Q&39 W83 Wolff - P. r. fnr> Southern Branch of the University of California Los Angeles Form L I QA39 33 This book is DUE on the last date stamped below V9-15m-8,'26 MATHEMATICS FOR AGRICULTURAL STUDENTS McGraw-Hill BookGompany Electrical World The Engineering andMining Journal Engineering Record Engineering News Railway Age Gazette Amei-ican Machinist Signal Engineer American Engneer Electric Railway Journal Coal Age Metallurgical and Chemical Engineering Power 2j_ f EDITED BY CHARLES S. SLIGHTER MATHEMATICS FOR AGRICULTURAL STUDENTS BY HENRY C. WOLFF, PH. D. ASSISTANT PROFESSOR OF MATHEMATICS UNIVERSITY OP WISCONSIN FIRST EDITION McGRAW-HILL BOOK COMPANY, INC. 239 WEST 39TH STREET, NEW YORK 6 BOUVERIE STREET, LONDON, E. C. 1914 COPYRIGHT, 1914, BY THE McGRAW-HiLL BOOK COMPANY, INC. THE. MAPLE. PKES3. YORK-PA 3 PREFACE The present book was designed as the text for a working course in elementary Mathematics given in the College of Agriculture of the University of Wisconsin. It is not a text book of Advanced Algebra, Trigonometry, Analytic Geometry, or "Practical Mathe- matics." Students pursuing a scientific course, in which but a year of college Mathematics is offered, receive little profit from a formal course in Advanced Algebra, Trigonometry, or Analytic Geometry. On the other hand, such students benefit little from a so-called "practical" course of a type which does little more than train them to substitute in given formulas, or to use formulas merely as a means to an end. Such a course does little to develop the habit or power of clear and logical thinking. This book is the outgrowth of mimeograph notes used for the past three or four years with undergraduate students of agricul- ture. With the exception of the illustrations and exercises, the book contains, nevertheless, nothing which is of exclusive in- terest to agricultural students. It may be used equally well with any class of scientific students who desire only a short course in mathematics beyond elementary Algebra and Geometry. The plan of the book is: First, to select material primarily on the basis of its usefulness to scientific students. Second, to illustrate the principles with problems of interest to the agricultural student, or with problems having direct applica- tion to his work. Third, to give minute and detailed explanations of all new work, and to assume the minimum attainments in mathematical preparation on the part of the student. On the other hand, details are often omitted from work that is not new to the student, in order that they may be supplied by the student himself. Fourth, to give detailed directions for doing work "at home" so vi PREFACE that the student of moderate mathematical attainments may, nevertheless, "learn by doing." Fifth, to include certain material not elsewhere readily acces- sible to the student of science, for the purpose of rendering the book useful for reference throughout the four years of the college course. Sixth, to select the topics and the amount of material under each topic so that either a half year or a full year may be devoted to the work. In writing this book the author had in mind the preparation of a text which would : First, train the student to do neat and careful work. Second, encourage the student to make further use of elemen- tary Algebra and Geometry. Third, develop in the student the habit of careful and logical thinking. Fourth, train the student to study a problem with a view of discovering the shortest and easiest method of handling it, rather than attacking it by the "first thought-of " method. Fifth, show the student, by illustrations and exercises, how mathematics may be helpful in pursuing other subjects of study and useful in a "practical" way. While this book includes enough material for a year's course of study, it is believed that it does not contain too much material for a shorter course. The inclusion of additional material in the text, will at least let the student know of the existence of sub- jects in mathematics not covered by him in the class room. Even this superficial knowledge may be very helpful to him later in connection with other scientific work, especially if he, knows he can find brief discussions in a familiar book. The author takes this opportunity to thank Professor C. S. Slichter for assistance in the preparation of the entire manuscript; and to acknowledge his indebtedness to Professor E. V. Hunting- ton for valuable suggestions and criticisms. Acknowledgments are due Mr. E. Taylor and Mr. T. C. Fry for suggestions based upon their use of the preliminary notes in the class room. A very brief review of elementary Algebra is given in the intro- duction, besides a list of materials and instruments. In the PREFACE vii appendix is given, for reference, a list of common mathematical symbols and a few formulas of mensuration. Material in the book which may be omitted when less than a full year is devoted to the course, is indicated by enclosing exercise and section numbers, and chapter headings within brackets. HENRY C. WOLFF. UNIVERSITY OP WISCONSIN September, 18, 1914 CONTENTS PAGE PREFACE v INTRODUCTION 1 CHAPTER I GRAPHIC REPRESENTATION 23 II LOGARITHMS 60 III THE CIRCULAR FUNCTIONS: THE TRIANGLE 80 IV THE ELLIPSE 128 V THE SLIDE RULE 143 VI STATICS 155 VII PERMUTATIONS, COMBINATIONS, AND THE BINOMIAL EX- PANSION. 189 VIII PROGRESSION 200 IX PROBABILITY 207 X SMALL ERRORS 231 XI POINT, PLANE, AND LINE IN SPACE 239 XII MAXIMA AND MINIMA 252 XIII EMPIRICAL EQUATIONS 264 APPENDIX 286 INDEX . 303 MATHEMATICS FOR AGRICULTURAL STUDENTS INTRODUCTION 1. Instruments and Materials. A large number of problems in this book are to be solved graphically or solved both graphically and analytically, and a large number of the exercises are exercises in graphic representation of observed or computed data. In order that this work may be performed neatly and accur- ately, it is necessary that the student have at least a few simple drawing instruments. The indispensable instruments, together with other material required for the work, are here listed and briefly described. 1. Two 4H Drawing Pencils. One, sharp- ened to a fine point, is used for marking points upon paper, or for sketching free-hand. The other, sharpened to a chisel-point, is used for drawing straight lines. Two views of a chisel- pointed pencil are given in Fig. 1. When drawing lines, a flat side of the chisel-point is held against the straight edge of the guide, thus causing the pencil always to move in a direction with the chisel edge and to produce a fine and distinct line. For fine and accurate work, both pencil points must be kept sharp. Neither point ought to become so rounded or flat that any appreciable metallic luster will appear at the wearing surface of the graphite. FIG. 1. Two views of a chisel- pointed pencil. A good way of keeping pencil points sharp is by occasionally whetting upon fine 1 2 MATHEMATICS [1 sand-paper. Small pads of sand-paper for this purpose may be procured from stationers or dealers in drawing materials. Hex- agonal pencils should always be used. The student may prefer to have one pencil sharpened at both ends; one end with a round point, and the other end with a chisel-point. 2. One 3H Pencil, sharpened to a round point for lettering. 3. A small drawing board made of soft wood to which the draw- ing paper may be fastened with thumb-tacks. A drawing board should have at least one straight edge along which the T-square may slide. A board 12 by 18 inches will be large enough. FIG. 2. T-square. 4. A small T-square about 15 inches long. A T-square, Fig. 2, consists simply of a thin, straight-edged blade, A, screwed to a heavier block of wood, the head, B. The inside of the head has a smooth, straight edge. The T-square is placed upon the drawing board with its head held against the left-hand edge of the drawing board. By sliding it up and down with the left hand, the draughts- man is enabled to draw with it as guide a series of parallel lines running from left to right. From this use of the T-square, it is seen that the edge of the drawing board along which it slides must be smooth and straight, otherwise the lines drawn along the blade of the square would not be parallel. It is hot at all necessary that the blade of a T-square be exactly at right angles to the inner edge of the head. 5. A 60 and a 45 Transparent Celluloid Triangle. Drawing triangles are thin triangular guides, usually made of wood or celluloid, one angle of which is a right angle. (See Fig. 3.) 1] INTRODUCTION 3 In a 60 triangle, the acute angles are 60 and 30, and in a 45 triangle, the two acute angles are each 45. With these triangles straight lines may be drawn making angles of 15, 30, 45, 60, 75, and 90 with the lines drawn with the T- square as a guide. For the work in this book 6-inch triangles will be found large enough, although an additional larger 60 triangle will be found very convenient. 60 Triangle 45 Triangle FIG. 3. 60 and 45 triangles. 6. A protractor is an instrument for laying off or measuring angles upon a drawing. One, consisting of angular graduations upon a semicircular piece of transparent celluloid, about 6 inches in diameter, is recommended for this work. (See Fig. 4.) 7. A scale divided decimally (the unit of length divided into tenths) should be used rather than one in which the unit is divided into eighths, twelfths, or sixteenths. It is advised that a triangular box-wood scale about 4 inches long, or longer, be used in this work. (See Fig. 5.) 8. A pair of 6-inch pencil compasses used for drawing circles and arcs of circles. The lead in a cheap pair of compasses usually is not found satisfactory and should be replaced by a good piece of medium hardness, sharpened to a narrow chisel-point. The lead should be so adjusted that when the two legs of the compasses are brought together the two points will be together or directly opposite each other. 4 MATHEMATICS [1 9. A good rubber eraser for erasing lead pencil marks. 10. Twelve small thumb-tacks. 11. Fifty sheets of squared paper, form Ml, used for plotting data. FIG. 4. Protractor. 12. Twelve sheets of paper, form M7, used for numerical calculations and for tables. 13. Six sheets of polar coordinate paper, form M3. 14. Twenty-four sheets of heavy bond paper, letter size, 8 by 11 inches; to be used for drawings. Y \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\V FIG. 5. Triangular scale. \ 15. Two heavy paper note covers suitable for holding paper 85 by 11 inches without folding. These covers are to be used for submitting daily work to the instructor. 16. One note cover large enough for filing 8J by 11-inch sheets 2] INTRODUCTION 5 without folding. This cover is to be used for filing the written work after having been examined and returned by the instructor. 17. A set of "Slichter's four-place logarithmic and trigono- metric tables" used in connection with numerical computations. There are other logarithmic and trigonometric tables published, but in order to do efficient classroom work without waste of time, there must be complete uniformity in the tables used by the mem- bers of the class. If the student wishes larger tables for his own private use, the two following are among the best: Albrecht's five-place, and Bremiker's six-place. While the above list of instruments and materials includes about the minimum in number and smallest in size for efficient work, more and better instruments may be purchased if the student desires. 2. General Directions. All work must be done neatly. In general, accuracy in graphic work can only be acquired by neat work. At first, perhaps, much time will be sacrificed, but after the habit of doing neat work has been established much time will be saved both in avoiding errors and in the ease with which errors may be located or the work checked. The student at the begin- ning must not be discouraged if a large number of his drawings and exercises are returned marked " redraw" or "rewrite," for neatness of work will always be insisted upon as one of the means o* obtain- ing accurate work. Unless special directions are given, all work submitted to the instructor is to be upon paper of letter size, 8 by 11 inches, and enclosed without folding in a paper folder. These exercises when returned by the instructor are to be corrected in accordance with instructions, and kept in order in a note cover. Exercises marked "rewrite" or "redraw "are to be done over, and both the original and the corrected sheets filed. This file of exercises may be called for at any time by the instructor. All drawings are to be done in pencil, unless the student has had training in the use of the ruling pen, in which case he may, if he desires, "ink in" his drawings with drawing ink. 6 MATHEMATICS [3 All descriptive work upon a drawing is to be lettered, not written. 1 Each drawing or exercise handed in must be numbered in the upper right-hand corner. It must have a complete yet brief title, and contain a sufficient number of descriptive words in the body of the drawing so that anyone familiar with the prob- lem will be able to read the drawing; i.e., to understand the meaning of every line or figure in the drawing. ABCDEFGHIJK a b cd efghijklmno pqrstuv wxyz 1 234 567890 FIG. 6. Set of letters and figures. All exercises not drawings are to be done in ink. Use any good writing fluid. It is suggested that the student acquire the habit of working numerical problems, and the like, with pen and ink. 3. To draw a straight line through a point with the T-square or triangle as guide, place the triangle or T-square upon the paper, so that the point is just visible at one of its edges; press the tri- angle firmly against the paper with the fingers of the left hand and with the pencil in the right hand hold a flat side of the chisel-point against the straight edge and move the pencil from left to right. The pencil should be held with its upper end tipped slightly away 1 A set of letters and figures is given in Fig. 6. 3] INTRODUCTION from the triangle in order to bring the point of the pencil in contact with the straight edge. (See Fig. 7.) As far as possible, the whole right hand should be given a straight-line motion in order that the pencil may be made to move parallel to itself. (Pencil Pencil Triangle Triangle FIG. 7. a, Incorrect way of holding a pencil; b, correct way of holding a pencil. Care must be taken that the triangle is well pressed down upon the paper at the point where the line is being drawn. This is partly accomplished by the pressure of the fingers of the left hand, and partly by pressing down upon the triangle with the third and fourth fingers of the right hand. This pressing down with the fingers of the right hand not only helps to hold the triangle in place, but, at the same time, gives a rest to the hand. 8 MATHEMATICS [4 4. Parallel Lines. Horizontal parallel lines may be drawn with the aid of the T-square. Parallel lines making angles of 30, 45, 60 or 90 with the horizontal may be drawn with the aid of triangles placed upon the edge of the T-square. Parallels other than those mentioned above may be drawn as follows: Suppose that through the point P, Fig. 8, we wish to draw a line parallel to the line AB. Place two triangles together, as shown in the figure, one edge of No. 2 coinciding with the given line. Hold triangle No. 1 fixed and slide No. 2 along it in the direction of the arrow until its edge which originally coincided with AB passes through the point P. Along this edge draw a line, PQ. PQ is parallel to AB, for, by geometry, if two straight lines are cut by a transversal, and if the interior exterior angles on the same side of the transversal are equal, the lines are parallel. Exercises 1. Fasten a sheet of paper upon the drawing board with its longer dimension extending from left to right. About 2 inches from the lower edge of the sheet draw, with the aid of the T-square, a horizontal line 6 inches long. Divide this line into six equal parts. At the points of division draw vertical lines with the aid of the T-square and triangle. Beginning with the horizontal line mark off four equal spaces, each 1 inch long, upon the left-hand vertical line. From the- points of division just located draw horizontal lines, thus forming a rectangle 4 by 6 inches, divided into twenty-four equal squares. Through the lower left-hand corner draw lines making angles of 30, 45, and 60 with the horizontal. Divide the upper left-hand square into 100 equal squares. A drawing (reduced) of this exercise is given in Fig. 9. Complete the exercise by giving, to two decimal places, the dimensions and quotients omitted in Fig. 9. 2. In a 45 triangle, prove by geometry that the hypotenuse is \/2 times as long as a leg. 3. In a 60 triangle, prove that the hypotenusejs 2 times as long as the shorter leg. Prove that the longer leg is V3 times as long as the shorter leg. The hypotenuse is how many times as long as the longer leg? 4. Find, correct to three decimal places, V2; V3; V5. 6. If the hypotenuse of a 60 triangle is two units in length, what 4] INTRODUCTION 9 is the length of the shorter leg ? Of the longer leg ? If the hypotenuse is four units in length, what is the length of each leg? 6. What is the length of the hypotenuse of a 45 triangle, if each C 5 w ^ Ss fc) 3 ^ fc. i: ^ ^ ^ T^ fe] K| K leg is a unit in length? Two units in length? Three units in length? 7. What is the length of each leg of a 45 triangle, if the hypotenuse is one unit in length? Two units in length? Three units in length? 10 MATHEMATICS [4 8. Find the length in rods of a diagonal of a section of land, assum- ing that the section is a true square 1 mile on a side. 9. A, B, C, and D are the corners of a section of land. Through A, a line is drawn making an angle of 30 with the side AB, and inter- secting EC at the point E. Find the length, in rods, of BE, and of AE. Find the area in acres of the triangular plot ABE. 10. Continuing with exercise 9, let F be the mid-point of AB, and G the mid-point of CD. Draw the quarter section line FG, cutting the line AE at H. Find the length in rods of AH, and of FH. Find the area in acres of the plot AFH. FIG. 10. 11. The longer leg of a plot of land in the form of a 60 triangle is 80 rods. Find the area of the plot in acres. 12. A plot of land in the form of a 60 triangle contains 72 acres. Find the length in rods of each side of the triangle. HINT: Let x represent the number of rods in the length of the shorter leg. 13. The shorter side of a rectangle is 100 feet, the diagonal is 200 feet. Find the length of the longer side. 4] INTRODUCTION 11 14. The diagonal of a square is 100 feet. Find the length of a side. 16. A barn 36 feet wide has a gambrel roof. The lower rafters make an angle of 60 with the horizontal. The upper rafters make an angle of 60 with the vertical. The lower and upper rafters are equal in length. Find the length of the upper edge of the rafters. See Fig. 10. HINT: Let x be the number of feet in the desired length. 16. If the rafters referred to in exercise 15 are 2 X 6's, and if they run short 3/16 inch, find the dimensions of the block sawed from each end. (See Fig. 11.) Find y and z. ./A 17. If the building referred to in exercise 15 is 84 feet long, find the number of cubic yards 'space under the roof above the plates. 18. A circular concrete silo, 12 feet outside diameter, is 20 feet tall. The wall is 8 inches thick. Find the number of cubic yards of con- crete in the wall, no allowance being made for openings. After the problem has been solved by the student, the instructor will show how algebraic operations may be used in eliminating considerable numerical work. 19. A man walks 3/4 mile up an incline making an angle of 30 with the horizontal. How many feet does he move in a hori- zontal direction? 12 MATHEMATICS [5 20. A ball rolls down a smooth inclined plane making an angle of 45 with the horizontal. If the distance s measured in feet along the incline is given by the formula where t is time in seconds, find the vertical distance through which the ball moves in one second; in two seconds; in three seconds. 6. Review of Elementary Algebra. The course for which this book is intended presupposes a knowledge of elementary algebra, and of plane and solid geometry, as usually covered in two years of High School work. Many students, however, having had their algebra and geometry some years since, may feel a need of review in these subjects. When this is the case, the student should take a hasty review touching only the essentials, and then study those subjects or theorems which he wishes to apply, at the time he wishes to apply them. For this purpose, a very brief review of algebra, consisting mostly of examples and illustrations, is here given. 6. Addition. To 3x + 4y - 6z Add 2x - 5y + 2z Sum 5x y 4z Exercises Find the sum of the following: 1. 3a - 66 + 7c and 5a - 2c - 76. 2. 3x 2 + 6xy + y z and x 2 5xy + y*. 3. 2xy - 3x 2 y + 3xy* + y 2 and 2x 2 y - 3y* + 3xy - 7. 4. a + b + c - d and 2a - 7c - 2d + 66 - 7. 6. a - 26 + 6, 3a + 7c - 6 + 26, and 5a + 6 - 7c. 6. x 2 + 2xy + y*, x 2 - 2xy + y 2 , and x 2 - 2j/ 2 . 7. 3x 3 + 6x 2 y + 6xy 2 + y 3 , 5x 3 Qy 3 + 6x 2 y, and 3x 3 - y 3 8. a + b + c + 3d, 2a - c + 76 - 5d + 6, and c - 7 + 2d - a. 9. a 3 - 6 3 , a 2 6 + a& 2 , and - 4o 2 6 + 2a6 2 . 10. 25a 3 - 276 3 , 5a 2 6 - 6a6 2 , and 266 3 - 24a 3 + a6 2 . 7. Subtraction. From 3a - 66 + 7c - 6 Subtract 2a + 56 - 2c - 6 Difference a 116 + 9c 8] INTRODUCTION 13 Exercises 1. From 3o 2 - 26 2 - Sab subtract a 2 - 3a6 + 6 2 . 2. From x 2 - 2xy + y 2 subtract x 2 + 2xy + y 2 . 3. From x + 2x z + 3x 3 - 2 subtract 2x + x 2 + 7x 3 . 4. From x 3 + 3x 2 y + 3xy 2 + y 3 substract Qx 2 y + 2y 3 . 5. From 3x 2 - 2y + y 2 subtract 2x 2 - 2y + 6 + 7z 2 . 8. Multiplication. The product of two numbers having like signs, either positive or negative, is positive. The product of two numbers having unlike signs is negative. Thus, (a) (6) = ab (_a)(-6) = ab (a)(- 6) = - ab (-a) (6) = -ab Multiply 3z 2 - 2xy + y 2 By 2x 2 - sy + y 2 - 2xt/ 3 ___ _ Product 6a; 4 - 7x 3 7/ + 7x 2 y 2 - 3xy 3 + y* Exercises 1. Multiply x 3 + x?/ + y 3 by x y. 2. Multiply x 3 xy + y 3 by x + y- 3. Multiply x 2 + xy + y 2 by x 2 - xy + y 2 . 4. Multiply x 7y by x 2y. 5. Multiply x + y by x y. 6. Multiply 3x + 7y by 3x + 8y. 7. Multiply a + b+c+dbya + b-c-d. 8. Multiply 2x - 3y + 6 by 2x + 7y + 2. 9. Multiply 3xy - 2x 2 - 2x by 2xy - 2x - 3y - 6. 10. Multiply 9 - 3x + x 2 by 3 + x. A few simple multiplications may be performed mentally. We see at once that (a + 6) (a 6) = a 2 6 2 , where a and b are any two numbers. From this we have (3x - 2y)(3x + 2y) = 9x 2 - 4?/ 2 14 MATHEMATICS Exercises Multiply the following mentally: 1. (3z -y)(3x+y). 2. (2x -y)(2x + y). 3. (a - 9) (a + 9). 4. (x*y -3a)(x 2 ?/ + 3o). 6. (a +36) (a - 36). 6. (29) (31), or (30 - 1)(30 + 1) = 900 - 1 = 899. 7. (51)(49). 8. (52) (48), or (50 + 2) (50 - 2). 9. (103) (97). 10. (25) (35). A few powers of binomials are: (a + 6) 2 = a 2 + 2ab + b\ (a - 6)2 = a 2 - 2ab + 6 2 . (a + 6)> = a 3 + 3o 2 6 + 3a6 2 + b 3 . (a - 6) 3 = a 3 - 3a 2 6 + 3a& 2 - b 3 . (a + &) 4 = a 4 + 4a 3 6 + 6a 2 6 2 + 4a6 3 + 6 4 . (a - 6) 4 = a 4 - 4a 3 6 + 6a 2 6 2 - 4a6 3 4 Thus, (3 - a) 3 = 27 - 27a + 9a 2 and (x + ?/ 2 ) 4 = x 4 Exercises Expand the following mentally: 1. (2o - a;) 2 . 4. (x - 3) 4 . 2. (x + 3y) 2 . 5. (1 - z) 3 . 3. (2z - I) 3 . 6. (2 + ?/) 4 . 7. (52) 2 , or (50 + 2) 2 , or 2500 + 200 + 4 = 2704. 8. (31) 2 , or (30 + I) 2 . 9. (29) 2 , or (30 - I) 2 . 10. (98) 2 . The square of a polynomial is illustrated by the following: (a + 6 + c) 2 = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc. (a + 6 + c + d) 2 = a 2 + 6 2 + c 2 + rf 2 + 2o6 + 2ac + 2 + 2bc + 26rf + 2cd. (3 - x + ?/ 2 ) 2 = 9 + x z + y* - 6x + 6i/ 2 - 2^. 9] INTRODUCTION 15 Exercises Expand the following mentally: 1. (o + b + 2) 2 . 4. (2o - x + 3) 2 . 2. (a + b - 2) 2 . 6. (x 2 - 2?/ 2 + 4) 2 . 3. (a - b - c) 2 . The product of two binomials having a common term may be found by inspection. Illustrations: (a + 6) (a + c) = a 2 + (b + c)o + be. (x + 7)(x + 2) = z 2 + 9z + 14. (x - 5)(z + 3) = x 2 - 2x - 15. (x 2 - 2y)(x* - By) = x* - 5x 2 y + 6?/ 2 . Exercises Find mentally the value of each of the following: 1. (x + 2)(x + 3). 6. (3x + 2y)(3x - 7y). 2. (x - 2)(x + 3). 7. (x 2 - 3)(x 2 - 4). 3. (x - 2)(x - 3). 8. (3xy - z)(3xy + 7z). 4. (x + 2)(x - 3). 9. (xV - 3)(xV - 10). 5. (x 2 + 5y)(x 2 - 5y). 10. (x - 2y)(2x - 2y). 9. Division. The quotient of two numbers having like signs is positive. The quotient of two numbers having unlike signs is negative. Thus, (a) -(&)=- f \ f M a (0) T ( _ 6) = _ _ (-a)-*- (6)= '- Divide a; 3 - Go; 2 - 19x + 84 by x - 7 x - 7(x* - 6z 2 - 19x + 84) z 2 + a; - 12 Quotient x 2 -19z + 84 a 2 - 7x - 12x + 84 -12x+ 84 16 MATHEMATICS [10 Exercises 1. Divide 15x 2 - llx - 14 by 3x + 2. 2. Divide 12a 2 - 28a + 15 by 6a - 5. 3. Divide 8m 5 - 14m 2 - 18m + 21 by 4m 3 + 6m - 7. 4. Divide 3x 5 - 7x 3 - 2x - 36 by x - 2. 5. Divide x 8 + 2x 6 - 3x 4 + 7x 3 - 2x + 69 by x - 3. 6. Divide x 4 + x 2 y 2 + y* by x 2 + xy + y 2 . 7. Divide 8x 4 - 22x 3 y + 43x 2 t/ 2 - 38xy 3 + 24y* by 2x 2 - 3xy + 4y 2 . 8. Divide x 3 y 3 by x y. 9. Divide x 4 y 4 by x + y. 10. Divide x 4 y 4 by x y. 10. Symbols of Aggregation. If a sign of aggregation is pre- ceded by the negative sign, change all signs within when the sign of aggregation is removed. If the sign of aggregation is preceded by the positive sign, all signs within remain the same when the sign of aggregation is removed. 5x 2 - [3y 2 + {2x 2 - (?/ 2 + 3x 2 ) + 5y*} - x*] = 5x 2 - [3y 2 + {2x 2 - y 2 - 3x 2 + 5r/ 2 } - x 2 ] = 5x* - [3y 2 + {4t/ 2 -x"'} - x 2 ] = 5x 2 - [3y 2 + 4y 2 - x z - x 2 ] = 5x 2 - [7y 2 - 2x 2 ] = 5x 2 - 7y 2 + 2a; 2 = 7x 2 - 7y 2 Exercises Simplify the following by removing the signs of aggregation : 1. ab - 46 2 - (2a 2 - 6 2 ) - { - 5a 2 + 2ab - 36 2 } . 2. x - { y + z - [x - ( - x - y) + z}} + [z - (2x - y)]. 3. a - { - a - [ - a - ( -a - 1)]}. 4. 3yz - [2yz + (9z - 2yz}]. 5. - { - 1 -[-1 - ( -1)]}. 11. Factoring. Since (a + 6) 2 = a 2 2ab + b 2 , any expres- sion of the form of the right-hand side can be factored by inspec- tion. Thus, x 2 - 6xy + Qy 2 = (x - 3y) 2 and :. 4 + 4(o + 6) + (a + 6) 2 - (2 + a + 6) 2 11] INTRODUCTION 17 Exercises Factor the following by inspection : 1. 9x 2 - 30xy + 25y 2 . 2. 4 + IQt + 16< 2 . 3. zV + lOxVz 2 + 25z 4 . 4. 9 + 6(x 2 + t/ 2 ) + (x 2 + y 2 ) 2 . 5. a 4 + 4a 2 6 2 + 46 4 . Since (a + 6) (a 6) = a 2 6 2 , any expression of the form of the right-hand side can be factored by inspection. Thus, 4a 2 - 96 2 = (2a + 36) (2o - 36) Exercises Factor the following by inspection : 1. xV - 2 2 . 4. 25 - 3x 2 . 2. (a + 6) 2 - c 2 . 5. 81 - 625x 4 . 3. c 2 - (a + 6) 2 . Since (a + 6) (a + c) = a 2 + (6 + c)a + &c, any expression of the form of the right-hand side can be factored by inspection. Thus, x 2 - 5x - 14 = (x - 7)(x + 2) Exercises Factor the following by inspection : 1. x 2 + 7x + 10. 4. 9x 2 - 18x - 27. 2. a 2 + 4ay - 21y 2 . 5. 25 + 30a - 27a 2 . 3. 4x 2 - ISxy + 18j/ 2 . Since (a + 6) (a 2 ab -f- 6 2 ) = a 3 + & 3 , any expression of the form of the right-hand side can be factored by inspection. Thus, 27 + 125x 3 = (3 + 5x)(9 - I5x + 25z 2 ) Exercises Factor the following by inspection: 1. zV + 1- 4- 125 + xy. 2. x 6 + y\ 6. x 3 + 8y 6 . 3. 8 + 27x 3 . Since (a 6) (a 2 + 06 + 6 2 ) = a 3 b 3 , any expression of the form of the right-hand side can be factored by inspection. Thus, 27 - 125z 3 = (3 - 5s) (9 + 15* + 25z 2 ) 18 MATHEMATICS [12 Exercises Factor the following by inspection : 1. x 3 y 3 - 1. 4. 125 - x 3 y 3 . 2. x 6 - y 6 , or (x 3 + y 3 )(x 3 - y 3 ). 5. x 3 - 8y*. 3. 8 - 9x 3 . The following may be factored by grouping the terms. Thus, a 3 m + a 3 n m n = a 3 (m -\- n) (m -\- n) = (a 3 - l)(m + n) = (a - l)(a 2 + a + l)(m + n) Exercises Factor the following : 1. ax ay + bx by. 4. x 5 xy* x*y + y b . 2. x 3 + 3x 2 + 3x + 1. 6. x 4 - x s y - xy 3 + y*. 3. ax 2 - 2axy + ay 2 + 6x 2 - 2bxy + by*. 12. Fractions. Multiplying or dividing both numerator and denominator of a fraction by the same number, excepting zero, does not change the value of the fraction. To reduce a fraction to its lowest terms, factor both numerator and denominator and then divide each by the common factors if there are any. Thus, ax* + ax 2 y 2 + ay* a 2 z 3 a 2 y 3 a(x* + xy + t/ 2 )(a; 2 - xy + y z ] a(x y)(x 2 -\-xy-\- y 2 ) x 2 xy + y 2 (x - y) Exercises Reduce the following to lower terms: ^ ox 2 ay 2 > ax + ay x y. x 3 y 3 ' x 3 + y 3 x 2 + x - 6_ 27 - x 3 2 * x 2 - 4 12- 7x + x 2 . x 4 - y\ x 6 - 6 ?/ 12] INTRODUCTION 19 To multiply fractions together, multiply the numerators together for the numerator of the product, and multiply the denominators together for the denominator of the product. To simplify the result, divide both numerator and denominator by common factors if there are any. Thus, a 3 - b 3 b a 2 6 - ab 2 ab X a 2 + ab + X ( a -b) z (a 3 - 6 3 )(6)(a 2 6 - a& 2 ) " ab(a 2 + ab + 6 2 )(a - by _ (a - 6) (a 2 + ab + 6 2 )(a6 2 )(a - 6) a&(a 2 + ab + 6 2 )(a - 6) 2 = 6 In working this problem, the second expression should not be written down, and possibly the third could be omitted. Exercises Simplify the following : a 4 - y* a 2 - 2ay + y 2 a z (x + a) x 2 + lax + a a 2 + 2a.b + b 2 - 4c 2 (a - b) 2 - 4c" (2a - b) 2 - 4c 2 . 4a 2 - (b + 2c) 2 (a + 6)' - 4c 2 a 2 - (b - 2c) 2 x 3 - y 3 x* xy + y* x + y % s\ . , , X 7*3 _L_ 4/3 /72 _! TII 1 -i/2 /> _ _ 91 *' \ y * \ -Ly i y * j/ _ oa; + ay bx 36 x 2 xy + y 2 b < x + j/ 3 X x 2 - 5x +~6~' To add fractions reduce each to the same denominator by multi- plying numerator and denominator of each by a suitable number, then add the numerators and place the sum over the common denominator. Thus, x -y _J_ xy I -v I .. I _3 I x 2 - zt/ + 7/ 2 ' a: + y ' x 3 + y 3 x 2 - y* x 2 - xy + y 2 xy x 3 + y 3 "* a; 3 + y 3 + a; 3 + y 3 2a: 2 20 MATHEMATICS [13 Exercises Simplify the following : i n I o I o f* 2 i v-2 + 3). a; 5 x 1 . o x- 5 x - te + 5 y) ( + y)*' y (*-) 13. Exponents, Definitions: a n = a a a . . . to n factors, if n is a positive integer ap/ = v/a p , if p and g are positive integers a = 1 a~ 8 = -, if s is positive ft 8 Laws of indices: a m a n = a m +" (a m ) = a mn Exercises 1. Express the following using positive exponents only: (a) x~ 2 ab~ 3 . ar^-'x-*. (6) a-^b^. W c2 Sa;6 a~ l bc~* (c) (x-^y-^). (e) ^7 c ^i- 2. Perform the indicated operations: (a) 25 W . (6) 8 ?4 . (c) (9/16)-*. (d) 27. 3. Simplify: 14] INTRODUCTION 21 4. Multiply Sa % - 4o6 w + 2o w 6 - & ? * by 2o w - 36^. 6. Divide a* - 3a 2 + a^ - 4 + 12o w - 4o % by a** - 4. 14. Radicals. Simplify ^24 + 5^81 - 6^192 - 10^1029 + = 2 i/3 + 15^3 - 24i/3 - 70^3 = - 77^/3 Rationalize the denominator of (\/2 + \/3)(>/2 + \/3) \/2 - \/3 (\/2 - 2-3 = 2 + 2\/6 +3 - 1 = - (5 + 2V6) Exercises 1. Simplify Vl2 - 2\/27 + 9 A/48 - Vl08 - 6\/75. 2. Simplify V (a - b)*x - V(o + 6) 2 z - Va 2 x + 2 3. Rationalize the denominator of 4. Rationalize the denominator of V3 - V2 2 - \/5 6. Rationalize the denominator of 7= 7=' V3 - V 5 16. Quadratic Equations. A quadratic equation may be solved by completing the square, or by using the formula, or somtimes by factoring. Solve for x: x 2 - 6x + 8 = 0. By completing the square we have x 2 - 6z + 9 = 1, or x - 3 = 1, or x = 3 + 1, or x = 4 or 2. 22 MATHEMATICS [15 By factoring the left-hand side we have (x - 4)(x - 2) = 0. This equation is equivalent to the two equations. x - 4 = and z - 2 = from which we have x = 4 and x = 2 If ax z + bx + c = is the general equation of the second degree, its solutions are - b + V'b 2 - 4ac 2a In the above equation, a = 1, b = 6, andc = 8. Substitut- ing these values in the formula we have 6 + V31T-4-8 x ' 2. = 3 1, or x = 4 or 2 Exercises Solve the following equations for x: 1. x 2 - 9x + 20 = 0. 2. 4z 2 + 7x - 2 = 0. 3. x 2 - 7z + 6 = 0. 4. 8x 2 - 4x + 3 = 0. 6. 9x 2 + 7x - 20 = 0. CHAPTER I GRAPHIC REPRESENTATION The Straight Line. The Parabola. The Equilateral Hyperbola. 16. The Scale. In solving problems graphically the funda- mental instrument used- is the scale. Unless otherwise stated, the term scale, as used in this book, means what is commonly understood by a scale, an instrument for measuring short linear distances. The term scale, in a broader sense, may be defined as any curve, including a straight line, the points along which are supposed to correspond in order to the numbers of a sequence. We are all familiar with various scales, such as the foot-rule, divided into inches; the meter, divided into centimeters and milli- meters; the thermometer scale, divided to correspond to degrees and tenths of degrees of temperature; the scale on a burette, divided to correspond to cubic centimeters and tenths of cubic centimeters; and scores of other scales used for measuring different quantities. If a scale is marked so that the subdivisions are all of equal length, as the foot-rule, we say the scale is divided uni- formly, or that we have a uniform scale; otherwise, the scale is a non-uniform scale. For the solution of problems graphically, the student should have for measuring lengths a scale divided into centimeters and millimeters or into inches and tenths of inches. Illustrations of non-uniform scales will be given later. 17. Drawing to Scale. A large number of graphic problems consist in nothing more than drawing to scale, i.e., making a picture of some object in which a definite length, as the inch, on the drawing represents a definite length, as the foot, on the given object. If the dimensions on such a drawing were to be given in feet and inches, or if dimensions in feet and inches were to be scaled from the drawing, a scale divided into tenths of inches would apparently be very ill-adapted. For this kind of work, scales may be procured on which the unit of length is sub- 23 24 MATHEMATICS [18 divided into twelve equal parts. If each unit throughout the entire length of the scale is subdivided, the scale is called a full _ divided or chain scale. If only one unit, usually the left-hand unit, is subdivided, it is called an open divided scale. These scales are made with the unit J, \, f, , \, 1, 2, 2, 3, etc., inches long, used, respectively, when |, j, f , |, $, 1, 2, 2, 3, etc., inches on the drawing are to represent 1 foot on the object. Fig. 12 represents an 18- ^ inch, flat, open divided scale. % The unit on one edge is f ^ inch, and on the other edge ]> it is j inch. This scale may ^ be used in drawing the plans g, of a barn. The small scale 9 is used in drawing the plan ^ and elevations ? inch to the foot, while the larger scale is 2 used in drawing, on a larger scale, the details of the plank frame truss bents. The method of measuring lines is illustrated in the figure. The line AB is found to be 22 feet, 7 J inches long. 18. Positive and Negative Directions. If upon a scale some point is chosen as zero, or origin, distances measured in one direction, or the num- bers represented by the points on one side of it, are called positive, and are designated with the positive, or plus, sign +. Distances measured in the other 19] GRAPHIC REPRESENTATION 25 direction, or numbers represented by points on the other side of the origin, are called negative, and are designated by the nega- tive, or minus, sign . 19. Rectangular coordinate paper, or squared paper, is paper ruled in both directions, horizontally and vertically. These rulings are generally spaced uniformly, dividing the page into a number of equal squares or rectangles. The size of these squares or rectangles is entirely arbitrary. The length of a side may be an inch, a tenth of an inch, a centimeter, a millimeter, etc., x- Y' FIG. 13. Coordinate, or squared paper. depending, to some extent, upon the use for which the paper is designed. For convenience, every fifth, tenth, fourth, eighth, or twelfth line is made somewhat heavier than the others. A sheet of squared paper is represented in Fig. 13. Upon this sheet two lines, mutually at right angles, are chosen as axes. The horizontal, X'OX, is called the axis of abscissas; the vertical, Y'OY, the axis of ordinates. The point of intersection, 0, of these axes is called the origin. A use of squared paper is illustrated by the following: In Table I are given the maximum and minimum air temperatures at Madison, Wisconsin, for each day of the month of October, 1910. 26 MATHEMATICS [19 TABLE I. MAXIMUM AND MINIMUM TEMPERATURES AT MADISON WISCONSIN, FOR OCTOBER, 1910 Date Max. temp., F. Min. temp., F. Date Max. temp., F. Min. temp. op 1 66 55 17 81 53 2 71 48 18 81 57 3 75 58 19 69 45 4 68 55 20 45 40 5 62 53 21 49 41 6 58 45 22 52 34 7 66 43 23 60 37 8 68 47 24 60 49 9 60 44 25 52 44 10 67 42 26 60 40 11 75 49 27 42 32 12 61 46 28 35 30 13 69 45 29 38 26 14 73 52 30 60 31 15 76 50 31 63 39 16 80 56 These data are said to be represented graphically in Fig. 14, or the data are said to be plotted upon squared paper. To do this: first, points equidistant along the axis of abscissas are chosen to repre- sent the days of the month. These points are numbered con- secutively from 1 to 3 1. 1 Distances from this horizontal line represent temperatures. If the temperature is above zero, or (+), it is represented by a distance above the horizontal axis; if below zero, or ( ), by a distance below. The distance above or below is proportional to the magnitude of the temperature reading. Since distances from the horizontal axis represent tem- perature readings, points along the axis of ordinates are numbered to correspond to degrees of temperature. In the figure, the verti- cal distance between two adjacent horizontal lines represents 2, but only every fifth line is 'actually numbered, since numbering each line would not aid in any way in reading the drawing. This vertical axis is in reality a uniform scale, for equal 1 Only the even numbers are placed in the drawing; otherwise the numbers would be so near together as to be confusing. Omitting the numbers from every other point does not make the drawing harder to read. 19] GRAPHIC REPRESENTATION 27 -^ <^ ^ \D V v 1 <<; > > > H 3 p * -^^ V, is. I E -1 *v ^ S i-^~ "^N (X- 1 k^ \ .a 50 fa ^ ^ & \ X 1 *> 1111 ^^ | 40 if i >. ''/ ^ 5 s w 30 - - - | 20 Hourly Air Te at n V r, l an< t.. I nperatures Hm Ma< liso Ma Oe Wisconsin 4.1910 i 0, 1910 i 2 4 6 8 10 12^ 2 4 ^6 8 10 1 < A.M; > | < F.M. * Time of Day Hours FIG. 15. TABLE II. HOURLY AIR TEMPERATURES AT MADISON, WISCONSIN. MAY 14, 1910; OCTOBER, 10, 1910 Time May 14, 1910, temp., F. Oct. 10, 1910, temp., F. Time May 14, 1910, temp., F. Oct. 10, 1910, Temp., F 1 a. m. 37 44 1 p. m. 58 65 2 a. m. 35 44 2 p. m. 61 66 3 a. m. 35 44 3 p. m. 63 67 4 a. m. 34 43 4 p. m. 62 67 5 a. m. 34 43 5 p. m. 62 65 6 a. m. 35 42 6 p. m. 61 62 7 a. m. 37 43 7 p. m. 59 57 8 a. m. 42 45 8 p. m. 56 55 9 a. m. 46 51 9 p. m. 53 55 10 a. m. 49 55 10 p. m. 50 55 11 a. m. 54 60 llp.m. 47 52 12 a. m. 56 ! 62 12p.m. 45 51 30 MATHEMATICS [21 21. Precipitation Charts. In Fig. 16 we have represented diagrammatically, by the usual method, the mean monthly pre- cipitation at Tacoma, Washington; Madison, Wisconsin; Tampa, Florida; and Yuma, Arizona. In these charts the amount of pre- cipitation is represented by the height of a shaded rectangle. This method of representation is similar to that used in Fig. 14 and enables one to see at a glance when the rainy and the dry periods occur at these localities. J FMAM J JASOND Tacoma, Wash. J FMAM J JASOND Tampa, Fla. . 6- a> -S5 a 4. 13 .2 3 3 2- TIM Ml I I I I I I I I IN J F MAM J J A S O N D Madison, Wis. P-iU J FMAM J JASOND Yuma, Ariz. FIG. 16. Precipitation charts. 22. Marking Axes. Whenever data are represented upon squared paper, both horizontal and vertical axes should be marked in such a way as to make perfectly clear what quantities are represented and what length represents a unit of measure of that quantity. If two or more curves are drawn upon the same set of axes, they should be so marked as to show definitely what each curve represents. Occasions may arise when it is desirable to draw upon the same 22] GRAPHIC REPRESENTATION 31 sheet of paper, for comparison, two curves having in common only one element, such as time. A scheme for doing this is illustrated by exercise 12, below. The length chosen to represent a unit is entirely arbitrary and need not be the same along both axes. This is already illustrated in Fig. 15. Further, the numbering in the horizontal and vertical directions need not be placed upon the axes, as will be illustrated by exercise 12, below. Exercises 1. At what hour was the temperature a maximum on May 14, 1910? At what hour was the temperature a maximum on Oct. 10, 1910? 2. At what time was the temperature a minimum on May 14, 1910? At what time was the temperature a minimum on Oct. 10, 1910? 3. At what hour was the temperature increasing the fastest? At what hour was the temperature decreasing the fastest? 4. During what hours of the day was the temperature decreasing? During what hours of the day was the temperature increasing? 5. What was the temperature at 10:30 a. m.? 10:30 p. m.? 6. What was the temperature at 6: 15 a. m.? 6: 15 p. m.? 7. At what time was the temperature on May 14 the same as on October 10? 8. For what portion of the day was it warmer on May 14 than on October 10? 9. During what hours was the temperature below the freezing point? 10. Plot data given in Table III. 11. Plot data given in Table IV. 12. Plot the data given in Table V upon squared paper. Use the, same horizontal axis for all three curves. Put the temperature curves above the discharge curve, using the same horizontal (time) scale for both. Let 1 cm. (or 1/2 inch) on the vertical scale represent 0.1 second-foot 1 discharge, and 10 temperature. Start the temperature scale with 60, and place the 60, 6 cm. (or 3 inches) above the horizontal scale. Start the discharge scale with 3.4 placed on the horizontal scale. 1 Second-foot (or sec-ft.), when applied to the measurement of flow of water means one cubic foot per second. 32 MATHEMATICS [22 TABLE III MAXIMUM AND MINIMUM TEMPERATURES FOR OCTOBER, 1911, AT MADISON, WISCONSIN Date Max. temp., F. Min. temp., F. Date Max. temp., F. Min. temp., F. 1 55 50 17 61 51 2 56 50 18 67 47 3 72 50 19 57 45 4 61 48 20 54 39 5 51 42 21 46 38 6 63 47 22 48 35 7 53 45 23 42 34 8 61 40 24 46 30 9 62 45 25 47 37 10 61 49 26 38 30 11 65 48 27 42 28 12 69 49 28 43 29 13 57 47 29 53 30 14 61 50 30 45 37 15 57 49 31 40 32 16 65 52 TABLE IV. HOURLY TEMPERATURES AT MADISON, WISCONSIN, FOR APRIL 11, 1911, AND NOVEMBER 10, 1911 Hour Temp., F., Nov. 10, 1911 Temp., F., April 11, 1911 Hour Temp., F., Nov. 10, 1911 Temp., F., April 11, 1911 a. m. 1:00 59 39 p. m. 1:00 62 59 2:00 59 40 2:00 60 61 3:00 59 40 3:00 59 59 4:00 59 40 4:00 51 58 5:00 60 40 5:00 44 56 6:00 60 40 6:00 39 53 7:00 61 41 7:00 35 53 8:00 62 44 8:00 31 52 9:00 64 47 9:00 28 51 10:00 66 50 10:00 25 47 11:00 68 55 11 :00 22 46 12:00 70 57 12 :00 20 45 23] GRAPHIC REPRESENTATION 33 13. What was the discharge at 7 : 00 p. m.? At 10 : 00 a. m.? 14. What was the minimum rate of discharge? At what time? 16. What was the water temperature when the discharge was a minimum? 16. What was the maximum water temperature? At what time? 17. What was the maximum air temperature? At what time? 18. What was the greatest difference between the air temperature and the water temperature? 