UC-NRLF $B ME3 5MD i ATICS '.,-.-. -,.,.. /#*X IN MEMORIAM FLOR1AN CAJOR1 ^.QIUl~ tsv ^L PRACTICAL MATHEMATICS FOR BEGINNERS PRACTICAL MATHEMATICS FOR BEGINNERS BY FRANK (^ASTLE, M.I.M.E. MECHANICAL LABORATORY, ROYAL COLLEGE OF SCIENCE, SOUTH KENSINGTON ; LECTURER IN MATHEMATICS, PRACTICAL GEOMETRY, MECHANICS, ETC., AT THE MORLEY COLLEGE, LONDON MACMILLAN AND CO., Limited NEW YORK : THE MACMILLAN COMPANY 1905 All rights reserved First Edition 1901. Reprinted 1902. New Edition 1903, 1904, 1905 (twice). CAJORI GLASGOW : PRINTED AT THE UNIVERSITY PRESS BY ROBERT MACl.EHOSE AND CO. LTD. PREFACE. The view that engineers and skilled artizans can be given a mathematical training through the agency of the calculations they are actually called upon to make at their work, steadily gains in popularity. The ordinary method of spending many years upon the formal study of algebra, geometry, trigonometry, and the calculus may be of value in the development of the logical faculty, but it is unsuitable for the practical man, because he has neither the time nor the inclination to study along academic lines. But though Practical Mathematics secures more and more adherents, the subject is still in a tentative stage. The recent revision of the syllabus issued by the Board of Education only two years after its first appearance, is evidence of this. The present volume is designed to help students in classes where the new course of work issued from South Kensington forms the basis of the lessons of the winter session. Such students are supposed to be familiar with the simple rules of arithmetic, including vulgar fractions, hence the present volume commences with the decimal system of notation. The modern contracted methods of calculation, which are so useful in practical problems, are not taught in many schools and they are therefore introduced at an early stage. In the extensive range of subjects included in the present volume care has been taken to avoid all work that partakes of the mere puzzle order, and only those processes of constant practical value have been introduced. Since, in mathematical teaching especially, " example is better than precept," a promi- nent place is given to typical worked out examples. In nearly ail cases these are such as occur very frequently in the work- shop or drawing office. 911250 PREFACE. The order in which the subjects are presented here merely represents that which has been found suitable for ordinary students. Teachers will have no difficulty in taking the different chapters in any order they prefer. Any student working without the aid of a teacher is recommended to skip judiciously during the first reading any part which presents exceptional difficulty to him. So many practical examples of a technical kind, not usually to be found in mathematical books, have been included in this volume that some errors may have crept into the answers, but in view of the careful method of checking results which has been adopted, these will in all probability prove to be small in number. I desire again strongly to emphasize what I have already said in another volume of somewhat similar scope. "Readers familiar with the published works of Prof. Perry, and those who have attended his lectures, will at once perceive how much of the plan of the book is due to his inspiration. But while claiming little originality, the writer has certainly endeavoured to give teachers of the subject the results of a long experience in instructing practical men how to apply the methods of the mathematician to their everyday work." Mr. A. Hall, A.R.C.S., has read through some of the proof sheets, and I am indebted to him for this kindness. I also gratefully acknowledge my obligations to Prof. R. A. Gregory and to Mr. A. T. Simmons, B.Sc, not only for many useful suggestions in the preparation of my MSS., but also for their care and attention in reading through the whole of the proof sheets. F. CASTLE. London, August, 1901. PEEFACE TO NEW EDITION. Several important additions have been made in this edition. Sections dealing with Square Root, Quadratic Equations, and Problems leading to Quadratic Equations, have been added, and, where possible, more exercises have been introduced. Some corrections in the Answers have been made, and I am indebted to many teachers for calling my attention to the necessity for them ; as it is too much to hope that there are no more mistakes in so large a number of figures, I shall be grateful to anyone who may call my attention to other inaccuracies. In its present form the book is not only suitable for students of classes in connection with the Board of Education, but for candidates for the Matriculation examination of the London University under the new regulations ; it will also assist students preparing for the Army and Navy Entrance examina- tions to answer the new type of questions recently introduced into the mathematical papers at these examinations. F. C. London, November, 1902. CONTENTS. CHAPTER I. PAcm Arithmetic : Decimal Fractions. Addition. Subtraction. Multiplication and Division. Contracted Methods of Multiplication and Division, 1 CHAPTER II. Arithmetic : Ratio, Proportion, Percentages, - 16 CHAPTER in. Arithmetic : Powers and Roots, 24 CHAPTER IV. Plane Geometry. - - 31 CHAPTER V. Algebra : Evaluation. Addition. Subtraction, 57 CHAPTER VI. Algebra : Multiplication. Division. Use of Brackets, - 66 CHAPTER VII. Algebra : Factors. Fractions. Surds, .... 73 CHAPTER VIII. Algebra: Simple Equations, ...... 31 CONTENTS. CHAPTER IX. Algebra : Simultaneous Equations and Problems Involving Them, 90 CHAPTER X. Algebra : Ratio, Proportion, and Variation, - - - 100 CHAPTER XI. Algebra : Indices. Approximations, 107 CHAPTER Xn. British and Metric Units of Length, Area, and Volume. Density and Specific Gravity, - - * - - 114 CHAPTER XIII. Logarithms : Multiplication and Division by Logarithms, - 125 CHAPTER XIV. Logarithms : Involution and Evolution by Logarithms, - 133 CHAPTER XV. Slide Rule, 143 CHAPTER XVI. Ratios : Sine, Cosine, and Tangent, 151 CHAPTER XVII. Use of Squared Paper. Equation of a Line, 171 CHAPTER XVIII. Use of Squared Paper : Plotting Functions, - - 189 CHAPTER XIX. Mensuration. Area of Parallelogram. Triangle. Circum- ference of Circle. Area of a Circle, - - 216 CONTENTS. xi PAGE CHAPTER XX. Mensuration : Area of an Irregular Figure. Simpson's Rule. Planimeter, 229 CHAPTER XXI. Mensuration. Volume and Surface of a Prism, Cylinder, Cone, Sphere, and Anchor Ring. Average Cross Section and Volume of an Irregular Solid, 241 CHAPTER XXII. Position of a Point or Line in Space, .... 262 CHAPTER XXHI. Angular Velocity. Scalar and Vector Quantities, - 275 CHAPTER XXIV. Algebra (continued) ; Square Root ; Quadratic Equations ; Arithmetical, Geometrical, and Harmonical Pro- gressions, 290 Mathematical Tables, 311 Examination Questions, 317 Answers, 334 Index, 346 PRACTICAL MATHEMATICS FOR BEGINNERS. In a similar manner, 8 - 073 would be read as eight, point, nought, seven, three. The relative values of the digits to the left and right of the decimal point can be easily understood by tabulating the number 432 1-2345 as, follows : 02 02 -5 P 1 I P 4 C P. P w 02 j5 02 -5 1 B w "T3 P i P o & o P 3 d o 1 p 4 3 2 l 2 3 4 5 Also it will be obvious that in multiplying a decimal by 10 it is only necessary to move the decimal point one place to the right ; in multiplying by 100 to move it two places to the right, and so on. Similarly, the decimal point is moved one place to the left when dividing by 10, and two places when dividing by 100. Addition and subtraction of decimal fractions. When decimal fractions are to be added or subtracted, the rules of simple Arithmetic can be applied. The addition and subtraction of decimal fractions are performed exactly as in the ordinary addition and subtraction of whole numbers ; the only pre- caution necessary to prevent mistakes is- to keep the decimal points under each other. For instance : Ex. 2. Subtract 578 9345 from 702-387. 702-387 578 9345 123-4525 Ex. 1. Add together 36 053. 0079, -00095, 417-0, 85-5803, and -00005. 36-053 0079 00095 4170 85 5S03 00005 538-64220 The decimal points are placed under each other, and the addition and subtraction are carried out as in the familiar methods for whole numbers. MULTIPLICATION OF DECIMALS. EXERCISES. I. Add together 1. 47'001, 2 1101 16, -0401, and 75 8 1983. 2. 23-018706, 1907, '07831, and 1 006785. 3. 4715132, 17-927, 800704, and 20898. 4. 32-98764, 5-0946, -087259, and -56273. 5. 65-095, -63874, 214 89, and -0568. 6. 3720647, 41 62835, 964738, and 876. 7. -7055, 324-88, 7*08213, and -0621. Subtract 8. 15-01853 from 47'06. 9. 708*960403 from 816'021. 10. 28-306703 from 501-28601. 11. 39765496 from 140 3762. 12. 27*9876543 from 126*0123. 13. 13'9463 from 15*10485. 14. 23*872592 from 35 073 16. 15. 22 94756 from 23*002. 16. 11-72013 from 113*408. Multiplication of decimal fractions. The process of the multiplication of decimal fractions is carried out in the same manner as in that of whole numbers. When the product has been obtained, then : The decimal point is inserted in a position such that as many digits are to the right of it as there are digits following the decimal points in the multiplier and the multiplicand added together. Ex. 1. 36-42x4-7. Multiplying 3642 by 47, we obtain the product 171174. As there are two digits following the decimal point in the multiplicand and one digit following the decimal point in the multiplier, we point off three digits from the right of the product, giving as a result 171-174. Ex. 2. -000025 x 005. Here 25x5 = 125. In the multiplicand there are six digits following the decimal point, and in the multiplier three. Hence the product is -000000125. The positions to the right of the decimal point, occupied by the six digits and the three digits referred to, are often spoken of as "decimal places"; thus -000025 would be said to consist of six decimal places. A similar method is used when three or more quantities have 4 PRACTICAL MATHEMATICS FOR BEGINNERS. to be multiplied together, as the following example will make clear : Ex. 3. 2-75 x -275x27-5. The continued product of 275x275x275 will be found to be 20796875. Now, there are two decimal places in the first multiplier, three in the second, and one in the last. This gives a total of six decimal places to be marked off from the right of the product. Hence, the required product is 20 "796875. In addition to applying this rule for determining the number of decimal places in the way shown, the student should mentally verify the work wherever possible. Thus, by inspection, it is seen that *275 is nearly \, and ^ of 27 is 9. This result mul- tiplied by 2 shows that the final product will contain two figures, followed by decimal places. Ex. 4. 730214 x -05031. The product obtained as in previous cases is 3*673706634. In practice, instead of using the nine decimal places in such an answer as this, an approximate result is, as a rule, more valuable than the accurate one. The approximation consists in leaving out, or, as it is called, rejecting decimals, and the result is then said to be true to one, two, three, or more significant figures, depending upon the number of figures which are retained in the result. The rule adopted is as follows : If the rejected figure is greater than 5, or, five followed by other figures, add one to the preceding figure on the left ; if the rejected figure is less than 5, the preced- ing figure remains unaltered. When only one figure is to be rejected andthat figure is 5, it is doubtful whether to increase the last figure or to leave it unaltered. An excellent rule is in such a case to leave the last figure as an even number, thus using this rule we should express 35*15 and 36*85 as 35*2 and 36*8 respectively. In this manner a result may be stated to two, three, four, or more significant figures ; the last figure, although it may not be the actual one obtained in the working, is assumed to be the nearest to the true result. Thus in Example 4, above, the result true to one decimal place is 3*7 ; the rejected figure 7 being greater than 5, the SIGNIFICANT FIGURES. 5 preceding figure 6 is increased by unity. The result, true to two places, is 3*67 ; -the rejected figure 3 is less than 5, and the preceding figure is therefore unaltered. The result true to three and four decimal places would be 3*674 and 3*6737 respec- tively. Applying a rough check, in the way previously mentioned, it is easily seen, that as the multiplier lies between T J(j and y# 3, the result lies between 73 x jfa and 73 x y^. In other words the result lies between 3*65 and 4*38. In simple examples of this kind it may at first sight seem to be unnecessary to use a check, but, if in all cases the result is verified, the common mistakes of sending up, in examinations or on other occasions, results 10, 100, or more times, too great or too small (which the exercise of a little common sense would show to be inaccurate) would be avoided. Significant figures. When, as in Ex. 2 (p. 3), the result is a decimal fraction in which the point is followed by a number of cyphers, the result must include a sufficient number of significant figures to ensure that the result is sufficiently accurate. The term significant figure indicates the first figure to the right of the decimal point which is not a cypher. Thus, if the result of a calculation be "0000026 this includes seven decimal figures ; but an error of 1 in the last figure would mean an error of 1 in 26, or nearly 4 per cent. (p. 21). If the result were 78*6726, then an error of 1 in the last figure would simply denote an error of 1 in 780,000, or, -00013 per cent. Again, in Ex. 2 (p. 3), the result -000000125 must include the three significant figures 125, for an error of 1 in the last figure would mean an error of 1 in 125 or *8 per cent. Some common values. There are many decimal fractions of such frequent occurrence in practice that it may be advisable to commit them and their equivalent vulgar fractions to memory. Thus '125 = ^5 = 4 ; -25 = ^ = 1; -375 = ^ = 1; * = &=}; *-flif=!- It will be noticed that by remembering the first of the above results the other fractions can be obtained by multiplying it by 2, 3, etc., or in each case the result is obtained by mentally dividing the numerator by the denominator. 6 PRACTICAL MATHEMATICS FOR BEGINNERS. Conversion of a vulgar to a decimal fraction. To convert a vulgar fraction to a decimal fraction, reduce the vulgar fraction to its lowest terms and then divide its numerator by its denominator. Ex. 1. 2 9 3- = f = 3-^8= -375; | = 7 4- 8 = -875. Ex.2. T ijj = '00625. Ex. 3. ^ = '432. In many cases it will be found simpler and easier to reduce a fraction to its equivalent decimal if the numerator and denomin- ator are first multiplied by some suitable number. Ex. 4. Reduce ^^ to a decimal. Multiplying by 4 we get t|-o~ = '028. In a similar manner j^T = To"o~ = ^4. Other examples can be worked in like manner. In some cases the figures in the quotient do not stop, and we obtain what are called recurring (they are also called repeating, and sometimes circulating) decimals. Ex. 5. }=-333.... The result of the division is shown by as many threes as we care to write. The notation '3 is used to denote this unending row. Ex. 6. Again = -666 = -6. In each of these, and in similar cases, the equivalent vulgar fractions are obtained by writing 9 instead of 10 in the denominator, thus 3 = |- = ^, etc. In a similar manner y= '142857, and these figures again recur over and over again as the division proceeds, hence j= 142857. When it is necessary to add or subtract recurring decimals, as many of the recurring figures as are necessary for the purpose in hand are written, and the addition or subtraction performed in the usual manner. With a little practice the student soon becomes familiar with the more common recurring decimals. Any decimal fraction, such as 3, 142857 in which all the figures recur is called a pure recurring decimal ; the equivalent vulgar fraction is obtained by writing for a numerator the figures that recur, and for the denominator as many nines as there are figures in the recurring decimal. DECIMALS OF CONCRETE QUANTITIES. 7 When the decimal point is followed by some figures which do not recur and also by some which do recur, the fraction is called a mixed recurring decimal, and the equivalent fraction is obtained by subtracting the non-recurring figures from all the figures to obtain the numerator, and by writing as many nines as there are recurring figures, followed by as many cyphers as there are non-recurring figures for the denominator. Ex. 7 Express as a vulgar fraction the recurring decimal *123. Here there are two recurring figures and one not recurring, . .lOQ_12 3-l_122_ 61 .. LZ6 -g- IJ - 9IO-49 5" 7?V 8 Wfltf _32fi57-32_32825_145 jX. O. HZKiDi qq yo "9" 9 "9 _ 4~4 T' Decimals of concrete quantities. It is often necessary to express a given quantity as a fraction of another given quantity of the same kind. Thus, in the case of 1. 15s., it is obvious that 15s. = $ of 20 shillings, and 1. 15s. may be written lf ; or, f = '75, we may also write 1. 15s. as 1'75. Ex. 1. To reduce lOd. to the decimal of a pound. As there are 240 pence in 1, .". required fraction is -^To" = T = '04167 .... Ex. 2. Express 7s. 6|d. as the decimal of a pound. Here 7s. 6d.=90'5d. . 90 5 ' 240 " 611 ' And 1. 7s. 6|d. may be written 1\377. Ex. 3. Express 6 days 8 hours as the decimal of a week. As there are 24 hours in a day, 6 days 8 hours = 6^ = 6 J days, . .*. 6 days 8 hours = -=- = -90476i week. Ex. 4. Reduce 5d. to the decimal of Is. T 5 2 = -416s. Ex. 5. Express in furlongs and poles the value of '325 miles. Here, multiplying by 8, the number of furlongs in a mile, '325 we obtain 2*6, and multiplying the decimal -6 by 40 (the 8 number of poles in a furlong) we get 24 poles. 2*600 Hence '325 mile = 2 fur. 24 po. 40 24 8 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 6. Reduce 9 inches to the decimal of a foot. There are 12 in. in a foot. Hence the question is to reduce y^- to a decimal. .*. 9 in. = -75 ft. Given a decimal of a quantity, its value can be obtained by the converse operation to that described. Ex. 7. Find the value of *329 of 1. The process is as follows : First multiplying by 20 we obtain the product 6580, and marking off three decimals -329 we get the value 6 '580 shillings. In a similar manner 20 multiplying by 12 and 4 as shown, we obtain the value of 6*580 "329 of 1, which is read as 6 shillings 6 pence 3 farthings 12 and -84 of a farthing. 6 960 The result could be obtained also by multiplying \329 4 by 240, the number of pence in 1, giving 78*96d. and 3*840 afterwards expressing in shillings, etc. Ex. 8. Find the number of feet and inches in '75 yard. Here '75 x 3=2*25 feet, and -25 ft. = *25 x 12 in. = 3 in. .*. *75 yard = 2 f t. 3 in. Contracted methods. The results of all measurements are at best only an approximation to the truth. Their accuracy depends upon the mode of measurement, and also, to some extent, on the quantity measured. All that is requisite is to be sure that the magnitude of the error is small compared with the quantity measured. It is clear that in a dimension involving several feet and inches, an error of a fraction of an inch would probably be quite unimportant. But such an error would obviously not be allow- able in a small dimension not itself exceeding a fraction of an inch. By means of instruments such as verniers, screw gauges, etc., measurements may be made with some approach to accuracy. But these, or any scientific appliances, rarely give data correct beyond three or four decimal places. Thus, if the diameter of a circle has been measured to *001 inch, then, since no result can be more exact than the data, there is no gain in calculating the circumference of such a circle to more than three decimal CONTRACTED MULTIPLICATION. 9 places. Hence 3'1416 is a better value to use for the ratio of the diameter of a circle to its circumference than 3*14159. In such cases, too, the practical contracted methods of calculation are the best. In a similar manner when areas and volumes are obtained by the multiplication of linear measured distances the arithmetical accuracy to any desired extent may be ensured by extending the number of significant figures in the result, but it should be remembered that the accuracy of any result does not depend on the number of significant figures to which the result is cal- culated, but on the accuracy with which the measurements or observations are made. In any result obtained the last significant figure may not be accurate, but the figure preceding should be as accurate as possible. It is therefore advisable to carry the result to one place more than is required in the result. It is evident that loss of time will be experienced if we multiply together two numbers in each of which several decimal figures occur, and after the product is obtained reject several decimals. Especially is this the case in practical questions in which the result is only required to be true to two or more significant figures. In all such cases what is known as Contracted Multiplication may be used. Contracted multiplication. In this method the multiplication by the highest figure of the multiplier is first performed. By this means the first partial product obtained is the most important one. The method can be shown, and best understood by an example. Ex. 1. Multiply -006914 by 8*652. The product of the two numbers can of course be found by the ordinary methods; and to compare the two methods, "ordinary" and " contracted," the product is obtained by both processes : Ordinary Method. Contracted Method. 6914 6914 8652 2568 13828 55312 34570 4148^ 41484 346^ 55312 14$$ 059819928 059820 10 PRACTICAL MATHEMATICS FOR BEGINNERS. The ordinary method will be easily made out. In the contracted method the figures in the multiplier may be reversed, and the process continued as follows : Multiply first by 8, so obtaining 55312 ; next by 6 this step we will follow in detail 6 x 4 = 24, the 4 need not be written down (but if written it is cancelled as indicated), and the 2 is carried on. Continuing, 6x1=6, and adding on 2 gives 8. Next, 6x9 = 54, the 4 is entered ; and 6x6 gives 36, this with the 5 from the preceding figure gives 41, hence the four figures are 4148. In the next line, multiplying by 5, we can obtain the two figures and 7, but as these are not required unless there is some number to be carried, it is only necessary to obtain 69 x 5, and write down the product 345, add 1 for the figure rejected (because it is greater than 5) thus making 346. Finally, as 2x9 will give 18, we have to carry 1, and therefore we obtain 2x6 = 12, together with the one carried from the preceding figure which gives 13, add 1 for the figure (8) rejected, which gives 14. Adding all these partial products together we obtain the final product required. Thus, in the second row one figure is rejected, in the next row two figures, and in the last row three figures are left out. It may be noticed again, with advantage, that when the rejected figure is 5 or greater, the preceding figure is increased by 1, also that the last figure of the product is not trustworthy. Having noted (or cancelled) the rejected figures, as will be seen from the example, the decimal point is inserted as in the ordinary method, i.e. marking off in the product as many decimal places as there are in the multiplier and multi- plicand together. Though the multiplier is very often reversed, this is not necessary, except to avoid mistakes. The multiplier may be written in the usual way, and the work will then proceed from the left hand figure of the multiplier, i.e. the work is commenced by multiplying by 8 and not by 2. Ex. 2. The circumference of a circle is obtained by multiplying the diameter of the circle by 3*1416. Find the circumference of a circle 13-25 inches diameter. CONTRACTED MULTIPLICATION. 11 Here, we require the product of 13*25 and 3 1416. .-. 13-25 61413 3975 132$ l$ft 41-63 Hence the required circumference is 41 '63. EXERCISES. II. 1. Multiply 6-234 by '05473, leaving out all unnecessary figures in the work. 2. 4-326 by '003457. 3. 8 09325 by 62-0091. 4. -72465 by '04306. 5. 5 '80446 by '10765. 6. 21 -0021 by '0098765. 7. 24 9735 by 30-307. 8. 73001 by 7'30121. 9. '053076 by 98 '0035. 10. 3-12105 by 905008. 11. '0435075 by 3*40604. 12. 76-035 by '0580079. 13. 5'61023 by '597001. 14. 59-6159 by 30807. 15. -020476 by 2-406. 16. 43-7246 by "24805. 17. -01785 by 87 "29. 18. 40-637 by 028403. 19. 2 030758 by 36 409. 20. 82 5604 by 08425. 21. 6 04 by 35. 22. 8-0327 by -00698. 23. 390-086 by -00598. 24. 4-327615 by -003248. 25. Add together five-sevenths, three-sixteenths, and eleven- fourteenths of a cwt. , and express the sum in lbs. 26. Express 9s. 4|d. as the decimal of 1. 7s. 27. Subtract '035 of a guinea from 1 '427 of a shilling. 28. Subtract 3 '062 of an hour from 1'5347 of a day. 29. Add together 0029 of a ton and '273 cwts. 30. Reduce '87525 of a mile to feet. 31. Find the sum of 2 35 of 2s. Id. and 0*03 of 6. 3s. 9d. 32. Add together ^ of a guinea, -|-g- of a half-crown, I-gnj shilling, and ^ of a penny, and reduce the whole to the decimal fraction of a pound. 33. Express 3s. 3d. as the decimal of 10s. 34. Add together -| of 7s. 6d., 2*07 of 1. 8s. 2d., and f of '0671 of 16s. 8d. Express the answer in pence. 12 PRACTICAL MATHEMATICS FOR BEGINNERS. Division of Decimal Fractions. The division of one quantity by another when decimals enter into the operation, is performed exactly as in the case of whole numbers. The process can be best explained by an example as follows : Ex. 1. Divide '7 by -176. This may be described as finding a number, which, when multi- plied by "176, gives a product equal to '7. Though decimals may be divided as in the case of whole numbers, care is necessary in marking off the decimal point. In the present, and in all simple cases, the position of the decimal point is evident on inspection. Practically, it is often convenient to multiply both terms by 10, or some multiple of 10 100, etc. and so obtain at once, without error, the position of the unit's figure, and hence of the decimal point. Thus, in the above example, multiplying 1*76) 7 00 (3*97 both terms by 10, we have to divide 7 by 5 28 1*76, and it is evident that the number 1 720 required lies between 3 and 4. This deter- * 584 mines the position of the unit's figure. As 1360 7'0 is unaltered by adding any number of \22>2 ciphers to the right, we add two for the 1280 purpose of the division. Multiplying 1 # 76 by 3 we obtain 5*28, which, subtracted from 7 "00, gives a remainder 1*72; to this we affix a cipher and carry on the division as far as necessary ; when this is done, we find *7-=- "176 = 3 9772727. It will be seen that the ordinary method of performing division necessarily requires considerable space, especially when there are several figures in the quotient. Italian Method. Another method, referred to as the Italian method, in which only the results of the several subtractions are written down, is often used ; the method of procedure is as follows : Note, as before, that 1 '76 will divide into 7 ; then, since 3x6 = 18, the 8 is not written down T76 ) 7 00 ( 3 '97 but is instead mentally subtracted from 10, 1~720 leaving 2. Next 3x7 = 21 and 1 carried ~1S60 makes 22 ; the 2 is again not written down, 1 280 but instead, after the addition of unity (from the multiplication of 6 by 3), we say 3 from 10 = 7. In a similar manner the remaining figure is obtained; the next row of figures is arrived at by a like method and so on. CONTRACTED DIVISION. 13 Comparing the two examples it will be seen, that as at each step of the work one line of figures is dispensed with, the working takes up far less room than is the case in the ordinary method. It is obviously bad in principle to use more figures than are essential for the work in hand ; these are not only unnecessary, but give additional trouble, and also increase the risk of making mistakes. In many cases, students are found to work with ten or more decimal figures, when, owing to errors of observation, or measurement, or to slightly incorrect data, even the first decimal place may not be trustworthy. It is, of course, in- advisable to add an error of arithmetic to an uncertainty of measurement or data, but even a slight error is preferable to working out ten, or fifteen, places of decimals to a practical question, and when the result is arrived at, to proceed to reject the greater part of the figures obtained, leaving only two or three decimal places. To avoid this, what is known as con- tracted division is often adopted. Contracted Division. It is assumed that the student is familiar with the ordinary method of obtaining the quotient in the case of division. The long process of division can, how- ever, also be advantageously contracted. The method of doing this will be clear from the following worked example. Ex. 1. Divide -03168 by 4 '208. We shall work this example by the contracted method alone. To begin with, the number 7 is obtained by the usual process of division. By multiplying the divisor by 7 the product 29456 is arrived at. When this is subtracted from 31680 the remainder 2224 is left. It is seen that if we drop or cancel the 8 from the divisor 4208, thus obtaining 420, it can be divided into the remainder 2224, five times. 4208 ) 31680 ( 7529 In multiplying by five we take account of the 29456 8, thus, as 5x8 is 40, the is not entered 2224 but the 4 is carried. Proceeding we have 2104 0x5=0, and adding 4, we see this is the 120 'figure to be entered. Now proceed to the 84 next and the following figures, obtaining in 36 the usual way 2104 ; subtract this from 36 2224, and the remainder 120 is obtained. Proceeding in like manner with the multiplier 2, we obtain 84, which, subtracted from 120 leaves 36, and our last figure in the 14 PRACTICAL MATHEMATICS FOR BEGINNERS. quotient is 9. By the method described on p. 12 the answer is written -007529. As the product of the divisor and quotient, when there is no remainder, is equal to the dividend, it follows that the dividend may be multiplied by any number if the quotient is divided by the same number. Thus, in the last example, if 03168 is multiplied by 1000, then, 31 '68 divided by 4'208 gives the result 7'259. Dividing this by 1000 we obtain the answer 007259. This process of multiplying and dividing by 1000 simply means shifting the decimal point three places to the right in the divisor, and three places to the left in the quotient. The above example shows that the method of contracted division consists in leaving out or, as it is called, rejecting a figure at each operation. Any number which would be added on to the next figure by the multiplication of the rejected figure is carried forward in the usual way. To avoid mistakes it may be convenient either to draw a line through each rejected figure of the divisor, or to place a dot under it. Ex. 2. When the circumference of a circle is given, the diameter is obtained by dividing the circumference by 3-1416. The circumference of a circle is 41 "63 inches ; find the diameter of the circle. 3-1416) 41 630 (13-25 31-416 10214 9-424 790 628 162 157 5 EXERCISES. III. Divide the following numbers, leaving out all unnecessary figures in the work. 1. -43524 by 2197962. 2. -00729 by -2735. 3. 24-495 by -0426. 4. 13195 by 4 '375. 5. 33-511 by '0713. 6. -414 by 34 '5. CONTRACTED DIVISION. 15 7. 32-121 by 498. 8. 166*648 by -000563. 9. 1-6023 by 294. 10. 7'3by584. 11. -292262 by 32 7648. 12. Find the value of 09735 -f 5*617 to four significant figures. 13. How many lengths of '0375 of a foot are contained in 31 '7297 feet? 14. If sound travels at the rate of 1125 feet per second, in what time would the report of a gun be heard when fired at a distance of 1-375 miles? 15. Find the value to four significant figures of 6 234 x '05473, also -09735^-5-617. Divide 16. 19-305 by '65. 17. 325 '46 by 0187. 18. 172 9 by 0'142. Find the value of 19 i of 8 ' 236 20 12-4+ -064- -06 6 ' T 9 21 Now put the double of 3, that is 6, on the left 784 ) 3136 of the number 652, and by trial find that 6 will 3136 divide into 65 nine times. Put the 9 with the first figure of the square root on the right, and also on the left with the 6, and multiply 69 by 9 obtaining 621, which when subtracted from 652 gives a remainder of 31. Bring down the next two figures, thus obtaining 3136. Double the number 39, the part of the root already found, and put the result 78 on the left, as shown. By trial, find that 78 will divide into 313 four times. Put the 4 on the right with the other numbers, 39, of the square root which is being obtained, and also with the 78, making the number 784 on the left ; this last number multiplied by 4, the figure just added, gives 3136, which subtracted, leaves no remainder. Hence 394 is the square root required. If we proceed to extract the square root of 394 we obtain 19*85, and this is the fourth root of 155236; .-. 155236*= >/l55236 = 19-85. The student should always begin to point at the unit's place, whether the given number consists of integers, or decimals, or both. Ex. 2. Find the square root of 1481-4801. rru u 4. *u >*.> i 1481-4801 (38-49 The pointing begins at the unit s place, q v and every alternate figure to the right and , r left of the unit's place is marked as indicated ' * in the adjoining example. As there are two , - dots to the left of the unit's place, the square ' " ' oAr root consists of the whole number 38 and ' the decimal ; the working is exactly the same ' "" / "xH9, . .. . % * 69201 as in the previous example. It should be observed that, to obtain the square root of a decimal fraction, the pointing should commence from the second figure of the decimal place. 28 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 3. Find the square root of '9216. 9216 ( -96 81 186)1116 1116 The method adopted will be evident from the working shown. As examples, obtain the square roots of the following fre- quently occurring numbers ; these should be worked out care- fully, and the first two at least committed to memory. x/2=l'414..., x/3=1732... , x/5=2'236..., x/6=2*449.... The square root of each of these numbers is an unending decimal. Thus, the square root of 3 can be carried to any number of decimal places, but the operation will not terminate. Such a square root is often called a surd, or an incommensurable number. In any practical calculation in which surds occur, the value is usually not required to more than two or three decimal places. If a number can be easily separated into factors, the square root can be obtained more readily. The method adopted is to try in succession if the number is divisible by 4, 9, 16, and other numbers of which the square roots are known. Ex. 4. To find the square root of 1296. 1296 = 4x324=4x4x81, .*. Vl296 = \/l6x81 = 4x9 = 36. A similar method may be employed in the case of numbers the roots of which cannot be expressed as whole numbers. Ex. 5. Vl28=\/64x2 =8\/2; and remembering that the \/2 is 1*414 approximately, the value 8 x 1-414 = 11 '312 can be found. Ex. 6. \/243=\/8T>r3 =9^3. SQUARE ROOT. 29 In many cases where a surd quantity occurs in the denominator of a fraction, it will be found advisable, before proceeding to find the numerical value of the fraction, to transfer the surd from the denominator to the numerator. This is readily effected by multiplication. Thus, if as a result to a given question we obtain the fraction 100 r=, we may proceed to divide the numerator by cr. p w a a o 5 <> ^ u >=> _ similar manner, if the drawing were made so that every length of 3 inches on the drawing represented an actual length of 12 inches, the scale would be said to be \. The fraction of ^ or , etc., is called the representative fraction of the scale. Hence, Representative fraction of a scale number of units in any line on the drawing number of units the line represents The term representative fraction is not always used, but, more shortly, the drawing is said to be made to a scale of \ or ^. When dimensions are inserted on a drawing a convenient notation is to use one dash ' to denote feet and two dashes * for inches, thus a dimension of 1 ft. 3 in. could be written 1' 3". Scales of boxwood or ivory are readily obtainable ; the former are cheaper than the latter, and the student should possess at least one good boxwood scale about 12 inches long. What is called an open divided scale will be found most useful. These can be obtained with the following scales : 1", J", J", and J" on one side, all divided in eighths. The same scales in tenths are found on the obverse side. Such a scale is shown in Fig. 10. These scales are divided up to the edge, which is made thin, as shown in the sections a and b, and so allows dimensions to be marked off direct from the scale with a fine-pointed pencil or pricker. It is not advisable to use compasses or dividers, if it can be avoided, when transferring dimensions from scales. The frequent use of dividers soon wears away the divisions on the scale, and renders them useless for accurate measurements. *S^^^S2*"* ^^m^^^s DIVISION OF LINES. Division of a line into equal parts Given any line AB (Fig. 11), to divide it into a number of equal parts is comparatively an easy task when an even number of parts are given, such as 2, 4, 8, etc. In such a case the line would be bisected by using the dividers, each part so obtained again bisected, etc. When an odd number of parts are required, such as 5, etc., a length may be taken representing about one-fifth of AB (Fig. 11). This, on trial, may prove to be than the necessary length. By A Fig 11. Division of a line into five equal parts. slightly longer or shorter alteration of the dimension in the required direction, and by repeated trials, a length is ultimately found which is exactly one-fifth. Much unnecessary time and labour may be spent in this way. A better method is to set off a line AC (Fig. 11) at any con- venient angle to AB, and to mark off" any five equal lengths along AG from A to 5, and join 5 to B. If lines are drawn through the successive points 1, 2, 3, 4, parallel to the line 5B, then AB will be divided into the required number of equal parts. It will be obvious that the process of marking off a given number of equal distances along the line A may be carried out by using the edge of a strip of squared paper, or a piece of tracing paper or celluloid on which a number of D parallel lines have been drawn. Conversely, given a line denoting a num- ber of units, then the length of the unit adopted can be found. Division of a line into three segments in a given proportion (say 1*5, 2*5, 3). Draw a line AB making any convenient acute angle with AB. Set off 7 units (equal to the sum of three ACE B Pig. 12. Division of a line into three segments in a given proportion. 40 PRACTICAL MATHEMATICS FOR BEGINNERS. segments) along AD. Join point 7 to B. Through points 1*5 and 25 draw lines parallel to IB. AB is thus divided at C and E in the required proportion. Ex. 1. To cut off a fraction of a given line. To cut off a fraction say f of AB (Fig. 12). Draw AD at an acute angle to AB, and along AD mark off 7 equal divisions. Join 7 to 5. Through 4 draw 4# parallel to IB; then ^4#= f J.B. Construction of scales. The cheaper kinds of scales are often very inaccurate, those which are machine divided of the type shown in Fig. 10 are expensive, and it sometimes becomes necessary to substitute some simple form which can be readily made for oneself. For this purpose good cartridge paper, thin cardboard, or thin celluloid may be used. If the latter is em- ployed the lines may be scratched on the surface by using a small needle mounted in the end of a penholder and projecting about a J in. or ^ in. 9 8 7 6 4 3 2 \ II 1 1 1 1 1 . 2 \ A B Fig. 13. Construction of a simple scale. To make a scale, two lines are drawn about a j in. or h in. apart. A number of divisions are then marked off along A B, Fig. 13, each one inch in length. The end division is sub- divided into 10 equal parts. The lines denoting inches are made slightly longer than those indicating half inches, and these in turn longer than the remaining divisions. Finally, numbers are inserted as shown, the larger divisions being numbered from left to right, the smaller from right to left. When this notation is adopted any dimension such as 1*7" can be estimated without risk of error by counting. Other similar scales may be made as required. In the preceding scale, although a dimension such as 1 # 7" involving only one decimal place can be made accurately, yet to obtain a dimension such as 1 "78 it would be necessary to further divide mentally the space between the 7th and 8th division into 10 equal parts, and to estimate as nearly as possible a length PROPORTION. 41 equal to 8 of such parts. Such a method is a mere approxima- tion. When distances involving two or more decimal places have to be estimated other measuring instruments, such as diagonal scales, verniers, screw-gauges, etc., are used. Diagonal scale. A diagonal scale of boxwood or ivory is usually supplied with sets of mathematical instruments. They can be purchased separately at a small outlay. To make such a E D C o B Pig. 14. Diagonal scale. scale, set off AB (Fig. 14) equal to 1 inch, draw BC perpendicular to A B, and divide AB and BC each into ten equal parts ; join the point B to the first division on CE and draw the remaining lines parallel to it as in the figure. A dimension 1*78 is the distance from the point b to a, the point of intersection of a sloping line through 7 and the horizontal line through 8. Proportion. It has been shown (p. 19) that - C when four quantities are proportional we may write them as A : B= C : D. Given A, B, and C, we proceed to find the fourth proportional geo- metrically as follows : Draw two lines at any convenient angle to each other. In Fig. 15 the lines are at right angles. -----D=Ans--- Fia. 15. Proportion. Set off a distance oa = A along the vertical line, and a distance ob~B along the horizontal 42 PRACTICAL MATHEMATICS FOR BEGINNERS. line. Join a to b. Set off the third quantity G along the vertical line, making oc = G ; draw a line cc? parallel to ab, and meeting 06 produced at d. Then od=D is the fourth propor- tional, or answer required. Denoting the fourth proportional by x, A : B= G : x ; or, multi- plying extremes and means, A xx = Bx G ; BxG :. x= 3 . In many cases the value of a complicated fraction can be found by proportion by a similar geometrical method. Ex. 1 . Find the value of i 3 1^ Arvs Pig. 16. Simplification of a fraction. In Fig. 16, on a convenient scale, ob is made = If, oa=lf, and oc = 2|-. Join a to b and through c draw cd parallel to ab, meeting ob pro- duced at d. Then od is the required result. When measured, od will be found to be 3*7 units. Mean proportional. given lines AB and AC. To find a mean proportional to two Draw the two lines, as in Fig. 1 7, so that they form together one line A C. On A C describe a semicircle, and at B draw a perpendicular BD meeting the semicircle in D. Then BD is the mean proportional required. If the line AB to a given scale represent a certain num- ber of units and BG one unit on the same scale, then BD is the square root of A B. Square root. The square root of a number is often required in practical calculations, and may be calculated as already Fig. 17. Mean proportional PLANE FIGURES. 43 Fig. 18. Square root. explained on p. 27, or obtained by means of the slide rule (p. 149), or by graphical construction, as follows : Ex. 1. Find the square root of 4f- Using any convenient scale, mark off ab = 4%, and be = unity on same scale (Fig. 18). On ac describe a semicircle, and at b draw bd perpendicular to ac, and meeting the semicircle ind. Then bd is the square root required. The construction shown in Fig. 18 is the same as that of finding a Geometrical Mean or the mean proportional of the two numbers 4f and unity. Ex. 2. To obtain the fourth root of 4^ or J4' Having obtained, as in the previous example, the square root bd, make be (Fig. 19) equal to bd. This is effected by using b as centre, bd as radius, and describ- ing the arc be, meeting ac in e. On ec describe a semicircle. Let/ be the point of intersection of bd with the semicircle. Then bf is the fourth root required. In a similar manner the 8th, 16th, etc., roots, can be obtained. Plane figures. A triangle is a figure enclosed by three as ^5, BC, an CA. These lines form at their points of in- tersection three angles. The three lines are called the sides of the triangle. The angle formed at the point of intersection of the sides AB and BC may be called the angle ABC, but more simply the angle B. The two remaining angles are called A and C. Any one of its three angular points A, B, or C (Fig. 20) may be looked upon as the vertex and the opposite side is then called the base of e b Fig. 19. Fourth root. A triangle. 44 PRACTICAL MATHEMATICS FOR BEGINNERS. tjie triangle. The altitude of a triangle is the perpendicular distance of the vertex from the base. Equilateral triangle. When the three sides of a triangle are equal, the triangle is an equilateral triangle ; the angles of the triangle are equal, each being 60. Isosceles triangle. When two sides of a triangle are equal, the triangle is an isosceles triangle. A right-angled triangle (Fig. 21) is a triangle one angle (C) of which is a right angle ; the side (AB) opposite the right angle is called the hypotenuse. A parallelogram is a four-sided figure, the opposite sides of which are equal and parallel. A rectangle is a parallelogram having each of its angles a right angle, or, in other words, each side is not only equal in length to the opposite side, but is also perpendicular to the two adjacent sides. a c Fig. 21. A right-angled triangle. Fig. 22. A parallelogram. Fig. 23. A rectangle. A square is a parallelogram which has all its sides equal, and all its angles right angles. Fig. 24. A square. Rhombus. A rhombus is a parallelo- gram in which all the sides are equal but the angles are not right angles. Fig. 25. A rhombus. The altitude of a parallelogram is the perpendicular distance between one of the sides assumed as a base and the opposite side. The circle. The curved line ABED (Fig. 26) which encloses a circle is called the circumference. Any straight line such as 0A, OB, etc., drawn from the centre to the circumference is a radius, and a line such as AD passing through the centre and GEOMETRICAL TRUTHS SHOWN BY INSTRUMENTS. 45 terminated by the circumference is a diameter of the circle. A poi'tion of a circle as OBEC, cut off by two radii, is a sector of a circle. A line such as BC which does not pass through the centre is a chord, and the portion of the circle BEC cut off by it is called a segment of a circle. A line touching the circle is a tangent, the line joining the point of contact to the centre is at right angles to the tangent. Perimeter. The term perimeter of a figure is used to denote the sum of the lengths of all its Pia. 26. A circle. thus the perimeter of a parallelogram is the sum of the lengths of its four sides. Geometrical truths illustrated by means of instruments. The following important geometrical truths may be verified by means of drawing instruments. Lengths are measured by a scale ; angles by a protractor or a scale of chords ; any necessary calculations are made arithmetically, and tracing paper may be used to show the equality of angles. Parallel lines. When two parallel straight lines are crossed by a third straight line, the alternate angles are equal. Draw any two parallel F lines (Fig. 27), and a third line EF crossing them. Show, by tracing the angles on a sheet of paper, or by measuring the angles, that the alternate pairs of . angles marked x and are in each case equal to each other. Also show by measurement that the four angles formed by the intersection of the third line with each of the parallel lines are equal to 360. Parallelogram. Draw a parallelogram ABCD (Fig. 28), the longer sides being 3", and the shorter 2" long. Verify by Parallel lines. 46 PRACTICAL MATHEMATICS FOR BEGINNERS. measurement that* the angle at B is equal to the angle at Z), and the angle at A equal to the angle at C. Join the points A and C and B and D. The B lines AC and BD are called the diagonals of the parallelogram. 