 a; — 3. Solution. D^ V2^-3 = ^ D^(2x-S) = . 2V2X-3 ' V27:r3 Example 2. Find the derivative of x Vl — x^. Solution. D^{x Vl - x^) = Vl - x" Z»^x + xD^ VT = Vl - a;2 + : x" -2x l_2a 2Vl-a;2 Vl-a;2 EXERCISES Find the derivatives of each of the following functions : 1. V5x. 4. Vx^ - 4. 7. VlOO - 25 a?. ,„ b 10. - Va^ - 3?. a 2. Va;2 + 1. Ia;+1 8. ■ ^- \^Z^- Va;+1 11. a:Wx+l. -I a; „ 3a; 1 Draw the graphs of the functions in Exs. 1-9, locating any maximum or minimum values. 178. Differentiation of implicit functions. When we have an equation in the variables x and y which determines y as an im- plicit function of x, it is easy to write down the value of B^y without first solving the equation for y. Eules I-V (pp. 214-217) are all that is needed. Example. a;^ + 4 / = 4 (1) (1) being true, the derivative of the left-hand side must equal the derivative of the right-hand side, „ , „ „ . ^ Z)^(3:2-f-4y2) = Z)^(4); that is, D^ (^2) + D^ (4 y^ = 0. Therefore 2 z -I- 4 • 2 ^ D^y = 0. Therefore D^y^ — -. — 42/ (The above work assumes that the function y has a derivative. This assumption is justified in every case with which we shall have to do.) 224 THE ELEMENTARY FUNCTIONS EXERCISES Find D^y from each of the following equations : 1. 0^ + %/= 9. Ans. D^ij=--- 2. 4x^ + 2/' = 4. ^ 3. 9x2 -42/= =36. 4. xy -\- y'^= 4. Ans. D^y = — _^ , ^ ^ ■ 6. x''-2xy + y'' + ix-Sy=0. 6. x' + / = 65. 7. x"y + a;?/" = 26. 8. (x-«f +(y_^/ = rl 9. 6V + ay = a^'Sl a? ^ b' ~ 12. ax" 4- 6x2/ + cy" + c?x + ey +/= 0. 13. Prove that the equation of the tangent to the parabola 2/" = 4 ax at the point (Xj, y^) is yy^= 2 a(x + x^). 14. Obtain the equation of the tangent to the hyperbola xy =1 at the point (Xj, y^) in the form y^x + x^y = 2. 15. Show that the hyperbola of Ex. 14 and the circle x' + y'' = 2 have the same tangent at their common points ; that is, that they are tangent to each other. 16. At what point on the ellipse ix^ + 9 y^ = 36 is the tangent parallel to the line y =— x? 17. Verify the results of Ex. 17, § 109, and Ex. 2, § 110, by finding the value of D^y in each case. 18. Prove that the tangents to a parabola at the ends of the latus' rectum intersect at right angles on the directrix. 179. Summary. We have now seen how to find the derivative of any rational function, and of irrational functions of degree not higher than the second. Since the derivative of y with respect to X gives the slope of the tangent at a point, we can find the equa- tion of the tangent to any curve that is the graph of one of these functions. INTEODUCTION TO DIFFERENTIAL CALCULUS 225 The derivative gives also the rate of change of the function per unit change of the independent variable, so that this impor- tant problem is solved for all the algebraic functions mentioned. We have accordingly reached the point where we can consider the applications of this work to problems from various fields. Example 1. The Law of Falling Bodies. If s = 16 fi (1) states the relation between distance and time in the case of a falling body, then the instantaneous rate of change of the distance with respect to the time, that is, the velocity of falling, is given by Dts. Since DfS = 32 t, we have at once the formula ti on < /ox V = Dts = 32 t, (2) which is well known from elementary physios, where it appears as a second formula, as if it were independent of (1). But we now see that it is not, since, if (1) is true, (2) must be true also. Example 2. If a body moves according to the law s = <^ — 10 < + 2, find its velocity at any instant t. When is it moving in the positive direction, and when in the negative ? Solution. The velocity is given by D(S = 2t — 10, which is negative until t — 5, when it equals zero ; for <> 5 it is positive. Hence the body is moving in