19. At what hour did this "maximum" difference occur? 20. What was the minimum difference between the air temperature and the water temperature? ' 21. At what hour did this "minimum" difference occur? 22. Can you think of any cause which would produce the indenture in the air temperature curve at 1:30 p. m., at which time a normal temperature curve should have a maximum? 23. Can you think of any reason why the discharge curve should have a minimum when the temperature curves have maxima? TABLE V. DISCHARGE OF A SEEPAGE DITCH Time, Aug. 24, 1905 Discharge of ditch, sec.-ft. Temp, of water, F. Temp, of air, F. 8 :00 a. m. 3.72 65 9:00 a. m. 11 :00 a. m. 1 :30p. m. 2 : 15 p. m. 3.70 3.66 3.52 3.49 70 79 83 74 84 85 3 : 30 p.m. 5: 30 p. m. 6 :00 p. m. 3.52 3.66 3.73 84 78 90 88 8: 00 p.m. 3.84 67 76 23. The Point. The location of a point upon a sheet of rec- tangular coordinate paper is fixed by giving its distances from the vertical and horizontal axe^. The distance from the vertical axis is called the abscissa, and the distance from the horizontal axis is called the ordinate of the point. These two distances are called the coordinates of the point. The horizontal axis is commonly called the X-axis, or the axis of x; and the vertical axis, the Y-axis, or the axis of y. The abscissa and ordinate are then 3 MATHEMATICS [23 called, respectively, the x and y coordinates. When writing the coordinates of a point, the value of the abscissa is placed before the value of the ordinate, and separated from it by a comma. Thus, "the point (3, 4)," (read, "three, four"), is a point three units to the right of the F-axis, and four units above the X-axis. The point ( 2, 6) is two units to the left of the F-axis and six 1 r 1 (-2,6 ) 6 5 4 (3,4) 3 n 2 I 1 7 - 6 - 5 - 4 - 3 - 2 - 3 (-2, 1 O 1) - 1 l 2 3 4 5 6 7 . 2 (3,-: S) III _ 3 IV 4 _ 5 . 6 7 FIG. r' 17. units above the X-axis. The point (2, 1) is two units to the left of the F-axis and one unit below the X-axis. The point (3, 2) is three units to the right of the F-axis and two units below the X-axis. These four points are shown on the sheet of squared paper in Fig. 17. The two axes divide the plane into four parts, called quadrants. They are numbered from "one" to "four" as shown in Fig. 17. 23] GRAPHIC REPRESENTATION 35 Exercises 1. Plot the following points: (a) (1, 1); (6) (2, 3); (c) ( - 3, 2); (d) ( - 1, - 1); (e) ( - 2, - 3); (J) (3, - 4); (g) (V2, 1); (ft) C - V3, V5). 2. Find the distance of each of the following points from the origin: (a) (1, 1); (6) ( - V2, V2); (c) ( - 3, - 2). 3. Find the distances between the following pairs of points: (a) (1, 1) and (2, 2); (&) ( - 2, 3) and ( - 5, - 6); (c) (3, - 5) and (- 1, 0). 4. What is the y-coordinate of any point upon the axis of a;? 6. What is the x-coordinate of any point upon the axis of y? 6. What is the y-coordinate of any point upon a straight line parallel to the axis of x, three units above it? 7. What is the ^-coordinate of any point upon a straight line parallel to the F-axis, five units to its left? 8. What relation exists between the coordinates of any (every) point of a line bisecting the angle between the positive directions of the two axes? Between the negative directions of the two axes? 9. What relation exists between the coordinates of any (every) point of a line bisecting the angles having the positive direction of the F-axis and the negative direction of the ^T-axis as sides? 10. What relation would exist between the x- and ^-coordinates of any (every) point of the line in exercise 9, if it were raised five units parallel to itself? If it were lowered six units ? 11. What relation would exist between the coordinates of any (every) point of the line in exercise 9, if the line were moved up three units parallel to itself? If it were lowered five units? 12. What are the coordinates of the origin? 13. Compute for every 10 temperature (between and 200) read upon a Fahrenheit scale, the corresponding reading upon a centigrade scale. Put your results in tabular form, using a sheet of paper, form M7. 14. Compute for every 10 temperature (between 20 and 100) read upon a centigrade scale, the corresponding readings upon the Fahrenheit scale. Put your results in tabular form, using a sheet of paper, form Ml. 16. From the table constructed in exercise 13, convert the following readings Fahrenheit into readings centigrade: (a) 12.6; (6) 21.7; (e) 28.55; (d) 116.75; (e) 151.86. 16. From the table constructed in exercise 14, convert the following 36 MATHEMATICS [24 readings centigrade into readings Fahrenheit: (a) 2; (6) 22.2; (c) 61.3; (d) 62.6; (e) 75.4. 17. If x represents the number of degrees temperature, as read upon a Fahrenheit scale, and if y represents the number of degrees of the same temperature as read upon a centigrade scale, give the equation connecting y and x. 18. Plot upon squared paper the numbers in the tables constructed in exercises 13 and 14. Plot degrees Fahrenheit along the horizontal axis and degrees centigrade along the vertical axis. Recompute numbers, if any, whose plotted points do not fall upon a straight line connecting the points (32, 0) and (212, 100). 19. Construct a table, for every sixty-fourth of an inch, con- verting sixty-fourths into decimal equivalents. Compute to three decimal places. 20. Plot the corresponding numbers found in exercise 19 upon squared paper. Do all the points plotted fall upon a straight line passing through the origin? 21. Construct a table converting centimeters, from to 100, into inches. One meter = 39.37079 inches. Compute to three decimal places. Plot the numbers of the table upon squared paper. Do all the points fall upon a straight line passing through the origin? 24. The Double Scale. From the preceding exercises we have seen that relations between numbers may be represented by tables, by equations, and by curves upon squared paper. A fourth method of representing a relation between numbers will be ex- plained by means of an illustration. If x represents a pressure read in feet of water, and if y represents the same pressure read in inches of mercury (since mercury is 13.55 times as heavy as water), the relation between x and y is 12x 13.55 or y = 0.886 x. In Fig. 18, the scale on the under side of the horizontal line represents pressure measured in feet of water, the scale on the upper side represents pressure measured in inches of mercury. The scales are so drawn that readings directly opposite the horizontal line represent the same pressure, i.e., the length representing a unit 25] GRAPHIC REPRESENTATION 37 on the upper scale is 1.13 times as long as the length represent- ing a unit on the lower scale. From this double scale, one can easily convert pressure readings in feet of water into inches of mercury , or vice versa. Inches of Mercury, y 1 2 3 4 5 l lll ll | l|ill||l.lj I 11111111.(1111 ,,|l I ll|lll,|lll.|l , 123456789 10 Feet of Water, a; FIG. 18. Double scale showing relation between pressure measured in inches of mercury and in feet of water. Exercises 1. Draw a double scale connecting centimeters and inches. 2. Draw a double scale connecting inches and tenths of a foot. 3. Draw a double scale connecting pressure measurements in feet of water and pounds per square inch. A cubic foot of water weighs 62.5 pounds. 25. The Equation of a Curve, the Graph of an Equation. Any (every) point of a line parallel to the axis of x and three units above it has an ordinate three, and no point whose ordinate is three is found off the line. Since the equation y = 3 is satisfied by the coordinates of every point of this line and by the coordinates of no other point, it is said to be the equation of the line. The equa- tion of a line parallel to the X-axis and ten units below it is y = 10. The equation of the X-axis is y = 0. The equation of a line parallel to the F-axis, five units to its right, is x 5. The equation of a curve is an equation satisfied by the coordi- nates of every point of the curve and by the coordinates of no other point. The graph of an equation is the locus of a point whose co- ordinates satisfy the equation. To illustrate: if we have a circle, 1 radius 5 and center at the origin of coordinates, its equation is x z -+- y 2 = 5 2 . For, if we draw from any point of this circle a line to the origin and a perpendicular to the X-axis we have formed a right triangle whose legs are x and y, and whose hypotenuse is 5; 1 The term "circle" as used in this book means the locus of a point at a constant distance from a fixed point. 38 MATHEMATICS [25 and if the point is chosen off the circle, the hypotenuse of the right triangle is no longer 5. Thus, for points of the circle we have x 2 + y 2 = 25, and for points off of the circle, x 2 + y 2 ** 25. * The graph of the equation z 2 + ?/ 2 = 25 is a circle, radius 5 and center at the origin. The equation of a straight line passing through the origin and making an angle of 45 what the positive direction of the axis of x is y = x. The graph of the equation y = x is a straight line passing through the origin, making an angle of 45 with the positive direc- tion of the axis of x. For brevity, instead of saying "the circle," or "the straight line," or "the curve" naming the equation "representing the equation" we shall simply say the circle, or the straight line, or the curve naming the curve. Thus, the circle x 2 + y 2 = 25; the straight line y = x, and the curve x 3 3xy y z = 1. Exercises 1. What is the equation of the axis of x? 2. What is the equation of the axis of y? 3. What is the equation of a line parallel to the X-axis, three units above? Five units above? One hundred units below? Five units below? Fifty units below? 4. What is the equation of a line parallel to the axis of y, five units to the right? Ten units to the right? Six units to the left? Eighty units to the left? 5. What is the equation of a right line passing through the origin and bisecting the angle between the positive directions of the coordi- nate axes? 6. What is the equation of a straight line passing through the origin and bisecting the angle between the positive direction of the axis of y and the negative direction of the axis of a;? 7. What is the graph of y = z? Of y = - xl Of y = x + 2? Of y = x - 3? 8. What is the graph of y = - z? Of y = - x + 1? Of y = - x + 3? Of y = - z - 6? 9. What is the equation of a straight line making an angle of 45 with the positive direction of the axis of x and cutting the F-axis five units above the origin? Ten units above the origin? Six units below the origin? Twenty units below the origin? i "^ " is read "is not equal to," or "is different from." 25] GRAPHIC REPRESENTATION 39 10. What is the equation of a straight line making an angle of 135 with the positive direction of the axis of x and cutting the F-axis one unit above the origin? Three units above the origin? Five units below the origin? 11. What is the equation of the circle, center at the origin and radius 1? Radius 2? Radius 3? 12. Show that the equation of a circle, radius 5 and center at the point (2, 3), is (x - 2) 2 + (y - 3) 2 = 5 2 , or z 2 + y* - 4x - 6y = 12. HINT: Take any point, P, of the circle. Its coordinates are x and y. Draw PA perpendicular to the axis of x. Call the center of the* circle C, and draw CB perpendicular to PA. Draw PC. PCB is a right triangle, and CB 2 + BP 2 = CP 2 , or . (x - 2) 2 + (y - 3) 2 = 5 2 , x 2 - 4x + 4 + y 2 - 6y + 9 = 25, or or The student will show that the same equation is obtained when the point P is to the right of, and below C', to the left of the F-axis ; below the X-axis. When the point P is to the left of the F-axis the student must remember that a; is a negative num- ber; when below the x-axis the y is a negative number. 13. Show that the equa- tion of a circle, radius 5, center at the point (3, 2), is x 2 + y 2 - 6x - 4y = 12. 14. Show that the equa- tion of a circle, radius 5, center at the point ( 2, 3). is x 2 + y 2 + 4x - &y = 12. 15. Show that the equation of a circle, radius 5, center at the point ( - 3, 2), is x 2 + y 2 + 6z - 4y = 12. 16. Find the equation of a straight line passing through the origin and making an angle of 30 with the positive direction of the X-axis. Let T'OT, Fig. 19, be the line. Let P = (x, y) be any (every} point of 40 MATHEMATICS [26 the line. Let PA be the ordinate, y; and let OA be the abscissa, x. OAP is a 30 triangle, and (see exercise 3, page 8) is the equation of the line. For, the coordinates of every point of the line satisfy the equation and the coordinates of any point not of the line do not satisfy the equation. If the line weie raised three units, x its equation would be y = 7= + 3; if it were lowered ten units, its V3 equation would be y = - 10. 17. Show that the equation of a right line passing through the origin and making an angle of 60 with the positive direction of the axis of x is y = \3x. Show that if this line were raised six units its equation would be y = V 3x + 6; and if it were lowered five units, its equation would be y = \3x 5. 18. Find the equation of a straight line cutting the F-axis one unit above the origin and the X-axis one unit to the left of the origin; cutting the F-axis one unit above the origin and the X-axis one unit to the right of the origin; cutting the F-axis one unit below the origin and the X-axis one unit to the right of the origin; cutting the F-axis one unit below the origin and the X-axis one unit to the left of the origin. 26. The Equation of the Straight Line. Let T'OT, Fig. 20, be any straight line passing through the origin. If, in addition to stating that the line passes through the origin, its direction is given, the line is fixed. Its direction may be given by the angle XOT. Let Q and R be any two points upon the line. Suppose a de- scribing point slide along the line from Q to R, its ^-coordinate FIG. 20. 26] GRAPHIC REPRESENTATION 41 will change by the amount QS and its ^-coordinate will change by the amount SR. If the line extends in the first and third quadrants, both QS and SR will be positive, or both negative, de- pending upon whether R is above and to the right of Q, or below and to the left of Q. If the line extends in the second and fourth quadrants, QS and SR will have opposite signs. If R is above and to the left of Q, QS will be negative, and SR positive; if R is below and to the right of Q, QS will be positive and SR negative. It is SR readily seen, however, that for any fixed line the ratio is con- Qb stant (remains unchanged as the positions of Q and R are changed) . SR Let us abbreviate this ratio, , by a. If the line extends into the QS first and third quadrants, a is positive; if into the second and fourth quadrants, a is negative. If the line is very nearly hori- zontal, a is numerically small. If the line is very nearly vertical, a is numerically very large. For a line making an angle of 30 with the positive direction of the axis of x, a is 1 / "V3. For a line making an angle of 45 with the positive direction of the axis of x, a is 1. For a line making an angle of 60 with the positive direction of the axis of x, a is >3. For lines making angles of 120, 135, and 150 with the positive direction of the axis of x, a is, respectively, Vs, 1, and r _- Vs If the line were to rotate counter-clockwise about the origin from a horizontal position toward a vertical position, a' would in- crease continuously from zero through all positive real values. If the line were to rotate clockwise from a horizontal position to- ward a vertical position, a would decrease continuously from zero through all negative real values. From this, we see that every line passing through the origin has an a, and for every value assigned to a there corresponds one and only one straight line through the origin. The direction of a line is fixed by the algebraic value of a as well as by the number of degrees in the angle between the line and the positive direction of the axis of x. The numerical value of a is a measure of the steepness of the line. It represents the number 42 MATHEMATICS [26 of times faster a point moves in a vertical direction than in a horizontal direction as it slides along the line, a is called the slope of the line. The term slope is analogous to the term percent grade in railroad work. A 1 percent grade means a rise of 1 foot for every 100 feet of horizontal distance, or a slope of 1 /1 00. Again we speak of the hydraulic gradient of a stream as 2 feet per 1000, or as 8 feet per mile; these mean, re- spectively, a slope equal to 2/1000 and 8/5280. AP Let P, Fig. 20, be any point on the line. Then -~ . = a, or y/x = a,ory = ax. This, then, is the equation of a straight line passing through the origin. 1 If the line is raised two units, its equation becomes y = ax + 2; if r.aised five units, y = ax + 5; if lowered ten units, y = ax 10. The equation of a line parallel to the line y = ax, and cutting the F-axis b units from the origin, is y = ax + b. If the line cuts the F-axis above the origin, b is positive. If the line cuts the F-axis below the origin, b is negative. y = ax + b is called the slope equation of the straight line, a is called the slope, and b the y-intercept. Exercises 1. Give the slope and y- intercept for each of the following lines: (a) y = 2x + 1; (g) y = - Bx + 2; (ro) y = \x + 3; (b) y = 2x - 3; (h) y = x + 1; (n) y = \x - 3; (c) y = 3z - 6; (i) y = x - 1; (o) y = - %x; (d) y = 2x - 1; (j) y =x+ 2; (p) y = - fx - 8; (e) y = - 2z - 2; (fc) y = - a; + 6; (?) y = - fz; (/) y = - 3x + 4; (I) y= -x + 2; (r) y = - O.Ola: - 6. 2. In exercise 1, what lines are parallel to (a)? To (c)? To (e)? To (/)? To (A)? To (fc)? To (TO)? To (p)? 3. If a point were to move along the lines of exercise 1, so that its x-coordinate is decreased by one unit, would the y-coordinate be increased or decreased and how much? 4. Find the equation of a straight line passing through the points P, (3, 2), and Q, (1, 5). To solve the problem is to find the value of a and of 6 and substitute in the equation y = ax + b. As we pass from the point P to the point Q, the x-coordinate decreases two units 1 The student must guard against getting the idea that the slope of a line is, in general, y over x. This is only true when the line passes through the origin. 27] GRAPHIC REPRESENTATION 43 and the ^-coordinate increases three units. Then the slope, a, of the line is 3/2. Let R be the point where the line through P and Q cuts the F-axis. Draw PT perpendicular to the F-axis, meeting it at the point T. Draw QS perpendicular to PT, meeting it at S. PSQ and PTR are similar triangles, and we have TP _ TR SP ~ SQ* or 3 _ TR 2 ~ 3 ' or TB-\ and b = L 2 3 , since b = OT + TR. The equation of the line is _ 3 13 ? or 3z + 2y = 13. This work can be checked, for, substituting for x and y the co- ordinates of the points P and Q, the equation is satisfied in each case. 6. Does the line of exercise 4 pass through the point (2, 7/2)? The answer is yes, for substituting 2 for x and 7/2 for y, in the equation of the line, the equation is satisfied. 6. Find the equation of the straight line passing through the points (a) (-3, 2) and (2, 1); (6) (1, 1) and (2, 2); (c) (-1, 3) and (2, - 3); (rf) (- 3, -1) and (-1, 3); (e) (-1,2) and (6, - 1). 7. Which lines of exercise 6 pass through the point ( 2, 6)? 8. Find the equation of a straight line passing through the origin and the point (3, 5); through the points (3, 1) and ( 3, 1); through the points ( 3, 5) and ( 3, 2). 27. The Graph of an Equation of the First Degree between x and y. In 26 it was shown that every equation -of the form y = ax represents a straight line passing through the origin; and further, when the line was moved up or down its equation became y = ax + b, where 6 represents the algebraic distance the line was translated in the y direction. If the line were moved from a very low position to a very high position, the b would take on all real values between its very small initial value and its very large ter- minal value. Thus, by moving the line over a greater and greater 44 MATHEMATICS [27 Vertical distance, b can be made to take any real value whatever. Any equation then of the form y = ax + b, for any values given to a and b, represents a straight line. It is readily seen that this form of the equation includes all lines excepting those parallel to the F-axis. The equation of a line parallel to the F-axis is, however, x = k, where the constant k is the number of units the line is distant from the F-axis. The equation of any straight line is then either of the form y = ax + 6 or of the form x = k. Since any equation of the first degree between x and y, Ax + By = C, may be put in one of these forms, it is the equation of a straight line, and is called a linear equation. To illustrate: 3x + % + 6 = may be written y = fx 3, the equation of a straight line having the slope f, and a y-intercept of 3. Exercises Find the slope and intercept of each of the following lines: 1. 2z - 3y = 1. 4. 3x - 2y + 6 = 0. 2. 3x - 2y - 6 = 0. 5. y - 3x = 5. 3. x - y = 1. 6. y + x = 0. 7. Find the point of intersection of each line given above with the axis of x. (The distance from the origin to the point of intersection of a line with the X-axis is called the re-intercept.) HINT: This is done by putt'ng y equal to zero, and solving the re- sulting equation for x. Why does this give the rc-intercept? 8. Draw the straight lines of exercises 1 to 6 upon one sheet of squared paper. Use 1 inch (or 2 cm.) to represent the unit and place the origin near the center of the sheet. 9. Find the coordinates of all points of intersection of these lines in so far as they intersect upon the sheet of squared paper. The coordinates of the points of intersection are to be read to two decimal places. To make this possible great care must be exercised in drawing the lines. Tabulate your results. In the first column place the numbers of the lines, as 2-5, meaning the line in exercise 2, and the line in exercise 5. In the second column place the x-coordinate and in the third column place the y- coordinate of the point of intersection of the two lines. A fourth and a fifth column are to be filled in with the results obtained from the next exercise. 10. Consider all pairs of equations given in exercises 1 to 6 as simultaneous equations and solve for x and y. Record the results in the table of the preceding exercise. Compare the corresponding x v GRAPHIC REPRESENTATION 45 values and the corresponding y values found by the two methods. Should the corresponding values be the same? Why? 28. Analytic Method of Finding a and b. In exercise 4, 26, there was described a geometric method for finding the a and the b in the equation of a straight line passing through two points. An analytic method of finding a and b will now be illustrated. Let (3, 2) and (1, 5) be the points through which the line passes. The coordinates of each point when substituted for x and y in the gen- eral equation of the line, y = ax + b, must satisfy that equation. Thus we have, upon substitution, 2 = 3a + b, and 5 = a + b. These two equations are called conditional equations, for they are, in mathematical language, statements of the two and only two conditions imposed upon the line. 1 The first equation is the equation of condition which says that the point (3, 2) is upon the line y = ax + b; the second equation is the equation of condi- tion which says that the point (1, 5) is upon the line y = ax + b. Solving these two conditional equations for a and b we ob- tain fl = f, and b = V. Substituting these values in y = ax + b, the equation of the line through the two given points becomes y = lx -f V, or 3z + 2y = 13. Exercises 1. Find, by the analytic method, the equation of the line through each of the following pairs of points: (a) (1, 2) and (- 3, 6); (c) (0, 0) and (- 1, - 1); (b) (-2, 6) and (3, - 2); (d) (0, 0) and (- 2, - 3). 2. Plot upon squared paper the data given in Table VI. Plot hydraulic gradient 2 along the F-axis using 1 cm. (or 1/2 inch) 1 A conditional equation may be denned as an equation expressing some condition given in the problem. To illustrate: If two numbers x and y are to be found such that their sum is 9, and the sum of their squares is 25, we have x + y = 9 and x 2 + j/ 2 = 25. The first equation expresses the condition imposed upon x and y, that their sum is 9. The second equation expresses the condition that the sum of the squares of the numbers is 25. 2 By hydraulic gradient is meant the drop of the surface of water in moving down stream a given distance. Thus in the table the hydraulic gradient of 10 feet per thousand feet means a drop of the water plane (water surface) at the rate of 10 feet for every 1000 feet measured along the horizontal in the direction of greatest decline of the plane. 46 MATHEMATICS [28 to represent 10 feet. Plot velocity along the X-axis, using 2 cm. (or 1 inch) to represent 10 feet. From the plotted points it appears as if the points would fall (or would nearly fall) along a straight line if there were no errors of observation, or if all the readings were taken under the same conditions. The best that can be done in this case is to assume a linear relation between hydraulic gradient and velocity. Draw with the aid of a transparent triangle a straight line among the points which in your judgment is the best line to repre- sent this relation. Does the line pass through the origin? Compute a by taking two points at or near the ends of the line. The student must remember that a is the ratio of the number of units in the change in the ^-coordinate to the number of units in the change in the x-coordinate. 3. Plot upon squared paper the data for the 8-inch well given in Table VII. Plot head 1 along the F-axis using 1 cm. (or 1/2 inch) to represent 1 foot. Plot yield along the X-axis, using 1 cm. (or 1/2 inch) to represent 10 gallons. As in the preceding exercise draw a straight line among the points and determine the a. TABLE VI. VELOCITY OF WATER THROUGH SAND Hydraulic gradient, feet per 1000 Velocity of water, feet per 24 hours Hydraulic gradient, feet per 1000 Velocity of water, feet per 24 hours 31.9 16.9 21.0 10.1 20.8 11.4 119.1 58.8 54.1 10 24.5 4 3 55.8 22.7 TABLE VII. YIELD OF AN 8-INCH AND OF A 14-INCH WELL Size of well, inches Head, feet Yield, gallons per minute Size of well, inches Head, feet Yield, gallons per minute 14 2.02 31.4 8 2.50 39.5 14 4.00 59.7 8 3.85 49.4 14 6.40 93.2 8 6.47 89.3 14 8.68 121.1 8 8.28 116.3 14 11 73 151.1 1 By "head" in this problem is meant the distance the water was lowered in the well below its normal position. Thus, a head of 2.5 feet in the table means that when 39.5 gallons of water per minute were pumped from the well the elevation of the sur- face of the water in the well was 2.5 feet below its normal elevation. 29] GRAPHIC REPRESENTATION 47 4. Plot upon squared paper the data for the 14-inch well given in Table VII. Draw a straight line among the points and find its a. 1 29. Cost of Concrete. Suppose concrete mixture is made by mixing cement, sand, and crushed rock. The expression " mixture (ra, n, fe) " means the relative amounts, by volume, of the three materials. Thus, a mixture (1, 2, 3) means that for each volume of cement there are two volumes of sand and three volumes of rock. Curves will now be drawn from which the cost of a cubic yard of concrete may be determined. The method of drawing the curves will be illustrated by an example. The student is to con- struct the drawing as explained. Suppose the percent voids of rock is 45 and that of sand 33. Let us take the mixture (1, 2, 3). For each cubic yard of cement, 2 cubic yards of sand and 3 cubic yards of rock are used. In the 3 cubic yards of rock there are 3-0.45 = 1.35 cubic yards of voids. Then 1.35 cubic yards of the 2 cubic yards of sand will be taken up by the voids of the rock leaving 0.65 cubic yard. Thus mixing 3 cubic yards of rock and 2 cubic yards of sand will give but 3 + 0.65 = 3.65 cubic yards of mixture. In the 2 cubic yards of sand there are 2-0.33 = 0.66 cubic yard of voids. Thus there is an excess of 1 0.66 = 0.34 cubic yard of cement above filling the voids of the sand. Hence in mixing the rock, sand, and cement there will be 3.65 + 0.34 = 3.99 cubic yards of mixture. If Xi is the cost (say in. dollars) of a cubic yard of stone, the cost, yi, of the stone in 1 cubic yard of concrete is 3 yi ~ 3^9 Xl or yi = 0.752 x, (1) Upon a sheet of squared paper draw a straight line represent- ing equation (1). Let 10 cm. represent $1. Then each centi- meter will represent 10 cents. To draw this line join the points (0, 0) and (2, 1.504). 1 The equations found in exercises 2, 3, and 4 very probably do not represent the true relations among the measured numbers, and are at best only approximate relations. Such equations are called empirical equations, or empirical laws. 48 MATHEMATICS [29 If x 2 represents the cost of 1 cubic yard of sand, the cost, y 2 , of the sand in 1 cubic yard of concrete is 2/2 = 3799 2 or ?/ 2 = 0.501 x 2 (2) Plot equation (2) upon the same sheet of paper upon which equation (1) was plotted, using the same scales. A barrel (four sacks) of cement contains 3| cubic feet, or 5/36 cubic yard. If x s represents the cost of cement per barrel, the cost, y 3 , of cement in 1 cubic yard of concrete is 36 z ~ 5-3.99 3 or 7/3 = 1.805 x 3 (3) Draw this line upon the same sheet of paper upon which equations (1) and (2) were drawn. Mark the first line drawn "rock;" the second "sand;" and the third "cement." From these three lines the cost of material in 1 cubic yard of concrete can be scaled off, if the costs of stone, sand, and cement are given. Thus, suppose stone worth $1.40 per cubic yard, sand $1.25 per cubic yard and cement $1.10 per barrel. From the point 1.4 on the X-axis go up to the rock line, thence to the left to the F-axis and obtain the reading $1.05, the cost of the stone in 1 cubic yard of concrete. From the point 1.25 on the X-axis go up to the sand line, thence to the left to the F-axis and obtain the reading $0.63, the cost of the sand in 1 cubic yard of concrete. From the point 1.1 on the X-axis go up to the cement line, thence to the left to the F-axis and obtain $1.99, the cost' of the cement in 1 cubic yard of concrete. The total cost of material in 1 cubic yard of concrete is then $1.05 + $0.63 + $1.99 = $3.67. The following table gives the number of cubic yards of rock, cubic yards of sand, and barrels of cement in 1 cubic yard of concrete based on 45 percent voids in rock and 33 percent voids in sand. 30] GRAPHIC REPRESENTATION 49 Mixture Rock, cu. yds . Sand, cu. yds. Cement, barrels 1:2 :3 0.752 0.501 1.805 1:2 :4 0.881 0.446 1.586 1:2^:5 0.922 0.461 1.327 1:3 :5 0.951 0.475 1.125 Exercises 1. Verify the above table. 2. Construct curves from which to scale off the cost of material used in a 1, 3, 5 mixture of concrete. 3. From the drawing of exercise 2, find the cost of the material in 1 cubic yard of concrete, if rock is worth $1.50 a cubic yard, sand $1.10 a cubic yard and cement $1.30 per barrel. 30. The Graph of y = px 2 . Exercises 1. Upon a sheet of paper, form M7, construct a table per the following instructions. Head the first vertical column x, the second x 2 , the third x, the fourth |x 2 , and the fifth ?x 2 . In the horizontal spaces in the first column place the numbers 0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, . . ., 4.6, 4.8, 5.O. 1 In the second column place the squares of these numbers, as 0, 0.04, 0.16, 0.36, 0.64, 1.0, . . ., 21.16, 23.04, and 25.00. In the third, fourth and fifth columns place, respectively, the halves, the thirds and the fourths of the num- bers in the second column. Head this table "Table of Squares." 2. Plot upon squared paper a curve for y = x 2 . Let the X-axis run the shorter and the F-axis the longer dimension of the sheet of paper. Choose the origin near the lower center of the page. Plot points for negative values of x as well as for positive values. Make use of your "Table of Squares." Let 2 cm. (or 1 inch) on both axes represent one unit. Use a very sharp pointed pencil in plot- ting these points, and exercise great care in locating them as accurately as possible. Do not draw a curve through the plotted points until all exercises of this list have been completed. 3. Plot a curve for y = \x*. Follow directions given in exercise 2. 4. Plot a curve for y = $x 2 . Follow directions given in exercise 2. 6. Plot a curve for y = Jx 2 . Follow directions given in exercise 2. 6. Plot a curve for y = ix 2 . Follow directions given in exercise 2. 1 The marks " . . . " mean "and so on," indicating all corresponding inter- mediate numbers. 4 50 MATHEMATICS [30 7. Prom the coordinates of the points of the curve y = z 2 compute the slope of lines passing through the consecutive plotted points. The slope of the secant lines will be approximately the slope of the tangent lines drawn to the curve midway between the points. Thus, if we find a slope of 4.2 for the secant line drawn through the points whose ^-coordinates are 2.0 and 2.2, we shall have the approximate slope of the tangent line drawn to the curve at the point whose x-coordinate is 2.1 Tabulate your results. 8. Same as exercise 7 but use curve y = z 2 . 9. Same as exercise 7 but use curve y = |z 2 . 10. Same as exercise 7 but use curve y = |x 2 . 11. Same as exercise 7 but use curve y = z 2 . 12. Plot upon squared paper the results of exercise 8. Plot x along the horizontal axis, and slope, a, along the vertical axis. Draw a straight line through the origin and among the plotted points. Measure the slope of this line; call it A\. 13. Same as exercise 12 but use results of exercise 9. Call the slope Az. 14. Same as exercise 12 but use results of exercise 10. Call the slope A*. 15. Same as exercise 12 but use results of exercise 11. Call the slope At. 16. Divide A 2 by f; As by |; and A^ by j. These quotients and AI should all be equal to 2. It can be shown by the method of the Calculus that the slope of a tangent line drawn to the curve y = px 2 (p being a constant) is 2px, where x is the abscissa of the point of tangency. 1 1 A formal proof is given here. Let P, (x, y), Fig. 21, be any point, and Q any second point on the curve y = px*. Let k, (PR), be the difference of the two z-coordinates, and let h, (RQ), be the difference of the two ^-coordinates. The coordinates of Q are then (x + k, y + h) . For the point P we have y = px*; and for the point Q we have (y + h) = p(x + k) t . Subtracting, we have h = 2pkx + pk*. Dividing by k we have h/k = 2px + pk. This shows that the slope, (h/k), of the secant line drawn through the points P and Q is 30] GRAPHIC REPRESENTATION 51 This is only saying that if a point move along the curve its upward motion (downward motion if the point is to the left of the y-axis) is at any instant 2px times its horizontal motion, or its y-coordinate is changing 2px times as fast as its x-coordinate. Here x represents the x-coordinate of the position of the moving point. Experiment shows that a body falling freely from rest gives the law s = 16. li 2 , where s represents the distance, in feet, passed over in the time t, in seconds. If s be plotted along the vertical axis and t along the horizontal axis the slope of the tangent line drawn to the curve will be 32.2t. Distance s, then, is changing 32.2t times as fast as t, i.e., the speed of a body falling freely from rest is 32.2J, when t is the time, measured in seconds, elapsed from the instant the body began to move. If v represents speed, in feet per second, we have v = 32. 2t. Plotting v along the vertical axis and t along the horizontal axis we obtain a straight line representing the relation between speed and time. The slope of his line is 32.2, i.e., v changes 32.2 times as fast as t, or the acceleration of gravity is 32.2 feet per second per second. 17. Plot a curve for y = x 2 2x + 1 = (x I) 2 . How is this curve related to the curve y = x 2 ? 18. Plot a curve for y = x* - 2x + 2 = (x - I) 2 + 1. How is this curve related to the one plotted in exercise 17? 19. Plot a curve for 3y = x* - 4x + 13 = (x - 2) 2 + 9, or y = 3(2 2) 2 + 3. How is this curve related to the curve for y = *x 2 ? 20. Without plotting, compare the following curves: (a) y = x 2 and y = x 2 . (6) y = |x 2 and y = - x 2 . (c) y = |x 2 and y = \x l . (d) y = lx 2 and y = }x 2 . (e) y = (x - I) 2 and y = - (x - I) 2 . 2px + pk. Aa the point Q slides along the curve and approaches P, k, (PR), grows smaller and smaller and approaches zero; the slope of the secant line then ap- proaches 2px. But, as Q approaches P the secant line rotates about the point P and approaches the tangent line PT. Therefore, the slope of the tangent line is the limit of the slope of the secant line, i.e., 2px. 52 MATHEMATICS [31 21. Have the curves considered in exercise 20 lines of symmetry? Give the equation of the line of symmetry for each curve. 1 22. Change each of the following equations to the form y = p(x q) z + r, where p, q, and r are constants, and find the coor- dinates of the highest or lowest point on the curve: (a) 4x 2 + By = 3x + 5, or 3y = - 4x 2 + 3x + 5, or - | y = x 2 f x I , or - f y = x 2 - f x + V* - f - &, or - \y = (x - f) 2 -ff, or y = - f (x - f ) 2 + ||. P = - I; 9 = 1; r = If. The coordinates of the highest point are f and If. (6) y = x* + 4x - 6. (f) x*+y = 6. (c) 3y = x 2 + 6x + 10. (0) x 2 + 2z + y + 1 = 0. (d) 3y = 4x 2 + 6x + 10. (h) 2x 2 - 5x - 6y + 10 = 0. (e) 4x 2 = y - 2x - 6. (i) 5x 2 + 2x - y = 0. 31. The Parabola. All curves drawn and discussed in the preceding section have equations reducible to the form y = p(x q) 2 + r. They have a line of symmetry, x = q, and change from an increasing to a decreasing curve or vice versa at the point (q, r). 2 Each curve is the curve y = px 2 translated horizontally a distance q, and vertically a distance r. 3 Any curve which is of the shape of y = px 2 is called a parabola. The line of symmetry of the curve is called the axis of the parabola, and the point of intersection of the curve with its axis is called the vertex. 4 1 A line such that line segments drawn perpendicular to it and terminated by the curve, are bisected by it is called a line of symmetry of the curve. Thus any line passing through the center of a circle is a line of symmetry for the circle. A point such that line segments drawn through it atid terminated by the curve, are bisected by it, is called a point of symmetry. Thus, the center of a circle is a point of symmetry. 2 A curve is said to be increasing when its ordinate increases as its abscissa in- creases; and decreasing when its ordinate decreases as its abscissa increases. The increase in each case being understood to be the algebraic increase. 8 Distance here means algebraic distance. When q is positive the curve is moved to the right, when negative to the left. When r is positive the curve is moved up, when negative down. 4 It can be shown that the parabola is the locus of a point moving in a plane so that its distance from a fixed line, the directrix, is always equal to its distance from a fixed point, the focus. The parabola is also the curve of intersection of a right cir- 31] GRAPHIC REPRESENTATION 53 Any equation of the form y = ax 1 + ftx + 7, where a, /3 and 7 are constants, positive or negative, represents a parabola with its axis parallel to the Y-axis; for ax 2 + 0x + y = a which is of the form p(x q) 2 + r, where p = a, q = - > and 2a r = -- - ---- The expresson ax 2 + fix + 7 is the most general expression of the second degree in x, or it is a general quadratic. Thus we see that the equation y = ax 2 + 0x + 7 represents a parabola with axis parallel to the Y-axis, and by completing the square in the x terms we are able to locate its vertex. Problems illustrating this point have already been given in exercise 22, 30. We have already seen that the slope of the tangent line drawn to the curve y = px 2 is 2px, i.e., 2p times the x-coordinate of the point of contact of the tangent line. Another way of saying this is that the y is changing 2px times as fast as the x. From this and what was said in the preceding paragraph we see that the slope of the tangent line drawn to the curve for y = ax 2 + /3x -\- y is 2p(x q), or 2p times the distance of the point of contact of the tangent line from the axis of the parabola. Conversely, it may be shown that if a number y changes 2p(x q) times as fast as the number x, x and y are connected by the parabolic law. Exercises 1. Plot data given in Table VIII. Plot temperature along the horizontal axis and weight along the vertical axis. cular cone and a plane parallel to an element of the cone. The parabola has many useful and interesting properties, one of which is: If a parabola be rotated about its axis forming a surface of revolution, the paraboloid of revolution, rays of light passing out from its focus are reflected by the surface into lines parallel to the axis of the paraboloid. This property is made use of in the construction of reflectors. 54 MATHEMATICS [31 2. An empirical equation of the second degree in T for the data given in Table VIII is W = 0.06T 2 - 0.367 7 + 0.47. Here W is used in order to distinguish between weight computed by formula and observed weight, W. Copy Table VIII upon a sheet of paper, form M7. In a third column headed W place weights computed by the empirical formula. In a third column headed W W place the differences between the observed and computed weights. Plot W upon the drawing of exercise 1. Mark the points representing W so that they are readily distinguished from the points representing W. TABLE VIII. RELATION BETWEEN TEMPERATURE AND WEIGHT OF 10 LITERS OF WATER T = temperature in degrees centigrade. W = loss in weight, measured in grams, of 10 liters of water, as temperature differs from 4. T w T w 1.3 16 10.3 2 0.3 18 13.8 4 0.0 20 17.7 6 0.3 22 22.0 8 1.2 24 26.8 10 2.7 26 31.9 12 4.8 28 37.1 14 7.3 30 43.3 3. Plot data given in Table IX. 1 Plot ^ along the horizontal axis using 1 cm. to represent one-tenth unit. Plot velocities along the vertical axis, using 1 cm. to represent two one-hun- dred ths of 1 foot per second. Sketch a smooth curve among the points. The curve may not pass through all the plotted points. 2 Begin to number the vertical axis with 2.90. The true origin will be far below the sheet of paper. 1 The velocity at any point of a moving stream is determined by a current meter placed at that point. 2 This curve is called a vertical velocity curve. In practical work, however, velocities are plotted along the horizontal axis and depths along the vertical axis, and down from the origin. Your drawing gives the usual form of plotting if it is turned 90 in a clockwise direction. Vertical velocity curves are parabolic in shape, with the axis of the parabola parallel to the surface of the water. 32] GRAPHIC REPRESENTATION 55 TABLE IX. RELATION BETWEEN VELOCITY AND DEPTH AT A POINT IN THE LOWER MISSISSIPPI Depth at observed point = d; whole depth = D. d D Velocity, feet per second d D Velocity, feet per second 0.0 3.195 0.5 3.228 0.1 3.230 0.6 3.181 0.2 3.253 0.7 3.127 0.3 3.261 0.8 3.059 0.4 3.252 0.9 2.976 4. An empirical formula connecting velocity, V, and depth, x, is V' = - 0.78(x - 0.296) 2 + 3.261. Compute V from this formula and tabulate your results following, in a way, directions given in exercise 2. 