0^ D C Fig. 28. Opposite angles of a parallelo- gram are equal. and the altitude of each is the gram is double that of the triangle. B Fig. Verify that the two diagonals are bisected at 0, their point of intersection. /. A0 = 0C, and D0= OB. If a triangle and a parallelogram are on the same or equal bases same, the area of the parallelo- Draw any parallelogram A BCD; join A to C C (Fig. 29). Cut the paper along A C and make the triangle ABC coincide with the triangle ADC. Or, using a piece of tracing paper, trace care- fully the triangle A BC, then place it on ABC with B at D. Note that the lines forming the triangles are coincident. Triangles. The angles at the base of an isoceles triangle are equal. Draw to scale an isoceles triangle A BC, that is, make the side AB = side A C (Fig. 30). Show by measuring that the angle at C is equal to the angle at B. Also prove the equality ^ by cutting out the angle tf C & 9 C a t C and placing it on B. ~^\Tgletre^r ' ^ iS SCeleS S ma 7 be effec t ed by marking off a length not greater than half BC, and drawing gf perpendicular to Fig. 29. The area of a parallelogram is double that of a triangle on the same base and the same altitude. GEOMETRICAL TRUTHS SHOWN BY INSTRUMENTS. 47 BC, meeting AG inf. If Gfg be placed as shown in Fig. 30 with the angle G on and Gg coinciding with Bg', then the line Gf will be found to coincide with the line BA. Or, using a piece of tracing paper, trace the triangle, fold the paper and see that the angles are equal. If a line be drawn at right angles to the base of a triangle, and passing through the vertex it will bisect the base. Draw a tri- angle ABC with AD at right angles to the base. Make a tracing of the triangle AGD, then place it on the triangle ABD, with the point G coincident with j5, all the other lines of the triangles can be made to coincide. Hence verify that the triangles AGD and ABD are equal, and D is the middle point of BG Equilateral triangle. Make a triangle having all its three sides equal, (a) Measure by means of a protractor any one of the three angles and write down its magnitude ; (b) carefully trace one of the angles on a piece of tracing paper, and placing the paper on each of the other two angles, verify that all the angles are equal and that the sum of the three angles is 1 80. Each side may be made equal to 3" ; draw a line perpendicular to the base and passing through the vertex of the triangle. Verify by measurement that the line so drawn bisects the base, and also the vertical angle at A. The three angles of a triangle are together equal to 180 Draw the triangle ABG (Fig. 31), making the two sides AG and BG respectively equal to 4 and 3 units of length. Join AB, measure by the protractor the ^^-^USr^afto ^ angles at A, B, and G ; add the values of the angles measured in degrees together, and ascertain if the angles A, B and G make 180. It will be found that =53 7', ^ = 36 53', and G=90. Complement of an angle. When the sum of two angles A, B, is equal to 90, one angle is said to be the complement of the other. That is, A is the complement of B, or, B is the complement of A. 48 PRACTICAL MATHEMATICS FOR BEGINNERS. Right angled triangle. The side opposite the right angle is called the hypotenuse and in a right-angled triangle the follow- ing relation always holds : The square on the hypotenuse is equal to the sum of the squares on the other two sides. Thus in Fig. 31, 3 2 + 4 2 = 25 = 5 2 . As the two lines BG and GA in each case represent a certain number of units of length we can write the above statement simply as ab=Jbc*+ca*. Various values for BG and GA should be taken, and in each case it will be found that the relation holds good. This property of a triangle, that when the three sides are proportional to the numbers 3, 4, and 5, the angle opposite the greater side is a right angle, is largely used by builders and others to obtain one line at right angles to another ; instead of 3, 4, and 5, any multiples of these numbers such as 6, 8, 10, etc., may be used ; also it is obvious that the unit used is not necessarily either an inch or foot, it may if necessary be a yard, or a chain, etc. The greater angle of every triangle is subtended by the greater side. Draw a triangle having its sides 9, 7, and 3 units, measure the angles with a protractor, verify that the sum of the three angles is equal to 180, and care- fully observe that the greatest angle is opposite the greatest side 9, and the smallest angle is opposite the smallest side 3. If a line be drawn parallel to the base of a triangle, cutting the side or sides produced, the segments of the sides are proportional. Draw a triangle ABG, and a line DE B C parallel to the base (Fig. 32) ; show by Fig. 32. If a line is drawn measurement that parallel to the base of a triangle, the segments of the J$(J J) J sides are proportional. at* ~T~ri '> An AD :. if DE is one-half BG, then AD is one-half of AB. Similar figures. Similar figures may be defined as exactly SIMILAR FIGURES. 49 alike in form or shape although of different size; or, perhaps better, as figures having the same shape but drawn to different scales. Two triangles are similar when the three angles of one are respectively equal to the three angles of the other. The student Similar triangles. is already familiar with similar triangles in the form of set squares which may be obtained of various sizes. Obviously all the three angles of a 45 or 60 set square are the same whatever be the lengths of the sides. As a simple illustration suppose that one side of a 45 set square be twice that of a corresponding side in another 45 set square, then the remaining sides of the second square are each twice those of the former. Thus if EF (Fig. 33) be twice A C, then it follows that ED is twice AB and DF is twice BC. It also follows that if one or more lines be drawn parallel to one side of a triangle the two sides are divided in the same proportion. Thus if in Fig. 34 BC be drawn parallel to the base DF, thenAB:AD=AC:AF= BC.DF. Thisissufficiently clear from Fig. 34, in which AD and A Fare each divided into a number of equal parts and the ratio of AB to A D is seen to be equal to the ratio of AC to A F or BC to DF. These important relations may be verified by drawing various triangles to scale. Fig. 34. When two or more lines are drawn parallel to one side of a triangle, the two sides are divided in the same proportion. P.M. B. 50 PRACTICAL MATHEMATICS FOR BEGINNERS. Similar figures as in Fig. 35 may be divided into the same number of similar triangles. If each side of the figure ABODE is three times the corresponding side of the other, then the line AG is three times A'C and AD is three times A'D'. Fig. 35. Similar figures. Similar figures are not necessarily bounded by straight lines, the boundaries may consist of curved lines, as for example two maps of the same country may be drawn one to a scale of 1 inch to a mile, the other to a scale of j inch to a mile, and any straight or curved line on the one will be four times the corresponding line on the other. Circles. The angle in a semicircle is a right angle. Draw a line AB to any convenient length, say 3 inches. On AB describe a semicircle, Fig. 36 and from any , - - ,P 2 point P on the semicircle draw / \ "\ lines to A and B. Measure the angle APB, or test it by inserting the right angle of a set square. It will be found by taking several positions, and in each case joining to A and B, that the angle at P is always a right angle. In a similar manner it may be proved that in a segment less than a semicircle such as AP X B Fig. 36, the angle formed is greater than a right angle, and when as at A P 2 B the segment is greater than a semicircle the angle is less than a right angle. -Angle in a segment of a circle. THE CIRCLE. 51 -Construction of a right-angled triangle. Another important result is shown, where a line is drawn from P to the centre of the circle G, then as GA, GP, and CB are all radii of the same circle, they are obviously equal. Hence the line joining the middle point of the hypotenuse of a right angled triangle to the opposite angle is equal in length to half the hypotenuse. Ex. 1. Construct a right- angled triangle in which the hypotenuse is 3 "75" and one side is 1 -97". Draw two lines at right angles intersecting at a point B (Fig. 37) ; measure offAB = 1 -97", with A as centre and radius 3*75" describe an arc cutting BC at G. Or, make AG equal to 3*75", and describe a semicircle on it. Then with A as centre and with a radius 1*97" describe an arc intersecting the semicircle at B. Join B to A and G, then ABC is the triangle required. If two chords in a circle cut one another, the rectangle on the segments of one of them is equal to the rectangle on the segments of the other. Thus, if A G and BD be two chords in a circle cutting each other at a point E, the rectangle AE . EG= rect- angle BE . ED. If, as shown in Fig. 38, the lines are perpendicular to each other, and one passes through the centre, then BE=ED; :. AE.EG=ED 2 . From this the graphic method of Fig. 38. If two chords in a circle cut one another, the rect- angle on the segments of one is obtaining square~root as shown on ?%&$.%>,* on the p. 43 is obtained. If a quadrilateral ABCD he inscribed in a circle the sum of the opposite angles equals 180. Thus angle ABG+ angle .42X7=180 (Fig. 39) and similarly angle A + angle G= 180. 52 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 1. Draw a circle 4" diameter, take any four points on the circumference and join by straight lines as in Fig. 39. Verify that the sum of the opposite angles in each case is 180. In any circle the angle at the centre is double the angle at the circumference, on the same or on equal arcs as bases. Draw a circle of 3 or 4 inches diameter, select any two points, A and C (Fig. 40) on the circumference, and join to centre 0. D Pig. 39. Quadrilateral in a circle. Fig. 40. On the same or on equal arcs the angle at the centre is double the angle at the circumference. Also join A and G to any point B on the circumference. Measure the angles A OC and ABC by the protractor and prove the theorem. Make AD=AC ; join A and D to any point E on the circum- ference ; show that the angle AED = ABC=\AOC. Angles on the same, or on equal arcs, are equal to one another. Prove this by joining points D, A and C to different points on the circumference. The products of parts of chords of a circle cutting one another are equal. Draw a circle of 3 or 4 inches diameter and draw any diameter such as DC (Fig. 41) from any convenient point in DC produced, draw a line PAB cutting the circle in points A and B. Measure the lengths, PC, PD, PA, and PB. Show thai PD x PC=PB x PA. If from a given point P (Fig. 41) a line PT be drawn touching the circle, and a line PAB cutting the circle in points A and B, then the rectangle contained by PB . PA, is equal to the square on PT. Using the previous construction draw from P a tangent to the CONSTRUCTION OF TRIANGLES. 53 Fig. il.PTi=PB.PA. circle, measure the line FT, and verify that PT**='PB. PA, and therefore from previous result we have also PT 2 = FCxPD. It will be noticed thcit as the angle DPB is increased the points A and B approach each other, thus taking some position as PA'B', then the chord A'B' is less than AB. When the two points of intersection are coincident we obtain the tangent FT. Hence the tangent may be taken to be a special case of a chord in which the two points of intersection are coincident. To construct a triangle having given the length of its three sides. Draw a line AB (Fig. 42) equal in length to one of the given sides ; with one of the remaining lengths as radius, and A as centre, describe an arc ; with B as centre and remaining length obtain an arc intersecting the former in C. Join C to A and to B. A CB is the required triangle. Ex. 1. Three sides of a triangle measure 2 '5, 1*83, and 2*24 inches respectively. Construct the triangle. Draw AB (Fig. 42) equal to 2 5 inches. With B as centre, radius 2*24 inches, j m , .,, i . Fig 42. To construct a triangle describe an arc, and with A as centre havil]g giveu the i cngt h of its three and a radius of 1*83 describe an arc sides. intersecting the former in G. Join GA and CB. ABG is the required triangle. The construction is obviously impossible when the sum of two sides is less than the third. The angles may be measured by using the scale of chords. Do this and show the angles are as follows : A is 60, B is 45, and C is 75. To construct a triangle having given two sides and the angle included between the two sides. Ex. 2. Two sides of a triangle each measure 5*4 in. and the angle 54 PRACTICAL MATHEMATICS FOR BEGINNERS. included between these sides is 40. Construct the triangle and find the length of the remaining side of the triangle. By means of the table of chords (p. 36), or by a protractor, set out at G (Fig. 43) an angle of 40. Make GB and GA each equal to 5 "4 in. , join B to A. Then BGA is the triangle required. The length Fig. 43. To construct a triangle given two sides and included angle. Fig. 44. To construct a triangle given one side and two ad- jacent angles. of BA will be found to be 3 69, and this is the length of the third side. If the angles be measured it will be found that A and B are each 70. To construct a triangle given one side and two adjacent angles. Ex. 3. Construct a triangle having one side equal to 4 78 in. and the two adjacent angles equal to 35 and 63 respectively. Make the base AB (Fig. 44) 478 in. in length. At A and B set out, by the scale of chords or protractor, the angles BAG =63 and the angle ABG=35. If C is the point of intersection of the two lines, then AGB is the required triangle. Measuring the sides we find AG to be 2-77 in., and BG to be 43 in. Given two sides and the angle opposite one of them to construct the triangle. Ex. 4. Construct a triangle having two sides 2*7 in. and 2 '5 in. respectively and the angle opposite the latter equal to 60. At any con- venient point B (Fig. 45) set out an angle of 60. Make BA =2*7 in. With A as centre and a radius equal to 25 in. describe Pig. 45. To construct a triangle given two sides and the angle opposite one of them. EXERCISES. 55 an arc. The arc so drawn may intersect the base in two points D and C. Join D and C to A. Then each of the triangles DBA or CBA satisfies the required conditions, and the remaining side may be either BD or BG This is usually called the ambiguous case. If the arc just touches the line BC, one triangle only is possible ; if the radius is less than A P the problem is impossible. EXERCISES VIL 1. The side of an equilateral triangle is 10 inches ; find the length of the perpendicular from the vertex to the base. 2. Two sides of a triangle are 12 feet and 20 feet respectively, and include an angle of 120 ; find the length of the third side. 3. (i) Construct a right-angled triangle, base 1'75", hypothenuse 325. (ii) One side of a right-angled triangle is 29 ft. 6 in. and the adjacent acute angle 27 ; find the hypothenuse. 4. Two sides of a triangle are 5 and 7, base 4 feet ; find the length of the perpendicular drawn to the base from the opposite vertex, also find the area of the triangle. 5. Two sides of a triangle are 10*47 and 9*8 miles respectively, the included angle is 30 ; find the third side. 6. Measure as accurately as you can the given angle BOA, also the length OA (Fig. 46). From A draw AM per- pendicular to OB, measure ()M and A M, divide OM by OA and AM by OA, and in each case give the quotient. 7. Draw a circle of 2 '25" radius. In this circle inscribe a quadrilateral A BCD having given : Sides ^5 = 2-87", DC = 2 5", Angle BCD = 16-5. Measure the angle BAD. Draw the tangent to the circle at A. Join A C, and measure the angles which AC makes with the tangent. Also measure the angles ABC and ADC. 8. (i) Prove that the sides of a quadrilateral figure are together greater than the sum of its diagonals. (ii) A quadrilateral A BCD is made from the following measure- ments : The diagonals AC and BD cut in at right angles, CM =3 in., 0^ = 4 in., 0C=8in., OD = Qin. Show that a circle may be described about the quadrilateral. 9. In a triangle one side is 11-14 ft. and the two adjacent angles are 38 and 45. Find the length of the side opposite the former angle. 56 PRACTICAL MATHEMATICS FOR BEGINNERS. 10. A quadrilateral framework is made of rods loosely jointed together, and has its opposite sides equal. Show that when one side is held fast, all positions of the opposite side are parallel to one another. 11. Divide a line 6 in. long into nine equal parts. Construct a triangle with sides equal to 2, 3, and 4 of these parts respectively. Measure the angles of the triangle. Circumscribe the triangle with a circle and measure its radius. 12. Construct a triangle having its sides in the ratio 5:4:2, the longest side being 2f " long. 13. In a triangle given that a =1-56', B = 57, 0=63, find the two remaining sides. 14. The angles of a triangle being 150, 18, and 12, and the longest side 10 ft. long, find the length of the shortest. 15. Find the least angle of the triangle whose sides are 7 "22, 7, and 9-3. 16. Two sides of a triangle are 1*75 ft. and 1'03 ft., included angle 37 ; find the remaining parts of the triangle. 17. The perimeter of a right-angled triangle is 24 feet and its base is 8 feet ; find the other sides. 18. Find the least angle of the triangle whose sides are 5, 9, and 10 ft. respectively. 19. Determine the least angle and the area of the triangle whose sides are 72'7 feet, 129 feet, and 113*7 feet. 20. The two sides AB and BC of a triangle are 447 ft. and 96*8 ft. respectively, the angle ABC being 32. Find (i) the length of the perpendicular drawn from A to BC ; (ii) the area of the triangle ABG ; (iii) the angles A and C. 21. In a triangle ABC, AD is the perpendicular on BC ; AB is 3-25 feet ; the angle B is 55. Find the length of AD. If BC is 4*67 feet, what is the area of the triangle ? Find also BD and DC and AC. Your answers must be right to three significant figures. 22. The sides of a triangle are 3*5, 4 '81 and 5*95 respectively; find the three angles. 23. Construct a triangle, two sides 5 and 6 inches respectively, and having an angle of 30 opposite the former side ; find the remaining side. 24. Construct a triangle, one side 2 - 5 inches long and adjacent angles 68 and 45 respectively. What are the lengths of the remaining sides ? Also find the area of the triangle. 25. Two parallel chords of a circle 12 in. diameter lie on the same side of the centre, and subtend 72 and 144 at the centre. Show that the distance between the chords is 3", , SECTION II. ALGEBRA, CHAPTER V. EVALUATION. ADDITION. SUBTRACTION. Explanation of symbols. In dealing with numbers, or digits, as the numerals 1, 2, 3 ... are called, accurate results can be obtained whatever be the unit employed. Thus the digit 7 may refer to 7 shillings, ounces, yards, or other units. In adding two digits, such as 7 and 5, together, we obtain the sum 12, whatever the unit employed may be. The signs already made use of in Arithmetic are also em- ployed in Algebra, but in Algebra representations of quantities * 1 1 1 A BCD Fig. 47. are utilised which have a further generality. Both letters and figures are used as symbols for numbers, or quantities. These numbers may be knowyi numbers, and are then usually repre- sented by the first letters of the alphabet, a, b, c, etc. ; or, they may be numbers which have to be found, called unknown numbers, and these are often denoted by a?, y, z. A more general meaning is given to the signs + and in algebraical expressions than in arithmetic. If a distance AC measured along a line AD (Fig. 47) from left to right is said to be positive, the distance CA measured in the opposite direction from right to left would be negative. 58 PRACTICAL MATHEMATICS FOR BEGINNERS. The result of the first measurement A G could be indicated by +a, while the same distance CA, but measured in the opposite direction, would be indicated by a. Again, if a length CD be measured in the same direction from left to right and be denoted by + b, the length DC measured from right to left would be indicated by b. Hence + a+b would mean the sum, or addition, of the two lines A C, CD, and so a line of length equal to AD is obtained, where AD=AC+CD. Similarly, + a-b would denote the length AB obtained by measuring a length a in the positive, and a length b in the negative direction. The beginner will probably experience some difficulty in the consideration of these positive and negative quantities. In Arithmetic the difficulty is avoided, for the only use that is made of the sign (minus) is to denote the operation of subtraction, and in this idea the assumption is made that a quantity cannot be subtracted from another smaller than itself. Moreover in Arithmetic we are apt to assume that no quantity is less than 0. In Algebra, on the contrary, we must get the conception of a quantity less than 0, in other words, of a negative quantity. Thus, as an illustration, consider the case of a person who neither owes nor possesses anything ; his wealth may be represented by 0. Another person, who not only possesses nothing but owes 10, is worse off than the first, in fact he is worse off to the extent of 10 compared with the first person. His wealth may, therefore, be denoted by 10. Again, in what is called the Centigrade Thermometer the temperature at which water freezes is marked 0, and that at which water boils 100, and any temperature between these may be at once written down. But it is often necessary to refer to temperatures below the freezing point. To do this we represent the reading in degrees, but prefix a negative sign to indicate that we measure downwards instead of upwards. Thus + 5 C. or, as it is usually written, 5 C. indicates five degrees above freezing point, whereas 5C. indicates five degrees below zero. ALGEBRAICAL SUM AND PRODUCT. 59 If AG denote a distance of 4 miles in an easterly direction, and AB a distance of 2^ miles (Fig. 47), then a person starting from A and walking 4 miles in an easterly direction will arrive at G. If when he arrives at G he then proceeds due west a distance equal to 1| miles he will arrive at B, and his distance from A will be 2^ miles ; or, if as before, a denote the distance AG, and c the distance AB, then if BG be denoted by 6, the distance from A would be expressed by + a b=+c. Algebraical sum. When writing down an expression it is usual, where possible, to place the positive quantity first and to dispense with the + sign. The above expressions would, there- fore, always be written as a + b and a b. The signs placed between the numbers indicate in the first case the sum of two positive quantities, and in the second case the subtraction of one positive quantity from another. In the latter case the quantity a b could also be described as the addition of a minus quantity b to a positive quantity a, by which what is called the algebraical sum of the two quantities is obtained. The algebraical sum of two or more quantities is, therefore, the result after carrying out the operations indicated by the signs before the quantities. The algebraical sum of +10 and 18= 8. In the quantity a - b, if a represents a sum of money received, then b may represent a sum of money paid away. The algebraic sum is represented by the balance a b. It will be seen that in Algebra the word sum is used in a different and wider sense than in Arithmetic. Thus, in Arithmetic 7-3 indicates that 3 is to be subtracted from 7, but in Algebra a similar expression means the sum of the two quantities whether the expression is in numbers, as 7-3, or in letters, as in a b. How a product is expressed. The arithmetical symbols of operation, +, , x, and -4-, are used in Algebra, but are varied according to circumstances. The general sign for the multi- plication of quantities is x ; but the product of single letters may be expressed by placing the letters one after another ; thus the product of a and b may be written a x b or a . b, but is usually written as ab. In a similar manner the product of 4, a, x, and y is expressed by Aaxy. It will be obvious that this method is not applicable in Arithmetic. Thus 5x7 cannot be written as 57. 60 PRACTICAL MATHEMATICS FOR BEGINNERS. The product of two quantities such as a+5 and c + d may be expressed as (a + b) x (c + d), or, and usually, as (a + b)(c + d). Expression. Quantity. When, as in a + b-c, or 4axy, several terms are joined together with or without signs, they together form what is called an algebraic expression or quantity. Other names used in Algebra. Any quantity, such as 4a, indicates that a quantity a must be taken four times ; the multi- plier, 4, of the letter is called a coefficient ; and 4a, containing a coefficient and a letter, is called a term. Multiples of the quantities a, b, c, etc., may be expressed by placing numbers before them as, 2a, 36, 5x ; the numbers 2, 3, and 5 thus prefixed are called the coefficients of the letters a, 6, and x. When no coefficient is used the coefficient must be taken to be 1, thus x means 1 x x, be indicates 1 x b x c, etc. The product of a quantity multiplied by itself any number of times is called a power of that quantity, and is indicated by writing the number of factors on the right of the quantity and above it. Thus : a x a is called the square of a and is written a 2 ; b x b x b is called the cube of b and is written 6 3 . Similarly, the multiplication of any number of the same letters together may be represented; thus c n indicates c to the power n, where n represents a given number. The number denoting the power of a given quantity is called its index (plural, indices) or exponent. It is very important that the distinction between coefficient and index be clearly understood. Thus 4a and a 4 are quite different terms. Let a = 2, then 4a = 8 ; buta 4 = 2 4 = 16. The use of signs may be exemplified in the following manner : Ex. 1. In the expression a 2 + b - c, Let a = 4, 6 = 7, and c = S. Then a 2 + 6-c = 4 2 + 7-3 = 23-3 = 20. Ex. 2. Find the value of d from the equation d - t = fy ; (ii) when t = %. - 1 "2*Jt, (i) when (i)d=l-2jjL=l-2x%=-9. (iD^lVl^^^ 1,2 ^ 898 -^. EXERCISES. 61 Ex. 3. Find the value of aar 2 + 6 2 ^ when a = 3 h = ^ c = ^ X = Q bx-a z -c Here ax i + 6 2 = 3 x6 2 + 5 2 =133, Also te-a 2 -c=5x6-3 2 -2= 19, " 6-a 2 -c~ 19 ~ / * .Sir. 4. Find the value of c\/l0a6 + bs/Sac + a*s/45bc, when a = 8, 6 = 5, e=l. Substituting the given values in the expression given we obtain \/l0x8x~5 + 5\/8 x 8 x 1 + 8\/45 x 5 x i = \/400 + 5x 8 + 8x15 = 20 + 40 + 120 = 180. ^c. 5. Find the value of (ac - bd)s/a?bc + b 2 cd + c 2 ad - 2, when a = l, 6 = 2, c = 3, d = 0. Substituting these values in the given expression we obtain (3 - 0)\/lx2x 3 + 4x3x0 + 9x1x0-2 = 3\/6^2 = 6. In Ex. 5 it should be carefully noticed that, as one of the given terms d is equal to 0, any term containing that letter must be 0. Hence, we may either omit all the terms containing that letter or, by writing them as in the above example, the terms in which the letter occurs are each seen to be equal to zero. EXERCISES. VIII. If a = 4, 6 = 3, c = 2, d = 0, find the value of 1. 3a + 26 + c. 2. 2a-2b-c. 3. a6W. 4. a -b 2 + c 3 -d 4 . Find the value of 5. a? + 3a 2 6 + 3a6 2 + 6 3 when a = 1, b = 2. 6. The resistance of a wire is given by R=-k. Given =100, a= -002, and k= -00002117 find the value of B. 7. The heating effect of a current is given by H= 'I^CPIlt. Find H given 0=20, = 30, = 60. GV 8. HP==jj,. This is the relation between horse power, current in amperes, and volts. Given C = 30, F=100. Find HP. 62 PRACTICAL MATHEMATICS FOR BEGINNERS. If o=l, 6 = - 1, c = 2, d = 0, find the value of Q a + b c + d ad -be c 2 -d? iJ. 1 1 a-b c-d bd + ac a 2 + b 2 When a=10, 6 = 3, c = 7 of, 10. 6 + c 11 3c 4& <" 2c -- 36 " 4a + 2~ r 10a-16" 12. J^IL JSEI* \c-6 \ c-6 When a = 5, 6 = 4, c=3 of, 13 76 + c 14 7a 116 JOc 3a + 46' ' 116 -3c 86 -7c 7a -56* Addition. The addition of algebraical quantities denotes the expression in one sum of all like quantities, regard being had to their signs. When like quantities have the same sign, their sum is found by adding the coefficients of similar terms and annexing the common letters. Thus 7a + 4a = 11a. Also, 7a + 3a + 36 + 5a = 15a + 36. Ex. 1. Add together: 7a + 56, -5a + 46, 3a -26. When several such quantities have to be added together, they may be arranged so that all terms having the same letter, or letters, and the same powers of the letter, or letters, are in columns as shown ; the positive and negative coefficients in each 7a + 56 column are then added separately, and the sign of the -5a + 46 greater value is prefixed to the common letters. The 3a -26 operation would proceed as follows. Arrange in 5a + 76 columns, placing the letters in alphabetical order. Commencing with the row on the left-hand side, we have 7a + 3a = 10a. Now add to this - 5a ; or, in other words, from 10a subtract 5a, and the result is 5a. Again 56 + 46 = 96; and 96-26 = 76. Hence the sum required is 5a + 76. Ex. 2. Add together : lx 2 - 9y 2 + 20xy, - 11a; 2 + 10y 2 - xy, 8x 2 - 7y 2 - 4xy. First adding together the coefficients of the terms in x 2 we get 7-11 + 8 = 4. In a similar manner the coefficients of y 2 are -9+10-7= -6; and of xy are 20 - 6 - 4 = 10. Hence, the required sum is Ax 2 - 6y 2 + lOxy. It is not necessary to separate the coefficients and to write them down. It is much better to perform the addition mentally. ALGEBRAICAL SUBTRACTION. 63 EXERCISES. IX. Add together 1. b + lc-^a, c + ^a-^b, a + ^b-^c. 2. ax 2 - bx 2 + cy 2 - ab 2 c, U ax 2 - 3$ bx 2 + cy 2 + 4ab 2 c, 5 \ a x 2 - 5j bx 2 -3cy 2 - 2a6 2 c, -lax 2 + Ibx 2 + cy 2 - ab 2 c. 3. 6m-l3n + 5p, Sm-9p + n, -p + m, n + 5p + m. 4. 7a + 56-13c, -6 + 4c + a, 36 + 3c -3a, and find the value of the result when a =1, 6 = 2, c = 3. 5. 2x + 4y-z, 2z-3y-2x, 3x-z, 2y-x. 6. a-26 + 2c, 6-3c + 3a, e-4a + 46. 7. ax 2 -2dx 2 -2x + 2c-Sf, ax 2 + 2dx 2 -bx + cx-l, ax 2 - dx 2 -bx-cx-c+1, -x 2 + 3bx -c + 2j. Find the sum of 8. 2{a + b + c), 3(a + 6-c), 4(a + c-6), b + c-a; and obtain its numerical value if a = J, 6 = ^, c=] 3 2. 9- i* + ? y 4 + *% " 5 a-'V and far 4 + f y 4 - J a% - a% 2 - Simplify 10. 4a: 2 + 8a^ + 4 t y 2 -9a: 2 +18a^-92/ 2 . 11. a; 6 + 3x* + lx 4 + 15a; 3 + 10a; 2 - 3a; 5 - 9a; 4 - 21 x 3 - 45a; 2 - 30a; + 2a; 4 + 6ar> + 14a; 2 + 30a; + 20. 12. Simplify and find the value of x 2 + xy + x - x 2 + xy x 2 + y 2 + 66, when a; =100, y = 50. 13. 4a?-16a; 2 +162/ 2 + 4a;+16a; 2 - I6y 2 -4y whena;=l, y=0. Subtraction. In Algebra, to perform the operation of sub- traction, arrange like terms together as in addition, change the signs of all the terms to be subtracted, and then add to the other expression. Thus, to subtract la from 13a, we reverse the sign of la and so make it minus ; for 13a la is only another way of expressing that la is to be subtracted from 13a. Thus 13a - la = 6a. Ex. 1. From 5a + 3a; -26 subtract 2c -4y. The quantity to be subtracted, when its signs are changed, is -2c + 4y, .'. the remainder is 5a + 3a; - 26 - 2c + 4y. Ex. 2. Subtract a 2 -2b- 2c from 3a 2 - 46 + 6c. Here, after arranging as in addition and changing g a 2 _ 4^ . g c the signs, we proceed as in addition, thus : _ a. 2 + 26 + 2c 3a 2 - a 2 = 2a 2 ; 26 - 46 = - 26 ; and, finally, 2a 2 -26 + 8c 6c + 2c = 8c. Hence, the result is 2a 2 - 26 + 8c. It is not necessary to perform on paper the actual operation of changing the signs of all the terms in the expression to be subtracted. The operation should be carried out mentally. 64 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 3. From 7a? 2 - 2x + 5 subtract 3x 2 + 5x - 1. Here we may, as in Ex. 2, write the terms under each other. Then, after mentally altering the sign of Sx 2 , we obtain by addition 4a: 2 . Again mentally altering 729 the sign of 5x and adding to - 2x we obtain - lx ; , e i and, finally, repeating the operation for the last . 2 _n x ,a figure we get the number 6. Hence the result is 4* 2 -7* + 6. It will be noticed that the process of subtraction in the last Example, where we subtract Zd? from 7#*, simply means to find a quantity such that when added to Zx l will give 7a 2 . So that in (Ex. 3) adding the second and third rows together the sum is equal to the first row. This affords a ready check and should always be used. The subtraction of a negative quantity is equivalent to adding a corresponding positive quantity. If a length AB y 4. a -*-- b -> i t 1 < 4 .... ai * A D B C Fig. 48. Fig. 48, be denoted by or, and another length BG by 6, then a + b would be represented by AG, a line equal in length to AB + BC, both being measured in the positive direction (from left to right). Also a b would be a quantity obtained by subtracting b from a, and could be obtained by measuring off a length BD in a negative direction from B, so that a b is appropriately represented by AD. As BG is positive, the reversal of direction indicated by CB is negative, and would be indicated by -b. Thus, a minus sign before a quantity reverses the direction in which the quantity is measured. Now, to subtract b from a, we reversed the direction of b and added it on to a. If, then, we have to sub- tract a negative quantity, b or GB, from a positive quantity a, by reversing the direction we obtain BG or + b, and adding on to a we get AG or a + b. We could indicate this by a ( 6), the negative sign outside the bracket indicating that the quantity EXERCISES. 65 inside the bracket has to be subtracted from a. The change in sign is true whether the quantity subtracted be positive or negative. Hence, the diagram illustrates the rule already given, namely, to subtract one quantity from another change the sign of the quantity to be subtracted and proceed to add the two together. EXERCISES. X. 1. From 6a -26 + 2c take 3a -36 + 3c. 2. 8a + x-6b take 5a + a + 46. 3. 9x 2 -3x + 5 take 6x 2 + bx-3. 4 Subtract 2a - 2b - 36 + 3c from 2a + 26 + 36 + 3c. 5. ax -bx-yd + yc from 2ax -bx + 2yc - yd. 6. xc-xd + ya + yb from xa - xd - yc + yb. 7. 3p-3m + 2m-2n from 2m - 2p - 3n + 3p. 8. 3yz - 3xz - 4xy from 3yz - 3xy - 4xz. 9. a-b-d-c-e from 2d + 2b + 2a + 3e-2c. 10. x 2 - 3y 2 + Qxz - 3xy from x 2 + 4xy - 5xz + z 2 . 11. Add together a-26-c, 4a-36 + c-2d, 4d-3c + 2h, c-5a-b-d, a + 66 + 5c + 3d; and subtract c-a + 2d from the sum. 12. What must be added to a + 6-c to make c + d, and what must be taken away from x 2 (l +2y) -y 2 (l+2x) to give as remainder 2xy{x-y) t > 13. Add together 4a 3 + 36c 2 - 6a 2 6, 5a 2 6 - c 3 - 3>.c 2 , c 3 -2a 3 -2a 2 6. 14. From 6x?y + 10x 2 y 2 + 13xy* + I9y 4 take 5x*y - 2x 2 y 2 + 3xy 3 - 2y 4 . 15. Subtract 2a 4 + 3a 3 6 + 5a 2 6 2 + 8a6 3 + 1 16 4 from 4a 4 + 6a 3 6 + 8a 2 6 2 + 10a6 3 + 126 4 . 16. Find the sum of 4a~> - 5ax 2 + 6a 2 x - 5a 3 , 3a 3 + 4ax 2 + 2a 2 x + 6a 3 , - 17a 3 + I9ax 2 - \5a 2 x + 8a 3 , I3ax 2 - 27a 2 a + 18a 3 , 12a 3 + 3a 2 x - 20a 3 . 17. Find the sum of ab + 4ax + 3cy + 2ez, 14aa + 20ez + 19a6 + 8cy, \3cy + 21ez+15ax + 24ab. P.M.B. CHAPTER VI. MULTIPLICATION. DIVISION. USE OF BRACKETS. Multiplication. As in Arithmetic, multiplication may be considered as a concise method of finding the sum of any quantity when repeated any number of times. The sum. thus obtained is called the prodioct. In multiplying, what is called the Rule of Signs must be observed, i.e. The product of two terms with like signs is positive ; the product of two terms with unlike signs is negative. If a is to be multiplied by b, it means that a has to be added to itself as often as there are units in b ; hence, the product is ab. If a is to be multiplied by - b, it means that a is to be subtracted as often as there are units in b, and the product is ab. Again, if a is to be multiplied by b, it means that a is to be added to itself as often as there are units in b, hence the product is - ab. The same result would be obtained by multi- plying a by -b. At the present stage it is necessary to be able to apply the rules of multiplication readily and accurately. The proofs will come later. Rule. To multiply two simple expressions together, first multiply the coefficients, then add the indices of like letters. Re- member that like signs produce + (plus), unlike signs produce (minus). Ex. 1. Multiply 4a& 3 by 3a 2 6 2 . Here the product of the coefficients is 4x3 = 12. Adding the indices of like letters we have a x a 2 = a i+ * = a 3 , and b 3 x b 2 =b 3+2 =b 5 . Hence the product is 12a 3 & 5 : .-. 4a6 3 x3a 2 fi 2 =12a 3 & 5 . Ex. 2. 4a 3 6 2 c 5 e 4 x 6a 4 6 3 c 4 e 3 = 24a 7 6 W. ALGEBRAICAL MULTIPLICATION. 67 When the expressions each consist of two terms, the process of multiplication may be arranged as follows : Ex. 3. Multiply x + 5 by x + 6. Write down the two expressions as shown, one under the other ; multiply each term of the first expression by each term of the second, and arrange the results as here indicated ; begin at the left-hand side, thus, x x x = x 2 . Write M both a+h and a b when n is even. The cases where n equals 2, 3, 4, 6 have already been taken. Other values of n should be used and more complete veri- fications be obtained of the rules given. In finding the factors of any given expression any letter or letters common to two or more terms may be written as a multiplier, thus, given ac + ad we can write this as a(c + d). Again, ac + bc + ad+bd=a(c + d) + b(c + d) = (c + d) (a + b). By multiplying x + 2 by x + 3 we obtain xP + bx + Q. ;. (x + 2)(x + Z)=x 2 + 5x + 6. Hence, given the expression x* + 5x +6, to find the quantities x + % and # + 3, or the factors of the given expression, we find that The first term is the product of x and x, or x 2 . last 2 and 3, or 6. middle the first term, and the sum of 2 and 3, or bx. Proceeding in this manner the factors of a given expression are readily obtained. Ex. 7. Resolve into factors x 2 + Sx + 12. Here, the two numbers required must have a sum of 8 and a product equal to 12. Of such pairs of numbers, the sum of which is 8, are 4 and 4, 7 and 1, and 6 and 2, but only the last pair have a product of 12. Hence, the factors are {x + 2){x + 6). A convenient method is to arrange the possible factors in vertical ,, x + 4 x + 7 x + 2 rows, thus x + x + l x + Q These may be multiplied together as in ordinary multi- plication, but it is much better to perform the process mentally, obtaining first the product of the two first terms, then the pro- duct of the two last terms, and finally the sum of the diagonal products. Thus, in the second group, the product of the two first terms is x 2 , of the two last is 7 ; and the sum of the diagonal products 7#+#=8.z\ Hence, these are the factors of x 2 -\-8x + 7. 76 PRACTICAL MATHEMATICS FOR BEGINNERS. Proceeding in like manner the product of the terms in the last is x 2 + 8x + 12. Similarly (x - 6) (x - 2) = x 2 - 8x + 1 2, (x+6)(x-2)=x 2 + 4x-l2. Also, (.r-6)0+2)=.2? 2 - 4^-12. All these products should be verified and in each case the process should be carried out mentally. Or, we could write the given expression x 2 + 8#+12 as x 2 + 2x+6x + \2, Taking out the quantity common to two terms we obtain .27 (# + 2) + 6 (#+2). This shows that x + 2 is common to both terms, hence we may write x 2 + 8x+12 = (x + 2)(x + 6). Ex. 8. In a similar manner, x 2 - 9x + 20 = x 2 - 5x - 4x + 20 =x{x - 5) - 4{x - 5) = {x - 4) (x - 5). Ex. 9. a; 2 +lla; + 30=a; 2 + 5:c + 6a; + 30=(a; + 5)(a; + 6). Factors obtained by substitution. The factors in the pre- ceding, and in other, examples may also be found by substituting for x some quantity which will reduce the given expression to zero. Thus, in x 2 9# + 20 the last term suggests that two of the following, 4 and 5, 10 and 2, or 20 and 1, are terms of the factors, but the middle term of our expression denoting the sum of the numbers selects 4 and - 5. To ascertain if 4 and 5 are terms of the factors, put # = 4 ; then 16-36 + 20=0; thus x 4 is a factor. Similarly putting ^=5we obtain 25-45 + 20 = ; .'. x 5 is a factor. Hence x 2 - 9x + 20 = (#-4) (#-5). Ex. 9. # 2 + 6a;-55. Put x= - 11 ; this reduces the given expression to zero ; .*. x + 1 1 is a factor. Next put x= +5 ; a;-5is found to be a factor ; /. x 2 + 6z-55=(a; + ll)(a;-5). FRACTIONS. 77 EXERCISES. XIV. Resolve into factorfa 1. x 2 -7x + 10. 2. a; 2 -#-90. 3. a; 2 -3a;-4. 4. a?+2a;-15. 5. 27a 3 + 86 3 . 6. 8a^-27. 7. a; 2 -a: -30. 8. x 1 + 12a; -85. 9. x 2 - 2xy - xz + 2yz. 10. 3a; 2 -27y 2 . 11. x 2 + 18a; -175. 12. a; 2 -3a;2;-2a;y + 62/z. 13. 625a^-^. 14. 10^ + 79a; -8. 15. ar J -13a; 2 y + 42a;y 2 . 16. (a 2 + 6 2 -c 2 ) 2 -' ia 2 b 2 . 17. {x-2yf + y 3 . 18. (i)a 2 -6 2 + c 2 -d?-2(ac-bd); (ii) (p 2 + q 2 -^) 2 -4p 2 q 2 ; (iii) 1 -m^-m + rn^. Fractions. The rules and methods adopted in dealing with fractions in Algebra are almost identical with those in Arith- metic. In both cases fractions are of frequent occurrence and their consideration is of the utmost importance. Some little practice is necessary before even a simple fraction can be re- duced to its lowest terms. Perhaps the best method in the simplification of fractions is to write out the given expressions in factors wherever possible. To do this easily the factors already referred to on pp. 73 and 74 should be learnt by heart. When proper fractions have to be added, subtracted, or com- pared, it is necessary to reduce them to a common denominator, and to lessen the work it is desirable that this denominator shall be as small as possible. Ex. 1. Addi+i-. 2 3a; First reduce to a common denominator 6a; ; mentally multiply both numerator and denominator of the first fraction by 3a;, and obtain -? ; and similarly, by multiplying by 2, get . ba; ox bar . 1 1 _3a; + 2 2 + 3a;~ 6a; Jte& Simplify (1 + ^)^9,-1). 1 1 3a: + 2 2 + 3a; 6a; a;(3a; + 2) 4 9a; 2 - 4 6a;(3a; + 2)(3a;-2) 6(3a;-2) x x The factors x (3a; + 2), which are common to both numerator and denominator, have been cancelled. 78 PRACTICAL MATHEMATICS FOR BEGINNERS. r o ct i:* ,-\ x* + x 2 y 2 r -v x 2 -\-3x-\-2 Ex. 3. Simplify (i) __, (11) ^^ Here, (i) ^ a V = * 2 W) * T y \i* + y*)\7*-y*) tf 2 -y 2 * 1 \ x 2 + x-2~(x-l)(x + 2)~x-V Ex.4. Simplify ^2 xSLJ. _1 _ 1 , _a a 6 a: x-a a-b x-a a-b 1 1 . a~b -a x-a a b x ab x (x-a)ab (a-b)x - = 1 r J x = -- abx. b-a x-a The terms common to numerator and denominator are cancelled ; the term b-a being for this purpose written in the form - (a - b). Highest Common Factor. When the denominators of two or more fractions can be written in the form of factors, the reduction of the fractions to their simplest form can be readily effected. But the process of factorisation cannot in all cases be easily carried out, and in such cases we may proceed to find the Highest Common Factor (h.c.f.). The process is analogous to that of finding the g.c.m. in arithmetic. The h.c.f. of two or more given expressions may be defined as the expression of highest dimensions which can be divided into each of the given expressions without a remainder. Ex. 5. Simplify the fraction ^ . t^^" 4 . X s + Sx 2 - 4 To find the h.c.f. we proceed as follows : x 3 + Sx 2 -4)x 4 + x s + 2x -(x-2 -2X* +Gx- 4 -2^-63? + 8 6^ + 62:- 12 = 6{x 2 + x-2); x 2 + x-2)x s + 3x 2 -4(x + 2 x* + x 2 -2x 2x 2 + 2x-4 2x 2 + 2x-4: Therefore the h. c. f. = x 2 + x - 2. SURDS. 79 Hence a? 4 + a? 3 + 2a;-4 _ (a; 2 + a;-2)(a; 2 + 2) _ x* + 2 .x 3 + Sx 2 -4: ~ {x 2 + x-2){x + 2)~ x + 2' Least Common Multiple. When required to add, subtract, or compare two fractions, it is often necessary to obtain the Least Common Multiple (l.c.m.) of the denominators, i.e. the expression of least dimensions into which each of the given ex- pressions can he divided without a remainder. To find the l.c.m. we may find the h.c.f. of two given ex- pressions, divide one expression by it and multiply the quotient by the other. Thus the h.c.f. of the two expressions x 3 -3x 2 -l5x + 2b and # 3 + 7# 2 + 5j?-25 is ^7* 2 + 2^7 5 ; dividing the first expression by this h.c.f. the quotient is x- 5. Hence the l.c.m. is (x - 5) (x 3 + 7x 2 + 5^-25) or (x - 5) (x + 5) (x 2 + 2a? - 5). Ex. 6. Simplify the following : 1 1 X s -3a; 2 -15a; + 25 X s + 7a; 2 + 5a; - 25* The common denominator will be the l.c.m. of the two denomin- ators, and the fractions become a; + 5 a;-5 (a; 2 + 2a; - 5) (x - 5) (x + 5)" (a; 2 + 2a; -5) (a? -5) (x + 5)' 10 ~{x-5){x + 5){x 2 + 2x-5)' Surds. As already explained on p. 29, when surd quantities occur in the denominator of a fraction it is desirable to simplify before proceeding to find the numerical values of the fraction. 20 Ex. 1. Find the value of -=. s/2 Unless some process of simplification is adopted it would be necessary to divide 20 by 1*4142..., a troublesome operation. If, however, we multiply both numerator and denominator by s[2 we obtain ,- 2 P= l(k/2 = 10x1-414..., a result easily obtained. A similar method is applicable when the numerator and de- nominator of a fraction each contain two terms. Thus, Ex. 2. Find the value of 2+ *l . 2-n/3 SO PRACTICAL MATHEMATICS FOR BEGINNERS. Here, as ^3 = 1732, if we proceed to insert the value of the root 2 + V3 2 + 1-73205 373205 we get -pz = , 2-\/3 2-1-73205 -26795 and it would be necessary to divide 3 73205 by 26795. Instead of doing this we may rationalise the denominator, i.e. multiply both numerator and denominator by 2 + s/Z. The fraction then becomes (2 + x/3)(2 + \/3) ^ (2 + \/3) 2 =: 4 + 3 + W3 ^ (2-n/3)(2+n/3) 4-3 ' 1 ~ 7 + 4V In this form the necessary calculation can readily be carried out. EXERCISES. XV. Simplify 1 20abx 2 a?-x 2 3 x? + a s 4 4 + 12a; + 9a; 2 15a 2 ' ' a + x' ' x 2 + 2ax + a 2 ' ' 2+13a; + 15a; 2 " K 4a- 5 + 4a; 2 -7* + 2 6 2a~ J -lla; + 15 a; 2 + 5a;-6 4ar* + 5a; 2 -7a;-2 a^ + 3a;-18 2* 2 -3a:-5 7 x + y X ~V 8 rr4_a4 . x^ + ax x-y x + y ' (x-af x-a' 9 a; 2 + 4a; + 3 tf + Gx + S 1Q a; 2 + 6a;-7 . a; 2 + 4a;-21 a> + 5a; + 6 X a; 2 + 5a; + 4' ' a; 2 + 3a;-4 * 2a; + 8 11. Express as the difference of two squares 1 + x 2 + x*, and thence factorise the expression. 12 168a 3 6 2 c 13 a?+{a + b)ax + bx 2 14 a; 4 + a; 2 +l ' 48a 2 6c 3 ' ' tf-bW ' ' x 3 -!. ' 1K a; 2 -a;-6 3 3 -2a?-8 1C 2a; 2 -a;-15 17. a; 2 + 4a; + 4 x*-lx+\2 5a; 2 -13a;-6 1 1 3x 2{'Sx-2y) 2{Sx + 2y) 9x 2 -4y 18 l 6y 1 19 3a; 3 -6a? + a;-2 ' x + 3y + x 2 -9y 2 3y-x ' ar*-7a; + 6. 20. (i-^U + ^L). \ x+yj\ x-y) 1 1 x+S 4 21. (x-1) 2(a;+l) 2(a; 2 +l) a?*-l 1 2b 1 a + b + a?-b 2 + a-b' CHAPTER VIII. SIMPLE EQUATIONS. Symbolical expression. One of the greatest difficulties experienced by a beginner in Algebra is to express the condi- tions of a problem by means of algebraical symbols, and considerable practice is necessary before even the simplest problem can be stated. The few examples which follow are typical of a great number. Let x denote a quantity ; then 5 times that quantity would be bx ; the square of the quantity would be x 2 ; and a fourth part of it would be indicated by -. If a sum of 50 were equally divided among x persons, then each would receive . x If the difference of two numbers is 7, and the smaller number is denoted by x, the other will be represented by x + 7. If the larger is denoted by x, then the smaller would be represented by x -7. If the distance between two towns is a miles, the time taken by a train travelling at x miles an hour would be - ; when the x numerical values of a and x are known, the time taken can be obtained. Thus, let the distance a be 200 miles, and x the velocity, or speed, be 50 miles an hour, then the time taken to complete the journey is = 4 hours. Although the letters a, x, etc., are used in algebraical opera- tions, symbols are often employed which at once, by the letters used, express clearly the quantities indicated. P.M.B. f 82 PRACTICAL MATHEMATICS FOR BEGINNERS. Thus, space could be denoted by s ; the velocity by v ; and the time taken by t ; then instead of - in the last example we use - ; or, the relation between s, v, and t is given by s = vt. From this, when any two of the three terms are given, the remaining one may be obtained. In the case of a body falling vertically, the relation between space described and time of falling is given by s=^gt 2 ; where s denotes the space described in feet, t the time in seconds, and g denotes 32"2 feet per second in a second, or the amount by which the velocity of a body falling freely is increased in each second of its motion. In this case, given either s or t, the remaining term may be calculated. Equations. An equation may in Arithmetic, or Algebra, be considered simply as a statement that two quantities are equal. Thus, the statement that 2 added to 7 is 9, may be expressed as an equation thus 2 + 7 = 9. In a similar manner, other state- ments of equality, or, briefly, other equations, could be formed ; indeed, the greater part of the student's work in Arithmetic has been concerned with such equations. All such equations, involving only simple arithmetical opera- tions, may be called Arithmetical Equations, to distinguish them from such equations as 2^ + 7 = 9, which are called Algebraical Equations. As in Arithmetic, the answer to any given question remains unknown until the calculation is completed. So in Algebra the solution of an equation consists in finding a value, or values, which at the outset are unknown. Simple equations. When two algebraical expressions are connected together by the sign of equality, the whole expression thus formed is called an equation. The use of an equation con- sists in this, that from the relations expressed between certain known and unknown quantities we are able under proper conditions to find the unknown quantity in terms of the known. The process of finding the value of the unknown quantity is called solving the equation; the value so found is the solution or the root of the equation. This root, or solution, when sub- . stituted in the given expression makes the two sides identical. SIMPLE EQUATIONS. An equation which involves the unknown quantity only to the first power, or degree, is called a simple equation ; if it contains the square of the unknown quantity it is called a quadratic equation ; if the cube of the unknown quantity, a cubic equation. Thus, the degree of an expression is the power of the highest term contained in it. If an equality involving only an algebraic operation exists between two quantities the expression is called an identity, thus {x+y) 2 x 2 -\-2xy+y 2, is an identity. In the equation 2# + 7 = 9, x represents an unknown number such that twice that number increased by 7 is equal to 9. It is of course clear that x l y but we may with advantage use this simple example to explain the operation of solving an equation. Before doing so, it is necessary to note that as an equation con- sists of two equal members or sides, one on the left, the other on the right-hand side of the sign of equality, the results will still be equal when both sides of the equation are : (i) equally increased or diminished, which is the same in effect as taking any quantity from one side of an equation and placing it on the other side with a contrary sign ; (ii) equally multiplied, or equally divided ; (iii) raised to the same power, or, the same root of each side of the equation is extracted. And also, if (iv.) the signs of all the terms in the equated expressions are changed from + to - , both sides of the equation being altered similarly, the result will still be the same. Thus, in the equation 2.27+7 = 9, subtracting 7 from each side we get 2^+7-7 = 9-7, or 2#=2. Dividing by 2, then, .27=1. Ex. I. Solve the equation 4x + 9 = 37. Subtracting 9 from each side we get 4x = 28 ... *= T =7. To prove this, put 7 for x. Then each side is equal to 28. Instead of subtracting we can transpose the 7 in the preced- ing example from one side of the equation to the other by changing its sign ; thus 4^7 = 37-9 = 28. 