the negative direction until t = 5, when it has zero velocity, after which it moves in the positive direction. Example 8. A ladder 40 ft. long rests against the side of a house. If its foot A is pulled away from the house at the rate of 5 ft. per second, how fast is the top of the ladder descending when the bottom is 10 ft. from the house? Solution. Let x = AC, the distance from the house to the foot of the ladder ; and let y = CB, the height of the top of the ladder. Then y is a function of x, and the problem is to find the rate of change of y, knowing that of x. From the right triangle ABC ^2 ^ igoo - x^ Therefore 2 yD^y = -2x, X or DxV = ~ " ' that is, y is changing at any instant times as fast as x is. We are to find the rate of change of 2/ when :r =K). When x= 10, y = VI6OO - 100 = Vl500 = 10 Vlo ; W - J_ hence ^^^ = " I^VW ~ VlS' 226 THE ELEMENTAEY FUNCTIONS Therefore y is at that instant changing ■r= times as fast as x. But, by vl5 the conditions of the problem, x is changing at the rate of 5 ft. per second, and accordingly the rate of change of y is p= • 5 = — - vl5 ft. per Vl5 ^ second. The negative sign shows that y is diminishing. EXERCISES 1. Two persons start from the same point and walk in directions at right angles to each other at the rates of 3 mi. per hour and 4 mi. per hour respectively. How fast are they separating after 15 min. has elapsed ? 2. A light is 10 ft. above a street crossing, and a man 6 ft. tall walks away from it in a straight line at the rate of 4 mi. per hour. How fast is the tip of his shadow moving after 10 sec? after 1 min.? How fast is the shadow lengthening in each case ? 3. A point moves along the parabola j/" = 8 a;. How does the rate of change of the ordinate at the point (^, 2) compare with that of the abscissa ? 4. At what point on the parabola of Ex. 3 are the ordinate and the abscissa changing at the same rate ? 5. When a stone is thrown into still water, how fast does the area of the circle formed by the ripples change in comparison with the radius ? 6. If the radius of a spherical soap bubble is 3 in. and is increasing at the rate of ^ in. per second, at what rate is the volume increasing ? 7. If a point moves along the curve a? + 2y^ = 2, how does the rate of change of the ordinate compare with that of the abscissa at the point (I, ^)? 8. An automobile track is in the form of an ellipse whose major axis is ^ mi. long and whose minor axis is \ mi. long. If a car moves around the track at the rate of 40 mi. per hour, how fast is it mov- ing toward the north, and how fast toward the east when it is at the end of the latus rectum ? (Suppose the length of the track to be from north to south.) 9. A man on a wharf is pulling in his boat by means of a rope fastened to its prow. If his hand is 10 ft. above the boat and he pulls in the rope at the rate of 3 ft. per second, how fast is the boat coming in when 20 ft. away ? when 6 ft. away ? INTRODUCTION TO DIFFERENTIAL CALCULUS 227 10. A railroad crosses a road at right angles by an overhead crossing 30 ft. high. If a train and an auto cross at the same time, the one at the rate of 30 mi. per hour, the other at the rate of 15 mi. per hour, how fast will they be separating after 1 min.? 11. One ship sails east at the rate of 10 mi. per hour, another north at the rate of 12 mi. per hour. If the second crosses the track of the first at noon, the first having passed the same point 2 hr. before, how is the distance between the ships changing at 1 p.m.? How was it at 11a.m.? When was the distance between them not changing? 