6. A strip of tin L feet long and 40 inches wide is made into a trough with rectangular cross section by bending up an equal portion of each side. Find the width of each portion bent up such that the volume of the trough is a maximum. HINT: The volume is a maximum when the area of the cross sec- tion is a maximum. Let y be the area of the cross section. Let x be the width of the portion bent up. The dimensions of the cross section are then x and 40 2z. Hence y = x(40 - 2x) y = - 2(z 2 - 20x) y = - 2(x 2 - 2Qx + 100) + 200 y = - 2( - 10) 2 + 200 The equation is that of a parabola with its high point at (10, 200). Thus the maximum area, 200, occurs when 10 inches of tin are turned up. 6. Same as exercise 5, but with the width 60 inches and length equal to 10 feet. Find the volume of the trough. 32. The Equilateral Hyperbola. Exercise 1. Plot upon squared paper a curve for 123 (a) y = - (6) y = ~ (c) y - XXX 56 MATHEMATICS [32 Choose the origin near the center of the sheet and plot for negative as well as for positive values of x. Let 1 cm. (or 1/2 inch) rep- resent the unit along each axis. When x is numerically small the interval between successive values assigned to it must be small in order to obtain points near enough together for sketching the curve. Compare curves (a), (6), and (c). These curves are equilateral hyperbolas. They are special cases of the more general equilateral hyperbola, y = ' where a X is a constant. When one quantity, y, varies with another, x, so that the relation connecting the two is represented by the equa- tion y = , y is said to vary inversely with x. tC If points plotted upon squared paper arrange themselves along a curve which seems to be an equilateral hyperbola, y = - X may, as a first assumption, be taken as an empirical equation connecting the variable numbers plotted. The problem now becomes one of selecting a value for the constant a, and of deter- mining the agreement between the values of the numbers plotted and those called for by the equation. If the a cannot be selected giving a fairly close agreement between observed and calculated numbers, it means that the law we are dealing with in nature is not that of one number varying inversely as the other. An easy elementary method for determining the a, as well as showing whether or not the law is that of the inverse power, will be illustrated by means of an example. In Table X are given the relative volumes of a mass of air subjected to pressures ranging from to 100 pounds per square inch measured above atmospheric pressure. The temperature of the air is constant, 60 Fahrenheit. Plot the data upon a sheet of squared paper. Plot pressure along the F-axis, using 2 cm. (or 1 inch) to represent 10 pounds. Plot volume along the X-axis, using 10 cm. (or 5 inches) to rep- resent the unit volume. This curve seems to be of the general shape of y = excepting that it crosses or touches the .X-axis X at unity instead of remaining above the X-axis and becoming nearer and nearer to it. 32] GRAPHIC REPRESENTATION 57 To find an equation connecting x (volume) and y (pressure) proceed as follows. Construct a third column of numbers for Table X, by taking reciprocals of the numbers representing rela- tive volumes given in the second column. Call these values the x' values. (The prime is put upon the x to distinguish it from the x which represents the true volume reading.) Plot a curve be- 100 90 8 70 #60 " 3 40 30 20 10 Reciprocal I of Volume A 8 0.2 0.4 0.6 Volume 0.8 1.0 FIG. 22. tween pressure and x'. Let 2 cm. (or 1 inch) along the Y- axis represent 10 pounds; let 2 cm. (or 1 inch) along the hori- zontal axis represent one unit of x' '. Such a curve, reduced, is given in Fig. 22. This curve is a straight line with slope equal to 14.7 and y-intercept equal to 14.7. The equation of the 58 MATHEMATICS [32 line is y = 14. 7o/ 14.7. Since x' , the relation connect- ing volume and pressure is represented by the equation 14.7 y = - 14.7. It is readily seen that this is the equation of u/ 14.7 the curve y = - - lowered 14.7 units; that the first curve *c plotted between pressure and volume is an equilateral hyperbola; and that the pressure readings given in Table X, when increased by 14.7, vary inversely as the volume. TABLE X. AIR COMPRESSION FROM ONE ATMOSPHERE AT SEA-LEVEL Pressure, Ib. per sq. in. Volume Pressure, Ib. per sq. in. Volume 1,000 45 0.246 1 0.936 50 0.227 2 0.880 55 0.210 3 0.830 60 0.196 4 0.786 65 0.184 5 0.746 70 0.173 10 0.595 75 ' 0.163 15 0.495 80 0.155 20 0.423 85 0.147 25 0.370 90 0.140 30 0.328 95 0.134 35 0.295 100 0.128 40 0.268 The pressure readings given in the table were taken by means of a pressure gauge which registers the difference in pressure between the outside and inside of the receptacle containing the air. If we assume Boyle's law for the compressibility of a gas (the volume varies inversely as the pressure) the y-intercept, 14.7, found above, shows that when the experiment was conducted the pres- sure gauge was registering 14.7 pounds less than the true pressure exerted upon the air, or that the atmospheric pressure was 14.7 pounds per square inch. If, in Fig. 22, the plotted points had not arranged themselves along a straight line the method outlined in 28 could have been 32] GRAPHIC REPRESENTATION 59 used in finding the slope and intercept. When the plotted points arrange themselves -along a curve which is not a straight line, it may mean that the original variables are not connected by the equilateral hyperbolic law. Exercise Find the equation connecting I and W, using the data given in Table XI, 1 where 7 represents the indicated horse power of a steam engine, and W the number of pounds of steam consumed per hour. TABLE XI I W / W 36.8 460.0 15.8 222.8 21.5 406.2 12.6 182.7 26.3 344.5 8.4 137.0 21 279 3 1 Taken from Elementary Practical Mathematics by John Perry. CHAPTER II LOGARITHMS 33. Logarithm Defined. The equation y = x z 1 is solved for y. If values are assigned to ' x, the corresponding values of y may be calculated. The same equation may be written x = \/y -f- 1. Here the equation is solved for x, and if values are assigned to y, 1;he corresponding values ofz may be calculated. The two equations represent the same law connecting x and y. Exercises 1. Solve each of the following equations for x: (a)y = x; (i) y = 3x 2 - 2x + 1; (6) y = x - 1; (j) y = x s ; (c)t/ = 2x+3; (k)y=x*-l; (d)y = |x -6; (I) y = 3z -3; (e) y = x 2 ; (m) y = x 3 + 3x 2 + 3x + 1; (/) y = 2x^5 (n) y = X 3 - 3x 2 + 3x; (0) y = x 2 - 2x + 1; (o) y = Vx* - 25. (h) y = 3x 2 + 6x + 3; 2. Solve each of the following equations for x and also for y: (a) x 2 + y 2 = 25; (i) xy = 2; (6) x 2 -0 = 25; (j) zy +y = 1; (c) x 2 + 2x + y 2 = 25; (jfe) xy 2 = 3; (d) x 2 - 2x + y 2 = 25; (I) x*y = 5; (e) x 2 + y 2 - 2y = 25; (m) x 2 y + xy 2 = 3; (/) x 2 + 2x + y 2 - 2y = 25; (n) x 2 ?/ + xy + xy 2 = 0; (g) x 2 + x + y 2 - 2y = 16; (o) xy 3 = 1; (A) xy = 1; (p) x=y = - 1. All equations of the preceding exercise can be solved for x by means of the symbols and operations of elementary algebra. The equation y = a x , where a is a known positive constant different 60 33] LOGARITHMS 61 from zero, cannot, however, be solved for x by means of these elementary operations. The best we can do is to say, x = the exponent of the power to which a must be raised to produce y. In this verbal sense, the equation is now solved for x. The words "exponent of the power to which a must be raised to pro- duce y" are abbreviated by "logarithm of y to the base a," and in writing they are further abbreviated by "Iog y." Thus, the equation y = a x , solved for x, gives x = log a y- Logo is then a new symbol; considered as an operator, operating upon y, it gives the number, x, such that when a has x for an exponent, the power equals y. a is called the base. The definition then of the logarithm of a number x to the base a is: "the exponent of the power to which a must be raised to produce the number in question." Of the two equivalent equations, y = a* and x = logo y, the first is called the exponential form, and the second the logarithmic form. Exercises 1. Write the following equations in logarithmic form : (a) 100 = 10 2 ; (e) 25 = 5 2 ; (i) 1 = 10; (6)10 = 10'; (/) 9 = (V3) 4 ; 0')2 = (V2) 2 ; \C) TC == *w j \Q ') -LD ^ = ~ *j \fo) *5 ^~ \'\' *J ) j (d) 8 = 2 3 ; (h) 125 = 5 3 ; (1) 8 = (V2) 6 ; (m) 2 = 1; (n) 5 = 1. 2. Write the following equations in exponential form: (a) Iog 2 4= 2; (/) logio 10 = 1; (ft) logs 9 = 2; (b) logio 100 = 2; (0) logs 3 = 1; (I) logio 1000 = 3; (c) logio 1 = 0; (h) Iog 5 5 = 1; (m) logio 10,000= 4; (d) Iog 2 1 = 0; (i) logs 25 = 2; (n) logs 1 = 0; (e) logs 1=0; (j) Iog 2 32 =5; (o) Iog 99 1 = 0. 3. Give the value of each of the following: (a) Iog 2 4; (h) log, 100; (o) logio 0.1; (6) logs 9; (i) logio 1000; (p) logio 0.01; (c) logs 27; (j) logio 10,000; (a) logio 0.001 ; (d) logt 25; (ft) logio 1; (r) logio 100,000,000; (e)log 4 4; (I) logs 1; () logio 0.000000001. (/) log 6; (m) logs 1; (o) logio 10; (n) logo 1; 62 MATHEMATICS [34 4. Evaluate the following: (a) Iog 2 2 + Iog 4 4 + log B 5; (6) logs 9 + Iog 5 25 + logs 125 + logs 36; (c) logio 1000 + Slogs 36-5 Iog 2 16+2 log B 625; (d) Iog 2 1 - logs 1 + logs 3 - logio 10 + logs 27; (e) \ logio 100 - | logio 10-| logio 0.1-2 log, 0.01; (/) logo a; (g) Iog l 34. Theorem : The logarithm of the product of two numbers is the sum of their separate logarithms. Proof: Let logo u = x and Iog v = y Then a x = u and a" = v By multiplication a*+v = uv, or Iog uv = x + y, or Iog uv = logo u + logo v Thus the theorem : The logarithm of the product of two numbers is the sum of their separate logarithms. It is easily seen that the logarithm of the product of three or more numbers is the sum of their separate logarithms. If no base is indicated, the base 10 is understood. Exercises If log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990, and log 7 = 0.8451, find the logarithms of the following: 1 (a) 6 (g} 35 (m) 9 (6) 15 (h) 40 (n) 27 (c) 10 (t) 4 (o) 18 (d) 20 (j) 8 (p) 24 (e) 45 (k) 16 (q) 30 (/) 21 (i) 32 (r) 12. 1 The logarithms used in this book are taken from a four-place logarithmic table and are correct to the nearest fourth decimal place. 35] LOGARITHMS 63 35. Theorem : The logarithm of the quotient of two numbers is the logarithm of the dividend diminished by the logarithm of the divisor. Proof: Let logo u = x and Iog v = y. Then a* = u and a = v. By division _ u v or logo (-) = x - y, or logo y-j = 10g U logo V. Thus the theroem: The logarithm of a quotient is equal to the logarithm of the dividend diminished by the logarithm of the divisor. Exercises Using the logarithms given in preceding Exercise, find the loga- rithms of the following: (a) f ; (6) f- ; (c) f ; (d) I; (e) J; (/) |. 36. Theorem : The logarithm of the power of a number is the logarithm of the number multiplied by the index of the power. Proof: Let logo u = x. Then u a 1 , Then or logo u n = nx, or Iog u n = n logo u. Thus the theorem : The logarithm of the power of a number is equal to the logarithm of the number multiplied by the index of the power. The theorem is true for fractional exponents as well as for integral exponents, and from it then follows the law that the logarithm of the 64 MATHEMATICS [37 root of a number is equal to the logarithm of the number divided by the index of the root. ILLUSTRATION : If log 2_= 0.3010, log 4 = 2 log 2 = 2 X 0.3010 = 0.6020, and log V2 = J log 2 = } X 0.3010 = 0.1505. Exercises 1. Using the logarithms of the number given for the exercises) 34, find the logarithms of the following: (a) 3 2 ; (6) 3 3 ; (c) 5 2 ; (d)27; (e) 125; (f)V5; (g) 3?; (h) 7*; (i) 1251; (j) 625; (k) 2. If the log 2 is 0.3010, what is the (a) log 20? (6) log 200? (c) log 2000? (d) log 20,000? (e) log 200,000? How do these loga- rithms differ? What part of these logarithms is the same? Why? 3. If log 3 = 0.4771, what is the (a) log 30? (6) log 300? (c) log 3000? (d) log 30,000? What part of these logarithms is the same? Why? 4. If log 373 = 2.5717, what is the log 37.3? log 3.73? log 3730? log 37,300? What part of these logarithms is the same? Why? If a number is greater than unity, how does moving the decimal point affect its logarithm? 37. Characteristic and Mantissa. If 10 is used as base, the following equations are true: log 10,000 = 4 log 1,000 = 3 log 100 = 2 log 10 = 1 log 1 = log 0.1 = log 1 - log 10 = - 1 = 9 - 10 log 0.01 = log 1 - log 100 = - 2 = 8 - 10 log 0.001 = log 1 - log 1,000 = - 3 = 7 - 10 log 0.0001 = log 1 - log 10,000 = - 4 = 6 - 10 From the above table, which can be extended indefinitely in both directions, we see that any number greater than 1000 and less than 10,000 has a logarithm lying between 3 and 4, i.e., the logarithm is 3 plus some (decimal) fraction. All numbers between 37] LOGARITHMS 65 1000 and 10,000 have four digits preceding the decimal point, and all numbers, excepting 1000, having four digits preceding the decimal point lie between 1000 and 10,000. Therefore, the loga- rithm of any number, excepting 1000, having four digits preceding the decimal point is 3 plus a (decimal )fraction. In a similar way it may be shown that the logarithm of any number which is not an integral power of 10, consists of an integral part plus a positive decimal fraction. For fractional numbers the integral part of the logarithm is usually written in the form of a binomial, as 9 10, 8 10, 7 10, etc. The integral part of a logarithm is called the characteristic. The positive decimal part is called the mantissa. The mantissa of a positive integral power of 10 or of the reciprocal of a posi- tive integral power of 10 is zero. If the log 373 = 2.5717: log 3.73 = log 373 - log 10 = 2.5717 - 1 = 1.5717 log 3.73 = log 37.3 - log 10 = 1.5717 - 1 = 0.5717 log 0.373 = log 3.73 - log 10 = 0.5717 - 1 = 9.5717 - 10 log 0.0373 = log 0.373 - log 10 = 9.5717 - 10 - 1 = 8.5717 - 10 log 0.00373 = log 0.0373- log 10 = 8.5717 - 10 - 1 = 7.5717 - 10 The logarithms of numbers between 1 and are negative. It will be noticed from the above illustration of the method of writing negative logarithms that the mantissa is always the same for the same sequence of digits, and is entirely independent of the position of the decimal point. The characteristic of the logarithm of a number depends only upon the position of the decimal point in the number, and is entirely independent of the sequence of the significant digits. A rule for determining the characteristic of the logarithm of a number is: the characteristic of the logarithm of a number, using 10 as a base, equals the number of places the first significant figure is removed from the units' place, and is positive if the first significant figure is to the left of the units' place, and is negative if it is to the right of units' place. Thus the characteristic of the logarithm of 37.3 is + 1; of 3.73 is 0; of 0.373 is - 1, or 9 - 10; of 0.0373 is - 2, or 8 - 10, etc. 5 66 MATHEMATICS [37 Exercises 1. Give the characteristic of the logarithm of each the following numbers : (a) 375; (e) 87.36; (ft) 0.6127; (k) 17.31; (6) 172.6; (/) 8.276; (t) 0.08217; (I) 0.00082; (c) 37.62; (g) 0.561; (j) 0.00756; (m) 0.000002. (d) 186.2; 2. Locate the decimal point in the numbers for which the follow- ing are logarithms: (a) 3.7283; (d) 1.8725; (g) 3.1786; (j) 7.6862 - 10; (6) 3.6872; (e) 0.8271; (ft) 8.1268 - 10; (/c) 9.3621 - 10. (c) 2.7826; (/) 9.1267- 10; (i) 0.3675; 3. Do you see any reason for using 10 as a base rather than any other number? 1 1 When 10 is used as base the system of logarithms is called the common system. The system of logarithms employed in theoretical investigation has for its base the sum of the infinite series 2 I 1 I 1 I 1 1 . T 2 T 2-3 ^ 2-3-4 This number is represented by the letter e, and to seven decimal places is 2.7182818. When the base e is used the logarithm is called the natural, the hyperbolic, or the Naperian logarithm. The number e enters into a great many formulas rep- resenting laws of nature. To illustrate: if the number of times faster y is changing than x is proportional to y, it can be shown that the law connecting y and x is y = ce kx where candk are constants. This law is sometimes called the compound interest law. The rate of growth of" bacteria is proportional to the number of bacteria. The formula giving the number at any time is then the compound interest formula. It is an easy matter to find the logarithm of a number to the base e, if we know its logarithm to the base 10. For, let log t N = x, where N is the number whose natural logarithm is desired. Then e x = N i_ ,,x e = N . Taking the common logarithm, we obtain: log e = ~ log N, solving for x log AT = ~ log e ' whence i w log ^ log e N = -. - . log e The log c = 0.434294. Then Iog e N = 2.302585 log N. This formula converts the common logarithm of the number N into its natural logarithm. To find the common logarithm of N if its natural logarithm is given we use the formula, log AT = 0.434294 log e N. 38] LOGARITHMS 67 4. If the log 2 = 0.3010, log 3 = 0.4771, log 7 = 0.8451, and log 11 = 1.0414, find the logarithms of the following: . 121 . 70 . 84V24 (a) -v^-- (o) T-j-p '"* ' 15 144 007 I o\ I I I f\ 21 1.25 (c) 16 / w ' 0.064 5. Fill in the blanks of the following table: 55 /500\ i I 6/ ' Number Logarithm Base Number Logarithm Base 100 2 32 V2 125 5 81 5 81 4 16 2.0 5 2 17 5.0 27 3 V3 V2 5 10 3V3 1.5 625 5 38. Logarithmic Tables. A table of common logarithms con- sists of the mantissas of the logarithms of numbers put in tabular form. Since the characteristic of the common logarithm of a numbed is dependent only upon the position of the decimal point, it is not published in the tables, but is supplied by the computer; and since the mantissa is independent of the position of the decimal point, no decimal points are placed in the numbers in the table. Thus, if one were to find the logarithm of 27.6 from a table of common logarithms, he would enter the table with 276, take out the corresponding mantissa, which is a decimal fraction, and prefix the characteristic 1. The tables commonly used are known as four-place, five- place, and six-place; meaning by this that the mantissas are given, respectively, to four, five, and six decimal places. In a four- place table the number interval is usually 1 in the third place; in a five-place table, 1 in the fourth place; and in a six-place 68 MATHEMATICS [39 table, 1 in the fifth place. If one wishes to compute accurately to four, five, or six places, he should use, respectively, four-, five-, or six-place tables. Do not use a five- or six-place table if a four-place table will suffice to give accurately the required number of places. The use of logarithms and logarithmic tables is primarily to shorten the work of numerical computation, and to relieve the computer of as much mental work as possible. In the explanations of the use of logarithmic tables, it is as- sumed that the student is using Slichter's four-place tables. 1 All numerical work, unless otherwise stated, is to be done accu- rately to four figures. 39. To Find the Logarithm of a Number from the Tables. On the first page of the tables will be found the logarithms (i.e., the mantissas) of numbers from 1 to 1000. There are nine sets of vertical columns. The first column of each set is headed No. and contains numbers whose logarithms are given directly op- posite in the second column, headed log. In the third column, headed d, are found the differences (called tabular differences) of the consecutive mantissas given in the second column. From the table one finds the mantissa for the sequence of digits 376 to be 0.5752. Hence log 3760 = 3.5752 log 376 = 2.5752 log 37.6 = 1.5752 log 3.76 = 0.5752 log 0.376 = 9.5752 - 10 log 0.0376 = 8.5752 - 10 log 0.00376 = 7.5752 - 10 Exercises Find the logarithm of each of the following numbers : 1. (a) 365. (6) 36.5. (c) 3.65. (d) 3650. 2. (a) 725. (6) 72.6. (c) 7.26. (d) 7260. *" Slichter's logarithmic and trigonometric tables" are printed both in pamphlet form and upon four leaves of cardboard bound together along the longer edge. For desk use the cardboard form should be used. If a student does any considerable amount of numerical work he cannot afford to use poorly arranged and slow tables, such as are frequently found in text-books on trigonometry and algebra. 39] LOGARITHMS 69 3. (a) 816. (6) 81.6. (c) 8.16. (d) 8160. 4. (a) 261. (b) 26.1. (c) 2:61. (d) 2610. 6. (a) 113. (b) 1.13. (c) 0.113. (d) 0.0113. 6. (a) 216. (6) 0.216. (c) 0.00216. (d) 216,000. 7. (a) 101. (6) 1010. (c) 0.0101. (d) 0.000101 8. (a) 2. (6) 0.2. (c) 0.02. (d) 0.00002. 9. (a) 9.99. (6) 99.9. (c) 0.999. (d) 9,990,000. 10. (a) 0.0136. (6) 0.000136. (c) 136,000. (d) 136,000,000. In writing a decimal fraction without an integral part before the decimal point, always place a zero before the decimal point; as, 0.1276; 0.0281; 0.1201; 0.0026. In writing and reading the mantissas of logarithms do not omit the last one or more digits, even though they be zero. Thus, write: log 399 as 2.6010 not 2.601 log 414 as 2.6170 not 2.617 log 455 as 2.6580 not 2.658 In reading the mantissas of the logarithms of numbers, read two digits at a time, not one at a time. Thus, read 2.7162, "two, (pause) seventy-one, sixty-two;" not, "two, point, seven, one, six, two." If the number consists of more than three digits the method of finding the mantissa of its logarithm from the table is illustrated by the following examples. Find the log 376.8. From the table log 376 = 2.5752 log 377 = 2.5763 The difference between these logarithms is 1 1 (disregarding the decimal point). This difference is given in the table in the column headed d. We see that a difference of 1 in the number pro- duces a difference of 11 in the logarithm. A difference of 0.8 in the number will then produce (approximately) a difference of 11 X 0.8, or 8.8 (use 9) in the logarithm. Then, since the log 376 = 2.5752, the log 376.8 = 2.5752 + 0.0009 = 2.5761. The 9 added to the logarithm of 376 is called the correction for 8. Find log 521.7. log 521 = 2.7168 70 MATHEMATICS [40 Correction: 9 X 0.7 6 log 521.7 = 2.7174 Find log 8618. log 8610 = 3.9350 Correction for 8 =4 log 8618 = 3.9354 Exercises Find the logarithms of the following numbers : 1. (a) 172.6. 2. (a) 276.5. 3. (a) 365.8. 4. (a) 812.6. 6. (a) 8.887. (6) 17.26. (6) 27.65. (6) 3.658. (6) 0.8126. (b) 8887. (c) 1.726. (c) 2.765. (c) 0.3658. (c) 0.08126 (c) 0.008887. (d) 17,260. (d) 27,650. (d) 3,658,000. (d) 812,600. (d) 0.8887. If there are five digits in the number the correction may be found by multiplying the tabular difference by the last two digits of the number. But, with the exception of some of the numbers oc- curring near the beginning of the table where the tabular differ- ences are large, no error will be introduced in the logarithm if the number is fist reduced to four significant figures. Thus, to four decimal places, the log 86326 is the same as log 86330. 40. Tables of Proportional Parts. In finding corrections the tabular difference is multiplied by the digit in the fourth place of the number, or by the last two digits of a five-place number. This multiplication is performed by means of the tables of pro- portional parts, the tables headed p. p. Thus, to find the correction for the log 37,267, turn to the p. p. table headed 16, 16 being the tabular difference. Opposite 6 we find 9.6, the correction for 6 in the fourth place; opposite 7 we find 11.2, the correction for 7 in the fourth place. But 7 is in the fifth place, therefore the correction for it is 1.12. Add mentally 9.6 and 1.12, and obtain for the complete correction 10.72 (use 11). In taking out the mantissa from the tables, the addition of corrections should be performed mentally. A good method to follow is illustrated by example. To find log 13.78. First write down the characteristic, 1. Then, with the table at your left, find 137 in the No. column and mark the corresponding mantissa by placing the thumb nail above, or the finger nail below it. 41] LOGARITHMS 71 Do not read this mantissa, but read the tabular difference, 32. From the p. p. tables find the correction, 26, for 8. Now, go back to the mantissa marked by the finger nail, and read it increased by 26, i.e., 1393. Then place 1393 after the characteristic 1 previously written down. 41. Arrangement of Work. All work should be arranged in a vertical column and done with pen and ink. Make the digits about 1/8 inch tall and space them horizontally about eight to the inch. Study the formula in which you are substituting and decide upon some arrangement of your work in the vertical column which will make the additions, subtractions, etc., of logarithms as systematic and easy as possible. Fill out the vertical column as far as possible before turning to the table of logarithms. This is called blocking out the work. The method of block- ing out the work is illustrated by the following example: Find the area, in acres, of a triangular piece of land, the sides a, b and c being, respectively, 127.6, 183.7, and 201.3 rods. The formula giving the number of acres, A, is 1 A = Vs(s a)(s b)(s c) 160 1 The proof of this formula, as generally given, presupposes a knowledge of trigonometry. A proof involving only algebra and geometry is given here. C FIG. 23. Let ABC, Fig. 23, be the triangle whose sides are a, b, and c. Let CD be drawn perpendicular to AB. Then CD* - a* - (c - AD)* (1) AD = \/b* - CD* (2) Substituting in (1), CD* = a* - c* + 2c\/b* - CD* - b* + CD*. (3) and 72 MATHEMATICS [41 where s is one-half the sum of the three sides. We may block out the work as follows: s = 256.3 a = 127.6 6 = 183.7 c = 201.3 2s = 512.6 s - a = 128.7 s - b = 72.6 s-c= 55.0 logs = 2.4087 log (s -a) = 2.1096 log (s -6) = 1.8609 log (s - c) = 1.7404 sum = 8.1196 l/2sum= 4.0598 log 160= 2.2041 log A = 1.8557 A = 71.73 In performing the work the expressions on the left of the equality signs were all written down first. Then the numerical -' - -' and - c c b a b a From a property of similar triangles it is seen that these ratios are entirely independent of the size of the triangle; i.e., independent of the position of the point P upon the terminal side, provided, how- ever, that the angle at the origin does not change. If, however, the angle is made to increase or decrease by any amount, no matter how small, each of the six ratios will change in value. 1 Since these ratios depend for their values upon the size of the angle and are entirely independent of the size of the triangle of reference, they are functions 2 of the angle, and are called the trigonometric, or circular, functions. In the triangle of reference the side opposite the given angle (the angle whose vertex is at the origin) is called the opposite (opp.) side. The side adjacent the angle is called the adjacent (adj.) side. The six trigonometric ratios are defined as follows, a is the angle in question: opp. The ratio . is called sine a; abbreviated by sin a. hyp. 1 The increase or decrease in the size of the angle should not be so large that the angle will not remain in the first quadrant. Later on it will be seen that the above statement may not be true if the terminal side of the angle were to move to another quadrant. 2 One quantity, y, is said to be a function of another quantity x, if, when x is known, y is determined. Thus, the area of a circle is a function of its radius; for if the radius is known, the area is determined. In this case, since we know the formula connecting the radius and the area of a circle, we can actually compute the area. If an angle in the first quadrant is known, the six ratios among the lengths of the three sides of the triangle of reference are determined. 45] THE CIRCULAR FUNCTIONS 85 The ratio , ' is called cosine a; abbreviated by cos a. hyp. opp. The ratio .. is called tangent a; abbreviated by tan a. adj. The ratio , or the reciprocal of tan a, is called cotangent opp. a; abbreviated by cot a. hi/D The ratio ~TT, or the reciprocal of cos a, is called secant a; duj . abbreviated by sec a. hwo . The ratio ', or the reciprocal of sin a. is called cosecant a; opp. abbreviated by esc a. It is to be remembered that in the order given above, the first and last, the second and fifth, and the third and fourth ratios are reciprocals. The side opposite and the side adjacent are, respectively, the ^-coordinate and the z-coordinate of P, the point of the terminal side from which the perpendicular was let fall upon the X-axis. Since the point P is in the first quadrant, so that its x- and ^-coordinates are both positive, the side opposite and the side adjacent of the triangle of reference are considered positive. The length of the hypotenuse is the distance of the point P from the origin, and is always considered positive. If the angle is in the second quadrant, a triangle of reference is formed as above by dropping a perpendicular from any point P of the terminal side upon the X-axis (the initial side produced through the vertex) . The six ratios of the lengths of the sides of this triangle are called the trigonometric functions of the angle and are named the same as if the angle were in the first quadrant. In this case, however, the side adjacent, the z-coordinate of the point P, is negative; hence the cosine, the tangent, the cotangent, and the secant are negative. In a similar manner, if the angle is in the third quadrant, we see that both the adjacent and opposite sides of the triangle of refer- ence are negative, and the sine, the cosine, the secant, and the 86 MATHEMATICS [45 cosecant are negative. If the angle is in the fourth quadrant, the sine, the tangent, the cotangent, and the cosecant are negative. Exercises 1. In what quadrant or quadrants is the angle if: (a) The sine negative? (6) The tangent positive? (c) The secant negative? (d) The cosine positive? (e) The cosecant positive? (/) The sine negative and the tangent positive? (g) The secant positive? (h) The sine positive? (i) The tangent positive and the cosine negative? (j) The cosine negative and the sine positive? (fc) .The cosine negative? (I) The sine negative and the cosecant negative? (m) The cosine negative and the cosecant negative? (ri) The cosine positive and the tangent negative? (0) The secant negative and the cotangent positive? . (p) The cotangent positive and the tangent positive? (q) The sine negative and the tangent positive? (r) The cotangent positive and the cosecant negative? (s) The secant positive and the cosecant negative? (t) The cosecant positive and the tangent positive. (u) The cosine negative and the tangent positive? (v) The cotangent negative and the secant positive? (w) The sine negative and the secant positive? (x) The sine positive and the cosecant negative? (y) The tangent negative and the cotangent positive? (z) The cosine negative and the secant positive? 2. Give the values of the six trigonometric functions of the following angles. Tabulate your results. Do not reduce radicals to approximate decimal fractions, (a) 30; (6) 45; (c) 60; (d) 120; (e) 135; (/) 150; (g) 210; (h) 225; (t) 240; (j) 300; (k) 315; (1) 330. 3. At 117 feet from the base of a flagstaff standing on level ground, the angle of elevation 1 of the top of the staff is 45; what is 1 If a point A is at a higher elevation than a point B, the angle of elevation of the point A at B, is the angle between the horizontal and a line drawn from B to A, If a point A is at a lower elevation than a point B, the angle of depression of the point A at B, is the angle between the horizontal and a line drawn from B to A. 46] THE CIRCULAR FUNCTIONS 87 the height of the staff? What is the height if the angle of eleva- tion is 60? If 30? 4. In exercise 15, page 11 how high above the plates is the point of meeting of the upper edges of the upper and the lower rafters? How far from the outside of the building will a plumb-line dropped from this point strike the floor? How high above the plates is the ridge of the barn? 6. The diagonal of a rectangular plot of ground is 82 rods. One side is 41 rods. What is the length of the other side? 6. The sides of a rectangular plot of ground are 100 and 173.2 rods. What is the approximate length of the diagonal? 7. The sine a = f ; the cosine is positive. Find the values of the remaining five functions. HINT : Since the sine and cosine are both positive the angle is in the first quadrant, and all functions are positive. A triangle whose hypotenuse is five units in length and whose opposite side is three units in length (hence the adjacent side four units in length) is a triangle of reference for a. Hence we have, cos a = -5-, tan a = f , cot a = %, sec a = -f , and esc a = f . 8. Find the values of the trigonometric functions, construct to scale a triangle of reference, and measure the angle with a protractor, for each of the following: (a) sin a = f , and the cosine is negative. (6) cos a = -f , and the tangent is negative. (c) tan a = %, and the sine is negative. (d) sin a = ^ and the tangent is negative. (e) sec a. = 2, and the tangent is negative. (/) sin a = |, and the cosine is negative. (00 cos =, and the sine is negative. (h) sec a = -f, and the sine is negative. 9. At two points, A and B, 100 feet apart and in line with the base of a tower standing on level ground, the angles of elevation of the top are 45 and 30, respectively. Find the height of the tower. 10. In exercise 9, if the angles of elevation are 60 and 30, what is the height of the tower, and how far are the points A and B from its base? 46. Functions of 0, 90, 180, and 270. The functions of 0, 90, 180 and 270 are not included in the above definition of trigonometric functions, since for these angles there are no tri- angles of reference. We shall define the functions of 0, 90, 180, 88 MATHEMATICS [46 and 270, as the limits toward which the functions of any angle, say a, approach as the angle a approaches 0, 90, 180, or 270. Thus, if the angle a is in the first quadrant, and if the hypotenuse of the triangle of reference is kept constant as the angle becomes smaller and smaller, the opposite side will become smaller and smaller and approach zero. The sine of this variable angle will then approach zero as the angle approaches zero, or the sin = 0. The length of the adjacent side approaches the length of the hypotenuse as the angle approaches zero, or the limit of the cosine is 1, or cos = 1. As the angle approaches zero, the cotangent and the cosecant increase without limit. 1 By considering a a small negative angle (hence in the fourth quadrant) the cotangent and cosecant are both negative and increase numerically without limit as a approaches zero. These facts together with what was said above regarding cotangent and cosecant are expressed by writing cot + and esc = + oo . Thus we have : sin = 0, tan = 0, sec = 1, cos = 1, cot = oo esc = By allowing the angle to approach 90, the student will show that: sin 90 = 1, tan 90 = oo sec 90 = 00, cos 90 = 0, cot 90 = 0, esc 90 = 1. 1 If 6 is a variable which approaches zero as its limit, and if a is a constant or a a variable whic approaches a limit c, the fraction - increases without limit. Thus, 6 if a = 1, and if b approaches zero by taking successively the values 1, 1, J, J, . . ., the fraction has successively the values 1, 2, 3, 4, . . . By giving to 6 a value sufficiently small, the value of the fraction, v t may be made larger, and for subse- quent values of 6 remains larger, than any assigned number. We then say that increases without limit as 6 approaches zero. The expression, "r increases without limit," or "increases indefinitely," as 6 approaches zero is usually written r = oo The student is not to infer that (infinity) is a number or the limit of a number- The expression, r= oo,is a symbol representing, "r increases without limit," or "r increases indefinitely." CO 46] THE CIRCULAR FUNCTIONS 89 The student will further show that: sin 180 =0, tan 180 = 0, sec 180 = -1, cos 180 = -1, cot 180 = co, esc 180 = <, and that: sin 270 = -1, tan 270 = + oo , sec 270 = cos 270 = 0, cot 270 - 0, esc 270 = - 1 Exercises 1. Tabulate the functions of 0, 90, 180 and 270. 2. Find the values of the functions of 360; of 450; of 540; of 630; of 720; of 810. 3. Is it possible for an angle to have a sine equal to 1/2? to 3/4? to - 2? 4. Is it possible for an angle to have a cosine equal to 1/2? to 3/4? to - 2? 5. Is it possible for an angle to have a tangent equal to 1/2? to 3/4? to - 2? 6. Is it possible for an angle to have a cotangent equal to 1/2? to 3/4? to - 2? 7. Is it possible for an angle to have a secant equal to 1/2? to 3/4? to - 2? 8. Is it possible for an angle to have a cosecant equal to 1/2? to 3/4? to - 2? 9. As the angle varies from to 360 in a positive sense, how do the values of the sine, the cosine, the tangent, the cotangent, the secant and the cosecant change? 10. In the equation of the straight line, y = ax + b, a is what trigonometric function of what angle? 11. Compare the sin 30 with sin ( 30) ; sin 45 with sin ( 45) ; sin 60 with sin (- 60); sin 135 with sin (- 135); sin 300 with sin ( 300); sin a with sin (a). 12. Compare cos 30 with cos (- 30) ; cos 60 with cos (-60); cos 210 with cos (- 210); cos a with cos (- a). 13. Compare tan 45 with tan ( - 45) ; tan 150 with tan ( - 150) ; tan 330 with tan (- 330); tan a with tan (-a). 14. Compare cos 30 with sin 60; sin 30 with cos 60; tan 30 with cot 60; sec 30 with esc 60; sin 45 with cos 45; tan 45 with cot 45; cos a with cos (90 a); sin a with cos (90 a); tan a with cot (90 a); sec a with esc (90 a); esc a with sec (90 a), where a is an angle in the first quadrant. 90 MATHEMATICS [47 16. Compare sin 135 with sin 45; sin 120 with sin 60; cos 120 with cos 60; tan 120 with tan 60; sin 150 with sin 30; cos 150 with cos 30; tan 150 with tan 30; sin 240 with sin 60; cos 240 with cos 60; tan 240 with tan 60; sin 300 with sin 60; cos 300 with cos 60; tan 300 with tan 60; sin (180 - a) with sin ; cos (180 - a) with cos a; tan (180 - a) with tan ; sin (180 + a) with sin a; cos (180 + a) with cos a; tan (180 + a) with tan a; sin (360 a) with sin a; cos (360 a) with cos a; tan (360 a) with tan a. 16. Express the following as functions of some positive angle less than 90: sin 112; cos 127; tan 171; sin 92; cos 170; tan 131; cot 175. HINT: Construct an angle of 112. Its terminal side falls in the second quadrant. Draw a triangle of reference. Construct an angle of 22. Draw a triangle of reference for 22. These two triangles of reference are similar, and their homologous sides are in proportion. Whence: sin 112 = + cos 22; cos 112 = - sin 22 (because cosine is positive in the first quadrant, and negative in the second), and tan 112 = - cot 22. 17. Show that sin a tan a COS a and that cos a cot a = -s 47. Even and Odd Functions. Let us consider the function x 2 (i.e., y = x 2 ). If we give to x any two values numerically equal, one positive and the other negative, the two corresponding values of x 2 (i.e., oiy) are equal. The curve for x 2 (i.e., y = x 2 ), plotted upon squared paper is symmetrical with respect to the F-axis. The same can be said of the functions x 4 , x 6 , x 6 , x 2 + 3a; 4 , and x 2 + 5x 4 6z 6 7. These functions are said to be even. An even function is one whose value remains unchanged when the sign of the argument is changed. Let us next consider the function x 3 (i.e., y = x 3 ). If we give to x two values numerically equal but with opposite signs, the two corresponding values of x 3 (i.e., of y) are numerically equal but of opposite signs. The curve for x 3 (i.e., y = x 3 ), plotted upon squared paper is symmetrical with respect to the origin. The 48] THE CIRCULAR FUNCTIONS 91 same can be said of the functions x 5 , re 7 , x, x + 3x 3 7x & , and x 3 2z 5 + 6z 7 . These functions are said to be odd. An odd function is one whose value 'remains numerically the same but changes sign when the sign of the argument is changed. In previous exercises the fact was developed that cos a and sec a are even functions, and that sin a, esc a, tan a and cot a are odd functions. These facts are shown graphically in 49. 48. Circular Measure. The circular unit of angular magnitude is defined as an angle whose sides intercept an arc equal in length to the radius of a circle when the vertex of the angle is placed at the center of the circle. This unit is called the radian. 1 This sys- tem of angular measure is fundamental in mechanics, physics, and pure mathematics. It must be thoroughly mastered by the student. Since the circumference of a circle is 2irr, where r is the radius of the circle, there are 2ir radians in one complete revolution; or 360 = 2-ir radians 180 = TT radians 90 = K radians, etc. 6 To find the number of radians in an angle expressed in degrees, divide the number of degrees by 180 (this gives the number of straight angles in the given angle) and multiply the quotient by TT, the number of radians in a straight angle. Thus, to change 176 to 1 7 ft radians, we have T^TT = 3.07: or 176 = 3.07 radians. loU To find the number of degrees in an angle expressed in radians, divide the number of radians by TT (this gives the number of straight angles in the given angle) and multiply this quotient by 180 (the number of degrees in a straight angle). Thus, to change 4.6 radians to degrees we have '- = 263. 6, or 4. 6 radians 7T equals 263.6. 1 If is the angle measured in radians at the center of a circle of radius r, the length of the arc intercepted by the angle is rO, and the area of the circular sector is *r 2 0. 92 MATHEMATICS \ \ \0 y 7 Exercises 1 . Express in radians the following: (a) 135; (6) 45; (c) 161; (d) 275. 