84 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 2. Solve Ax + 5 = 3x + 8. Subtract 3x from both sides of the equation, and we get Ax-Sx + 5 = S; next subtract 5 from each side ; .-. # = 8-5 = 3. It is sufficiently clear that +Zx and +5 on the right- and left-hand sides of the equation respectively may be removed from one side to the other (or transposed) and appear on the opposite side with changed sign. Hence the rule for the solution of equations is : Transpose all the unknown quantities to the left-hand side, and all the known quantities to the right-hand side ; simplify, if necessary, and divide by the coefficient of the unknown quantity. Ex. 3. Solve 5x -1 2 Ix - 2 33 x 10 " 5 2* Multiply both sides of the equation by 10. .'. 25a; - 5-7x + 2 = ffi-5x; 23a; =69; .-. x=3. Eractional equations. If the attempt is made to solve all equations by fixed methods or rules, much unnecessary labour will often be entailed. Thus, in equations containing fractions, or, as they are called, fractional equations, the rule usually given would be to first clear of fractions by using the l.c.m. of the denominators ; but, if this is done in all cases the multiplier may be a large number, troublesome to use. In such cases it is better, where possible, to simplify two or more terms before proceeding to deal with the remaining part of the equation. 25 21 7 '4 We may with advantage simplify three of the given terms, using 21 as a multiplier, thus : 21 (11a; -13) in /, n SM 21 (5a? -254) .*. v og -+17a; + 4-f57a; + 9 = 591 + l - -. % 25 4 FRACTIONAL EQUATIONS. 85 Multiplying by 100 we obtain 84 ( 1 la; - 13) + 7400a; - 525 (5a; - 7 ^) = 57800, or 924a: - 1092 + 7400a; - 2625a; + 13300 m 57800. .-. 5699a; - 45592. When decimal fractions occur in an equation it is often desirable to clear of fractions by multiplying both sides of the equation by a suitable power of ten. Ex. 5. Solve -015a; + -1575 - -0875a? = -00625a;. We can clear of fractions by multiplying every term by 100000. .-. 1500a; + 15750 - 8750a; = 625a;, or 15750= 7875a;. /. x =2. Ex. 6. Solve ^--a? = b 2 -. ox ax First remove the fractions by multiplying all through by abx :. a 2 - a?bx = ab 3 x - b 2 , transposing, - a?bx - ab 3 x = - a 2 - b 2 , changing sign or multiplying by - 1, x{a 2 + b 2 )ab = a 2 + b 2 ; a 2 + b 2 1 ab' Ex. 7. Solve V a + x- sja - x Equations of this kind are simplified by adding the numerator and the denominator to obtain a new numerator, and then subtracting in order to find the new denominator as on p. 101. 'a + x + v a - x si a + x + si a -x + si a + x- sla-x 2 + 1 Hence s/a + x-s/a-x s/a + x+s/a-x-s/a + x + s/a-x 2- 2s/a + x_ " 2s/a-x~ a + a;=9(a-a:) .'. x=^a. EXERCISES. XVI. Solve the equations 1. 18a; + 13 = 59 -5a;. 2. 4a? + 16=10a;-5. 3. 3(a:-2)=4(3-a;)-4. 4. 7a;-3=5a; + 13. 5. 3,-1=42-2,. 6. | + ^-|-| = ll. 86 PRACTICAL MATHEMATICS FOR BEGINNERS. 7. 3Oc+12 + 32a;-8 = 500. 8. 2x + 3= 16 -(2a: -3). 9. x - 7 (4as - 11) = 14 {x - 5) - 19 (8 - x) - 61. 10. 3(aJ-5)-5(aj-4)=21a;-41. 11. 21a? + 7 = 4(a;-3) + 3a: + 6]. 12. 5^-^zi^iz^. 13. 6a: + 4(2x-7)-9(7-2a;) = 645. 14. 5a?-7(a;-8)-20(8-a;) = 10(2a;-19). M 7a; + 17 2*+l If , . 3. , lox ) 15. _ Ig _= g _ +i | a . + 6- i (3a: + 19)|. 16 - rH(r* : 4>}+> 17. * + ^=&. 18. *=* = - 6 + *<+I>. a o-a b a ab 10 x-a ax+l A OA (x-b\ ,(x-a\ a 2 + b 2 a-6 ab+1 \a + b/ \a-bj 2{a + b) Problems involving simple equations with one unknown quantity. When a question or problem is to be solved, its true meaning ought in the first place to be perfectly understood, and its conditions exhibited by algebraical symbols in the clearest manner possible. When this has been done the equation can be written down and the solution obtained. Ex. 1. If 3 be added to half a certain number the result is equal to 7. Find the number. Let x denote the number ; then, one-half the number is - ; and, x ^ 3 added to this gives the expression ^ + 3 ; but the sum is equal to 7- x Hence we have - + 3 = 7 as the required equation. Subtracting 3 from each side of the equation it becomes Next multiplying the equation throughout by 2 Thus the required number is 8. The result in this and in all equations should be substituted in both sides. When this is done the left-hand side is seen to be equal to the right, or, the equation is said to be satisfied. The beginner will find that simple exercises of the type shown in Ex. 1, are easily made and tend to give clear notions how to express arithmetical processes by algebraical symbols. PROBLEMS LEADING TO SIMPLE EQUATIONS. 87 Ex. 2. The sum of two numbers is 100 ; 8 times the greater exceeds 1 1 times the smaller part by 2 ; find the numbers. Let x denote the smaller part. Then 100 - x = greater part, and 8 times the greater = 8 ( 100 - x). Hence 8 (100 - a) = 1 la: + 2, or 800-8a;=lla: + 2; .-. 19a; = 798, x = 42. Also (100 -x) = 58. Hence the two numbers are 58 and 42. Ex. 3. A post which projects 7 feet above the surface of water is found to have ^ its length in the water and J its length in the mud at the bottom ; find its total length. Let x denote its total length in feet. Then ? is the length in the water. And - is the length in the mud. But the length in the mud, the length in the water, together with 7, is equal to the total length. Hence -^ + -7 + 7 =x, o 4 or 4a; + 3a; + 84 = 12a;; /. 5a; = 84, ora;=16f feet. Ex. 4. A rectangle is 6 feet long ; if it were 1 foot wider its area would be 30 square feet. Find the width. Let x denote the width in feet. Then x+ 1 is the width when one foot wider. The area is 6{x+ 1), but the area is 30 square feet ; .-. 6(jc+1) = 30, or 6a; + 6 = 30. Transposing, 6x = 24 ; .'. x = 4. A practical application. In electrical work equations are of the utmost importance. Asa simple case we may consider what is known as Ohm's Law. This law in its simplest form may be expressed by the equation w s-4 (i) where R denotes the resistance of an electric circuit in certain units called ohms, E the electromotive force in volts, and C the 88 PRACTICAL MATHEMATICS FOR BEGINNERS. current in amperes. An explanation of the law may be obtained from any book on electricity, and need not be given here. Our purpose is only to show that in (1), and in all such equations involving three terms, when two of the terms are given, the remaining one (or unknown quantity) may be found. Ex. 5. A battery contains 30 Grove's cells united in series ; a wire is used to complete the circuit. Find the strength of the current, assuming the electromotive force of a Grove's cell to be 1*8 volts, the resistance of each cell # 3 ohm, and the resistance of the wire 16 ohms. Here Electromotive Force = 30 x 1 *8 = 54 volts. Resistance = (30 x 3) + 16 = 25 .". G jf = 2-16 amperes. Falling bodies. The space s described by a body falling freely from rest in a time t is given by the formula s\gt 2 . It should be noticed that as g has the value 32'2 ft. per sec. per sec, if either s or t be given the remaining term can be obtained. Such equations, which involve three, four, or more terms, are of frequent occurrence. In all cases the substitution of numerical values for all the terms except one enables the remaining term to be obtained. Ex. 6. Lets = 128-8. Find t. Here 128 '8 = J x 322 x t\ . , 2 _ 128-8x2 _ g ' l " 322 ~ Hence *=V8 = 2*8 sec. EXERCISES. XVII. 1. Divide 75 into two parts, so that 3 times the greater shall exceed 7 times the lesser by 15. 2. Divide 25 into two parts, such that one-quarter of one part may exceed one- third of the other part by 1. 3. The sum of the fifth and sixth parts of a certain number exceeds the difference between its fourth and seventh parts by 109 ; find the number. 4. At what times between the hours of 2 and 4 o'clock are the hands of a watch at right angles to each other ? 5. There are three balls, of which the largest weighs one-third as much again as the second, and the second one-third as much again EXERCISES. as the third : the three together weigh 2 lbs. 5 oz. How much do they each weigh ? 6. Five years ago A was 7 times as old as B ; nine years hence he will be thrice as old. Find the present ages of both. 7. Divide 111 between A, B, and G, so that A may have 10 more than B, and B 20 less than C. 8. A broker bought as many railway shares as cost him 1875 ; he reserved 15, and sold the remainder for 1740, gaining 4 a share on the cost price. How many shares did he buy ? 9. Two pedestrians start at the same time from two towns, and each walks at a uniform rate towards the other town, when they meet ; one has travelled 96 miles more than the other, and if they proceed at the same rate they will finish their journeys in 4 and 9 days respectively. Find the distance between the towns and the rates of walking per day in miles. 10. A man gives a boy 20 yards start in 100 yards, and loses the race by 10 yards. What would have been a fair start to give ? v ^_i 11. A father leaves 14,000 to be divided amongst his three ,* children, that the eldest may have 1000 more than the second, and ^v^ twice as much as the third. What is the share of each ? 12. Divide 700 between A, B, and C, so that G may have one- a V fourth of what A and B have together, and that A's share may be 2^ times that of B. ' 13. A cistern can be filled by two taps, A and B, in 12 hours, and by B alone in 20 hours. In what time can it be filled by A alone ? 14. Two cyclists, A and B, ride a mile race. In the first heat A (j-j. |r\_ r < wins by 6 seconds. In the second heat A gives B a start of 58f yards and wins by 1 second. Find the rates of A and B in miles per hour. &f* jLjf- < 15. A slow train takes 5 hours longer in journeying between two given termini than an express, and the two trains when started at the same time, one from each terminus, meet 6 hours afterwards. Find how long each takes in travelling the whole journey. 16. A man walks a certain distance in a certain time. If he had gone half a mile an hour faster he would have walked the distance in 4 of the time ; if he had gone half a mile an hour slower he would have been 2^ hours longer on the road. Find the distance. 17. Two pipes, A and B, can fill a cistern in 12 and 20 minutes respectively, and a pipe G can carry off 15 gallons per minute. If all the pipes are opened together the cistern fills in two hours. How many gallons does it hold ? 18. A man walks at the rate of 3^ miles an hour to catch a train, but is 5 minutes late. If he had walked at the rate of 4 miles an hour he would have been 2^ minutes too soon. Find how far he has to walk. 19. Two trains take 3 seconds to clear each other when passing in opposite directions, and 35 seconds when passing in the same direction. Find the ratio of their velocities. nt- V 1 ~/-*\ Z_^-i- (a: -3) (a? -4) #-3 a; -4 SIMULTANEOUS EQUATIONS 95 n qu **-* 9 a; + 20 2 7 Ex.9. Show that ^ + 5x + Q = ^2 + ^r S - Fractions of a more complicated character may be reduced to partial fractions by an extension of the previous methods. Reference must be made to more advanced works for these cases, and also for the theory of the subject. EXERCISES. XVIII. Solve the equations : L 3 * + f = 42.) 2 - 9 * + |=' f + % = 27. 7x 5)/ = 65.1 -} 3 *.y *' 5 4 ' ~J 10a;-6> 3x-4y=10.J 5. 7x + 3y = \0.\ 6. 2-15y=3-24y=a. 35cc-6?/ = l.J 7. 6*-12y=l, 8^ + 9?/= 18. 8. \6x-y=4x + 2y = Q. 9. 3a?-7y = 7, ll* + 52/ = 87. 10. 3^-42/ = 25.) 5rc + 2y=: 7. J x-y^x + y_ 9l \ 12. 3* + 2y=118.1 2 + 3 ~ 2 * a; + 5y=191.J 2 + 3 and their difference 3. Let x denote one number and y the other. Then, the sum of the numbers is x + y ; but this, by the question, is equal to 19. Hence x + y = \9 (i) Also x~y= 3 (ii) Adding (i) to (ii) 2x = 22; Subtracting (ii) from (i) 2y=16 ; .'. y = 8. Hence, the two numbers are 11 and 8. It is easy by inspection to see that when these are inserted in the equations both are satisfied. /. 11 + 8 = 19 and 11-8 = 3. Ex. 2. If 3 be added to the numerator of a certain fraction, its value will be J, and if 1 be subtracted from the denominator, its value will be ^. What is the fraction ? Let x be the numerator and y the denominator of the fraction. Add 3 to the numerator, then = -. V 3 x 1 Subtract 1 from the denominator, and ^ = ^=; y-1 5 x+3 1 , x 1 y 3' y-l 5 .: Sx + 9=y, (i) and 5x=y-l (ii) Transposing we get y-3x=9 (iii) y-5x=l .(iv) Subtracting (iv) from (iii), 2a; = 8; .". x = 4. Substituting this value of x in (iii), y- 12 = 9; .*. y=21. Hence the fraction is ^-. SIMULTANEOUS EQUATIONS. 97 Ex. 3. A number consisting of two digits is equal in value to double the product of its digits, and also equal to twelve times the excess of the unit's digit over the digit in the ten's place ; find the number. If we denote the digits by x and y, and y denote the digit in the unit's place, then the number may be represented by lOx + y. But this is equal to double the product of the digits ; ,\ \x + y = 2xy (i) The excess of the unit's digit over the digit in the ten's place is (y - x), and we are given that I2(y-x) = 2xy (ii) Hence 10# + y = 12y - \2x ; .'. 22af=ll3/ or 2x = y (iii) Substituting this value in (i) we get 5y+y=y 2 ; ,\ 6y = y 2 or y = Q ; and from (iii), x=S. Hence the number is 36. Ex. 4. Find two numbers in the ratio of 2 to 3, but which are in the ratio of 5 to 7 when 4 is added to each. Let x and y denote the two numbers. Then the first condition that the two numbers are in the ratio of 2 to 3 is expressed by -=% (i) y 3 Similarly, the latter condition, that when 4 is added to each of them the two numbers are in the ratio of 5 to 7, is expressed by x l\4 : (ID y + 4 7 From (i) %x = 2y (iii) From (ii) 7^ + 28 = 5^ + 20 or 7x + 8 = 5y (iv) Multiplying (iii) by 5 and we obtain 15#=10?/ (iv)by2 14a;+16 = 10y Subtracting, x -16= .\ a; = 16. From (iii), y=f# = 24. Hence the two numbers are 16 and 24. The unknown quantities to be found from a simultaneous equation are not necessarily expressed as x and y. It is fre- quently much more convenient to use other letters. Thus P.M.B. G 98 PRACTICAL MATHEMATICS FOR BEGINNERS. pressure, volume, and temperature may be denoted by p, v, and t respectively. Also, effort and resistance may be indicated by the letters E and R. It will be obvious that letters consistently used in this manner at once suggest, by mere inspection, the quantities to which they refer. Some applications. It is often necessary to express the relation between two variable quantities by means of a formula, or equation. The methods by which such variable quantities are plotted and the law obtained have already been explained but practice in solving a simultaneous equation is necessary before any such law can be determined. Ex. 5. The law of a machine is given by R = aE+b, (i) and it is found that when R is 40, E is 10, and when R is 220, E is 50 ; find a and b. Substituting the given values in (i) we get 220 = 50a + 6 (ii) 40=10a + 6 (in) Subtracting, 1 80 = 40a 180 A K Substituting this value in (iii), 6 = 40- 10x4-5 =-5. Hence the required law is R = 4:'5E-5. EXERCISES. XIX. 1. Find the fraction to the numerator of which, if 16 be added, the fraction becomes equal to 4, and if 11 be added to the denomin- ator the fraction becomes l. 2. The difference of two numbers is 14, their quotient is 8. Find them. 3. What fraction is that which, if the denominator is increased by 4, becomes ^ ; but, if the numerator is increased by 27, becomes 2 ? 4. Find that number of two digits which is 8 times the sum of its digits, and the half of which exceeds by 9 the same number with its digits reversed. 5. Find a fraction which will become 1 if 1 is added to its de- nominator, and ^ if 3 is taken from its numerator. EXERCISES. 99 6. Two numbers differ by 3, and the difference of their squares is 69. Find them. 7. If 1 be added to the numerator and 1 subtracted from the denominator of a certain fraction, the value of the fraction becomes 2 ; if 2 be added to the numerator and 2 subtracted from the de- nominator, the value becomes ^. What is the fraction ? 8. Find two numbers such that the first increased by 15 is twice the other when diminished by 3 ; while a half of the remainder when the former is subtracted from the latter, is an eighth of that sum. 9. A, B, and G travel from the same place at the rates of 4, 5, and 6 miles an hour respectively ; and B starts 2 hours after A . How long after B must G start in order that they may overtake A at the same instant ? 10. If six horses and seven cows cost in all 276, while five horses and three cows cost 179, what is the cost of a horse and what is the cost of a cow ? 11. There is a fraction such that when its numerator is increased by 8 the value of the fraction becomes 2, and if the denominator is doubled, its value becomes ; find the fraction. 12. Twenty one years ago A was six times as old as B ; three years hence the ratio of their ages will be 6 : 5 ; how old is each at present ? 13. There are two coins such that 15 of the first and 14 of the second have the same value as 35 of the first and 6 of the second. What is the ratio of the value of the first coin to that of the second ? 14. Each of two vessels, A and B, contains a mixture of wine and water, A in the ratio of 7 to 3, and B in the ratio of 3 to 1 ; find how many gallons from B must be put with 5 gallons from A in order to give a mixture of wine and water in the ratio of 1 1 to 4. 15. Eliminate t from the equations v = u +/L 8=ut + ft\ 16. A racecourse is 3000 ft. long ; A gives B a start of 50 ft., and loses the race by a certain number of seconds ; if the course had been 6000 ft. long, and they had both kept up the same speed as in the actual race, A would have won by the same number of seconds. Compare the speed of A with that of B. 17. The receipts of a railway company are apportioned as follows : 49 per cent, for working expenses, 10 per cent, for the reserved fund, a guaranteed dividend of 5 per cent, on one-fifth of the capital, and the remainder, 40,000 for division amongst the holders of the rest of the stock, being a dividend at the rate of 4 per cent, per annum. Find the capital and the receipts. CHAPTEE X. RATIO, PROPORTION, AND VARIATION. Ratio and proportion. It has already been seen that ratio may be defined as the relation with respect to magnitude which one quantity hears to another of the same kind. By means of algebraical symbols the ratio between two quantities can be expressed in a more general manner than is possible by the methods of Arithmetic. Thus, the ratio between two quantities a and b may be expressed by a : b or ^ ; and the ratio is unaltered by multiply- ing or dividing both terms by the same quantity. Proportion. When two ratios a : b and c : d are equal, then the four quantities a, 6, c, d are said to be in proportion or are Hence a : b = c : d or - = - (i) b a The two terms b and c are called the means, and a and d the extremes. When four quantities are in proportion the product of the means is equal to the product of the extremes. Thus if a : b = c : d then b x c=a x d. This important rule can be proved as follows : As the value of a ratio is unaltered by multiplying both terms by the same quantity we may, in Eq. (i), multiply the first ratio by d and the second by b. Then, we have ^=^ 7 ; hence bc=ad. bd bd PROPORTION. 101 In the proportion \ %\ by adding unity to each side we get b d -' ! In a similar manner subtracting 1 from each side we obtain a b c d /...v T T (m) Dividing (ii) by (iii) then ^| = C -^L CL O C Ci a result often required in both Algebra and Trigonometry. The most general form of the above may be written ma + nb _mc + nd pa + qb pc + qd y whatever m, n, p, and q may be ; this can also be obtained as follows : In Eq. (i) we have |=f. Multiplying both sides by p we obtain " ' * t ' pa_pc T~~d' Again dividing both sides by q, P a _pc qb qd Adding 1 to each side, pa + qb _pc + qd qb qd ' or pa + qb ^qbj) pc + qd qd d In a similar manner we can obtain ma + n mc + nd d Hence pa + qb _ ma + nb . pc + qd mc + nd' . ma+nb _mc+nd pa + qb pc + qd' This important proposition should be tested by substituting simple numbers for the letters. Thus, let a = 3, 6 = 4, c = l'5, d=2, Then %= c - becomes ?=!?. b d 4 2 102 PRACTICAL MATHEMATICS FOR BEGINNERS. Now let m = 5, n = 6, p = 9, q 10. pa + qb pc + qd becomes 5x3 + 6x4 = 5x1-5 + 6x2 9x3 + 10x4 9x1-5 + 10x2' 39 = 19-5 '* 67~33'5 And the ratio of the first two numbers is equal to the ratio of the second. Other simple numbers should be inserted in each case, when it will be found that the two ratios remain equal to each other, or, in other words, the four quantities are proportionals. Mean proportional. When the second term of a proportion is equal to the third, each is said to be a mean proportional to the other'two.* *Thas "6 is said to be a mean proportional to 4 and 9. Geometrical' mean (written G.M.). The mean proportional between two ouar-tities is also called the geometrical mean, and is etyuaVto *jne square rdot of the product of the quantities. Thus the g.m. of 4 and 9 is \^4x9 = 6. Similarly if a : b = b : c then b = Jac. Arithmetical mean (written A.M.) is half the sum of two quantities. Thus, the a.m. of 4 and 9 is - = 6'5. The arithmetical mean of a and c is -^I_. 2 Third proportional. When three quantities are in proportion and are such that the ratio of the first to the second is the same as the second to the third, then the latter is called a third proportional to the other two. Variation. When two quantities are related to each other in such a manner that any change in one produces a corresponding change in the other, then one of the quantities is said to vary directly as the other. The symbol cc is used to denote variation. Thus, the state- ment that x is proportional to y, or, that x varies as y, may be written x qc y. For many purposes, especially to obtain numerical values, it is necessary to replace the sign of variation by that of equality, VARIATION. 103 hence we may write that y multiplied by some constant (k) is equal to x, :. x = ky. The value of h can be obtained when x and y are known. Having obtained the value of k, then, given either x or y, the value of the other can be found. Nearly all the formulae required by the engineer are con- cerned with the sign of variation. As there are so many applications to choose from it is a difficult matter to make a selection. The following are a few typical cases : Ex. 1. The space described by a falling body varies as the square of the time. If a falling body describes a distance of 64 "4 feet in 2 seconds, find the distance moved through in 5 seconds. Here, denoting the space by a 1 and the time by t, then s Strain = j. fl 7$ = ^ = -0002, original length 12 x 12 '' ^ = ^^ = 25 ' 000 ' 000 or 2 ' 5 x 10? lbs ' P er S< 1- in - Ex. 3. The heat H, in calories, generated by a current of G amperes in a circuit, varies as the square of the current, the resist- ance of the circuit R and the time t in seconds during which the current passes, .-. H = 100 lbs. per square inch v is 2 cubic feet, find the pressure when v is 8 cubic feet. 6. It is known that x varies directly as y and inversely as z ; it is also known that x is 500 when y is 300 and z is 14 ; find the value of z when x is 574 and y is 369. 7. If x varies as the square of y, and if x equals 144 when y equals 3, find the value of y when x = 324. 8. A person contracts to do a piece of work in 30 days, and employs 15 men upon it. At the end of 12 days one-fourth only of the work is finished. How many additional hands must be engaged in order to perform the contract ? 9. If 30 men working 9 hours a day can build a certain length of wall in 16 days, find how many youths must be employed to build a similar wall of half that length in 20 days, working 8 hours a day, the work of 4 youths being equal to that of 3 men. 10. If 360 men working 10J hours a day can construct a road 1089 yards long in 35 days ; how long would the job take 420 men working 9 hours a day ? ^.V-Ql. A garrison of 1500 men has provisions for 12 weeks at the rate of 20 ounces for each man per day ; how many men would the same provisions maintain for 20 weeks, each man being allowed 18 ounces per day ? 12. If 20 men can build a wall 70 ft. long, 8 ft. high, and 4 ft. thick in five days, working 7 hours a day, how many hours per day must 30 men work to build a wall 120 ft. long, 12 ft. high, and 3 ft. y* thick in the same time ? 13. The volume of a sphere varies as the cube of the diameter. If a solid sphere of glass 1 -2 inches in diameter is blown into a shell bounded by two concentric spheres, the diameter of the outer sphere being 3*6 inches, show that the thickness of the shell is 0225 inches (nearly). 14. The expenses of a certain public school are partly fixed and /partly vary as the number of boys. In a certain year the number of boys was 650 and the expenses were 13,600 ; in another year the number of boys was 820 and the expenses were 16,000. Find the expenses for a year in which there were 750 boys in the school. CHAPTER XI. INDICES. APPROXIMATIONS. Indices- The letter or number placed near the top and to the right of a quantity which expresses the power of a quantity is called the index. Thus in a 5 , a 7 , a 9 , the numbers 5, 7, and 9 are called the indices of a, and are read as " a to the power five," " a to the power seven," etc. Similarly a b denotes a to the power b. There are three so-called index rules or laws. First index rule. To multiply together different powers of the same quantity add the index of one to the index of the other. To divide different powers of the same quantity subtract the index of the divisor from the index of the dividend. Thus a 3 x a 2 = (a x ax a) (ax a) = a 3+2 =a 5 . Ex. 1. a 3 xa 5 = a 3+5 = a 8 . Ex. 2. a*xa 3 x a*=a 2 + 3 + 4 =a 9 . This may be expressed in a more general manner as follows : a m =(a x a x a. . .to m factors) and a n = (axaxa. .. to n factors), .'. a m xa n = (ax ax a... to m factors) (ax ax a... to n factors) = (axaxa...to m+n factors) = a m+n . This most important rule has been shown to be true when m = 3 and n = 2. Other values of m and n should be assumed, and a further verification obtained. A1 a 5 axaxaxaxa , Also -7= = a 5 ~ 3 = a 2 . a 6 ax ax a . ., , a m a x a x a to m factors Similarly = , x = a- n . a n ax ax a to n factors 108 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 3. Explain why the product is a? when a 3 is multiplied by a 4 , and why the quotient is a when a 4 is divided by a 3 . a 4 xa 3 =(axaxaxa) x (ax ax a) = a 4+3 =a 7 . A1 ataxaxaxa Also ,= a. a 6 ax ax a It is often found convenient to use both fractional and negative indices in addition to those just described. The meaning attached to fractional and negative indices is such that the previous rule holds for them also. When one fractional power of a quantity is multiplied by another fractional power the fractional indices are added, and when one fractional power is divided by another the fractional indices are subtracted. a 2 xa 2 =a *+i =a L =a, Hence, the meaning to attach to a? is the square root of a ; to a 5 is the cube root of a squared, and to a 3 the cube root of a. Thus si a can be written as a^, %/a can be written as d 3 . Also and sja va Again -r=rxa a* _-_- Also Similarly axaxa axaxa = 3-3^0 Generally, since a m xa n =a m+n is true for all values of m and If n be 0, then a m xa = a m+0 o a 1 a = = 1. INDICES. 109 The second index rule. To obtain a power of a power multiply the two indices. Ex. 1. Thus to obtain the cube of a 2 we have (a?) s = (a xa){ax a) (a x a) = a 2x3 = a 6 , where the index is the product of the indices 2 and 3. Ex. 2. Find the value of (2'15 2 ) 3 . (2-15 2 ) 3 =2-15 2X3 = 215 6 =98-72, or, expressing this rule as a formula, we have (a m ) n =a mn , .'. a quantity a m may be raised to a power n by using as an index the product run. The third index rule. To raise a product to any power raise each factor to that power. Ex. 1. (abcd) m = a m xb m xc m xd m . Ex. 2. Let a=l, 6=2, c=3, d = 4, and m = 2. Then {abcd) m = (1 x 2 x 3 x 4) 2 = l 2 x 2? x 3 2 x 4 2 = 24 2 =576. In fractional indices the index may be written either in a fractional form or the root symbol may be used. The general form is a n . This may be written in the form v a m , which is read as the n th root of a to the power m. Ex. 6. 2*= Z/2 5 = 4/32=3-174. Ex. 7. Find the values of 8*, 64~^, 4~^* Here 8*= W=$M=4. i 1 1 4 -f-J_-_L__i 4 I v/64 8* Ex. 8. Find the value of 64* + 4 l * + 2 s 5 + 27** Here 64* = 8, 4 1<5 = 4* = 64*=8, 22' 5 =2^=32* = 5-656, 27^=3. Hence 64* + 4 1 -5 + 2 2,5 + 27*=24'656. Ex. 9. Find to two place's of decimals the value of x 2 - 5x* + x~ 2 when x=5. 110 PRACTICAL MATHEMATICS FOR BEGINNERS. Here x 2 - 5x^ + x~ 2 = 25- W5 + ^ = 25 - y x 2236 + -04 = 13-86. Powers of 10. Reference has already, on p. 25, been made to a convenient method of writing numbers consisting of several figures. Thus the number 6340000 is 6-34x1000000, or, 6*34 xlO 6 . Similarly 6340- 6'34 x 10 3 , 634 = ?2~ -6-34X10" 1 , . 000634 =-^- = 6-34 xlO" 4 , etc. lOOOO Suffix. A small number or letter placed at the right of a letter but near the bottom is called a suffix and it is important to notice the difference between an index and a suffix. Thus P 2 means PxP, but P 2 is merely a convenient notation to avoid the use of a number of letters, each of which may refer to different magnitudes of similar quantities. In this manner the letters P , P lt P 2 ..., etc., may each refer to forces, etc., of different magnitudes, and in different directions. Binomial theorem. We have already found that (a + b) 2 = a 2 + 2ab + b 2 , and by multiplying again by a + b we obtain (a + bf = a 3 + 3a 2 5 + Zab 2 + b 3 . It is seen at once that some definite arrangement of the coefficients and indices of such expressions may be made so that another power, say (a + by, can be written down : the method used, and called the Binomial Theorem, is very important. The rule should be applied to the operation of expanding several simple expressions, such as (a + b) 3 , (a + b)*, etc., and afterwards committed to memory. / 7\ wa M-1 6 n(n 1) 07 , (a+b) n = a n + + v x 2 ' a n - 2 b 2 +.... Take n=2, then and :; = 2ab. Hence (a + b) 2 = a 2 + ^ + bK ( 2 ' 1) = a 2 + 2ab + < l l . z APPROXIMATIONS. 1 1 1 Take ?i = 3 ; here a n =a? ; ^ =3a?b, etc. *b 1 , , ,., Sl 3a 2 6 , 3.2a& 2 , 3.2.1 0/3 = a 3 + 3a 2 6 + 3a& 2 + 3 . As a handy check, the reader should notice, that in each term the sum of the powers of a and b is equal to n. Thus, when % = 3, in the second term a is raised to the power 2, and b to the power 1. Therefore sum of powers = 3. Also, each coefficient has for its denominator a series of factors 1 . 2 . 3 . . . r, where r has the same numerical value as the power of b in that term. Thus, in the term containing 6 3 , a must be raised to the power T&3, and the coefficient must be n(n-l)(7i-2) 1.2.3 "Writing down terms in the numerator to be afterwards can- celled by corresponding numbers in the denominator, may appear to the beginner to be an unnecessary process, but to avoid mistakes it is better to write out in full, as above, and afterwards to cancel any common factors in the numerator and denominator. Approximation. The expansion of /, x . , na n(n l)a 2 (l +a ) ls l + _ + _L__^ + ... when a is a very small quantity, the two first terms are for all practical purposes sufficient ; thus, when a is small (1 + a) n = 1 + na (approximately). Similarly when a and b are small quantities (1 + a) n (1 + b) m = 1 + na + mb (approximately). Thus if a= "01 and 71 = 2, then (l + -01) 2 = l+2x'01 = l-02. Ex. 1. (1 + -05)2 = 1 + 2 x -05= 1-1; more accurately (1*05) 2 = 1*1025. Ex. 2. (l + -05) 3 = l + 3x -05 = 1*15. Ex. 3. 4/( 1*05) = (1 + -05)*= 1 + Jx -05 = 1-0167. Ex. 4. 3 J = ( 1 + -05)"^ = 1 - J x 05 = 1 - -0167 = '9833. 112 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 5. Find the superficial and cubical expansion of iron, taking a, the coefficient of linear expansion, as '000012, or 1*2 x 10 " 5 . If the side of a square be of unit length, then when the tempera- ture is increased by 1 C. , the length of each side becomes 1 + a, and area of square is ( 1 + a) 2 = 1 + 2a + a 2 . .'. (l + a) 2 =l+2x'000012 + ('000012) 2 . As a is a very small quantity its square is negligible. Hence the coefficient of superficial expansion is 2a -000024, or 2*4 x 10 " 5 . Again, (l + a) 3 may be written as 1 + 3a, neglecting the terms in a 2 and a 3 . :. coefficient of cubical expansion = 3a = '000036 = 3 '6 x 10" 5 . W/ EXERCISES. XXI. 4 1. Multiply a? + b* + c 2 - dfi - ca? - r*M by a* + b% + c. 4 2. Find the value of (i) >/64 + n/4 5 " + 2* + #27 - 9*. (\i)J' + sl2 E + y&. Simplify fi (pl>-g)g+g x (qy)+ a m-n a m- S n a m-6n g _ (a* > )* > ~* ' ' a n_TO a w ~ 3m a n " 6w ' 9. If a glass rod 1 inch long at C. is 1 '000008 inches long at 1 C, find the increase in the volume of 1 cubic inch of the glass when heated from C. to 1 C. 10. How much error per cent, is there in the assumption (l+a)(l + 6) = l + a + 6 when a =-003, 6='005? 11. Using the rule (1 + a) n = 1 + na, find the values of VI '003 and (*996) 2 , and find the error per cent, in the latter case. 12. How much error per cent, is there in the assumption that (l+a)(l + /3) = l+a + j8, when a = - '002, )8= -'004? 13. Having given 10^=3*1623, and 10 = 1 '3336, find the values of 10 and 10 to five significant figures. 14. If x = 1 '002 and y=0'997, write down the values of x 3 and y* correct to three places of decimals. EXERCISES. 113 15. Given that 10*=3'1623, 10^ = 1-1548, 10^ = 17783, 10^=1-0746, 10*= 1-336, find to five significant figures the values of 10 7!r , 10 1 **, 10*. Explain how you would illustrate that 10= 1. lfr. Divide (x 3 y mn )m by (x 2 y mn )n. 17. Raise {a?b{a?bc) T }^ to the 7 th power. 18. Find the m th root of 2a m b 2m c 2 . / 20 {(oho)V r (a\*y ' \Vb" . a/by}* ' Iw /' 21. (ilX'W^vlJAr 1 )- (ii) ax~ l + g- l x + 2 , a s x ^ + a % T -1 22. Find the value of a; 2 - f x* + ar 1 when a? =3. 23. Va'^'Hv^- 1 . 24. Find the value of 2-W + 2" 3 3 _ y - 10(27#)~* when a; =64. P.M.B. CHAPTER XII. BRITISH AND METRIC UNITS OF LENGTH, AREA, AND VOLUME. DENSITY AND SPECIFIC GRAVITY. Measurement. The measurement of a quantity is known when we have obtained a number which indicates its magnitude. It is necessary, therefore, to select some definite quantity of the same kind, as a unit, and then to proceed to find how many times the unit is contained in the quantity to be measured. The number of times that the unit is contained in the given quantity is the numerical value of the quantity. Units of length. In order that length may be measured there must be both a unit and a standard. The unit is a certain definite distance with which all other distances can be com- pared ; and a standard is a bar on which the unit is clearly, accurately, and permanently marked. The two units most generally adopted are the yard and the metre. The British System. In this system the unit of length is the yard. It may be defined as the distance between two lines on a particular bronze bar when the bar is at a certain temper- ature (62 F.). The bar is deposited at the Standards Office of the Board of Trade. British Measures of Length. [The unit is divided by 3 and 36, etc. ; also multiplied by 2, 51 220, and 1760.] 12 inches = 1 foot. 40 poles, or 220 yards = 1 furlong. 3 feet =1 yard (unit). 8 furlongs^ 2.yards = 1 fathom. 1760 yards \ = 1 mile. 5| yards = 1 rod, or pole. or 5280 feet J 6080 feet = l knot, or nautical mile. MEASUREMENT OF LENGTH. 115 The French or Metric System. The Metric System is extensively used for all scientific, and in many cases for com- mercial purposes, and for many purposes is better and simpler than the British method. The metre is divided into 10 equal parts called decimetres ; the decimetre is divided into 10 equal parts each called a centimetre : hence a centimetre is one hundredth of a metre, and this sub- multiple of the unit is the most commonly used of the metric measures of length. The centimetre is divided into 10 equal parts each known as a millimetre. The metre is equal in length to 39*37 inches, and is thus slightly longer than our yard. Its length is roughly 3 feet 3J inches, which number can be easily remembered as it consists throughout of threes. The foot is equal in length to 30'48 centimetres. It will be seen on reference to Fig. 49, which represents one end of a steel scale, that a length of 10 cm. is approximately CENTIMETRE Inch 1 2 3 4 1 i ill! ill III c b J i i~ * Fig. 49. Comparison of inches and centimetres. The inches and centimetres are not drawn full size, but their comparative dimensions may be seen. equal to 4 inches. A more accurate relation to remember is that a length of 25*4 centimetres, or 254 millimetres, is equal to the length of 10 inches. Thus, the distance from a to b may be expressed as 1 inch, 2 '54 centimetres, or 25*4 millimetres. The following approximate relations are worth remembering : 35 yards = 32 metres. 10 metres = 11 yards, or 20 metres equals the length of a cricket pitch = 1 chain. 5 miles = 8 kilometres. 116 PRACTICAL MATHEMATICS FOR BEGINNERS. British to Metric Measures Metric to British. of Length. 1 millimetre = 039 inch. 1 inch = 2-54 centimetres. l centimetre = 394 inch. 1 foot= 30'48 centimetres. r 39*3/1 inches. 1 yard = 0*914 metre. 1 metre = -j 3*28 feet. 1 mile = 160933 metres. I 1094 yards. 1 kilometre = 0'621 mile. Abbreviations. The following abbreviations are generally used, and should be carefully remembered ; this may be easily effected by taking the precaution to use the abbreviations on all possible occasions. Length. in. is used to denote inch or inches. ft. 55 55 feet. kilom. 55 55 kilometres. dcm. 55 55 decimetre or decimetres. cm. 55 55 centimetre or centimetres. mm. 55 55 millimetre or millimetres. gm. 55 55 gram or grams. kilog. 55 55 kilogram. Unit of Area. Measurement of area, or square measure, is derived from, and calculated by means of, measures of length. Thus, the unit of area is the area of a square the side of which is the unit length. Area of a square yard, or unit area. If the unit length be a yard proceed as follows : Make AB equal to 3 feet, as in Fig. 50, and upon AB construct a square. Divide AB and BC each into 3 equal parts, and draw lines parallel to AB and BC, as in the figure. The unit area is thus seen to consist of 9 smaller squares, every side of which represents a foot ; thus, the unit area, the square yard, contains 9 square feet. The smaller measures of length, the foot and the inch, are much more generally used than the yard. If the unit of length AE (Fig. 51) be 1 foot, the unit of area AEF is 1 square foot. In a similar manner, when the unit of length is 1 inch, the unit of area is 1 square inch. If the unit of length be 1 centimetre, the unit of area is 1 square centimetre (Fig. 51). MEASUREMENT OF AREA. 117 If the side of the square on AE (Fig. 50) represent, on some convenient scale, 1 foot, then by dividing AE and A F each into D C A E B Pig. 50. 1 square yard equals 9 square feet, or 9x144 square inches. 12 equal parts, the distance between consecutive divisions would denote an inch. If through these points lines be drawn parallel to AE and AF respectively, it will be found that there are 12 rows of squares parallel to AE, and 12 squares in each of these 12 rows. Hence, the area of a square foot represents 144 square inches Pig. 51. Square inch and square centimetre. British Measures of Area or Surface. [ Unit area= 1 square yard. Larger and smaller units obtained by multiplying by 4840 and dividing by 9 and 1296.] 144 square inches = 1 square foot. 1296 square inches or 9 sq. ft. = 1 square yard. 4840 square yards = 1 acre. 640 acres = 1 square mile. * 118 PRACTICAL MATHEMATICS FOR BEGINNERS. When comparatively large areas, such as the areas of fields, have to be estimated, the measurements of length, or linear measurements, are made by using a chain 22 yards long. Such a chain is subdivided into 100 links. The square measurements, or areas, are estimated by the square chain, or 484 (22x22) square yards in area. Or the area of a square, the length of one side of which is 22 yards, is 100x100 = 10000 sq. links ; for each chain consists of 100 links. Hence we have the relation : 1 chain 1 square chain = 10 square chains = square inches (sq. , square feet 30J square yards 40 square poles 4 roods 640 acres 144 9 22 yards = 100 links. = 484 square yards = 10000 sq. links. = 4840 square yards = 1 acre. in.) = l square foot (sq. foot). = 1 square yard (sq. yd.). = 1 square perch, rod, or pole (sq. po.). = 1 rood (r.). = 1 acre (ac.) = 4840 square yards. = 1 square mile (sq. m.). Metric measures of area. As the metric unit of length is the metre, the unit of area (Fig. 52) is a square ABDB, having the length of its edge equal to 1 metre, and its area consequently equal to 1 square metre. If AB and BD are each divided into 10 equal parts and lines drawn parallel to AB and BD, as shown, the unit area is divided into 100 equal squares, each of which is a square decimetre. In scientific work the centimetre is the unit of length usually selected, and the unit of area is one square centimetre (Fig. 51). D 101 9 J i 7 e 1 _s 2 B Fig. 52. Representing a square metre divided into 10 decimetres. Scale ^, Metric Measures of Area. 100 square millimetres = 1 square centimetre. 10000 = 100 sq. cm. m 1 sq. decim. 100 decimetres = 1 square metre. MEASUREMENT OF VOLUME. 119 Conversion Table. British to Metric. 1 sq. in. = 6 '451 sq. cm. 1 sq. ft. = 929 sq. cm. 1 sq. yard = 8361*13 sq. cm. 1 acre =4046*7 sq. metres, 1 sq. mile= 2*59 sq. km. = 2 59 x 10 10 sq.cm. Metric to British. 1 sq. cm. = 0*155 sq. m. 1 sq. m. =10*764 sq. ft. lsq.m. = 1*196 sq. yard. 1 sq. km.= 0*3861 sq. mile. EXERCISES. XXII. 1. Find the number of square metres in (i) 10 square feet, (ii) 10 square yards. 2. Find the number of square metres in a quarter of an acre. 3. Find the number of square metres in 1000 square yards. 4. Express 2 sq. ft. 25 sq. in. as the decimal of a square metre. 5. Reduce 1000 square inches to square metres. 6. Find the number of square miles in 25,898,945 square metres. 7. Find which is greater, 10 sq. metres or 12 sq. yards, and express the difference between these areas as a decimal of a square metre. Units of capacity and volume. In the British system an arbitrary unit, the gallon, is the standard unit of capacity and volume, and is defined as the volume occupied by 10 lbs. of pure water. A larger unit is the volume of a cube on a square base of which the length of each side is 1 foot and the height also 1 foot. The volume of such a cube is one cubic foot. A good average value for the weight of a cubic foot of water is 62 -3 lbs. For convenience in calculations, a cubic foot is some- ^ iS\,^ ^ "TOE ;h]]G4^ B times taken to be 61 gallons, and its weight 1000 oz., or 62*5 lbs. Hence the weight of a pint is 1^ lbs. The connection between length, area, and volume, may be shown by a diagram as in Fig. 53. Let A BCD represent a square having its edge 1 yard, the area of the square is 9 square feet. If the vertical sides, one of which Fig. 53. Showing a cubic yard and a cubic foot. 120 PRACTICAL MATHEMATICS FOR BEGINNERS. is shown at DE, be divided into three equal parts, and the remaining lines be drawn parallel to BE and the base respec- tively. Then, as will be seen from the figure, there are nine perpendicular rows of small cubes, the sides being 1 foot in length, area of base 1 square foot, and volume 1 cubic foot. Also there are three of these cubes in each row, making in all 3x9 = 27. Thus, 1 cubic yard = 27 cubic feet, i.e. 3x3x3 = 27, and the weight of a cubic yard of pure water would therefore be 27 x 62-3 lbs. = 1682' 1 lbs. In this example, and also in considering the weight of a gallon, the student should notice that the specification " pure " water is necessary, for if the water contains matter either in solution or mixed with it, its weight would be altered. Thus, the weight of a cubic foot of salt water is usually taken to be 64 lbs., and the weight of a gallon of muddy water may be 11 or 12 lbs. in- stead of 10 lbs. Units of Volume and Weight. 1728 cubic inches = I cubic foot. 27 cubic feet = 1 cubic yard. 1 gallon = -1605 cub. ft. =277 '3 cub. in. One fourth part of a gallon is a quart and an eighth part is a pint. Metric measures of volume. We proceed in a similar way when we wish to measure volumes by the metric system. A block built up with cubes re- presenting cubic centimetres is shown in Fig. 54. Each side of the cube measures 10 centimetres, and its volume is therefore a cubic decimetre. There are 10 centimetres in a decimetre, so the edge of the decimetre cube is 10 centimetres in length ; the area of one of its faces is 10x10 = 100 square centimetres ; and its volume is 10x10x10 = 100x10 = 1000 cubic centimetres. This unit of volume is caned a Litre. At ordinary temperature / 2 .'i J G 7 S CI "InM 9 Iff s ry 7Uh 6 m '' rr ' Uj/ ' 1 * m 1 1 Fig. 54. Cubic decimetre (1000 cubic centimetres) or 1 litre holds 1 kilogram or 1000 grams of water at 4C. DENSITY. 121 it is very nearly a cubic decimetre, or 1000 cubic centimetres (Fig. 54), and is equal to 176 English pints. We have found that the unit of area is, for convenience, taken to be one square centimetre, the corresponding unit of volume is the cubic centimetre (c.c). For all practical purposes a litre of pure water weighs 1 kilo- gram or 1000 grams. Thus we have the relation 1 gram = weight of 1 cubic centimetre of water. 1 litre = 1 cubic decimetre = 1000 grams. It is advisable to remember that there are 453'59 grams in a pound ; that 1 gram = 15'432 grains and that a kilogram = 21 lbs. The unit of weight is one pound. A smaller unit is obtained by dividing by 7000 and larger units by multiplying by 14, 112, and 2240, as follows : 7000 grains = 1 lb. 14 lbs. =1 stone. 112 lbs. =1 hundred- weight, 1 cwt. 20 cwts. = 1 ton = 2240 lbs. Conversion Table. 1 cub. in. = 16*387 cub. cm. 1 ft. =28316 1 yard = 764535 1 pint =567*63 1 gallon =4541 1 grain ='0648 gm. 1 ounce avoirdupois = 28*35 gm. 7000 grains 1 1 pound (lb.)} =453-59 gm. 1 ton =1-01605 xl0 6 gm. 10 milligrams = 1 centigram. 10 centigrams = 1 decigram. 1000 grams = 1 kilogram. Density. The density of a substance is the weight of unit volume. Assuming the density to be uniform, the density of a substance, when the unit of weight is one pound and the unit of volume one cubic foot, is the number of pounds in a cubic foot of substance. 