180. Problems involving maxima and minima. As we have seen, the use of the derivative of a function enables us to discover for what values of the independent variable the function will have a maximum or a minimum value. This fact is of use iu solving a great variety of problems. Example 1. Find the rectangle of greatest area which can be inscribed in a given circle. Solution. Let r be the radius of the given circle, and 2 x and 2 y the dimen- sions of the rectangle (Fig. 131). Then ^2 _ ^ _ J.2 or y = Vr" — x^. The area of the rectangle is 4 xy, which is to be made the greatest possible by choosing the proper value of x (and hence of y, wh ich is a function of x). Now i xy = i x Vc^ — x- =f{x), so that the question is, For what value of x will f(x) have its maximum value ? For that value of X, as we have seen (p. 220), we may expect to find D^f^x) = ; that is, Fig. 131 Therefore But Da;[x^r^-x^] Therefore : Vr2 - !» - ■ ■2x^ ■■-2x^ = 0, Vr" - x'' V2 When X < -^ > Dxf(x) > 0, and the function is increasing as x increases ; V2 when a;>-7=' Dxf(^)<0> ^.i^ t^® function is decreasing as x decreases. V 2 228 THE ELEMENTARY FUNCTIONS That is, the function is increasing before x reaches the value V2 and decreasing after x passes the value V2 Hence that value of x gives a maximum value of /(x) ; that is, of the area of the rectangle. Geometrically it is also evident that this value is a maximum rather than a minimum, because as X increases from very small values the area of the rectangle also increases, but beyond a certain point it will diminish again, approaching as i approaches r. The locating of this "certain point" as being the value is made possible by the use of the derivative. r Notice that when x = — = , V2 a square. : — -= , so that the maximum rectangle is V2 Example 2. What would be the dimensions of a cylindrical steel tank, open at the top, to require the least material possible and still contain 50 gallons ? Solution. The quantity of material used will be least when the area of the surface is least. The surface con- sists of the side, whose area is 2 wxy, and the bottom, whose area is ■itx'^ ; total, S = 2 ■n'xy + ■rrx^ (x represent- ing the radius of the base and y the altitude, as in Fig. 132). In this expression y is a function of j, whose value is determined by the fact that the volume is to be 50 gal.; that is, 6.684 cu. ft. The volume of the cylinder = irx^y. Fig. 132 Therefore nx^y = 6.684, 6.684 2' = -Tir- Therefore .. „ ■ 6.684 , 2 13-368 ^ , This function of x is to be made a minimum. Therefore Da:S = 0. T^ C-, 13.368 . „ DxS = ;— -I- 2 w = 0. 13.368 3/6.684 X = -»/ = 1.286, approximately. From the nature of the problem this value makes the function S a minimum, not a maximum (as can also be verified by noting that Dq.S INTRODUCTION TO DIFFERENTIAL CALCULUS 229 changes sign from — to + as i increases through the value 1.286), so that the value x — 1.286 gives the radius of the base that will require the least amount of material for the tank. When X = 1.286, y — — — -- = 1.286, approximately ; that is, y = X. The exact value of x is -* / — > which equals V^, so that x = Vifiy- Therefore x = y exactly, in fact. Thus the height of the tank should equal the diameter of the base in order to require the least amount of material. EXERCISES 1. Work through Ex. 2 above, using V as the volume of the tank, instead of 50 gal. Show that y = x will give the proportions to require the least material. 2. Divide a length AB into two parts so that the product of the lengths of the segments shall be the __;; __^_ greatest possible (Fig. 133.) Fig. 133 3. Divide a length AB into two parts so that the sum of the squares of the lengths of the segments shall be the least possible. 