2. Express in degrees the following angles measured in radians: (a) 7r/2; (6) ir/4; (c) Tr/6; (d) 37T/2; (e) 1; (/) 2; fo) 3.6; (ft) 11.6. [49.] Graphs of the Tri- gonometric Functions. In Fig. 26 let OPiPzPa be a circle with unit radius. Let OCPi be any angle at the center of the circle. Then PiAi represents the sine of the angle 0CP,. Take OB equal in length to the circum- ference of the circle, i.e., equal to 2-n-. Divide OB into thirty- six equal parts, and erect per- pendiculars at each point of division. Each of these equal distances of the X-axis, OB, represents 10, orT/18 radian. Upon each vertical line plot a point whose ordinate repre- sents the sine of the corre- sponding angle. To do this divide the circumference of the circle into thirty-six equal parts, and join each to C, the center of the circle. (Only a few of these points, as PI, Pz, PZ, PI, P&, are lettered in the drawing.) Project these points to the right upon the vertical line corresponding to the angle at the center of the 49] THE CIRCULAR FUNCTIONS 93 circle. Thus, the angle 50 determines the point P 2 . P 2 pro- jected upon the line passing through the point of the X-axis repre- senting 50 gives P' 2 . ThenP' 2 A' 2 = P 2 A 2 = sin 50. The curve OKMNB (called the sine curve) is the graph of the equation y = sin x, as x varies from to 2?r. It is readily seen that as x continues beyond this interval the curve is repeated for every addition of 2ir to the X-axis. The length of any portion of the X-axis represents an angle at the center of the circle expressed in circular units. The graph of y = 2 sin x may be obtained from the curve in Fig. 26 by making all ordinates twice as long; the curve y = 3 sin by making all ordinates three times as long; the curve y = f sin x by making all the ordinates one-half as long; the curve y = % sinx by making all ordinates one-third as long; etc. FIG. 27. If the sine curve in Fig. 27 were translated to the left a distance ir/2, its equation would be y = sin (x + ir/2). Since the sin (x + 7T/2) = cos x, the translated curve in Fig. 27 is the graph for y = cos x. To construct the graph y = tan x we proceed by a method similar to that used in the construction of the sine curve. Let C, Fig. 28, be the center of a circle of unit radius. Upon the X-axis lay off the distance OD equal to 2ir, the circumference of the circle. Divide OD into thirty-six equal parts, and at each point of division erect a perpendicular. Let the F-axis, Y'OY, be tangent to the circle. Beginning with the horizontal line draw thirty-six radii at intervals of 10, extending them to the tangent 94 MATHEMATICS t50 line Y'OY. Project the points of intersection with the tangent line upon the vertical line whose distance from the origin repre- sents the particular angle at the center of the circle. Thus, OCPi is an angle of 30. Project PI upon the third vertical line to the right of the F-axis, and obtain P'i, a point upon the graph of y = tan x. When the angle is in the second or third quadrants the radius must be extended back through the center in order to intersect with the tangent Y'OY. We see that as the angle FIG. 28. The tangent curve. passes through 90 and 270 the tangent curve jumps from + to . The tangent curve extends to infinity in both y direc- tions, and only a portion of it near the X-axis is represented in the figure. Curves for y = cot x, y = sec x, and y = esc x may be plotted from the tangent, cosine, and sine curves, respectively, by taking reciprocals of the lengths of the ordinates. [50.] The Line Representation of the Trigonometric Functions. Let C, Fig. 29, be the center of a circle with unit radius. Let CO be a horizontal line and CD a vertical line. Let OY and DQ be [50 THE CIRCULAR FUNCTIONS 95 tangents. Let OCP be an angle a, at the center of the circle. Then, AP represents the sine of a; CA represents the cosine of a; OP i represents the tangent of a; DP 2 represents the cotangent of a; CPi represents the secant of a; and CPz represents the cose- cant of a. AO, which is equal to 1 cos a, is sometimes called the versed sine of a (vers a). FIG. 29. Exercises 1. Draw a curve for y = sin x. Use 1.15 inches as the unit of length. This gives" 2?r = 7.2 inches, approximately, or each 0.2 inch of the X-axis represents 10. 2. Draw a curve for y = cos x. 3. Draw a curve for y = tan x. 4. Prove (sin a) 2 + (cos a) 2 = 1; or, as it is usually written, sin 2 a + cos 2 a = 1, where a is any angle. HINT: In a triangle of reference for any angle, excluding 0, 90, 180, and 270, the square of the adjacent side plus the square of the opposite side equals the square of the hypotenuse. Note that this is true even though we consider the sides negative. For the angles 96 MATHEMATICS [51 excepted above, prove the truth of the formula by direct sub- stitution. 6. Prove sec 2 a = 1 + tan 2 a. 6. Prove esc 2 a. = 1 + cot 2 . 51. Trigonometric Tables. In preceding exercises we have calculated the numerical values of the trigonometric functions of a few angles; viz., 30, 45, 60, etc. If the values of the functions of other angles were desired they could be found by drawing the angles with a protractor, constructing *a triangle of reference, and from it scaling off the lengths of the sides and computing the ratios. This method of finding the functions would not only be slow, but the degree of accuracy would usually be unsatisfactory. A method has been devised for calculating the trigonometric func- tions to any number of decimal places, but it cannot be explained in this place. A table of natural trigonometric functions, for every ten minutes from to 90, is given in Table XXII, pages 300-302. Since the cosecant and secant are, respectively, reciprocals of the sine and cosine, they need not be used in formulas, and their values are not usually published in a trigonometric table. If the angle is less than 45 the degrees and minutes of the angle are printed in the left-hand column and the names of the functions are printed at the top of the page. If the angle is greater than 45 the degrees and minutes are printed in the right-hand column and the names of the functions are printed at the bottom of the page. This arrangement of the table is possible through the fact that the cosine of an angle is the sine of its complement, and the cotan- gent is the tangent of its complement. 52. Graphic Table of Trigonometric Functions. From a sheet of polar coordinate paper, form MS, the sine and the cosine of any angle, and the tangent of an angle less than 45, may be scaled off accurately to within two or three points in the third decimal place. Upon this sheet concentric circles, 2/100 of a unit apart, are drawn. Every fifth circle is shown by a heavy line. Running from their common center radiating lines for every degree are drawn. Every tenth line is drawn heavy. Two additional circles are drawn each passing through the origin, the center of the con- 52] THE CIRCULAR FUNCTIONS 97 centric circles, and each having a diameter equal to unity. The circle with its center upon the F-axis is called the sine circle, the other with its center upon the X-axis is called the cosine circle. At the right is a vertical scale from which may be read the tan- gent of any angle up to 45. This scale is called the tangent scale, and its use is apparent at once. In explaining the method of scaling off the sines and cosines from the sine and cosine circles, FIG. 30. reference is made to Fig. 30. Let be the center of the system of concentric circles, and let CABD be the circle with radius equal to unity, and center at 0. Let OEA be the cosine circle, and OFB the sine circle. Let AOE be any angle, say a. Draw AE and FB. OAE and OFB are right triangles. Then OE = OA cos a. But OA = 1. Therefore OE = cos a. Upon form MS the length OD may be read off at once by means of the system of concentric circles. Similarly, OF = OB sin OBF. But, since OB = 1 and the angle OBF = angle a, we have OF = sin a. 7 98 MATHEMATICS [53 63. The Graphic Solution of the Right Triangle. Let A, B, and C be the vertices of any triangle; and let a, /3, and y be, re- spectively, the interior angles of the triangle at these vertices. Let the sides opposite the vertices A, B, and C be, respectively, a, b, and c. In a right triangle the right angle is marked 7, and the hypotenuse c. The three sides together with the three angles are called the elements of the triangle. If any three of the six elements, except- ing the three angles, of a possible triangle are given, the triangle may be constructed to scale, and the remaining three elements measured. The triangle is then said to be solved graphically. Exercises Solve graphically the following right triangles; record the time spent upon each problem. l.o = 173, b = 216. 4. a = 175, a = 36. 2. b = 136, c = 527. 5. c = 516, = 73. 3. a = 210, c = 728. 6. b = 172, a = 47. 54. Solution of Right Triangles Analytically. Instead of solv- ing a right triangle graphically we may solve it analytically. This method will be illustrated by solving two right triangles. ILLUSTRATION 1 : Given, a = 216 and a = 36; to find b, c, and /3. We see at once that /3 = 54, the complement of 36. To find c a, a use the relation c = , and to find b use the relation b = sin a tan a From the table of the natural trigonometric functions, sin a = 0.5878, and tan a = 0.7265. 216 216 Substituting, c = -- = 367.5; and b = ~ - 297.3. To check the result, make use of the relation c 2 - a 2 = 6 2 , or \/(c a)(c + a) = b. c - a = 151.4. c + a = 583.4. V'(c^-~aXc +"a) = 297.3 = b 54] THE CIRCULAR FUNCTIONS 99 This last value of 6 agrees with the value computed above. This not only checks the value of b, but also the value of a, which is used in computing the second value of 6. ILLUSTRATION 2: Given c = 516.2 and a = 176.5; to find b, a, andjS. a 176.5 Solution: sin a = - = r< . _ = 0.3419 c 51o.2 From the table a = 19.994 whence = 70.006 b = csin/3 = (516.2) (0.9397) = 485.1 Check: \/(c a)(c + a) = b, or \/235,310.19 = 6, or 485.1 = 6, which checks all work. Exercises Solve the following right triangles. Use natural values of trigono- metric functions taken from Table XXII. Perform the necessary multiplications and divisions by the arithmetical process, and hand in all numerical work. Be sure to check your work. Keep a record of time spent upon each problem. [1.] a = 173, b = 216. [4.] a = 175, a = 36. [2.] b = 136, c = 527. [5.] c = 516, ft = 73. [3.] a = 210, c = 728. [6.] b = 172, a = 47. Instead of performing the multiplications and divisions by the lengthy arithmetical method, logarithms may be used. In the logarithmic solutions the natural values of the trigonometric func- tions are not used, because their logarithms may be taken directly from a logarithmic table. The two problems solved below illus- trate the arrangement of the work. The student should go over the solutions and checks of these two problems very carefully, in order to become familiar with the use of a table of the logarithms of the trigonometric functions. Especial study should be given to the finding of the angles when the logarithms of the functions are given. 100 MATHEMATICS [54 ILLUSTRATION 1: Given a = 175.6 and 6 = 216.1; to find a /?, and c. a 6 Solution: tan a = ~ , /3 = 180 a, c = - 6 sm /S log a = 2.2443 log 6 = 2.3347 log tan a = 9.9096 - 10 a = 39 4' .6 = 50 55' .4 log sin ]8 = 9.8900 - 10 log c = 2 . 4447 c = 278.44 Check: c - a = 102.84 c + a = 454.04 log (c - a) = 2.0122 log(c + a) = 2.6571 logfc 2 = 4.6693 log 6 = 2.3347 The computed log 6 agrees with log b used above, which checks all of the work. ILLUSTRATION 2: Given, a = 176.3, a = 37 17'.4; to find b, c, and j9. Solution: /3 (the complement of a) = 52 42'. 6. b = a tan /3, a and c = - smp log a = 2.2463 log tan /3 = 0.1183 logfe = 2.3646 6 = 231.53 log sin |8 = 9.7824 - 10 logc = 2.4639 c = 291 . 00 Check: c- a = 114.7 c + a =467.3 log (c - a) = 2,0596 54] THE CIRCULAR FUNCTIONS 101 log (c + o) = 2.6696 log& 2 = 4.7292 log 6 = 2.3646 which agrees with log b above and thus checks the entire work. This illustrative example was chosen so that the calculated value of the logarithm of b would agree with the value found in the table. This absolute agreement will, in general, not occur. For, in finding the second logarithm of b, calculated numbers are used which are correct only to four places, and the errors in these numbers, especially in c-a will effect the calculated logarithm of 6. If the calculated logarithm differs from that taken from the table by only one or two in the fourth decimal place, the student may consider his work checked. In checking the work it will be noticed that it is immaterial whether the log a or log b is recomputed. If the hypotenuse and a leg are given, the logarithm of the unknown leg should be recom- puted. Otherwise the logarithm of the longer leg should be recomputed by the check formula. The reason for this is that it keeps the percent of error in the difference between the hypotenuse and the leg as small as possible. Exercises Solve the following right triangles using logarithms. Keep a record of the time spent upon each problem. Check the work. l.o = 173, b = 216. 4. a = 175, a = 36. 2. 6 = 136, c = 527. 5. c = 516, ft = 73. 3. a = 210, c = 728. 6. b = 172, a = 47. 7. Having measured a distance of 281.6 feet in a direct horizontal line from the bottom of a tower, the angle of elevation of the top was found to be 24 16'.6. Find the height of the tower. 8. A person on top of a tower 75 feet high observes the angle of depression of two objects on the horizontal plane, which are in line with the base of the tower, to be 37 27'.6 and 25 17'.8. Find the distance of each object from the base of the tower. 9. A tower stands by a river. A person on the opposite bank finds the angle of elevation of its top to be 50. He recedes 40 yards 102 MATHEMATICS [54 in a direct line from the tower, and finds the angle of elevation of the top to be 40. Find the breadth of the river and the height of the tower. 10. A rope 100 feet long is fastened to the top of a building 40 feet high. Find the angle the rope is inclined, if its lower end just touches the ground. 11. From a balloon, which is directly over one town, is observed the angle of depression of another town, 12 16'. The towns are 8.5 miles apart. Find the height of the balloon. 13. From a station A at the base of a mountain, its summit S is seen at an ele- vation of 43 17'. After walking 5000 feet toward the summit, up a plane making an angle of 28 13' with the horizontal, to another station, B, the angle ABS was found to be 137 17'. Find the height of the mountain. HINT: In solving any problem, first build up a formula giving the unknown number. As far as possible put this formula in a form suitable for logarithmic computation. This will be illustrated by building up a formula for x, the height of the mountain. In Fig. 31, SE = x is the required height. BA = 5000 feet; a = 28 13' and the angles /3, y, and 5 can be computed from the given data. BC is perpendicular to AS. BC = 5000 sin /3 BC 5000 sin ft BS= . sin 5 sin 8 DS = BS sin y = 5000 sin -7 sin 5 DE = BF = 5000 sin a SUIT x = DS + DE = 5000 sin /3 - sin 5 5000 sin S + 5000 sin a + sin a THE CIRCULAR FUNCTIONS 103 Compute /3, y, and 5, and substitute in the formula. Compute the value of the first term within the parenthesis by using logarithms. 13. From a point in a window in the same horizontal plane with the bottom of a steeple, the angle of elevation of the top of the steeple is 43 16'. From another window, 18 feet below the first, the angle of elevation of the top of the steeple is 52 8'. Build up a formula giving the height of the steeple. Find the height. A c B A c B FIG. 32. 55. The Law of Sines. Let ABC, Fig. 32, be any triangle. Let p be a perpendicular let fall from C upon c. Then p = a sin /3 p = b sin a. a sin j8 = b sin a, a _b_ sin a ~ sin In a similar way we may show that or a Hence sin a sin /3 sin 7 which is known as the law of sines. Exercises 1. Suppose we are required to determine the distance from a given point A to an inaccessible object B, Fig. 33. Let C be any other 104 MATHEMATICS [55 convenient accessible point. Set a transit over A and measure the angle B AC. Set the transit over C and measure the angle BCA. Measure the distance AC. Solve for AB, using the law of sines. Data: b = 500 feet a = 101 13' 6 7 = 4228'.7 FIG. 33. To find c = AB, use the formula sin 7 sin /3 or b sin 7 sin /3 The value of /3 may be found by subtracting the sum of a and 7 from 180, for, from geometry, the sum of the three interior angles of a triangle is equal to 180. 2. Same as exercise 1, with the following data: b = 375.6 feet a = 27 42'. 7 7 = 113 17'.6. HINT: Sin (113 17'.6) = sin (66 4.2' A.) = cos (23 17'.6) 56] THE CIRCULAR FUNCTIONS 105 56. The Law of Cosines. Let ABC, Fig. 34, be any triangle. Let p be a perpendicular let fall from B upon 6. C 2 = p2 + (6 - k)* if 7 is acute (Fig. 34, a). Since p z + # 2 = a 2 , we have, C 2 = 2 + 52 and since k = a cos 7, we have 6 2 2a& cos 7. FIG. 34. If 7 is obtuse (Fig. 34, 6), c 2 = p 2 + (6 + A;) 2 c 2 = p 2 + 6 2 + 26fc + A; 2 c 2 = a 2 + 6 2 + 2kb: But, since k = a cos D(7# = a cos 7, we obtain the same expres- sion for c as above, c 2 = a 2 + b 2 - 2ab cos 7, which is known as the law of cosines. Exercises 1. Suppose we wish to determine the distance between two points A and B on opposite sides of a building, Fig. 35. Choose any con- venient point, C, and measure AC, CB, and the angle ACS. Data : a = 137.6 feet b = 186.5 feet 7 = 76 54'.6. To find c use formula C 2 = a 2 + b 2 - 2abcos7. 106 MATHEMATICS [57 2. Same as exercise 1, with the following data: a = 378.6 feet. b = 276.5 feet. 7 = 11227'.8. 57. The Law of Tangents. Let ABC, Fig. 36, be any triangle. With C as center and with CA as radius describe an arc cutting BC at E and BC produced at D. Draw AE and AD. BAD is a right triangle with the angle AEC equal to The angle EAB is equal to Draw EK perpendicular to AE. K FIG. 36. The triangles BEK an4 BAD are similar. Hence AD AE a + - = tan AE EK 2 BD _ AD BE ~ EK Since BD = a + b, and BE = a b, a + b _ tan J (a + a b tan \ (a cot (D 57] THE CIRCULAR FUNCTIONS 107 which is known as the law of tangents. Since a + /3 + 7 = 180, 2 2' The law of tangents may be used instead of the law of cosines for solving triangles in which two sides and the included angle are given. ILLUSTRATION : Given a = 337. 6 feet b = 213. 5 feet 7 = 27 16' to find c, a and |8. Solution: tan $( - |8) = CI ^- tan \(a + j8) a + 6 a - 6 = 124.1 a + b = 551.1 J(a + j8) = 76 22' log (a - 6) = 2.0938 log tan |(a + 0) = 0.6152 log (o + 6) = 2.7413 log tan |(a - 0) = 9.9677 - 10 i( _ 0) = 42 52' a = 119 14' /8 = 33 30' Find the value of c by using the law of sines, b sin 7 c = sin |8 log 6 = 2.3294 log sin 7 = 9.6610 - 10 log sin /3 = 9.7419 - 10 log c = 2.2485 c = 177.2 108 MATHEMATICS [58 Exercises Solve exercises 1 and 2, 66, by using the law of tangents. 58. The Addition Formulas. Let ABC, Fig. 37, be any tri- angle. Let ACD be the exterior angle formed by producing the side BC. Draw CK perpendicular to AB. The law of sines gives, AB sin a smy = ~BC (1 > Since a + ft = 180 7, sin (a + ft) = sin 7. Equation (1) may then be written . , _. AB sin a sm (a + ft) = or sin (a + /3) = sin a cos /? + cos a sin ft. (2) This formula gives the sine of the sum of two angles, provided the sum is less than 180. We shall show that this limitation may, however, be removed. Let a + ft be greater than 180 but less than 360. We then have a + ft = 180 + + ft, where e + ft is less than 180. sin ( a + ft) = sin [180 + ( e+ ft)] = -sin (e + ft) By formula (2) sin (e + ft) = sin e cos ft -f- cos e sin /3 58] THE CIRCULAR FUNCTIONS 109 and since e = a 180, sin ( a + ft) = - [sin (a - 180) cos/3 + cos (a - 180) sin ft], or sin ( a + j8) = sin a cos /3 + cos a sin /3, which shows that formula (2) is true if a. + ft is less than 360. Similarly it may be shown that formula (2) is true for any value of the sum, a + ft- To prove a similar formula for sin (8 a), drop a perpen- dicular, BM, Fig. 37, from B upon AC: sin ft = sin (5 a) By the law of sines s * n a n/ * = BC AM + MC . _^_-Bin BMAM , MC . or sin (5 a) = sin 7 cos a + cos 7 sin a (3) Since sin 7 = sin 5 and cos 7 = cos 5, equation (3) becomes sin (d a) = sin 5 cos a cos 6 sin a (4) Formula (4) gives the sine of the difference of two angles pro- vided neither angle exceeds 180. The student will show that this limitation can be removed in a way similar to the way in which it was removed from formula (2) above. Formulas (2) and (4) are usually written together as, sin (a ft) = sin a cos ft cos as in /3. (5) Since cos (a ft) = sin [(90 - a) + ft], formula (5) gives cos (a ft) = sin (90 - a) cos ft + cos (90 - a) sin ft, or cos (a ft) = cos a cos ft + sin a sin /3. (6) Formulas (5) and (6) are called the addition formulas for the sine and cosine. If ft = a, formula (6) gives cos (+) = cos a cos a sin a sin a, 110 MATHEMATICS [59 or cos 2 a = cos 2 a sin 2 a, (7) or cos 2a = 1 2 sin 2 a, (8) since cos 2 a = 1 sin 2 a. cos 2a = 2 cos 2 a 1 (9) When /? = a, formula (2) becomes sin (a + a) = sin a cos a + cos a sin a sin 2a = 2 sin a cos a. (10) Exercises 1. Find without using the tables, the exact value of sin 75, and of cos 75. HINT: 75 = 45 + 30. 2. Find without using the tables, the exact value of sin 15; and of cos 15. HINT: 15 = 45 - 30. X I 1 COS X 3. Show that sin = -\ HINT: Let 2 a = x in formula (8). 4. Show that cos ~ = \ I Zi HINT: Let 2 a = x in formula (9). 6. Show that x 11 cos x tan - = \ 2 \ 1 + cos x 1 cos x sin x sin x 1 + cos x HINT: Use results of exercises 4 and 5. [59.] Area of Triangle by Drawing to Scale. In 41, the areas of triangles, with the three sides given, were found both graphically and analytically. 60] THE CIRCULAR FUNCTIONS 111 In the following exercises construct the triangle to scale. Drop a perpendicular from any vertex upon the opposite side. Measure the length of this perpendicular, and multiply this length by the length of the side upon which it was dropped. One-half of this prod- uct is the area of the triangle. In drawing the triangle use a convenient scale, just small enough to admit drawing upon a sheet of paper, 8J X 11 inches. Always mark in words upon the draw- ing the scale used, as "scale 1 inch = 1 foot"; "scale 1/2 inch = 10 rods"; "scale 1/4 inch = 20 rods," etc. Exercises Find the area of each of the following triangles. Record time spent upon each problem. Place but one problem upon a sheet of paper. l.o = 175.6 rods; b = 276.5 rods; /3 = 27 35'. 2. a = 267.75 feet; ft = 37 45'; 7 = 43 17'.6. 3. b = 262.5 feet; a = 178.4 feet; /3 = 34 47'.5. 4. b = 175.37 rods; c = 216.82 rods; a = 27 47'. 6. b = 73.78 feet; c = 53.62 feet; a = 136 43'.6. 6. b = 100.37 feet; a. = 27 31'.6; ft = 73 6'.7. 60. Area of Triangle by Formula. Case I. Two sides and the included angle given. Let ABC be any triangle with b, c and a given. From C draw p perpendicular to the side AB. Then area = %pc. Since p = b sin a, area = i&c sin a. ILLUSTRATIONS: Find the area of the triangle having given b = 172. 36 feet c = 103. 27 feet a = 27 17'.6. log b = 2.2364 logc = 2.0140 log sin a = 9.6614 - 10 log = 9.6990 - 10 log area = 3.6108 area = 4081 square feet 112 MATHEMATICS [60 Exercises Find the area of each of the following triangles: l.o = 173.6 rods b = 263.8 rods y = 73 47'.6. 2. a = 278.3 rods c = 172.7 rods = 21 21'.7. 3. 6 = 100.1 rods c = 216.5 rods a = 127 26'.3. Case II: One side and two angles given. If two angles are given the third can be computed by sub- tracting their sum from 180. Let ABC be any triangle with c, a and /3 given. From C drop a perpendicular, p, upon the side c. Let M be the foot of this perpendicular. Then AM = p cot a BM = p cot ft and by adding c = p(cot a + cot ft) 'cos a cos j8"j . sin a sin J = P = P = P sin ft cos a + cos ft sin sin a sin ft sin (a + 0) in al sin a sin /3 Then _ c sin a sm p P ~ sin (a + 0) ' Since area, A = %pc, _ c 2 sin a sin / = 2sin(a:+p') ILLUSTRATION: Find the area of the triangle having given c = 263.75 feet, a = 33 17'. Q, and ft = 73 21'.6. a + /3 = 106 39'. 2 logc = 2.4212 logc 2 = 4.8424 log sin a = 9.7395 - 10 log sin ft = 9.9814 - 10 log J = 9.6990 - 10 log sin (a + ft) = 9 . 9816 - 10 log area = 4 . 2807 area = 1909 square feet 60] THE CIRCULAR FUNCTIONS 113 The log sin (a + /3), or the log cos 16 39'.2, is subtracted from the- sum of the four preceding logarithms. The subtrac- tion is performed digit by digit, as the addition progresses from right to left. Exercises Find the area of each of the following triangles. Record time spent upon each exercise. 1. c = 216.3 feet; = 24 16'.7; ft = 61 15'.6. 2. a = 372.6 feet; a = 37 27'.7; /3 = 81 21'.6. 3. b = 87.75 feet; ft = 16 37'.6; 7 = 63 21'.7. 4. c = 217.75 feet; a = 93 17'.5; = 21 37'.6. 5. a = 126 17'.6; 7 = 25 57'.3; b = 200.37 feet. [Case III:] Given two sides and an angle opposite one of them. Let ABC be any triangle. Let a, c, and a be given. Drop a perpendicular, p, from B upon AC. Let M be the foot of the perpendicular. MA = c cos a, p = c sin a MC = Va? - p 2 = A/a 2 - c 2 sin 2 ~a AC = AM MC The minus sign is used if 7 is obtuse. Substituting for AM and MC, ^_____ AC = c cos a \/a 2 c 2 sin 2 a. Then, since area = %pA C, area = fc sin a [c cos a A/a 2 c 2 sin 2 a] area = |c 2 sin a [cos a \/(a/c sin a)(a/c + sin a)] If a /c < sin a, the quantity under the radical sign is negative, which means that in this problem there is no real area. From a figure it will be seen that this is the impossible case studied in geometry. If a/c = sin a we have a right triangle, and the formula reduces to: area = ^c 2 sin a cos a, which is as it should be. If 7 is acute use the + sign before the radical; if obtuse use the - sign. If a > c, and if a is the interior angle of the triangle, there is but one solution given by the formula with the + sign before the radical. 114 MATHEMATICS [61 ILLUSTKATION: Find the area of the'triangle if c = 216.7 feet, a = 187.56 feet, and a = 37 15'.6, and 7 obtuse, log a = 2 . 2730 logc = 2.3359 loga/c = 9.9371 - 10 a/c = 0.8652 sin a = 0.6054 a/c sin a = 0.2598 a /c + sin a = 1 . 4706 log(a/c - sin a) = 9.4147 - 10 log (a/c + sin a) =0. 1675 sum = 9.5822 - 10 sum = 9.7911 - 10 \/(a/c) 2 - sin 2 a = 0.6181 cos a = 0.7958 1 cos a \/(a/c) 2 sin 2 a = 0. 1777 log 0.1777 = 9.2497- 10 log sin a = 9.7821 - 10 log c 2 = 4.6718 log = 9.6990 - 10 log area = 3 . 4026 area = 2527 square feet. Exercises Find the areas in acres, of each of the following triangles. Record time spent upon each problem. 1. c = 317.62 rods, a = 217.75 rods, a = 27 41'.7, and y is obtuse. 2. c = 267.7 rods, a = 298.6 rods, a = 87 8'. 6, and y is acute. Case IV. Three sides given. This case has already been dis- cussed in 41. 61. Graphic Method of Constructing the Curve of Intersection of a Right Circular Cylinder with a Plane. In exercise 3, 42, there was described a method of constructing the curve of inter- section of a right circular cylinder and a plane making an angle of 60 with the axis of the cylinder. A graphic method will now be 1 In taking out the cosine or the logarithm of the cosine of an sngle, the correc- tion, if any, must be subtracted, because, in the first quadrant, the cosine decreases as the angle increases. 61] THE CIRCULAR FUNCTIONS 115 described for the construction of this curve, not only when the angle is 60, but for any given angle. Let BQCB', Fig. 47, be the normal cross section of the cylinder, 1 and let BPAB' be the section made by a plane making an angle a with the normal cutting plane. From the explanation given in exercise 3, 42, it is readily seen that RP = RQ cos a = RQ sec a. Draw the circle BQCB', Fig. 38, with radius OB equal to the radius of the cylinder. Extend the diameter BOB' to some con- venient point 0'. Consider O'B and O'Y' a set of rectangular R B FIG. 38. coordinate axes. Through 0' draw the 45 line O'K and the line O'M whose slope is equal to the value of sec a. Draw QN par- allel to O'B intersecting O'K at N. Draw NS parallel to O'Y' intersecting O'M at S. Draw SP parallel to O'B intersecting RQ, produced, at P. P is a point on the curve to be drawn. For, since O'E = EN = RQ, and since ES = RP = O'E sec a, RP = RQ sec a. In this manner any number of points may be located upon the curve. In the above construction the horizontal and vertical lines need not actually be drawn; it is sufficient to locate only the points N and S. If the drawing is done on squared paper the horizontal 1 A normal cross section is a section made by a cutting plane perpendicular to the axis of the cylinder. 116 MATHEMATICS [62 and vertical lines of the paper will be of assistance; if the drawing is done on plain paper the T-square and triangle should be used. Exercises Construct the curve called for in exercise 3, 42, by the method of this section. Construct the complete curve on a heavy sheet of paper, cut out the inner portion and fold to form a model of the saddle. 62. The Pattern for a Right Circular Cylindrical Surface Cut by a Plane. In the preceding section a method was given for laying out the saddle for a circular ventilator. A method will now be FIG. 39. described for laying out a pattern for cutting the sheet iron which when rolled up into a cylindrical surface will give the flue fitting upon this saddle. The length of this pattern will be equal to the circumference of the cylinder. Draw the line BQC, Fig. 39, equal in length to one-fourth the circumference of the cylinder. This line corresponds to the arc BQC, Fig. 47, i.e., if the sheet of paper upon which BC is drawn is rolled up into a cylindrical surface such that lines perpendicular to BC become the elements of the cylinder, BC becomes the arc BQC of Fig. 47. From points of BC, such as Q, perpendiculars must be laid off equal to QP, Fig. 47. From Fig. 47, it is appar- ent that the distance QP = RQ tan a where a is the angle between the normal and inclined cutting planes. Extend CB to some convenient point 0. With as center and with a radius equal to the radius of the cylinder, describe the arc 63] THE CIRCULAR FUNCTIONS 117 BQC. Through draw the 45 line OK, and the line OM making the angle MOC equal to a. By means of a protractor divide the arc BQC into a number of equal parts (nine for convenience). Divide the line BC into the same number of equal parts. From each point of division of the arc, as Q, draw a horizontal line inter- secting OK at N. Through N draw a vertical line intersecting OM at S. Through S draw a horizontaliine intersecting at P the perpendicular erected at Q of the line BC. P is a point upon the desired curve. For, since QR = NE = OE, and since PQ = SE = OE tan a, PQ = RQ tan a. Any number of points may be located in this way and a smooth curve drawn through them. The pattern thus obtained may be used four times, giving the complete pattern. Exercises 1. Construct a pattern by the method explained above using the dimensions in the exercise of the preceding section. HINT: This drawing will require a sheet of paper larger than 8 X 11 inches. 2. An elbow is formed by joining two right circular cylinders of equal radii in such a way that their axes intersect at right angles. Lay out a pattern for the surfaces. 63. Pattern for a Conical Roof. A pattern is to be drawn for a conical roof, Fig. 40. It is apparent that the pattern may be made by cutting out a cir- cular sector from a circular sheet of paper of radius r. If R is given, r = R sec a. A > The circumference of the circular sheet of paper from FIG. 40. which the circular sector is cut, Fig. 41, is then 2irR sec a. The circumference of the lower edge of the roof is 2irR. Hence the length of the arc DE = 2irR sec a - 2-rrR = 2irR(sec a - 1). 118 MATHEMATICS Then the angle BOD expressed in radians is _27rR(sec a - 1) r 27rfi! (sec a 1) R sec a = 2ir(l cos a). Exercises 1. Construct a pattern for a conical roof if a, Fig. 40, is 30, and if R is 20 percent greater than the radius used in exercise 1 of the preceding section. 2. Build up suitable formulas for a pattern for the convex surface of a tank in the form of a frustum of a right circular cone, if the radius of the larger base is R, of the smaller base is r, and if the depth is H , Construct a pattern (reduced) if R = 5, r = 4, and H = 3. Miscellaneous Exercises 1. The roof rises from the adjacent sides of a rectangular house at an angle of 30. Find the angle which the corner of the roof makes with the horizontal. Find the length of this edge measured from the plate to the ridge, if the building is 24 X 32 feet. What is the length of the ridge of the roof? 2. The height of a butte is desired. At a point B, the angle of 63] THE CIRCULAR FUNCTIONS 119 elevation of the base of a tree, A, standing on the top of the butte was measured and found to be 63 18'. 7. At a second point, C, in the same vertical plane with AB, 'at the same elevation with, and 200 feet from, the point B, the angle of elevation of A was found to be 32 7'. 8. Build up a formula giving x, the elevation of A above B; and y, the horizontal distance of A from B. 3. Same as exercise 2, excepting' that instead of ABC being in the same vertical plane the angle between the vertical planes through A and B, and B and C is 163 5'.6. 4. Same as exercise 2, excepting that the point C is 10 feet higher than the point B. 5. Same as problem 3, excepting that instead of B and C being on a level, the point C is 10 feet lower than the point B. 6. A 100-foot tape hangs down the face of a vertical cliff. The zero mark is at A, and the 100 mark at B. At a point C in the plane below the cliff, the angle of elevation of A is 43 30'. 7, and of B is 36 20'. 7. Find the elevation of the point A above C. 7. A and B are two points on the face of a vertical cliff. At a point C the vertical line AB subtends an angle of 7 10'; i.e., the angle BCA = 7 10'. At a point D it subtends an angle of 6 7'. If A, B, C and D are in the same vertical plane and if C and D are 200 feet apart at the same elevation, find the elevation of A above C, and the horizontal distance of A from C. 8. Same as exercise 7, but let D be 5 feet lower than C. 9. Same as exercise 7, but let the vertical plane through CD make an angle of 137 10'. 7 with the vertical plane through AC. 10. Same as exercise 9, but let D be 6 feet lower than C. 11. A, B, C, D, E, and F are the vertices of a six-sided field. Find the area in acres if: AB = 162.75 rods. BC = 96.63 rods. CD = 86.73 rods. DE = 100.75 rods. EF = 101.62 rods. FA = 98.76 rods. AC = 200.60 rods. AD = 156.73 rods. AE = 200.00 rods. 12. The line AB runs north and south. The line AC makes an angle of 52 8'. 6 with AB. Locate the line BC perpendicular to AB 30 that the area ABC shall be 1 acre. 120 MATHEMATICS [63 13. Same as exercise 12, excepting that the angle ABC is to be 29 17'. 6 instead of 90. 14. The area, in acres, of a four-sided plot of land with vertices M, N, 0, and P, is desired. The side MP crosses a low swamp so that it can not be chained. The sides MN, NO, and OP are measured and found to be, respectively, 798.8 feet, 7103.7 feet and 2181.1 feet. The angles PMN and NOP are measured and found to be, respectively, 76 47'.6, and 126 10'.2. HINT: Draw the diagonal NP and find the area of each triangle. 15. Same as exercise 14, but instead of measuring the angle PMN the angle MNP was measured, and found to be 63 42'.6. HINT: Compute the length of NP as a side of the triangle NOP. 16. Same as exercise 14, but angle ONM was measured instead of angle, PMN and found to be 127 0'.3. HINT: Compute the size of the angle ONP as an angle of the triangle NOP. 17. M, N, 0, and P are the vertices of a four-sided field. NO = 3172.6 feet, Z MNP = 43 21'.3, ZPNO = 51 2'.1, Z NOM = 55 6'.5, and Z MOP = 47 52'.6. Find the area of the field in acres. HINT: Calculate MN and angle MNO. Find the area of the triangles NOP and MNP. 18. Find the volume, in cubic yards, of a right circular cylindrical shell of uniform thickness. To make the problem specific, let us find the number of yards of concrete in the vertical wall of a circular con- crete silo, assuming, first, that there are no openings. Let R be the outer radius, d the thickness of the wall and H the height, all measured in feet. From the horizontal cross section of the silo, it is seen that the area of the cross section of the wall is the difference of the areas of the two circles, one with radius R and the other with ra- dius R - d. This area A = irR z - ir(R - d) 2 = TrlR 2 - (R - d) 2 ]. The quantity within the parentheses is the difference between two squares, and may be resolved into the product of two factors, the one the sum and the other the difference of the numbers. Then A = ir(2R d~)(d). This area multiplied by the height and divided by 27 gives the volume, V, of the wall in cubic yards. Then ir(2R - 27 a formula suitable for logarithmic computation. ILLUSTRATION : Suppose the outside diameter of a silo is 13 feet, 63] THE CIRCULAR FUNCTIONS 121 the thickness of the wall 7 inches (0.5833 feet, from Table XVII), and heighl; 35 feet. Then H = 35 d = 0.5833 2R - d = 12.4167 log H = 1.5441 log d = 9.7659 - 10 log (2R - d) = 1.0940 log TT = 0.4971 log 27 = 1.4314 log V = 1.4697 V = 29.49 cubic yards. If only a single computation were to be made, the formula for volume could be built up using the data directly. For the above measurements : 7r(6.5) 2 = area of larger circle 7r(6.5 0.5833) 2 = area of inner circle 7r(6.5 2 - 5.9167 2 ) = 7r(6.5 + 5.9167) (6.5 - 5.9167) = 7r(12.4167) (0.5833) = area of cross section Then V = 35rr (12.4167) (0.5833) -h 27 = volume in cubic yards. If several sets of measurements were given, and if volumes cor- responding to all were to be computed, it would be desirable to derive first the general literal formula and then substitute the given nu- merical values. The example explained above illustrates how much easier it is to effect the transformations when letters rather than numerical values are used. The student is reminded that in general it is advisable to represent the known and unknown numbers by letters and build up, if possible, a formula for each unknown in terms of known numbers before mak- ing numerical substitutions. 19. Find the volume in cubic yards of a right circular cylindrical shell, if the outer radius is 7.25 feet, the thickness 8 inches, and the length 28.5 feet. 20. In exercise 18, the volume of the opening must be deducte from the total volume of the wall. Let the sides of the opening be two vertical planes passing, if extended, through the axis of the 122 MATHEMATICS [63 cylinder. 1 Fig. 42 represents a cross section of the opening. Let S be the width of the opening on the outside, and s the width on the inside. These distances are to be measured along the arcs of circles and not along the chords. %RS = area of circular sector ABO. %(R d)s = area of circular sector CDO. %[RS - (R - d)s] = area ACDB. Before the last equation can be used we must find an expression for s, for this is, as yet, unknown. From geometry, s R - d = ? S R or _ S(R - d) R ~~' Substituting for s in the above formula for the area of ACDB we have K decreases from ? 2 to 0. When a = 90, the equation of the ellipse reduces to v^ = 1, or x 2 = b 2 , or z 2 6 2 = 0, or (x b) (x + b) = 0. The last equation is equiva- lent to the two equations, z 6 = 0, and z + 6 = 0, or x = b and z = b; for if the product of two expressions is equal to 9 130 MATHEMATICS [66 zero, the equation is satisfied by equating either factor to zero. 1 Each equation, x = b and x = b, represents a straight line parallel to the F-axis, 6 units from it, but upon opposite sides of the origin. This is exactly what sliould be expected, for when a = 90, the cutting plane becomes parallel to the elements of the cylinder and the curve of intersection reduces to two straight lines. 66. General Shape of the Ellipse. Let OA, Fig. 47, be repre- sented by a. Since OC = 6, and since / AOC = a, it follows that - = cos a or - - = a. Substituting this value for a cos a in the equation of the ellipse., it reduces 2 to cos a or, upon solving for y, y = 1 V& 2 - x 2 This equation shows that for every value assigned to x there cor- respond two values for y, numerically equal but with opposite 1 It will be recalled that the method of replacing one equation by two equations is that which may be used in solving quadratic equations. (See 15). Let x 2 x 6 = be the equation to be solved. By factoring the left-hand side, this equation may be written (x + 2)(z 3) = 0. To solve an equation for x is to find all values such that when they are substituted for x in the equation, the equation is satisfied, i.e., the left-hand side reduces identically equal to the right-hand side. To solve then the above equation, considering it written in the second form, means to find all values which when substituted for x reduce the left- hand side to zero. To make the left-hand member of this equation zero in all possible ways is to equate each factor to zero. Thus x + 2 = and x 3 = 0. But, for x + 2 to be equal to zero, x must equal 2, and for x 3 to be equal to zero, x must equal 3. Therefore, 2 and 3 are the only values of x which when substituted for x in the original equation satisfy that equation, i.e., 2 and 3 is the solution of the equation. In general, if the right-hand member of an equation is zero, and if the left-hand member breaks up into two or more factors, the equation is equivalent to the system of equations formed by equating each factor to zero. Finding all solutions of the system of equations gives all solutions of the original equation. 1 It must be remembered that Fig. 47 is only a perspective drawing of the cylinder and the two cuttng planes, and that we are at present concerned with the curve AB'A'BA as it appears upon the plane of the paper. We must imagine that we are looking down upon the inclined cutting plane in such a way that OB extends to our right and that OA is the upward vertical direction. The outline would then appear to us as in Fig. 48. 66] THE ELLIPSE 131 signs. Geometrically this means that the curve is symmetrical with respect to the axis of x. For if the curve were plotted point by point, the two numerically equal values of y, corresponding to a single value of x, show that there are two points upon the curve at the same distance from the .XT-axis; and the opposite signs of the two values of y show that one point is above the X-axis while the other is below. These two points, since they correspond to a single value of x, are located upon the same side of, and at the same distance from, the F-axis. The line joining the two points is then perpendicular to, and bisected by, the X-axis. Thus the two points are symmetrical with respect to that axis. Since all points of the curve occur in pairs symmetrically situated with respect to the X-axis, the curve is said to be symmetrical with respect to that axis. As an illustration: let the radius, 6, of the cylinder be 2 feet, let the angle between the normal and the in- clined cutting planes, a, be 30. Then __ _ " " ~ ~ ~ ' a . , = V6 2 x 2 = -- V 4 x 2 . When x = 1 the last equation gives . 2V3 , Thus, (1, 2) and (1, 2) are two points on the curve, each, one unit to the right of the 7-axis. One point is two units above the X-axis, and the other is the same distance below. From the equation y = -r V& 2 a; 2 it is seen that y increases in numerical value as the quantity under the radical increases, and decreases as the quantity under the radical decreases. The quantity under the radical, & 2 x z , increases as x decreases in numerical value, and decreases as x increases. The value of 6 2 x 2 , which is 6 2 when x is zero, decreases to zero as x increases to 6, and becomes negative when the value of x exceeds that of b. 132 MATHEMATICS [66 The numerical value of y, which has a maximum when x equals zero, decreases to zero as x increases to b. When the value of x exceeds b, y becomes imaginary, 1 since the number under the radical sign is then negative. This says, geometrically that for all values of x greater than 6, there is no real value of y and no point can s A T be found whose coordi- nates satisfy the given equation; or the curve does not extend to the right of a vertical line, b units to the right of the F-axis. The curve is symmetrical with respect to the F-axis. This fact is shown by its equation, since x occurs only as the square. If two values are assigned to x, numerically equal but with opposite signs, the corresponding pairs of values for y are the same. Which says, geometrically that for any point of the curve there is another point symmetrical to the first point with respect to the F-axis. 2 From the preceding discussion it is seen that the curve does not extend to the right of the line T'T, Fig. 48, nor to the left of the line SS'; that the curve does not extend above the line ST nor below the line S'T'. Therefore the curve is entirely within the rectangle of dimensions 2a and 26. These conclusions are geometrically apparent from Fig. 47. 