1 c.cm. = -061 cub. in. 1 litre = 61-027 = 1*76 pint or = 22 gallon. 1 gram = 15*43 grains. 1 kilo =2*2 lbs. 122 PRACTICAL MATHEMATICS FOR BEGINNERS. In the cases where metric units are adopted, the density is the number of grams in a cubic centimetre of the substance. Density of water. The weight of a cubic foot of water is 62*3 lbs., of a cubic centimetre 1 gram, and of a litre 1 kilogram. Relative density. The relative density of a substance is the ratio of its weight to the weight of an equal volume of some sub- stance assumed as a standard. It is necessary that the standard substance should be easily obtainable at any place in a pure state. Pure water fulfils these conditions. Specific gravity. The relative density of a substance is usually called its specific gravity. The specific gravities of various sub- stances are tabulated in Table I. If s = specific gravity of a body, then the weight of 1 cub. ft. = sx62-3 lbs. If the substance is a liquid, then 1 gallon = s x 10 lbs. Again, as a litre of water weighs 1 kilogram, weight in kilograms = 5 x volume in litres. A vessel containing 1 cub. ft. of water would when the water is replaced by mercury weigh 13*596 x 62 3 lbs. If for cast iron s is 7*2, then the weight of a cub. ft. = 7-2 x 62-3 lbs. =448*56 lbs. The weight of a cub. centimetre will be 7*2 grams. The weight of V cubic feet of water will be Fx623 lbs. or Vw, where w is the weight of unit volume of water. Hence if V denote the volume of a body in cub. ft. its weight will be Vws. In this manner it is customary to define specific gravity as the ratio of the weight of a given volume of a substance to the weight of the same volume of water. If the volume of the body is obtained in cubic inches then w will denote the weight of one cubic inch (the weight of one cubic inch of water = 62 -3 -=-1728= *036 lbs.). Principle of Archimedes. The method of obtaining the specific gravity of a solid (not soluble in water) depends on what is known as the " Principle of Archimedes " : When a body is immersed in a liquid it loses weight equal to the weight of the liquid which it displaces ; that is, if the weight of a body is obtained, first in air, and next when immersed in water, the PRINCIPLES OF ARCHIMEDES. 123 difference in the weights is the weight of an equal volume of water : . . e , , weight in air .'. specific gravity of body= . t , . s- 2 . , , . r . r J J weight in air -= weight in water Ex. 1. A piece of metal weighs 62*63 grains in air and 56 grains in water. Find its specific gravity. 62-63 SG, "62-63-56" y Ex. 2. A piece of metal of specific gravity 9*8 weighs in water 56 grains. What is its true weight ? Let w denote its true weight. w Then 0-8= w-5G ;. 9-$w-9'8x56 = w; 9-8x56 w = 8-8 62-36 grains. Ex. 3. If 28 cubic inches of water weigh a pound, what will be the specific gravity of a substance, 20 cubic inches of which weigh 3 lbs. ? As 20 cubic inches weigh 3 lbs., 1 cub. in. =$j ; .-. 28 cub. in. = 28 x ^ lbs. = 4 lbs. But the same volume of water weighs 1 lb. ; /. specific gravity = -y = 4-. TABLE I. RELATIVE WEIGHTS. Name. Weight of Unit Volume in pounds. Relative Density , or Specific Gravity. Cub. ft. Cub. in. Water, .... Cast Iron, Wrought Iron, Steel, .... Brass, .... Copper, .... Lead, .... Tin, .... Antimony, 62-3 450 480 490 515 552 712 462 418 036 26 28 29 298 3192 4121 267 242 1 7-2 7 698 7-85 8-2 8-9 11-418 7'4 6 72 124 PRACTICAL MATHEMATICS FOR BEGINNERS. EXERCISES. XXIII. 1. What is the specific gravity of a substance, 20 cubic inches of which weigh 3 lbs. ? 2. A body A has a volume 1 *35 cub. ft. and a specific gravity of 4*4, a second body B has a volume of 10'8 cub. in. and a specific gravity of 19 '8 ; what ratio does the quantity of matter in A bear to that in B ? 3. If the specific gravity of iron be 7 '6, what will be the apparent weight of 1 cwt. of iron when weighed in water, and what weight of wood, of specific gravity 0*6, must be attached to the iron so as just to float it ? 4. A piece of iron weighing 275 grams floats in mercury of density 13*59 with of its volume immersed. Determine the volume and density of the iron. 5. A ship weighing 1000 tons goes from fresh water to salt water. If the area of the section of the ship at the water line be 15,000 sq. feet, and the sides vertical where they cut the water, find how much the ship will rise, taking the specific gravity of sea water as 1-026. 6. A cubic cm. of mercury weighs 13'6 grams ; obtain the equi- valent of a pressure of 760 mm. of mercury in inches of mercury, in feet of water, in pounds per square inch and per square foot, and in kilograms per square cm. 7. A cubical vessel, each side of which is a decimetre, is filled to one-fourth of its height with mercury, the remaining three-fourths with water, find the total weight of the water and the mercury. 8. Find the number of kilograms in '7068 of a ton. 9. The area of a pond is half an acre when frozen over ; find the weight of all the ice if the mean thickness be assumed to be 2 inches. Specific gravity of ice *92. 10. A body weighs 80 lbs. in air, its apparent weight in water is 56 lbs. and 46 lbs. in another liquid. Find the specific gravity of the liquid. 11. Three pints of a liquid whose specific gravity is 0*6 are mixed with four pints of a liquid specific gravity 0'81, and there is no con- traction ; find the specific gravity of the mixture. CHAPTER XIII. MULTIPLICATION AND DIVISION BY LOGARITHMS. Logarithms. By the use of logarithms computations involving multiplication, division, involution, and evolution are made much more rapidly than by ordinary arithmetical processes. Many calculations which are very difficult, or altogether im- possible, by arithmetical methods are, moreover, readily made by the help of logarithms. Logarithms of numbers consist of an integral part called the index or characteristic, and a decimal part called the mantissa. If the reader will refer to Table III., he will find that opposite each of the numbers from 10 to 99 four figures are placed. These four figures are positive numbers, and each set of four is called a mantissa: the characteristic, which may be either positive or negative, has to be supplied in a way to be presently described when writing down the logarithm of any given number. Logarithmic tables of all numbers from 1 to 100000 have been calculated with seven figures in the mantissa, but for ordinary purposes, and where only approximate calculations are required, such a table as that shown in Table III., at the end of this book, and known as four-figure logarithms, is very con- venient. By means of the numbers 10 to 99, with (a) those at the top of the table, and (b) those in the difference column on the right, the logarithm of any number consisting of four significant figures can be written down. In logarithms all numbers are expressed by the powers of some number called the base. The logarithm of a number to a given base is the index showing the power to which that base must be raised to give the number. 126 PRACTICAL MATHEMATICS FOR BEGINNERS. If N denote any number and a the given base, then by raising a to some power a; we can get N. This is expressed by the equation #=a*. Any number can be used as the base, but, as we shall find, the system of logarithms in which the base is 10 is that commonly used. Thus, if the base be 2, then as 8 = 2 3 , 3 is the logarithm of 8 to the base 2. This can also be expressed by writing log 2 8 = 3. In a similar manner, if the base be 5, then 3 is the logarithm of 125 to the base 5 ; .'. log 5 125 = 3. Also 64 = 2 6 = 4 3 = 8 2 . Similarly, 6 is the log of 64 to the base 2 ; 3 is the log of 64 to the base 4 ; 2 is the log of 64 to the base 8, etc. ; .-. log 2 64 = 6; log 4 64 = 3; log 8 64 = 2, etc., using in each case the abbreviation log for logarithm. Logarithms to the base 10. It is most convenient to use 10 as the base for a system of logarithms. It is then only neces- sary to print in a table of such logarithms the decimal part or mantissa ; the characteristic can, we shall see, be determined by inspection. The tables are in this way less bulky than would otherwise be the case. When calculated to a base 10, logarithms are known as Common Logarithms. Since N= 10* ; ' log 10 iV=#. Or by definition, substituting positive numbers for A 7 , as 1 = 10 Also, 10 = 10 1 Again, 100 = 10 2 .'. log 1=0. .*. log 10 = 1. /. log 100 = 2, etc. In the chapter on Indices (p. 110) we found that "1, or j^, can be written as 10 -1 ; also '01, or t q, can be written as 10~ 2 . Hence log *1 = log ^ = - 1, and log *01 = - 2, etc. The mantissa is always positive, and instead of writing the LOGARITHMS. 127 negative sign in front of the number, it is customary in logarithms to place it over the top ; thus log '1 is not written - 1 but as 1, and log '01 = 2. In the preceding logarithms we have only inserted the characteristic ; each mantissa consists of a series of ciphers. Thus, log 1=0-0000, log 10 = 1-0000, log 100 = 2-0000, and so on. As the logarithm of 1 is zero, and log 10 is 1, it is evident that the logarithms of all numbers between 1 and 10 will consist only of a certain number of decimals. Thus, log 2 = -3010 indicates, that if we raise 10 to the power 3010 we shall obtain 2, or 10 3010 =2. In a similar manner, the logarithm of 200 = 2 x 100 might be written as 10 2 xl0 3010 , /. 200 = 10 2 ' 3010 . Hence we write log 200 = 2 '30 1 0. Also -0002 = j^tjo o = 2 x 10 ~ 4 - Hence -0002 = 10 T3010 . :. log -0002 = 4-3010. The characteristic of a logarithm. Eef erring to Table III., opposite the number 47 we find the mantissa -6721, and as 47 lies between 10 and 100 the characteristic is 1. Hence the log of 47 is 1-6721. Again, the number 470 lies between 100 and 1000, and there- fore the characteristic is in this case 2 ; .-. log 470 = 2-6721. In a similar manner, the. logarithms of 4700 and 47000 are 3'6721 and 4'6721 respectively ; in each case the mantissa is the same, but the characteristic is different. The rule by which the characteristic is found may be stated as follows : The characteristic of any number greater than unity is positive, and is less by one than the number of figures to the left of the decimal point. The characteristic of a number less than unity is negative, and is greater by one than the number of zeros which follow the decimal point. X28 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 1. To write down log -047 Here the two significant figures are 47, and the mantissa is the same as before._ As one zero follows the decimal point, the characteristic is 2 ; .-. log -047 = 2-6721. Again, to obtain the log of '00047. There are three zeros following the decimal point, the characteristic is 4, and .-. log -00047 = 4-6721. Similarly, in the case of log -47. Here the rule will give I for the characteristic ; .-. log -47 = 1-6721. Another method of determining the characteristic is to treat any given number as follows : 470 = 47 x 100 = 4-7 x 10 2 . Hence as before the characteristic is 2. Similarly, 4700 = 4*7 x 10 3 , '47 = 4*7 x 10" 1 , 047 = 4-7 x lO- 2 , -0047 = 4*7 x 10~ 3 . If all numbers are written in this convenient form the characteristic is the index of the multiplier 10. If this method be applied to all numbers it will save the trouble of remembering rules. To obtain the logarithm of a number consisting of four figures. Ex. 1. Find the log of 3768. First look in Table III. for the number 37, then the next figure 6 is found at the top of table, so that the mantissa of log 376= -5752. At the extreme right of the table will be seen a column of differ- ences, as they are called ; thus, under the figure 8 on a horizontal line with 37 is found the number 9. This must be added to the mantissa previously obtained. Hence we have mantissa of log 376 = 5752 Add difference, 9 .-. mantissa for log 3768 = 5761 Hence log 3768 = 3-5761, also log -003768 = 3-5761, and log -3768 = 1-5761, etc. LOGARITHMS. 129 To find the number corresponding to a given logarithm or the antilogarithm of a number. Ex. 2. Given the log 2*4725, to find the number. From Table IV. of antilogarithms. Opposite the mantissa -472 we have 2965. In the difference column under the number 5, and on the horizontal line 47, we have the figure 3. Hence the corresponding mantissa = 2968, and as the characteristic is 2, the number required is 296*8. If the given logarithm had been 2*4725 the required number would be -02968. Multiplication by logarithms. Add the logarithms of the multiplier and multiplicand together : the sum is the logarithm of their product. The number corresponding to this logarithm is the product required. Let a and b be the numbers. Let log a = x and log b =y ; .; a = 10% 6 = 10*. or log 10 a& = x + y = log a + log 5. Ex. 1. Multiply 2*784 by 6*85. ' From Table III. log 278= 4440 Diff. col. for 4= _6 /. log 2*784= -4446 Also log 6 '85 = -8357 .*. logarithm of product = 1 *2803 From Table IV. antilog280= 1905 Diff. col. for 3 = 1 antilog 2803 = 1906 Hence 2 *784 x 6 *85 = 1 9 06. Ex. 2. Multiply -002885 by *0915. log -002885 = 3*4602 *0915 = 2*9614 4-4216 Hence -002885 x -0915 = -000264. P.M.B. I 130 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 3. Find the numerical value of a x b when a 32 '4, b = '000467. log 32 -4 =15105 '000467 = 4 '6^93 21798 .'. ax b= -01513. Using the data of Ex. 1, to prove the rule, then by the definition of a logarithm '4446 is the index of the power of 10, which is equal to 2-784 .-. 10' 4446 = 2-7S4. Similarly 10 8357 = 6'85, .-. 2-784 x 6 -85 = 10 -4446 x 10 8357 = 10 1 ' 2803 . EXERCISES. XXIV. Multiply 1. -000257 by 3-01. 2. -000215 by -0732. 3. -0032 by 23 45. 4. 3-413 by 10 16. 5. 05234 by 3 87. 6. 4 132 by -625 and -1324 by -00562. 7. 4*017 by -00342. 8. 003 x 17 x 004 x 20000. 9. 76 05 by 1 036. 10. Find the numerical value of a x b when (i) a =14-95, b =00734. (ii) a = 420-3, 6 = 2'317. (iii) a = 5-617, 6= '01738. (iv) a =-01342, b= -0055. H. Calculate (i) 23-51 x 6 71. (ii) 168-3x2-476. 12. Why do we add the logarithms of numbers to obtain the logarithm of their product ? Division by logarithms. Subtract the logarithm of the divisor from the logarithm of the dividend and the result is the logarithm of the quotient of the two numbers. The number corre- sponding to this logarithm is the quotient required. Let a and b be the two numbers. Let log a = x and log b =y ; .-. a = l0*, b = 10P. Hence hw= 10X - y > or log 10 |=^-y = loga-log6. LOGARITHMS. 131 Using this rule for division, it is an easy matter to write down the logarithm of a number less than unity, and to verify the rule given on p. 127. Thus, log 047 = 2-6721. 4 "7 This may be verified by noting that *047 = y^.; :. log -047 = log^5 =log 4-7 - log 100= -6721 - 2 = 2-6721. In a similar manner *47 = -y^r " :. log -47 = log^=log 4-7 -log 10 = '6721 - 1 = 1-6721. Ex. 1. Divide 3-048 by -00525. From Table III. log 304 = 4829 Diff. col. for 8, 11 /. log 3-048= -4840 (i) Also log -00525 = 3-7202 .'....(ii) Subtracting (ii) from (i), 2-7638 From Table IV. antilog 763 = 5794 Diff. col. for 8, 11 .-. antilog of 7638 = 5805 Hence log 2 -7638 = 580-5 ; .-. 3 048 -r -00525 = 580-5. Ex. 2. Divide -00525 by 3 048. Here as in Ex. 1 subtracting the log of 3*048 from log '00525 we obtain log 3 '2362. From Table IV. antilog corresponding to this is 1723. /. 005254-3-048= 001723. In some cases it should be noticed that when four-figure logarithms are used the fourth significant figure, although not always quite exact, is usually not far wrong : three significant figures are necessarily accurate. Thus, in Ex. 13 048 4- '00525 = 580 '57 1 ... , and thus, as on p. 4, the fourth figure should be 6, not 5. Again, -00525 -"- 3 "048 = -0017234..., and the four significant figures are correct. Evaluation by logarithms involving multiplication and division. It is easily possible to evaluate any arithmetical calculation true to three significant figures. 132 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 1. Evaluate ^^ when a = 1986, b = -1188, c = '5046. c Substituting the given numbers we obtain 1 '986 x '1188 5046 ' .-. log 1 -986 + log -1 188 - log -5046 = '2980+1-0749 - T'7029 =1-6700. antilog 6700-4677; .'. I^^? = -4677, *Y ^ ^='4677. EXERCISES. XXV. Divide / 1. 30 by 6'25. n/ 2. '325 by 1300 and 3250 by '01 3.J 3. '00062 by 64. 4. Why is it that we subtract the logarithms of two numbers to obtain the logarithm of their quotient ? Divide . 5. 429 by '0026. ^ 6. (i) ('02- 002+ -305) by 016 x -016. , (ii) '05675 by 0705. I 7. 05344 by 83'5. 8. '00729 by '2735. i 9. '0009481 by '0157. J 10. Calculate axb + c when (i) a = 619'3, b = '117, c = l'43. (ii) a = 6'234, b ='05473, c = 756'3. 11. Calculate a + b when (i) a= '0004692, b= -000365. (ii) a = 94'7S, 6=2-847. (iii) a = 907 '9, & = 17'03. 12. Calculate (i) 23 '51 + 6 '78. (ii) 23 51 +0678. 13. Compute (i) 16'83 + 24'76. (ii) 1613+ '002476. 14. If = cxjd?\/2gh, find the numerical value of c, given = 1-811, d='642, = 32-2, ft=l'249. 15. Calculate ab and a + & when a = '5642, 6= '2471. 16. (i) Compute 4-326 x -003457. (ii) -01584 + 2-104. 17. (i) Compute 30-56 + 4-105. (ii) '03056x0-4105. CHAPTER XIV. INVOLUTION AND EVOLUTION BY LOGARITHMS Involution by logarithms. To obtain the power of a number, multiply the logarithm of the number by the index representing the power required ; the product is the logarithm of the number required. Let log a = x. Then a = 10*. And a n = (10*) n ; .'. log 10 a n = nx=n log a. Ex. 1. Find log a 3 , also log cfi. In the first case the index is 3, and, hence log a 3 is three times the logarithm of a. Similarly, log a? is one half the logarithm of a. These examples are illustrations of the general rule, viz. : loga w =wloga, where n is any number, positive or negative, integral or fractional. Ex. 2. Find by logarithms the value of ( 05) 3 . The process is as follows : Write down the log of the number as is shown on the 2*6990 right ; multiply the log by the index 3, and in this way 3 obtain for the mantissa "0970, and for the characteristic 4 "0970 4. This result is arrived at by saying when obtaining the characteristic 3 x 6=18 plus 2 carried from last figure gives 20, and we write down 0. Next 3 x - 2 = - 6, but - 6 added to +2 carried from the previous figure gives - 4, which is written 4. .'. log('05) 3 = 4-0970. It is seen from Table IV. that antilog -097 = 1250. The characteristic 4 indicates that three cyphers precede the first significant figure. Hence the required number is '000125. .-. -05 3 = -000125. 134 PRACTICAL MATHEMATICS FOR BEGINNERS. Contracted multiplication may be used with advantage when the given index consists of three or more figures. Ex. 3. Calculate the value of (9) 3 ' 76 . log 9 =-9542; .-. 9542 673 2-8626 6679^ 572^ 3-5877 antilog 587 = 3864 Diff. col. for 7= 6 /. antilog 5877 = 3870 Hence (9) 376 = 3870. When the index of a number not only consists of several figures, but the number itself is less than unity, so that the characteristic of the logarithm of the number is negative, it is advisable to convert the whole logarithm into a negative num- ber before proceeding to multiply by the index. Ex. 4. Calculate (-578)- 376 . log -578 = 1-7619, or, - 1 + -7619= - -2381. The product of - -2381 and -3 76 is -8952. antilog 8952=7856; .-. (-578)- 376 = 7-856. When the mantissa of a logarithm is positive, and the index a negative number, the resulting product is negative. If such a result occurs the mantissa must be made positive before reference is made to Table IV. Ex. 5. Calculate the value of (8'4)- 1-97 . log 8-4= 9243. -l-97x -9243= -1-8208. As the mantissa -8208 is negative, it must be made positive, by subtracting it from unity and prefixing I for the characteristic, i.e., - -8208=1 1792. Hence -1-8208 = 2-1792. antilog 1792=1511; .-. (8-4)- lfl7 = -01511. This may be verified, if necessary, by writing (8-4)" 197 in its equivalent form, (8-4) 197' LOGARITHMS. 135 In any given expression when the signs + and occur it is necessary to calculate the terms separately and afterwards to proceed to add or subtract the separate terms. The method adopted may be shown by the following example : Ex. 6. Evaluate 2a 3 + (6 2 ) 376 + 2c " 3 76 - d~ lin . When a = -07, b = 3, c = -578, d = 8 '4. Let x denote the value required, then x = 2a 3 + (6 2 ) 3 ' 76 + 2c - 3 ^ _ d ~ . Here the four terms must be separately calculated. 07 3 is found to be '000343 .'. 2a 3 = -000686. From Ex. 3. (3 2 ) 376 3870. From Ex. 4. -578- 376 7*856 ,\ 2c" 376 = 15712. ; From.Ek. 5. 8-4" 197 01511. Hence x = -000686 + 3870+ 15-712- -01511 = 3885-727696. Evolution by logarithms. The logarithm of the number, the root of which is required, is divided by the number which indicates the root. No difficulty will be experienced when the characteristic and mantissa are both positive. But, although the characteristic of the logarithm may be negative, the mantissa remains positive. Hence the characteristic, when negative, usually requires a little alteration in form before dividing by the number, in order to make such logarithm exactly divisible by the number. The methods adopted can best be shown by examples. Ex. 1. Find the cube root of 475. From Table III., mantissa of log 475 = 6767 ; /. log 475 = 2-6767. To obtain the cube root it is necessary to divide the logarithm by 3 ; we thus obtain 2^67= -8922. 3 From Table IV. we get antilog 892 = 7798 Diff. col. for 2, 4 /. antilog 8922 = 7802 Hence ^475 = 7 "802. 136 PRACTICAL MATHEMATICS FOR BEGINNERS. When the given number is less than unity, the characteristic of its logarithm is negative, and a slight adjustment must be made before the division is performed. Ex. 2. Find the value of 4/-I75. log -475 =1-6767. To obtain the cube root it is necessary to divide 1 *6767 by 3 ; before doing so the negative characteristic is, by adding - 2, made into - 3, so as to be exactly divisible by 3. Also + 2 is added to the mantissa, thus 1'6767 becomes 3 + 2'6767. Hence 4(3 + 2-6767) - 1 '8922. As in the preceding example, the corresponding antilog is 7802 : ' 4^475= -7802. The adjustment indicated in the preceding example should be performed mentally, although at the outset the beginner may find it advisable to write down the numbers. In dividing a logarithm by a given number it is necessary, when the divisor is greater than the first term in the mantissa, to prefix a cipher. Ex. 3. Find the fifth root of 3. log 3= -4771, and K -4771) = -0954. antilog 0954= 1246; .-..3* =1-246. In this example, since the divisor 5 is greater than the first term 4 in the mantissa, a cipher is prefixed. Then, by ordinary division, we have 5 into 47 gives 9 ; the remaining two figures 5 and 4 are obtained in a similar manner. Ex. 4. Find the fourth root of 0'007 or (-007)*. log -007 = 3-8451. I (3 -8451)=| (4 + 1-8451) = 1-46127; .-. log (-007)* = 1-4613. Corresponding to the mantissa 461 we find the antilogarithm = 2891 Diff. col. for 3= 2 -. the antilogarithm corresponding to the logarithm 1*4613 is 2893. Hence (0-007)*= '2893. LOGARITHMS. 137 Ex. 5. Find the seventh root and the seventh power of 0*9306. log -9306 = 1-9688, log of seventh root =\ (7 + 6 '9688) =1 -9955. antilog 995 = 9886 Diff. col. for 5, 11 .-. antilog of -9955 = 9897 The characteristic 1 indicates that the number is less than unity. Hence seventh root= *9897. Let x denote the seventh power of '9306. Then ic = (0'9306) 7 ; .\ log x =7 log -9306 = 7x1 -9688 = 1-7816. antilog 781=6039 Diff. col. for 6, 8 6047 Hence x =-6047. Ex. 6. Evaluate ah* (a + b)~^ x (a - &)* when a=3 142, & = 2-718. In this example the signs + and - must be eliminated before logarithms can be used. This elimination is effected by first finding the values of a + b and a -b. Thus a + & = 3-142 + 2-718 = 5*86 and a -6 = 3-142 -2-718 ='424. Hence denoting the given expression by x we have x= (3-142p x (2-718)* x (5 -86)~* x ( -424)* log x=% log 3-142 + f log 2-718 + log -424 -{ log 5-86 = x -4972 + f x 4343 + 1 x 1-6274- J x -7679 = 3315+ -3619+1-9255- 1-7917 = 2-8272. antilog 8272 = 6717; .-. x= -06717. Miscellaneous examples. As logarithms enable calcula- tions involving the arithmetical processes of multiplication, division, involution, and evolution to be readily performed, a few miscellaneous examples involving formulae frequently required are here given. 138 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 1. The collapsing pressure of a furnace flue (in lbs. per sq. in. ) may be found from the formula : p _ 174000 xfi dxs/l Given d=S3 ; t=% ; and 1= 15 ; find P. Substituting the given values we have p _ 174000 x ( |) 2 33x\/l5 .-. log P = log 1 74000 + log 3 2 - (log 33 + Mog 15 + log 8 2 ) ; .-. log P = log 1 74000 + 2 log 3 - (log 33 + 1 log 1 5 + 2 log 8) = 5-2405 + -9542 - (1 '5185 + -5880+ 1 -8062) = 61947- 3-9127 = 2-2820. antilog 282= 1914; .-. P=191'4. Ex.2. If 2 -718* =148 -4, find a:. # log 2'718 = log 148-4; .-. xx -4343 = 2-1715. Hence x= 77^= 5. 4345 The numerical values of equations in which the ratio of one variable to another is required, can also be obtained by logarithms. Ex. 3. Given ^s^ ^^-y 6 ) find ^ yalue of y x x Multiplying both sides by x, then x 6 = 7-4x 6 -7'4y 6 , or 7*4i/ 6 = 6-4a^; x \7'4' log^ = l(log6-4-log7-4) x = h -8062 --8692) 6 =1-9895 ; /. ^=-9761. x Ex. 4. Given 7* = 3 X+1 + 2 X ~ 2 , find x. Here 7*x 2*~ 2 =3* +1 ; .-. a: log 7 + (a - 2) log -2= (a: +1) log 3. From which we find x = 1 614. EXERCISES. 139 Ex. 5. If Q denotes the quantity of water passing over a V- shaped notch per second and h the height of the water above the bottom of the notch (Fig. 55) then Q oc ifi. If Q is 7*26 when h is 1-5 find h when Q is 5*68. Q=kht; :. 7^6=*x(l-5)*i ., K _ * d T x * ! " '"' log k= log 7 -26 -flog 1-5 =-4207; *rl\ z W :. k = 2 -634. ^^T ? Hence when Q is 5 "68 we have 5-68 = 2-634 x A* ; ^f . h = 1-359. Ex. 6. If V is the speed of a steam vessel in knots, D the dis- placement in tons, and HP the horse-power, then HP cc V 3 D* or HP = JcV s D%. Given F= 17, D= 19700, and #P= 13300, find k. Here 13300 = k x 17 3 x 19700 ; .-. log *= log 13300- 3 log 17 - 1 log 19700 = 3-5697 ; .-. k= -003713. Hence the equation becomes HP= -003713 V 3 D%. Napierian Logarithms. The system of logarithms em- ployed by Napier the discoverer of logarithms, and called the Napierian or Hyperbolic system, is used in all theoretical investi- gations and very largely in practical calculations. The base of this system is the number which is the sum of the series this sum to five figures is 2-7183. Usually the letter e is used to denote a hyperbolic logarithm, as for example log 2 to base 10 would be written log 10 2 or more simply as log 2, but the hyperbolic logarithm of 2 is written as log e 2. Transformation of logarithms. A system of logarithms calculated to a base a may be transformed into another system in which the base is b. Let N be a number. Its logarithms in the first system we may denote by x and in the second system by y. X Then N=a x = b y or b = a& J .-. ~ = \og b and *-=* *. y &* 29. From the two given formulae jr ; , , ? D = l-860 2 + 1-08, N=^, find the values of iV when (7 has the values T23, 2*89, 4 63, 6 '48, 32. 142 PRACTICAL MATHEMATICS FOR BEGINNERS. and 8 "06. Arrange the work so as to carry out the computation with the least trouble. 30. Calculate the values of (i) 0-25 2 - 19 . (ii) #'00054 x 3-6. (iii) If m = r- 116 , find r when m = 2 -263 and a= '4086. "* 31. In a horse-shoe magnet the following relation is found to hold very nearly, P = cxd?x 10 ~ 7 , find P. Where P is the pull in lbs. per sq. in., c is a constant = 5*77, d is the density in the air gap = 6000 per square centimetre. Find the value of x from the equations, / 1-03 xlQ- 5 x 9300x1 -05 xM \* ~-\ 240 )' M 11 *6 x -4785 33 - * = n^r MISCELLANEOUS EXAMPLES. XXVII. Calculate 1. #23-51. 2. 6*78 234 . 3. *678- 1301 . ^ 4. Work out the values when s = '95 and r = 1 *75 of (i) (sr _1 -r-*)-f (.s-1). (ii) ( 1 + log e r ) -*- r. 5. If 2>m 10646 = 479 find p when w is 12*12, and find u when p is 60*4. Compute 6. 1-683 365 . 7. *01683-^. s 8. If His proportional to ZW, and if 77 is 871 when D is 1330 and v is 12, find H when D is 1200 and i; is 15. 9. Calculate the ratio of d to d 1 from the equation : 10. Find the ratio of a to 6 when a?= -. 11. Find the ratio of y to x from the equation x s = 12. Find the ratio of x to y from y 4 = ' 8 ( x5 + y 5 \ 13. If Q oc m and when # is 8 '5, Q is 557 1 ; find Q when i7is4 25. 14. (i) If H is proportional to D^tf, and if D is 1810 and v is 10 when H is 620, find H ii D is 2100 and t? is 13. (ii) If y^axv+bxz*. And y 62S when #=4 and s=2. Also y 187*2 when #=1 and z = l*46 ; find a and b, and find the value of y when # is 9 and z is 0*5. CHAPTER XV. SLIDE RULE. Slide rule It will be clear already to the reader who has followed the section dealing with logarithms, that by their use the multiplication of two or more numbers is affected by adding the logarithms of the factors, and their division by the subtrac- tion of the logarithms of the factors. Or, shortly, by the use of logarithms multiplication is replaced by addition, and division by subtraction. Hence, if instead of the equal divisions of a scale (Fig. 56), unequal divisions corresponding to logarithms are employed, then, when performed graphically, in the manner to be immedi- ately described, multiplication will correspond to addition and division to subtraction. Fig. 56. Tt is an easy matter to add together two linear dimensions by means of an ordinary scale or rule. Thus, to add 2 and 3 units together. Assume the scale B (Fig. 56) to slide along the edge of the scale A, then the addition of the numbers 2 and 3 is made when the 2 on B is coincident with on A ; the sum of the two numbers is found to be 5 opposite the number 3 on the scale A. If the scales on A and B are not divided in the proportion of the numbers, but of the logarithms of the numbers, then, using this graphic method, we can by sliding one scale along 144 PRACTICAL MATHEMATICS FOR BEGINNERS. 3 r the other perform the operation of addition ; but, as the scales are logarithmic, the result would correspond to the product of the numbers added. Similarly, the number corresponding to the difference would be a quotient. Construction of slide rule. The object of the slide rule is to perform arithmetical calculations in a simple manner. There is a great saving of time and labour effected by its use, as it solves at sight all questions depending on ratio. It consists of a fixed part or rule con- taining a groove in which a smaller rule slides. Reference to Fig. 57 shows that the upper part of the rule contains two scales exactly alike, while the lower part of the rule con- tains only one scale, its length being double that of the upper one. As the upper part contains two scales, it will be convenient to refer to the division 1 in the centre of the rule, shown at E as the left-hand 1, the other to the right of it as the right-hand 1. There are two scales on the smaller rule or slide, as we may call it, at B ; and at C one double the length. These scales on the slide correspond to those on the rule. It will be convenient to refer to the four scales by the letters A, B, C, D. There is in addition to the parts mentioned a movable frame or thin metal runner, held in position on the face of the rule by sliding in two grooves. This is shown both at E and in the end view. Although it slides freely along the instrument, any shake which might otherwise occur is prevented by a small steel spring placed at the upper part of the carrier. SLIDE RULE. 145 The principle of action is the same in all slide rules, although the arrangement of the lines depends upon the purpose to which the rule ^ is to be applied. The modified form of the calculating rule, which we propose to explain, is one of the most accurate instruments of the kind that can be obtained. The instrument, with the exception of the runner B, is usually made of boxwood or mahogany. The wood may be faced with white celluloid, the black division lines showing more clearly on the white back- ground. Graduation of slide rules. In Fig. 58 it will be seen that the distance apart of the divisions are by no means equal. The divisions and subdivisions are not equidistant as in an ordinary scale, but are pro- portional to the logarithms of the numbers and are set off from the left or commencing unit. In studying Indices it was found, p. 107, that if 10 3 be multiplied by 10 4 the result is 10 3 * 4 or 10 7 . From the definition of a loga- rithm, 2 is the logarithm of 100, since 10 2 =100. Or, as 10 raised to the power 2 gives 100, the logarithm of 100 is 2. In a similar manner if 10 be raised to a power '4771 we obtain the result 3 ; .-. log 3= -4771. P.M.R K NifjfflB r* J jpj jj 146 PRACTICAL MATHEMATICS FOR BEGINNERS. Also since 10 3010 = 2 ; .*. log 2 = '3010. Hence '7781 is the log of 6 = 10 3010+4m . With the rule this addition is effected by drawing the slide to the right until the left-hand index of scale B is coincident with 2 on scale A. Then over 3 on the scale B is found 6 on the scale A. 107781 So also jpMo^ 107781 "' 8010 ^ 101 * 771 - Or more simply, to divide 6 by 2. log 6 - log 2 = 7781 - -3010 = '4771; and -4771 is the log of 3. This is obviously only the converse operation to that already described. Thus, set 3 on B to 6 on A ; then, coincident with the index 1 on B is the answer 2 on A. A model slide rule. Simple exercises similar to the above will be found very useful as a first step, and such practice will enable the student to deal with numbers with certainty and ease. It is an excellent practice to make a slide rule, using two strips of cardboard or thick paper. Assuming any length, such as from 1 to E, scale A, to be 10 inches long and to be divided into 10 parts, then the distance from 1 of any intermediate number (from 1 to 10)' is made pro- portional to its logarithm. Or, as the length of the scale is to be 10 inches, the distance in inches of any number from 1 is equal to the logarithm of the number multiplied by 10. To find the position of the 2nd division, since log2 = *301, '301 parts, or 3"01 inches from 1, would indicate its position. In like manner the 3rd division would be '477 parts, or 4'77 inches ; the 4th, "602 parts, or 6 02 inches ; the 5th, 6 99 inches, etc. Denoting the distance of any division from point 1 by g, if I denote the length of the scale from 1 to E, and L the logarithm of the number indicating the division required, then x = I . L. When the upper scale A is set out, the scales B and C on the slide and the scale D may be similarly marked from it. SLIDE RULE. 147 The excellence of any slide rule depends upon the skill with which these division lines have been constructed. In good rules they are as accurate as it is possible to make them. In dealing with a carefully made slide rule we deal with the effect of a considerable amount of labour and thought which have been expended in its construction. Although a knowledge of logarithms is not essential before a slide rule is used, any more than it is necessary that a man should be able to make a watch before he is allowed to use one, or that he should understand the nature of an electric current before using an electric bell, it is much better to clearly under- stand the principles underlying the construction of any instrument. Multiplication with a slide rule. - In (Fig. 58) putting the units' figure of the slide opposite the 2 on the fixed scale A, we get registered the products of all the numbers on the slide and 2 above. Thus 2x1 = 2, 2x1-75 = 3-5, 2x2 = 4, etc. Or, we may use the two lower scales C and D. In each case we make the index 1 on the slide coincide with either of the factors read on scales A or Z>, and the product will be found coincident with the other factor read on the slide. The use of the two upper scales enables a much larger series of values to be read with one motion of the slide, but as the scales on C and D are double the length of those on A and B it is obvious that the former are more suitable to obtain accurate results. It should be carefully noticed that the values attached to the various divisions on the scales depend entirely upon the value assumed for the left-hand index figure. Thus, the left-hand index, or units' figure, may denote Ol, 1, 10, 100, or any mul- tiple of 10 ; and, when in any calculation the initial value is assumed, it must be maintained throughout. Thus, in Fig. 58, the products may be read off as 2x1 = 2, etc. ; 2 x 10 = 20, etc. ; or 2 x 100 = 200, etc. If the product cannot be found when the left-hand index is used the right-hand index must be employed. Division with a slide rule. Set the divisor on B under the dividend on J, and read the quotient on A over the index of B ; 148 PRACTICAL MATHEMATICS FOR BEGINNERS. or, set the divisor on G over the dividend on D, and read the quotient on D under the units' figure of C. Ratio with a slide rule. This, as already indicated, is only a convenient method of expressing division. One of the simplest applications of ratio is to convert a vulgar fraction to a decimal fraction. Thus, the decimal equivalent of f is found by placing 8 on the scale B opposite to 3 on the scale A ; then, coincident with the index on B, is the result '375 on A The two lower scales (7 and D may, of course, be used instead of A and B. The circumference of a circle is obtained by multiplying its diameter by 3*1416. Hence if the index on the scale B be put into coincidence with this number 3*1416 (marked it on scale A), then against any division representing the diameter of a circle on B, on A the division indicating the circumference of the circle is found. Conversely, the diameter may be obtained when the circumference is given. As proportion is simply the equality of two ratios, the rules for performing proportion by the help of the slide rule follow at once from those already given for ratio. Proportion with a slide rule. Making use of the two upper scales A and B, operate so as to find the quotient, and without reading off the answer, look along the rule for the product of the quotient by the third factor in the proportion. Ex. 1. 3: 4 = 9: x. Read off the answer 12 by the process described, or put the pro- 3 9 portion in the form of ratios; thus -=-. 4 it- Place 4 on B under 3 on A, and under 9 on A read off the answer x=\2. Opposite the division 2 '8, on G, is found 9*52 on D. The line on the runner can be made to coincide with this. Next move the slide until the graduation 1*7 is coin- cident with the line. If necessary opposite 1 on G the result of 3*4 x 2*8 1 - could be read off on D ; but instead of doing so the line on the runner is made to coincide with the 1 on scale C, and the slide is moved until *3 is coincident with the line. The answer 18*6 is now read off opposite the index on G. In this manner in any com- plicated calculation the runner may be used to record any operation. CHAPTER XVI. RATIOS, SINE, COSINE, AND TANGENT. Measure of angle in radians = Measurement of an angle in radians. A very convenient method for measuring angles, which is especially useful when dealing with angular velocity, is obtained by estimating the arc subtended by a given angle, and dividing the length of the arc by the radius of the circle. Thus, in Fig. 59, the measure of the angle BOA in radians is the ratio of the length of arc A B to the radius OA, or length of an arc AB ... length of its radius OA ^ ' Evidently this measure of an angle will be unity when the arc measured along the curve is equal to the radius, or, the unit angle is that angle at the centre of a circle subtended by an arc equal in length to the radius. This unit is called a radian. The circle contains 4 right angles or 360. This may be expressed in radians by the ratio of circumference of circle radius of circle ' but the circumference of the circle as shown on p. 222 is 2irr. Where r is the radius OA and tt denotes the number of times that the diameter of a circle is contained in the circum- ference, the value of ir is 3*1416, or more accurately 3*14159. Fig. 59. Measurement of an angle in radians. 152 PRACTICAL MATHEMATICS FOR BEGINNERS. 27T7* Hence the measure of 4 right angles = =2ir. Thus 4 right angles may be expressed in two ways, viz., as 360, or as 2tt radians, and one radian = -5 = = 57*29578 (taking it = 3 - 141 6). For many purposes the approximate value 57'3 is used instead of the more accurate value. It will be seen from Fig. 59 that if any circle be drawn with centre and cutting the lines OA and OB in any two points such as D and E, then the angle E0D = angle AOB. It follows, therefore, that the unit angle is independent of the size of the circle and is an invariable unit, being, as already indicated, equal to , or 57-2958. IT It is advisable not only to be able to define the two units, the degree and the radian, but also to realise their relative magni- tudes. Thus an angle 5 denotes an angle of 5 degrees, but an 180 angle simply denoted by 5 contains 5 x degrees. /. From (i) it follows that if an arc of a circle is five times as long as the radius, the angle subtended at the centre is five radians, if the arc is one-third the radius, the angle is one-third of a radian, etc. To find the length of arc subtending a given angle it is only necessary to write (i) arc = angle x radius. Also when tt refers to a number it denotes 3"1416, but applied to an angle, then the angle contains it radians and is 180 degrees. Ex. 1. An angle is radians, what is its value in degrees ? In this example, as the angle is in circular measure, its value in degrees will be fxunit angle, or x 57 '2958 = 38 -1972 or 38 "2 using 57 '3. Ex. 2. (a) What is the numerical value of a right angle in radians? (6) Find the radian measure of an angle of 112 43'. (c) Find the length of an arc which subtends an angle of 112 43' at the centre of a circle whose radius is 153 feet. (a) The measure of 4 right angles is 2-rr radians, therefore the measure of 1 right angle is radians = t- radians == 1 '5708 radians. 4 I FUNCTIONS OF ANGLES. 153 (b) In each degree there are 60 minutes, hence 112 43' = 6763 minutes. Also 180 x 60 = 10800 minutes ; 6763x3-1416 10800 = 1 "967 radians. , v -r i length of arc (c) In radian measure, angle = ^r-. ; /. length of arc = angle = radius = 1*967 x 153 = 301 feet nearly. EXERCISES XXVIII. 1. If an arc of 12 feet subtend at the centre of a circle an angle of 50, what is the radius of the circle ? 2. Explain the different methods of measuring angles. Find which is greater, an angle of 132 or an angle whose radian measure is 2*3 radians. 3. Find the number of degrees in the angle whose radian measure is 1. 4. A train is travelling on a curve of half a mile radius at the rate of 20 miles an hour ; through what angle has it turned in 10 sees. ? Express the angle in radians and in degrees. 5 Define the radian measure of an angle. Find the measure of the angle subtended at the centre of a circle of radius 6^ inches by an arc of 1 ft. 6. The radius of a circle is 10 ft. Find the angle subtended at the centre by an arc 3 ft. in length. 7. If one of the acute angles of a right-angled triangle be 1*2 radians, what is the other acute angle ? 8. Two angles of a triangle are respectively J and |- of a radian ; determine the number of radians and degrees in the third angle. 9. A certain arc subtends an angle of 1 "5 radians at the centre of a circle whose radius is 2'5 feet ; what will be the radius of the circle at the centre of which an arc of equal length subtends an angle of 3*75 radians? Functions of angles. We have already explained the use of the protractor, and a table of chords in setting out and measuring angles. It is now necessary to refer to another useful method of forming an estimate of the magnitude of angles, that namely by means of the so-called functions of angles, the sine, cosine, tangent, etc. The values of these functions have been tabu- lated for every degree and every minute up to 90. It is possible, in addition, by the help of columns of differences to calculate these functions to decimal parts of a degree, or in 154 PRACTICAL MATHEMATICS FOR BEGINNERS. seconds and fractional parts of a second. It will be an advan- tage to understand what information is derivable from such tables, and how to use them with facility. If at any point B along a straight line AD (Fig. 60) a line BChe drawn perpendicular to AD, then ABC is a right-angled triangle. One angle at B, a right angle, is known ; the other two, and the three sides, can be determined by the data of any given question. We may denote the angle BAC by the letter A (the letter at the angular point), and the remaining two angles may be referred to as the angles B and C. If one of the sides be given, and one of the two remaining angles, the other sides and the remaining angle can be found either by construc- Fig. 60. tion or by calculation. Also any two of the three sides will give sufficient data to enable the triangle to be drawn and the two angles and the remaining side to be found. It is not necessary to give the actual lengths of two of the sides ; it will answer the same purpose if the ratio of AB to BC be known, for if any line such as DE be drawn parallel to BC (Fig. 61) it will cut off lengths AD and DE from the two sides AB and AC produced, such that the ratio of AD to DE is equal to the ratio of AB to BC. Also the equality is unaltered if BC and DE be replaced by AC and AE respectively. AB AD This statement can be written -j-p -rp- FUNCTIONS OF ANGLES. 155 AB To obtain the value of the ratio -j-~ when the angle A is 60. Draw any line AD as base (Fig. 61), make AC equal to 10 units and at 60 to AB. From C draw CB perpendicular to AD and meeting AD at B ; ascertain as accurately as possible the lengths of AB and BC. AB will be found to be 5 units, and BC to be 8*66 units. ., A . AB 5 1 , BC 866 Hence the ratio -tti^tk^-, and -j-*. ,a * ^4(7 10 2' ^46 10 r>/"Y This ratio of - is called the sine of the angle BAC or (as A Is only one angle is formed at A) sine A. It will be seen from the above that the sine of an angle (which is abbreviated into sin) is formed by the ratio of two sides of a right-angled triangle ; the side opposite the angle being the numer- ator, and the hypotenuse or longest side of the triangle (adjacent to the angle) the denominator. Let the three sides of the triangle be represented by the letters a, b, and c, where a denotes the side opposite the angle A, b the side opposite the angle B, and c the side opposite the angle C. ^ . _ AO side BC a 8-66 _._ Then, sine 60 =-r 1 ~= T = = -866. side AC b 10 Eef erring to Table V., opposite the angle 60 this value will be found, and the length of the side BC, or a, can be obtained bv calculation. Thus, in the right-angled triangle ABC (Fig. 61) we have a 2 = 6 2 -c 2 = 10 2 -5 2 =75, ;. a = V75 = 8 '66. The ratio of -jj, or j- is called the cosine of the angle BAC (cosine is abbreviated into cos), c 5 .'. cos A = cos 60 = T = - = *5. o 10 Angle of 30. The sum of the three angles of a triangle is 180. As one of the angles in Fig. 61 is 90 and the other 60, the remaining angle is 30. Also, sin30 = | = cos60 = -5 o 156 PRACTICAL MATHEMATICS FOR BEGINNERS. and cos 30 = t = sin 60 = '866. Again referring to Table V., these calculated values are found opposite sin 30 and cos 30 respectively. The tangent. Of the three sides of the triangle AB, BG, and GA (Fig. 62) we have already taken the ratio of -j~ and -r-~, the former is called the sine and the latter the cosine of 0. One other ratio, and a most important one, is the ratio of jtn -j-fr This ratio is called the tangent AB of BAG, or, denoting BAG by #, and using the abbreviation tan for tangent, we have tan 6= -j-~- Ex. 1. Construct angles of 30, 45, and 60. In each case make the hypotenuse AG =10 units on any convenient scale. Measure the lengths AB and BG to the same scale, and tabulate as follows : Angle of 30 Angle of 45 Angle of 60 Lengths of : Numerical values of : sin. cos. tan. AB BC BC AC AB AC BC AB' It has already been seen (p. 47) that two angles are said to be complementary when their sum is a right angle. Eef erring to Fig. 61, the sin of 30= cos 60, the ratio in each BG case being . Hence the sine of an angle is the cosine of the AG complement of that angle ; and the cosine of an angle is the sine of the complement of that angle. FUNCTIONS OF ANGLES. 