4. A piece of cardboard 10 in. square has a small square cut out at each corner, and the remainder is folded up so as to form a box. How large a square should be cut out so that the box may be the largest possible ? Ans. | in. is the side of the square cut out. 5. A rectangular piece of cardboard, 15 in. x 7 in., is to be made into a box by cutting out a square of equal size from each corner. How large should this square be in order ^ that the box may have the largest possible contents ? 6. A person in a boat, 6 miles from 6 miles 10 milea Q Beach the nearest point of the beach, wishes to i-m. 134 reach in the shortest possible time a place 10 miles from that point along the shore; if he can walk 5 mi. per hour, and row 4 mi. per hour, where should he land ? (Fig. 134.) 230 THE ELEMENTARY EUNCTIONS 7. A rectangular box, open at the top, is to be constructed to contain a certain volume V. What must be its dimensions in order to require the least amount of material ? 8. Find the largest rectangle that can be inscribed in an isosceles triangle whose base is 4 in. and whose slant height is 10 in. (Fig. 135). 9. Find the largest cylinder that can be inscribed in a given sphere. 10. Find the largest cone that can be inscribed in a given sphere. 11. What should be the dimensions of a cylindrical tin can (both ends being closed) in order to require the least material ? 12. The lower corner of a sheet of paper whose width is a is folded over so as just to reach the other edge of the paper. How wide should the part folded over be in order that the length of the crease may be the minimum ? 13. Find the shortest and the longest distance from the point (4, 6) to the circle x^+f= 25. 14. Find the shortest distance from the point (0, 1) to the hyperbola xy = 1. Show that this distance is measured along the normal to the curve through the given point. 15. If the cost of running a steamboat is proportional to the cube of the velocity generated, what is the most economical rate of steaming against a 3 mi. per hour current ? PORTION OF GREEK ALPHABET Alpha a Lambda Beta i8 Pi Gamma y Eho Delta A 8 Tau Epsilon e Phi Theta 6 Omega <^ 231 APPENDIX A PROOF THAT THE DIAGONAL OF A SQUARE IS IISrCOMMENSURABLE WITH ITS SIDE This theorem proves the fact asserted in the text (p. 3), that there exist segments which cannot be measured in terms of a specified unit. The fact to be proved may be stated in the following words : If the side of a square be taken as unit, there exists no number — n (m and n being integers) such that the length of the diagonal of the square is — n Proof. By the Pythagorean Theorem, d^ = 1^ + 12 = 2. (1) We can prove that the assumption rf = — , where m and n are integers, leads to a contradiction. Suppose — to be in its lowest terms; that is, n suppose m and n to have no common divisor. Then, since d^ = 2, ?n2 _ Therefore m^ = 2 n\ (2) Therefore rrfi is divisible by 2. This is only possible if m, is an even integer (since the square of an odd integer is odd) ; and since m is even and — is in ' n its lowest terms, n must be odd. Further, since m is even, we can write m = 'ik, where k is some integer. Therefore nfi = 4:B. , (3) Comparing (3) with (2), 2 ra^ = 4 P ; that is, n2 = 2F. (4) Hence n^, and therefore n, is even. 233 234 THE ELEMENTARY FUNCTIONS But we proved above that n is odd ; hence a contradiction results from the assumption that d = —■ Therefore the assumption is false, and there n wi does not exist any number — representing the diagonal of a square whose side is the unit. Note. This proof is of ancient origin, being given in almost exactly the above form by Euclid, in his famous " Elements of Geometry," in the third century b.c, and being referred to by Aristotle, as well known, much earlier. It is very probable that it was discovered by Pythagoras himself in connection with the study of his famous theorem (see p. 4, footnote). APPENDIX B LAWS OF OPERATION WITH RADICALS