1 The even root, as the square root, the fourth root, the sixth root, etc., of a nega- tive quantity is called imaginary. 2 That the curve is symmetrical with respect to the F-axis may also be shown by first solving the equation for x. For, in this form, it is readily seen that, for any value of y, numerically less than a, these are two numerically equal values of x, one positive and one negative. S' 67] THE ELLIPSE 133 OE,( = b) and OA,( = a) are called the semi-axes of the ellipse. 1 67. A Geometric Method of Locating Points upon an Ellipse. With as center, Fig. 49, and with OB, or b, and OA, or a, as radii, draw two circles. From draw any line making an angle 6 with the positive direction of the axis of x, cutting the smaller FIG. 49. circle at the point M and the larger circle at the point N. Through M draw a line parallel to the F-axis, and through N a line parallel to the .X"-axis. We shall show that P, their point of intersection, is a point upon the ellipse. The x-coordinate, OS, of the point P 1 In text-books on Analytic Geometry it is customary to take the major semi-axis of the ellipse coinciding with the axis of x. The ellipse would then appear as Fig. 48 rotated through 90. Its equation is where 6 > a. 134 MATHEMATICS [68 is OM cos 6, or x = b cos 6. The T/- coordinate, SP, or TN, of the point P is ON sin 6,ory = a sin 0. From these two equations , -T = cos 6 and V = sin a + i - cos 2 + sin 2 J or, since sin 2 + cos 2 0=1, This is the equation of the ellipse. Therefore, P is a point upon an ellipse. By the above method any number of points may be lo- cated upon the ellipse and the curve drawn through them. 68. The Foci; the Eccentricity; the Sum of the Focal Radii. At B, Fig. 49, draw a line tangent to the smaller circle cutting the larger circle at the points E and E'. From E and E' drop perpendiculars upon the F-axis meeting it at the points F and F'. F and F' are called the foci of the ellipse, and the lines FP and F'P, lines drawn from the foci to any point P of the ellipse, are called the focal radii for the point P. We shall show that FP + F'P = 2a (a constant for all points upon the ellipse). Let FP be abbreviated by n andP'Pby r t , and OF ( = OF') by c. Then or ri = x 2 + y 2 + c 2 - 2cy, or ri 2 = b 2 cos 2 + a 2 sin 2 + c 2 - 2ca sin 0. jt is easily seen from the rectangle OBEF that c 2 = a 2 b 2 , whence, ri 2 = b 2 cos 2 + a 2 sin 2 + a 2 - b 2 - 2ac sin 0, or since cos 2 0=1 sin 2 0, ri 2 = 6 2 (1 - sin 2 0) + a 2 sin 2 + a 2 - b 2 - 2ac sin 0, 68] THE ELLIPSE 135 or ri 2 = (a 2 - 6 2 ) sin 2 - 2ac sin + a 2 n 2 = a 2 - 2ac sin -f c 2 sin 2 ri = a c sin 0. Similarly it can be shown that TZ = a + c sin 0. Upon adding, ri + r 2 = 2a, or the sum of the lengths of the two focal radii is a constant. This property of the ellipse furnishes an easy method for its construction. From the relation c 2 = a 2 6 2 , locate the foci, F and F'. Fasten the ends of a string, whose length is 2a, at the points F and f". Place a pencil point in the loop of this string and draw taut. By moving the pencil, allowing the string to slip around it, an ellipse will be described. We shall prove the converse of the above theorem: The locus of a point, moving in a plane such that the sum of its distances from two fixed points remains constant, is an ellipse. Let F and F' be the fixed points, and P any point upon the curve. Let the F-axis pass through F and F'. Let 0, the mid- point of FF', be the origin. Let the distance OF ( = OF') be rep- resented by c. Let FP be represented by r t and F'P by r 2 . n 2 = x z + (c - 7/) 2 (1) r 2 2 = z 2 + (c + y) 2 . (2) By adding, n 2 + r 2 2 = 2(z 2 + ?/ 2 + c 2 ). (3) By subtracting, r 2 2 ri 2 = 4cy. (4) fi + r-i = 2a (constant sum) by hypothesis. (5) Dividing (4) by (5), By adding (5) and (6), r 2 = a+f- (7) By subtracting (6) from (5), * * (8) 136 MATHEMATICS [69 Squaring (7) and (8) and adding, ri 2 + r 2 2 = 2(a 2 + c 2 ?/ 2 /a 2 ). (9) Substituting in the first member of (3) and canceling the common factor 2, C 2 W 2 a 2 + = x 2 + y 2 + c 2 . Clearing of fractions, or a 2 z 2 + (a 2 - c% 2 = a 2 (a 2 - c 2 ). (10) Dividing (10) by the quantity a 2 (a 2 - c 2 ), -Y.2 /j;2 x i y_ _ j a 2 c 2 a 2 But, this equation is of the form fc 2 + a 2 = L which we have shown to be the equation of the ellipse. The distance OF = OF' = c is called the focal distance. The ratio, c/a, which is never greater than unity, is called the ec- centricity of the ellipse. Referring to Fig. 47 the focal distance is equal to AC, and the eccentricity is the sin a. A circle then is an ellipse with zero eccentricity. 69. The Area of the Ellipse. Let ABCD and ABC'D', Fig. 50, be two planes inclined to each other at an angle a. Let EFG'H' be a rectangle in the plane ABC'D' with its base EF in (or parallel to) the line of intersection of the two planes. Let EFGH be the orthographic projection of EFG'H' upon the plane ABCD. 1 Since EH = EH' cos a, the area of EFGH = (area of EFG'H') cos a. If, instead of a rectangle, an irregular area, as E'F'G'H', 1 If from a point a perpendicular is let fall upon a plane, the foot of the perpendicular is called the orthographic projection of the point upon the plane. If from every point of a curve a perpendicular is let fall upon a plane, the locus of the foot of this perpendicular upon the plane is called the orthographic projection of the curve upon the plane. If the bounding curve or curves of any surface be projected orthographic- ally upon a plane, the area bounded by the projection is called the orthographic projection of the area of the given surface upon the plane. 69] THE ELLIPSE 137 Fig. 51, on the plane ABC'D' be projected orthographically upon the plane ABCD, divide the projected area into strips by equidis- FIG. 50. FIG. 51. tant parallel lines drawn perpendicular to AB, and from the points of intersection of each of these lines with the boundary of 138 MATHEMATICS [70 the area draw lines to the right until they meet the next parallel. In this way, on the plane ABC'D', is formed a series of rectangles which project into another series of rectangles on the plane ABCD. From what was shown above it is seen that the sum of the areas of the projections of these rectangles upon the plane ABCD is equal to the sum of the areas of the projected rectangles multi- plied by the cos DAD'. As the distance between the parallel lines is made smaller and smaller, i.e., as the number of strips is increased indefinitely, the sum of the areas of the rectangles in the plane ABC'D' ap- proaches nearer and nearer the irregular area E'F'G'H' ; and the sum of the areas of the rectangles in the plane ABCD approaches nearer and nearer the area EFGH, the proj ection of E'F'G'H' . Therefore, area EFGH = (area E'F'G'H') cos DAD'. Then (Fig. 47) the area of the circle BCB'D = (area of the ellipse ABDB') cos a, or, since cos a = b/a, and since area of the circle = irb 2 , jrb 2 = (area of ellipse)' or area of ellipse = Trba 70. The Orthographic Projection of a Circle. Let ABCD, Fig. 52a, be a circle with radius a. Let AOC, Fig. 526, be an end elevation of the same circle. Rotate this circle about BOD as an axis through an angle, /3, to the position A"OC". Project the rotated circle upon its original plane, into the curve A'BC'D. We shall show that A'BC'D is an ellipse. Take any point P upon the original circle. It rotates into the point P", and P" projects intoP'. The equation of the circle is x* + y 2 = a 2 , where x = MP. To get the equation for the curve A'BC'D replace MP by its equal MP' /cos 0. (See Fig. 526.) Whence, MP' 2 - 4- v 2 = a 2 cos 2 ,3 +y Since MP' is the x-coordinate of P', 71] or THE ELLIPSE 139 a 2 cos 2 /3 "" a 2 Replacing a cos /3 by &( = OC"), a; 2 v 2 the equation of an ellipse. The focal distance, Va 2 ft 2 , is equal to C'C", and the eccentricity c/a = sin ft. 1 FIG. 52. [71.] Tangent to an Ellipse. Let PTQ, Fig. 47, be a tangent plane to the cylinder with PQ the line of tangency. Let QT and PT be, respectively, the lines of intersection of the tangent plane with the horizontal and the inclined cutting planes. TQ is tangent to the circle DBQCB' at the point Q, and TP is tangent to the el- lipse BPAB'A' at the point P. These tangent lines meet at the point T upon the shorter axis of the ellipse produced. This fact will be made use of in drawing the tangent to an ellipse. Let ABA'B', Fig. 53, be an ellipse to which a tangent line is to be drawn at the point P. Draw the circle with B'B as diameter. From P drop a perpendicular upon B'B cutting the circle at Q. 1 It can be shown that the section of a right circular cone made by a plane cutting all of its elements on the same side of the vertex is an ellipse. 140 MATHEMATICS [71 Through Q draw QT tangent to the circle (perpendicular to OQ). Draw TP, which will be tangent to the ellipse at the point P. FIG. 53. FIG. 54. It is left for the student to show that the above statement is correct. We shall now show how to draw a tangent to an ellipse through 71] THE ELLIPSE 141 a point S, not upon the curve. Let S, Fig. 47, be a point in the plane of the ellipse, through which the tangent line is to be drawn. Draw SS' parallel to PQ, intersecting the plane OCQ at S'. Draw SK and S'K perpendicular to BB' From the similar right triangles SS'K and AGO it follows that: S'K/SK = CO/AO Let ABA'B', Fig. 54, be the ellipse to which the tangent is to be drawn from the point S. Draw ASW. Draw CW. Draw SK perpendicular to BOW, cutting CW at the point S'. From S' draw the two tangents to the circle BCB'D. Let Qi and Q 2 be the points of tangency. Draw RiQiPi and RiQ^Pz perpendicular to B'OB cutting the ellipse in the points PI and P 2 . Draw PiS and PzS. PiS and PyS are the tangents to the ellipse drawn from the points S. PiS and QiS' intersect upon B'BW ; and P*S and Q 2 S' intersect upon B'BW. It is left for the student to show that the above statements are correct. Exercises 1. Find the value of each semi-axis, the focal distance, and the eccentricity for each of the following: (a) + = 1- (c) lOOz 2 + 4i/ 2 = 400. 4 y (b) ^ + !^ = 1. (d) 7x* + 3*/ 2 = 11. 2. Construct an ellipse by the method given in 67 if a =5 inches, and 6=4 inches. 3. Construct an ellipse by the method given in 68 if a = 4 and a = 30. 4. Find the weight of metal removed in cutting the opening in the saddle for a circular ventilator 85 feet in diameter, if the rafters make an angle of 30 with the horizontal, and if the ventilator is made of 20 gauge iron. Twenty gauge iron is 0.038 inch thick, U. S. standard. The specific gravity of steel is 7.83. A cubic foot of water weighs 62.5 pounds. [5.] Show that a right elliptical cylinder, has a circular cross section. If, x - + y- = i 16 ^ 9 142 MATHEMATICS [71 M is the equation of its normal cross section, find the angle between the axis of the cylinder and the plane giving the circular section. [6.] Show that every section of a right elliptical cylinder is an ellipse. [7.] A circular window in the east wall of a building is 4.6 feet in diameter. The wall runs true north and south. The rays of light from the sun pass through the window and fall upon a horizontal A floor. Find the area of the floor in the sunlight if the angle of elevation of the sun is 57 42'. 6, and if it is 37 27'.7 east of the meridian. [8.] Construct an ellipse with semi-axes equal to 1^ and 2^ inches. Draw a line bisecting the angle between the axes. At the point of intersection of this line with the ellipse, draw a tangent line. Measure the distances from the center of the ellipse to the points of intersection of the tangent line with the axes of the ellipse produced. [9.] Construct an ellipse with semi-axes equal to 1\ and 3| inches. Through its center draw a line making an angle of 30 with the minor axis. From a point of this line 2 inches from the center draw two tangents to the ellipse. [10.] Through the point of tangency of the tangent line drawn in exercise 8 draw a line perpendicular to the tangent line. With the protractor measure the angles between this perpendicular and the focal radii. [11.] Let OA, Fig. 55, be the vertical wall of a building; OB a hori- zontal pavement; MN a ladder of length a + b. P is a point of the ladder such that MP = b and NP = a. Show that as the foot of the ladder is drawn out along the pavement, P moves along the arc of an ellipse. HINT: Consider OB as the X-axis and OA the F-axis of a rec- tangular coordinate system. OT ( = x) and TP ( = y) are the co- ordinates of P. - = sin a, and r = cos a. Square these equations d and add. N FIG. 55. CHAPTER V [ THE SLIDE RULE ] 72. Accuracy of Logarithmic Computation. The logarithms given in tables are only approximations. In a four-place table, they are correct to the nearest fourth decimal place; in a five- place table to the nearest fifth decimal place; in a six-place table to the nearest sixth decimal place; etc. The log ir = 0.497 to three places, 0.4971 to four places, 0.49715 to five places, 0.497150 to six places, 0.4971499 to seven places, and 0.49714987 to eight places. The log TT is given correct to three, four, five, six, seven and eight places of decimals; but none of these values is absolutely correct. Since the logarithms of numbers taken from tables and used in logarithmic calculation are not absolutely correct, the results of the calculations are not absolutely correct. But in nearly all numerical calculation, results are required accurate only to a certain number of places; and the number of accurate places re- quired is the fact that determines the table to be used. If a set of logarithms are added (corresponding to multiplication) or added algebraically (corresponding to multiplication and division) the errors in the logarithms will be, in most cases, compensating in nature. That is to say, the positive errors and negative errors tend to balance, producing only a small error in the sum. Below are written ten numbers selected at random. In the sec- ond column are given their logarithms to six places; in the third column their logarithms to four places; and in the fourth column the errors in the four-place logarithms as compared with the six-place logarithms. 143 144 MATHEMATICS [72 Number Logarithm Logarithm Error 3.72 0.570543 0.5705 -0.000043 9.81 0.991669 0.9917 +0.000031 7.15 0.854306 0.8543 -0.000006 1.23 0.089905 . 0899 -0.000005 5.06 0.704151 0.7042 +0.000049 7.17 0.855519 0.8555 -0.000019 8.23 0.915400 0.9154 0.000000 3.56 0.551450 0.5514 -0.000050 1.17 0.068186 . 0682 +0.000014 8.75 0.942008 0.9420 -0.000008 Sum 6 4431 -0.000037 The sum of the errors is 0.00004, which represents approximately the error in the sum of the ten logarithms. The mantissa 4431 is found to correspond to the number 2774. Since the tabular dif- ference at this point of the table is 15, the above error, 0.00004 makes only a difference of between 2 and 3 in the next (fifth) place in the number. Thus by using a four-place table in multi- plying these ten numbers together, the product is found correct to the nearest fourth place. If the sum of the ten logarithms had been 6.9872, the tabular difference at this point of the table would have been 5, the error of 0.00004 would have produced an error in the corresponding number of about 7 in the fifth place. Thus, the result of the prod- uct of the ten numbers would have been one unit too small in the fourth place. We see from this illustration that an error in the logarithm, the sum, may effect the fourth digit one, or possibly two points, if the logarithm stands near the end of the table; while the fourth digit is given correctly if the logarithm stands nearer the begin- ning of the table. Multiplying and dividing with a four-place table gives re- sults correct to four places, or correct to three places while the fourth place is in error one or two points only. The error in the product or quotient will usually be less than 2/100 of 1 percent, and in the majority of cases it will be very much less than this. 72] THE SLIDE RULE 145 This is what is meant when we say a four-place table computes ac- curately to four places. In taking out the number corresponding to a logarithm, do not, in general, write more than four significant digits if a four-place table is used. If there are more than four digits preceding the decimal point, determine the first four, and suffix ciphers between the fourth digit and the decimal point. Thus, if log x = 6.1738, x = 1,493,000. The fourth digit is determined by interpolation, and three ciphers are suffixed. This number, however, occurring near the beginning of the table, could have been corrected to another place, in which case the fifth digit would have been correct to within three or four points. In numerical work, use the smallest table which will insure the required degree of accuracy in the result. If a five-place table will suffice, do not use a six-place table. If a four-place table will suffice, do not use a five-place table. If the numerical work of a problem is to be done with four-place logarithms, the student who has had but little practice in logarithmic computation is very apt to believe that by using a five-place table and discarding the last digit he will save the time and trouble required in apply- ing corrections. This method, however, does not gain the end sought; for, by using a well-arranged four-place table, corrections can be applied, after a very little practice, more expeditiously and easily than the logarithms can be found from the larger ta- ble containing ten times as many numbers. Another oversight on the part of a large number of students, is the attempt to compute more accurately than the observed data warrant. Thus, to illustrate, suppose a and b are two readings. They are not absolutely correct. The first has a possible error ei and the second a possible error e 2 . The two readings are to be multiplied together. What is the possible error of the product? How many places in the product are reliable? (a ci) (b e 2 ) = ab ae 2 + 6ei + ci2. The sum of the last three terms in this product, taken with the upper signs, is the largest possible error in the product due to the errors ei and e 2 in the factors a and b, respectively. eie 2 is 10 146 MATHEMATICS [73 small compared with ae 2 + &ei, and may be neglected. The pos- sible error in the product is then + (ac 2 + &ei). Suppose 325 and 52 1 are the readings. Suppose that in each case 1 in the third place is the least count of the instrument, i.e., there is a possible error of 0.5 in each reading. Then, a = 325 b = 521 i = 6z = 0.5 ab = 169,325 aei + 6e 2 = 423 Thus we see that the product of the two numbers whose measure- ments were attempted lies between 169,325 + 423 and 169,325 423. This shows that the product may be correct to three places only, and should be written 169,000. The digits following the 9 in the product ab have no significance since all may be incorrect; hence the digits should not be written, but ciphers substituted in their places. In the above illustration, even a four-place table carries the accuracy of the work beyond what is required. In this prob- lem the calculation could have been done with a three-place logarithmic table, or with a slide rule. 73. The Principle of the Slide Ride. Let C and D, Fig. 56, represent two uniform scales. The two scales are so placed that the zero mark of C stands opposite the 3 mark of D. Any number on D to the right of 3 is the sum of the number directly above it on C, and 3. Thus, 5(read on scale D) is the sum of 2 and 3; 6 is the sum of 3 and 3; 4| is the sum of 1J and 3; etc. In this way any number may be added mechanically to the number 3. By placing the of C above any point of D, any other number may be added mechanically to the number corresponding to this point of the D scale. It is immaterial whether the of C or the of D is set to cor- respond to one of the numbers to be added. If the of D is set, the sum is read from the C scale. By a similar process, subtraction may be performed mechanic- ally. Thus, if 3 is to be subtracted from 6, move scale C so that 73] THE SLIDE RULE 147 3 upon it stands above 6 of D, the number of D below the of C, or 3, is the difference. In exercise 10, 41, there were constructed two logarithmic scales, each in juxtaposition with a uniform scale. 1 These two pairs of scales are represented (reduced) in Fig. 57. Bend scales C and D along the horizontal lines; turn the lower part, the uniform scale, of C back behind the non-uniform scale; and turn the upper part of D back behind the lower part. Now place these two scales together as shown in Fig. 58. The lines marked 1 and 10 of these scales are called indices. If we bring the left-hand index of C opposite 2 of D, any number to the right of 2 on the D scale will be the product of the corre- sponding number of C, and 2. Thus, in the figure, 6 of D stands beneath 3 of C; 6 is the product of 3 and 2. To multiply 2 by 3 mechanically, bring the index of the upper scale over 2 of the lower scale, and read the number on the lower scale which stands directly under 3 of the upper scale. This number is the product. For, upon the back of each scale, C and D is a uniform scale, and by sliding C and D upon each other we are adding me- chanically, as was shown above, the numbers of these uniform scales. Thus, on the back of D is a number, which is the sum of the number back of 2 on D and the number back of 3 on C. But, the number back of 2 on D is the logarithm of 2, and the number back of 3 on C is the logarithm of 3; then on the back of D opposite 3 on C is the logarithm of 2 plus the logarithm of 3, or the logarithm of 6, since the sum of the logarithms of two numbers is the logarithm of their product. On the face of D are numbers whose logarithms are on the back. Therefore, opposite the logarithm of 6 upon the back of D, i.e., opposite 3 of C, we have 6, or the product of 2 and 3. In constructing scales C and D, one unit of length only, of the uniform scale was used. By extending these double scales indefi- nitely in both directions, the uniform parts will include all al- gebraic numbers, i.e., all numbers from to + < ; the non-uni- form parts will include all positive arithmetical numbers, i.e., all numbers from to + . A port : on of one of these scales (the C i If this exercise was not performed when the work on logarithms was taken up, it should be performed now, or at least the description read carefully. 148 MATHEMATICS [73 O - : : - 73] THE SLIDE RULE 149 scale) is represented in Fig. 59. The extended D scale is the same, but inverted. For convenience of description, different segments of these scales will be referred to as "block I," "block II," etc. By block II is meant that portion of the double scale between 1 and 2 of the uniform part, or between 10 and 100 of the non-uniform part; by block I is meant that portion of the uniform part between 1 and 0, or between 0.1 and 1 of the non-uniform part, etc. Following directly from the facts that the mantissas of the com- mon logarithms are independent of the position of the decimal point in the number, and that by moving the decimal point one place to the right or left increases or decreases, respectively, the characteristic of the logarithm by 1, it is seen that all blocks of the double scale, Fig. 59, are identical. To illustrate: the line on the non-uniform part representing 20 is in the same position relative to 10 as the line representing 2 is to 1. This must be so, for the logarithm of 20 equals the logarithm of 10 plus the loga- rithm of 2; and, since the logarithms are represented on the uniform part, the logarithm of 20 is as far to the right of 1 on the uniform part as the logarithm of 2 is to the right of 0. Since all blocks are the same, one block will suffice for the entire indefinite scale. In using scales C and D, Fig. 58, for mechanical multiplication the operator may disregard the positions of the decimal points in the factors, and point off the proper number, of places in the product after the sequence of digits has been found. Since the index at the right-hand end of C may be looked upon as 1 as well as 10, the right-hand index as well as the left-hand index may be set above one of the factors of the product. It will be seen, however, that in multiplying two numbers together, using only one block each of the C and D scales, only one of the two indices of C can be set over the.mutiplicand, for if the other index were set it would throw the product off the D scale into the next block. By interchanging the multiplicand and multiplier the other index of C may be set. To illustrate: let 3 be multi- plied by 6. Set the left-hand index of C over 3, then 6 of C falls beyond the D scale. In this case the product may be found either by setting the right-hand index of C over 3 and reading 150 MATHEMATICS [74 cq O C) under 6, or by setting the right-hand index of C over 6 and reading under 3. By means of scales C and D any two num- bers may be multiplied together mechan- ically. The principle involved, i.e., of adding logarithms mechanically, is the un- derlying principle of the construction of the (Mannheim) slide rule. 1 The opera- tions performed by the mechanism may be compared with logarithmic work as follows. Locating the numbers on the scales may be compared with taking out the logarithms of the numbers and the anti-logarithm of their product, while sliding the C scale above the D scale may be compared to the addition ^ of the logarithms. 2 Since the logarithms of the factors and 3 the logarithm of their product are not read "" they need not be printed upon the back of ^ the scales C and D. | 74. The Slide Rule. The slide rule con- sists of three parts: the body, the slide, and e the indicator (Fig. 60). The body of the rule consists of a wooden base about 10 inches long, 2 about 1 inch wide, and about 1/8 inch thick, upon one side of which are fastened two parallel strips, each carrying 1 Many mechanical devices have been constructed for performing special numerical operations. Often they are in the form of one scale sliding over another, and for this reason are also termed, with a proper qualifying adjective, -slide rules; as, stadia slide rule, sewer slide rule, horse-power slide rule, etc. The slide rule is also constructed by placing the logarithmic scales upon two concentric circular arcs, one rotating within the other. ' Slide rules are made in three sizes: viz., the 5-inch, about 5 inches long; the 10-inch, about 10 inches long; and the 20-inch, about 20 inches long. The 10-inch is the size commonly used. The 5-inch is less accurate, but is very convenient for carrying in the pocket and for rough estimates. 75] THE SLIDE RULE 151 a logarithmic scale engraved upon celluloid. These two scales are, in Fig. 60a, marked A and D. A transverse section of the rule is shown in Fig. 606. The two upper parallel strips of the body are separated about 1/2 inch, and between them there is placed the slide, a thin strip of wood carrying two logarithmic scales, B and C. The scales C and D are the same as those explained in the preceding section, and by means of them, multiplications may be performed. The indicator /, Fig. 60a, consists of a piece of glass, about an inch square, mounted in an aluminum frame, which slides in grooves lengthwise the rule. On the under side of the glass, very nearly in contact with the surfaces of the four scales, is etched a fine vertical line, or, "cross line" as it will be called. Scales A and B are logarithmic scales exactly one-half as long as scales C and D, i.e., upon A and B are two blocks of the in- definite logarithmic scale. Scales A and B, then, have each three indices, one at each end and one at the mid-point. When we come to the explanation of the use of the A and B scales we shall see that the end indices are considered to correspond to the num- bers . . . 0.0001,0.01,1,100,10,000,. . . , while the central in- dex corresponds to the numbers . . . 0.001, 0.1, 10, 1000, 100,000. . . The scales A and B are placed upon the body of the rule so that the left-hand index of A is directly over the left-hand index of D, and so that the right-hand index of A is directly over the right-hand index of D. When the cross line of the indi- cator is set over the left-hand index of A it will also be over the left-hand index of D; when set over the right-hand index of A it will also be over the right-hand index of D. 75. The multiplication of two or more numbers will be il- lustrated by an example. To multiply (37.5) (212) (11.2) (0.315) proceed as follows: Set left-hand index of C over 375 of D. Set cross line of indicator over 212 of C. (Directly under the cross line on the D scale is the product of the first two factors, but do not read it.) Set the left-hand index of C under cross line (i.e., over the product of the first two factors on Z>). Set cross line over 112 on C. Bring the right-hand index of C under the cross line. Move cross line over 315 of C, and under it on the D scale read 152 MATHEMATICS [76 the product of the four factors. The digits read are 2805. To locate the decimal point, find mentally a very rough estimate of the product; thus, the product of the first two factors is about i of 21,000, or about 7000; the next multiplier (about 10) makes the product of the first three factors about 70,000; the next multiplier is about ^, which makes the product of four factors about 23,000. This shows that there are five digits be- fore the decimal point. The product is then 28,050. It is to be noticed that in the above operations no readings are taken from the rule excepting the final product. In a similar way any number of factors may be multiplied together. If several numbers are to be multiplied by the same number, set the index of C over the common factor on D. Slide in turn the cross line over the given numbers on C, and read off the various products from D. In this way all the products can be read with one, or at most two, settings of the slide. 76. Division. The division of one number by another may be performed upon a slide rule in two ways. The principle of one method is that subtracting the logarithm of the divisor from the logarithm of the dividend gives the logarithm of the quotient. The principle of the other method is that multiplying the quotient by the divisor gives the dividend. Both methods will be illus- trated by example. To divide 372 by 176, set the slide so that 176 of C is directly above 372 of D. (To do this easily, first set the cross line over 372 of D and then bring 176 of C under it.) Read the quotient from D, the number under the index of C. The quotient is 2.113. By inspection it is seen that the decimal point follows the 2. This method of division corresponds to the subtraction of logarithms. For, the distance from the index to 372 (measured upon the uniform scale on the back of D) is the logarithm of 372. The distance from the index of C to the number 176 is the logarithm of 176. Then the distance from the index of D to the index of C is the difference of these two logarithms, or the logarithm of the quotient; and the number on the face of D is then the quotient. To divide by the second method, set the index of C over the divisor, 176, on D. Set the cross line over the dividend, 372, on 77] THE SLIDE RULE 153 D. Under the cross line on the C scale read the quotient, 2. 1 13 It is easily seen that this method gives the quotient, for it is the reverse of that used in multiplication. 1 The second method of division is useful if several numbers are to be divided by a common number; for example, to find the per- centages which a set of numbers are of a given number, place the index of C over the common divisor on D, bring the cross line over the dividend on D and read the quotient from C under the cross line. In this way a series of quotients may be found with one, or at most two, settings of the slide. 77. The A and B Scales. Since the "block" on A is just one- half as long as the "block" on D, the numbers (in imagination) on the back of A are just double those directly opposite on the back of D. The numbers on the face of A are then the squares of the numbers directly opposite on the face of D. This follows from the fact that the logarithm of the square of a number is two times the logarithm of the number. ILUSTRATIONS : To find the square of 3.77, either multiply 3.77 by 3.77 or set the cross line over 3.77 on the D scale and read its square, 14.2, from the A scale. To extract the square root of 7.13, set the cross line over 713 of the left-hand block of the A scale, read off its square root, 2.67, from the D scale. To extract the square root of 71.3, set the cross line over 713 of the right-hand block of A, read off its square root, 8.44, from the D scale. In extracting square roots care must be exercised to enter the number in the proper block of the A scale. Scales A and B may be used for multiplication and division in the same way that scales C and D are used. But on account of the A and B scales being only half as long, the results are not as accurate as when C and D are used. 2 78. The Cube Root. 3 The method of finding cube roots of numbers will be illustrated by examples. 1 The first method may also be looked upon as the reverse of multiplication. 1 The Polyphase Slide Rule is one having two additional scales. One of these scales gives cubes (and cube roots) directly; the other gives reciprocals of numbers directly. * Some rules are provided with a cube scale by means of which cubes and cube roots may be obtained directly. 154 MATHEMATICS [79 Example 1 : Find the cube root of 8. Set the cross line over 8 on the A scale (block I). Move the slide so that the number on the B scale under the cross line is the same as the number on the D scale under the left-hand index of the C scale. This is found to be 2, the cube root of 8. Example 2: Find the cube root of 80. Set the cross line over 8 on the A scale (block I). Move the slide so that the number on the B scale under the cross line is the same as the number on the D scale under the right-hand index of the C scale. This is found to be 4.31, the cube root of 80. Example 3: Find the cube root of 800. Set the cross line over 8 on the A scale (block II). Move the slide so that the number on the B scale under the cross line is the same as the number on the D scale under the right-hand index of the C scale. This is found to be 9.28, the cube root of 800. The student will explain why these operations give the cube roots. 79. Other Operations with the Slide Rule. It was the inten- tion to give above only the underlying principle of the slide rule; and to explain its use in connection with the simple opera- tions of multipli cation, division, extracting square root, and squaring. It is believed that the student will, after some prac- tice with the slide rule, devise methods of procedure which will involve the fewest number of settings of the slide, and the fewest number of intermediate readings. Pamphlets and books may be procured giving in considerable detail the methods of perform- ing numerical calculation with the slide rule. In performing numerical calculations with the slide rule, the work should be blocked out as when calculating with logarithms. 1 Exercises The student will practice on simple multiplications, divisions, squares, square roots, and cube roots. 1 Paper, form M 8, is ruled especially for slide rule calculations. CHAPTER VI STATICS 80. Vector and Scalar Quantities. In Physics and in Me- chanics we have to deal with two kinds of quantities: vector quantities and scalar quantities. Quantities, such as velocity, force, acceleration, etc., which require for their complete definition direction as well as magnitude, are called vector quantities. When we say a body has a uniform velocity of 10 feet per second, we mean that in every second of time the body moves a distance of 10 feet and in some given direction. Velocity, then, is a vector quantity. Quantities, such as mass, time, area, volume, etc., having only magnitude, are called scalar quantities. In order to convert a gram of ice at zero degrees centigrade into water at the same temperature, 79 units of heat must be added to the mass. This quantity, 79 calories, is a scalar quantity. 81. Graphic Representation of Vector Quantities. In order to represent a vector quantity graphically, choose some unit of length, as the inch, the half-inch, the centimeter, to represent the unit of magnitude; with this unit length, draw a line whose length measures the magnitude of the quantity, and whose direc- tion is that of the vector quantity. To illustrate: a body moves FIG. 61. from west to east with a speed of 3 miles per hour. Draw the line AB, Fig. 61,3 cm. long and from left to right. The line drawn 3 cm. long represents the magnitude, the speed, 3 miles per hour, and drawn from A to B represents the motion from left to right, or, from west to east. An arrow is placed upon the line indicat- ing its sense, i.e., from left to right. A directed line, as AB, Fig. 61, is called a vector. 155 156 MATHEMATICS [82 A vector quantity is also termed, for brevity, simply a vector. In what follows it is assumed that all problems are of such a nature that all vector quantities act in the same plane. The vec- tors representing these vector quantities may then be drawn upon a plane, the plane of the paper. 82. Vector Addition. In elementary algebra real numbers are usually represented graphically by points along a straight line. Thus, if 0, Fig. 62, be any point upon the straight line X'X, FIG. 62. points to the right of represent the positive numbers, while points to the left represent the negative numbers. To every real number there corresponds some point upon the line, and for every point of the line there exists a real number. If the points upon the line are chosen such that distances between them are propor- tional to the steps between the numbers which they represent, we have what was previously called a uniform scale. In what follows this uniform scale is assumed as the graphic representation of the real number system. The points A,B,C, D, etc., or the vectors OA, OB, OC, OD, etc., represent, respectively, the numbers 2, 5, -3, -4, etc. o A B c X - 1 - 1 ____ i ___ >! ___ >! ____ i ___ J - X FIG. 63. The addition of two real numbers, say 2 and 3, may be looked upon as consisting of the following operations: Move the vector OA, Fig. 63, parallel to itself; place it so that its initial point, which was originally at 0, coincides with the terminal point, B, of the vector OB. Its new position is indicated by the dotted line BC. Draw the vector OC. It is easily seen, since the scale X'X is uniform, that the vector OC represents the number 5, the sum of 3 and 2. It is readily seen that this process of addition holds for any set of real numbers (negative as well as positive), and that, in the process, either of the two vectors may be moved. 82] STATICS 157 In general, let OB and OA, Fig. 64, be any two vectors. Move OA parallel to itself so that its initial point coincides with the terminal point, B, of the vector OB. It then takes up the po- sition BC. Draw the vector OC. The vector OC is defined as the vector sum of the vectors OB and OA. The negative of a vector is a vector of the same length, but extending in the opposite direction (having opposite sense). FIG. 64. To subtract the vector OA from the vector OB means to find a third vector OC, Fig. 65, such that the sum of the vectors OC and OA is the vector OB. If OA is moved parallel to itself, so that its terminal point, A, falls upon the point B, and its initial point at C, the vector OC is such that the vector OC plus the vector OA equals the vector OB. Therefore, vector OC = vector OB vector OA FIG. 65. From the figure it is readily seen that, to subtract the vector OA from the vector OB is to add the negative of the vector OA , that is, AO, to the vector OB. To find graphically the vector sum of any three vectors, find the sum of any two and to this sum add the third. 158 MATHEMATICS [83 A vector is designated by giving its length, and the angle which it makes with the positive direction of the axis of x. Thus (10, 30) when referring to a vector means that the vector is ten units long and its direction makes an angle of 30 with the positive direction of the axis of x. Exercises Find graphically the sum of the following vectors: 1. (2, 10) and (3, 20). 3. (6, 70) and (5, 217). 2. (5, 30) and (10, 135). 4. (5, 110) and (3, - 71). 6. (3, 30), (7, 121) and (11, - 75). 83. Resultant Velocity. Let a raft move along a straight path with uniform speed of 2 feet per second; suppose a man Direction of Raft FlG. 66. upon the raft walks at the rate of 3 feet per second in the direc- tion indicated in Fig. 66. Let A be the point of the raft occupied by the man at some particular instant. In one second of time the point A "will be translated by the movement of the raft to the point B. The length of the line AB represents 2 feet, and the direction of AB is the direction of motion of the raft. If 84] STATICS 159 the raft were not moving, the man would walk along the line A C, and at the end of one second would be at the point C. The motion of the raft, however, has carried the point C to the point D ; at the end of one second of time the man is at the point D instead of at the point C. ABDC is a parallelogram. The sides AB and AC are vectors representing, respectively, the velocity of the raft relative to the earth and the velocity of the man relative to the raft. By the same reasoning it is seen that at the end of the second second the man will be at the point G. The points A, D, and G are upon a straight line. At the end of the first half second the man was at the point K, the center of the parallelogram ABDC, and at the end of any other fractional part of one second he was at some point X on the line AD. From similar triangles, it is easily shown that X is not only on the line AD, but at a position such that AX is to A D as the time interval is to one secohd. Thus, the man moves, relative to the earth, in the direction AD, with a uniform speed, and over a distance AD in one second of time. This is only another way of saying that the velocity of the man may be represented by the vector sum of the vector representing the velocity of the raft relative to the earth, and the vector representing the velocity of the man relative to the raft. In general, then, if a body is subjected to several velocities, its resultant velocity may be found by taking the vector sum of the vectors representing the component velocities. 84. Resultant of Concurrent Coplanar Forces Found Graphic- ally. Experimentally, it is shown that the resultant of two concurrent 1 forces is such that it may be found in magnitude and direction by considering it the vector sum of the two vectors representing the two given forces. If a force of 10 pounds acts horizontally to the right, and if a second force of 6 pounds acts in a direction making an angle of 60 with the horizontal, the result- ant of the two may be found graphically in the following manner. Lay off a line AB 5 inches long, Fig. 67. The vector AB repre- 1 Forces acting upon a body are concurrent if they have the same point of applica- tion; otherwise they are non-concurrent. Forces acting upon a body are coplanar if their lines of action lie in a plane; other- wise they are non-coplanar. A single force, if it exists, which is equivalent to a system of forces is called the resultant of the system. 