157 Prove these statements by reference to the tabulated values obtained by measurement. The supplement of an angle is the angle by which it falls short of two right angles (180) ; thus, the supplement of an angle of 60 is 120 ; the supplement of an angle of 30 is 150 ; or two angles are said to be supplementary when the sum of the two angles is 180. It is important to be able to readily write down the values of the sine, cosine, etc., of angles of 60 and 30. To do this, the best method is to draw, or mentally picture, an equilateral triangle ABC (Fig. 63) each side of which is 2 units in length. From the vertex C let fall a perpendicular CD on the base A B. As shown on p. 47, the point D bisects A B, and AD=DB. Also the angle ACD is equal to DCS ; each of these equal angles is one- half the angle ACB, and is there- fore 30 ; or, as the angle at D is 90 r and the angle at C is equal to ACD must be 30. From the right-angled triangle ADC, we have DC 2 = AC 2 -AD* = 4-l=3; :. dc=sJz. D<0_JI AC~ 2 ' AD_\ AC~2 ; 60 c B . Equilateral triangle. the remaining angle Thus sin 60 cos 60 Hence . AO DC Jz ,- sin 60 tan 60 = J3 = T =^-.-- sW . sin30 = -T^ = AD AC~~ 1 = 2~ cos 60 ; DC AC = V3 * 2 = sin 60; cos 30 AD 1 tan 30 = ^ = -j, = cot 60. 158 PRACTICAL MATHEMATICS FOR BEGINNERS. It should be noticed that sin60_Z)<7 AD DC _, RCV> ^60- AC+ AC~ AD- 1 W ' In a similar manner, for any angle A, sin A 4 T = tan^i. cos A Instead of attempting to remember the important numerical r* values for the ratios, it will be found much better to use the triangle, as de- scribed in Fig. 63, its angles 90, 60, and 30, and its sides in the proportion of 2, 1, and V3. To ascertain the numerical values of the sine, cosine, and tangent of 45, a similar method may be used. Thus, if FlG *angTe I dTianglI ight * AB aild BC ( Fi S 64 ) f rm tW0 sideS f a right-angled triangle in which BA=BC and each is one unit in length, the angle BAC*=BCA, and as the sum of the two angles is 90 each angle is 45. Length of AC=JAB 2 + BC 2 = J2. Hence the three sides of the triangle ABC are in the ratio ofl, l,and/2; . ... BC 1 . KO AB 1 sin 45 = = -j= ; cos 45 = = ^ ', AC s/2 AC >J2 or sin 45 = cos 45 ; tan 45 =-^=1. An Angles greater than 90. On p. 33 we have found that an angle is expressed by the amount of turning of a line such as AB (Fig. 65). If the movable radius, or line, occupies the posi- tions AC, AC, AD, and AE, then it is seen that as BC' BC and the remaining sides of one triangle are equal to the corre- sponding sides of the other that the triangle BAC is equal to BAC. Hence angle J e'4C=(180 o -30j = 150, or sin 150 = sin 30; or, generally, sin (180 -A) = sin A. If the line AB be assumed to rotate in a negative direction until FUNCTIONS OF ANGLES. 159 it reaches a point E, a negative angle equal to - 30 is described ; thus, the angle BAE may be written either as 330 or -30. In addition to the convention that all angles are measured in an anti-clockwise manner, the following rules are adopted : All lines measured in an upward direction from BB' are positive, and all lines measured from A' A towards B are positive ; those in the opposite directions, i.e. downwards, or from A' A to B' are negative. The movable radius, or line AC, or AC, is always positive. Hence, if BA C denote any angle A, then, since BC and A* B'C' are both in an upward direction, we have, as before sin(180-J) = sin,4. But AB and AB' are lines drawn in opposite directions ; /. cos (180-^)= -cos A. In the case of the angle formed by producing C'B' to D, both B'D and AB are negative, hence the sine and cosine of the angle are nega- tive ; when the angle is formed by producing OB to E, the sine of the angle is negative, the cosine is positive. On reference to Table V., it will be found that the functions of an angle sine, cosine, etc. are only tabulated for values from to 90 ; but from these the value of any angle can be obtained by means of the above conventions. Thus, the numerical value of the sine of 30 is J or "5, and this is also the value of sin 150. The cosine of 30 is n/3 or -866, and cos 150 is - * As the tangent is _ . , its sign, positive or negative, will depend upon the signs of the sine and cosine of the angle ; when these are alike the tangent is positive, and negative when they are unlike. Further, when the numerator is zero, the value of the tangent is ; the value is indefinitely great when the denominator is : this is written as cc. 160 PRACTICAL MATHEMATICS FOR BEGINNERS. The following important relations should be proved by draw- ing a right-angled triangle to scale, as already described : sin60 = ^ = sinl20; 2 cos 60= - cos 120 = 2' 2 tan 60 = */3, tan 1 20 = - \/3. These results and those previously arrived at are collected in the following table, the table should be extended to include angles up to 360 ; each result should be expressed as a decimal fraction, and afterwards verified by reference to Table V. 30 45 60 90 120 135 150 180 sin 1 2 ] v/2 2 1 v/3 2 1 \/2 1 2 cos 1 n/S 2 1 s/2 1 2 1 2 1 s/2 2 -1 tan 1 v/3 1 v/3 QO -s/S -1 1 From the figures already drawn and also from these tabulated values, it will be seen that as an angle increases from to 90, the sine of the angle increases from to 1, but the cosine decreases from 1 to 0. Conversely from 90 to 180, the sine of an angle decreases from 1 to 0, the cosine increases from to 1. Other ratios of an angle. In any right-angled triangle, ABC (Fig. 66) the angle BAG is denoted Pig. 66. in the usual manner (by A). FUNCTIONS OF ANGLES. 161 Then, 1 I AC, .. 1 1 AC cosecant A = . 7=7577 = 1577 secant A = 7=TB=^r^; sin A BG BG cos A AB AB AC AC 1 1 AB COtaneentA = t^A = Bd == BC' AB The above are usually designated as cosec A, sec A, and cot A respectively. Some important relations. (i) sin 2 A + cos 2 A = l. a BG A AB For sin.4 = -j~; cos.4=-j^. / ,Na,/ am BG 2 AB 2 BC 2 + AW J Then (sin ,4) 2 + (cos^) 2 = 3+= _^__ = i ; or (sin^) 2 +(cos^) 2 =l. Usually, (sin A) 2 is written sinM. And in a similar manner (cos^4) 2 = cos 2 J. (ii) sec 2 A=l+tan 2 A. A AC secA= AB' aecA- AB2 - AB2 -1+^^-1 + tan 4. Aiso cosec 2 A=l+cot 2 A. _, tA AC 2 BC 2 + AB 2 AB* -- 3. For cosecM = ^r 2 = ^ ~ = x + ^2 = 1 + cot ^ Construction of an angle from one of its functions. Given the sine of an angle \ ~ to construct the angle. ""- " Oiven sin 6=%. .'>^^ Draw a line AB (Fig. \>^^ 67), and at point Z? erect ' -j^* - ^ a perpendicular i? (7 to any *^^ convenient scale 3 units a-^^ in length, with C as centre A B and radius 7 units, de- Fig. 67. Given the sine of an angle to .. . . . _ construct the triangle. scribe an arc cutting AB in A. Join A to C; then BAG is the angle required. Measure the angle, and verify by referring to Table V. p.m. a l 162 PRACTICAL MATHEMATICS FOR BEGINNERS. Given the cosine of an angle to construct the angle. Given cos # = f. . Set out AB = 4: units (Fig. 68), and erect a perpendicular BG. With A as centre, radius 7 units, describe an arc cutting BG in C; BAG is the angle required. Measure the angle by the pro- tractor or the table of chords, p. 36, and verify by Table V. Given the tangent of an angle to construct the angle. Given tan # = . G "4 -+--B Fig. 68. Given the cosine of an angle to construct the triangle. Fig. 69. Given the tangent of an angle to construct the triangle. Draw a line AB, 5 units in length (Fig. 69), and at B draw BC perpendicular to AB and equal to 7 units ; join A to G ; then BAG is the angle required. Verify as in the preceding cases. Inverse ratios. A very convenient method of writing sin 0=f is to write it as #=sin 1- f, which is read as the angle the sine of which is ^. In a similar manner we may write cos # = f and tan 0=J, as #=cos _1 f and 0=tan -1 f respectively. Small angles. Referring to Table V., and also to Fig. 84, it will be seen that, when the size of an angle is small, the values of the sine, tangent, and the radian measure of an angle differ but little from each other. EXERCISES. XXIX. 1. Draw an angle of 35, and make the hypotenuse 10 units. Find by measurement the values of the sine, cosine, and tangent of the angle ; compare the values obtained with those in Table V. USE OF TABLES. 163 Using the values, ascertain if the following statements are true : sin 2 35 + cos 2 35 = 1 ; ^| = tan 35. cos 35 sec 2 35 = 1 + tan 2 35 ; cosec 2 35 = 1 + cot 2 35. 2. If sin 6 = f , find cos 6 and tan 6. 3. The cosine of an angle is y 2 ^, find the sine and tangent of the angle. 2 4. If tana = t=, find sin a and cos a. s/b 5. If tan A=\, find the value of the following : (i) cosM -sinM. (ii) cosecM - secM. (iii) cot 2 ^+sinM. (iv) Show that cosecM - cosM =cot 2 A + ain 2 A. 6. If A =90, = 60, (7=30, Z) = 45, show that sin B cos C+sin G cos B = smA, and that cos 2 D - sin 2 Z> = cos A . 7. The tangent of an angle is 675 ; draw the angle, without using tables, and explain your construction. Along the lines forming the angle set off lengths OA=4:'23" and OB = 376". Find the length AB, either by measurement or by calculation. 8. The lengths of two sides of a triangle are 3*8 and 4*6 inches, and the angle between them is 35 ; determine by drawing or in any way you please (1) the length of the third side, and (2) the area of the triangle. 9. In a triangle A BO, A is 35, G is 55, and AC is 3*47 ft. Find A B and BG. 10. The sides a, b, c of a triangle are 1 2, 1 '6, and 2 feet respec- tively. Find the number of degrees in the angle A, and determine the area of the triangle ABG. Use of tables. Having explained how the sine, cosine, tangent, etc., of such angles as 30, 45, and 60, can be obtained, it remains now only to indicate how the trigonometrical ratios of any angle can be found. The method may be understood by a reference to Table V., in which the values of the sine, cosine, etc., of various angles are tabulated. On the extreme left of the table angles to 45 are found, and from this column and the columns marked at the top by the words sine, tangent, etc., the value of any of these ratios for a given angle may be seen. At the extreme right the angles are continued from 45 to 90. The ratios for these angles are indicated at the bottom of each column. 164 PRACTICAL MATHEMATICS FOR BEGINNERS. Thus, given an angle of 25, we find 25 in the column marked angle, and corresponding to this in the column marked sine we have the value '4226, hence, sin 25 = *4226. By referring to the columns marked tangent and cosine we obtain tan 25 = '4663, cos 25 = -9063. To find the value of sin 65, look out 65 in the right-hand column, and in the column, marked sine (at the bottom) we find corresponding to an angle 65 the value '9063. Hence sin 65 = cos 25. This result agrees with that on p. 156, that the sine of an angle is equal to the cosine of the complement of the angle. Tables are obtainable in which the values of ratios consisting of degrees, minutes, and seconds, or degrees and decimal parts of a degree, are to be found, but Table V. may also be used for such a purpose. Ex. 1. Find the value of sin 25 12' and cos 25 12'. We find sin 26 =4384 sin 25 =-4226 Difference for 1 or 60'= '0158 Hence difference for 12 = 0158 x 12-f60= '0032. .-. sin 25 12' = '4226 + '0032 = -4258. As an angle increases the value of the cosine of the angle decreases (p. 160). Thus cos 25 ='9063 cos 26 = -8988 Difference for 60'= -0075 Hence diff. for 12'= 0075 x 12-r60 = -0015. .-. cos 25 12' = -9063 - 0015 = 9048. Ex. 2. Take out from Table V. the tangent of 3 15' and calcu- late the cube root of the tangent. tan 4 = -0699. tan 3 = -0524. Hence difference for 1 or 60'= '0175. .-. diff. for 15'= -0175 x 15 -f 60= '0044 tan 3 15' = ;0524 + 0044 = 0568 log -0568 = 2-7543 2 7543 -r 3= 1-5847. antilog 1-5847 ='3843 /. #tan315'=-3843. USE OF TABLES. 165 Ex. 3. Find the value of sin .4 cosl?-cos-4 sini? (i) when A is 65 and B is 20. Substituting values from Table V., in (i) we have 9063 x -9397 - "4226 x -3420= -8516 - '1445= '7071. In a similar manner sin 65 cos 20 + cos 65 sin 20 is found to be 9960. The reader familiar with elementary trigonometry will see that sin (65 - 20) = '7071 = sin 45, and sin (65 + 20) = *9962 = sin 85. In like manner the sum, or difference, of two cosines can be obtained. When an angle is given in radians, it is necessary either to multiply the given angle by 57 "3 or, from Table V., to ascertain the magnitude of the angle in degrees before proceeding to use the ratios referred to. Ex. 4. y = 2-3 sin ( /9618a? + ? Y find y when (i) x=l, (ii) find x when y = \ '9716. 7T 6 : number of radians in the angle, to find the number of degrees we multiply by 57 3. .*. 7854x57 "3 = 45, or, from Table V., in the column marked radians corresponding to *7854, we have 45, and sin 45= "7071. .-. 2-3 x sin 45 =2-3 x -7071 = 1-6263. (ii) l-9716=2-3sin(-2618x + ^ or sin(-2618x+'5236)=i^^=-8572. Referring to Table V. this is found to correspond to 1 '0297 radians, /. -2618 x = 1-0297 -'5236 5061 2618 1-934. Functions of angles by slide rule. Given the numerical value of the sine or tangent of an angle, the number of degrees in the angle can be found ; or, conversely, given the number of degrees in the angle, the numerical value can be ascertained : 166 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 1. To find the numerical value of sin 30. Reverse the rule as shown in Fig. 70, placing sin 30 opposite the upper mark. Fig. 70. Slide rule reversed. Again reverse the rule, and opposite the right-hand 1 on scale A the numerical value '5 on slide is obtained on scale B. Similarly, placing sin 60 opposite the mark, the value *866 is obtained. Also, sin 50= '766 may be read off. For practice other values should be selected, and their numeri- cal values written down and afterwards verified. Ex. 2. Obtain and write down the numerical values of the sines of angles of 15, 20, 25, 30, and for intervals of 5 up to sin 70. Conversely, when the numerical value of the sine is given, the corresponding angle can be found. Tangents. At the opposite end of the rule a gap similar to the one just described is to be found ; this may be used to find the numerical value of tangents. Ex. 3. Find the value of tan 30 and tan 20. Move slide until 30 is opposite the mark, then on upper scale coincident with 1 on the slide the value -577 is obtained. In like manner tan 20 '364. Ex. 4. Obtain and tabulate the numerical values of the tangents of angles from 10 to 40. Ex. 5. Write down the values of the sine and cosine of 5, 10, etc. , up to and including 45. Verify by reference to Table V. Logarithms of numbers. Logarithms can be obtained by using the transverse mark on the lower edge of the gap. To obtain log 2, set the index on scale C to coincide with 2 on lower scale D, reverse the rule, and opposite the lower mark, the log of 2 = '301 is read off ; also setting it opposite 3, log 3 = "477, etc. Ex. 1. Obtain and write down the logarithms of all numbers from 1 to 10. Verify the results by reference to Table III. EXERCISES. 167 EXERCISES. XXX. 1. Take out from Table V. log tan 16 6' and calculate the square root of the tangent. 2. Find log tan 81 12'. 3. Calculate the numerical value of (tan 50 tan 22 30') 4. Find log tan 35 15' and the numerical value of %/{ sin 44). 5. Evaluate Vtan40-f65. 6. Find the numerical value of the seventh root of tan 53 30' -f 32. 7. Find log tan 58 5', also the value of the cube root of tan 52 30' 4- 15. 8. Find the logarithms of (sin 26 13') _T . 9. (sin 18 37'P*- 10. Evaluate Vsin 50 tan 2 38 20'. 11. If y = 23 sin ( *2618a* + |Y find y when x = 0, 2, 3, 5,6, 7. Also find x when y = 2*049. 12. Find the value of e bt sin {at) when 6 = -0-7, t = l% and a =3*927. 13. Find the numerical value of e^(a 2 - 6 2 ) tan 0, where c = 25*2, a = 90, 6 = 49*6, = sin- 1 (-528). (0 is less than 90.) 14. Find to the nearest integer the value of the expression 700s/n{| S i- M 0-1426) + l.} when the given angle is positive and less than ^. 15. If ?? = 11 *78, q = 5'61, = 4712 radians, calculate the value of p sin (p 2 + q 2 )~% ; also the value of sjp q . 16. In the following formula, a = 25, 6 = 8*432, c = 0*345, 0=0*4226 radians ; find the value of a ii57 b~ * (c 3 + a log e b . tan 0). 17. sin 162 tan 2 140 -r Vsec 105. 18. Find the value of a * (a 2 - 6 2 )^ -*- sin log e 6 If a is 9*632, 6 = 2*087, is 0*384 radians. 168 PRACTICAL MATHEMATICS FOR BEGINNERS. / 19. Find the value of ah+ \/c 2 + shrU when a =4-268, 6=0249, c = 3*142, ^4=26. Applications to problems on heights and distances. It is not always convenient, or possible, to measure directly the height of a given object, nor to find the distance of two objects apart. Instruments are used for measuring purposes by which the angle between any two straight lines which meet at the eye of the observer can be measured. For this purpose what are called Sextants and Theodolites are used. By means of these the cross-wires of a telescope can be made to coincide with considerable accuracy with the image of an observed object, and by means of a vernier attached, the readings of the observed angles can be made to a fraction of a minute. The angle contained between a horizontal line and the line which meets a given object is called the angle of elevation when the object is above the point of observation ; and the angle of depression when the object is below. Thus, if B be the point of observation (Fig. 71) and A the given object, the angle made by the line joining B to A, with a the horizontal line BC, is called the angle of elevation. In a similar manner if A be the point of observation and B an object, then the angle be- tween the horizontal line (DA, drawn through A) and the line AB is called the angle of de- pression. The following problems will show the methods adopted in Fig. 71. Angles of elevation and depression. working examples involving these angles : Ex. 1. At a distance of 100 feet from the foot of a tower the angle of elevation of top of tower is found to be 60. Find the height of the tower. *To any convenient scale make AB (Fig. 72) the base of a right- angled triangle equal 100 units. HEIGHTS AND DISTANCES. 169 Draw the line AG, making an angle of 60 with AB and inter- secting BG at G. Then BG is the required height. 45 = tan 60 ; but tan 60 = J3 ; AH .: J BC=45tan60 o =10 3 s^v ' % M -s> t \ 9 - / % 2 f S :: zzfr. :: ^ :: : : 7 ^ t J- V 1 re N ^ t 6^ 1 S t -7 X. o 7 i 2 3 4 Fig. 77. Solution of simultaneous equation. Thus, when a cord is passed round a fixed cylinder and a force N is applied at one end and a force M at the other, the cord remains at rest not only when N is equal to J/, but also when N is increased. If the increase in N is made gradually a value is obtained at which the cord just begins to slip on the cylinder. The amount by which N must be greater than M when slipping occurs is readily found by experiment, and depends not only on the surfaces in contact, but also on the fractional part of the circumference of the cylinder embraced by the cord. Ex. 3. Denoting by n the fractional part of the circumference of a cylinder embraced by a cord, then the following table gives a 180 PRACTICAL MATHEMATICS FOR BEGINNERS. series of values of n and corresponding values of N. Find the relation between n and iV. n 25 "5 75 1 1-25 1-5 1-75 2 2-25 2-5 2-75 N 150 195 295 375 515 615 755 1045 1435 1735 2335 LogN 21761 2'29 2 4698 2-5740 2-7118 27887 2-8779 3-0191 3-1568 3-2392 3-3683 When simultaneous values of n and iVare plotted, a curve lying evenly among the points (Fig. 78) can be found ; but by plotting n and logiV the points lie approximately on a straight line. The relation between n and log N may be expressed in the form n=a\ogN+b. Fig. 78. To find the numerical values of the constants a and b it is only necessary to substitute for two points on the line simultaneous values of n and log N, thus obtaining two equations from which a and b can be obtained. Thus at c (Fig 78), n = l, logJ\r=2-54, and at d, n = 2 '6, log N= 3 -28. PLOTTING A LINE. 181 Hence By subtraction, 26: 1: :ax 3*28 + 6 :x2-54 + 6 (i) (ii) l-6 = ax-74 a =^? = 2-162. And substituting this value for a in (i) we have 6 = 2-6 -2-162x3-28= -4'49. Hence the relation between the variables is expressed by 71=2-162^^-4-49. y ' s / : s / C $ J* 1L 111 A5 . S 5. . S ? 5 j? %' " 'it l Z2"l \ *? 8* S 7^ S y S *.- - 3p -os I IS 2 '25 v3 Deflection* Fig. 79. Ex. 4. The depths d and deflections 5, when loaded with the same load, of a series of beams of varying depths and constant breadths are given in the annexed table. Find the equation con- necting d and 5. d 1 75 625 5 375 25 8 02 033 06 118 27 934 1 1 238 41 8 18-9 64 When the variables d and 5 are plotted a curve is obtained ; but by plotting 5 and -^ a straight line, lying evenly among the points, 182 PRACTICAL MATHEMATICS FOR BEGINNERS. can be drawn as in Fig. 79. The line passes through the origin, 1 and its equation may be written 8 = ax From Fig. 79 at c, Hence -25 = ax 17; 1 8= -25 and ^= 17. a 6 a=^? = -0147. 17 The relation is therefore 8= 0147 cF i: !:_:::: : cz / y! 3-6 / - ^ 7 - z i*t '- I 3y -/ f^. - Z ,N v! 2 T- 1 , * ' /- V -/ - s 7 a z I 1 U /- t / - k / 7 : z . ft /: 7 z ^ : ' s z : ^ ~r -W / < 42 CZi -03 * Deflectiorvs. Fig. 80. iik. 5. The following table gives a series of values of the breadths b and deflections 5 of a series of beams of constant depths and vari- able breadths ; find the equation connecting b and 5. b 25 375 5 625 7 1 5 03 017 014 on 009 007 b 4 267 2 1-6 1-33 1 If the first two columns are plotted the points lie on a curve, but SLOPE OF A LINE. 183 by plotting the second and last columns 5 and 7 (Fig. 80), a straight line through the points and passing through the origin may be drawn. At the point c, 5=*024 and r = 3'2. Substituting these 1 values in the equation 5 axj, the relation between the variables is found to be 5='007x r . Slope of a line. The ratio of increase of one variable quantity relatively to that of one another is of fundamental importance. Simple cases are furnished as in the preceding examples, when on plotting two variable quantities on squared paper a straight line connecting the plotted points can be obtained. The rate of increase which is constant is denoted by the inclination or slope of the line. Care should be /\ taken to clearly distin- guish between the usual meaning attached to the term " slope of a line " and the meaning given to the same words in Mathe- matics. LEVEL What is usually meant jmmmw//////sW^^ by the statement that a p Ia 8 i s hill rises 1 in 20 is that for every 20 feet along the hill there is a vertical rise of 1 foot. To indicate the slope of a railway line a post may be placed along the side of the line and a projecting arm indicates roughly the slope by the angle which it makes with the horizontal, and in addition the actual amount is marked on it. As in Fig. 81, the termination of the slope and the commencement of a level line may in like manner be denoted by a horizontal arm with the word level on it. This so-called slope of a line which is largely used by engineers and others is not the slope used in Mathematics . The slope of a line such as A B (Fig . 82) is denoted by drawing a horizontal line at any convenient point A and at any other point B a perpendicular BC meeting the former line OB in 0. The slope of the line is then measured by the ratio of -jjy- 184 PRACTICAL MATHEMATICS FOR BEGINNERS. If at any point in AB a point B! be taken and a perpendicular B'C be drawn the ratio remains unaltered ; C'B' CB ; ,_, Denoting the coordinates of the point A by (x, y\ then, if B' is a point near to A, the distance AC may be called the increment ofx, and the distance C'B' the increment of y. Instead of using the word increment it is better to introduce a symbol for it, this is usually the symbol 8 ; hence, 8x is read as " increment of x," and does not mean 8xx. Similarly increment of y is written 8y. CB 8y AG 8x is the tangent of the angle BAG, or the tangent of the angle of slope. It will be obvious that the former method would give the sine of the angle of slope. Rate of increase. To find the rate at which a quantity is increasing at any given point we find the rate of increase of y compared with the increase of x at the point. Let the equation of the line AB (Fig. 82) be y = ax + b (i) and let A be the point (x y y\ then the coordinates of B' a point near to A may be written as x + 8x, and y+8y, substi- tuting these values in (i) then we obtain y + 8y = a (x + 8x) + b (ii). 8y Subtract (i) from (ii), :. 8y=a8x, or #-=. Thus we find, as on p. 176, that the slope or inclination of the line depends on the term a. In Ex. 3, the line y = x + 2 has been plotted ; proceeding as in 81/ the preceding example we find that ~- = 1. As the slope or tan- gent of the angle made by the line is 1, we know that the inclination of the line to the axis of x is 45. **- Ex-*- Fig. 82. Slope of a line EXERCISES. 185 J EXERCISES. XXXII. 1. The following numbers refer to the test of a crane. Resistance just overcome, R lbs. Effort just able to overcome re- sistance, E lbs. } 100 200 300 400 500 600 700 800 U-5 128 170 214 25 6 29 9 34 2 38-5 Try whether the relation between E and R is fairly well repre- sented by the equation E=aR + b, and if so, find the best values of a and b. What effort would be required to lift a ton with this crane ? V 2. In the following examples a series of observed values of E, R, and F are given. In each case they are known to follow laws approximately represented by E=aR + b, F=cR + d ; but there are errors of observation. Plot the given values on squared paper, and determine in each case the most probable values of a, b, c, and d. E 35 5 6 75 8-25 9*75 11-5 1325 14-78 R 14 28 42 56 70 84 98 112 F 2-86 3-83 5-00 5 92 6-83 8-00 917 101 T^ii E 5 1 ii 2 2-5 3 3-5 4 R 4 15 28 40 52 64 76 88 F 32 57 80 104 128 162 176 200 (iii) B 3 25 4-25 5 5-75 35 6-75 7'5 8-5 9-25 10 R 14 21 28 42 49 56 63 70 ( ' 2-68 3-39 3-86 4-32 5 04 5-5 6 22 6 68 714 186 PRACTICAL MATHEMATICS FOR BEGINNERS 3. A series of observed values of n and N are given. Find the relation in each case between n and log N. (i) n 25 5 75 1 1-25 1-5 1-75 2 2-25 2-5 2-75 N 154 180 265 375 485 635 835 1135 1535 1835 2435 (ii) n 25 145 5 186 75 235 1 1-25 1-5 1-75 2-0 2 5 3 N 296 385 495 558 683 1115 1515 (iii) n 25 *5 75 1 235 1-25 1-5 1-75 2-0 N 115 145 185 300 385 490 605 4. An electric light station when making its maximum output of 600 kilowatts uses 1920 lbs. of coal per hour. When its load factor is 30 per cent, (that is, when its output is 600 x 30 -r 100) it uses 1026 lbs. of coal per hour. What will be the probable consumption of coal per hour when the load factor is 12 per cent. ? 5. Plot on squared paper the following observed values of A and B, and determine the most probable law connecting A and B. Find the percentage error in the observed value of B when A is 150. A 50 100 150 200 250 300 350 400 B 6 2 7 4 8-3 9 5 103 11-6 12-4 13-6 14-5 6. The following observed values of M and N are supposed to be related by a linear law M=a + bN, but there are errors of observa- tion. Find by plotting the values of M and N the most probable values of a and b. N 25 35 44 5-8 7-5 9-6 12-0 151 18-3 M 13-6 176 222 28-0 35 5 47-4 561 74-6 84-9 EXERCISES. 187 7. (i) The following values, which we may call x and y, were measured. Thus when x was found to be 1, y was found to be *223. X i l 1-8 2-8 39 5-1 60 y 223 327 525 730 910 1-095 It is known that there is a law like y = a + bx connecting these quantities, but the observed values are slightly- wrong. Plot the values of x and y on squared paper, find the most likely values of a and 6, and write down the law of the line. (ii) X 05 1-7 3 47 5-7 71 8-7 9 9 106 11-8 652 y 148 186 265 326 388 436 529 562 611 State the probable error in the measured value of y when x = S'7. 8. In the annexed table, values of L, the length of a liquid column, and T, its time of vibration, are given. The relation between L and T 2 is given by L = aT 2 + b ; find a and h. L 2-4 2-8 3-0 3-2 3-4 3-6 T 1-06 1-23 1-29 1-34 1-38 1-42 9. It is known that the following values of x and y are connected by an equation of the form xy = ax + by, but there are slight errors in the given values. Determine the most probable values of a and b. X 18 28 54 133 -455 -111 -65 y 5 6 7 8 9 10 11 10. The following measurements were made at an Electric Light Station under steady conditions of output : W is the weight in pounds of feed water per hour, and P the electric power, in kilowatts, given out by the station. When P was 50, W was found to be 3800 ; and when P was 100, W was found to be 5100. If it is known that the following law is nearly true W=a + bP, find a and b, also find W when P is 70 kilowatts. State the value of W -r P in each of the three cases. 188 PRACTICAL MATHEMATICS FOR BEGINNERS. 11. Some particulars of riveted lap joints are given in the following table : t= Thickness of plate, 1 * 1 t 1 1" d= Diameter of rivet, f 1 *t H 1 1 p 1 = Pitch of Rivets \ (single riveted), - / 2-06 2-25 2-3 2-37 2-40 2-63 p 2 = Pitch of Rivets \ (double riveted),-) 3 33 3-58 3-60 3 63 3 63 395 x-" (i) Plot d and t and obtain the values of the constants a and b in the relation d = at + b for plates from " to " thick. ^ (ii) Plot d and V7and obtain for the whole series of values given in the table the value of c in the relation d c \/t. (iii) Find values of d when t is yg-, j 7 ^-, and ^-. (iv) Plot d and p x and d and p 2 and obtain the constants in the relations p x = d + b ; p 2 = d + c. 12. The following table gives some standard sizes of Whitworth bolts and nuts. All the dimensions being in inches. d = Diameter of bolt, i 1 1 1 1 H H 2 8| W = Width of nut\ across the corners, / 605 818 1-06 1'50 1-93 2'36 2-77 3-63 4 50 A = Area of bolt at) bottom of thread, l" 027 068 112 304 554 894 1-30 2-31 3-73 (i) Plot d and W and obtain a relation in the form W=ad + b. (ii) Plot A and d? and obtain a relation in the form A =ad?. (iii) Obtain a more accurate relation in the form A = ad? + b for bolts from ^" to l" diameter. 13. In the following table a series of values of the pull P lbs. necessary to tow a canal-boat at speeds V miles per hour are given. If the relation between P and V can be expressed in the form P=CV n , what is the numerical values of the constants G and nt p 10 1-82 2-77 3-73 4 4 V 1-82 253 3 24 3 86 427 CHAPTER XVIII. PLOTTING FUNCTIONS. In the preceding chapter the student will have noticed that when the numerical values of two variables are obtained from a simple formula, the curve passes through the plotted points. When, however, the given numerical values are experimental numbers involving errors of observation the curve is made to lie evenly among the points, in this manner errors of experiment or observation may be corrected, and by interpolation any intermediate value can be obtained. The applications of squared paper are so numerous and varied that it becomes a difficult matter to make a suitable selection. The following examples may serve to illustrate some of the uses to which squared paper can be applied. Ex. 1. In a price list of oil engines the prices for engines of a given brake horse power are as follows : Brake horse power, 11 3* 6* H m 16 Price in pounds (), 75 110 160 200 225 250 Plot the given values on squared paper and find the probable prices of engines of 5 and of 8 horse power. In Fig. 83 the given values are plotted and a curve is drawn, passing through the points. The coordinates of any point on the curve shows the horse power and probable price of an engine. Cor- responding to sizes 5 and 8, we obtain the probable prices as 135 and 180 respectively. 190 PRACTICAL MATHEMATICS FOR BEGINNERS. The calculation of logarithms. The following method described by Prof. Perry, and also devised independently by Mr. E. Edser, may be used to calculate a table of logarithms to three or four significant figures. The square root of 10 or 10* = 3*162. Referring to Table III. it will be found that log 3*1 62 = '5000. Again \/3*162 or 1(F = 1 '778, and log 1778 = -2500. Now 10 ' 5 x 10" 25 = 3*162 x 1-778 = 5-623 ; .-. 10 075 = 5-623. In a similar manner we obtain 10 = 1*336, 10 T?r = 1*1548, and 10^ = 1*0746; .*. 10^ x 10^ = 10^ = 3*162 x 1*0746=3*398. Also 10* = 1000* = 31*62, and 10 1 * = 10^ =56*23. 10 = 1 ; .*. log 1=0. 10 1 = 10; .*. log 10 = 1. When a series of values have been obtained by calculation the logarithms may be plotted soL y on squared paper as or- dinates and the numbers as abscissae. By drawing the logarithmic curve through the plotted points any intermediate value can be read off. Even with the cheapest squared paper, tables of logarithms and antilogarithms can be made fairly accurate in this manner. Using better paper and with care, a table of logarithms Fig. 83. Price list of oil engines. accurately giving logarithms to four figures can be obtained. Ex. 2. By means of squared paper shew the values of the sine, cosine, tangent, and radian measure of all angles from to 90. EQUATIONS. 191 Find from the curves the values of the sines, cosines, and tangents of 15, 30, 45, 75. Here as in Fig. 84 we may denote degrees as abscissa and numerical values as ordinates ; these are obtained from Table V. Having drawn curves through the plotted points the values for 15, etc., can be read off. Notice carefully that when the angle is 45 the sine and cosine curve cross, i.e. the values of the sine and cosine are equal and the curve denoting values of the tangent has an ordinate unity at this point. Again it will be obvious that for small angles not exceeding 20 the values of the sine, radian and tangent are approximately the same. Ex. 3. In the following table some population sta- tistics of a certain country are given. Let P denote the population and t the time in years. Show the relation between P and t by a curve, and find from the curve the probable population in 1845 and in 1877. 1 V /, / J u \ / 1 K? / W/ /\ 4 t \ f \ // f \ / '/ Fig. 84. Values of the sine, cosine, tangent and radian measure of angles from to 90. year. 1821 1831 1841 1851 1861 1871 1881 1891 1901 Pi in millions. 10-2 12-8 15 4 18-4 216 25 6 30-0 38-0 When as in Fig. 85 the given values are plotted and a curve drawn, the probable populations in 1845 at a, and in 1877 at 6, can be read off, and are found to be 16 5 millions and 28*2 millions respectively. Equations. On p. 75 a method has been indicated by which in a given expression such as x 2 - Ax -f 3 the factors {x - 1) (x - 3) can be obtained by substitution; these values x=l, x3 are 192 PRACTICAL MATHEMATICS FOR BEGINNERS. called the roots of the given equation. In equations which are more complicated such a method may become very troublesome ~6? a/ Fig. 85. and laborious ; the roots of an equation, or better, the solution of an equation, which would be difficult by algebraical methods, may in many cases be obtained by the use of squared paper. To gain confidence the method may be applied to any simple equation such as the one above. Ex. 4. To solve the equation x 2 - 5x + 5*25=0, let y= a; 2 -5a; + 5 "25. Substitute values 0, 1, 2, etc., for x, and find corresponding values ofy. Thus, when x = 0, y = 525 ; when x=\, y=l -5 + 525= 1*25; the values so obtained may be tabulated as follows : X y 1 2 3 4 5 5 25 1-25 -75 -75 1-25 5-25 Plotting these values on squared paper a curve of the form shown in Fig. 86 is obtained. The curve crosses the axis of x in two points, A and B ; the two values of x given by 0^4 and OB make y=0 t and therefore are the two roots required; 0.4 is 1*5 and EQUATIONS. 193 OB is 3*5. When these values are substituted they are found to satisfy the given equation. Hence x = 1*5 and # = 3 # 5 are the two roots required. Ex. 5. Find the roots of the equation x 3 - Sx - 1 = 0. Let y=x 3 -Sx- 1. As before, put x=0, 1, 2, etc., and calculate corresponding values of y as follows : X y -2 -3 -1 1 -1 1 -3 2 1 3 17 Plotting these values as in Fig. 87 the curve cuts the axis of x in three points, G, B, and A. At each of these points the value of x V \ \ / / .... / / V / V / 1 \ 2 \ 3 / 4 / 5 X \ y 2 ~ -A. \b % ill 3 Fig. 86. - Graph of .r 2 - 5x+5'25 =0. Pig. 87. Graph of x* - 3x - 1=0. makes y = 0, and hence is a solution of the given equation. At A the value of x is seen to be between -1*5 and -1*7, and at B between - "3 and - *5, by plotting this part of the curve to a larger scale, the more accurate values are found to be - 1 "532, and - - 347. In a similar manner the value at C is found to be 1 879. Probably a simpler method than the one described, and which may be shown by an example, is as follows : Ex. 6. Solve the equation X s - 6x + 4=0. , Write the given equation in the form of two equations. y=x 3 (i), y = 6x-4:{n). P.M. B. N 194 PRACTICAL MATHEMATICS FOR BEGINNERS. From (i) we shall by plotting obtain a curve, and from (ii) a line. The points of intersection of two lines, as in p. 179, give values which satisfy the equations, and in like manner the points of inter- section of the line and curve will give the required values of x. Thus in (i), by giving x various values 0, 1, 2, etc., we can calculate corresponding values of y as follows : X i 2 3 4 5 y l 8 27 64 125 By plotting these values we obtain the curve shown in Fig. 88. Positive values of x have been assumed, but if negative values are used the values of y are of the same magnitude but with altered sign. Hence the corresponding part of the curve, below the axis of x, can be obtained. In (ii), if x = 0, y -4, and if x = 5, y = 26, the line drawn through these plotted points will give at their points of in- tersection A and B the required values. As all equations of this kind can be reduced to the forms shown at (i) and (ii) the curve indicated by (i) may be used for all equations of this form. Plotting of functions. Functions of the form y=ax M , y=ae 6 *, y = sinax, where a, b, and n may have all sorts of values, are easily dealt with by using squared paper. Thus in the equation y = aa?, when a and n are known, for various values of x corresponding values of y can be obtained. Ex. 7. Let a ='25 and n = 2. The equation y = ax n becomes y=0'25x 2 x By giving a series of values tox, 1, 2, 3, etc., we can obtain from Eq. (i) corresponding values of y. Thus, when x=0, y=0, also when x= 1, y = *25. H - / / / / / / / / / / CD / ? 1 \ ** ' ! 1 S >5 Fig. 88. Graph of x*-6x+i=0. PLOTTING OF FUNCTIONS. 195 It will be convenient to arrange the two sets of values of x and y as follows : Values of x, 1 2 3 4 5 Corresponding \ values of y, / 25 1 2 25 4 6-25 As y is when x is 0, the curve passes through the origin (or point of intersection of the axes). Plotting the values of x and y from the two columns, as shown in Fig. 89, a series of points are obtained. The figure obtained is a parabola. iiniiiiiiiiiiiJiiiiiiiiiiit/iM -i A 2 \ s * V / ^ t X 4 \ T 1^7 ^ -i z S / \ I / _s: :2_ *^ --r s - -j "' jg^-y^T 1 3 n i > 7 % ,/ "' \ / \ Z . S J. - V- t -/ + - J - N i z x i ** ^ ^ 7 3 V 3 -2 "* 5 L ._.._! 5 Fig. 89. Graph of y = -25a;2 It is sometimes difficult to draw a fairly uniform curve through plotted points, but when a curve has been drawn improvements may be made, or faults detected, by simply holding the paper on which the curve is drawn at the level of the eye, and looking along the curve. Some such simple device should always be used. As the square of either a positive or a negative number is necessarily positive, it follows that two values of x, equal in magnitude but opposite in sign, correspond to each value found 196 PRACTICAL MATHEMATICS FOR BEGINNERS. for y. By using positive values of x, the curve shown on the right of the line oy is obtained. The negative values give the corresponding curve on the left. If the constant a be negative, its numerical value remaining the same, then the equation becomes y= '2bx* ; this when plotted will be found to be another parabola below the axis of x (Fig. 89). The equation y = ax M becomes when a = l, y=x n . Giving various values 2, 3, ^, J, 1, etc., to the index n then functions of the form y=x 3 , y=x*, etc., are obtained. Assuming values 0, 1, 2 . . . for x corresponding values of y can be found. The curves can be plotted, and are shown in Fig. 90. It will be seen that the curves y = x 3 ,y = x^, and the straight line y = x all intersect at the same point. O 0-1 0-2 0-3 0-4 0-5 0-6 0-7 0-8 0-9 VOX Fig. 90. Graph of y-ax^. The hyperbolic curve is of great importance, more especially to an engineer, and is obtained from the general equation y = ax n by making n 1 ; the equation then becomes y = ax~ 1 or y ~ x - ocy^a (i) The curve is shown in Fig. 91, and should be carefully plotted. The rectangular hyperbola is the curve of expansion for a gas PLOTTING OF FUNCTIONS. 197 such as air, at constant temperature, and is often taken to represent the curve of expansion of superheated or saturated steam. If p and v denote the pressure and volume respectively of a gas, instead of the form shown by (i), the equation is usually E A- > 23 45 6 789 Fig. 91. Graph of xy =9. written, pv = constant = c*, and is known as Boyle's Law; e is a constant, this is either given, or may be obtained from simul- taneous values of p and v. Ex. 8. Plot the curve xy = 9. 9 y =x (ii) From (ii), when x=l, y = 9. x = 2, 2/ = 4-5. x=Tjhr<), y=9000. .*. when x is very small y is very great. Thus let *=TITO then y= 9000000. When x=0, then y=$, or is infinite in value. In other words the curve gets nearer and nearer to the axis oy as the value of x is diminished, but does not reach the axis at any finite distance from the origin. This is expressed by the symbols y oo when x = 0. 9 As Eq. (ii) can be written x=- it follows as before that when y=0> y 198 PRACTICAL MATHEMATICS FOR BEGINNERS. The two lines or axes ox and oy are called asymptotes and are said to meet (or touch) the curve at an infinite distance. Arranging in two columns a series of values of x and corresponding values of y obtained from Eq. (ii) we obtain. Values of x, ! o 1 l 2 3 4 5 6 7 8 9 Corresponding values of y, 1 1 9 4-5 3 2-25 i-s 1-33 1-3 113 1 Plotting these values of x and y on squared paper then the curve or graph passing through the plotted points is a hyperbola as in Fig. 91. One of the most important curves with which an engineer is concerned is given by the equation pv n = c, where p denotes the pressure and v the volume of a given quantity of gas. The constant c and index n depend upon the substance used ; i.e. steam, air, etc. When, as in the preceding example, the values of c and n are known, for various values of one variable, corresponding values of the other can be obtained, and plotted. The converse problem would be, given various simultaneous values of p and v calculate the numerical values of c and n. To do this it is necessary to write the equation pv n = c in the form log p + n log v = log c. Plotting logp and logv a straight line may be drawn lying evenly among the plotted points, and from two simultaneous values of p and v the values of c and n may be found. EXERCISES. XXXIII. 1. A man sells kettles. He has only made them of three sizes as yet, and he has fixed on the following as fair list prices : 12 pint kettle, price 68 pence. 6 50 2 22 He knows that other sizes will be wanted, and he wishes to publish at once a price list for many sizes. State the probably correct list prices of his 4 and 8 pint kettles. EXERCISES. 199 2. Plot the following values of D and 0, and determine (i) The value of D when 6 is 0. (ii) The value of 6 when D is 0. (iii) The maximum value of D. 6 D inches -45 -15 15 45 75 105 -0-25 98 1-80 2 24 2-05 1-32 3. Plot the corresponding values of x and y given below, and determine the mean value of y. 11-5 25 40-5 58 5 96-4 109 120 y i 7'6 102 12-6 14-4 15-6 16 15 2 13'8 112 v 4. Plot the curve given by the equation y = 0'le x where e = 2*718. 5. The population of a country is as follows : Year. 1830 1840 1850 1860 1870 1 1880 i 1890 Population ) (millions), J 20 23 5 29 34-2 41 49-4 57-7 Find by plotting the probable population in 1835, 1865, and in 1895. Find the probable population at the beginning of 1848 and the rate of increase of population then. 6. A manufacturer finds that to make a certain type of cast-iron pump the cost is 45 shillings for a pump of 3 inches diameter, and 115 shillings for a pump of 6 inches diameter. Estimate the probable cost of pumps of 4 inches and 5 inches diameters so far as you can from these data. If the actual cost of a 5-inch pump when made is found to be 82 shillings, what would now be the estimate of the probable cost of a 4-inch pump ? Solve the following equations : 7. x 2 - 5 -45a; + 7 '181 = 0. 8. 0-24x 2 -4'37- 8'97 = 0. 9. 2 3a; 2 -6'72a;- 13-6 = 0. 10. x 3 -7x 2 +Ux-8 = 0. Find in each of the following a value of x which satisfies the equations : 11. 2.x- 31 -Sx- 16 = 0. 12. 2'42ar 5 -3-151og e a--20 5 = 0. 200 PRACTICAL MATHEMATICS FOR BEGINNERS. 13. e*-e-* + 0-4:r-i0 = 0. The answer to be given correctly to three significant figures. 14. The following values of p and u, the pressure and specific volume of water-steam, are taken from steam tables : p 15 20 30 40 50 65 80 100 u 25-87 19 72 13-48 10-29 8-34 6 52 5-37 4 36 Find by plotting log p and log u whether an equation of the form pu 11 constant represents the law connecting p and u, and if so, find the best average value of the index n for the range of values given. / 15. Find a value of x which satisfies each of the equations 4i) x s +4'73x- 1-746=0. *(ii) ^ + 9a; -16 = 0. 16. Find a value for x for which tan x = 2 -75a;. 17. Given y=l -4-818^ + 7*514^, calculate and enter the values of y in the following table : n X i 2 3 4 5 6 7 8 9 10 y Plot the curve and find one root. Solve the equations : 18. x*- 13a? -12 = 0. 19. a- 3 -237a;- 884= :0. 20. x s -27x-4Q=0. Slope of a curve. The slope of a curve at any point is that of the tangent to the curve at the point. The tangent to the curve is the straight line which touches the curve at the point. If, in Fig. 92, the tangent at P makes an angle 42 with the axis of x, then slope of curve at P=tan 42 = 0*90. It is an easy matter to draw a line touching a given curve at a point when the inclination of the line is known, but if the direction is not known, then at any point P several lines apparently touching the curve could be drawn, but it would be difficult by mere inspection to draw a tangent at the point. Before this can be done it is necessary to know the direction of the line with some approach to accuracy. This may be effected by taking the values of x and y at a given point and the values x' y' of another point close to the former ; from these values SLOPE OF A CURVE. 201 x'-x, and/-y can be obtained, the former may be denoted by 8x and the latter by 8y. The ratio -M- gives the average rate of increase of one variable compared with the other, and also approximately the slope of the tangent at P. If Q and P (Fig. 92) be two points on a curve, the former the point (x', y') the latter the V \ f / k. / ' J Ov M M, 12345678 X Fig. 92. Slope of a curve. point (x, y\ then x'-x or 8x is 7-5 = 2, and y'-y or 8y is 5-1-5 = 3-5. o^ _ 2 ~ 175 * Draw Q M and PM parallel to oy and ox respectively, then if points P and Q be joined by a straight line ^=tan QPM=6Z'3. This is obviously only a very rough approximation, a better result is obtained by using point Q x (Fig. 92), its co- ordinates (5*4, 1*88). The increments 8y and 8x become - 38 and 0*4 respectively. /. ^=^='9499; /. slope of line = 43-3. If the numbers on the vertical axis denote distances in feet, and along the horizontal axis time in seconds, then the slope of the curve at any point gives rate of change of position or velocity at the point. Thus at P the velocity would be '9 ft. per sec. If the ordinates denote velocities and the abscissae times, then the slope of the curve at any point gives rate of change of velocity, or the acceleration at the point. When the increments 8y and 8x are made smaller and smaller the slope given by -~ becomes more and more nearly the actual slope at the point. Finally, when each increment is made indefinitely small the ratio is written -~ and is the tangent or the actual rate at the point. 202 PRACTICAL MATHEMATICS FOR BEGINNERS. The slope of a curve at a given point may be indicated by a simple example as follows : Ex. 10. Plot the curve y = x 2 and find the slope of the curve at the point x = 2. The square of a negative and a positive quantity are alike posi- tive, so that for each value of y there are two values of x. Thus, when a;=2 or -2, or x=2, y=4, etc. By substituting values 0, 1, 2, 3, etc., for x corresponding values of y are obtained as in the following table : x 1 2 3 4 5 y 0' 1 4 9 16 25 To obtain the slope of the curve at the point 2, if we take the two points x = 2 and x' = S, then y=4, y' = 9 ; . y'-y_5. a/ -x 1 5. I 7C \ K I 7 \ SO xit t \ \ 40 j_ \ ' \ X ~J_ l2_ K / \i / \ K) ~v \ 9>Q 6 7 1 3 ? I 4t Fig. 93. Graph of y=x 2 . This is obviously only a rough approximation. The line PQ (Fig. 93) joining the two points cuts the curve, and the slope of the line is 7/ -y ^5 x' -x r given by SLOPE OF A CURVE. 203 Assuming a point Q' nearer to P, then a better approximation is obtained. Thus if x = 2 5, then 2/' = (2-5) 2 =6'25 ; .*. x'-x = *5, y' -y = 2'25. Hence =.