The following are the laws of operation with irrational numbers that are in the form of radicals : I. Multiplication. Va • a/* = -Vab. Special case. Va ■ Vfi = Va6. Examples. V2 • Vs = Vg ; 712 = Vi • VI = 2 Vs. II. Division. —= = ->M-- ; but in no ease may the denominator equal 0, since division by is an impossible operation. Examples. J| = ^; ./! = Je^lVG; Jl = JI = 1 Vs. V4 2'\3 \9 3 '\5 \25 5 III. Addition and subtraction. Only similar radicals can be added or subtracted; that is, radicals that involve the same root of the sarne numher. Examples. V2 + Vs cannot be simplified, nor can V2 + V^, but V2 + Vl8 =V2 + 3V2 = 4V2= V32. (It is a very common error to write v a + ft = V a + Vj, whicli violates this law.) EXERCISES Give the value of each of the following in the simplest form : 1. VI. 4. (2-V3)(2 + VI). ^ V6 + V2 o 2-V3 ^ V^^_V^3-3 ■ 2 + V3 2 + Va _ ■ V^TjTi" + Va;-3 „ 4 - V6 - V2 3. 2-3V2 e. ^^ ■ V6-V2 3 + 2 V2 1 - Vl - 3; Y Hint. A fraction containing Va + Vft in the denominator is simplified by multiplying both numerator and denominator by Vo — V6. Accordingly, in Ex. 3, we should multiply both terms of the fraction by 2 — V3. 9. Prove: (a) -^ = ^. (b)-;^— = 1+V2. (c) ^^^E^li^ = 9 - 4 Vs. ■v^ 2 ^ W2-I 5V5+IO (d) V3 - 2 V2 = (V2 - 1). 235 APPENDIX C TO CONSTRUCT A SEGMENT HAVING A RATIONAL LENGTH 1. To divide a given segment AB into any number of equal parts : Draw any line AE making an angle with AB. On AE lay off any length, as AD, n times, and let the last division point be E. Draw EB, and from all the other division points draw lines parallel to EB. These parallels intersect AB in the points of division required, and one of the resulting segments is - oi AB. (Why ?) EXERCISES Take a random segment as unit, and construct J, J, J, \, J, J, If, 2J, 0.6, 3.1, 5.3. Take two random .segments, a and b. Construct Ja + J 6, 3a + 4i, l|a + 2ii. 2. To divide a given segm,ent AB into two segments having any given ratio, as m : n. This problem is solved by a method very similar to that above, and it is accordingly left to the student as an exercise. 236 APPENDIX D TO CONSTRUCT THE SQUARE ROOT OF ANY GIVEN SEGMENT This important problem of construetion is solved by means of the following Theorem. In a right triangle the altitude drawn from the vertex of the right angle is the mean proportional between the segments into which it divides the hypotenuse. The proof is found in any textbook of elementary geometry. By using this theorem we can construct the mean proportional between any two given segments, as follows : Let a and b be the two segments. Draw AB = a, and produce it to C so that BC = b. Draw a semi-circle on 4 C as diameter. At B erect the perpendicular to A C, meeting the semi-circle in D. Then BD is the mean proportional between AB and BC ; that is, between the given segments a and b. The proof follows at once from the fact that the angle ADC is a right angle. Now, to construct the square root of any given segment m, construct as above the mean pro portio nal to m and 1. This mean proportional then has the length Vm ■ 1 ; that is, it is the required Vro. EXERCISE Take any random segment and construct its square root. 237 APPENDIX E SIMULTANEOUS LINEAR EQUATIONS IN TWO UNKNOWN QUANTITIES If we have the two equations r3x-72/ = l, (1) {; I2x+ 2/ = 12, (2) which are both to be satisfied by a certain pair of values (x, y), we proceed to combine the two equations in such a way as to obtain a new equation which shall contain only one unknown quantity. To do this we may add (1) and (2) as they stand or after they have been multiplied by any number we please. If we multiply (2) by 7, and add the result to (1), it is clear that we shall get rid of the y-term altogether; we call this "eliminating y" from (1) and (2). Thus, (2) multiplied by 7 gives 14a; + 7?