160 MATHEMATICS sents the 10-pound force. One inch in length represents a magni- tude of two pounds. Draw the line AC 3 inches long making the angle BAG 60. The vector AC represents the 6-pound force. Complete the parallelogram ABDC, or the triangle ABD, or the Reduced B FIG. 67. triangle ACD. AD is a vector representing the resultant of the 10- and the 6-pound forces. Measuring the length of AD it is found to be 7 inches, which shows that the magnitude of the resultant force is 14 pounds. Measuring the angle BAD it is found to be approximately 21 45', the direction angle of the resultant. ABDC is called the parallelogram of forces. Reduced FlG. 68. The resultant of three or more concurrent coplanar forces may be found by repeated application of the parallelogram of forces. Thus, to find the resultant of the five forces, PI, P%, Pz, Pt, PS, whose magnitudes are, respectively, 3, 6, 2, 2, and 4 pounds, and 85] STATICS 161 whose directions make, respectively, the angles 0, 30, 70, 210 and 170 with the positive direction of the axis of x, proceed as follows: Draw the vector AB, Fig. 68, 3 inches long in a hori- zontal direction. Draw BC 6 inches long making an angle of 30 with the horizontal. AC is a vector representing the result- ant of the forces PI and P 2 . Draw a vector CD representing the force P 3 . AD is a vector representing the resultant of PI, P 2 and P 3 . Draw the vector DE representing the force P 4 . AE is the vector representing the resultant of PI, P 2 , PS, and P 4 . Draw EF, a vector representing the force P 6 . AF is the vector repre- senting the resultant of PI, P 2 , PS, P 4 , and P 5 . AF is 5.59 inches in length, which shows that the magnitude of the resultant force is 5.59 pounds. The angle BAF measured with a protractor is found to be approximately 55. It is seen that in the construction, the lines AC, AD, and AE are entirely unnecessary and should be omitted. The figure ABCDEFA is called the polygon of forces. AF, the resultant, is called the closing side. It will be noticed that the arrows on the vectors representing the given forces all run in the same sense around the polygon, while the arrow of the closing vector runs in the opposite sense. If the vectors representing the given concurrent forces of them- selves form a closed polygon, the resultant is zero, or the system of given forces is in equilibrium. Exercises Find graphically the resultant of the following concurrent, coplanar forces : 1. (3, 30), (5, 45), and (7, 190). 2. (2, 0), (3, 40), (5, 170), and (7, 135). 3. (2, - 10), (6, 90), (5, 0), and (7, 135). 4. (3, 180), (2, 270), (5, 45), and (1, 150). 5. (1, 270), (5, - 30), (7, 90), and (2, 30). 85. Resultant of Concurrent Coplanar Forces found Analytic- ally. Let AD, Fig. 69, represent a force both in magnitude and direction. If AB, be a line of any length, drawn in any direction, and if ABiDCi be a parallelogram so constructed that ABi is a side and AD a diagonal, AD may be considered the resultant of two 11 162 MATHEMATICS [85 forces represented by AE\ and ACi. Then in a process of analysis the force represented by AD may be replaced by the two forces represented by ABi and AC\. The force AD is said to be resolved into the components AB\ and ACi. It is apparent that AD can be resolved into an infinite number of pairs of components; and if it were desirable, either, or both, of these components could be resolved further into two or A BI B p IG gg more components. If the resolution is such that the components are parallel one to each axis of coordi- nates, the force is said to be resolved into its x- and ^-com- ponents, or into its horizontal and vertical components. It is seldom desirable to resolve a force into other than its x- and ^-components, and this form of resolution is very helpful at times in finding the resultant force acting upon a body. If OD, Fig. 70, represents a force acting in a direction making an angle 6 with the positive direction of the axis of x, the magni- tudes of the horizontal and vertical components are, respectively, R cos 6 and R sin B, where R is v the magnitude of OD. The vector representing the force in Fig. 70 is in the first quadrant; but the student will easily see, by sketching other figures, that the algebraic sign of the sin 6 and cos 6 will give to the component force the proper algebraic sign, i.e., positive when up or to the right, and negative when down or to the left. What is here called the analytic method of finding the resultant of a set of given forces, consists of the following process: resolve each force into its horizontal and vertical components, add alge- braically the horizontal components, add algebraically the vertical components, and from these two sums find the resultant as in- dicated below. FIG. 70. 85] STATICS 163 It is required to find analytically the resultant of the five forces given in 84. The work is put in tabular form: Magnitude Angle Cosine Sine Jf-component F-component Pi 3 1.000 0.000 3.000 0.000 P 2 6 30 0.866 0.500 5.196 3.000 P 3 2 70 0.342 0.940 0.684 1.880 p< 2 210 - 0.866 - 0.500 - 1.732 - 1.000 Pi 4 170 - 0.985 0.174 -3.940 0.696 Sum 3.208 4 576 In the first column are placed the letters indicating the forces. In columns two and three are placed, respectively, the magnitudes and angles of these forces. In the fourth and fifth columns are placed, respectively, the cosines and sines of these angles. In the sixth column are placed the products of the magnitudes of the forces by the cosines of the angles. In the seventh column are placed the products of the magnitudes of the forces by the sines of the angles. In columns six and seven, then, are given, respectively, the x- and ^-components of the five forces. Below are given the algebraic sum of the z-components and the algebraic sum of the ^/-components. The five forces may be considered as being re- placed by two; one of 3.208 pounds acting to the right along the X-axis, the other of 4.576 pounds acting up along the 7-axis. It now remains to unite these two forces into one resultant, the resultant of the five forces. Since the two components of 3.208 and 4.576 pounds are mutually at right angles, the parallelogram of forces (See Fig. 70) is a rectangle; hence, the magnitude of the resultant R = ^(3.208) 2 + (4.576) 2 = 5.59. The direc- tion of OC is given by tan 6 = ^~ = 1.426, or, 6 = 54 58'. Of the two values of the angle corresponding to tan 6, the student must be careful to select the correct value for 6, by noting the quadrant within which the resultant lies. Exercises 1. Solve, analytically, placing work in tabular form, the exercises given in 84. 164 MATHEMATICS 86. The Simple Crane. To illustrate the application of the polygon of forces to mechanical problems, let us consider a simple crane. AB, Fig. 71, is the post, CE is the jib, and DE the rope attached to the jib at the point E. At the end of the jib is sus- pended a weight of 10 tons. It is required to find the tension in the rope and the compression in the jib. The point E is acted upon by three forces: one, of magnitude 10, acts down; one, of unknown JD magnitude, acts in the direction CE, and the third, of unknown magnitude, acts in the direction ED. Suppose CD = 12 feet, \OT FIG. 71. FIG. 72. CE = 8 feet, and DE = 7 feet. Draw a triangle to scale with sides proportional to 12, 8, and 7, Fig. 72. Draw the side propor- tional to 12 vertical. Draw the side proportional to 7 through the upper end, and the side proportional to 8 through the lower end, of the vertical side. The triangle CED is drawn to scale. A vec- tor representing the 10- ton force will point in the direction DC; a vector representing the compression (call it PI) in the jib will STATICS 165 point in the direction CE, and a vector representing the tension (call it P 2 ) in the rope will point in the direction ED. Draw the polygon of forces, DC'E', for the three forces, 10, PI, and P 2 . This polygon must close if the point E is in equi- librium. Scale off the length C'E', and the length E'D, which give, respectively, the magnitude of the compression in the jib and of the tension in the rope. If DC' is made 10 inches long, or 1 inch representing 1 ton, C'E' is found to be approximately 6.67 inches and DE' approximately 5.83 inches. This gives a compres- sion in the jib of 6.67 tons, and a tension in the rope of 5.83 tons. Exercises Find graphically the compression in the jib, and the tension in the rope, Fig. 71, if CD = 12 feet, CE = 8 feet, and ED = (a) 8 feet, (6) 9 feet, (c) 10 feet, (d) 11 feet, (e) 12 feet, (/) 13 feet, and (g) 14 feet. Plot two curves upon a sheet of rectangular coordinate paper, one representing the compression in the jib, the other representing the tension in the rope. Plot forces as ordinates, and lengths of DE as abscissas. 1 87. Parallel Forces; Moment of Force. A force applied to a body may act throughout the whole, or a part, of its mass; or it may act over the whole, or a part, of its surface. A book lies upon the table. Gravitational force acts throughout its entire mass, whereas the force exerted upon it by the table acts only over that portion of its surface in contact with the table. If the sur- face over which the force acts is small compared with other dimen- sions, the force is treated as though it acted at a single point of the body. Thus, in Fig. 71, the post of the crane presses to the right, 1 This problem may be solved more easily by making uae of the theorem in ge- ometry which says that homologous sides of similar triangles are in proportion. In Fig. 72, the triangles DCE and DC'E' are similar, and DC' C'E' DE' DC " CE " DE' or 10 C'E' DE' 12 * ~F " ~7~' or C'E' - 6J DE' - 5.833 Solve the above exercises by this method. 166 MATHEMATICS against the ceiling, and the ceiling presses against the crane with a force of equal magnitude, but in the opposite direction. This force, PS, is distributed, uniformly or non-uniformly, over the entire surface of the post in contact with the ceiling; but this bearing surface is so small compared with other dimensions of the crane that the force P$ is considered acting, and is said to act, at a point. A weight of 10 tons is suspended from the end of the jib. Its line of action is vertical and passes through the point E. If the cable suspending the weight were two, three, or any number of times as long, the weight would still exert the same influence upon the crane. For the cable still pulls upon the crane with a force of 10 tons and in a downward direction. The line of action remains the same, i.e., vertical and through the point E. In considering forces acting upon a body, the position of the line of action, and not the exact point of application of the force, is what enters into the solution of the problem. It is readily seen that the line of action of the resultant of two forces passes through the point of intersection of their lines of action. In finding the resultant of two parallel forces it becomes neces- sary to consider two cases: when the forces act in the same direc- tion and when the forces act in opposite directions. 88. Two Parallel Forces Acting in the Same Direction. Let AP, Fig. 73, be a vector representing a force, P, whose line of action passes through the point A. Let BQ be a vector represent- ing a force, Q, whose line of action passes through the point B. P and Q are parallel forces acting upon a body in the same direc- tion. The resultant of P and Q together with its line of action is to be found. At the points A and B insert two forces, S and S', having the same line of action, equal in magnitude but acting in opposite directions. The resultant of these two forces is zero, hence their introduction in no way changes the resultant force acting upon the body. If AS and BS' represent the forces S and iS', AP' represents P', the resultant of P and S; and BQ' represents Q', the resultant of Q and S'. Since S and S' balance, the result- ant of P' and Q' is the resultant of P and Q. The lines of ac- tion of P' and Q' intersect at C. At the point C resolve the force P' (represented by CP") into its original components P STATICS 167 and S, represented by CP"' and CS", respectively. At the point C resolve Q' (represented by CQ") into its original components Q and S f , represented by CQ'" and CS'", respectively. The forces S and S' again balance, leaving the resultant of P and Q equal to the vector sum of CQ'" and CP'". From similar triangles it is seen that CQ'" is equal and parallel to BQ, and that CP'" is -V-S' equal and parallel to AP. Hence, the resultant of P and Q is P + Q, whose line of action passes through the point C and is par- allel to the lines of action of P and Q. From similar triangles: AP _ CK BQ _ CK AS ~ AK a BS' ~ BK' From these two equations, and since AS = BS', it follows that _ BQ ~ AK' If a is the distance of K from the line of action of P, and if 6 is 168 MATHEMATICS [88 the distance of K from the line of action of Q, it follows from similar triangles that b BK a~ AK' Uniting this equation with the next preceding, AP _b BQ ~ a' or P _ b Q ~ a' or Pa = Qb. The product of the magnitude of a force by the distance of its line of action from a fixed point is called the moment of the force about the point. The point about which moments are taken is called the origin of moments, and the perpendicular distance from the origin to the line of action of the force is called the arm of the force. If the relation of the direction of the force to the origin of moments is such as to suggest positive rotation about the origin, the FIG. 74. moment is considered positive. If the relation is such as to sug- gest negative rotation, the moment is considered negative. Thus, the moment of the force P about the point 0, Fig. 74, is positive; the moment of the force Q about is negative. The moment of the force Q about the point K, Fig. 73, is Q 6; the moment of the force P about the same point is P a. The (algebraic) sum of the moments is Q b P a, but since in magni- tude P a = Q b, this sum is zero. In other words, the sum of the STATICS 169 moments of two parallel forces acting in the same direction about any point in the line of action of their resultant is zero. Let 0, Fig. 75, be any point in the plane. Let c be its perpen- dicular distance from the line of action of R, the resultant of the parallel forces P and Q; and d its perpendicular distance from the FIG. 75. line of action of the force Q; and e its perpendicular distance from the line of action of the force P. Then and since R = P + Q, and since Pa = Qb, c = e a Re = Re Ra Re = Pe + Qe- Pa - Qa Re = Pe + Qe - Qb - Qa Re = Pe + Q(e - b - a) and since e b a = d, Re = Pe + Qd. This equation shows that the moment, about any point, of the resultant of two parallel forces acting in the same direction is equal to the sum of the moments of the two forces about the same point. The proof given above considers the lines of action of the two given forces, and the line of action of their resultant, to be upon one side of the point 0. The student may easily show that the same law holds when the point is taken between the lines of action of the two given forces, if by sum is meant algebraic sum. 170 MATHEMATICS [89 89. Two Parallel Forces Acting in Opposite Directions. Let P and Q, Fig. 76, be two parallel forces acting in opposite direc- tions. By analysis similar to that of the preceding section, the student will show: that the resultant of the two forces is the alge- braic sum of P and Q; that its line of action is parallel to that of P and Q, and so placed that the moment of the resultant about any point is the algebraic sum of the moments of P and Q about the same point. The above proof fails when P and Q are equal in magnitude, for then AP' and BQ' are parallel and the point C is at infinity. In FIG. 76. this case the algebraic sum of the two forces is zero, and they can- not be replaced by a single force whose effect upon the body is equivalent to that of the two forces. A system of two non-collinear parallel forces, equal in magnitude, and acting in opposite directions, is called a couple. The product of the magnitude of the forces by the perpendicular distance between their lines of action is called the moment of the couple. If the couple tends to produce positive rotation its moment is posi- tive; if negative rotation, its moment is negative. 90] STATICS 171 The student will show that the moment of a couple is equal to the algebraic sum of the moments of the two forces of the couple about any point. He will also show that the algebraic sum of the moments of two equal parallel forces acting in opposite directions is the same for all origins of the moments. 90. A System of Parallel Forces is Reducible Either to a Single Force or to a Single Couple. In a system of three or more parallel forces there must be at least two that do not form a couple. These two forces may be replaced by their resultant force, as was shown in 88 and 89. The magnitude of the resultant is the algebraic sum of the two components and its moment about any point is the algebraic sum of the moments of the two forces about the same point. In this way the system of forces is replaced by another system having a number of forces less by one. The algebraic sum of the forces and the algebraic sum of thier moments about any point remain, however, the same. By continuing this process of reducing the number of forces, the system is eventually replaced by one consisting of only two parallel forces. This system, if not a couple, can be replaced by a single force, the resultant of the two forces. Thus, the resultant of a system of parallel forces is either a couple or a single force. The resultant can be a single couple only when the algebraic sum of the given forces is zero, and when the algebraic sum of the moments of all the given forces about any point is different from zero. For, by the above method of reduction of the number of forces of the system one at a tune, the algebraic sum of the magni- tudes of the forces remains unchanged, and since this sum is zero when a couple is obtained it must have been zero at the beginning; the algebraic sum of the moments about any point remains un- changed, and since the moment of the couple cannot be zero, the sum of the moments of the given forces cannot be zero. 91. System of Parallel Forces in Equilibrium. A system of forces applied to a body are said to be in equilibrium if the state of motion of the body is unaltered by the application of the forces. If the algebraic sum of a system of parallel forces is not zero, the resultant of the system is a single force and the system is not in equilibrium. One of the necessary conditions for equilibrium of a system of parallel forces is, then, that the algebraic sum of the 172 MATHEMATICS [91 forces be zero. If the algebraic sum of the moments of the forces about any point is not zero while the algebraic sum of the forces is zero, the resultant is a couple and the body is not in equilibrium. The necessary and sufficient conditions for a system of parallel forces to be in equilibrium are: The algebraic sum of the forces must be zero; and the algebraic sum of the moments of the forces about any point must be zero. The above law, it will be noticed, includes, as a special case, the law of the lever. Exercises 1. A lever 14 feet long is used for raising a weight of 500 pounds. Where may the fulcrum be placed if a downward pressure of 100 pounds is available at the other end of the lever? 2. The two end clevis holes of a double-tree are 14 inches and 20 inches from the middle clevis hole. The team is pulling 300 pounds. How much does each horse pull? 3. What must be the ratio of the longer arm to the shorter arm of a double-tree: (a) if one horse is to pull 5 percent more than the other horse ; (6) if one horse is to pull 5 percent less than the other horse? If the team pulls 300 pounds, how much does each horse pull in each of the above cases ? 4. A horizontal bar, AB, 30 B feet long weighs 30 pounds per linear foot. At the point A is suspended a weight of 10 tons, and at the point B is suspended a weight of 3 5 tons. Find the magnitude, the direction, and the line of action of the resultant of the two weights and the gravi- tational forces acting upon the bar. HINT: The line of action of the gravitational forces acting upon the particles of the bar passes through the geometric center of the bar. 5. Let AB, Fig. 77, be a plank hinged at the point A, and held in a horizontal position by a vertical rope. The plank is 20 feet long and weighs 110 pounds. The rope is fastened to the plank 15 feet from the hinge. Find the tension in the rope when a man weighing 150 pounds stands upon the plank x feet from the hinge. FIG. 77. 91] STATICS 173 HINT: If PI represents the tension in the rope, taking mo- ments about A as origin gives 15Pi (110)(10) 150x = 0, or PI = 10.r + 220/3. For, the system of parallel forces is in equilibrium, and the algebraic sum of the moments about any point must be zero. The above equation, which is linear, expresses the relation between the tension in the rope and the distance of the man from the point A. When the man is at the point A, the tension is 73J pounds. When he is at the point B, the tension is 273 J pounds. P 2 + PI HO 150 = 0, or P 2 = 360 - Pi, or P 2 = 286 - 10z, which gives the reaction at the hinge. For, by 90 the sum of the parallel forces must be zero. When the man is at the point B, P 2 = 865 pounds. 200^ FIG. 78. 6. A weight of 200 pounds is to be raised by a lever x feet long, Fig. 78. The distance from the weight to the fulcrum is 3 feet and the lever weighs 4 pounds to the foot. If the lever is very short there is but little advantage gained by its use, whereas if it is very long its increased weight will more than compensate the gain in leverage. Evidently there is some length which will make the force P a minimum. 2(300 + x 2 ) Show that P = . Find the values of P when x equals x 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 17, 18, 19, and 20. Plot results upon a sheet of rectangular coordinate paper. HINT: When substituting in the formula, place it in the form (?+) 7. The end holes A and B of a double-tree, Fig. 79, are 2a inches apart. The center hole C is placed d inches in front of the mid-point of AB. If one horse is pulling ahead of the other as indicated in Fig. 79, find what percent more the rear horse is pulling than the forward horse. HINT: Pi (a cos a d sin a) = P 2 (a cos a + d sin a) 100(Pi - P 2 ) _ 20Qd sin Pj a cos a d sin at 174 MATHEMATICS [92 8. Same as exercise 7, but with a = 13 inches, d = 3 inches, and = 30. 9. With the statement of exercise 7, the difference between PI and z is what percent of the load? FIG. 80. 92. The Sum of the Moments of Two Non-parallel Forces. Let AP and AQ, Fig. 80, represent two forces. Let AR represent their resultant R. The moment of P about is the magnitude of 92] STATICS 175 P, (AP), multiplied by the perpendicular distance from to the line of action of P. This moment is then represented by twice the area of the triangle OAP. Similarly, the moment of Q about is represented by twice the area of the triangle OAQ; and the mo- ment of R by twice the area of the triangle OAR. Draw PP', RR', and QQ' perpendicular to AO; and QB perpendicular to RR'. The area of the triangle OAP = %(PP' -AO). The area of the triangle OAQ = %(QQ' -AO). The area of the triangle OAR = %(RR' -AO). RR = RB -j- BR , or RR' = PP' + QQ'. By uniting these equations 2 (area of A OAP) + 2 (area of A OAQ) - 2 (area of A OAR) = 2(PP' + QQ' - RR') = 0, which shows that the moment of the resultant, R, is equal to the sum of the moments of P and Q. In the above proof, the point was so located that the algebraic signs of the three moments were the same. The student will prove the formula for the case in which is taken between the sides of the angle QAP. When this has been done it will be seen that "sum" in the above law means algebraic sum. From the above theorem and from what has been proved regard- ing moments of parallel forces, it will be seen that the algebraic sum of the moments of any number of forces is equal to the moment of their resultant, and the algebraic sum of the moments of any number of forces in equilibrium is zero. A restatement of the above conclusions follows: The resultant of a system of coplanar forces (not in equilibrium) is either a single force or a couple. If the resultant is a single force, its magnitude and direction may be found graphically by finding the sum of the vectors representing the given forces; its line of action is such that its moment about any point is the algebraic sum of the moments of the several forces about the same point. If the resultant is a couple, the vector sum of the vectors representing the forces is zero, and the algebraic sum of the moments of the several forces about any point is equal to the moment of the couple. For 176 MATHEMATICS [93 a system of coplanar forces to be in equilibrium, the vector sum of the vectors representing the several forces must be zero, and the alge- braic sum of the moments of the several forces about any point must be zero. The first condition throws out the possibility of the resultant being a single force. The second condition throws out the possibility of the resultant being a couple. 93. Equations of Equilibrium. In the preceding sections it was shown that if a system of coplanar forces is in equilibrium, the algebraic sum of the moments of the forces about any point must be zero. This law is generally written in the form of an equation: 2Pa = P represents the magnitude of force, a its moment arm, and 2 (the Greek letter Sigma) is a symbol meaning to take the alge- braic sum. Another condition for equilibrium is that the vector sum of the vectors representing the given forces must be zero. Another way of stating this law is that if the given forces were con- sidered concurrent, their resultant would be zero. It was shown, however ( 85), that one method of finding the resultant of a system of concurrent coplanar forces is first to resolve each force into its x- and ^-components, and from the algebraic sum (X) of the z-components and the algebraic sum (F) of the ^/-components compute the magnitude of the resultant (R) by the formula R = \''X 2 + F 2 . If R is zero, X and Y separately equal zero. This law is generally written in the form of two equations: 2X = and 27 = The symbol 2 (the Greek letter Sigma) means to take the alge- braic sum. X in the above equation means z-component. 2Z then means the algebraic sum of the z-components. 2F means the algebraic sum of the y-components. The equations 2Z = 2F = and 2Pa = 93] STATICS 177 are called the equations of equilibrium for a system of coplanar forces. The application of the equilibrium equations will be illustrated by examples: EXAMPLE 1: The crane, Fig. 71, considered as a whole, is acted upon by four forces: viz., the downward force of 10 tons, the upward push of the floor, P^, the push of the floor to the right, PS; and the push of the ceiling to the left, P 3 . P 4 , P s , and PS are unknown in magnitude, and are to be found. Let AB = 12 feet and FE = 4 feet. Then 2Z = P 5 - P 3 = ZF =P 4 - 10 = and by taking moments about the point A, ZPa = 12P 3 - 10-4 = 0, or PS = 10/3. The forces Pt and P 5 contribute nothing to the moment equation, for they act through the point A about which moments are taken; hence their moment arms and moments are zero. The second equation gives P* = 10 tons. The first equation gives P 3 = PS = 10/3 tons. This means that the post of the crane pushes to the left against the flange of the floor with a force of 31 tons. The construction at this point, as well as at the ceil- ing, must be such as to with- stand this pressure. EXAMPLE. 2: Let AB, Fig. 81, be a ladder 16 feet long, resting against a smooth vertical wall, CB. The lower end of the ladder rests upon the horizontal ground, 6 feet from the building. CB is 14.83 feet long. The ladder weighs 35 pounds, and a man 12 FIG. 81. 178 MATHEMATICS [93 who ascends weighs 150 pounds. Find the pressure of the upper end of the ladder against the wall when the man stands on a rung 5 feet from the point A. By taking moments about the point A, SPa = +35-3 + 150-15/8 - 14.83 Pi = 0, or PI = 26.04 pounds. (The weight of the ladder is considered as a single force acting through the middle point.) FIG. 82. Taking moments about the point A, letting x be the distance (measured along AB) the man stands from the point A, SPa = 35-3 + 150?| - 14.83 Pi = 0, or Pi = _ f 225 - L 4 x + 105 -s- 14.83. From this equation PI can be calculated for all values of x, i.e., for any position of the man upon the ladder. The equation is linear and hence its graphic representation upon a sheet of squared paper is a straight line. In order to draw the straight line all that is needed are the coordinates of two points. These may be found by giving to x two values, as and 16, and calcu- lating the corresponding values of PI. EXAMPLE 3 : Let all members of the bridge truss, Fig. 82, be equal in length, k feet. The bridge sustains loads as indicated. Find the reactions, PI and P 2 , at the points A and B. By taking moments about the point A, it follows that 94] STATICS 179 k 3k *>k 3k P 2 - 1000 2 - 4000 k - 1000 y - 5000 2fc - 1000 ^ - 3000 -3k = 0, or 3P 2 - 500 - 4000 - 1500 - 10,000 - 2500 - 9000 = 0, or 3P 2 - 27,500 = 0, or P 2 = 9166f pounds. SF = Pi + P 2 - 2000 - 1000 - 4000 - 1000 - 5000 - 1000 - 3000 = 0, or P! + P 2 = 17,000, or Pi + 9166| = 17,000, or PX = 7833i Exercises 1. Find the magnitudes of the forces PI, P 2 , and Pa, Fig. 83, neglect- ing the weight of the crane. 2. Find the magnitudes of the forces PI and P 2 , Fig. 84, also the tension in the rope. HINT: Draw ABC to scale, using 1/2 inch to represent a foot. Neglect the weight of AC. [94.] Tension and Compression. The tension or compression in members of a construction may be easily computed, if the weights of the members are neglected, if the loads are applied at the joints, and if the members are discontinuous at the joints. By discon- tinuous is meant that a single piece does not form a side of more than one polygon. Thus, in the roof truss, Fig. 85, the tie AB con- sists of three pieces, AD, DE, and EB, bolted together at the points D and E. If the loads are applied at the joints, the tension or compression in every member is the same throughout the entire length of that member, and the line of action of the compressive or tensile force within the member is that of the longitudinal dimension of that member. 180 MATHEMATICS [94 The method of finding tension and compression will be illus- trated by solving the problem represented by Fig. 85. Let AD = DE = EB = lOf feet /\ 10 Tons Let and FIG. 83. CK = 8 feet AC = BC = 17.88 feet CD = CE = 9.61 feet 94] STATICS 181 To find the reactions, PI and Pt, of the supports use 2 Y = 0, and 2Po = 0, taking moments about the point A. FIG. 84. /\ 10 Tons or FIG. 85. 27 = Pi + P 2 - 500 - 2000 - 1000 - 1000 - 500 = 0, P! + P 2 = 5000. 182 MATHEMATICS [94 SPa = 2000 lOf + 1000 16 + 1000-2H + 500-32 - P 2 -32 = 0, or P 2 -32 = 74,667, or P 2 = 2333 pounds. Then PI = 2667 pounds. Imagine the members AC and AD cut by the section a and the right-hand portion removed. In order to keep the left-hand por- tion in equilibrium, Fig. 86, a, forces P 3 andP 4, equal in magnitude to the tension or compression in AC and AD, must be introduced c' A' FIG. 86. upon the cut ends. To express this in another way: suppose an imaginary plane through the rafter AC divide it into an upper and a lower part. The part below pushes against the part above, and the part above pushes against the part below; the two compressive forces are equal in magnitude and this magnitude is called the compression within the rafter. If the upper part of the member were removed, and if a force equal to the compression were introduced acting upon the remaining cut end oi AC and in a direc- tion toward A, the lower part of the rafter would be acted upon by the same force as before the upper portion was removed. The four forces acting upon the remaining portion of the truss, Fig. 86, a, are in equilibrium. The two vertical forces acting at the point A may be considered as one force acting up, of magni- tude 2167 pounds. Draw the line C'K', Fig. 86, 6, to scale, representing the upward 2167-pound force. C'K' is parallel to CK, Fig. 85. Through C' draw a line parallel to AC and through K' draw a line parallel to AK. These two lines intersect at A'. 94] STATICS 183 K'C'A' is then the force polygon for the three forces, 2167, Pa and P\. C'A' is the vector representing the force PS, and A'K' is the vector representing the force Pi. The magnitudes of PS and P* may be scaled off from this triangle. Since K'C'A' is a closed polygon, the arrows run around the polygon in the same sense. The arrow of C'A' then points from C" toward A', which says that the arrow of PS, Fig. 86, a, was chosen to point in the proper direction, or the member AC is in compression. Similarly, the arrow of P\ was chosen in the proper direction, or the member AD is in tension. The magnitudes of PS and Pi may be found by another method which does not require drawing the triangle K'C'A' to scale. The triangles K'C'A' and KG A, Fig. 85, are similar, and C'A' _ K'C' t CA == KG or C'A 1 2167 17.88 " 8 ' or C'A' = 4842, PS = 4842 pounds. Again : A'K' _ K'C^ ~AK '' = KG' or A'K^ _ 2167 16 8 ' or A'K' = 4333, P 4 = 4333 pounds. To find the tension or compression in the members CD and DE, consider the portion about the point D removed by the cut 6, Fig. 85. 184 MATHEMATICS At the cut ends insert the forces PS and P 6 , Fig. 87, a, equivalent to the tension or compression in the cut members. The force polygon for the four forces is represented by the triangle K'C'D', Fig. 87, b. This triangle is similar to triangle KCD, Fig. 85. 4333 Hence or or Again: or 4333 7) 2000^ FIG. 87. D'C' C'K' DC D'C' 9.61 CK' 2000 D' , ,2000^ K' D'C' = 2402.5, P 5 = 2402.5 pounds tension. K'D' 2000 KD M'D' + 4333 8 ' 2000 or M'D' = - 3000 pounds. The negative sign shows that the arrow of P& points in the wrong direction. P& = 3000 pounds tension In a similar way by means of cuts c and d, Fig. 85, the tension or compression in EB, BC, and EC may be found. 95] STATICS Exercises 185 1. Find the tension or compression in the members BC, BE, EC, and ED, of the truss given in Fig. 85. 2. Find the tension or compression in each member of the truss given in Fig. 82. 95. Rope and Pulley. Let PI and P 2 , Fig. 88, be the tensions in a rope upon opposite sides of a pulley over which it passes. If FIG. 88. there be no friction between the pulley and the axle, the line of action of the force exerted by the axle upon the pulley will pass through the center of the pulley. Taking moments about the center of the wheel (calling a the radius of the wheel) Pia-P 2 a = 0, orPi=P 2 FIG. 89. This says that if a flexible rope passes over a series of friction- less pulleys, the tension at every point of the rope is the same. In Fig. 89, the pull on each end of the rope is the same; and if the 186 MATHEMATICS [95 rope be cut at any point, K, a force equal to P must be introduced upon each free end of the rope in order to hold the system in equilibrium. Fig. 90 represents a system of pulleys by which a pull, P, holds a weight of 10 tons in equilibrium. To find the tension in the rope passing around pulley E, imagine the rope cut by the section /\ 10 Tons FIG. 90. a, and forces introduced upon the cut ends equal to the magnitude of the tension in the rope. Represent these forces by PI. The forces acting upon the part of the body below the section are in equilibrium, and PI + PI 10 = 0, or PI = 5. The tension in the rope passing about pulley E is 5 tons. If a second cut 6 be made, and if P 2 represent the tension in the rope passing about pulley F, P 2 + P 2 P : = 0, or P 2 + P 2 5= 0, or P 2 = 2J. The tension in the rope passing about pul- ley F is 2 tons. 95] STATICS 187 If a third cut c be made, and if P represents the tension in the rope passing around pulleys G and H , P + P P 2 = 0, or P + P - 2f = 0, or P = l. The pull, P, necessary to hold the weight 10 tons in equilibrium is lj tons. Exercises 1. Find the pull P which will hold, by means of the system of pulleys shown in Fig. 91, the weight of 10 tons in equilibrium. Find the pull on the hooks A, B, and C. 10 Tons FIG. 91. 2. Find the pull P which will hold, by means of the system of pulleys shown in Fig. 92, the weight of 10 tons in equilibrium. Find the pull on the hook A. 3. The differential wheel and axle shown in Fig. 93 consists of three drums of radii a, b, and c. A force P is applied to a rope which unwinds from the largest drum. As the drums rotate a rope unwinds from the 188 MATHEMATICS [95 smallest and winds up upon the medium sized drum, lifting a weight of 10 tons. Find P which will hold the weight in equilibrium. As P moves down x feet, how far does the weight rise? y' ~\10 Tons FIG. 92. /\ 10 Tons FIG. 93. CHAPTER VII PERMUTATIONS, COMBINATIONS, AND THE BINOMIAL EXPANSION 96. A Fundamental Principle. From a recitation room there are two doors leading into the hall; from the hall there are three exits. In how many ways can a person leave the building from the room? Let the doors of the room be numbered 1 and 2, and let the exits from the hall be marked A, B, and C. A person may pass through door 1 and exit A; door 1 and exit B; door 1 and exit C; door 2 and exit A ; door 2 and exit B; or door 2 and exit C. After passing through door 1 there are three possible paths: one through exit A; one through exit B; and one through exit C. There are exactly the same number of paths passing through door 2. In all there are 2 X 3 or 6 ways of passing from the building. If there were ra doors leading from the room into the hall, and n exits from the hall, obviously there would be m X n ways of leaving the building. The following is a fundamental principle: // one thing can be done in m ways, and if a second can be done in n ways, the two things can be done in the order indicated in m X n ways. It is obvious that the order in which the two things are done does not affect the number of ways in which they may be done. They may even be done simultaneously. To illustrate: suppose a penny and a die are thrown simultaneously; the penny will, fall in one of two ways; the die will fall in one of six ways; then, the two will fall simultaneously in one of 2 X 6, or 12 ways. This illustration shows that the fundamental principle applies not only to the number of ways of doing two things but also to the number of ways two events may occur. It will be seen at once that if one thing can be done in m\ ways, a second in m* ways, a third in m$ ways, and so on, the num- ber of ways they can be done together is m\ X m^ X m$ X . . . 189 190 MATHEMATICS [97 97. Permutation and Combination Defined. Suppose we have five cubes, one painted white, one black, one red, one blue, and one green. From these five a selection of one is made. This can be done in five ways. Each time a selection of one is made, a selection of four is left consisting of a different combination of colors. A selection of two cubes is made. This may be done in ten ways, as: white and red, black and blue, white and black, black and green, white and blue, red and blue, white and green, red and green, black and red, blue and green. Each time a selection of two cubes is made, a selection of three cubes remains. Then a selection of three cubes from the five cubes may be made in ten ways. A selection of four cubes may be made in five ways, and a selection of five cubes may be made in one way. If from n things, a group of r (r ^ ri) are selected, the selection is called a combination of the n things taken r at a time. The symbol representing the number of combinations that can be made of the n things taking r at a time is n C T . Thus, from the previous illustrations, 5 (7i = 5; 2 = 10; 5 Ca 10; 4 = 5, and 5^5 = 1. Consider again the five colored cubes. Select two by first pick- ing out one and then a second. This may be done in twenty ways if the order of the selection is considered (the selection first black, then white, is considered different from the selection first white, then black). For, by the fundamental principle, the first selec- tion may be made in five ways, and the second selection in four ways; the total number of selections possible is 4 X 5 = 20. Any selection of r (r ^ n) things taken from n things which takes into consideration the order of selection is called a permu- tation of n things taken r at a time. The symbol representing the number of permutations of n things taken r at a time is n P r . It is to be remembered that with permutations we consider the order of selection, or the arrangement, of the objects as well as the PERMUTATIONS, COMBINATIONS, ETC. 191 individuals which go to make up the selection, while with combina- tions we consider only the individuals of the selection without reference to the order of their selection or arrangement. To illustrate: If in a plane there are five points, no three of which are upon one straight line, the problem of finding the number of lines determined by the points is a problem in combinations. For it is immaterial whether we think of the line drawn, for ex- ample, from the point A to the point B, or from the point B to the point A. From the five digits 1,2, 3, 4, and 5, how many products may be formed, using two and only two digits at a time? This is a problem in combinations, for example, the product 3 X 2 is the same as the product 2 X 3. From the five digits 1, 2, 3, 4, and 5, how many two-figured numbers may be formed? This is a problem in permutations; for example, the number 32 is different from the number 23. Exercises 1. Now many products can be formed from the digits 1, 2, 3, 4, and 5 taking (a) two and only two digits to form a product; (6) three and only three digits to form a product; (c) four and only four digits to form a product; and (d) five digits to form a product? 2. How many lines may be drawn through five points in a plane, no three of which are upon one straight line? 3. How many planes are determined by five points in space, no four of which are on one plane? 4. How many points of intersection are determined by five straight lines on a plane, if no three intersect in the same point? 5. How many numbers can be formed from the digits 1, 2, 3, 4, and 5 (a) if two and only two digits are used in each number; (6) if three and only three digits are used in each number; (c) if four and only four digits are used; (d) if five and only five digits are used? 98. Formula for n Pr. Let a, b, c, d, e, f be permuted two at a time. The first letter may be selected in six ways, i.e., the first letter selected may be the a, the b, the c, the d, the e, or the /. When the first letter is selected, the second letter may be selected in five ways from the remaining five letters. Then, by the funda- mental principle, the number of ways two letters may be selected from six letters is 6 X 5, or 30. Thus, ^ = 6 X 5 = 30. 