**.4* . *1 = VK = M x'-x~ '5 ' " 5x PM' 1 As the magnitude of x' - x is diminished the corresponding values obtained approach nearer and nearer to the actual value. Thus when x' = 2 -05, y' = 4 '2025, &c # 05 When x' = 2 -005, y' = 4 '020025, In each case we obtain the average rate of increase at the point P, the average rate approaching nearer and nearer to the actual rate as the increments get smaller and smaller. The actual rate at P will be 4 when the increments 5y and dx are made small enough, and it is easy to show this by using algebraic symbols as follows : If the equation to a curve be V = x 2 (i) and {x, y) the coordinates of a point on the curve, the coordinates of a point close to the former may be written x + 5x and y + dy (p. 184). Substituting these values in (i) we get y + 8y = (x + dxf = x 2 + 2x5x + {5x) 2 (ii) Subtracting (i) from (ii), .-. dy = 2x5x+{8x) 2 . Dividing both sides by 8x, 8v j L = 2x+dx (in) Equation (iii) is true whatever value is given to 8x, and values of ~ corresponding to values for 8x of '5, *05, etc., have been obtained, in each case giving an approximation to the tangent at the point and also giving the average rate of increase of y with respect to x. If we imagine the increments 6y and 6> to get smaller and 204 PRACTICAL MATHEMATICS FOR BEGINNERS. smaller without limit, then the ratio ^ is denoted by -J~. and '7 ox J dor equation (iii) becomes -h = 2x. This is the actual rate at the point P, or in other words is the tangent at P. Hence the slope of a curve at a given point is represented by the tangent of the angle which the tangent to the curve makes with the axis of x. The symbol ~- is read as the differential coefficient of y with respect to x, and simply denotes a rate of increase. Its numerical value can be ascertained when the law or the relation connect- ing two variables x and y is known. The beginner should notice that the differential of a variable quantity denoting the difference between two consecutive values is an indefinitely small quantity, and is expressed by writing the letter d before the variable x or y. When this is clearly understood the symbol dx (which is read as the differential of x) will not be taken to mean dxx, nor dy as dx y. It is obvious that there cannot be any rate of change of a constant quantity, hence the differential of a constant quantity is zero. Ex. 11. Find the slope of the curve y = x? at the point x=25. As before, we may write y + dy (x + dx) 3 ; .-. y + dy = x 3 + 3x 2 dx + Sx{dx) 2 + {dx) 3 . Subtracting y = x 3 , dy = 3x 2 dx + Zx(dx) 2 + (dx) 3 , or ^- = 3x* + 3x5x + (dx)*. ox When the increments become indefinitely small the ratio on the left is -^, and on the right all terms involving 5# disappear. dx Hence < & = 3x 2 . dx The slope of the curve at the point x =25 is 3x2-5 2 = 18-75. On p. 176 we have found that the equation y = ax + b represents the equation to a straight line in which the slope or inclination of the line to the axis of x depends on a. SIMPLE DIFFERENTIATION. 205 Let y = ax + b, (i) then y + 8y = a(x + 5x) + b = ax + a5x + b, (ii) Subtracting (i) from (ii), .'. by = adx ; . fy dy ..... Hence a is the slope, or the tangent of the angle which the line makes with the axis of x. Generally if y = ax n then ~=nax n ~ 1 , (iv) where a is a constant and n is any number positive or negative. Simple differentiation. The process of finding the value of -A the rate of change from a given expression, is called differen- tiation, and in simple cases, which are all that are required at the present stage, it is only necessary to apply the rule given by Eq. (iv). Any constant which is a multiplier or divisor of a term will be a multiplier or divisor after differentiation, but as the differ- ential of a constant is zero, any constant connected to a variable by the signs + or disappears during differentiation. The process may be seen from the following examples : Ex. 12. y = 3x*. Ex. 13. y = 5x*. ^ = 2x3xV-V = 6x. ax dy = 4:x5xU-V = < 2 = '53e- 52 ; .-. log y = log -53 + '52 log 2 -7 1 8 = 1-7243+ -52 x -4343 = T-9501. .-. y=-8915. 208 PRACTICAL MATHEMATICS FOR BEGINNERS. Other values of x can be assumed and values of as in the following table : calculated Values of x. 1 2 3 4 5 6 7 8 Corresponding ) values of y. ) 53 6874 8915 1-156 1*500 1945 2-522 3-271 4-242 To obtain the average value of y from x = to x = 8 we can apply Simpson's Rule, p. 233. Thus Sum of end ordinates=4 - 772, ,, even = 7 0594, odd =4-9135. Area of curve from x=0 to x = 8 is 4-772 + 7 0594 x 4 + 4-9135x2=42-8366; 42-8366 .-. area of cur ve = ^ But average value of y multiplied by length of base = area ; , , 42-8366 . _ Q _ .'. average value of y = 1 "785. 3x8 Proceeding as in preceding problems, If y=ae bx , -^- = abe bx dx dx = -53 x -26x2 -7 18 7 8 =-3006. .-. slope of curve at point #=3 is '3006. Maxima and minima. If a quantity varies in such a way that its value increases to a certain point and then diminishes, the decrease continuing until another point is reached, after which it begins to increase ; then the former point is called a maximum and the latter a minimum value of the quantity. Ex. 20. Given Zx 3 - I5x 2 + 24# + 25 = ; determine values of x so that the expression may be a maximum or a minimum. Denoting by y the value of the left-hand side of the equation, and substituting various values for x, corresponding values of y can be obtained, as shown in the annexed table : X 5 1 2 3 4 5 y 25 33 6 36 29 16 9 20 MAXIMA AND MINIMA. 209 Plotting these values on squared paper and joining the points, the curve ABGD (Fig. 94) is obtained. From the tabulated values B 30 A / N^ / \ \ D C I 2 3 4- S Fig. 94. Curve showing maximum and minimum values. the expression seems to be a maximum when x=l, and a minimum when x = ; this is confirmed by the curve, the former value being located at B, the latter at C. Also, as explained on p. 205, we have, y = 2ar 5 -15x 2 + 24# + 25, ^| = 6x 2 - 30a; + 24 ; this gives the slope at any point, or the angle which the tangent to the curve at any point makes with the axis of x. At the points B and G the angle is zero, i.e. the tangent is horizontal, hence dy dx :. 6x* -30a; + 24 = 0, or x 2 -5x + 4 = (x-l){x-4), and the values of x which satisfy this equation will give the points at which the tangent to the curve is horizontal. The required values are therefore x=l, x = \. It will be seen from the above example that the values of x corresponding to -^=0 may give either a maximum or mini- mum value ; in many practical questions the conditions of the question will at once suggest whether the value obtained is a maximum or minimum. From Fig. 94, for instance, it is obvious that at point C, where y is a minimum, an increase in x gives an increase in y, i.e. ~ is positive, at B where y is a maximum, for dy an increase in x we obtain a decrease in y, hence -# is negative. P.M. B. O 210 PRACTICAL MATHEMATICS FOR BEGINNERS. As there may be several loops in a curve similar to those indicated, the terms maocimum and minimum may in each case be taken to denote a point such that a point near to, and on the left of it, has a different slope to a similar point near to, and on the right of it. Ex. 21. Divide the number 8 into two parts, such that their product is a maximum. Let x denote one part, then 8 - x is the other ; x{8 - x), or 8a; - x 2 is the product ; denote this by y. Substitute values 0, 1,2, 3, 4, 5, 6 for x, and find the corresponding values 0, 7, 12, 15, 16, 15, 12 for y. Plot on squared paper, and at x 4 a point corresponding to B (Fig. 94) is obtained ; or let y = 8x - x 2 , then -~ = %-1x; .". 8 -2a; = 0, for a maximum, gives x = 4. Average velocity. Probably every one has a more or less clear idea of what is meant by saying that a railway train, which may be continually varying its speed, is at any given instant moving at the rate of so many miles per hour. Suppose that in t hours the train has gone over a distance 5 miles ; then if the rate were uniform, the rate - would denote the number of miles per hour. As a simple numerical example suppose the distance between two places to be 150 miles and the time taken by a railway train from one place to the other is 5 hours ; the uniform rate 150 per hour or average velocity would be -=30 miles per hour. 5 Such an average would include all the variations of speed, including stoppages on the journey, and is clearly not what is meant by the statement that at a given moment the train is going at one particular speed. To obtain the numerical value of such a speed it is necessary to recognise that as the speed is variable the value of - is continually changing, and can only give a good approximate value of the average velocity when the time interval is very short, i.e. when t and therefore 5 are both small quantities. Such small intervals or increments may be denoted by 8s and 8t, .'. Average velocity = -*-. AVERAGE VELOCITY. 211 Ex. 22. Suppose a body to fall from rest according to the law s=16* 2 (i) where s is the space in feet and t the time in seconds. Find the actual velocity of the body when t is one second. In this example, if s and t are plotted the curve is of the form shown in Fig. 93. To find the velocity at time 1, we can, from the given equation, find the space described in a fractional part of a second ; by dividing the space described by the time, the average velocity is obtained. We may take values of t such as 1 and 125, 1 and 1*1, and 1 and l'Ol, the approximation becoming closer and closer to the actual value as the interval is diminished. When the points are 1 and 1*1, then from (i) s=16x(l\L) 2 = 19-36 feet; /. space in 1 second=16{(M) 2 - 1 2 }=3'36 ; .'. average velocity during *1 second = ---=33 *6. T"tf Hence the average velocity obtained is too great, and its inaccuracy becomes greater as the interval of time is increased. Thus space is one quarter of a second = 16 {(1) 2 - l 2 } = 16(M-1) = 9; 9 /. average velocity during "25 second = ^ = 36. If the interval be from 1 to 1*01, space-16(101 2 -l 2 ) = 16(1'0201-1)=-3216; "3216 .*. average velocity during '01 second = -^-=32 '16. Other values for t may be assumed, the average value obtained becoming closer and closer to the actual value as the interval of time is diminished. Thus, the intervals of time may be '001, 0001 of a second, etc. These small intervals of time and corresponding small space described may be indicated in a convenient manner by the symbols 8s and 8t. The actual value is obtained when the increments are made indefinitely small, and 8t , ds ^becomes^-. The preceding results are easily obtained by means of Algebra. The coordinates of any point on the curve s = 16tf 2 ... (i) may 212 PRACTICAL MATHEMATICS FOR BEGINNERS. be denoted by (s, *), and those of a point near to it by s + 8s and * + 8*. Substituting these values in (i) we get s + 8s=16(*+8*) 2 =16{* 2 + 2*(8*) + (8*) 2 } (ii) Subtracting (i) from (ii) we have 8s = 32*(8*) + 16(8*) 2 . Dividing by 8t, 8s St = 32* + 168* (iii) When 8t is made smaller and smaller without limit, then the last term 168* is zero and (iii) becomes ds dt 32*. Hence the actual value when * is 1 is 32. Ex. 23. At the end of a time * it is observed that a body has passed over a distance s. Given that s=10 + 16* + 7* 2 , (i) find s when * is 5. Taking a slightly greater value for *, say * = 5 - 01, calculate the new value of s and find the average velocity during the '01 second. Also find the exact velocity at the instant when * is 5. Assuming values 0, 1, 2 ... for * values for s can be found from (i) as follows : * 1 2 3 4 5 s 10 33 70 121 186 265 When* is 5; s=10 + 80 + 7 x 25 = 265. * is 5-01; 8=10 + 16x5-01 + 7x(5-01) a = 265'8607. space in '01 sec. = 265 '8607 - 265 = '8607. Average velocity or 8s '8607 = 86-07. St -01 When * is 5'001, proceeding as before 8s = 086007. . Ss_ -086007 5*~ 001 86-007. Again when * is 5*0001, te= -008600007. . 8s _ -008600007 " 8t~ 0001 86-00007. ACCELERATION. 213 It will be seen that the average velocities approach a certain value during smaller and smaller intervals of time, and the limiting value is the actual velocity at the point ; or, by differentiation, s=10 + Wt + W; dt 16 + 14*; and when t = 5 this gives 86 as the actual velocity. Acceleration. When the ordinates of a curve denote the space or distance passed over by a moving body and the abscissae the time, the slope of the curve at any point gives the velocity at that point. If the ordinates are made to represent the velocity and the abscissae, time, then the slope of the curve, or the tangent to the curve at any point, gives the rate at which the velocity of the moving body is increasing or diminishing. In the former case the rate of increase is called acceleration, if the latter then the rate of decrease is called the retardation. Thus the velocity of a body falling freely is known to be g feet per sec. at the end of one second (where g denotes 32 '2 ft. per sec), the velocity at the end of the next second would be 2g. Hence if we proceed to plot velocities and times we should obtain a straight line, indicating that the " slope " is constant. The body is said to move with uniform acceleration. M -4- / ! f v \\ / ts v \ / <: 1 \ \ Q / r A 1 ' \ ^ / $ r / f 1 A V X 4 10 I? Time Fig. 95. The slope of a curve PQMN (Fig. 95) at a point such as P, gives the rate of increase of the velocity, or acceleration at the point. At M the tangent to the curve is horizontal, the slope is 0, the acceleration is zero, and the body is moving with uniform 214 PRACTICAL MATHEMATICS FOR BEGINNERS. velocity. At a point such as Q the slope of the curve gives the rate of decrease or retardation, and at N the body is again mov- ing with uniform velocity. The points M and N correspond to maximum and minimum. EXERCISES. XXXIV, 1. What is meant by the slope of a curve at a point on the curve ? How is this measured ? If the co-ordinates of points on the curve represent two varying quantities, say, distance and time, what does the slope of the curve at any place represent ? Obtain an expression for the slope if the distance s and time t are connected by the equation s = 5t + 2'lt 2 and give the numerical value at the instant when t is 5. 2. At the end of a time t it is observed that a body has passed over a distance s reckoned from some starting point. If it is known that ,9 = 20 + 12^ + It 2 , find s when t is 5, and by taking a slightly greater value of t, say 5 '001, calculate the new value of s and find the average velocity during the "001 second. How would you proceed to find the exact velocity at the instant when t is 5, and how much is this velocity ? 3. A body is first observed at the instant when it is passing a point A. The time t hours (measured from this instant) and the distance s miles (measured from A) are connected by the equation s = 20t 2 : find the average speed of the body during the interval between t = 2 and t = 2l, between t = 2 and = 2'001, and between t = 2 and = 2*00001. Deduce the actual speed at the instant when t is exactly 2. How could you otherwise determine this speed, and what symbol is used to denote it ? 4. How do we measure (1) the slope of a straight line, (2) the slope of a curve at any point on it ? There are two quantities denoted by v and r which vary in such a way that v = 4*2r*. Explain what is meant by "the rate of increase of v relatively to the increase of r." How may the value of this rate of increase be exhibited graphically for any value of r? Calculate its value when r = 0'5. 5. A body weighing 1610 lbs. was lifted vertically by a rope, there being a damped spring balance to indicate the pulling force F lbs. of the rope When the body had been lifted x feet from its position of rest, the pulling force was automatically recorded as follows : X 11 ! 20 34 45 55 66 76 F 4010 3915 3763 | 3532 3366 3208 3100 3007 EXERCISES. 215 Find approximately the work done on the body when it has risen 70 feet. 6. A body is observed at the instant when it is passing a point P. From subsequent observations it is found that in any time t seconds, measured from this instant, the body has described s feet (measured from P), where s and i are connected by the equation s = 2t + 4t 2 . Find the average speed of the body between the interval 2=1 and t=l'\; between 2=1 and 2=1 001, and between 2 = 1 and 2=1 0001, and deduce the actual speed when 2 is exactly 1. 1 '5x 7. Plot the curve y=, jvk~- Determine the average value of 1 -f- KJ'OX y between x = and x= 10. Solve the equations : 8. 35a: 2 -5 '23a; -7 "86 = 0. 9. 2-065- 048 * = '826. 10. Find, correctly to three significant figures, a value of x which will satisfy this equation : 9ar* - 41x 08 + -5e 2 * - 92 = 0. 11. Divide the number 12 into two parts so that the square on one part together with twice the square on the other shall be a minimum. 12. Plot the curve y=x 2 -5'45x + 7'lSl between the points a; =0 and # = 4, and determine the average value of y between the points a:=325 and a; = 4. Plot the following curves from #=0 to #=8. 13. y = 4a;0-7. 14. y =2-3 sin ( -2618a: + ^Y 15. y=0-53e Q *. In each case find the rate of increase of y with regard to x where X=S; also find the average value of y from a: = to a; = 8. i -1 16. Given (i) y = ax n , (ii) y = ax 5 + bx~* + ex* + dx q y write down the value of ^. ax 17. By using squared paper, or by any other method, divide the number 420 into two parts such that their product is a maximum. Describe your method. 18. A certain quantity y depends upon x in such a way that y = a + bx + ex 2 , where a, b, and c are given constant numbers. Prove that the rate of increase of y with regard to x is b + lex. 19. Divide the number 20 into two parts, such that the square of one, together with three times the square of the other, shall be a minimum. Use any method you please. CHAPTER XIX. MENSURATION. AREA OF PARALLELOGRAM. TRI- ANGLE. CIRCUMFERENCE OF CIRCLE. AREA OF A CIRCLE. Areas of plane figures. When the numbers of units of length in two lines at right angles to each other are multiplied together, the product obtained is said to be a quantity of two dimensions, and is referred to as so many square inches, square feet, square centimetres, etc., depending upon the units in which the measures of the lengths are taken. The result of the multi- plication gives what is called an area. Or, briefly, the area of a surface is the number of square units (square inches, etc.) contained in the surface. It is obvious that although square inches or units of area are derived from, and calculated by means of, linear measure, those quantities only which are of the same dimensions can be added, subtracted, or equated to each other. Thus, we cannot add or subtract a line and an area, or an area and a volume. Results obtained in such cases would be meaningless. It must be observed, too, that the two lengths multiplied together to obtain the area are perpendicular to each other. This applies to the calculation of all areas. Area of a rectangle. The number of units of area in a rectangular figure is found by multiplying together the numbers of units of length in two adjacent sides. Thus, if A B and BG (Fig. 96) are two adjacent sides of a rectangle, its area is the product of the number of units of length in AB and the number of units of length in BG. Let a be the number of units of length in the lowest line of the MEASUREMENT OF AREA. 217 figure AB usually called its base, or length, and the units in line BC, perpendicular to this, its altitude, or breadth b. 1 2 3 4 5 6 7 8 9 10 11 12 B Hence or Fig. 96. Area of a rectangle. area = base X altitude, = length x breadth = ab. Ex. 1. If the base AB and height BC are 6 and 2 units of length respectively, the area is 12 square units. If AB be divided into 6 equal parts and BC into 2, then by drawing lines through the points of division parallel to AB and BC, the rectangle is seen to be divided into 12 equal squares (Fig. 96). The area is obtained in a similar manner when the two given numbers denoting the lengths of the sides are not whole numbers. Ex. 2. Obtain the area of a rectangle when the two adjacent sides are 5 ft. 9 in. and 2 ft. 6 in. in length respectively. We may reduce to inches before multiplying. Thus 5 ft. 9 in. =69 inches and 2 ft. 6 in. = 30 inches ; .*. area of rectangle = 69 x 30 square inches = 2070 square inches =14*375 square feet ; Or, instead of first reducing the feet to inches and afterwards multiplying, we may proceed as follows : 5 ft. 9 in. =5f feet and 2 ft. 6 in. =2j feet ; .'. area of rectangle = 5f x 2i square feet =^xf square feet = 14f square feet. If a rectangle is divided into three, four, or more rect- angles, the area of the whole is equal to the sum of the areas of the several parts. In Fig. 97 the rectangle A BCD is divided into four rectangles. 218 PRACTICAL MATHEMATICS FOR BEGINNERS. Area of AEFK=Sx 4 = 12 sq. in. ; area of BEFG=6 sq. in. ; area of GFHC=2 sq. in. ; area of ffFKD=4 sq. in. .-. Total area is 12 + 6 + 2 + 4 = 24 sq. -4- *H T>< >-r- *c B Fig. y" in., and this is equal to the area of A BCD. Using the letters a, b, c, d to denote the respective sides of the four rectangles, we have a verifi- cation of the formula (a + b)(c + d) = ac + bc + ad+bd. Area of a parallelogram. The rectangle is a par- and Fig. 98. Area of a parallelogram. ticular case of the parallelogram, area of pa?'allelogra?n = base x altitude. This may also be shown as follows : Let ABCD (Fig. 98) be the given paral- lelogram length of sides a and b respectively. Draw AF and BE perpendicular to AB, and meeting CD at F, and CD pro- duced at F, then the rectangle A BFE is equal in area to the parallelogram ABCD. Hence, area of parallelogram = base X altitude = ato ; or, the vertical distance between a pair of parallel sides multiplied by one of them. As b=AD sin ED A the area=.4Z? x AD sin EDA = ab sin 6 ; or, the product of two adjacent sides and the sine of the included angle gives the area of a parallelogram. As sin 90 = 1, this formula immediately reduces to that given for a rectangle when the included angle is 90. Of the three terms, area, base, and altitude, any two being given, the remaining term may be found. Similarly, if the area, one side, and included angle be given, the remaining side fcan be found. AREA OF PARALLELOGRAM. 219 Ex. 1. If the altitude be l ft. and the area 6 sq. ft., then the base is Ex. 2. The area of a parallelogram is 12 sq. ft., one side is 6 ft. and included angle is 30. Find the remaining side. Let a denote the side. Then we have ax6sin30=12; .\ a = 4 ft. EXERCISES. XXXV. 1. The length of a rectangle is 6 25 ft., its breadth 1 74 ft. Find its area. 2. The length of a room, the sides of which are at right angles, is 31 \ ft. and the area 46 sq. yds. What is the breadth ? 3. The length and width of a rectangular enclosure are 386 and 300 ft. respectively. Find the length of the diagonal. 4. Show by a figure that the area of a rectangle 8 in. long and 2 in. broad is the same as that of 16 squares each of them measur- ing one inch in the side. 5. Show that any parallelogram in which two opposite sides are each 15 in. long, while the shortest distance between them is 3 in. has an area of 15 sq. in. Write down an expression for the area of a parallelogram whose base is a inches, and altitude 6 inches. 6. The foot of a ladder is at a distance of 36 ft. from a vertical wall, the top is 48 ft. from the ground. Find the length of the ladder. 7. The side of a square is 24 ft. 6 in. What is its area ? 8. The sides of a rectangle are as 4 : 3 and the difference between the longer sides and the diagonal is 2. Find the sides. 9. Two sides of a parallelogram are 4*5 ft. and 5*6 ft. respectively, the included angle is 60. Find the area. 10. How many persons can stand on a bridge measuring 90 ft. in length by 18 ft. in width, assuming each person to require a space of 27 in. by 18 in. ? 11. What will it cost to cover with gravel a court 31 ft. 6 in. long and 18 ft. 9 in. broad at the rate of Is. 4d. per square yard? 12. The length of a rectangular board is 12 ft. 6 in. , its area is 18'75 sq. ft. Find its width. 13. The diagonal of a square is 3362 ft. Find the length of a side of the square. 220 PRACTICAL MATHEMATICS FOR BEGINNERS. Area of a triangle. In Fig. 99 the rectangle ABCD&nd the parallelogram ABEF on the same base AB and of the same altitude, are equal in area. a >B Fig. 99. Area of a triangle. When A is joined to C and E it is easily seen that the triangle ACB is half the rectangle ABOB, and the triangle AEB is half the parallelogram ABEF. Hence, the two triangles are equal in area, and the area in each case is equal to half the product of the base and the altitude. . \ area of a triangle = \ (base x altitude ) = ^ab. From this rule the area of any triangle can be obtained by measuring the length of any side assumed as a base and the perpendicular on it from the opposite angular point. The area is one-half the product of the base and perpendicular ; or, as the perpendicular is the product of the sine of the angle opposite and an adjacent side, the following rule is obtained. Multiply half the product of two sides by the sine of the included angle. This result may be shown graphically by drawing a rectangle on the same, or an equal base, and half the height of the triangle ABC. Ex. 1. The base of a triangle is 3 '5 in., the height 6 '25. Find the area. Area = \ x 3 '5 x 6 25 = 10 94 sq. in. Ex. 2. The sides of an equilateral triangle are 10 ft. in length. Find the area. As the included angle is 60 the area is given by 50y/3 2 \ x 10 x 10 sin 60 43-29 sq. ft. AREA OF TRIANGLE. 221 When the three sides of a triangle are given : If a, 5, c be the three sides and s= 5 ; or s = half the sum of the three sides, then the area of the triangle is given by the formula area = ^/s (s a) (s b) (s c), or, find half the sum of the length of the sides, subtract from this half sum the length of each side separately; multiply the three remainders and the half sum together ; the square root of the product is the area of the triangle required. Ex. 3. We may use this rule to find the area of a right-angled triangle, sides 3, 4, and 5 units respectively. The area can be determined by the method used in Ex. 1. Here s = J(3 + 4 + 5) = 6. Subtract from this the length of each side separately, i. e. 6-3=3, 6-4 = 2, 6-5=1. .*. Area of triangle = s/Q x 3 x 2 x 1 =\/36 = 6 sq. units. Ex. 4. Find the area of a triangle, the lengths of the three sides being 3*27, 436, and 5'45 respectively. ,s = |(3 27 + 4-36 + 5 -45) = 6 -54. /. Area=^6-54(6'54-3'27)(6-54- 436) (6-54 -5'45) =\/6-54 x 3-27 x 218 x 1 -09=^50 8 169 = 7*129 sq. ft. Ex. 5. The sides of a triangular field are 500, 600, and 700 links respectively. Find its area. J (500 + 600 + 700) = 900, area=\/900 x 400 x 300 x 200= 146969 sq. links = 1 ac. 1 r. 35 "15 poles. The area of any rectilineal figure is the sum of the areas of all the parts into which the figure can be divided. Usually the most convenient method is to divide the figure into a number of triangles, then, as the area of each triangle can be found, the sum of the areas will give the area of the figure. 222 PRACTICAL MATHEMATICS FOR BEGINNERS. EXERCISES. XXXVI. 1. The base of a triangle is 4-9 ft. and the height 2 525 ft. Find its area ? 2. Find the area of a triangle in which the sides are 13, 14, and 15 ft. respectively. 3. Find the area of a triangle sides 21, 20, and 29 in. respectively. 4. Make an equilateral triangle on a base 3 in. long and construct a parallelogram equal to it in area. 5. On a base of 10 yards a right-angled triangle is formed with one side two yards longer than the other. Find its area. 6. The sides of a triangle are 101 '5, 80*5, and 59*4. Find the area. 7. The span of a roof is 40 feet, the rise 15 feet. Find the total area covered by slating if the length of the roof is 60 ft. 8. The sides of a triangular field are 300, 400, and 500 yds. If a belt 50 yds. wide is cut off the field, what are the sides of the interior triangle, and what is the area of the belt ? 9. Find the area of a triangle, the sides being 15, 36, and 39 ft. respectively. 10. The sides of a triangle are 1 75, 1'03, and I'll ft. respectively e Find the area. 11. Find the area of a triangle whose sides are 25, 20, and 15 chains respectively. 12. In a triangle ABC the angle C is 53 p , the sides AC and AB are 523 and *942 mile respectively. Find the side CB and the area of the triangle. 13. The sides of a triangle have lengths a, b, c inches. State (1) which of the following relations are true for all triangles, (2) which untrue for all triangles, (3) which are true for some triangles and untrue for others : (a, > b denotes a is greater than b ; ab, a = b, ac, a + b = c, a + b < c. a 2 + b 2 >c 2 , a 2 + b 2 = c\ a 2 + b 2 p, and the sum of all the terms each of which is represented by the letter a. As the number of sides in the polygon is increased, its area becomes nearer and nearer that of the circle, and when the number of sides is indefinitely increased, the perimeter (or sum of the sides) of the polygon becomes equal to the circumference AKEA OF CIRCLE. 225 of the circle = 27rr; the perpendicular referred to above also becomes the radius of the circle. Hence the area of a circle = \ (2irr xr) = irr 2 . If d is the diameter of a circle, then as d2r the formula 7tt 2 becomes d 2 . 4 By dividing a circle into a large number of sectors, the bases may be made to differ as little as possible from straight lines. Each of the sectors forming the lower half of the circumference could be placed along a horizontal line A B (Fig. 100). A corre- sponding number of sectors from the upper half of the circum- ference could be placed along the upper line CD, completing the parallelogram ABCD. The length of the base AB will then be half the length of the circumference of the circle and the height of the parallelogram is equal to the radius of the circle, r. :. Area of circle ABxr = rXTrr = irr 2 . If a thin circular disc of wood be divided into narrow sectors, and a strip of tape glued to the circumference, then when the tape is straightened the sectors will stand upon it as a series of triangles. By cutting the tape in halves the two portions may be fitted together as in Fig. 100. Area of sector of a circle. To find the area of the sector AE (Fig. 101) the angle being known. As the whole circle consists of 360 degrees, or 27r radians, the area of the sector will be the same fractional part of the whole area that the angle 6 is of 360, or of 2tt. Denoting the angle in degrees by N, then f <. ' N 2 6 dr 2 area of sector = irr 8 - ^-Trr 2 = . Fia 10 i._ A rea of sector of a circle. Thus the area of a sector is given by half the product of the angle and the radius squared. Ex. 1. Find the area of the sector of a circle containing an angle of 42, the radius of the circle being 15 feet ; area of circle = tr x 1 5 2 ; area of sector = -foir x 15 2 =82'47 square feet. P.M. B. P 226 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 2. The length of the diameter of a circle is 25 feet. Find the area of a sector in the circle, the length of the arc being 13 "09 feet. The area of the sector will be the same fraction of the whole area that 13 09 is of the circumference ; 1300 7T _ : 81*79 square feet. 25 x XyX25 2 : 4 In a similar manner the length of an arc subtending a given angle 6 can be obtained from the relation : length of an arc is the same fractional part of the whole circumference that 6 is of 360 or 27r, or if r is radius of circle . length of arc _ ~~ &rr ~3W Area of an annulus. If R (Fig. 102) denote the radius of the outer circle, and r the radius of the inner, the area of the annulus is the difference of the two areas ; = tt(R 2 - r 2 ) = ir(R + r) (R-r) ; .'. Multiply the sum and difference of the two radii by 3f to obtain the area of an annulus. Segment of a circle. Any chord, not a diameter, such as AB (Fig. 103), divides the circle into two parts, one greater than, and the other less than, a semi- circle. If C is the centre of the circle of which the given arc ABB forms a part, then the area of the segment ABB is equal to the difference between the area of the sector CADB and the triangle ABC. Length of arc ABB.~ To find the length of the arc ABB (Fig. 103), we may proceed to find the centre of the circle of which ABB is a part. Then by joining A and B to C, the angle subtended by the given arc is known and its length can be obtained. To find the area enclosed by the arc and the chord ABwe can find the area of the sector CABB and subtract the area of the Fig. 102. Area of an annulus. SEGMENT OF CIRCLE. 227 triangle ABC from it ; this gives the area required. To avoid the construction necessary in the preceding cases several Fig. 103. Segment of a circle. approximate rules have been devised. Of these the following give fairly good results : t ti. f a 8.AB-AB Length of arc ABB o or in words, subtract the chord of the arc from eight times the chord of half the arc and divide the remainder by 3. Area of segment. The area of the segment may be obtained from the rule, A 3 2 _ area== 2^ + 3 cA > where c denotes the chord AB and h denotes the height ED. EXERCISES. XXXVIII. 1. Find the diameter of a circle containing 3217 sq. in. 2. The diameter of a circle is 69*75 in. Find its area. 3. The circumference of a circle is 247 in. Find its area. 4. Find the diameter of a circle when the area in square inches is (i) -7, (ii) 0000126, (iii) '00031, (iv) -0314. 228 PRACTICAL MATHEMATICS FOR BEGINNERS. 5. Find the area of a circle when its diameter in inches is (i) 064, (ii) -109, (iii) 3\3. 6. A pond 25 feet diameter is surrounded by a path 5 feet wide. Find the cost of making the path at Is. \^d. per square yard. 7. The perimeter of a circle is the same as that of a triangle the sides of which are 13, 14, and 15 ft. Find the area of the circle. 8. If the two perpendicular sides of a right-angled triangle are 70 and 98 ft. respectively ; find the area of a circle described on the hypothenuse as a diameter. 9. Find the area of the annulus enclosed between two circles, the outer 9 in. and the inner 8 in. diameter. 10. The inner and outer diameters of an annulus are 9^ and 10 in. respectively. Find the area. 11. The area of a piston is 5944*7 square inches. What is the diameter of the air-pump which is one-half that of the piston ? 12. A sector contains 42, the radius of the circle is 15 ft. Find the area of the sector. 13. The length of the arc of a sector of a given circle is 16 ft. and the angle J of a right angle. Find the area of the sector ; find also the length of the arc subtending the same angle in a circle whose radius is four times that of the given circle. 14. The diameter of a circle is 5 ft. Find the area of a sector which contains 18. 15. Find the area of the sector of the end of a boiler supported by a gusset-stay, the radius of the boiler being 42 inches, length of arc 25 inches. 16. A sector of a circle contains 270. Find its area when the radius of the circle is 25 ft. 17. In an arc of a circle the chord of the arc is 30 ft. and the chord of half the arc 25 ft. Find the length of the arc. 18. The circular arch of a bridge is 50 feet long and the chord of half the arc is 26'9 ft. Find the length of the chord or " span." 19. The length of a circular arc is 136 ft. and the chord of half the arc is 75*5 ft. Find the length of the chord. 20. Find the area of a segment in which the chord is 30 ft. and height 5 ft. 21. Find the area of a segment of a circle when the chord is 120 ft. and height 25 ft. CHAPTER XX. AREA OF AN IRREGULAR FIGURE. SIMPSON'S RULE. PLANIMETER. Area of an irregular figure. When the periphery of an irregular figure ABODE (Fig. 104) consists of a series of straight lines, the area may be obtained by dividing the figure into a number of triangles, and the area of each triangle may be obtained separately. The sum of the areas of all the triangles into which the figure has been divided will give the area of the figure. When the ordinates of an irregular figure, in which one or more of the boundaries may consist of curved lines, are given, the area may be obtained by drawing the figure on squared paper and counting the squares en- y^ B closed by the periphery. Fig. 104. Area of an irregular figure. In this method there will usually be a number of complete squares enclosed by the periphery and a number cut by it. To estimate the value of any square cut by the outline it is convenient to neglect any square obviously less than one-half and to reckon as a whole square any one cut which is equal to, or greater than, one-half. 230 PRACTICAL MATHEMATICS FOR BEGINNERS. One defect of this method is that large errors are likely to occur when portions of the periphery are nearly parallel to the lines of ruling. To avoid the errors likely to be introduced in the preceding method, other methods depending upon calculation are preferable. Of these the two in general use are known as the Mid-ordinate Rule and Simpson's Rule. The latter is usually the more accurate of the two. Mid-ordinate rule. A common method of estimating the area of an irregular figure, Q Z*~ --51 lE ,-s2 N . -2Z \ ? \ * N ^ 7 \ _2 \ 7 y Z 3 1 \ t - \ L 3 _ Fig. 105. Area obtained by using squared paper. such as GFED (Fig. 106), in which one of the boundaries is a curved line, is to divide the base OF into a number of equal parts, and at the centre of each of the equal parts G m to erect ordinates. The length of each ordinate, mn, pq, rs, etc., from the base GF to the point where the vertical cuts the curve, is carefully measured, and all these ordinates are added together. The sum so obtained, divided by the number of ordinates, gives approximately the mean height, A, or mean ordinate, GJV. A convenient method of adding the ordinates is to mark them on a slip of paper, adding one to the end of the other until the total length is obtained. MID-ORDINATE RULE. 231 The degree of approximation depends upon the number of ordinates taken. The approximation more closely approaches the actual value the greater the number of ordinates used. The product of the mean ordinate and the base is the area required. For comparatively small diagrams, such as an indi- cator diagram (Fig. 107), ten strips are usually taken. This Fig. 107. Area of an indicator diagram. number is sufficiently large to give a fair average, and, moreover, dividing by 10 can be effected by merely shifting the decimal point. The length GF (Fig. 107) may correspond on a reduced scale to the travel of the piston in a cylinder, and the ordinates of the curve represent, to a known scale, the pressure per square inch of the steam in the cylinder at the various points of the stroke. Hence, the mean height ON indicates the mean pressure P of the steam, in pounds per sq. inch, throughout the stroke (the stroke being the term applied to the distance moved through by the piston in moving from its extreme position at one end of the cylinder to a corresponding position at the other end). If A denote the area of the piston in square inches, then the total force exerted by the steam on the piston is FxA, and the work done by this force in acting through a length of stroke L is P x A x L. If N denote the number of strokes per minute, the work done per minute by the steam = PALN. 232 PRACTICAL MATHEMATICS FOR BEGINNERS. But the unit of power used by engineers, and called a Horse- power, is 33000 ft. lbs. per minute. Hence Horse-power of the engine =- 33000 Ex. 1. In Fig. 107 the indicator card of an engine is shown ; the diameter of the piston is 23^ inches, length of stroke 3 ft., and revolutions 100 per minute. Find the mean pressure of the steam, also the horse -power of the engine. Adding together the ten ordinates shown by dotted lines, we have 66 6 + 73-0 + 72-4 + 64 -8 + 53 -6 + 44 -4 + 38 -0 + 34-8 + 31 4 + 23 = 502. As there are 10 ordinates, 502 .-. mean pressure = -y^- = 50*2 lbs. per sq. inch. Area of piston = 420 sq. inches ; number of strokes per minute = 200. 50-18x3x420x200 OOD .-. Horse-powers 33^- =383'2. Simpson's rule. By means of what is called Simpson's rule the area of an irregular figure GFED (Fig. 108) can usually be ascertained more accurately than by the mid-ordinate rule. Fig. 108. The base GF is divided into a number of equal parts. This ensures that the number of ordinates is an odd number, 3, 5, 7, 9, etc. In Fig. 108 the base GF is divided into 6 equal parts, and the number of ordinates is therefore 7. SIMPSON'S RULE. Denoting, as before, the lengths of the ordinates GD, pm, nrj etc., by h u A 2 , h 3 ...h 7 ; then, if s denotes the common distance or, space between the ordinates, we have Area of GFFD^{h 1 + h 7 -^4(h 2 +h i +h 6 )-h2(h 3 +h 5 )} = S -(A+4B + 2C), where A denotes the sum of the first and last ordinates. B even ordinates. C odd ordinates. .-. Add together the extreme ordinates, four times the sum of the even ordinates, and twice the sum of the odd ordinates (omitting the first and the last). Multiply the result by one- third the common interval between two consecutive ordinates. Pig. 109. The end ordinates at G and i^may both be zero, the curve commencing from the line GF (Fig. 109). In this case A is zero, and the formula for the area becomes ^(0 + 4 + 2C). Or, using the given values in Fig. 109, where the length of the ordinates are expressed in feet, we have Area=|{0 + 4(9-0 + 10-4 + 6-8) + 2(9'7 + 8-8)} o = 2 (104-8 + 37) = 283-6 sq. ft. Comparison of methods. It will be found a good exercise to compare the various methods of obtaining the area of a plane figure by using them to obtain the area of a figure such as a quadrant of a circle. Ex. 2. Draw a quadrant of a circle of 4 in. radius and divide the figure into eight strips each in. wide. Measure all the ordinates 234 PRACTICAL MATHEMATICS FOR BEGINNERS. (including the mid-ordinate) and find the area by (i) the mid- ordinate rule, (ii) by Simpson's Rule, (iii) the ordinary rule ^-- Compare the results and find the percentage errors. Find also the mean ordinate in each case. In Fig. 110 a quadrant of a circle is shown in which the two radii OF and GD are horizontal and vertical. Divide the base into eight equal parts ; then, if the figure is drawn on squared paper, the lengths of the ordinates and also the mid- ordinate can be read off and marked as in Fig. 110. As the distance between each ordinate is \ in., we have by Simpson's Rule Areai^(?i)=^{0 + 4 + 4(3-98 + 3-7 + 3-12 + l-92) + 2(3-88 + 3-46 + 2-64)} = 12-61 sq. in. 4 3 j 2 o o o CO 4 CO to 9\ 1 G Fio. 110, -Area of quadrant of a circle. Using the formula r-, area : 22 The mean ordinate is 12-57 314 in. 12*57 when ==%- , area= 12*575 sq. in. when tt = 3-1416. Accepting the latter as the more accurate value the difference is 12 61 -12-575= 035; , . -035x100 00 . /. Percentage error is ^ = *3 %. 126 Q.IK- --; =3'15 n.. . 4 4 D E In a similar manner the and percentage error by using the mid-ordinate rule can be obtained. When the ar.ea is not sym- metrical about a line, and its boundary is an irregular curve, lines are drawn touching the curve ; two of these, FG and ED (Fig. Ill) may be made parallel r to each other and GD, FE drawn perpendicular to the former. GF is divided into a number of equal \ Pro. 111. As before, the COMPARISON OF METHODS. 235 parts and the ordinates of the curve measured ; from these values, proceeding as before, the area can be obtained. As the area of an irregular figure is the product of the length of the base GF and the mean ordinate, it follows that when the area is obtained, the mean ordinate may be found by dividing the area by the length of the base. Thus in Fig. 107, p. 231, where GF is 6 inches, the division into 10 equal parts will give the common distance between each ordinate to be '6 inch. On p. 232 a rough result for the mean ordinate has been obtained. A more accurate result can be found by Simpson's rule, as follows : Extreme ordinates = 55'8 + 13'6 = 69*4, Even ordinates = 71-2 + 70 + 48*2 + 36-2 + 28*4 = 254, Odd ordinates = 72*8 + 58'4 + 40*8 + 33 = 205, /. Area of figure = ^(69*4 + 4 x 254 + 2 x 205) o = 299-08 sq. in. j- 299*08 AnOK . :. Mean ordinate = = = 49'85 in. o In the preceding examples the given ordinates are equidistant ; when this is not the case, points corresponding to the given ordinates can be plotted on squared paper and a fair curve drawn through the plotted points. The area enclosed by the curve the two end ordinates, and the base, is the area required. This value may be obtained by counting the enclosed squares ; or, better, by dividing the base into an even number of parts and reading off the values of the ordinates at each point of division. The area may then be obtained either by the application of Simpson' 's or Mid- ordinate rule. Ex. 4. The following table gives the values of the ordinates of a curve and their distances from one end. Find the mean ordinate and the area enclosed by the curve. Distances I in feet. j 2-3 4 5 7-0 12-2 18-0 24-0 30-0 Ordinates. j 7 6-3 5-89 5-48 4-67 3 96 3 39 2-9 236 PRACTICAL MATHEMATICS FOR BEGINNERS. Plotting the given values on squared paper as in Fig. 112, we obtain a series of points through which a fair curve is drawn. Next, dividing the base into six equal parts, seven ordinates are drawn. The values of these ordinates are shown in Fig. 112. By Simpson's Rule, as the common distance is 5 feet, Area=|J7 + 2-9 + 4(5'8 + 4-3 + 3-3) + 2(5-0 + 3-75)| = 135sq. ft. Mean ordinate x 30: 135; Mean ordinate: 135 30 = 4-5 ft. Other methods of finding the area of an irregular figure, instead of those which have now been studied, are by means of weighing, and by using a planimeter. By weighing. Draw the figure to some convenient scale, or, if possible, full size, on thick paper or cardboard of uniform thickness. Cut it out carefully. Also cut out a rectangular piece from the same sheet ; find the weight of the rectangular piece, and hence deduce the weight of a square inch. Then, knowing the weight of the irregular figure and the weight of unit area, the area of the figure can be calculated. Planimeter. The planimeter is an instrument for estimating the areas of irregular figures. There are many forms of the instrument to which various names Hatchet, Amsler, etc. are applied. Of these the more expensive and accurate forms are mostly modifications of the Amsler planimeter. Hatchet planimeter. A hatchet planimeter in its simplest form may consist of a f"l -shaped piece of metal wire (Fig. 113)> one end terminating in a round point, the other in a knife edge. This knife edge is rounded or hatchet-shaped, the distance between the centre of the edge K and the point T may be made 7 6 i s s S \ k 1 ' \, r^ 3 2 s 9 1 * ot I ' s 10 ij 20 a X Fig. 112. PLANIMETER. 237 5, 10, or some such convenient number. This length may be denoted by TK. To determine the area of a figure we proceed as follows : (a) Estimate approximately the centre of area, and through this point draw a straight line across the figure. (b) Set the instrument so that it is roughly at right angles to this line, with the point T at the centre of gravity. When in this position a mark is made on the paper by a knife edge K. ^ ill IIIIHm j:IIIHIIH 2S Fig. 113. Hatchet planimeter. Holding the instrument in a vertical position, the point T is made to pass from the centre to some point in the periphery of the figure, and then to trace once round the outline of the figure until the point is again reached, thence to the centre again. In this position a mark is again made with the edge K. The distance between the two marks is measured, the product of this length and the constant length TK gives approximately the area of the figure. To obtain the result more accurately, it is advisable when the point T (after tracing the outline of the figure) arrives at the centre to turn the figure on point T as a pivot through about 180, and trace the periphery as before, but in the opposite direction. This should, with care, bring the edge K either to the first mark or near to it. The nearness of these marks depends to some extent on the accuracy with which the centre of area has been estimated. The area of the figure is the product of TK, and the mean distance between the first and third marks. To prevent the knife edge K from slipping, a small weight W (Fig. 113) is usually threaded on to the arm BK ; the portion of the arm on which the weight is placed is flattened to receive it. 238 PRACTICAL MATHEMATICS FOR BEGINNERS. The arm BA is usually adjustable, and this enables the instru- ment to be used, not only for small, but also for comparatively large diagrams. Amsler planimeter. One form of the instrument is shown in Fig. 114 and consists of two arms A and C, pivoted together at a point B. The arm BA is fixed at some convenient point s. The other arm BG carries a tracing point T. This is passed round the outline of the figure, the area of which is required. Fig. 114. Amsler planimeter. The arm BC carries a wheel 2), the rim of which is usually divided into 100 equal parts. When the instrument is in use the rim of the wheel rests on the paper, and as the point T is carried round the outline of the figure, the wheel, by means of a spindle rotating on pivots at a and 5, gives motion to a small worm F, which in turn rotates the dial W. One rotation of the wheel corresponds to one-tenth of a revolution of the dial. A vernier, V, is fixed to the frame of the instrument, and a distance equal to 9 scale divisions on the rim of the wheel is divided into ten on this vernier. The read- ings on the dial are indicated by means of a small finger or pointer shown in Fig. 114. If the figures on the dial indicate units, those on the wheel will be xoths ; as each of these is subdivided into 10, the subdivisions indicate y^tus. Finally, the vernier, V, in which T ^j of the wheel is divided into 10 parts, enable a reading to be made to three places of decimals. To obtain the area of a figure, the fixed point s may be set at some convenient point outside the area to be measured, and the PLANIMETER. 