/ = 84. Adding this to (1), 17 x = 85. Therefore x = 5. If we substitute this value of x in (1), we get 15-7y = l. 7y = 14. y = 2. Therefore the pair of values (6, 2) will satisfy both equations. We could also have eliminated x from (1) and (2) by multiplying (1) by 2 and (2) by 3 and subtracting : 6a;-14y = 2, 6a; + 3 2/ = 36. Therefore 17?/ = 34, y=2. 238 APPENDIX E 239 Substituting ?/ = 2 iu (1), 3x-14 = l. 3x = 15. X = 5. Therefore (5, 2) is the solution. This result should be checked by substituting x = 5, y = 2 in (1) and (2). It is clear that this procedure can always be adopted, giving a new equation in which one or the other of the two unknowns fails to appear. APPENDIX F THE QUADRATIC EQUATION IN ONE UNKNOWN QUANTITY Suppose we wish to solve the equation x'' + 6 r = 7. The method of solution most commonly employed in elementary algebra is that known as " completing the square," because it consists in adding to the left side of the equation such a number that that side becomes a perfect square. Success in using this method therefore depends upon familiarity with the form that an expression must have if it is a per- fect square. We know that (x + /.-y = x' + 2 kx + k", which shows that when the terms of a complete square are written in this order, the middle term (2 kx) is tirice the product of the square roots of the other two terms. As one of these square roots is x (the first term being x^), the other one must be half the coefficient of x in the middle term. Thus, in the equation x^ + &x = 7 above, a?-\-&x-\-k^ will be a perfect square if 6 = 2 A, that is, if /c = 3 ; hence the quantity to be added to x'' + 6 x to make it a perfect square is 2>^ ; x^+6x + 9=(x + 3)l Similarly, to x^ + 8 x we must add 4^ to complete the square, to x'^ -\- 5x we must add ( x ) , and to x^ -\- mx we must add ( -x- ) • In words, The quantity to be added is the square of half the coeffi- cient of X, when the coefficient of\c'^ is 1. EXERCISES Complete the square of each of the following : x^ + 10 X. z^ + (m + n) X. x^ + ^ x. x^ + 15 X. x^ + ax. 2 ab x^ + ^x. x^ + ^x. a + b ' Solve each of the following equations : 1. a;2 + 6 3; = 7. Solution. To complete the square of the left side we must add 9 : x2 + 6a; + 9 = 16. Extracting the square root, x + 3 = ± 4; that is, X = 1 or — 7. 240 APPENDIX F 241 2. x^ + bx = 6. 4. x'' + 7x = - 12. 6. x^ + 12 x = V- 3. x2-5x = 6. 5. x2 + 12x = -20. T. x^ - (a + l)x = - ab'. 8. x2 - 2 mx + m2 - «« = 0. 9. a6x2 = (a'-'-i2)a; + ah. 10. x2 - (a + i)x = 2 a2 + 2 i2 - 5 oi. 11. x-* — (a + c)x + = 0. INDEX Abscissa, defined, 10 Ambiguous case, 160 Angles, general definition, 61 ApoUonius, 139 n. Asymptotes, 87 Axis, of parabola, 115; of ellipse (major, minor), 124, 125 ; of hyper- bola (transverse, conjugate), 132 Center,of ellipse,125; of hyperbola, 138 Characteristic of logarithm, 193, 194 Circle, equation of, 110, 111 Component forces and velocities, 77 Conic sections, 139 Conjugate hyperbolas, 135 Constants, 20 Construction, of rational numbers, 4 ; of irrational numbers, 4-6 ; of pa- rabola, 120 ; of hyperbola, 138 Continuity of trigonometric functions, 181 Coordinate system, 9 Correspondence between numbers and points, 7 Cosines, Law ^f, 156, 157 Degree of function, 28 Derivative, 211 ; as slope of tangent, 212 ; rules for finding, 214-218 ; used in drawing graphs, 219 Descartes, 10 Determinants, 32-39 ; minors, 37 ; applied to solution of simultaneous linear equations, 33, 34, 38, 39 Diiference of two angles, trigonometric functions of, 170, 171 Difference of two sines or cosines, 174 Differentiation, of rational functions, 214-218 ; of irrational functions, 222 ; of implicit functions, 223 Directed