192 MATHEMATICS [98 If three letters are placed in each permutation, the first can be selected in six ways, the second in five (6 1) ways, and the third in four (6 - 2) ways. Thus, ePs = 6X5X4= 120. If four letters are placed in each permutation, the first can be selected in six ways, the second in five ways, the third in four ways, and the fourth in three ways. Thus, ^4 = 6X5X4X3 = 360. The student will show, by reasoning similar to the above, that ePs = 6X5X4X3X2 = 720, and ^6 = 6X5X4X3X2 X 1 = 720. If n things are permuted r at a time, the first selection is made from the n things, or it may be made in n ways; the second selection is made from the remaining n 1 things, or it may be made in n 1 ways. Then, by the fundamental principle, the first two things may be selected in n(n 1) ways. The third selection is made from the remaining n 2 things, or it may be made in n 2 ways. Thus, the first three things may be selected in n (n !)( 2) ways. Continuing in this way the first four things may be selected in n(n l)(n 2}(n 3) ways, the first five things in n(n \)(n 2)(n 3)(n 4) ways, the first six things in n(n l)(n 2)(n 3)(n 4)(n 5) ways, and so on. The r things may be selected in n(n - l)(n - 2)(n - 3) . [n - (r - 2)][n - (r - 1)] ways. Therefore nP r = n(n l)(n 2)(n 3) -to r factors, or n P r = n(n - l)(n - 2)(n - 3) (n - r + 1) (1) If r = n, formula (1) reduces to nP n = n(n l)(n 3) to n factors, or nP n = n(n - l)(n - 2)(n - 3) 3 X 2 X 1 nP n = n_ (2) where J^ (read "factorialn") means 1X2X3X4 (n - 2)(n - l)n, or the product of all the positive integral numbers from 1 up to and including n. The symbol nPn, the number of permutations of n things taken all (n) at a time, will be written P n . From formula (1), n -iPr-i = (n - l)(n - 2)(w - 3) (n - r + 1) 99] PERMUTATIONS, COMBINATIONS, ETC. 193 found by replacing n by n 1, and r by r 1. Then nP r = n( n _ 1 P,_ 1 ). (3) By multiplying both numerator and denominator of the right- hand side of formula (1) by nr , it reduces to ,P,= -l^-. (4) n r Exercises 1. Solve exercise 5 of the preceding section by formula (4). 2. How many different signals can be given with seven flags of different colors, by hoisting them in a vertical line, any number at a time? 3. How many different numbers can be formed from the digits 1, 2, 3, 4, 5, 6, if a digit cannot be used more than once in any one number? 4. From twenty men how many nines can be formed, assuming that every man can play any position? By nine is not meant merely the aggregate of nine men, but has reference as well to the positions played. 5. Same as exercise 4, excepting that only two can pitch and that only three can catch. 6. Same as exercise 5, excepting that the pitchers and catchers play no other positions. 99. Formula for n C r . Within each combination of n things taken r at a time there are __ permutations. The number of permutations of n things taken r at a time is equal to the product of the number of combinations of n things taken r at a time by the number of permutations within each combination taking r at a time, or n p r = or, or 1:3 n r 194 MATHEMATICS [100 Every time a combination of r things is selected from n things, a combination of n r things remains. Thus, the number of com- binations of n things taken r at a time is equal to the number of combinations of n things taken n r at a time. This fact can be shown from formula (1) by replacing r by n r, when it reduces to n Cn-r = n r n n r n-n + r a formula identical with (1). Exercises 1. Solve exercises 1, 2, 3, and 4, 97, by formula (1). 2. There are twenty points in a plane, no three of which are upon the same straight line, excepting five, which are upon one straight line. How many lines are determined by the points? 100. The Binomial Theorem. The product of the n binomials, (fli + bi)(az + 62) (as + 63) (a + bn), is the sum of the partial products formed by taking, in as many ways as possible, one and only one letter from each of the parentheses. When the binomials are all equal, the product becomes (a + &) (a -f 6) . . . to n factors. By taking the first letter from each of the parentheses the partial product a" is obtained. By taking a b from one of the parentheses, and an a from the remaining n 1 parentheses, a partial product, a n ~ l b, is obtained. Several par- tial products of this form may be obtained by selecting the b from different parentheses. In fact, it may be selected in n ways, giving n partial products of the form a n ~ l b. By taking the b from two of the parentheses, and the a from the remaining n 2 par- entheses, a partial product, a n ~ z b-, is obtained. Several partial products of this form may be obtained by selecting the b's from different pairs of parentheses. The number of partial products which may be formed in this way is equal to the number of ways in which two b's can be selected from n b's, or, n Cz. In the same way it may be shown that the number of partial products of the form a n-s 3 j s B (7 3 - ^0 number of partial products of the form a n ~ 4 , b* is n Ci, etc. Thus, (o + b) n = a n + na n - l b + n C z a n ~ 2 b 2 + n C 3 a n ~ 3 b 3 + n C t a n -W + . . . +b n , 101] or PERMUTATIONS, COMBINATIONS, ETC. 195 -(---JV^-C-lKn-Vv n(n - l)(n - 2)(n - 3) a n - 4 b 4 b n . This is known as the binomial expansion, or binomial theorem. The coefficients of the right-hand side of the equation are called the binomial coefficients. For n = 2, (a + 6) 2 = a 2 + 2ab + b\ For n = 3, (a + fe) 3 = a 3 + 3a 2 6 + 3a6 2 + b 3 . For n = 4, (a + 6) 4 = a 4 + 4a 3 6 + 6a 2 6 2 + 4a6 3 + 6 4 . For n = 5, (a + 6) 5 = a 5 + 5a 4 6 + 10a 3 6 2 + 10a 2 6 3 + 5a6 4 + & 5 . (2 6 X4 (2) 3 (3x 2 ) 6X5X4 6X5X4X3X2Xl (3x2)6> or (2 + 3x 2 ) 6 s 64 576x 2 4320x 6 + 4860a; 8 If 6 is negative, it will be seen that every alternate term of the expansion, beginning with the second, is negative. Exercises Expand the following : 1. (a - 6)*. 4. (x 2 - 3) 4 . 2. (2 + 6) 6 . 5. (x* + 3z 2 ) 6 . 3. (2 - 3x) 6 . 6. (x* + x- 1 ) 6 . 101. The Binomial Expansion for Negative and Fractional Exponents. In the preceding section the expansion was given for a binomial with a positive integral exponent. It is shown in higher mathematics that the expansion also holds for fractional 196 MATHEMATICS [102 and for negative exponents, providing that the numerical value of b is less than the numerical value of a. For a positive integral exponent the binomial expansion consists of a finite number of terms, n + 1. For fractional or negative exponents the number of terms is infinite, or the expansion is said to be an infinite series. ILLUSTRATION 1: Expand (1 + a;)" 1 . - ^ ^(_l)-4 (a .)S + or ILLUSTRATION 2: Expand (1 + a;)* . (1/2} (1/2 1} (I _i_ x y = (1) + (1/2) (lr a; 4- (1)~^ (a;) 2 2 or Exercises Expand the following to four terms: 1. (1 + x)U. 4. (1 - x)- 1 . 2. (1 - x)K. 5. (1 + x)~*. 3. (1 + x)H. 6. (1 - x)~*. 102. Approximation Formulas. If a; is numerically very small the expansion n(w 1) n(n l)(n 2) (1 + x) n ~ 1 + nx + ^^ - a; 2 + - I 3 - x 3 + is approximately equal to 1 + nx. For, a; 2 , a; 3 , . . . , are higher powers of a small fraction and hence 103] PERMUTATIONS, COMBINATIONS, ETC. 197 small compared with x. If the symbol = represents " approxi- mately equal," (l + x)" = l + nx (1) if x is numerically very small. Thus, to illustrate, (1 + 0.06) w = 1 + M0.06) = 1.02. The true value of this expression is, to five decimal places, 1.01961. Again: (1 - 0.06) * = 1 - 1(0.06) = 0.98. If x, y and z are numerically very small, the student will show by multiplication and division that the following are approxi- mate formulas: (2) (3) (1 + x)(l + y)(l + z) = 1 + x + y + z. (4) It is to be remembered that the above formulas hold for nega- tive values of x, y, and z as well as for positive values. Thus, by formula (2) (1 + 0.03) (1 - 0.05) = 1 + 0.03 - 0.05 = 0.98. Exercises Find the approximate value of each of the following: 1.03 2. (0.98) . '* 1.02' H _ 1.03 3. (0.93) . 7 - 0.97* 4. (l.Q3)(1.02). 8. (1.04) (1.06)(0.95). 103. The Compound Interest Law. If a dollars are loaned at r percent per annum, compound interest, at the end of one year the amount is a 1 + JQQ ' 198 MATHEMATICS [103 f r "12 at the end of two years the amount is a 1 + JQQ ' [r "is 1 + JQQ ' [r ~~\* 1 + TQQ If the interest is compounded semi-annually instead of annually the amount at the end of x years is [r "12* 1 + 20oJ if compounded monthly the amount at the end of the same period is [r "1 12* f I200J and if compounded n times a year the amount is [ 1 Now, if we find the limit of this expression as n increases indefi- nitely, we find the amount if the interest were compounded con- tinuously. For convenience let ,^,. be represented by Then the amount, y, is 1 lira TI ! 1^ y = a 1 + - 100 n = oo L u J , which may be written u - since a, r, and x are independent of n. By the binomial theorem, n u(u l)fl n2 wJ :m ! LwJ lim n = oo . C7 means the limit of U as n increases indefinitely. 103] PERMUTATIONS, COMBINATIONS, ETC. 199 Since u becomes oo as n becomes oo , equation (2) gives i r 1+ i> =1 + 1+ i + i + i_ + i + - - (3) u = oo L uj A A The sum of the infinite series on the right-hand side of (3) is represented by e and is approximately equal to 2.71828. It is the base of the natural or hyperbolic system of logarithms (see footnote, page 66). Equation (1) then reduces to TX y = ae Too. (4) Since r is any constant, r/100 is any constant, call it 6. Then equation (4) may be written y = ae>*, (5) where a and b are constants. It is the equation then which expresses the law: the rate of growth of a variable, y, is directly pro- portional to the variable itself. The multiplier a is the value of the variable when time, x, is zero, and 6 is the proportionality factor, i.e., it expresses the ratio of the rate of growth of y to y. The compound interest law is one of the very important laws of nature. ILLUSTRATION: Suppose it is found by experiment that 10 per- cent of light is absorbed in passing through a pane of glass 1 cm. thick. Then in passing through a second pane 1 cm. thick, 10 percent of the remaining 90 percent will be lost, and so on. The intensity of light in passing into the glass obeys then the com- pound interest law, y = ae bx where y represents intensity of light and x distance into the glass. When x = 0, y = a; thus, a is the intensity of the light as it falls upon the glass surface. When x = 1, y = 0.9a, hence 0.9a = ae b , or 0.9 = e b or b = log0.9 Then with the constants a and 6 determined, the intensity of the light at any depth below the surface is given by ae log(0 - 9 ^ = a(0.9) x CHAPTER VIII PROGRESSIONS 104. Arithmetical Progression Defined. An arithmetical pro- gression is a sequence of numbers such that any number of the sequence except the first may be obtained from the preceding by the addition (or subtraction) of a common number, called the common difference. The following are arithmetical progressions: 1. 2, 4, 6, 8, 10, 12. The common difference is 2. 2. 2, 5, 8, 11, 14. The common difference is 3. 3. 7, 2, 3, 8, 13. The common difference is 5. 4. 4, 2, 0, 2, 4, 6. The common difference is 2. 5. 2^, 2, If, 1, f, 0, f. The common difference is 1/2. Exercises From the following sequences select those which are arithmetical progressions, and give the common difference: 1. 5, 7, 9, 11, 13. 5. x - y, x, x + y. 2. 2, 4, 8, 16. 6. x, xy, x + y. 3. 10, 8, 6, 4, 2. 7. 0, - 3, - 6, - 9. 4. 3, 2f, 2i 2, If. 8. a, a + d, a + 2d. 105. The General Term; the Sum of an Arithmetical Pro- gression. Let a represent the first term, I the last term, d the common difference, n the number of terms, and s the sum of all the terms. The general series is then a, a + d, a + 2d, a + 3d, . . . , a + (w l)d. It is obvious that the coefficient of d in any term is one less than the number of the term. Thus, in the fourth term it is 3, in the twelfth term it is 11, and in the nth term it isn 1. A formula for the last term is then 1 = a + (n - l)d. (1) 200 105] PROGRESSIONS 201 The sum, s, is found by adding the n terms together, as s = a + (a + d) + (a + 2d) + + [a + (n - or s = na+d[l + 2 + 3 + 4+ . . +(ro-l)]. (2) The sum of the n 1 terms written within the brackets may be represented graphically as follows: Let AoA n -i, Fig. 94, be a line n 1 units long, divided into n 1 equal parts by the points A i, A 2 , As, etc. Upon the segments AoAi, A\Az, A 2 A Z , . . ., An-zAn-i, construct rectangles with altitudes respectively 1, 2, 3, Bo A n . 2 A n .\ FIG. 94. 4, . . . , n 1 units long. The areas of these rectangles are, respectively, 1, 2, 3, 4, . . . , n 1 square units. The areas of these rectangles may then be taken as representing the terms within the brackets of equation (2) , and the sum of the areas of the rectangles representing the sum of the terms within the brackets. Draw A C. An-lC = AfiAn-1 The angle CA A n . l = 45 The area of the triangle A^A n ^C = %(n I) 2 The area of each shaded triangle above the line A<>C is 1/2. 202 MATHEMATICS [105 The sum of the n 1 triangles is, then, %(n 1). The sum of the n 1 rectangles is then Kw- 1) 2 + K- 1) = *(- l)n. Therefore Substituting in (2) , TO (TO 1) s = na-\ ---- - d, or s = [2a+(n-l)d]- (3) From formula (3) the sum of an arithmetical progression may be calculated if the first term, the common- difference, and the number of terms are given. Formula (3) may be written. * * 2 a+ o+ (n l)d > or, since a + (n l)d = I, ' s= *(a+Z). (4) ILLUSTRATION 1: Find the sum of 11, 22, 33, . . . to twenty terms. This is an arithmetical progression, for any term is eleven greater than the next preceding term, o = 11, d = 11, TO = 20. Substituting in formula (3), s = jpJ2fl + (n - l)d\ ='~ [2-11 + (20 - l)ll] = 2310. ILLUSTRATION 2 : Find the sum of \/2 + 2 \/2 + 3 \/2 4- . . . to eleven terms. a = V% d = \/2, and n = 11. s = y[2A/2+ (11 - 1)V2J =66 V2. ILLUSTRATION 3: Find the sum of 6 + 5i + 4f + 4 + . . . to fifty terms. a = 6, d = f, and n = 50. 107] PROGRESSIONS 203 Exercises 1. Find the sum of 1 3 6 . . .to twelve terms. 2. Find the sum of - + -5- + \ + . . .to eighteen terms. 3. Find the sum of the first 100 odd numbers. 4. Find the sum of the first 100 even numbers. 5. A man saves $200 each year, and at the end of each year places this amount on simple interest at 6 percent. What is the amount of his savings at the end of ten years? 6. For drilling a well a contractor is to receive 10 cents for the first foot, and for each foot thereafter 1/2 cent more than for the pre- ceding foot. What does he receive if the well is 600 feet deep? 106. Geometrical Progression Defined. A geometrical pro- gression is a sequence of numbers such that any number of the sequence, except the first, may be obtained from the preceding term by multiplying by a common number, called the ratio. The following are geometrical progressions: 1. 2, 4, 8, 16, ... The ratio is 2. 2. 2, - 4, 8, - 16, . . . The ratio is - 2. 3. 2, 2^2, 4, 4 V2~ . . . The ratio is \/2. 4. "s/3, - 3, 3 -v/3", - 9, ... The ratio is - \/3, 5. a, ar, ar 2 , ar 3 , . . . The ratio is r. 107. The Last Term; the Sum of a Geometrical Progresssion. Let a represent the first term, I the last term, r the ratio, n the number of terms, and s the sum of the terms of a geometrical progression. The general sequence is, then, a, ar, ar 2 , ar 3 , . . . , ar*" 1 . It is obvious that the exponent of r in any term is one less than the number of the term. Thus, in the fourth term it is 3, in the twelfth term it is 11, and in the nth term it is n 1. A formula for the last term 's then 1 = ar"- 1 . (1) The sum, s, is found by adding the n terms together, as, s = a + ar + ar 2 + ar 3 + + ar n ~ l , or * = o(l + r + r 2 + r 3 + ' + r"- 1 ). (2) 204 MATHEMATICS (107 Since 1 + r + r 2 + r 3 + ' ' ' + r n ~ l = 1 ^ r and equation (1) becomes 3.(.l r n j ff.\ s = t _- (3) ILLUSTRATIONS: 1. Sum the progression 2 + 4 + 8 + to ten terms. a = 2, r = 2, and n = 10. d ~ r-) _ 2(1 - 2iQ) _ 1-r 1-2 2. Sum the series \/2 + 2 + 2 \/2 + to ten terms. a = -\/2, r = A/2, and n = 10. V2[l - ( V2) 10 ] = 3lV2 1 _ ^/2 V2 - 1 3. Sum the series \/3 - 3 + 3 V3 - to ten terms. a = A/3~, ^ = V3, and n = 10. Exercises 1. Sum 3 + 9 + 27 + to ten terms. 2. Sum 3+69+ -to ten terms. 3. Sum 3 1 + 1/3 to six terms. 4. Sum 3 + 6 + 15 + -to seven terms. 6. Sum \/3 + 2A/3 + 4 V3 + -to seven terms. 6. Sum \/3 + 2\/3~+ 3\/3 + to seven terms. 7. Sum 2 V2 + 1 to six terms. 8. Find the eleventh term in each of the above series. 1 An exception is made when r = 1, but in this case 1 + r + r 2 + r 3 + r"" 1 = n. 1()S] PROGRESSIONS 205 9. Show that the only right triangles whose sides are in arithmet- ical progression are those whose sides are proportional to 3, 4, and 5. HINT: Let x k, x, and x + k represent the sides. Find k in terms of x, and build up the progression. 10. A debt of $10,000 is to be paid in ten years. An equal amount is to be paid at the end of each year. Find this amount if the indebted- ness draws interest at 6 percent. 11. On a certain twenty-year life insurance policy $32 premium is paid annually. What do the premiums amount to by the end of the twentieth year, interest compounded annually at 3^ percent? 12. An equal amount of money is deposited at the end of each year for twenty years as a sinking fund to replace a piece of machinery valued at $10,000. How much must be deposited each year if the deposits draw 4 percent compound interest? 108. The Infinite Geometrical Progression. From the formula for the sum of a geometrical progression, s = ^ , it is evident that, if r is greater than unity in numerical value, the numerical value of s increases without limit as r increases without limit. For, as n grows very large, r n grows large without limit, causing s to grow large without limit. If r, however, is numerically less than unity, r n grows smaller and smaller as n increases, and approaches zero as n increases without limit. Thus, the quantity within the parentheses approaches 1, or the limit of s is a/(l r) as n increases without limit. The sum, s, of a geometrical progression approaches a definite finite limit as n increases without limit, when r is nu- merically less than unity. This limit, a /(I r), is called the sum of the infinite geometrical progression. ILLUSTRATIONS: 1. Find the sum of the infinite progression 1 + + i + i + ' to oo. a = 1 and r = %. * -o 1-r 1-t 2. Find the sum of the infinite progression \/3 + 1+1 \/3 ' to 0. a = V3 and r = I/ \/3. a \/3 3 = 3(V'3 + 1) < \/3 - 1 = 2 206 MATHEMATICS [108 3. Express the repeating decimal 0.237237237 ... in frac- tional form. 237 237 The decimal fraction 0.237237237 . . . means ^^ + (1QOQ) 2 237 + , an infinite geometrical progression, in which a = 237 /1000, and r = 1 /1000. Substituting in the formula for the sum of an infinite geometrical progression, 237 1000 237 237 1 1000 - 1 == 999' ~ 1000 Exercises 1. Find the sum ofl + i |+ . . . to < . 2. Find the sum of \/2 ?\/6 + i\/2 . . . to . 3. Find the sum of 7= -^ + 1 7=- ;= + . . to oo . \/2 + \/3 \/2 - \/3 4. Express 0.7261872618 ... in fractional form. 6. The middle points of the sides of a square, 10 inches on a side, are joined forming a smaller square. The middle points of the sides of this square are joined forming a third square. This process of form- ing smaller squares is continued indefinitely. Find the sum of the perimeters of the squares. 6. Same as exercise 5, but use an equilateral triangle 10 inches on a side. CHAPTER JX '[PROBABILITY] 109. Probability Defined. When a coin is thrown upon the flat surface of a table, it falls in one of two ways, heads or tails. If it is given a whirling motion when thrown, the chances of its falling heads are no greater than the chances of its falling tails, and the chances of its falling tails are no greater than the chances of its falling heads; that is, before the coin fell, we had no reason for expecting one rather than the other. All that could be said was that the coin must fall in one of two ways, both equally likely. When a common die is thrown it falls in one of six possible ways, all equally likely. One and only one of the six ways is that of the ace up. The chances then of throwing an ace with one throw of a die is one in favor and five against, or one to five, or one out of six. This same thought is expressed by saying that the probability of throwing an ace with one throw of a die is one to six, or 1 /6. If a thing can happen in n ways, and can fail to happen in m ways, it can happen or fail in just m -\- n ways. We say there are m + n possible cases, m favorable cases, and n unfavorable cases. If all these ways are equally likely to occur, ^p (the ratio of the favorable cases to the possible cases) is called the probability of the event happening. j- (the ratio of the unfavorable cases to the possible cases) is called the probability of it failing. When a coin is thrown, the probability of heads turning is 1 /2; for the number of favorable cases is one, and the number of possible cases is two. The probability of heads not turning is 1 /2. When a die is thrown, the number of possible cases is six, the number of favorable cases of the ace falling up is one. The proba- bility of throwing an ace is then 1/6. The probability of not throwing an ace is 5 /6. 207 208 MATHEMATICS [109 In a set of twenty-eight dominoes there are seven doublets. When one domino is selected at random the probability of its being a doublet is 7/28, or 1/4. For, the selection of one domino may be made is twenty-eight ways, or the number of possible cases is twenty-eight; and since a doublet may be selected in seven ways, the number of favorable cases is seven. When two dice are thrown the number of ways in which they may fall is 6 X 6 = 36. Of these thirty-six ways, one, and one only, is a pair of aces. The probability of throwing a pair of aces with two dice is then 1 /36. The probability of throwing a pair is 6/36, or 1/6. The probability of throwing a sum less than 5 is 1 /6. If an event is certain to happen, the number of favorable cases is equal to the number of possible cases, and the probability of its happening is one. If an event is certain to fail the number of favorable cases is zero, and the probability of its happening is zero. In all other cases the probability of an event happening is a positive proper fraction. If an event may happen in a ways, and fail in b ways, the Q probability of its happening is - r and the probability of its a. ~T~ D failing is 7 r- The sum of these two probabilities is 1. Exercises 1. Two dice are thrown. What is the probability of throwing a 5 and a 3? A sum greater than 7? A sum equal to 10? Conse- cutive numbers? One and only one 6? 2. A box contains three white, two black, and five red balls; if one is drawn, what is the probability it is white? Black? Red? If two are drawn, what is the probability that both are white? Both black? Both red? One white and one black? One white and one red? One black and one red? 3. Three coins are thrown. What is the probability that all turn heads? That two turn heads? That only one turns heads? 4. In a box there are five red, two white, and seven blue balls. If three are drawn one after the other, what is the probability of their being drawn in the order red, white, and blue? 110] PROBABILITY 209 HINT: The total number of ways, in order, in which three balls can be drawn from fourteen balls is 14 P 3 . 6. Four cards are drawn from a pack. What is the probability that there is one from each suit? 110. Mutually Exclusive Events. Two or more events are said to be mutually exclusive if the occurrence of one excludes the occurrence of the others. Thus, a 5 and a 6 cannot be thrown together with one die. The occurrence of one event excludes the occurrence of the other. The two events are mutually exclusive. Let A and B be two mutually exclusive events. There are three possible cases: (1) The event A may happen and the event B fail. (2) The event B may happen and the event A fail. (3) The event A and the event B may both faiJ. If a, b, and c are, respectively, the number of equally likely possible cases, the probability that either A or B happens is . For a + 6 + c is the number of possible cases, and a + o + c a + b is the number of favorable cases. The probability that A happens is , and that B happens is . Hence a + 6 + c' a + 6+c the probability of the occurrence of one or the other of two mutually exclusive events is the sum of the probabilities of the two separate events. It is easily seen that this law may be extended to any number of mutually exclusive events. The above law applies only to mutually exclusive events, and care must be exercised not to apply it to events not so related. Thus, if A's probability of winning a contest is 1 /3, and if B's probability of winning in the same contest is 1 /4, the probability of A or B winning is 5- + j- = o 4 7 ^. For the two events are mutually exclusive. If A's probability LZi of solving a problem is 1 /3, and B's probability of solving the same problem is 1 /4, the probability of the problem being solved is not the sum of the separate probabilities, for the events are not mutually exclusive. The solution of the problem by A does not prevent B from solving it. 14 210 MATHEMATICS [111 Exercises 1. The probability of throwing an ace with a common die is |; the probability of throwing a 6 is f . What is the probability of throwing an ace or a 6? 2. A ticket is drawn from a set numbered from 1 to 50. What is the probability that it is a multiple of either 7 or 9? 3. A die is thrown. What is the probability that the throw will be greater than 3? 4. Two dice are thrown. What is the probability that the throw will be greater then 7? 5. A bag contains four red, five white, two blue, and seven green balls. One is drawn. What is the probability that it is either red or blue? 111. Independent Events. The occurrence of several single events simultaneously or in succession constitutes a compound event. Events are said to be dependent or independent, if the occurrence of one does or does not affect the occurrence of the others. Thus, the drawing of a red and a white ball from a bag containing red, white, and blue balls is a compound event. Of two independent events, let a be the number of ways, all equally likely, the first may happen, and let b be the number of ways, all equally likely, it may fail; let ai be the number of ways, all equally likely, the second may happen, and let 61 be the number of ways, all equally likely, it may fail. The number of possible cases for the first event is a + 6, and the number of possible cases for the second event is a\ -\- b\. Each case of the first event may be associated with each case of the second event. The number of possible cases of the compound event is then (a + b)(a\ + &0, all equally likely to occur. In aa\ cases both events happen; in 661 cases both events fail; in 061 cases the first event happens and the second event fails; and in a\b cases the first event fails and the second event happens. Then the probability of both events happening is -, of both events failing is -. , ,., 1 , , . > (a + 6)(ai + 61) 111] PROBABILITY 211 of the first happening and _ 061 _ t second failing is (a + &)(fli + &i) of the first failing and second _ Oi& _ _ happening is (a From the first probability given above it is seen that the probability of the occurrence of two independent events is the product of the probabilities of the occurrence of the separate events. This law may be extended to include three or more single events. ILLUSTRATIONS: 1. A bag contains three white and four black balls. The probability of drawing a white ball is 3 /7. If a white ball is drawn and not replaced, the probability of drawing another white ball is 2/6. The probability of drawing two white u 11 3 ^ 1 l balls in succession is ^ X ^ = ^' lot 2. If A's probability of solving a problem is 1 /3, and if B's proba- bility of solving it is l'/4, the probability that A does not solve it is 2 /3, and the probability that B does not solve it is 3 /4. The 231 probability that both fail to solve the problem is ^ X 7- = " Since 1 /2 is the probability the problem is not solved, 1 1/2, or 1 /2, is the probability that the problem will be solved by A or B. 3. Five coins are thrown. What is the probability that they will all turn heads? The probability that a coin turns heads is 1 /2. Hence the probability that the five turn heads is o'o'o'o'o = Q2' Exercises 1. What is the probability that at least one ace will turn when three dice are thrown? 2. A bag contains six white and five red balls, and a second bag contains four white and three black balls. A ball is drawn from each, what is the probability that both are white? That one is white and one black? That one is white and one red? 3. What is the probability of holding four aces in a game of whist? 4. A sack contains three white and two black balls; a second sack contains four white and seven black balls. A ball is drawn from the first sack and placed in the second; then a ball is drawn from the 212 MATHEMATICS [112 second sack. What is the probability that the second ball drawn is white? 5. A bag contains three white and four black balls. Two persons draw alternately one ball without replacing it. If A draws first, what is the probability that he is the first to draw a white ball? 112. Expectation. If p denotes the probability of a person winning a sum of money, m, the product mp is called the value of his expectation, or his mathematical expectation. Thus, if a man is to receive $10 if he draw a doublet from a set of twenty-eight dominoes with one trial, his mathematical expec- tation is $2.50, found by multiplying $10 by 1/4, the proba- bility of drawing a doublet. 113. Successive Trials. On a single trial, let p be the proba- bility of an event happening and let q be the probability of the event not happening. Thus, if a die is thrown but once, the probability of throwing an ace is 1/6 = p, and the probability of not throwing an ace is 5/6 = q. If the die is thrown twice, the probability of throwing an ace each time is p 2 , or (1 /6) 2 = 1 /36. The probability of not throwing an ace is q 2 25 /36. The probability of throwing an ace on the first trial and not throwing 155 an ace on the second trial is pq, or X ^ = " The probability o o oo of not throwing an ace on the first trial and throwing an ace on 515 the second trial is qp, or _ X ^ = ~a' The probability of throw- ing an ace once and once only with two trials is + = j^- oo oo oo The probability of throwing an ace at least once with two trials ' l--l- = 18 36 + 36 ~ 36* The probability of the event happening on each of the two trials is p 2 . The probability of the event happening on the first trial and not happening on the second trial is pq. The proba- bility of the event not happening on the first trial and happening on the second trial is qp. The probability of the event happen- ing but once is 2pq. The probability of the event not happening upon either of the trials is q 2 . p* + 2pq + 2 is the sum of the squares of the deviations of the readings from the arithmetic mean, the approximate value, or 2/c 2 is the sum of the squares of the residuals as they are called. Thus the sum of the squares of the residuals must be a minimum. This principle is sometimes used in solving simultaneous observation equations in which there are more equation? (corresponding to a large number of readings) than unknown numbers. The method of uniting the equations is called the method of Least Squares. An observation equation, as here used, is an equation resulting from the substitu- tion of readings in an equation assumed, or known, to be of the form connecting the measured variables. In such an equation coefficients are the unknowns. Thus, from Table VI, if x represents the hydraulic gradient and if y represents velocity, and if it is assumed that the law connecting y and x is linear, i.e., y = ax + 6, there result, by substituting the measured values of x and j/, seven equations, viz.: 16.9 = 31. 9o + 6 11.4 = 20.8o + 6 24.5 = 54.1o + 6 4.3 = 10. Oo + b 10.1 = 21. Oo + 6 58.8 = 119.1o + b 22.7 - 55.8o + 6 a and b are two unknown numbers to be found, having given seven equations. The problem is not to find a set of values satisfying any two equations, but a set of values which will represent the most probable line upon which the points would fall if there were no errors of observation. 228 MATHEMATICS [119 118. The Probable Error of a Single Reading. The probable error of a single reading, E g . r ., is a number such that if it be sub- tracted from and then added to the arithmetic mean, the proba- bility of a reading falling within these limits is exactly equal to 1/2, i.e., the chance of a reading falling within these limits is exactly equal to the chance of its falling without. Graphically, the probable error determines two lines, as AB and A 'B', Fig. 99, parallel to and equidistant from the line of symmetry of the probability curve, such that the area between them, the curve, and the X-axis is one-half unit. (The area under the entire curve and the X-axis is unity.) It is found that 1 E s r = 0.6745 J = 0.6745 81, the corresponding values of z are im- aginary, which means that the surface does not extend to a point whose x and y are the values assigned to x and y in the equation. Thus, if x and y are each given the value 7 the equa- tion reduces to z = V 17, an imaginary. This says that the surface represented by the equation contains no point whose x- and y-coordinates are each equal to 7, or in other words, a line parallel to the z-axis seven units each from the XZ- and YZ-co- ordinate planes does not pierce nor touch the surface. The equation z = V 81 (z 2 + ?/ 2 ) may be put in the form x 2 + y z + z 2 = 81. The expression z 2 + y 2 + z 2 is the square of 132] POINT, PLANE, AND LINE IN SPACE 243 the distance from the origin to the point whose coordinates are x, y and z. (See 130.) This expression being a constant, 81, shows that all points whose coordinates satisfy the equation must be at a distance 9 from the origin. The locus of the equation is then a sphere with center at the origin and radius equal to 9. 132. The Equation of a Locus. The equation of the locus of a point is an equation satisfied by the coordinates of all points of the locus and by the coordinates of no other point. Thus, the equa- tion of a sphere, center at the origin and radius equal to 10, is x 2 + y* + z* = 100. The equation of the XZ-plane is y = 0. The equation of the XY-p\a,ne is z = 0. The equation of the YZ- plane is x = 0. The equation of the XY- and XZ-planes considered as a single locus is yz = 0. The equation of the three coordinate planes considered as a single locus is xyz = 0. Exercises 1. Find the distance between the following pairs of points : (a) (1, 1, 1) and (3, 2, 1). (6) (3, - 2, 6) and (- 2, 1, 7). (c) (-1, -2, -3) and (2, 6, -1). 2. What loci are represented by the following equations : (a) x = 3. (6) x = 0. (c) y = - 6. (d) y = 0. (e) z = 0. (/) z = - 10. (ff) x* + t/ 2 + z 2 = 1. (h) x* +y* +z* = 25. (i) xy = 0. (7) xyz = 0. (k) x = y. (I) x = z. (TO) y = z. 3. Find the equations of the locus of (a) a point ten units from the origin; (6) a point ten units from the point (3, 1, 2); (c) a point equi- distant from the XZ- and the FZ-coordinate planes. 133. The Direction Cosines of a Line. Let P, Fig. 103, be any point on the line OP passing through the origin. The projections of OP upon the three coordinate axes are the co- ordinates of the point P. Let a, ft, and 7 be the angles, called the direction angles of the line OP, between OP and the positive directions of the x-, y-, and z-axes, respectively. The cosines of these angles are called the direction cosines of the line OP. 244 MATHEMATICS [134 The direction angles of a line through the origin are positive and do not exceed 180. From the figure it is seen that x = OP cos a y = OP cos (8 z = OP cos 7 Squaring and adding these three equations, z 2 + y 2 + z 2 = OP 2 (cos 2 a + cos 2 + cos 2 7). But, since OP 2 = x 2 + ?/ + z\ it follows that cos 2 a + cos 2 /3 + cos 2 7 = 1. (1) The direction angles of a line not passing through the origin are defined as the direction angles of a parallel line passing through the origin. The sum of the squares of the direction cosines of a line is equal to unity. 134. The Projection of a Broken Line upon a Given Line. Let ABODE, Fig. 105, be any broken line in the plane or in A' B' a' FIG. 105. E' space. From the points A, B, C, D, and E perpendiculars are dropped upon the line OP meeting it at the points A', B', C', D', and E'. A'B', B'C', C'D', and D'E' are, respectively, the pro- jections of AB, BC, CD, and DE. Let a, p, 7, and 5 be the angles between the directed segments of AE and the line OP. A'B' = AB cos a, B'C' = BC cos 0, C'D' = CD cos 7 (which is negative, as 7 is obtuse) and D'E' = DE cos 6. Then A'E', 135] POINT, PLANE, AND LINE IN SPACE 245 the projection of a directed broken line upon any directed line, is the sum of the products of the length of each segment of the broken line by the cosine of the angle between it and the directed line upon which it is projected. 135. The Normal Equation of a Plane. A plane is determined if its normal distance, p, from the origin, and the direction angles of this normal are known. Let ON, or p, Fig. 106, be the normal FIG. 106. drawn from the origin to the plane ABC. Let a, (3, and 7 be the direction angles of this normal. Let P be any point in the plane. Draw PN, which is perpendicular to ON. The sum of the projections of x, y, z, and PN upon the normal is equal to p. The projection of x upon the normal is x cos a, of y is y.cos @, and of z is z cos 7. The projection of PN upon the normal is zero, since ONP is a right angle. It then follows that x cos a + y cos /3 + z cos 7 = p, (2) which is called the normal equation of the plane. In the equation Ax + By + Cz = D, let A, B, C, and D be any real constants. The equation is then the general equation of the 246 MATHEMATICS [135 first degree in x, y, and z. Upon dividing by V A 2 + B* + C' 2 , this equation reduces to = ^x + - B VA 2 + # 2 + c 2 " VA 2 + 2 + c 2 The sign before the radical is to be chosen so as to make the right-hand side of the equation positive. Since the numerical values of the coefficients of x, y, and z in equation (3) cannot exceed unity, they are each the cosine of some angle. Call these angles, respectively, a, /3, and 7. Then cos 2 + cos 2 ft + cos 2 7 = A2 + ^ 2 + c , + A . + %' +c . , h A 2 + 5 2 a, /3 and 7 are then, by 133, the direction cosines of some line. The left-hand side of (3) reduces to x cos a + y cos /3 + z cos 7 , the left-hand side of the normal equation of a plane; and since the right-hand side of (3) is a positive quantity, call it p, equation (3) reduces to x cos a + y cos j8 + z cos 7 = p, the normal equation of a plane. equation of a plane is of the first degree in x, y, and z; and, conversely, every equation of the first degree in x, y, and z is the equation of a plane. To reduce Ax + By + Cz = D to the normal form, divide the equation through by V A 2 + 5 2 + C 2 , choosing that sign before the radical which mil make the right-hand side positive. To illustrate: let x + 2y 3z = 2 be the given equation. To reduce it to the normal form divide by \S\4, which gives 1232 136] POINT, PLANE, AND LINE IN SPACE 247 2 1 The plane is 1= units from the origin. The cos a, j=. is V14 VI* negative, which says that the normal drawn from the origin to the plane extends into the third, fourth, seventh or eighth octant. 2 The cos /3, .^ is negative, which says that the normal from the origin to the plane extends into the fifth, sixth, seventh or eighth 3 octant. The cos y, + , > is positive, which says that the nor- mal drawn from the origin to the plane extends into the first, fourth, fifth or eighth octant. The three direction cosines, then, show that the normal drawn from the origin to the plane extends into the eighth octant; or the plane and the three coordinate planes form a tetrahedron in the eighth octant. Exercises Reduce each of the following equations of planes to the normal form, find the distance of each plane from the origin, and locate the octant in which the plane forms a tetrahedron with the coordinate planes. l.z+j/+z = l. 2. x - y + z = 6. 3. 3x - 2y + 5z - 6 = 0. 4. 2x + 7y - z + 7 = 0. 136. The Slope of a Plane, Given the Elevation of Three Points. In this section is described a graphic method for determining the slope, both in magnitude and direction, of a plane surface, if the relative elevations of any three of its points are known. A description and an illustration of the method will be given before the proof. Let A, B, and C, Fig. 107, be the three points whose elevations are known. Suppose A the highest. Compute the gradient (in feet per 100 feet, in feet per 1000 feet, in feet per mile, or in any other desired unit) from A to B and from A to C. From A toward B draw AB', whose length, drawn to scale, represents the gradient from A to B. From A toward C draw AC', whose length, to scale, represents the gradient from A to C. At the points B' and C' draw lines, respectively, perpendicular to AB and AC. 248 MATHEMATICS [136 These perpendiculars intersect at the point X. Draw AX, which represents in direction and magnitude the slope of the plane ABC. To illustrate: suppose B and C are, respectively, 2.26 and 3.20 feet lower than A. Suppose AB = 838 feet, and AC = 866 feet (measured along the horizontal). The gradient of the plane ABC in the direction AB is 2.7 feet per 1000 feet; and in the direction AC, 3.7 feet per 1000 feet. Lay off AB' and AC', respectively equal to 2.7 and 3.7 inches, thus making 1 inch represent a slope of 1 foot per 1000 feet. 1 AX is 3.79 inches long, which shows that any line on the plane parallel to AX has a gradient of 3.79 feet per 1000 feet. The direction AX is that of greatest slope. If A, B and C were points upon the plane surface of an uneroded sedimentary rock, the length of AX would represent the dip, and the direction perpendicular to AX would be the strike of the stratum. If A, B and C represent points upon the ground-water plane, AX would represent the direction in which the water would be seeping. 