239 point T at some point in the periphery of the figure. Note the reading of the dial and wheel. Carefully follow the outline of the figure until the tracing point T again reaches the starting- point, and again take the reading. The difference between the two readings multiplied by a constant will give the area of the figure, the value of the constant may be found by using the instrument to obtain a known area, such as a square, a rectangle, etc. EXERCISES. XXXIX. 1. Find the area of a quadrant of a circle of 5 inches radius by ordinary rule and by Simpson's Rule. Find the percentage 2. The transverse sections of a vessel are 15 feet apart and their areas in square feet up to the load water line are 4*8, 39 '4, 105*4, 159-1, 183*5, 173-3, 127*4, 57*2, and 6-0 respectively. Find the volume of water displaced by the ship between the two end sections given above. 3. The half ordinates of an irregular piece of steel plate of uniform thickness, and weighing 4 lbs. per sq. ft., are 0, 1*5, 2*5, 3, 5, 6 "75, 7-25, 9, 8 '75, 7, 6, 5-25, 3'5, 2, and ft. respectively, the common distance between the ordinates is 5 ft. Find the weight. 4. The ordinates of an irregular piece of land are 3 5, 4*75, 5*25, 7 "5, 8*25, 14-75, 6, 9'5, and 4 yards respectively, the common interval is 1 \ yds. Find the area in square yards. 5. The equidistant ordinates of an irregular piece of sheet lead weighing 6 lbs. per sq. ft. are respectively 2, 4, 9, 5, and 3 ft., the length of the base is 8 ft., find the weight. 6. The ordinates of a curve and the distances from one end are given in the following table. Find the area and the mean ordinate. Distances from one end (in inches). ! 20 35 56 72 95 110 140 156 Ordinates. 405 380 362 340 325 304 287 260 252 7. The girth or circumference of a tree at five equidistant places being 9'43, 7*92, 6*15, 4'74, and 3-16 ft. respectively, the length is \1\ ft. Find the volume, using the mid-ordinate and Simpson's Rule. 240 PRACTICAL MATHEMATICS FOR BEGINNERS. 8. Find the area of a curved figure when the distances and ordinates both in feet are as follows : X Distances from one end. 4 14 26 32 Ordinates. 20 16-5 12 8 7'5 9. Find the area of a half of a ship's water plane of which the curved form is defined by the following equidistant ordinates spaced 12 feet apart : 1, 51, 717, 875, 10-1, 9-17, 805, 64, -1 feet. 10. The half -ordinates of the load water-plane of a vessel are 13 ft. apart, and their lengths are -4, 33, 69, 10'5, 13*8, 163, 18-3, 19-5, 19-9, 20-0, 19*6, 190, 17 '8, 157, 11 8, 60, and -8 feet respectively. Calculate the area of the plane. 11. The load water-plane of a ship is 240 ft. long, and its half- ordinates, 20 ft. apart, are of the following lengths % 8'0, 12*4, 14-4, 156, 16*0, 18-0, 156, 142, 120, 9"2, 50 and '2 feet. What is the total area of the water plane ? 12. A river channel is 60 ft. wide, the depth (y) of the water at distances x ft. from one bank are given in the following table. Find the area of a cross-section and the average depth of the water. X 10 20 30 40 50 60 y 5 7 15 21 30 16 6 13. The work done by force is the product of the force and its displacement in the direction of the force, hence show that the work done by a variable force through a distance AB can be represented graphically by an area. If the distance AB be divided into two equal parts at O, and the magnitudes of the force at A , B, and G are P, Q, B respectively, show that the work is AB{P + 4:B + Q) + 6. Given P, Q, and B to be 50, 28, and 24 lbs. respectively, AB to be 12 ft., show that the work done is 372 ft. lbs. CHAPTER XXI. MENSURATION. VOLUME AND SURFACE OF A PRISM, CYLINDER, CONE, SPHERE, AND ANCHOR RING. AVERAGE CROSS SECTION AND VOLUME OF AN IRREGULAR SOLID. A solid figure or solid has the three dimensions of length, breadth, and thickness. When the surfaces bounding a solid are plane, they are called faces, and the edges of the solid are the lines of intersection of the planes forming its faces. What are called the regular solids are five in number, viz., the cube, tetrahedron, octahedron, dodecahedron, and icosahedron. The cube is a solid having six equal square faces. The tetrahedron has four equal faces, all equilateral triangles. The octahedron has eight faces, all equilateral triangles. The dodecahedron has twelve faces, all pentagons. The icosahedron has twenty faces, all equilateral triangles. Cylinder. If a rectangle ABCD (Fig. 115) be made to revolve about one side AB, as an axis, it will trace out a right cylinder. Thus, a door rotating on its hinges describes a por- tion of a cylinder. Or, a cylinder is traced by a straight line always moving parallel to itself round the boundary of a curve, called the guiding curve. Pyramid. If one end of the line AB always parses through a fixed point, and the other end be made to move round the boundary of a curve, a pyramid is traced out. P.M.B. Q Pig. 115 Cylinder. 242 PRACTICAL MATHEMATICS FOR BEGINNERS. Cone. If the curve be a circle and the fixed point is in the line passing through the centre of the circle, and at right angles to its plane, a right cone is obtained (Fig. 116) ; the fixed point is called the vertex of the cone ; an oblique cone results when the fixed point is not in a line at right angles to the plane of the base. Fig. 116. Cone. Fig. 117. Sphere. If a semicircle ACB (Fig. 117) revolve about diameter AB, the surface generated is a sphere. Fig. 118. Rectangular, Pentagonal, and Triangular Prisms. Prism. When the line remains parallel to itself and is made to pass round the boundary of any rectilinear polygon, the solid formed is called a prism. The ends of a prism and the base of a pyramid may be poly- gons of any number of sides, i.e. triangular, rectangular, penta- gonal, etc. MEASUREMENT OF VOLUME. 243 A prism is called rectangular, square, pentagonal, triangular, hexagonal, etc. (Fig. 118), according as the end or base is one or other of these polygons. A prism which has six faces all parallelograms is also called a parallelopiped. A right or rectangular prism has its side faces perpendicular to its ends. Other prisms are called oblique. b^ " ' c Fig. 119. - Volume of a right prism. In Fig. 119 a right prism, the ends of which are rectangles, is shown ; to find its volume, sometimes called the content, or solidity, it is necessary to find the area of one end DCGE, and multiply it by the length BC. Let v, I, b, and d denote the volume, length, breadth, and depth or altitude of the right prism respectively. Then area of one end = 6 x d. And volume of prism = bxdxl. As b x = area of base ; volume = area of base x altitude. As v = bdl it follows that if any three of the four terms be given the remaining one can be obtained. When the volume is obtained, the weight can be found by multiplying the volume by the weight of unit volume. Ex. 1. The length of a rectangular wrought iron slab is 8 ft., its depth is 3 ft. Find its breadth if its weight is 23040 lbs. (one cubic foot weighs 480 lbs. ). Here volume = 8 x 3 x b, Weight = 23040 = 8 x 3 x b x 480 ; 23040 8x3x480' :3 ft. 244 PRACTICAL MATHEMATICS FOR BEGINNERS. In Fig. 119, the length BO is divided into 8 equal parts, the breadth into 3, and the depth into 2. It is seen easily that there are 6 square units in the end DGGE of the slab, and these are faces of a row of six unit cubes. There are 8 such rows ; hence the volume is 8x6 = 48 cub. ft. Total surface of a right prism. The total surface is, from Fig. 119, seen to be twice the area of the face A BCD, and twice the area of A DBF, together with the area of the two ends ; .-. Surface = 2(ld+bl + bd) ; or, the total surface of a right prism is equal to the perimeter of base multiplied toy altitude together with areas of the two ends. Ex. 2. The internal dimensions of a box without the lid are : length 8 ft., breadth 3 ft., and depth 2 ft. Find the cost of lining it with zinc at 7d. per square foot. Area of base = 8 x 3 = 24 sq. ft. sides = 2(8x2) = 32sq. ft. ends = 2(3x2) = 12 ,, .\ Total area = 68 sq. ft. .;. Cost = ^y^ = l. 19s. 8d. EXERCISES. XL. 1. A cistern (without a lid) 6 feet long and 3 feet broad when two-thirds full of water is found to contain 187*5 gallons. Find the depth of the cistern, also the cost of lining it with zinc at 2\d. per square foot. 2. If the inside edge of a cubical tank is 4 ft. , find its volume ; also find the number of gallons it will hold when full. 3. The internal dimensions of a rectangular tank are 4 ft. 4 in., 2 ft. 8 in., and 1 ft. \\ in. Find its volume in cubic feet, the number of gallons it will hold when full, and the weight of the water. 4. A cistern measures 7 ft. in length, 3 ft. 4 in. in width. What is the depth of the water when the tank contains 900 gallons? 5. A tank is 4 metres long, *75 metres wide, and 1 metre deep. Find the weight of water it will hold. 6. A metal cistern is 12 ft. long, 8 ft. wide, and 4 ft. deep external measurements. If the average thickness of the metal is J in. , find the number of gallons of water it will hold. 7. Three edges of a rectangular prism are 3, 2*52, and 1*523 ft. respectively. Find its volume in cubic feet. Find also the cubic space inside a box of the same external dimensions made of wood one-tenth of a foot in thickness. I VOLUME AND SURFACE OF CYLINDER. 245 8. A Dantzic oak plank is 24 ft. long and 3f in. thick. It is 7 in. wide at one end and tapers gradually to 5f in. at the other. Find its volume and weight, the specific gravity being *93. 9. A Riga fir deck plank is 22 ft. long and 4 in. thick and tapers in width from 9 in. at one end to 6 in. at the other. If the specific gravity of the timber be "53, find the volume and weight of the plank. 10. Find what weight of lead will be required to cover a roof 48 ft. long, 32 ft. wide, with lead ^ in. thick, allowing 5 per cent, of weight for roll joints, etc. 11. A reservoir is 25 ft. 4 in. long, 6 ft. 4 in. wide. How many tons of water must be drawn off for the surface to fall 7 ft. 6. in. ? 12. If the surface of a cube be 491 '306 square inches, what is the length of its edge ? 13. A cistern is 9 ft. 4 in. long and 7 ft. 6 in. wide and contains 6 tons 5 cwt. of water. Find the depth of the water in the cistern. 14. Find the volume of a rectangular prism 3 ft. 4 in. long, 2 ft. wide, and 10 in. deep. Find also the increase in its volume when each side is increased by 8 in. 15. The internal dimensions of a rectangular tank are : length 2 metres, depth "75 metres, and width 1 metre. Find the weight of water it contains when full. Cylinder. It has been seen that the volume of a prism is equal to the area of the base multiplied by the length. In the case of a cylinder the base is a circle. If r denote the radius of the base and I the length of the cylinder (Fig. 120), Area of base = irr 2 ; .\ volume irr 2 x I. More accurately a cylinder of this kind in which the axis is perpendicular to the base should be called a right cylinder. This distinguishes it from an oblique cylinder in which the axis is not perpendicular, and from cylinders in which the base is not a circle. It is only necessary for practical purposes to consider a right cylinder. Surface of a cylinder. The surface of a cylinder consists of two parts, the curved surface and the two ends which are plane circles. If the cylinder were covered by a piece of thin paper this when unrolled would form a rectangle of length I and base 2irr. Thus, if the curved surface of a cylinder be conceived as unrolled and laid flat, it will form a rectangle of area %irrxl (Fig. 120). .*. Curved surface of cylinder = Zirrl. 246 PRACTICAL MATHEMATICS FOR BEGINNERS. To obtain the whole surface the areas of the two ends must be added to this . .*. Total surface of cylinder = 27rrl + 27rr 2 = 27rr(l + r). Fig. 120. Surface of a cylinder. In any problem in Mensuration it is advisable in all cases to express a rule to be employed as a formula. Thus, if V denote the volume and S the curved surface of a cylinder, then the preceding rules may be briefly written V=TrrH ; S^Zirrl. Ex. 1. Find the volume, weight, and surface of a cast-iron cylinder, 18 '5 inches diameter, 20 inches length. Area of base = tt x (9 *25) 2 = 268 *8 sq. in. Volume = 268 '8 x 20 =5376 cub. in. Weight = 5376 x -26 =1397 76 lbs. =2ttx ^5x20=1162-4 sq. in. Area of each end = 7r x ( * 8 ' 5) -=268-8 ; 4 .-. Total surface = 1699*99 or 1700 sq. in. Ex. 2. Find the effective heating surface of a boiler 6 ft. diameter, 18 ft. long, with 92 tubes 3| in. diameter, assuming the effective surface of the shell to be one-half the total surface. Effective heating surface of shell = ~ = 169 - 6 sq. ft. Heating surface of 92 tubes = 2 ^fn ~" ~" = 1517 ' 4 S< 1- ft - ; .-. Effective surface= 169'6 + 1517-4= 1687 sq. ft. CROSS-SECTION. 247 Cross-section. The term cross-section should be clearly understood. A section of a right cylinder by any plane perpen- dicular to the axis of the cylinder is a circle ; any oblique section gives an ellipse. Hence, the term area of cross-section is used to indicate the area of a section at right angles to the axis. Ex. 1. A piece of copper 4 inches long, 2 inches wide, and \ inch thick is drawn out into a wire of uniform thickness and 100 yards long. Find the diameter of the wire. Volume of copper = 4 x 2 x -^ =4 cubic inches. Length of wire = 100 x 3 x 12=3600 inches. Let d denote the diameter of the wire. Then Hence volume of wire = -(Z 2 x 3600. 4 ^2x3600 = 4 =4; . tf2_ 4x4 1 7TX3600" "225 xtt' .\ d= = -0376 inches Ex. 2. A piece of round steel wire 12 inches long weighs 65 lbs. If its specific gravity is 7*8, find the area of cross-section, also the diameter of the wire. Let a denote the area. Volume = 12 x a cubic inches. Also from Table I., weight of a cubic inch of water = *036 lbs. Weight = 12 x a x 7'8 x -036, but this is equal to *65 lbs. ; .-. 12a x 7*8 x -036= -65. *65 a = 12x7-8x-036 =' 193sq - in - .-. ^= -193. 4 Hence d=^ inch nearly. Hollow cylinder. The volume, V, is as before equal to area of base multi- plied by the altitude. If R and r denote the radii of the outer and inner circles re- spectively, D and d the corresponding diameters (Fig. 121), Fig. 121. Hollow cylinder. 248 PRACTICAL MATHEMATICS FOR BEGINNERS. Area of base = ttR 2 - irr 2 = tt(B? ~ r 2 ), And volume = tt(jR? - r-% ;; V='7854(B 2 -d 2 )l. To use logarithms, it is better to write this as 7854(D-d)(D + d)l. Also W= Vw, where W represents the weight of the cylinder and w denotes the weight of unit volume of the material. Ex. 1. The external diameter of a hollow steel shaft is 18 inches, its internal diameter 10 inches. Calculate the weight of the shaft if the length is 30 feet. Area of cross section = 7854 (18 2 - 10 2 ) = 7854(18 + 10) (18 -10) = -7854 x 28 x 8, volume = '7854 x 28 x 8 x 30 x 12 cubic inches, . , . -7854 x 28 x 8 x 360 x -29 A W61ght=_ 2240 t0nS = 8-2 tons. EXERCISES. XLI. 1. Let V denote the volume and 8 the curved surface of a cylinder of radius r and length I. (i) If V- 150 cub. in., =6 in., find r and S. (ii) If 7=100 cub. in., r=3 in., find I and S. (iii) Given r = 4 in., 110 in., find V and S. 2. The volume of a cylinder is 1608*5 cub. ft., the height is 8 ft. Find its diameter. 3. The curved surface of a cylinder is 402*124 sq. ft. If the height be 8 ft., what is the radius of the base? 4. 260 feet of round copper wire weighs 3 lb. ; find its diameter if a cubic inch of the copper weighs 0*32 lb. If the same weight of the copper is shaped like a hollow cylinder, 1 inch internal diameter and 2 inches long, what is its external diameter ? 5. A hollow cylinder is 4*32 inches long; its external and internal diameters are 3*150 and 1*724 inches. Find its volume and the sum of the areas of its two curved surfaces. 6. A portion of a cylindrical steel stern shaft casing is 12f ft. long, 1^ inches thick, and its external diameter is 14 inches. Find its weight. 7. What is the external curved surface and weight of a cast-iron pipe lj ft. internal diameter, 48 ft. long, and in. thick ? VOLUME AND SURFACE OF CONE. 249 8. The outer circumference of a cast-iron cylinder is 127 '2 in., and length 3 ft. 6 in. If the weight is 686 lbs., find its internal diameter. 9. If a cube of stone whose edge is 9 in. is immersed in a cylinder of 12 in. diameter half full of water, how far will it raise the surface of the water in the cylinder ? 10. Find the length of a coil of steel wire when the diameter is 025 inch and its weight 49 lbs. Cone. Volume of cone=^ {area of base x altitude) = i7rr 2 xA, where r = radius of base and h = altitude of cone. Or, the volume of a cone is one-tliird that of a cylinder on the same base and the same altitude. This result may be checked in a laboratory in ma-ny different ways. Thus, if a cone of brass and a cylinder of the same material, of equal heights, and with equal bases, be weighed, the weight of the cylinder will be found to be three times that of the cone. Or, the cone and cylinder may both be immersed in a graduated glass vessel, and the height to which the water rises measured. Or, if a cylindrical vessel of the same diameter and height as the cone is filled with water, it will be found, by inserting the cone point downwards, that one-third the water will be displaced by the cone, and will overflow. Curved surface of a cone. If the base of the cone be divided into a number of equal parts A B, BC, etc. (Fig. 122), then by joining A, B, C, etc., to the vertex V, the curved surface of the solid is divided into a number of triangles, VAB, VBC, etc. If a line be drawn perpendicular to BC, and passing through V; and its length be p, then Area of triangle VBC=\(BCxp). If n denote the number of triangles into w T hich the base is divided, and a the length BC, then Curved surface = - x ap approximately. 250 PRACTICAL MATHEMATICS FOR BEGINNERS. As the number of parts into which the base is divided is increased, the product na becomes more nearly equal to the cir- cumference of the base ; and becomes equal to the circumference when the number of parts is indefinitely increased, also p be- comes at the same time equal to I, the slant height. .*. Curved surface =\ 2irrl = irrl. Or, we may proceed as follows : Cut out a piece of thin paper to exactly cover the lateral surface Fig. ^.-Development of a cone. of & CQne When opened ^ .' will form a sector of a circle of radius I (Fig. 123). The length of arc CD circumference of base of cone = 27rr. But, as we have seen on p. 225, the area of a sector is equal to half the arc multiplied by the radius . :. Curved surface = J( CD x I) = J(27rr xl) = irrl, the curved surface of a cone equals half perimeter of base multi- plied by the slant height. Thus, if V denote the volume and S the curved surface of a cone, then Y=i 1T r 2 h ; S=Trrl. If h denote the height of the cone, then Ex. 1. Find the volume and curved surface of a right cone, diameter of base 67 in. , height 30 in. Area of base = 67 2 x ^=3525*66 sq. in. Volume of cone = ^(3525 '66 x 30) = 35256 '6 cub. in. Slant height = V33 '5 2 + 30 2 = 44 '98. .'. Surface = J > x 67 x 44*98) = 4733 -85 sq. in. EXERCISES. XLIL 1. From the two formulae V=^irr 2 h and S = ttH the volume and curved surface of a right cone can be obtained. (i) Given T = 200 cub. in., h = 8 in., find r. (ii) If F=200 r = 6 in., find h. (ill) If r = 6 in., ft = 8 in., find V and S. 2. The circumference of the base of a cone is 9 ft. Find the height when the volume of the cone is 22 '5 cub. ft. VOLUME AND SURFACE OF SPHERE. 251 3. Find the volume and weight of a cast-iron cone, diameter of base 4 in., height 12 in. 4. Find the volume and surface of a cone, radius of base 3 in., height 5 in. 5. If the weight of petroleum, specific gravity '87, which a conical vessel 8 inches in depth can hold is 3*22 lbs., what is the diameter of the base of the cone ? 6. If the volume of a cone 7 ft. high with a base whose radius is 3 ft. be 66 cubic feet, find that of a cone twice as high standing on a base whose radius is half as large as the former. 7. If the volume of a cone 7 ft. high with a base whose radius is 3 ft. be 66 cubic feet, find that of a cone half as high standing on a base whose radius is twice as large as the other one. 8. A right circular cone was measured. The method of measure- ment was such that we only know that the diameter of base is not less than 6 22 nor more than 6*24 inches, and the slant side is not less than 9*42 nor more than 9*44 inches. Find the slant area of the cone, taking (1) the lesser dimensions, (2) the greater dimensions. Express half the difference of the two answers as a percentage of the mean of the two. In calculating the area, if a man gives 10 significant figures in his answer, how many of these are unnecessary ? The sphere. A semicircle of radius r, if made to rotate about its diameter as an axis, will trace out a sphere. Any line such as AB or CD (Fig. 124) passing through the centre and terminated both ways by the surface is a diameter, and any line such as OA or OC passing from the centre to the circumference is a radius. By cutting an orange or a ball of soap it is easy to verify that any section of a sphere by a plane is a circle. The section by any plane which passes through the centre of the sphere is called a great circle. Surface and volume of a sphere. The following formulae for the surface and volume of a sphere of radius r should be carefully remembered. Surface of a sphere = 4?rr 2 (i) Volume of a sphere -firr 3 (ii) Fig. 124. Sphere. 252 PRACTICAL MATHEMATICS FOR BEGINNERS. The formula for the area of the curved surface may be easily remembered as follows : The area of a great circle CD (Fig. 124) is ttt 2 , where r is the radius of the sphere. The area of the curved surface or hemisphere DAC is twice that of the plane surface, and is therefore 27rr 2 . Hence the area of the surface of the sphere is 2 x 27r 2 = 47ir 2 . The area of the surface of a sphere is equal to that of the circum- scribing cylinder. Thus, in Fig. 125, the circumscribing cylinder or the cylinder which just encloses a sphere of radius r is shown. The curved ,, ^ surface of the cylinder will be the circum- ^ ^^^u '^ ference of the base 27rr multiplied by the height 2?' ; .'. Curved surface of cylinder = 27rrx2r = 4irr' 2 . The volume of the sphere is two-thirds that of the circumscribing cylinder. '* " - - y^^f-' * '"' Thus, area of base of cylinder = 7rr 2 , Fig. 125. Sphere and its and height of cylinder = 2r ; circumscribing cylinder. . volume of cylinder = 2^ ; two-thirds of 27rr 3 is Jttt 3 , and this is equal to the volume of the sphere. The formulae for the surface and volume of a sphere assume a much more convenient form when expressed in terms of the diameter of the sphere. d Let d denote the diameter, then r . A Surface of a sphere = 4ir dV 2 Volume of a sphere = -tt ( - j = ^ d 3 = -5236d 3 (iii) From Eq. (iii) (as *5 is one-half), the approximate method of quickly obtaining the volume of a sphere is seen to be, for the volume of a sphere, take half the volume of the cube on the diameter and add 5 per cent, to it. HOLLOW SPHERE. 253 Ex. 1. Find the surface, volume, and weight of a cast-iron ball ; radius 6 "25 in. Surface = ttx 12 5 2 sq, in. 2 log 12-5 = 2-1938 log 7T = -4972 antilog 6910= 4909 2-6910 /. Surface = 490-9 sq. in. Volume = 5236c? 3 cub. in. 3 log 12-5 =3-2907 log -5236 = 1-7190 antilog -0097 = 1023. 3-0097 .-. Volume =1023 cub. in. Weight of ball = (volume) x (weight of unit volume) = 1023 x -26 lbs. =266 lbs. Hollow sphere. If the external and internal diameters of a hollow sphere be denoted by r 2 and r x respectively, then the volume of the material forming the sphere would be i^ 3 -^ 3 , or -7r(r 2 3 - r*). This may be replaced by its equivalent 5236 (d*-d*). Ex. 1. Find the weight of a cast-iron ball, external diameter 9 inches, internal diameter 4 inches. Volume= -5236(9 3 -4 3 ) = -5236(729 - 64)= -5236 x 665. Weight of ball= -5236 x 665 x -26 = 90*53 lbs. EXERCISES. XLUI. 1. The external diameter of a cast-iron shell is 12 in. and its weight 150 lbs. Find the internal diameter ; also find the external surface of the sphere. 2 What is the weight of a hollow cast-iron sphere, internal diameter 18 in. and thickness 2 in. ? 3. Find the weight of a cast-iron sphere 8 in. diameter, coated with a uniform layer of lead 7 in. thick. 4. Determine (i) the radius of a sphere whose volume is 1 cub. ft., (ii) of a sphere whose surface is 1 sq. ft. 5. A sphere, whose diameter is 1 ft., is cut out of a cubic foot of lead, and the remainder is melted down into the form of another sphere. Find the diameter. 6. A leaden sphere one inch diameter is beaten out into a circular sheet of uniform thickness of ^q inch. Find the radius of the sheet. 7. Find the weight of a hollow cast-iron sphere, internal diameter 2 in. , thickness \ of an inch. 254 PRACTICAL MATHEMATICS FOR BEGINNERS. 8. The diameter of a cylindrical boiler is 4 ft., the ends are hemispherical, and the total length of the boiler is 8 ft. Find the weight of water which will fill the boiler. 9. The volume of a spherical balloon is 17974 cub. ft. Find its radius. 10. A solid metal sphere, 6 in. diameter, is formed into a tube 10 in. external diameter and 4 in. long. Find the thickness of the tube. 11. Two models of terrestrial globes are 2 35 ft. and 3*35 ft. in diameter respectively. If the area of a country is 20 sq. in. on the smaller globe, what will it be on the larger ? Solid ring. If a circle, with centre G, rotate about an axis such as AB (Fig. 126), the solid described is called a solid circular ring, or simply a solid ring. By bending a length of round solid indiarubber, a ring such as that shown in Fig. 127 may be ob- tained. The length of such a piece of rubber is the distance DC from the axis multiplied by 2tt. Examples of solid rings are found in curtain rings, in anchor rings, etc. Any cross-section of such a ring will be a circle. The ring may be considered as a cylinder, bent round in a circular arc until the ends meet. The mean length of the cylinder will be equal to 27rCD y or the circumference of a circle which passes through the centres of area of all the cross-sections. Area of a ring. The curved surface of a ring is equal to the circumference or perimeter of a cross-section multiplied by the mean length of the ring. If r denote the radius of the cylinder from which the ring may be imagined to be formed, R the mean radius of the ring, and A the area of the ring, then Fig. 127. Solid ring. VOLUME AND SURFACE OF RING. 255 Perimeter or circumference of cross-section = 2irr. Mean length = 2irR. Area of ring='2irr x2ttR (i) :\ A = 4ir 2 Rr (ii) Eq. (i) will probably be easier to remember than Eq. (ii). Volume of a ring. The volume of a ring is the area of a cross-section multiplied by the mean length. Area of cross-section = 7rr 2 . Mean length = 2ttR. Volume = irr 2 x 2irR. ;. V=2ir 2 Rr 2 (iii) In a similar manner the volume may be obtained when the cross-section is a rectangle (Fig. 128), by considering the ring to form a short hollow cylinder. Dividing (iii) by (ii) we get A 4ir 2 Rr '~2' from which when V and A are known r can be found, and by substitution in (ii) or (iii) the value of R can be obtained. Ex. 1. The cross-section of a solid wrought-iron ring, such as an anchor ring, is a circle of 5 inches radius, the inner radius of the ring is 3 ft. Find (a) the area of the curved surface, (b) the volume of the ring, (c) its weight. {a) Herer=5; i?=36 + 5 = 41. Area of curved surface = 4ir 2 x 41 x 5 sq. in. tt 2 x 20x41 ,. _ ' = yta sc l- ft. =56 '2 sq. ft. Volume. Area of cross-section = -k x 5 2 . Mean length = 2tt x 41. . Volume^ cub. ft. = 11 71 cub. ft. Ex. 2. The cross-section of the rim of a cast-iron fly wheel is a square of 5 inches side. If the inner diameter of the ring is 5 ft. , find (a) the area, (6) the volume, (c) the weight of the rim. As the inner diameter is 60 inches, the outer diameter will be 70. .*. Mean diameter =1(60 x 70) = 65 inches. The rim may be considered as a square prism, side of base 5 inches, length 7T x 65. 256 PRACTICAL MATHEMATICS FOR BEGINNERS. (a) Perimeter of square = 4 x 5 = 20 inches. .'. Total surface = 20 x -w x 65 sq. in. = 1300tt sq. in. (6) Volume = (area of base) x (length) = 5 2 xwx 65 = 16257T cub. in. (c) Weight = 1625tt x -26 lb. EXERCISES. XLIV. 1. The inner diameter of a wrought-iron anchor ring is 12 inches, the cross-section is a circle 4 inches diameter. Find the surface, volume, and weight of the ring. 2. The cross- section of the rim of a cast-iron fly wheel is a rect- angle 8 in. by 10 in. If the mean diameter is 10 ft. , find the weight of the rim. 3. The volume of a solid ring is 741*125 cub. in. and inner dia- meter 21 in. Find the diameter of the cross-section. 4. The outer diameter of a solid ring is 12 6 in. if the volume is 54*2 cub. in. Find the inner diameter of the ring. 5. Find the volume of a cylindrical ring whose thickness is 27 in. and inner diameter 96 in. 6. The section of the rim of a fly wheel is a rectangle 6 in. wide and 4 in. deep, the inner radius of the rim is 3 ft. 6 in. Find the volume and weight of the rim, the material being cast iron. 7. In a cast-iron wheel the inner diameter of the rim is 2 ft. and the cross-section of the rim is a circle of 6 in. radius. Find the weight of the rim. 8. Let V denote the volume and A the area of a ring, (i) If i?=6, r=l, find Fand A. (ii) If A =200 sq. in. and V= 100 cub. in., find the dimensions, (iii) If F=200 cub. in., i?=12 in., find r. 9. A circular anchor ring has a volume 930 cub. in. and an area 620 sq. in. Find its dimensions. 10. The cross-section of the rim of the fly wheel of a small gas engine is a rectangle 2 '33 in. by 2 5 in. If the mean diameter is 38 '4 in., find the volume of the rim in cubic inches and its weight, the material being cast iron. Similar solids. Solids which have the same form or shape, but the dimensions not necessarily the same, are called similar solids. All spheres and all cubes are similar solids. As a simple case we may consider two right prisms ; in one the length, breadth, and depth are 8, 3, and 4 respectively ; and in the other, 16, 6, and 8, i.e every linear dimension of the first is doubled in the second. These are similar solids. Further, if a SIMILAR SOLIDS. 257 drawing of the first is made to any scale it would answer for the second prism by simply using a scale twice the former. In other words two solids are similar when of the same shape or form but made to different scales. It will be seen that the area of any face of the second solid (as each linear dimension is doubled) is four times that of the first, and the volume of the second is 8 times that of the first. If a denote the area of the first and s the scale, then area of second is s 2 x a and volume s 3 x a. Ex. 1. The lengths of the edges of two cubes are 2 in. and 4 in. respectively. Compare the surfaces and volumes of the two solids. If the first cube weighs 2 lbs., what is the weight of the second ? The area of each face of a cube of 2 in. edge, is 2 2 . As there are 6 similar faces the surface is 6 x 2 2 =24 sq. in. In a similar manner the surface of the second cube is 6x 4 2 =96 sq. in. Thus the surface of the second is 4 times that of the first, The volume of the first cube is 2 3 =8. The volume of the second cube is 4 3 =64. Hence the volume of the second is 8 times that of the first As the weight of the first is 2 lbs., the weight of the second is 8x2=16 lbs. The definition that two solids are similar when a drawing of one to any convenient scale may by a mere alteration of the scale repre- sent the other, will be found to be a serviceable practical definition of similarity. And such a definition can be easily applied to cones, cylinders, and pyramids. Ex. 2. An engine and a small model are both made to the same drawings, but to different scales. If each linear dimension of the engine is 8 times that of the model, find its weight if the weight of the model is 100 lbs. If 1 lb. of paint is required to cover the surface of the model, what amount will probably be required for the engine ? Here volume of engine is 8 3 times that of model ; .-. weight = 512 x 100 = 51200 lbs. Area of surface is 8 2 times that of model ; .*. amount of paint required = 64 x 1 = 64 lbs. Irregular solids. When, as is often the case, the given cross- sections are not equidistant, as in Fig. 129, squared paper may be P.M. B. B 258 PRACTICAL MATHEMATICS FOR BEGINNERS. used with advantage. The given distances are set off along a horizontal axis, and the areas are plotted as ordinates. A series of plotted points are thus obtained. When a curve is drawn through the plotted points the dis- tance between the two end ordinates is divided into an even number of parts, and from the known values of the equidistant ordinates so obtained the area of the curve may be determined by any of the previous rules. Fig. 129. Ex. 2. The trunk of a tree (Fig. 129) 32 ft. long has a straight axis and has the following cross-sectional areas at the given dis- tances from one end. Find its volume. Distances (in feet) from one end. 4 14 26 32 Areas of cross- section. 20 16-5 12 8 7"5 Plotting the given values on squared paper a series of points are obtained, and through these points a curve is drawn as in CO IS 10 in 5 OO A> an( ^ l\ respectively. Join to P x , P 2 , and P 3 , then three angles will be found to be 51 3, 62'l, and 51'3. Ex. 2. Find the distance between the two points (3, 4, 5'3) (1, 2*5, 3) and the angles the line joining them makes with the axes. The solution of this problem can be made to depend on the pre- ceding example by taking as origin the point (1, 2'5, 3), then the co-ordinates of the remaining point will be (3-1), (4-2*5) and (5'3-3), or (2, 1*5, 2*3). Hence the true length, the projections and the angles made with the axes may be obtained as in Ex. 1. The manner in which the three axes are lettered should be noticed. It would appear at first sight to be more convenient to letter as the axis of x the line going from the origin to the right instead of y as in the diagram ; but when it becomes necessary to apply mathematics to mechanical or physical prob- lems the notation adopted in Fig. 136 is necessary, and therefore it is advisable to use it from the commencement. Calculation. In Fig. 136 let 6 denote the angle made by OP DIRECTION-COSINES. 267 with the axis of z, and (f> the angle which the projection Op makes with the axis of x ; then we have : x= OB = Op cos ; :. y=OPsin #sin, (ii) and z = OP cos (iii) Ex. 3. Let 0P= 100, = 25, = 70. Then x = 100 sin 25 cos 70 = 100 x -4226 x -3420-14-45 ; y= 100 sin 25 sin 70 = 100 x -4226 x -9397 = 39*71 ; z = 100 cos 25 = 100 x -9063 = 90-63. Ex. 4. Given the co-ordinates of a point x = 3, y = 4, z=5. Find OP or r, 0, and 0. OP 2 = r 2 = Op 2 + pP 2 also Op 2 = OB 2 + Bp 2 ; .-. r 2 =OB 2 + Bp 2 +pP 2 = x 2 + y 2 + z 2 = 9 + 16 + 25 = 50; .-. r=\/50 = 7'071. From (iii) z=rcos = 7*071 cos 0; .'. 008 = ^1^= -7071; /. = 45. From (i) 3 = r sin cos 0, CO8 = ^L= 7-071x-7071 ; ' = 53 6 ' Direction-cosines. "We have found that when the x, y, z co-ordinates of a point are given, its distance from the origin may be denoted by r where t- 2 ~x 2 -\-y 2 -\-z 2 . Hence we can i direction-cosines of the line. proceed to find the ratios -, *-, and -. These are called the r r r r Thus, if OP (Fig. 136) makes angles a, /3, and 6 with the axes of x, y, and z respectively, then x x C08o "5P = r 268 PRACTICAL MATHEMATICS FOR BEGINNERS. Similarly cos/3 = - and cos0 = -. Squaring each ratio and adding, we get cos 2 a + cos 2 ^ + cos 2 (9=^+f-2 + ^=^ = 1 - The letter I is usually used instead of cos a, and similarly m and n replace cos /3 and cos 6 respectively. Thus we get _x _ y _z I m n' where , ra, and n denote the direction-cosines of the line. From the relation cos 2 a + cos 2 /3 + cos 2 = l or its equivalent l 2 + 7n 2 + n 2 = l it is obvious that, if two of the angles which a given line OP makes with the axes are known, the remaining angle can be found. Also, as indicated in Ex. 1, the angles a, /3, and 6 can be obtained by construction, but more accurately by calculation. We may repeat Ex. 1 thus : Ex. 5. The co-ordinates of a point P are 2, 1*5, 2. Find the distance of the point from the origin O, and the angles made by the line OP with three axes. True distance, OP=\/2 2 +l'5 2 + 2 2 =3'2. Denoting the distance OP by r to find the angles a, /3, and 0, we have x=OB=r cosa ; /. cosa = - = .^='6250; r 3'2 .'. a = = 51 19'. y= = OG = OP cos , cos ^3 = = i|=-4688; .'. p = = 62 3'. z- = OD = OP cos 0, cosO- = 3-1=6250; :. Q-- = 51 19'. Ex. 6. A line OP makes an angle 60 with one axis, 45 with another. What angle does it make with the third ? LATITUDE AND LONGITUDE. Let 7 denote the required angle, then as cos60 = and cos 45 = -=, we have from the relation cosW + cos 2 45 + cos 2 = 1, J + l + cos 2 = l, or cos^^l -|=J; .'. cos 0= and = 60. A practical application. Some of the data we have assumed may perhaps be better expressed by the terms latitude and longitude of a place on the earth's surface. Thus, at regular S.Pola Pig. 138. Pig. 139. distances from the two poles a series of parallel circles are drawn (Fig. 138) and are called Parallels of Latitude. The parallel of latitude midway between the poles is called the Equator. These parallels are crossed by circles passing through the poles and called meridians of longitude. Selecting one meridian as a standard (the meridian passing through Green- wich), the position of any object on the earth's surface can be accurately determined. This information, together with the depth below the surface or the height above it, determines any point or place. The plane xoy may be taken to represent the equatorial plane 270 PRACTICAL MATHEMATICS FOR BEGINNERS. of the earth, and OZ the earth's axis. Then the position of a point P(Fig. 139) on the surface of the earth, or that of a point outside the surface moving with the earth, is known when we are given its distance OP (or r) from the centre, its latitude 6, or co-lati- tude (90 - 6), and its or east longitude, from some standard meridian plane, such as the plane passing through Greenwich. Assuming the earth to be a sphere of radius r, then the distance of a point on the surface can be obtained. If P be a point on the surface, the distance of P from the axis is the distance PM, but PM=r sin POM =r cos 6. Ex. 7. A point on the earth's surface is in latitude 40. Find its distance from the axis, assuming the earth to be a sphere of 4000 miles radius. Required distance = 4000 x cos 40 = 4000x -766 = 3064 miles. Having found the distance PM, the speed at which such a point is moving due to the rotation of the earth can be found. Ex. 8. Assuming the earth to be a sphere of 4000 miles radius, what is the linear velocity of a place in 40 north latitude ? The earth makes one revolution in 29*93 hours. Radius of circle of latitude = 4000 x cos 40 ; A _ 4000 x cos 40 x 2tt _ 4000 x -766 x 2ir ' Speed ~ 29^93 ~ 29^93 = 642*77 miles per hour. Line passing through two given points. If the co-ordinates of two given points P and Q be denoted by (x, y, z) and (a, b, c) the equation of the line passing through the two points is x a _y b _z c I m n Through P draw three lines Pp, Pp', Pp", parallel to the three axes respectively, and draw the remaining sides of the rectangular block as in Fig. 140. Complete a rectangular block having its sides parallel to the former one and q for an angular point. PL = Nq=NR-qR=Pp'-Lp'=x-a. PF=Mq = Md-dq=y-b. PS =Eq=Eq' -qq' = z -c. The line Pq is the diagonal of a rectangular block, the edges LINE THROUGH TWO GIVEN POINTS. 271 of which are x-a, y-b, z-c, and therefore to find the length of Pq we have p q = ^ - af + (y - bf + (z- cf. The angle between the line Pq and the axis of Z is the angle between Pq and qE a line parallel to the axis of Z. Hence denoting the angle by 6, Pq sl{x-af + {y-bf + {z-cf Fig. 140. Line passing through two points. Similarly, I x a _y b Pq ' ~ Pq ' It will be obvious that when the second point is the origin, a, 6, and c are each zero, and the equation x a _y b _z c I m n becomes x _y _z Ex. 9. If x=3, y = 4, z = 5, find r, I, m, and n. We have r 2 = x 2 + y 2 + z~ = 9 + 16 + 25 = 50; .-. r = \/50 = 7'07l, 1-2- 3 m r 7-071 y_ 4 r 7*071 2 5 / 7-071 4242, 5657, 7071. 72 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 10. Find the distance between the two points (3, 4, 5 3) (1, 25, 3) and the angles made by the line with the three axes. Distance=V(3-l) 2 + (4-2-5) 2 + (5-3-3) 2 =\/2 2 +l-5 2 + 2-3 2 =3-4. 3-1 = cosa: 3 4 = -5882 . 4-25 m = cos p 0.4. = 4413 '> ro=cosy= 0.4 '6764; a = 53 58'. /3=63 48'. = 47 24'. When the given point or points are in the plane of x, y, a resulting simplification occurs. Thus, denoting the co-ordinates of two points P and Q by (x, y) and (a, b respectively), and the angles made by the line PQ with the axes of x and y by a and fi. Then, if r be the distance between the points, r=J(x-ay+(y-by. Also f -,*-"*; cos a cos p , cos/?, . /. y-b= -(x-a); J cos a ' but /3 is the complement of a ; .*. cos/3 = sina. Hence we get y - 5 = tan a(# - a), and the equation of the line joining the two points may be written y-b = m'{x - a), where m! is the tangent of the angle made by the line with che axis of x. Thus, given x=3, y = 4, the point P (Fig. 141) is obtained by marking the points of intersection of the lines # = 3, y = 4. In a similar manner the point Q (1, 1'34) is obtained. Join P to Q, then PQ is the line through the points (3, 4), (1, 1-34), and PQ = ^(3 - l) 2 + (4 - 1 -34)2 = 3*33. Fig. 141. POLAR CO-ORDINATES. 273 and the equation of the line is y- 1-34 = |(^-1); ' y = ^ x or y = l*3&e. Polar co-ordinates. If from the point P a line be drawn to the origin, then if the length of OP be denoted by r, and the angle made by OP with the axis of x be 6, when r and are known, the position of the point can be determined, and the rectangular co-ordinates can be found. Conversely, given the x and y of a point, r and can be obtained. Ex. 11. Let r=20, = 35 ; find the co-ordinates x and y. Here x = r cos 35 = 20 x '8192 = 16 -384 ; y = r sin 35 = 20 x -5736 = 1 1 '472. Ex. 12. Given the co-ordinates of a point P (4, 3) ; find r and 0. tan 0=f = *75, = 36 54'. EXERCISES. XLVI. 1. A point P is situated in a room at a height of 3 ft. above the floor, 4 ft. from a side wall, and 5 ft. from an end wall. Determine the distance of P from the corner where the two walls and the floor meet. Scale, |"=1\ 2. Determine the length of a line which joins two opposite corners of a brick 9" x 4 J" x 3". Scale, |. 3. The floor A BCD of a room is rectangular. AB and CD are each 18 feet long, and AD, BC each 24 feet. A small object P in the room is 6 ft. above the floor, 10 ft. from the vertical wall through AB, and 8 ft. from the wall through BC. Find and measure the distances of P from A, B, C, and D, the four corners of the floor. Scale, 0-1"= 1'. 4. A small object P is situated in a room at a distance of 17" from a side wall, 24" from an end wall, and 33" above the floor. Find the distance of P from the corner of the room where these three mutually perpendicular planes meet. If a string were stretched from to P, find and measure the angles which OP would make with the floor, the end wall, and the side wall respectively. Scale, j^y. 5. The co-ordinates of two points A and B are (2, IV, \") and (J", 4", 2"). Determine the length of A B and the angles which AB makes with the planes XT, YZ. P.M. B. S 274 PRACTICAL MATHEMATICS FOR BEGINNERS. 6. The co-ordinates of a point are 1^", 2", 1". Draw and measure the three projections of the line OP on the planes of xy, yz, and zx. Find the true length of OP. 7. The three rectangular co-ordinates of a point P are x =1'5, y = 2'3, 3=1-8. Find (1) the length of the line joining P to the origin, (2) the cosines of the angles which OP makes with the three rectangular axes. 8. The polar co-ordinates of a point are r=20", = 32, = 70. Find the rectangular co-ordinates. 9. There are three lines OX, OY, and OZ mutually at right angles. The following lengths are set off along these lines : OA, of length 2 inches, along OX. OB, 34 or. 00, 295 OZ. A plane passes through A, B, and O. Determine and measure the angle between this plane and the plane which contains the lines OX and O Y. Also determine and measure the angle between the plane and the line OZ. 10. Describe any system which you know of that enables us to define exactly the position of a point in space. The three rectangular co-ordinates of a point P are 3, 4, and 5 ; determine (i) the length of the line joining P to 0, the origin of co-ordinates ; (ii) the cosines of the angles which OP makes with the three rectangular axes. 11. The polar co-ordinates of a point are r=S, = 65, = 50. Determine its rectangular co-ordinates. 12. The earth being supposed spherical and of 4000 miles radius, what is the linear velocity in miles per hour of a point in 36 North latitude? The earth makes one revolution in 23*93 hours. 13. A point is in latitude 52. If the earth be assumed to be a sphere of 3960 miles radius, how far is the point from the axis? Find the length of the circumference of a circle passing through the point called a parallel of latitude. What is the 360th part of this length, and what is it called ? 14. If the earth were a sphere of 3960 miles radius, what is the 360th part of a circle called a meridian ? What is it called ? 15. The polar co-ordinates of a point A are 3", 40, and 50 respectively. Find the rectangular co-ordinates. 16. The three rectangular co-ordinates of a point P are 2 5, 3 1, and 4. Find (1) the length of line joining P with the origin, (2) the cosines of the angles which OP makes with the three axes, and (3) the sum of the squares of the three cosines. CHAPTER XXIII. ANGULAR VELOCITY. SCALAR AND VECTOR QUANTITIES. Angular velocity. When a point moves in any manner in a plane, the straight line joining it to any fixed point continually changes its direction ; the rate at which such a straight line is rotating is called the angular velocity of the moving point about the fixed point. Angular velocity is uniform when the straight line connecting the moving and fixed points turns through equal angles in equal times, but variable when unequal angles are described in equal times. Measurement of angular velocity. The angular velocity of a rotating body is the angle through which it turns per second, expressed in radians. One of the most important cases of angular motion is when P (Fig. 142) is a point in a rigid body rotating about a fixed axis 0. All points of the body move in circles having their planes per- pendicular to and their centres in the axis. Hence, at any instant the angular velocity for all points of the body is the same. If a point P is describing the circle A PC, of radius r, with a uniform velocity of v feet per second, then, denoting the angular velocity by o>, the length of Pro. 142. 276 PRACTICAL MATHEMATICS FOR BEGINNERS. arc described in one second v, is the product of the angular velocity and the radius. - v :. v=ra), or 00=-. r In one revolution the moving point P describes a distance equal to the circumference of the circle. Hence, if t denote the time (in seconds) of a complete revolution rt ~ t' In one revolution angle turned through is 27r, and in n revolutions 2irn. From v = ior, v 2 ==(t) 2 r 2 = 2 2 7r 2 n 2 r 2 . If the number of revolutions per minute is given, it is necessary to divide by 60. Ex. 1. A wheel makes 100 turns a minute, what is its angular velocity ? Find the linear speed of a point on the wheel 7 feet from the axis. In one revolution the angle turned through is 2ir radians ; . . 100x2tt ._ . ., .-. angular velocity = ^r =10*47 radians per sec. Linear velocity = angular velocity x radius ; or, 10-47 x 7 = 73-29 feet per sec EXERCISES. XLVII. 1. A wheel diameter 5 ft. turns 40 times a minute. Find its angular velocity and the linear velocity of a point on the circum- ference. 