segments, 6, 7 Directrix, of parabola, 114 ; of ellipse, 122 ; of hyperbola, 131 Discontinuity of trigonometric func- tions, 181 Discriminant of quadratic equation, 47 Discriminant test graphically inter- preted, 44-53 Discriminant test used in obtaining tangents, 50 Distance between two points, 12 ; from a line to a point, 105-107 Division of segment in given ratio, 14-16 Eccentricity, of ellipse, 122 ; of hyper- bola, 131 Ellipse, 90 ; defined, 121 ; equation of, 122-128 Equation, linear, 29 ; quadratic, 41-47 ; of straight line, 97-109 ; of circle, 110, 111 ; of parabola, 114-118 ; of ellipse, 122-128 ; of hyperbola, 131- 136 ; of asymptotes to hyperbola, 133 Exponents, laws of, 188 ; fractional and negative, 189 Factor theorem, 55 Eocus, of parabola, 114 ; of ellipse, 122; of hyperbola, 131 Forces, problems on, 77 Functions, defined, 23; linear, 28; quad- ratic, 39 ; sign of quadratic, 56, 57 ; maximum and minimum values of quadratic, 58, 59 ; trigonometric, de- fined, 63-66 ; relations among trigo- nometric, 67, 69; trigonometric, of 30°, 45°, 60°, 68 ; fractional, 85-88 ; irrational, 88-90 ; logarithmic, 202 Galileo, 204 n. Graph, of linear function, 29 ; of quad- ratic function, 40 ; of fractional func- tions, 86-88 ; of irrational functions, 88-91 ; of trigonometric functions, 178-182 ; of logarithmic function, 202 Graphical representation, of number pairs, 9, 10 ; of equations, 19-22 ; of statistical data, 24-26 Graphical solution, of simultaneous linear equations, 30-32 ; of quad- 243 244 THE ELEMENTARY FUNCTIONS ratio equation, 40 ; of simultaneous quadratic equations, 141-150 "Greater than," 8 Half-angle formulas, 163 ; applied to plane triangles, 198, 199 Hyperbola, 87 ; asymptotes of, 87, 133, 134 ; defined, 131 ; equation of, 131- 136 I construction of, 138 Incommensurable segments, 3 Increment, 204 Independent variable, 23 Integer, 4 Intercepts, 29 Irrational functions, 85 Irrational numbers, 4 Kepler, 139 Latus rectum, of parabola, 116 ; of ellipse, 125 ; of hyperbola, 133 " Less than," 8 Limits, 208-210 ; theorems on, 210 Linear equation, graph of, 29 Locus, 19, 91-94 Logarithms, definition, 190 ; laws of, 191, 192 ; characteristic and man- tissa, 193, 194 Mantissa of logarithm, 193, 194 Maximum and minimum values, of quadratic function, 58, 59 ; with help of derivative, 220 ; problems involving, 227 Measurement, 1 Mid-point of segment, coordinates of, 12 Mollv?eide's Formulas, 201 Negative numbers, 6-8 Newton, 201 n. Normal form of equation of straight line, 101-105 Oblique triangle, solution of, 154, 157, 159 Ordinate, defined, 10 Parabola, 21 ; defined, 114 ; equation of, 114-118 ; construction of, 120 Periodicity of trigonometric functions, 180, 182 Polar coordinates, 183-187 Projections, Law of, 156 Ptolemy, 174 n. Pythagoras, 4 n. Pythagorean Theorem, 4 n. Quadratic equation, 41-47 ; formula for, 43 ; solution by factoring, 44 Quadratic function, 39 ; sign of, 56, 57 ; maximum and minimum values of, 58, 69 Radius vector, 65, 183 Rate of change, 203, 207 Rational functions, 185 Rational numbers, 4 Resultant forces or velocities, 77 Right triangle, solution of, 71-76 Roots of quadratic equation, 42 ; sum and product of, 53-55 Simultaneous linear equations, 30-39 Simultaneous quadratic equations, 141-152 Sines, Law of, 155 Slope of straight line, 79-82 Straight line, equation of, 97 ; normal form of, 101-105 Sum of two angles, trigonometric func- tions of, 168-171 Sum of two sines or cosines, 174 Tangent, to parabola, 50 ; to any conic, 143 ; to ellipse, 146 ; slope of, found by means of derivative, 211-214 Tangents, Law of, 201 Variables, 20 ; independent, 23 ; de- pendent, 23 Velocities, problems on, 77 Vertex, of parabola, 115; of ellipse, 125; of hyperbola, 132