1 The drawing here given is reduced. The student should redraw it to scale, and check the results given here. 137] POINT, PLANE, AND LINE IN SPACE 249 To lay out the direction of a line having a given slope, say 1 foot per 1000 feet, proceed as follows: Upon AX as diameter describe a circle. With A as center and with a radius equal to 1 inch describe arcs cutting the circle at the points M and N. AM and AN are then the directions having a drop of 1 foot per 1000 feet. To find the elevation of any point P, draw AP intersecting the circle AB'C'X at the point K. The length AK gives the slope of the plane in the direction AP. With this slope and the distance AP the elevation of P relative to A may be calculated. In our specific illustration, AK is 3.25 inches, indicating a drop of 3.25 feet per 1000 feet in the direction AP. The length AP is 2.18 inches, making P 436 feet distant from A. (The scale of ABC is 1 inch to 200 feet.) From these two measurements P is found to be 1.42 feet lower than A. Exercises 1. A, B and C are three points on a plane. AB = 500 feet, AC = 600 feet, BC = 400 feet, measured on the horizontal. The elevations of the plane at the points A, B, and C are, respectively, 137.6 feet, 132.4 feet, and 135.7 feet. Find graphically the magnitude and direction of the slope of the plane ABC. Find the elevations of two points each 300 feet from B and 200 feet from C. Find two directions having a slope of 3 feet per 1000 feet. Find the slope in the directions from A to two points each 300 feet from A and 400 feet from B. [137.] Proof of the Construction Given in the Preceding Section. In the proof for the above construction let A, Fig. 108, be taken as the origin of coordinates, and let the XY- plane pass through the point B. The coordinates of A are then 0, 0, and 0. Let the coordinates of B be a, b, and 0; and of C be a', b', and c'. The general equation of a plane, by 136, is Ax + By + Cz = D. (4) Since the plane passes through the origin, (0, 0, 0) must satisfy its equation, which shows that D is zero. Since the plane passes through the point B, its coordinates must satisfy the equation, thus giving aA-bB = 0. (5) 250 MATHEMATICS [137 Similarly, the coordinates of the point C must satisfy the equation, giving a' A - b'B + c'C = 0. (6) Solving equations (5) and (6) for A and B we find bc'C C. (a,-b, c) FlG. 108. With these values, equation (4) reduces to bc'x + ac'y + (ab 1 - a'b)z = 0. (7) The direction cosines of the upward drawn normal to this plane are, by 133, be' ac' . ab' a'b T=> T=> and j=~ Vi Vi Vi where / = 6V 2 + aV 2 + (ab'- a'6) 2 . 6 b' The gradient of the line AB is , and of A C is / ., , =j=- a v a * -f- c The equations of the lines B'X and C'X, Fig. 107, on the XZ- plane are, respectively, x = - and a'x + c'z = b'. The co- (Z ordinates of the point X are then x = andz = - , 137] POINT, PLANE, AND LINE IN SPACE 251 Let AT, Fig. 108, be the upward drawn normal to the plane. Take T such that AT = -y^-,. The coordinates of S, the .AC projection of T upon the .X"Z-plane, are , -- an d ac V7 ab' - a'b & a b' - a'b , . , r 7^ ' or, and - ; , which are seen to be the ac' \/I ' a ac' coordinates of the point X. This shows that the slope of the plane is in the direction AX. We shall now show that the length of AX represents the magnitude of the slope of the plane. The distance AX is ~ Vb*c' z +(ab'-a'b)*. By su b st i tut i ng etc the coordinates of X in the equation of the plane passing through the points A, B, and C, we find the plane to be units directly beneath X. Upon dividing this expression by the length of AX we have V&V 2 + (ab' - a'b) 2 , which is equal to the length of AX found above. This shows that AX represents the magnitude of the slope of the plane ABC. Since any angle inscribed within a semicircle is a right angle, a perpendicular drawn to AK, Fig. 107, at the point K will pass through X. From what was proved above, AK must represent the gradient of the plane ABC in the direction AP. For the same reason, AM represents the slope (given) in the direction AM. CHAPTER XII [MAXIMA AND MINIMA] 138. Maxima and Minima Defined. Let y be a function of x (See 45). If, as x increases, y increases to a certain value, M, and then begins to decrease, M is called a maximum value of y. If, as x increases, y decreases to a certain value, m, and then begins to increase, ra is called a minimum value of y. A maximum value of a function is not necessarily the largest value the function may have; and a minimum value is not neces- sarily its smallest value. A function may have more than one maximum, and more than one minimum, and a function may have neither maxima nor minima. >' FIG. 109. Graphically the maxima and minima values of a function are, respectively, the lengths of the ordinates of the high and low points of the graph of the function. Let ABODE, Fig. 109, be the graph of the function y = /(x). As x increases from zero to OA', y increases to the value repre- sented by the ordinate A' A. As x increases from OA' to OB', y decreases. Then by definition, A'A represents a maximum value of the function. Similarly C'C represents a second maxi- 252 139] MAXIMA AND MINIMA 253 mum value, and B'B and D'D represent two minimum values. The minimum value DD' is larger than the maximum value A' A. 139. Maximum and Minimum Values by Plotting. When observed data connecting two variables are plotted upon rec- tangular coordinate paper, and a smooth curve is drawn through the plotted points, the maximum and minimum values may be found approximately from the graph. Thus, from Fig. 15. the maximum air temperature was approximately 67.5 at 3 : 45, 13^'-- *- < --(13-2 *") > FIG. 110. If the equation of the curve be known it may be plotted point by point and the approximate values of the maximum and minimum values taken from the graph. To illustrate this method take the following problem: From a square piece of tin 13 inches on a side an equal square is cut from each corner. The edges are then turned lip forming a box. Find the size of the square cut from each corner such that the volume of the box shall be a maximum. Let y be the volume of the box. Let x be the side of the square cut out from each corner, Fig. 110. The volume of the box is then y = (13 2x) 2 x. By assigning to x values and calculat- ing the corresponding values of y the following table is con- structed: 254 MATHEMATICS [139 X 2/ || z y 7/2 126 1/2 72 8/2 100 2/2 121 9/2 72 3/2 150 10/2 45 4/2 162 11/2 22 5/2 160 12/2 6 6/2 147 13/2 From this table it is seen that a maximum value of y exists when x lies somewhere between f and f . By giving to x, values nearer together, the following table is constructed: X V x y 1.6 153.7 2.6 158.2 1.8 159.0 2.8 153.3 2.0 162.0 3.0 147.0 2.2 162.7 3.2 139.4 2.4 161.4 3.4 130.7 From this table it is seen that the maximum value of y occurs for some value of x between 2.0 and 2.4 In Fig. Ill is given the 160 150 1.6 1.8 2.0 X FIG. 111. 2.2 2.4 2.6 plot of the function for values of x ranging from f to f-. From this graph it is seen that an approximate maximum value of y is 162.7 A closer approximation may be obtained by assigning to x a larger number of values between 2.0 and 2.4. 140] MAXIMA AND MINIMA 255 Exercises 1. Same as the illustrative problem above, but with the side of the square of tin equal to, (a) 10 inches; (6) 15 inches; (c) 17 inches; (d) 20 inches. 2. If a represents the side of the square of tin in exercise 1, and x the side of the square cut out, calculate the ratio x/a for each case. 3. Find the dimensions of the greatest rectangle that can be in- scribed in a circle with radius 10. 140. The function ax 2 + &x + c. In 31 it was shown that the function ax z + bx + c has a maximum or a minimum value. -i -2 (*.-*) FIG. 112. Thus, to illustrate, the function y = 4x 2 + 8x 2 may be put in the form y = 4z 2 + Sx + 1 - 2, or y = (2x + I) 2 - 2, or 7/ = 4(x + l/2) 2 -2. The last equation shows that the vertex of the parabola is at the point ( TJ-, 2) , and that the axis of the parabola extends up from the vertex, as is represented in Fig. 1 12. The function then 256 MATHEMATICS [141 has a minimum value 2, which is the value corresponding to value x = ^-. Exercises Find the maximum or minimum value for each of the following functions : 1. x 2 +2x + 1. 4. 9x 2 + 3x - 2. 7. 2x - x 2 . 2. x 2 + 3x - 6. 5. 7x 2 _ 2x - 6. 8. 3 - 5x - 2x 2 . 3. 4x 2 + 16x. 6. 8x 2 - 3x - 7. 9. 3 + 5x - 2x 2 . 10. If a body is thrown vertically upward with an initial speed of a feet per second, its height, h, in feet, at the end of t seconds is given by the equation In = at 16. li 2 . To what height will the body rise if thrown with an initial speed of 32.2 feet per second? When will it reach this height? 141. Maxima and Minima Found by Limits. In some problems the maxima or minima values of a function may be found by a method illustrated by the following examples : EXAMPLE 1 : Find the maximum and minimum values, if any, of the f uncton 2*4-1 To find the maximum and minimum values of the function is to find the ordinates of the turning points of the graph of the equation a 2 + 6 Upon solving for x the equation . x = y V(y-2)(y + 3) (2) is obtained. If, in equation (2), values are assigned to y (y = k, where k is some constant) corresponding values of x are found; thus locating points upon the graph of the function. These values of x are given by x = k V(k - 2)(fc + 3). (3) The equation y = k represents a straight line parallel to the x-axis, k units distant, and the values of x found from equation 141] MAXIMA AND MINIMA 257 (3) are the z-coordinates of the points of intersection of this straight line with the graph of equation (2). The second factor, k + 3, under the radical of equation (3) is positive for all positive values of k, and for all negative values of k numerically less than 3, or positive for all values of k greater than 3. It is zero when k equals 3, and negative when k is less than 3. The first factor is positive when k is greater than 2, and zero when k equals 2, and negative when k is less than 2. When both factors are positive, i.e., when k is greater than 2 or less than 3, x is real, and the line y = k cuts the graph of equation (2) in two real points. The z-coordinates of these points are the values of x found from equation (3). 1 When k FIG. 113. is greater than 3 and less than 2, x is imaginary, for then the quantity under the radical is negative. This shows that all lines of the system y = k between the lines y = 2 and y = 3, fail to cut the curve. When k = 3 or 2 there is in each case but one value (or two equal values) of x. The lines y = 2 and y = 3 are tangent to the curve. From the above discussion it is seen that the graph of equa- tion (2) consists of two branches, one above and tangent to the 1 The student should plot the graph of equation (2). Give to y in (2), or to k in (3), values ranging from 10 to + 10. 17 258 MATHEMATICS l141 line y = 2, the other below and tangent to the line y = 3. These two branches extend to infinity, one in the positive, the other in the negative y direction, for x is real for all positive k values greater than 2, and for all negative k values less than 3. The graph has two turning points, one with a ^-coordinate 2, the other with a 7/-coordinate 3. The first is a minimum value, the second a maximum value of the given function. The mini- mum value corresponds to the value 2 for x, the maximum value corresponds to the value 3 for x. EXAMPLE 2: Find the maximum rectangle that can be in- scribed in a circle with radius 10. In Fig. 113, let x be one-half the length of the base, and z one- half the length of the altitude. The area, y, is y =.4z. (1) z = \/100 - z 2 . (2) Substituting in equation (1) y = 4z\/100 - x*. (3) Squaring equation (3) ?/ 2 = 16z 2 (100 - z 2 ). (4) Solving (4) for x z X2 = 800+ V(200-y)(200 + y) (fi) 16 In equation (5) the quantity under the radical is positive if y is less than 200 and greater than - 200, (- 200 < y < 200), and z 2 is real. Beyond these values for y, x 2 becomes imaginary. 200 then is the maximum value of y. When y equals 200, x 2 equals 50, or x = \/50, and from equation (2), z = \/50. The maximum rectangle is then a square with a side equal to 2 -x/50. Exercises 1. A gutter with rectangular cross section is made from a strip of tin 30 inches wide. Find the width of the portion of tin bent up upon each side so that the carrying capacity of the gutter shall be a maximum. HINT : Let y be the area of the cross section, and let x be the width of the tin bent up. Then y = x(3Q 2x). 2. Find the length of the lever, exercise 6, 91, which will make the upward pull a minimum. 142] MAXIMA AND MINIMA 259 142. Another Method of Finding Maxima and Minima. The method considered here of finding the maxima and minima of a function depends upon a study of the slope of the tangent line drawn to the curve representing the function. In Fig. 109, a tangent line drawn to the curve at any point between P and A makes an acute angle with the positive direction of the axis of x. The slope of any such tangent line is then positive. Any tangent line drawn to the curve between the points A and B makes an obtuse angle with the positive direction of the axis of x. The slope of any such tan- gent line is then negative. The slope of any tangent lines drawn to the curve between the points B and C, and between the points D and E, is positive. The slope of any tangent lines drawn to the curve between the points C and D is negative. When the slope of the tangent line is positive, the curve ascends; when it is negative FIG. 114. the curve descends. When the slope of the tangent line changes from positive to negative the curve changes from ascending to descending and possesses a high point. When the slope of the tangent line changes from negative to positive, the curve changes from descending to ascending and possesses a low point. EXAMPLE 1 : In 139 the volume, y, of a box formed from a square piece of tin 13 inches on a side was found to be y = (13 - 2z) 2 z = 169x - 52x 2 + 4z 3 . (1) Let us suppose that the curve in fig. 1 14 represents the graph of the function. Let P be any point on the curve whose coordinates 260 MATHEMATICS [142 are x and y. Let Q be any other point on the curve whose co- ordinates are x + h and y + k. Since these points are on the curve, their coordinates must satisfy equation (1), from which we obtain y = 169x - 52x 2 + 4x 3 (2) and y + k = 169(x + h) - 52(x + h) 2 + 4(x + ft) 3 . (3) Equation (3) reduces to y + k = 169x + 169ft - 52x 2 - 104ftx - 52ft 2 + 4x 3 + 12x 2 ft + 12xft 2 + 4ft 3 . (4) Subtracting equation (2) from equation (4) there results k = 169ft - 104ftx + 52ft 2 + 12ftx 2 + 12xft 2 + 4ft 3 , or | = 169 - 104z + 52ft + 12.x 2 + 12xft + 4ft 2 . (6) k/h represents the slope of the secant line drawn through the points P and Q. If the point Q is taken nearer and nearer the point P, the secant line rotates about the point P and approaches the tangent line drawn to the curve at that point. The slope of the tangent line is then the limit of the slope of the secant line as the point Q approaches the point P. But, as Q approaches P, h approaches zero, and the right-hand side of equation (6) ap- proaches 169 1042 + 12x 2 . Therefore, the slope of the tangent line drawn to the curve representing the volume of the box is r 1 04. 1 fiQ 12x 2 - 104x + 169, or 12 I x 2 ~ jrf * + ^ or slope =12(x-f)(x-f). (7) Since the side of the square of tin is 13 inches, x, the side of the square cut from each corner, must be less than 6j inches. Thus, / 13\ the last factor, ( x ^-j , of equation (7) is always negative. The second factor, (x -g-j, is negative if x is less than -^-, and 13 positive if x is greater than^-- The expression representing the 13 slope is then positive for values of x ranging from to - and 142] MAXIMA AND MINIMA 261 for these values of x the curve representing the function rises. 13 For all values of x ranging from -~- to 13 the slope is negative, and the curve descends. Thus, the function representing the 13 volume of the box has a maximum value when x = -^ To find o the maximum volume substitute this value for x in equation (1), which then gives y = 162.74. EXAMPLE 2: The area of a rectangle inscribed in a circle, radius 10, is given by equation (3), example 2, 141. y = 4x\/100 - x 2 , (8) or y 2 = 16x 2 (100 - a; 2 ) = IGOOx 2 - 16x 4 . (9) A value of x which makes y a maximum will at the same time make y 2 a maximum. For brevity call y 1 = u, and equation (9) becomes u = 1600x 2 - 16x 4 . (10) As in the preceding example, let P and Q be any two points upon the curve, Fig. 114, which represents the relation between u and x. If x and u are the coordinates of the point P, and x + h and u + k those of the point Q, u = IGOOx 2 - 16x 4 (11) and u + k = 1600(x + hY - 16(x + hY = IGOOx 2 + 3200/ix + IGOO/i 2 - 16x 4 - 64/ix 3 - 96/i 2 x 2 - 64^ 3 x - 16/i 4 . (12) Upon subtracting equation (11) from equation (12) there results k = 3200Ax + 1600/1 2 - 64/ix 3 - 96/i 2 x 2 - 64h 3 x - I6h*, (13) or k h = 3200* + 1600A - 64z 3 - 96/ix 2 - 64^ 2 x - 16A 3 . (14) As h approaches zero, the right-hand side of equation (14) ap- proaches 3200x 64x 3 , or slope = 64z(50 - x 2 ) = 64x( \/50 + x)( \/50 - x). (15) 262 MATHEMATICS [142 From the right-hand side of equation (15) it is seen that the slope is negative if is greater than \/5Q, and positive if x is positive and less than -y/50- The curve ascends for all values of x from to \/50, and descends for all values of x greater than \/50- Thus, u has a maximum value when x = \/50- When u is a maximum, y is a maximum; therefore the maximum rectangle is that for which x = \/50. The remaining discussion of this problem has been given in 141. EXAMPLE 3 : Find the most economical proportion of a closed cylindrical can. Let z be the altitude, x the radius of the base, v the volume, and y the surface of the can. The problem is to find the ratio x/z which will make y a minimum for any constant value of v. V = TTZX 2 (16) y = 2irzx + 2irx* (17) Upon eliminating z between equations (16) and (17) y = ~ + 2irx*. (18) U/ 2v y + k = + 27r(a; + h}\ (19) Subtracting equation (18) from equation (19) + - 2vh or K or k h As h approaches zero, the right-hand side of equation (20) ap- 2v preaches 4irx 2 - Then 2v slope = lirx -j x 2 or A.- / 11 \ (21) 142] MAXIMA AND MINIMA 263 3 / V If x is less than * / the quantity within the parentheses of 3 / V equation (21) is negative. If x is greater than -1/5 this quan- 3 / V tity is positive. For all positive values of x less than \IJT- the \ iir slope is negative, and the curve representing y is descending; for 3 / V values of x greater than \f^- the curve is ascending. Thus, \ 2?r 8/7 the surface is a minimum when x = \l^- From equation (16) = \ , \ 7T whence -- = 4-- Since the v does not appear in Z this equation, the ratio of x to z is constant, ^, for all values of v. The most economical proportions are such that the altitude is equal to the diameter of the base. The method used in the three above cases is applicable to any problem where the functional relation between the two variable quantities x and y is given by means of an equation. Exercises 1. Find the most economical dimensions of a pan with square base and vertical sides, if it is to hold 5 gallons (1 gallon = 231 cubic inches). 2. Find the most economical proportions for a cylindrical can, open at the top, if it is to hold 1 quart (1 quart = 57f cubic inches). 3. What is the ratio of the length of the legs of a right triangle when its area is a maximum, if the hypotenuse is a constant, 10? 4. The strength, y, of a rectangular beam varies as the product of its breadth, 2, by the square of its depth, x; find the dimensions of the strongest beam which can be cut from a log 20 inches in diameter. CHAPTER XIII [EMPIRICAL EQUATIONS] 143. Empirical Equation Defined. A series of readings con- necting two or more variable numbers should always be plotted upon rectangular coordinate paper (squared paper), thus rep- resenting to the eye the variation of one of the numbers with respect to the other. In Table VI, there are given seven readings of the velocity of water through a sample of sand. These seven velocities correspond each to a measured hydraulic gradient. This set of readings when plotted upon squared paper is shown in Fig. 115. Velocity- Feet 24 Hours i-. to co *>. 01 o .j OOOOOOOc C 4 + % ^ .'* ^ .," '"^ X ^ X 1 ' x* 3 X ' *r' -' ^ r + X- *' L 10 20 30 40 50 60 70 80 90 100 110 120 Hydraulic Gradient-Feet per 1000 Feet FIG. 115. The plotted points do not lie upon a smooth curve. The reason for this is that there are errors of observation errors both in the determinations of the velocities and in the hydraulic gradients. If there had been no errors of observation and if all pairs of read- ings h'ad been taken under exactly the same conditions, the plotted 264 143] MAXIMA AND MINIMA 265 points would have fallen upon a smooth curve. This smooth curve which we are trying to locate by the plotted points represents graphically the law of nature which we are trying to determine or study, viz., the law governing the flow of water through this particular sample of sand, relative to the hydraulic gradient. From Fig. 115 it is apparent that the plotted points do not justify the assumption of a locus other than a straight line along which the points would fall if there were no errors of observation. The true relation connecting the variable numbers measured may not be the linear law, but with the data at hand this is the only law which can be assumed as an approximation to the law of nature. With the aid of a transparent triangle the line QA Velocity- Feet per Second to CO CO CO CO 05 J- JO OS ~-~ ' r 1 > X f* x t I ) ^ \ > I \ \ \ 1 \ \ \ \ .1 .2 .3 .4 .5 .6 7 Depth -Part of Total FIG. 116. .8 .9 is drawn among the plotted points. The position of the line is located by the eye as the probable line upon which the points would fall if the true law were linear and if there were no errors of observation. The equation of this line is then y = ax -f- b, where x represents the hydraulic gradient and y the velocity. Since the line passes through the origin, b is zero, a is found to be 0.51. The equation of the line becomes y = 0.51x, and is called an empirical equation connecting velocity and hydraulic gradient. 266 MATHEMATICS [143 Frequently it is desirable to express the relation existing be- tween variable numbers in the form of an equation, even though this equation represents only an approximation to the true law. Such an approximate equation when found by means of experimental readings is called an empirical equation. As a second illustration of a method of finding the empirical equation, consider the plotted points in Fig. 116. From this figure it is seen that the law connecting velocity and depth is not linear, but appears to be parabolic. The vertex of the parabola is at, or nearly at, the point (0.29, 3.262) 1 . If the origin be moved to this point the readings (which are given in Table IX) become: X' y \\ x 1 y - 0.29 - 0.067 0.21 - 0.034 - 0.19 - 0.032 0.31 - . 082 - 0.09 - 0.009 0.41 - 0.135 0.01 - 0.001 0.51 - 0.203 0.11 - 0.010 0.61 - 0.286 and the equation connecting x' and y' is y' = a(x') 2 . If (x') 2 be replaced by X the equation becomes y' = aX (1) Hence if y"s are plotted against X's, the squares of the x"s, a straight line will be obtained if the curve in Fig. 116 is a parabola with its vertex at the point (0.29, 3.262). These plotted points are represented in Fig. 117. The points indicated by circles cor- respond to [positive values of x', while those indicated by dots correspond to negative values. A straight line drawn among the points with the aid of a transparent triangle gives the equation (2) (3) Since X= (x'}' Since and y ' = - 0.77Z. y ' = - 0.77(z') 2 . x' = x - 0.29 y' = y - 3.262, 1 From Fig. 116 it may at first appear that the vertex of the parabola is on the line x = 0.3; but by noting the vertical positions of the points plotted equidistant from this line it is seen that the line of symmetry of the curve is slightly to its left. 143] EMPIRICAL EQUATIONS 267 equation (3) becomes y - 3.262 = - 0.77(a; - 0.29) 2 , or y = - 0.77(z - 0.29) 2 + 3.262, or y = - 0.77 x z + 0.447 x + 3.197, an empirical equation connecting velocity with depth. 1 (4) (5) y -0.3 / / , / ' .A V / / / / < to / 0.64 0.08 0.12 0.16 0.20 0.24 0.28 O.S2 0.36 X FIG. 117. In the first illustration above the empirical equation is of the form y = ax + b, in the second illustration it is of the form 1 Compare equation (4) with the equation given in exercise 4, 31. 268 MATHEMATICS [144 y = ax 2 + bx + c. These equations are each a special case of the more general equation y = a x + aiZ"- 1 ^- a z x n - z + . . . + a n _iz + a n , (6) where a , 01, 02, . . . , a n are coefficients to be determined and where n is a positive integer. To determine the n + 1 coefficients of equation (6), n + 1 equations are required. Hence the curve corresponding to (6) may be made to pass through (n + 1) points of the curve drawn among the plotted points. Empirical equations of the form (6) are undesirable if more than three or four terms are required. Exercises 1. From the data given in Table VII, find empirical equations connecting yield and head, for the 8-inch and 14-inch wells. 2. From the data given in Table VIII, find an empirical equation connecting T and W. 144. Coefficients Determined by the Method of Least Squares. 1 In Fig. 115 a linear law was assumed connecting velocity and hydraulic gradient. A straight line was drawn among the points and the values of the coefficients a and b were determined from this line. If the plotted points had been so scattered that it would have been difficult to estimate by eye the position of the line, the coefficients a and 6 could have been determined by the method of least squares. This method will be illustrated by using the two illustrations of the preceding section. In the first illustration the equation is assumed to be y ax b = 0. Upon substituting the seven pairs of values given in Table VI there results: 16.9 - 31.9o -6 = 11.4 - 20. 8a -6 = 24.5 - 54. la - 6 = 4.3- 10.0a-6 = 10.1 - 21. Oa -6 = 58.8 - 119. la -6 = 22.7 - 55. 8a -6 = 1 The student will verify the numerical work of this section. 144] EMPIRICAL EQUATIONS 269 Here are seven equations from which to solve for a and &. This can, of course, not be done, since the number of equations is greater than the number of unknowns. The values of a and b, which are regarded as the best approximations, are found as follows: First multiply each member of each equation by the coefficient of a in that equation and add the resulting seven equations. In the illustration this gives 10627.52 - 22216.51a - 312.76 = 0. (1) Next multiply each member of each equation by the coefficient of 6 in that equation and add. In the illustration this gives 148.7 - 312.7a - 76 = 0. (2) Next solve equations (1) and (2) for a and b. This gives in the illustration a = 0.48 and 6 = - 0.2 Thus, the equation of the straight line drawn among the points determined by this method is y = 0.48z - 0.2 The equation found by the method of the preceding section is y = O.Slz In the second illustration (See Fig. 116) the empirical equation was assumed to be of the form y ax 1 7 = 0. Upon substituting the ten pairs of values given in Table IX there results; 3.195 - 0.00 a- 0.0/3 -7 = 3.230 - 0.01 a - 0.1/3 - 7 = 3.253 - 0.04 a - 0.2/3 -7 = 3.261 - 0.09 a - 0.30 -7 = 3.252 - 0.16 a- 0.40 -7 = 3.228 - 0.25 a - 0.50 - 7 = 3.181 - 0.36 a- 0.60 -7 = 3.127 - 0.49 a- 0.7/3 -7 = 3.059 - 0.64 a - 0.80 -7 = 2.976 - 0.81 a - 0.90 - 7 = 270 MATHEMATICS [145 To find the "best values" for a, /3 and 7 from these equations, first multiply each member of each equation by the coefficient of a in that equation and add the resulting ten equations. There results 8.8289 - 1.5333a - 2.025/3 - 2.867 = 0. (3) Next multiply each member of each equation by the coefficient of /3 in that equation and add the resulting equations. There results 14.09 - 2.025a - 2.85/3 - 4.57 = 0. (4) Next multiply each member of each equation by the coefficient of 7 in that equation and add the resulting ten equations. There results 31.762 - 2.85a - 4.5/3 - 107 = 0. (5) Next solve equations (3), (4), and (5) for a, /3, and 7, and obtain a = - 0.81 /3 = 0.48 7= 3.19 Hence an empirical equation found by this method is y = - 0.81x 2 + 0.48z + 3.19 This equation places the vertex of the parabola at the point (0.296, 3.26). These two illustrations show that a large amount of numerical work is required to find the coefficients by the method of least squares. If the plotted points fall upon, or nearly upon, a smooth curve the methods of the preceding section should be used. 1 145. Logarithmic Paper. The student will read 73 again. In Fig. 118 the square ADCB is one unit on a side. At the bottom and at the left-hand side of this square are logarithmic scales, each ranging from 1 to 10. Thus on the lower scale, AD, the numbers are placed opposite their logarithms measured on the uniform scale CB. For example, 1 is placed opposite 0; 2 opposite 0.301; 3 opposite 0.477; 4 opposite 0.602; 5 opposite 0.699; 6 opposite 0.778; 7 opposite 0.845; 8 opposite 0.903; 9 opposite 0.954, and 10 opposite 1. The distance from A to the position of any number on the scale AD is equal to the logarithm of that number. 1 The theory upon which the method of least squares is based may be found in texts on "Least Squares." 145] 271 Through the points of division of the AD and AC scales, vertical and horizontal lines are drawn forming what is called logarithmic coordinate paper. 1 To illustrate the use of logarithmic coordinate .1 1 .2 .3 I .< i .5 i . : 8 .9 LOg , _ 9 X> p ^f s" ^L x^ * ** 7 ^ ^ -^ (f),

is greater than. < is less than. is equal to or greater than, is equal to or less than. .^ angle. \n_ (read, "factorial n") means the product of all the integral num- bers from 1 up to and including n. Thus: |5_ = 1-2-3-4-5. n\ (1_^ is sometimes written n!). |a| the numerical, or absolute, value of a. /Or); (x); F(x)', ^(x), etc. function of x. logo x the logarithm of x to the base a. log x the logarithm of x to the base 10- 2 the sum of such terms as i ft the sum of such terms as, . . ., 2 asi varies from a to /3. TT the ratio of the length of the cir- cumference of a circle to the length of its diameter. An ap- proximate value of TT is 22/7. A better approximation is 3.1416. e the base of the natural (or hyper- bolic) logarithmic system, e = 2.71828 approximately, the number of permutations of n things taken r at a time. the number of permutations of n things taken all at a time. 288 CV = 00 lim _l!L In In u . m Y = (* MATHEMATICS the number of combinations of n things taken r at a time. infinite. limit. approaches. the limit of Y as x approaches a. FORMULAS Algebra 5. 6. 7. 8. 9. 10. ~~ a = d \a m ] n = a l" 1 J = a) (6) = - (06). o)(- 6) = 06. + 6) 2 = a 2 + 2a6 + b 2 . + b) 3 = a 3 + 3a 2 6 + 3a6 2 + b 3 . + 6) 4 = a 4 4a 3 6 + 6a 2 6 2 4a6 3 / \ - = a" + 11. a 2 + 2a6 + b 2 = (a + 6) 2 . 12. a 2 - 6 2 = (a + 6) (a - 6). 13. a 3 - b 3 = (a - 6)(a 2 + ab + 6 2 ). 14. a 3 +b 3 = (a + b) (a 2 - a& + 6 2 ). 6 4 . n(n - l)(n - 2) - 15. If c = 0, x = 16. logo (xy) = Iog x + logo y. 2C 17. logo - = logo x - logo y. 18. Iog x n = n logo x. 19. loga -- = - logo X. > 20. logo a; = (log b x)(log a 6). 22. log. x = log 10 logic x = 2.3026 logic x. 23. logic x = logic e log,, x = 0.4343 log e x. APPENDIX 289 24. a + [a + d] + [a + 2d] + + la + (n - l)d~] 25. a + ar + ar 2 + ar 3 + 26. a + ar + ar 2 + ar 3 + 27. 1+2+3+4 + 28. I 2 + 2 2 + 3 2 + 4 2 + + ar"" 1 = to oo = (- - r") 1 - r 1 - r > if |r|< 1. + = 2 (n + + n 2 = a(n + l)(2n + 1). 29. I 3 + 2 3 + 3 3 + 4' + - + n 3 = fj-(n +1)1* to ~ = Hm r i + 1? = * L n so. i + p + + + 31. e* = 1 + x + + ~ = 2.71828. to oo (for all values of x). 32. log, (1 + z) = + x - - + - + ... to oo (if - 1< x <+ 1). 33. e~ x = 1 x 2 + -jTj ^ + to oo (for all values of x). (n - r + 1) = 34. n P r = n(n - l)(n - 2) 35. n P B = P = n. 36. n P r = n(._iP,_i). 37. n C r = ~ n Pr = n In *r Trigonometry 38. sin a esc a = 1. 39. cos a sec a = 1. 40. tan a cot a = 1. 41. sin 2 a + cos 2 a = 1. 42. sec 2 a = 1 + tan 2 a. 43. esc 2 a = 1 + cot 2 a. , . 44. cos a 19 = tan a. 290 MATHEMATICS . COS a 45. - = cot a. sin a 46. sin (90 a) = cos a. 47. cos (90 + a) = + sin a. 48. tan (90 a) = +_cot a. 49. sin (180 a) = + sin a. 50. cos (180 a) = - cos a. 51. tan (180 + ) = tan a. 52. sin (270 a) = - cos a. 63. cos (270 a) = sin a. 54. tan (270 o) = + cot a. 56. sin (a + )3) = sin a cos j3 + cos a sin /3. 56. cos (a /3) = cos a cos /3 + sin a sin /3. tan a tan (3 67. tan (**)- 58. sin 2a = 2 sin a cos a. 59. cos 2a = cos 2 a sin 2 a. 60. =12 sin 2 a. 61. = 2 cos 2 a - 1. 2 tan or 62. tan 2a = z r 1 tan 2 a DO. sin 64. cos K RK tor, a 1 \ /I + COS a ' \ 2 VI COS a bo. tan o 66. CT 1 + COS a 1 COS a sin a sin a a + /3 a 68. sin a + sin /? = 2 sin s cos s- ^ ^ a + /3 a 69. sin a sin /3 = 2 cos o sm " a + 70. cos a + cos /3 = 2 cos o cos 71. cos a cos /3 = 2 sin ^ sin ^ & & 72. = a = ~ ' Law of sines. sm or sm /3 sm 7 APPENDIX 291 73. a 2 = 6 2 + c 2 26c cos a. Law of cosines. a- j8 tan 2 a b 74. -; ? Law of tangents. tan^ 3 a + b 75. Area of triangle = A/S(S a)(s 6)(s c), where 2s = a + 6 + c. cfe sin a 7b. = 2 c 2 sin a sin 8 77. 78. Radius of inscribed circle = A/ 79. sin 2 a = |(1 cos 2a). 80. cos 2 a = 5(1 + cos 2a). 1 cos 2a 1 + cos 2a (1 - cos 2a) 2 sin 2 2 a sin 2 2 a 81. tan 2 a = 82. 00 (1 + cos 2a) 2 84. sin 3a = 3 sin a 4 sin 3 a. 85. cos 3a = 4 cos 3 a 3 cos a. 3 tan a tan 3 a 86. tan 6 a = ^ o'+TTi 1 o tan 2 a a 87. a > sin a > a -- if a is expressed in radians. a 3 a 3 a & 88. a -^ < sin a < a -vr H ^ if a is expressed in radians. o o O If a is small and expressed in radians: 89. sin a = a approximately, or a 3 90. sin a = a r _ is a better approximation, or o 91. sin a = a - + ~=- is a still better approximation. a 2 92. 1 > cos a > 1 ~~- if a is expressed in radians. a 2 a 2 a 4 93. 1 ;- < cos a < 1 rfc- + r- if a is expressed in radians. I Le 4t 292 MATHEMATICS a* * 94. 1 *r if a is expressed in radians. If a is small and measured in radians: 95. cos a = 1 approximately, or 96. cos a = 1 TTT is a better approximation, or a 2 a 4 97. cos a = 1 pr- + nr is a still better approximation. \ L*. Analytical Geometry 98. y = az + b, slope equation of straight line. 99. x cos a + y sin a = p, normal equation of straight line. 100. x 2 + y 2 = r 2 , equation of circle, center at origin, radius r. 101. (x a) 2 + (y /3) 2 = r 2 , equation of circle, center at the point (a, /3), radius r. 102. y = px 2 , equation of parabola with axis coinciding with F-axis. 103. y = p(x a) 2 , equation of parabola with axis parallel to the F-axis, but a units to its right (to its left if a is negative). x 2 v 2 104. j + rj- = 1, equation of ellipse, axes coinciding with the co- ordinate axes, foci upon X-axis. 105. Area of the ellipse = irab. 2^2 4*2 106. j j-j = 1, equation of hyperbola, axes coinciding with co- ordinate axes. 107. x 2 y- = a 2 , equation of equilateral hyperbola, axes coinciding with coordinate axes. 108. xy = k, equation of equilateral hyperbola referred to asymptotes as axes. 109. Ax + By + Cz = D, equation of plane. 110. x cos a + y cos |8 + z cos y = p, normal equation of plane. 111. x 2 + y 2 + z 2 = r 2 , equation of sphere, center at origin, radius r. 112. (x a) 2 + (y b) 2 + (z c) 2 = r 2 , equation of sphere, cen- ter at the point (a, b, c). 113. cos 2 a + cos 2 ft + cos 2 7 = 1, where a, 0, and 7 are the direction angles of a line. APPENDIX 293 114. cos cos i cos at + cos 0i cos 02 + cos 71 cos 72, where is the angle between two lines whose direction angles are, respectively, a\, 0i, 71, and 2 , 2 , 72. MENSURATION General Triangle 116. Area = %bp. 116. a + + 7 = 180. 117. 5 = a + 0. ^T Right Triangle 118. 6 2 + a 2 = c 2 . 119. p 2 = mn. 120. b 2 = en. Circle r = radius. 121. Circumference = 2-irr. 122. Area = Trr 2 . s = length of arc ab. 123. Area of sector = %rs. Sphere radius = r. 124. Volume = f*-r 3 . 126. Area of surface = 4irr 2 . h = altitude of zone. 126. Area of zone abed = 2*rh. a, b = radii of bases of segment. 127. Volume of segment = 294 MATHEMATICS A.. Right Circular Cone r = radius of base. h = alitude. s = slant height. s = r 2 + h 2 . 128. Volume = \irr*h. 129. Area of convex = wrs. surface Frustum of Right Circular Cone R = radius of lower base. r = radius of upper base h = altitude. s = slant height. s = V/i 2 + (R - r)\ 130. Area of convex surface = irs(R - r). 131. Volume = ~(R^ + Rr + r*). 132. 3(R - r) Anchor Ring of generating r = radius circle. R = mean radius. 133. Volume = 2w*r*R. 134. Area of surface = 4v*rR. APPENDIX Theory of Probability 295 h "" 136. y = e" h * , probability curve. 136. 2 , probable error of mean. TABLE XVI. DECIMAL EQUIVALENTS OF COMMON FRACTIONS 8ths | 16ths 8ths | 16ths 1/8 =0.125 1/4 =0.250 3/8 = 0.375 1/2 =0.500 1/16 = 0.0625 3/16 = 0.1875 5/16 = 0.3125 7/16 = 0.4375 5/8 =0.625 3/4 =0.750 7/8 =0.875 9/16 = 0.5625 11/16 = 0.6875 13/16 = 0.8125 15/16 = 0.9375 TABLE XVII. CONVERTING INCHES INTO FEET Inches 1/8 1/4 | 3/8 1/2 | 5/8 | 3/4 7/8 0.000 0.010 0.021 0.031 0.042 0.052 0.062 0.073 1 0.083 0.094 0.104 0.115 0.125 0.135 0.146 0.156 2 0.167 0.177 0.188 0.198 0.208 0.219 0.229 0.239 3 0.250 0.260 0.271 0.281 0.292 0.302 0.312 0.323 4 0.333 0.344 0.354 0.365 0.375 0.385 0.396 0.406 5 0.417 0.427 0.437 0.448 0.458 0.469 0.479 0.490 6 0.500 0.510 0.521 0.531 0.542 0.552 0.562 0.573 7 0.583 0.594 0.604 0.615 0.625 0.635 0.646 0.656 8 0.667 0.677 0.688 0.698 0.708 0.719 0.729 0.740 9 0.750 0.760 0.771 0.781 0.792 V. 802 0.812 0.823 10 0.833 0.844 0.854 0.865 0.875 0.885 0.896 0.906 11 0.917 0.927 0.937 0.948 0.958 0.969 0.979 0.990 296 MATHEMATICS TABLE XVIII BOAED MEASURE Table giving the number of board feet in timbers of different dimensions; 1 board foot = 144 cubic inches. Size in inches Length in feet 8 10 | 12 | 14 16 18 20 1 X 4 2f 31 4 4! 5i 6 61 1 X 6 4 5 6 7 8 9 10 1 X 8 5| 61 8 9* lOf 12 13| 1 X 10 6f 8| 10 ill 13| 15 16 1 X 12 8 10 ; 12 14 16 18 20 2X4 51 6f 8 * 10| 12 13* 2X6 8 10 12 14 16 18 20 2X8 10f 13* 16 18f 21| 24 26f 2 X 10 is* 18f 20 23i 26| 30 33 2 X 12 16 20 24 28 32 36 40 3X4 8 10 12 14 16 18 20 3X6 12 15 18 21 24 27 30 3X8 16 20 24 28 32 36 40 3 X 10 20 25 30 35 40 45 50 3 X 12 24 30 36 42 48 54 60 4X4 10f 13* 16 18f 21| 24 26f 4X6 16 20 24 28 32 36 40 4X8 21* 26| 32 37| 42| 48 53i 4 X 10 26f 33| 40 46| 53| 60 66f 4 X 12 32 40 48 56 64 72 80 6X6 24 30 36 42 48 54 60 6X8 32 40 48 56 64 72 80 6 X 10 40 50 60 70 80 90 100 6 X 12 48 60 72 84 96 108 120 8X8 42f 53 * 64 74f 85| 96 106f 8 X 10 53 66| 80 93 106f 120 133^ 8 X 12 64 80 96 112 128 144 160 10 X 10 66f 83J 100 116f 133| 150 166! 10 X 12 80 100 120 140 160 180 200 12 X 12 96 120 144 168 192 216 240 APPENDIX 297 TABLE XIX WATER CONTAINED IN 1 FOOT LENGTH OF CIR- CULAR PIPE OF d INCHES INSIDE DIAMETER d Gallons Weight Iba. | d Gallons Weight Ibs. H 0.010 0.085 7 1.999 16.683 1 0.041 0.340 8 2.611 21.790 1M 0.092 0.766 10 4.080 34 . 048 2 0.163 1.362 12 5.875 49 . 028 2H 0.255 2.128 14 7.997 66.733 3 0.367 3.064 16 10.44 87.162 SM 0.500 4.171 18 13.22 110.314 4 0.653 5.448 . 20 16.32 136.190 5 1 . 020 8.512 22 19.75 164.790 6 1.469 12.257 24 23.50 196.114 TABLE XX TABLE OF HEADS OF WATER CORRESPONDING TO GIVEN PRESSURE AND OF PRESSURE CORRESPONDING TO GIVEN HEAD Pressure, Ibs. per sq. in.| Head, feet || Head, feet Pressure, Ibs. per sq. in. 1 2.307 1 0.434 2 4.614 2 0.867 3 6.920 3 1.301 4 9.227 4 1.734 5 11.534 5 2.168 6 13.841 6 2.601 7 16.147 7 3.035 8 18.454 8 3.468 9 20.761 9 3.902 10 23.068 10 4.335 20 46.135 20 8.670 30 69.203 30 13.005 40 92.271 40 17.340 50 115.338 50 21.675 60 138.406 60 26.010 70 161.474 70 30.346 80 184.541 80 34.681 90 207.609 90 39.016 100 230.677 100 43.351 298 MATHEMATICS TABLE XXI. CONVERSION TABLES Lengths 1 centimeter = 0.3937 inch. 1 meter = 39.37 inches. 1 kilometer = 0.62137 mile. 1 inch = 2.540 centimeters. 1 foot = 30.480 centimeters. 1 yard 1 mile (5280 feet) = 0.9144 = 1.609 meter, kilometers. Areas 1 square centimeter 1 square meter 1 square kilometer 1 square inch 1 square foot 1 square mile = 0.1550 = 10.764 = 0.3861 = 6.4516 = 0.0929 = 2.590 square inch, square feet, square mile, square centimeters, square meter, square kilometers. Volume 1 cubic centimeter = 0.0610 cubic inch. 1 cubic meter 1 liter 1 cubic inch = 1.308 = 61.023 = 16.387 cubic yards, cubic inches = 1 . 057 quarts (liquid), cubic centimeters. 1 cubic foot = 7.480 gallons (liquid) = 28.317 liters. 1 cubic yard 1 pint (liquid) 1 gallon = 0.7646 = 28.875 = 231 cubic meter, cubic inches = 0.4732 liter, cubic inches = 3. 785 liters. Weights 1 gram 1 kilogram 1 grain = 15.432 = 2.2046 = 0.0022857 grains = 0.03527 ounce (Av.). pounds (Av.). ounce (Av.) = 0.064799 1 ounce (Av.) 1 pound (Av.) = 437.5 = 0.45359 gram, grains = 28 . 3495 grams, kilogram = 7000 grains. APPENDIX 299 Pressure 1 foot of water column = 0.4335 = 22.419 = 0.8826 = 0.0295 1 inch of mercury column = 1 . 133 = 0.4912 = 0.03342 1 pound per square inch 1 atmosphere = 51.712 = 2.307 = 2.036 = 760 = 33,901 = 29.921 = 14.697 pounds per square inch. millimeters of mercury. inch of mercury. atmosphere. feet of water. pound per square inch. atmosphere. millimeters of mercury. feet of water. inches of mercury. millimeters of mercury. feet of water. inches of mercury. pounds per square inch. Weight and Volume of Water 1 cubic centimeter water 1 pint (liquid) water 1 cubic foot water 1 ounce (Av.) = 1 gram. = 1 . 043 pounds. = 62.428 pounds (Av.). = 28.35 cubic centimeters water. = 1.730 cubic inches water. 1 cubic foot per second 1 acre foot per day Flow of Water = 450 = 0.5042 gallons per minute (approxi- mately), cubic foot per second. 1 part per million 1 grain per gallon Salts in Solution = 0.05842 grain per gallon. = 17.118 parts per million. 300 MATHEMATICS oooooo oooooo oooooo oooooo oooooo WTTPOP1M WTfrPOPIM W^tPOPIM WfPOPlM W^-POPI" O *3" ^1" PO PO PI PI i- 1 M o Oi OlOO OO l*-O O W w *^ PO PO PI M M O OiOO 00 t^O W O\ O* O*. 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OlOO 00 00 t- t- N t o 9 z X O t- t- t-00 00 N d o o n d n d n o t o d O X .s z OOO t PI 000 00 M tt-O N d c -t 'i c x. c 1000 M tO Ol l~ t^OO 00 00 00 d t N Ol t* IO Pt Oi OiOiO O O d o Ooo ion MOO M t t^ O N M M M PI N n o ^c n - oo 10 n 1000 M no Oi PO d o N t o o U Z o o o o o o o H o o o o o o M n 1 10 H o o o o o o M PI n 1 10 o o o o o o HI M n 1 10 09 o o o o o o M n 1 10 o 302 MATHEMATICS 1 O O O O O O 10 *CO CM M en o o o o o o IO * CO N M CO o o o o o o t- o o o o o o 10*COCM M o o o o o o 10*CON H IO o e <3 O CM CO *10O O *CM OOOO f~ o r~oo Ov O O IH IO IO IO * * * r- o w N N CO CO CO CO M Ov t~ IO CO o' ****** IH Olt~lO CO M CO N CM p( Pi N o' CO CO CO CO CM CM Ovl^lO CO M Ov M M IH W M O d o d co o O Z M *r-O *t- OvOO t t^O IO M IH M M O M l* CO O O o *t- M looo IH O Ov OvOO r~* M O M M *" OvOO t~O P1O Ov CO t~n t^o 10 10 * * o 1010100 t-oo 10 Ov cot-M 10 CO N C1 M M O o M o o H t N.Tan 3 Z Ov*Oi*Ov* 00 d 00 o' Ov o Ov o Ov d o M q c oo o CM *r~ Ov CM IO t^ Ov IH CO M CO *O OO O O 00 O CM * t^ IH CO *O I"- Ov Ov IH coiO t^ Ov O M PI *IOO CM *O 00 O Pi t^-t-00 OvO O *o oo o co 10 r- Z o o' -o o o o o o o d o o o o o o M CM CO *1O o o o o o o o H o o o o o o H N CO *1O o o o o o o H CM C0*10 co O O O O O M Pi 10*10 <* o IO IO 000000 O O O O O co IO o o o o o o s 000000 TH 10 o o o o o o IO i CM IOOO M *t- O COO Ov IH * O Ov IH *o 00 O Pi *OOO O M CO10OOO Ov o a Z 00 e o o o o o o 00 o Ov Ov OvOv OvOO o' oo oo oo oo oo r- 6 o d 00 CO q * CO CO H CO N IH M r^o-o** M IH M Ov M IH 1 O *oo co r- PI O M 10 O * Ov "row pTSSo MO O 10 O 10 Ov*Ov*Ov* Ov o H o o o t^ 00 d d 00 o 00 o z g co Z 10 o o O O O O IH w o IH M Pi CM CM Pi >0 o Pi CO CO CO co * o N * q o o o o o o M N CO *1O CO O O O O O n co o o o o o o I- co O O O O O M Pi CO *>O CO CO o o o o o o IH Pi CO*IO * CO o i o o o o o o 10*COCM IH s o o o o o o 00 IO o o o o o o 10 *CO PI M t- o o o o o o 10*COCM M i o o o o o o 1O*CO Pi IH IO IO o O Z OOvOOvO 1O 00 o 00 o 00 00 o Pi N Pi Pi Pi Pi 00 o Ov 00 o w "o O Z M 10 O t~*CO co CM o Ovoo t- M M IH co O oo r~ r- t-* O Ov Ov Ov Ov Ov t*-O 10 * o 10 Ov M *00 CO Ov OvO O O M M CO CO CM M O Ov IO * O COM o O O Pi CO * 1OO t^ oo r^o 10 * co * 00 PI * m H * M o O O> t^ M IO Ov COO r~oo oo oo Ov Ov IO o OvOO 00 00 00 00 O O O IH IH Pi o OvOvO IH Pi CO *OO CO t- M 10 CM Pi CO CO * * vO o' *O t*- Ov M (O Ov co r- M o O o d *00 CO t- M IO