2. Explain what is meant by angular velocity of a rotating body ; knowing the angular velocity, how would you proceed to obtain the linear velocity ? P is a point of a body turning uniformly round a fixed axis, and PN is a line drawn from P at right angles to the axis. If PN describes an angle of 375 in 3 sec. , what is the angular velocity of the body ? If PN is 6 ft. long, what is the linear velocity of P ? 3. What is the numerical value of the angular velocity of a body which turns uniformly round a fixed axis 25 times per minute ? 4. The radius of a wheel is 14 feet, and it makes 42 revolutions a minute. Find its angular velocity and the linear velocity of the extremity of the radius. SCALAR QUANTITIES. 277 5. A wheel is 5 feet diameter, and a point on its circumference has a speed of 10 feet per second. Express in radians the angle turned through in second. How many revolutions will the wheel make per minute ? 6. Define angular velocity. A wheel makes 90 turns per minute. What is its angular velocity in radians per second ? If a point on the wheel is 6 feet from the axis, what is its linear speed ? 7. The diameter of a wheel is 3J feet, what is its angular velocity when it makes 120 revolutions per minute? What is the linear speed of a point in the rim of the wheel ? Scalar quantities. Those quantities which are known when their magnitudes (which are simply numbers) are given, such as masses, areas, volumes, etc., are called scalar quantities. Vector quantities. Quantities which require for their complete specification the enumeration of both magnitude and direction are called vector quantities, or shortly, vectors. Thus, forces, velocities, accelerations, displacements, etc., are vectors, and may in each case be represented by a straight line. To completely specify a vector we require to know (1) Its magnitude. (2) The direction in which it acts, or its line of action. (3) Its point of application. The term direction applied to vector quantities is not sufficiently explicit. For example, in the specification of a vector the direction may be given as vertical, but a vertical direction may be either upward or downward, hence what is called the sense of a vector must be known. Thus, if we include sense, four things require to be known before a vector is completely specified. The properties of a vector quantity may be represented by a straight line ; thus, for example, a vector acting at a point A can be fully represented by a straight line. The length x>i the line to some convenient scale may represent the magnitude of the force. One end of the line A (Fig. 143) will represent the point of application, while the direction in which the line is drawn as from to A will represent the direction or sense of the vector. Direction of a vector. The direction of a vector is specified when the angle made by it with a fixed line is known. When 278 PRACTICAL MATHEMATICS FOR BEGINNERS. two, or more, vectors are given the line referred to may be one of the vectors. In many cases the points of the compass are used. Thus, in Fig. 143, the vectors B, C, D, E, and F make angles of 30, 45, 90, 135, and 180 respectively with the line OX, or with the vector A. It is important to remember that all angles are measured in the opposite direction to the hands of a clock. Using the points of the compass A is said to be towards the East, B is 30 K of E., C is N.E., D is North, E is N. W., and F is W. The sense of a vector is indicated by an arrow-head on the line representing the vector ; the clinure of the line, or the p o A Fig. 143. Specification of vectors. direction of the line may be called the clinure or ort of the vector. Addition and subtraction of vectors. If A, B, C (Fig. 144) represent three vectors acting at a point 0, to find their resultant, or better, to add them, we make them form consecu- tive sides of a polygon. Thus, starting from any convenient point a, the line ab is drawn parallel to, and equal in magni- tude to, the vector A. In like manner be is made equal to, and parallel to, B, and cd to C. The last side of the polygon from a to d represents the resultant, or the sum of the three given vectors. The sides of the polygon, a, b, c, d, taken in order, indicate the magnitude and direction of each vector, but arrow-heads on each side of the polygon also indicate the sense VECTOR QUANTITIES. 279 of each vector. When taken in order, i.e. a to b, b to c, and c to d, as in Fig. 144, the vectors are said to be circuital, hence an arrow-head in a non-cir- cuital direction on the last side of the polygon represents the resultant or the sum of the given vectors. Thus, denoting the sum by D, we have ab + bc + cd=ad, or A + B+C=D. The result obtained is b the same if we begin Fig. 144. -Sum of three vectors, with B or C. In fact, taking them in quite a different order as the sides of a polygon, the same result follows : i.e. A + B+C-D=0, (i) or A+B = D-C. If at a fourth vector (shown by the dotted line) equal in magnitude and parallel to da, be inserted, the polygon is a closed figure having the arrow-heads on its sides circuital. The four vectors acting at are in equilibrium. Hence we can write Eq. (i) as A+B+C+D=0. Vector quantities may in fact be added or subtracted by the parallelogram, triangle, or polygon law. As a simple example consider two displacements A and B. The vector sum is at once obtained by setting off oa and ob (Fig. 145) equal in magnitude to A and B respectively. Com- pleting the parallelogram the diagonal oc is the resultant, or sum, of the given vectors. A negative sign prefixed to a vector indicates that the vector is to be reversed. Thus, if A and B are represented by oa and ob, then A- B will be represented by oa and the dotted line ob. Hence, one diagonal of the parallelogram gives A + B and the other gives A-B. Ex. 1. There are two vectors in one plane, A of amount 10 in 280 PRACTICAL MATHEMATICS FOR BEGINNERS. the direction towards the East, B of amount 15 in the direction towards 60 North of East. (i) Find the vector sum A+B. (ii) The vector difference A - B. (iii) Find A + B when B is in the direction towards the North. H/ Fig. 145. Resultant of two displacements, (i) Starting at any point a (Fig. 146), draw a line ab equal in magnitude to, and parallel to, A. From b draw be parallel and Fig. 146. Sum and difference of two vectors. equal to B. Then ac is the magnitude and direction of the vector sum, and its sense is denoted by an arrow-head, non-circuital with the rest ; ac measures 21 79 and is directed towards 36 46' N. of E. VECTOR QUANTITIES. 281 (ii) Again starting at a point d, draw de as before ; but, from e, draw e/in the opposite direction. Then, df, as before, is the required vector. Its magnitude is 13 '2, and its direction 10 15' E. of S. (iii) When the vector is in a direction towards N"., then the angle between the two vectors is 90, and G= gh? + hm? =s/A 2 + B 2 =18 '02. Its direction is 56*5 N. of E. {i.e. tan d = {). Ex, 2. A ship at sea is sailing apparently at 8 knots to the East, and there is an ocean current of 3 knots to the South-west. Find the actual velocity of the ship. We have to find the resultant of a velocity 8 in a direction E., and a velocity 3 in a direction S.W. If a velocity of 1 knot be represented by 1 inch, then 8 inches will represent 8 knots and 3 inches will represent 3 knots. Make op = 8 knots and oq = 3 knots (Fig. 147). On the two lines op and oq as sides complete the parallelogram oprq. The diagonal c Fio. 147. Resultant of two velocities. or is the resultant required. Measuring or we find it to be 6 '25 inches, therefore representing 6 *25 knots. Its direction is given by the angle por = M0% or it may be written as 19 '8 S. of E. We may obtain the same result by drawing from any point a the lines ab and be equal and parallel to op and oq respectively. The resultant is then given in direction and magnitude by the line ac. Resolution of vectors. We are able to replace two vectors acting at a point by a single vector which will produce the same effect. Thus, in Fig. 148, the two vectors A and B may be replaced by the vector C, 282 PRACTICAL MATHEMATICS FOR BEGINNERS. Fig. 148. Rectangular components of a vector. Conversely, we may replace a single vector by two vectors acting in different directions. The two directions are usually assumed at right angles to each other. Let OG (Fig. 148) represent in direction and magnitude a vector acting at a point 0. If two lines OX and OF at right angles to each other be drawn passing through 0, and BG and AG be drawn parallel to OY and OX respectively, we obtain two vectors OB and OA, which, acting simultaneously, produce the same effect on the point as the single vector OG. The two vectors OA and OB are called the rectangular components of OG, and the process of replacing a vector by its components is called resolution. The vector OG can be drawn to scale, and the components measured to the same scale. Or, they can, by means of a slide-rule or logarithm tables, be easily calculated as follows : Denoting the angle BOG by 0. By definition (p. 155)^7= cos 6 ; if and OG are known, then 0B= OG cos 6. In a similar manner, OA=OGx cos GO A = OGx cos (90 - 0) ' =0G sin 0. This important relation may be stated as follows : The resolved part of a vector in any given direction is equal to the magnitude of the vector multiplied by the cosine of the angle made by the vector with the given direction. If the vector is a given velocity V, then the resolved part of the velocity in any given direction making an angle 6 with the direction of the velocity is V cos 6. If a body is moving N.E. with a velocity of 10 feet per VECTOR QUANTITIES. 283 second, it has a velocity East of 10 cos 45 and a velocity North of 10 cos 45. As the angle made by line OC (Fig. 148) increases, the hori- zontal component diminishes, and the vertical component increases. When = 90, the vector is vertical and the vertical component is simply the magnitude of the vector, its horizontal component is 0. Conversely when the angle is the vertical component is 0. The process just described may be extended to two or more vectors acting at a point. The horizontal and vertical com- ponents of each vector are obtained, the sum of all the horizontal components is denoted by X, and the sum of all the vertical components by T ; X and Y are then made to form the base and perpendicular of a right-angled triangle, the hypotenuse of which will be the vector sum required. Denoting the vector sum by R and its inclination to the axis of x by 6, then R = s/T*TF\ and tan 0= By means of Table V. the values of the sine and cosine of any angle can be obtained and the calculations for R and are easily made. The results obtained from this and the graphical method may, if necessary, be used to check the result. The application of the rule can best be shown by an example : Ex. 3. The magnitudes and directions of three vectors in one plane are given in the following table. Find the vector sums and differences (i) A + B + G ; (ii) A+B-C. A. B. 0, Magnitude, 50 30 20 Direction, - 30 N. ofE. N. N.W. Graphically (i) Show the three vectors acting at a point O (Fig. 149) ; draw the polygon making ab, be, and cd to represent the three given vectors. The non-circuital side ad is the sum A + B + G=15% its inclination is 67 N. of E. (ii) In A + B - G the direction of the vector C is reversed ; hence, produce CO, and on the line produced put an arrow head indicating 284 PRACTICAL MATHEMATICS FOR BEGINNERS. a direction opposite to that of the vector C. Also in the polygon produce dc to d', making cd' = cd. Join ad' ; ad' represents A+B-C; the magnitude is 70 and the inclination 36. Fig. 149. Sum of three vectors. Set off a length OB on the vector A equal to 50 units, draw H M perpendicular to OX. Then OM is the horizontal component of A. Similarly, the horizontal component of C is the line ON. As ON is measured in a negative direction, the sum of the horizontal com- ponents is OM - ON. We may measure either OM and ON and subtract one from the other ; or, using as centre and a radius equal to ON, describe an arc of a circle to obtain N'. Then N'M=OM-ON, where N'M denotes the sum of the horizontal components ; .-. X = N'M=2916. In a similar manner, projecting on the vertical line Y, OP, is the vertical component of A, and OQ the vertical component of C. Hence, Y=OP + S0 + OQ = 69'U. Having found X and Y draw a right-angled triangle in which the base am is 29*16, and the perpendicular md equal to 69"14, then the VECTOR QUANTITIES. 285 hypotenuse gives the magnitude and direction of the resultant. It is equal fco 75 % and 67 N. of E. By calculation, X = 50 cos 30 -20 cos 45 =50 x -866 - 20 x 7071 =2916 Y = 50 sin 30 + 30 + 20 sin 45 = 50 x -5 + 30 + 20 x -7071 = 69-14 i?=^+ + (7=\/29-16 2 + 69-14 2 = 75 2. If 6 denote the inclination of R, then .-. 0=67. It will be obvious from the figure that the vertical line, or perpendicular, md, represents the sum of the vertical components, and the horizontal line, or base, am, represents the sum of the horizontal components of the polygon abed. The general case. In the preceding examples the given vectors have been taken to act in one plane. In the general case, in which the vectors may act in any specified directions in space, the sum or resultant of a number of vectors may be obtained by using, instead of two, the three co-ordinates, x, y, and z. In this manner the resolved parts of each vector may be obtained, and from these the magnitude and direction of the line representing their sum. The process may be seen from the following example : Ex. 4. In the following table r denotes the magnitudes of each of three vectors A, B, and C, and a and j3 the angles made by each vector with the axes of x and y respectively. Find for each vector the values of 0, x, y, and z, and tabulate as shown. Vector. r. a. 60 d. X. 35-35 y- 2. A 50 45 60 25 25 B 20 30( 100 61 21' 1731 -3-472 9-59 G 10 120 45 60 -5 7 071 5 286 PRACTICAL MATHEMATICS FOR BEGINNERS. From the given values of a and /3 the value of (where denotes the inclination to the axis of z) can be calculated from the relation cos 2 a + cos 2 /3 + cos 2 = 1. Thus, for vector A , we have cos 2 = 1 - cos 2 a - cos 2 /3 = 1 - i - t = T 5 ;. cos0=l and = 60. Similarly for B, cos 2 = l-(-866) 2 -(-1736) 2 ='23; .\ = 61 21'. And for G, cos 2 = l - J- J = J ; .'. = 60. To obtain the projections x, y, and z of each vector, we use the relations a;=rcosa, y = rcos/3, z = rcos0. Thus, for vector A , r=50, a = 45, /3 and are each 60; .. x = 50 cos 45 = 50 x '707 1 = 35 -35, y = 50 cos 60 = 50 x -50 = 25, z = 50 cos 60 = 25. For vector B we have a:=20 cos 30 = 17*31, y= -20 cos 80 = 3 472, z = 20cos6121' = 9-59. For G, x= -10cos60= -5, y = 10 cos 45 = 7 071, z = 10 cos 60 = 5. Adding all the terms in column x and denoting the sum by 2Ja?, 2^ = 35-35 + 17*31 -5 = 47 66. Similarly, Sy = 25 - 3 -472 + 7 *07 1 = 28 -6, Sz = 25 + 9-59 + 5 = 39-59. Hence the resultant, or sum of the three vectors, is ^+ + C=\/(47-66) 2 + (28-6) 2 + (39-59) 2 = (68-4). To find the angles made by the resultant vector with the three axes we have cosa = ^?=-6966; .'. a = 45 50'. 68 4 cosj8=H^=-4181; .'. 0=65 18'. cos *=? ='5788; .-. = 54 38'. bo "4 Multiplication of vectors. Addition, subtraction, and multiplication of scalar quantities involving magnitude and not direction may be carried out by any simple arithmetical process. VECTOR QUANTITIES. 287 In the case of vectors, addition and subtraction are performed by using a parallelogram or a polygon. In multiplication we may write the product of two vectors A and B as A, B, but it must be remembered that the letters indicate, not only magni- tude, but also direction. The process may be shown by the product of two vectors such as a displacement and a force. Ex. 5. The direction of the rails of a tramway is due N. , and a force A of 300 lbs. in a direction 60 N. of E. acts on the car. Find the work done by the force during a displacement of 100 ft. If 6 denote the angle between the direction of the force A and the direc- tion of the displacement ON, then the resolved part of A in the direction ON is A cos 6. The product of a force, or the resolved part of a force, and its displacement, or distance moved through, is the work done by the force. Thus, in Fig. 150, if B denote the displacement of the car a then the work done is A B cos d (i). As A is 300, = 100, and = 30. A B cos 30 = 300 x 100 x *866 = 25980 ft. -pds. Observe by way of verification that if 6 be 0, then cos = 1 ; the force A is acting in the direction ON, and hence work done =300 x 100=30,000 ft. -pds. When 6 is 90, then cos 90 = ; .'. work done = 0. This latter result is obvious from the fact that, when the angle is 90, the force is in a direction at right angles to the direction of motion, and hence no work is done by the force. Again, if the direction of the force were South, then negative work equal to - 300 x 10= - 3000 would be done. From Eq. (i) it follows that the product of two unit vectors such as unit force and unit displacement, is cos 0. In any diagram, when two vectors are shown acting at a point, care must be taken that the arrow-heads denoting the sense of each vector are made to go in a direction outwards from the point. When this is done 6 is the angle between the vectors. Fig. 150. 288 PRACTICAL MATHEMATICS FOR BEGINNERS. EXERCISES. XLVIIL 1. Two vectors A and B act at a point. The magnitude of A is 50, its direction E. B is 100, direction 30 N. of E. Find the resultant or sum A+B. 2. Two forces of 8 and 12 units respectively act at a point, the angle between them is 72. Find their resultant. 3. A ship is sailing apparently to the East, and there is an ocean current of 8*7 knots to the South-west. Find the actual velocity. 4. There are three vectors in one plane : A, of amount 2, in the direction towards the North-east. B, of the amount 3, in the direction towards the North. G, of the amount 2*5, in the direction towards 20 East of South. By drawing, or any methods of calculation, find the following vector sums and differences : (i) A+B + G, (ii) B + G-A, (iii) A-G. 5. There are three vectors in a horizontal plane : A, of amount 1*5, towards the South-east. B, of amount 3 *9, in the direction towards 20 West of South. C, of amount 2*7, towards the North. (a) Find the vector sums or differences : A+B + G, A-B + G, B-G. (b) Find the scalar products AB and AG. 6. Three horizontal vectors are defined as follows : Vector. Magnitude. Direction and Sense. A B G 25 20 14 Eastward. 25 North of East. 80 North of East. Determine (i) A + B + C, and (ii) A+B-G, and write down the results. Prove by drawing that A+B+G=A + G+B, and A-{B-G)=A-B + G. 7. You are given the following three vectors : A. B. a Magnitude, - - 21 15 12 Direction, - ' - 75 120 VECTOR QUANTITIES. 289 Determine and measure the magnitude and direction of the vector sum A + B+G. Also, verify by drawing, that A -{B - G) = A -B + C. 8. A force A acts on a tramcar, the direction of the rails being due north. If B denote the velocity of the car, find the vector product A x B (called the activity or the power). (i) A is 300 lbs. N. ; B is 20 ft. per sec. (ii) A is 250 lbs. N.E. ; B is 15 (iii) A is 200 lbs. E. ; B is 20 (iv) A is 150 lbs. S.E. ; B is 10 9. The following five vectors represent displacements : A. B. O. D. E. Magnitude, - 20 12 6-8 3 3 155 Direction, 75 310 225 120 Find the vector sums of (i) A+B + G+D + E. (ii) A+B+E+D+G (iii) A+B-G+D-E. (iv) A+B-E+D-O. 10. A cyclist is travelling at 10 miles per hour in a northerly direction and a south-west wind is blowing at 5 miles an hour. Determine the magnitude and direction of the wind which the rider experiences. 11. Three vectors A, B, and G act at a point. The magnitudes and inclinations of each vector to the axes of x and y are given in the following table. Find in each case the inclination to the axis of z. Also find the sum and the inclination which the line representing the sum makes with the three axes. r. a. 6. A 100 30 120 B 50 135 30 G 10 45 60 12. Let A a denote a vector, where A gives its magnitude, and a its direction. Find A and a in the following vector equation, that is, add the three given vectors, which are all in the plane of the paper : A a = 3 '7 3 o + 1 -4 8 2 + 2-6i57 Find also B and /3 from the equation /3 = 3-73o-l-4 82 <> + 2-6i57-. Use a scale of 1 inch to 1 unit. P.M.B. T CHAPTER XXIV. ALGEBRA {continued) ; SQUARE ROOT ; QUADRATIC EQUA- TIONS ; ARITHMETICAL, GEOMETRICAL, AND HAR- MONICAL PROGRESSIONS. Square root. In the process of division advantage is taken of the results obtained from multiplication. In like manner, the square root (p. 25) of an algebraical expression can often be obtained by comparing it with known forms of the squares of different expressions. Thus, the square of (a + b), or {a + b) 2 is a 2 + 2ab + b 2 . Hence, when any expression of this form is given, its square root can be seen at once and written down, e.g. sjx 2 + 2xy+y 2 =x+y. We may, from this example, proceed to derive a general rule for the extraction of the square root. a 2 + 2ab + b 2 (a + b a 2 2a + b)2ab + b 2 2ab + b 2 Thus, arrange the terms according to the dimensions of one term, as a. The square root of the first term is a ; taking its square from the whole expression, 2ab + b 2 remains ; mentally dividing 2ab by 2a, the double of the first term of the required square root, we find that it is contained b times in 2ab. Hence, adding b to the 2a previously obtained, we obtain the full trial divisor 2a + b. Multiply this result by the new term of the required root, b, and subtract the product from the first re- mainder. Then, as there is no remainder a + b is the root required. SQUARE ROOT. 291 Ex. 1. Find the square root of 4x 2 + 24xy + 36y 2 . Here 2x is clearly the root of the 4 * 2 + 24^ + 3% 2 ( 2* + 6y first term 4x 2 . Put 2x for the first 4^.2 term of the required root: square it, . Ta \^a , o* 9 4 ' M . ' 4:X + 6y)24:xy + 36y i and subtract its square from the given 2Axy + 36?/ 2 expression, Bring down the other two terms 24xy + S6y 2 . Multiply the first term of the root 2x by 2, giving 4a?, and using this as a trial divisor, the remaining term of the root is found to be Qy. Hence, put 6y as the second term in the root and multiply 4x + Qy by 6y, giving as a product 24xy + 36y 2 . subtract this from the two remaining terms of the given expression, and there is no remainder. The required root is 2x + 6y. Following the steps in the preceding worked out example the next will be readily made out. Ex. 2. Find the square root of 4x* + 4x 2 y 2 - \2xh 2 + y*- 6yh 2 + 9z 4 . 4a: 4 + 4x 2 y 2 - 1 2xh 2 + y*- Qy 2 z 2 + 9z 4 ( 2x 2 + y 2 -3z 2 4x* 4x 2 y 2 - \2xH 2 + y*- Qyh 2 + 9z 4 4x 2 + y 2 ) 4a;y jV 4a; 2 + 2y 2 - 3z 2 ) - \2x\ 2 - Qy 2 z 2 + 9z 4 -12x 2 z 2 -6y 2 z 2 + 9z 4 The expression for the expansion of (1 +a) n is given on p. 111. When n is \, and a is small compared with unity, the square root of (1 + a) can be obtained to any desired degree of accuracy. Ex. 3. Find the first five terms of the square root of 1 + x, and use them to find the value of \/l01. (1+x) =1+ - + ?L<^ 2+ ^J^W. _ j, this becomes When n = ^, this becomes 1 1 2 1 ^ 5 ^ ~ l+ 2 X 8 X + 16 128 + 292 PRACTICAL MATHEMATICS FOR BEGINNERS. Viol =n/(!oo+I) = 1( >V( 1+ iuo) = 10 ( 1+ 200~80000 + 16000000 ~ 12800000000 + etc 7 = 10-04988.... EXERCISES. XLIX. Find the square root of 1. 9a 4 - 42a 3 + 37a 2 + 28a + 4. 2. 4a 4 + 12a 3 - 1 la 2 -30a + 25. 3. a 4 + 4ax s + 2a 2 a 2 - 4a 3 a + a 4 . 4. a 6 - 22a; 4 + 34a 3 + 121a 2 - 347a + 289. 5. 25a 8 -60a 6 -34^ + 84a 2 + 49. 6. a 4 - 2a 3 + 9a 2 -8a + 16. 7. 25a 4 - 30a 3 + 49a 2 -24a +16. 8. a 4 + 8a 3 -26a 2 - 168a + 441. 9. 16a 4 -4005 + 89a 2 -80a + 64. 10. f^L + ^-S. y 2 a 2 11. 9a 4 - 30a 3 y + 31a 2 y 2 - lOay 3 + y 4 . 12. Show that the square of the sum of two quantities together with the square of their difference is double the sum of their squares. 13. Show that the sum of the squares of two quantities is greater than the square of their difference by twice the product of the quantities. 14. Find the square root of the difference of the squares of 5a 2 -8a +13 and 4a 2 + 2a -12. Quadratic equations. As already indicated (p. 83) when a given equation expressed in its simplest form involves the square of the unknown quantity it is called a quadratic equation. Such an equation may contain only the square of the unknown quantity, or it may include both the square and the first power. Ex. 1. Solve the equation a 2 -9 = 0. We have * 2 = 9 = 3 2 ; .\ a=3. It is necessary to insert the double sign before the value obtained for x as both +3 and -3, when squared give 9. The solution of a given quadratic equation containing both x 2 and x can be effected by one of the three following methods. First Method. The method most widely known, and generally used, may be stated as follows : Bring all the terms containing x 2 and x to the left hand side of the equation, and the remaining terms to the right hand side. Simplify, if necessary, and make the coefficient of x 2 unity. QUADRATIC EQUATIONS. Finally, add the square of one-half the coefficient of x to both sides of the equation and the required roots can be readily obtained. Ex. 2. Solve the equation x 2 + 4x - 21 = 0. We have # 2 + 4x=21 (i) Add the square of one-half the coefficient of x to each side ; .-. a 2 + 4# + (2) 2 =21+4=25; i.e. (a + 2) 2 = 5 2 ; .'. a? + 2=5; (ii) /. x= -25 = 3, or -7. It will be noticed that (ii) may be written x + 2= +5 and x + 2 = -5. From these equations the values # = 3 and x 7 are at once obtained. Second Method. The second and the third methods of solution are explained on p. 191, but it may be advisable to refer to them again here. Where the given equation can be resolved into factors, then the value of x which makes either of these factors vanish, is a value of x which satisfies the given equation. Ex. 3. Solve the equation x 2 + 4x - 21 =0. Since * 2 + 4a:-21 = (a;-3)(.r + 7) ; x - 3 = 0, when x = 3 ; and 07 + 7 = 0, when x=-l. Hence x = 3 or x= 1 is a solution of the equation and 3 and - 7 are the roots of the given equation. Third Method. Let y = .r 2 + 4# - 21. Substitute values 1, 2, 3 ... for x and calculate corresponding values of y. Plot the values of x and y on squared paper. The two points of intersection (of the curve passing through the plotted points) with the axis of x are the roots required. A quadratic equation in its general form may be written ax 1 + bx + c = 0. Then *+-#=-- a a adding to each side the square of half the coefficient of x, or \2a) 2 we have 294 PRACTICAL MATHEMATICS FOR BEGINNERS. a 2a/ 4a 2 2a a 4ac Aac 2a (i) The following important cases occur. If b 2 is greater than 4ac i.e. b 2 > 4ac, there are two values, or roots, satisfying the given equation. If b 2 = 4ac the two roots are equal ; each is - . If b 2 <4ac, there are no real values which satisfy the given equation, and the roots are said to be .imaginary. All these results may be clearly appreciated by using squared paper. Ex. 4. Solve the equations : (i) 2a; 2 -4a; + 1=0, (ii) 2a; 2 -4a; + 2 = 0, (iii) 2a; 2 -4a; + 3 = 0. (i) Let y = 2a; 2 - 4a; + 1. Assume a;=0, 1, 2, ... etc., and find cor- 1 responding values of y. Thus, when x = 0, y = 1 ; when x\, y=-l; when x = 2, ir=l. Plot these values on squared paper; then the curve passing through the plotted points in- tersects the axis of x at points A and B (Fig. 151) for values of x= -293, and 1 -707, and these are the roots required. It will be noticed each time the curve intersects the axis of x the value of y changes sign. Hence we know that one value lies between a;=0 and a?=l; and between x= 1 and a; = 2. (ii) Let y = 2a; 2 - 4a; + 2. Values of x and corresponding values of y are as follows : i // /// / / /// / i I V y V \ \ ' . 2 3 Pig. 151. -To illustrate Ex. 4. x 1 2 j 3 y 2 2 8 QUADRATIC EQUATIONS. 295 Plotting as before the curve (ii) (Fig. 181) is obtained and touches, or better is tangent to, the axis of x at the point x=\. Hence the two roots of the equation are equal. (iii) Proceed as before and obtain the following values : X 1 2 3 y 3 1 3 9 The curve joining the plotted points is shown by (iii) (Fig. 181) ; this does not intersect the axis of x, and the roots are imaginary. Much unnecessary labour will result if the attempt is made to obtain unity as the coefficient of x 1 in all equations. It may be found better to use another letter, such as y or 0, and then to proceed to solve the equation in the ordinary manner, finally solving the equation for x. The following examples will illustrate some of the methods which may be adopted. Ex. 5. Solve the equation ( j ) = 8 ( r ) - 15. By transposition, we obtain (a-xy Q f a-x \ \x^rs) ~ S \x^b) 15. If we write y for x- Hence x the equation becomes y 2 -8y + (4)2: .'. y=4l a-x x-b a-x x-b 3; 5; -15; -15 + 16=1 3, or 5. a + 3b 4 a + 5b Instead of using the letter y f the equation could be written as (H) 2 - 8 (H)+<*> 2 =-^=i> 41 = 3, or 5. Two simultaneous quadratics. Some methods which may be adopted to obtain the solution of simultaneous equations of the first degree are explained in Chap. IX., p. 91. Similar 296 PRACTICAL MATHEMATICS FOR BEGINNERS. processes are applicable in equations of the second degree. That is to say we can, by multiplication, division, or substitution, obtain an equation involving only one unknown quantity. From this equation the value of the unknown quantity can be deter- mined, and by substitution the value of the remaining unknown can be found. Ex. 6. Solve the equation x 2 + y = 8, 3x + 2y = r J. x 2 + y=8 (i) 3x + 2y = 7 (ii) Multiply (i) by 2 and subtract (ii) from it , .*. 2x 2 + 2y=lQ 3x + 2y~ 7 2x 2 -3#= 9 X 2 X+ \4j ~2 + 16 _ 16' q q * = 77 = 3, or -1-5. 4 4 From (ii), when x is 3 ; 2y = 7 - 9 ; .'. y = - 1 ; whenais -T5; 2?/ = 7 + 4'5; .*. y = 5'75. Ex. 7. Solve the equation (i) x 2 + xy = 84; (ii) xy + y 2 = Q0. Adding (ii) to (i) we get x 2 + 2xy + y 2 =U4; :. x + y=l2 (iii) From (i), x{x + y) = 84.. From (ii), y{x+y) = 60. Substituting from (iii), 12x=84 and 12*/ = 60. Hence x=7> y=5; therefore the four values are x = 7, x= -7, y = 5, y= -5. Equations reducible to quadratics. Equations of the fourth degree can in some cases be solved as two quadratic equations. Ex. 8. Solve x* - 17x 2 + 16 = 0. The equation may be written (x 4 -8x' 2 +16)-9x 2 =0, or (# 2 -4) 2 - (3#) 2 = ; :. {x 2 + 3x - 4) {x 2 - 3x - 4) = 0. Hence x 2 + 3x-4 = 0, (i) or x 2 -3x-4:=:0 (ii) From (i), x 2 + 3x-4: = {x + 4)(x-l) ; .-. x= -4, or 1. EQUATIONS REDUCIBLE TO QUADRATICS. 297 From (ii), x 2 -3x-4 = (x-4){x+l) ; .'. #=4, or - 1. Hence the values of x which satisfy the given equation are x=4, x=l. Relations between the coefficients and the roots of a quadratic equation. In the preceding examples we have been able, from a given quadratic equation, to find the roots, or the values which satisfy the given equation. The converse of this is often required, i.e. to form a quadratic equation with given roots. It has been already seen that if we can resolve the left-hand side of the given equation, when reduced to its simplest form, into factors, then the value of x which makes either of these factors zero, is a value of x which satisfies the given equation. Thus the roots of the equation (x-a)(x- /3) = are a and fi. Conversely, an equation having for its roots a and /3 is (x-a)(x-l3) = 0. Hence if a and /3 denote the roots of the equation ax 2 + bx + c=0. We have ax 2 + bx + c = a(x - a)(x - /5) ; \ ax 2 + bx + c = a(x 2 ax- /3x + a/3) = a(x 2 -(a + P)x + a/3). Comparing coefficients on both sides we have a(a + P)= -b and aa/3 c; .'. a + )8= and a/3 = -', therefore when the coefficient of x 2 is unity the sum of the roots is equal to the coefficient of x ; and the product of the roots is equal to the remaining term. Ex. 9. Form the quadratic equations having roots 1 and 4. Here (x- l){x-4) = x 2 -5x + 4:. Ex. 10. Form the quadratic equation having roots -3 + \/2 and -3-\/2. Here we have (x + 3- sfe) {x + 3 + s/2) = {x + 3) 2 - 2 ; .'. the required equation is x 2 + 6a? + 7=0, 298 PRACTICAL MATHEMATICS FOR BEGINNERS. Ex. 11. Form the quadratic equation having roots a and -. Here we have {x - a) I x J ; required equation is x 2 x + 1=0. EXERCISES. L. Solve the equations : 1. x 2 - 60 = 80 -Ax. 2. x 2 + S2x =320. 3. ?^-x + ~ = 0. 4. 2x 2 -4x-6 = 0. 5. 3# 2 -84 = 9x. 6. x 2 +'402x='\63. 7. 6x 2 -13x + 6 = 0. 8. x 2 -{a + b)x + ab = 0. \s/3^x~. 10. 19a: 2 -4x- 288 = 0. s/x + 2 2 11. 4# 2 + 4a:-3 = 0. 12. >J{5x + 9)-^3x+l=sj2(x -6). 13. x 3 -2x 2 -3# + 4 = 0. 14. (x 2 -4a: + 3) 2 -8(:r 2 -4a; + 3) = 0. 15. l+2xs/(T^x 2 j=9x 2 . 16. x + 1 = 2(1+^2). 1 1 1 1 17# l+# + 2 + a; _ l-a; + 2-a;' 18. 40^+~Y-286Ca; + ^ +493 = 0. 21. x s + y* = 72, xy{x + y)=A8. 22. * 2 -4y 2 = 8, 2{x + y) = 7. 23. 2x 2 -3y 2 = 5, 3x + y=15. 24. (i) Find the roots of the equation x 2 -2ax + (a-b)(a + b)=0. (ii) Form the equation the roots of which are the squares of the roots of the given equation. 25. Find the roots of the equation x 2 + 7 W2 = 60. Form the quad- ratic equation having roots a and -. a 26. If a and (3 are the roots of the equation ax 2 + bx + c = 0, show that a + 8= -- and aB = ~. r a r a Problems leading to quadratic equations. As already indicated on p. 81, one of the greatest difficulties experienced by PROBLEMS LEADING TO QUADRATIC EQUATIONS. 299 a beginner in Algebra is to express the conditions of a given problem by means of algebraical symbols. The equations themselves may be obtained more or less readily since the conditions are generally similar to those already explained, but some difficulty may be experienced in the interpretation of the results derived from such equations. Since a quadratic equation which involves one unknown quantity has two solutions, and simultaneous quadratics involving two unknown quantities may have four values, or solutions, it is clear that ambiguity may arise. It will be found, however, that although the equations may have general solutions only one solution may be applicable to the particular problem. The fact that several solutions can be found and only one applies to the problem is due to the circum- stance that algebraical language is far more general than ordinary methods of expression. Usually no difficulty will be experienced in deciding which of the solutions are applicable to the problem in hand. Ex. 1. A boat's crew can row at the rate of 9 miles an hour. What is the speed of the river's current if it takes them 2% hours to row 9 miles up stream and 9 miles down ? Let s denote the speed of the current in miles per hour. Then, 9 - s and 9 + s represent the crew's rate up and down stream respectively ; ... *+**mJL 9-s 9+s f 4 36 + 4s + 36-4s=81-s 2 . g*=9, s=S. Only the positive value is applicable to the problem. Ex. 2. A certain number of articles are bought for 1, and 1. Os. 7d. is made by selling all but one at Id. each more than they cost. How many are bought ? Let x denote the number bought. 240 Then = price per article in pence ; ,. ,.-l)(f + l)-M, .-. (x-l)(240 + x) = 247x; .-. x 2 + 239a; - 240 = 247*, o r x 2 - 8x -240 = 0; .% (a: -20) (a; + 12) =0. 300 PRACTICAL MATHEMATICS FOR BEGINNERS. The two values obtained are x=20 and x = -12. Obviously only the former is applicable to the problem, hence x = 20. Ex. 3. In the equation *= Vt + lflK Given s=80, F=64, and /=32, find t. Substituting the given values 80 = 64* + \ x 32* 2 = 64* + 16* 2 ; .'. * 2 + 4* + (2) 2 =5 + 4 = 9; ' /. *=-23=l, or -5. In the case of a body projected upwards with a vertical velocity 64, then, when / is 32, the body is at a distance 80 from the starting point when t = 1 and is moving upwards. The same conditions hold true again when t = - 5, and the body is moving in the opposite direction. EXERCISES. LI. 1. In the formula t Tr+. (i) given t=, g = 32, ir~ ij, find the numerical value of I. (ii) t = j^, 1=8, findgr. 2. In the formula s=Vt + \fp. (i) given F=12, . = 470, /=7, find t. (ii) T=172, s = 90, /=32, find*. 3. The area of a certain rectangle is equal to the area of a square whose side is six inches shorter than one of the sides of the rectangle. If the breadth of the rectangle be increased by one inch and its length diminished by two inches, its area is unaltered. Find lengths of its sides. 4. The perimeter of a rectangular field is to its diagonal as 34 to 13, and the length exceeds the breadth by 70 yards. What is its area ? 5. A traveller starts from A towards B at 12 o'clock, and another starts at the same time from B towards A. They meet at 2 o'clock at 24 miles from A, and the one arrives at A while the other is still 20 miles from B. What is the distance between A and B ? 6. From a catalogue it is found that the prices of two kinds of motors are such that seven of one kind and twelve of the other can be obtained for 250. Also that three more of the former can be purchased for 50 than can 4tee of the latter for 30. Find the price of each. 7. A boat's crew can row at the rate of 8 miles per hour. What is the speed of the river's current if it takes them 2 hours and 20 minutes to row 8 miles up stream and 8 miles down ? PROBLEMS LEADING TO QUADRATIC EQUATIONS. 301 8. A person lends 1500 in two separate sums at the same rate of interest. The first sum with interest is repaid at the end of eight months, and amounts to 936 ; the second sum with interest is repaid at the end of ten months, and amounts to 630. Find the separate sums lent and rate of interest. 9. Show that if the sum of two numbers be multiplied by the sum of their reciprocals the product cannot be less than 4. 10. Divide 490 among A, B, and G, so that B shall have 2 more than A, and C as many times i?'s share as there are shillings in A's share. 11. If in the equation ax 2 + bx + c=0 the relations between a, b, and c are such that a+b+S=0 and 2a -c + =0, what must be the value of a in order that one of the roots may be 5, and what is then the value of the other root ? _, x, -> a- - o Series. The term series is applied to any expression in which each term is formed according to some law. Thus, in the series 1 , 3, 5, 7 . . . each term is formed by adding 2 to the preceding term. In 1, 2, 4, 8 ... each term is formed by multiplying the preceding term by 2. Usually only a few terms are given sufficient to indicate the law which will produce the given terms. The first series is called an arithmetical progression, the con- stant quantity which is added to each term to produce the next is called the common difference. The letters a.p. are usually used to designate such a series. The second series is called a geometrical progression, the con- stant quotient obtained by dividing any term by the preceding term is called the common ratio or constant factor of the series. The letters g.p. are used to denote a geometrical progression. Arithmetical Progression. A series is said to be an arith- metical progression when the difference between any two con- secutive terms is always the same. Thus the series 1, 2, 3, 4 ... is an arithmetical series, the constant difference obtained by subtracting from any term the preceding term is unity. In the series 21, 18, 15, ... the constant difference is -3. Again in a, a + d, a + <2d, ... and a, a d, a-2d, ... the first increases and the second diminishes by a common difference d. In writing such a series it will be obvious that if a is the first term, a + d the second, a + 2d the third, etc., any term such as 302 PRACTICAL MATHEMATICS FOR BEGINNERS. the seventh is the first term a together with the addition of d repeated (7 - 1) times or is a + 6d. If I denote the last term, and n the number of terms, then l = a + (n-l)d (i) Let S denote the sum of n terms, then S = a + {a + d) + (a + 2d)+...+{l-2d) + (l-d) + l. Writing the series in the reverse order we obtain S=l + (l-d) + (l-2d) + ...(a + 2d) + (a + d) + a. Adding we obtain 2&=(a + l) + (a + l)+...to n terms = n(a + l); A S=%(a + l) (ii) From this when a and I are km wn the sum of n terms can be obtained. Again, substituting in (ii) the value of I from (i) we have S=^{2a + (n-l)d} (iii) From Eq. (iii) the sum of n terms can be obtained when the first term and the common difference are known. Arithmetical Mean. The middle term of any three quan- tities in an arithmetical progression is the arithmetical mean of the remaining two. Thus if a, A, and b form three quantities in arithmetical pro- gression, then A a = b- A ; or, the arithmetical means of two quantities is one-half their sum. Ex. 1. Find the 9 th term of the series 2, 4, 6 ... , also the sum of nine terms. Here, from (i), I = a + {n - 1 ) d. a = 2, n = 9, and d = 2; ;. J=2 + (9-l)2=18. From (ii), S=^(a + 1) =1(2+18) = 90. ARITHMETICAL MEAN. 303 Ex. 2. The second term of an a. p. is 24. The fifth term is 81. Find the series. Here a + d=24, also a + 4d = 81 ; .-. 3c? = 57, or d= 19. As the second term is 24, the first term is 24 - 19 = 5. Hence the series is 5, 24, 43 .... Ex. 3. The twentieth term of an a. p. is 15 and the thirtieth is 20. What is the sum of the first 25 terms ? Here a+\9d=\5 a + 29d = 20 By subtraction, \0d= 5; ;. d = \. By substitution, a=-^-; /. S=^{2a + (n-l)d} = |{ll + (25-lHH287^. EXERCISES. LIL Sum the following series : 1. 3, 3, 4 ... to 10 terms. 2. -2\ y -2, - 1 to 21 terms. 3. 7 + 32 + 57+... to 20 terms. 4. 2 + 3^ + 4+ .. to 10 terms. 5. i-i-l-...to20terms. 6. \-\- j ... to 21 terms. 3 3 4 4 4 7. 1-5-IZ-. ..to 12 terms. o o 8. Find the sum of 16 terms of the series 64 + 96 + 128 + . 9. Sum the series 9 + 5+1-3- to n terms. 10. The sum of n terms of the series 2, 5, 8 ... is 950. Find n. 11. The sum of n terms of an a. P. whose first term is 5 and common difference 36 is equal to the sum of 2n terms of another progression whose first term is 36 and common difference is 5. Find the value of n. 12. The first term of an a. p. is 50, the fifth term 42. What is the sum of 21 terms ? 13. The fourth term is 15 and the twentieth is 23|. Find the sum of the first 20 terms. 14. The sum of 20 terms is 500 and the last is 45. Find the first term. 15. The sum of three numbers is 21, and their product is 315. Find the numbers. 304 PRACTICAL MATHEMATICS FOR BEGINNERS. 16. If the sum of n terms be n? and common difference be 2, what is the first term ? 17. The sum of an a. p. is 1625, the second term is 21, and the seventh 41. Find the number of terms. 18. Find the sum of the first n natural numbers. 19. Find the sum of the first n odd natural numbers. 20. Show that if unity be added to the sum of any number of terms of the series 8, 40, 72 ... the result will be the square of an odd number. Geometrical progression. A series of terms are said to be in geometrical progression when the quotient obtained by divid- ing any term by the preceding term is always the same. The constant quotient is called the common ratio of the series. Let r denote the common ratio and a the first term. The series of terms a, ar, ar 2 , etc., form a geometrical pro- gression, and any term, such as the third, is equal to a multiplied by r raised to the power (3 1) or ar 2 . Thus, if I denote the last term and n the number of terms then we have Z = ar n_1 (i) Let S denote the sum of n terms then S = a + ar + ar 2 +... ar n ~ 2 + ar 71 ' 1 (ii) Multiplying every term by r Sr = ar + ar 2 + ar 3 + ... ar^ + ar" (iii) (Subtract ii) from (iii). .*. rS-S=ar n -a, or (r-l) = a(r w -l). a(r n -l) r . Ex. 1. The first term of a g.p. is 5 ; the third term is 20. Find the eighth term and the sum of eight terms. The third term will be ar 2 where a denotes the first term and r the common ratio ; .-. 5r 2 =20 or r = 2. From l^ar 71 ' 1 we get by substitution 1=5^ = 5 x 2 7 = 640. a a(r"-l) 5(2 8 -l) 05= ^ = ^ r- 1 1 = 5x255 = 1275. GEOMETRICAL PROGRESSION. 305 Ex. 2. The third term of a g.p. is 20. The eighth term is 640, and the sum of all the terms is 20475. Find the number of terms. Here ar 2 = 20 and ar 7 =640; ar 7 _640. '' ar 2- 20 ' or ^=32; .-. r=2, and a= - = 5. 4 r-1 _ 5(2-l ) " 2-1 J . 2M _ 1= 20475 = 4095 5 or 2 W =4096 = 2 12 ; /. w=12; or n log 2= log 4096; 36123 3010 = 12. Ex. 3. The sum of a g.p. is 728, common ratio 3, and last term 486. Find the first term. o = r ; r-\ but, r n ~ 1 = -; or ar n =lr; a s =7^' 728= 3x486-a. .*. a = 1458 -1456=2. By changing signs in both numerator and denominator Eq. (iv) becomes aJ=2 w \r When r is a proper fraction it is evident that r n decreases as n increases. Thus when r is ^, r 2 =Y^, ^ 3 = j , 5o"? e ^ c ') wnen ** is indefinitely great, r n is zero, and (v) becomes S~ r (vi) 17* P.M.B. U 306 PRACTICAL MATHEMATICS FOR BEGINNERS. Sum of a G.P. containing an infinite number of terms. Eq. (vi) is used to find the sum of an infinite number of terms, or as it is called the sum of a series of terms to infinity. Ex. 4. Find the sum of the series, 84, 14, 2J . . . to infinity. Here r = l -*J- ; 84 6' , S = JL " =* 100 . 8 . 1 ~ r i _i 5 6 Value of a recurring decimal. The arithmetical rules for finding the value of a recurring decimal depend on the formula for the sum of an infinite series in g.p. Ex. 5. Find the value of 3 '6. 3-6=3-666 ... --+$ 6 + 10 2 6 :+ 10 3 + r=i and a = '6; a ^ 1_ Io 6 9 10 6 2. 9~3' 8-4=4 Geometrical mean. The middle term of any three quantities in a geometrical progression is said to be a geometric mean between the other two. The two initial letters g.m. may be used to denote the geometric mean. Thus, if x and y denote two numbers, the a.m. is x ^ the g.m. is 4xy. In the progression 2, 4, 8 ... the middle term 4 is the g.m. of 2 and 8. In like manner in a, ar, ar 2 + ...ar is the g.m. of a (tr and It will be noticed that the g.m. of two quantities is the square root of their product. To insert a given number of geometric means between two given quantities. From l = ar n ~ 1 we obtain r n ~ 1 =- a from this when I and a are given r can be obtained. GEOMETRICAL MEAN. 307 Ex. 6. Insert four geometric means between 2 and 64. Including the two given terms the number of terms will be 6, the first term will be 2, and the last 64. " 2 ' r 5 = 32, or r=2. Hence the means are 4, 8, 16, 32. EXERCISES. LIII. Sum the following series : 1. 1 + 77 + 7^ to 12 terms. o oo 2. 1 - 1 2 + 1 -44 to 12 terms. 113 3. -- + -- 1+ .. to 10 terms. 4. The first term of a g.p. is 3, and the third term 12. Find the sum of the first 8 terms. 5. (i) What is the eighth term of the g.p. whose first and second terms are 2,-3 respectively, (ii) Find the sum of the first 12 terms of the series. 6. (i) What is the 6 th term of a g.p. whose first and second terms are 3, - 4 ? (ii) Find the sum of the first 10 terms. 7. Show that the arithmetical mean of two positive quantities is greater than the geometrical mean of the same quantities. 8. The arithmetical mean of two numbers is 15, and the geo- metrical mean 9. Find the numbers. Sum to infinity the series : 9. 14-4, 10*8, 8-1... . 10. (i) -32, (ii) -7, (iii) 2\ / 2-2>/3 + 3\/2 to 10 terms. 11. Find an a. p. first term 3, such that its second, fourth, and eighth terms may be in g.p. 12. The sum of the first 8 terms of a g.p. is 17 times the sum of the first four terms. Find the common ratio. 13. A series whose 1st, 2nd, and 3rd terms are respectively j_ 1 1 s/2 lW2 4 + 3^2 is either an a. p. or a g.p. Determine which it is and write down the fourth term. 14. If one geometrical mean O and two arithmetical means p and q be inserted between two given quantities show that G* = (2p-q)(2q-p). 308 PRACTICAL MATHEMATICS FOR BEGINNERS. 15. The continued product of three numbers in geometrical pro- gression is 216, and the sum of the products of them in pairs is 156. Find the numbers. Harmonical Progression. A series of terms are said to be in Harmonical Progression when the reciprocals of the terms are in Arithmetical Progression. Let the three quantities a, b, c be in h.p., then -, -, - are in a.p. a o c A 1111 ,.* b a c b we obtain the relation a\ ca-b \b c, or three quantities are in h.p. when the ratio of the first to the third is equal to the ratio of the first minus the second, to the second minus the third. Again from (i) the harmonical mean between two quantities , . , 2ac a and c is o = . a + c In problems in harmonical progression such as to find a number of harmonical means, to continue a given series, etc. ; it is only necessary to obtain the reciprocals of the given quantities and to proceed to deal with them as with quantities in arith- metical progression. Ex. 1. Find a harmonical mean between 42 and 7. , i, . . 2ac 2x42x7 10 1 -, 1 We may use the formula h. m. == = j~ s = 12, or as ^ and - a-\- c QcJi + / 4Z / are in \. p. J_ 1 42 + 7 1. .-. mean= = -^ Hence the required mean is 12, and 42, 12 and 7 are three terms in h.p. Ex. 2. Insert two harmonical means between 3 and 12. Inverting the given terms ^ and y^ are the first and last terms of an a.p. of four terms we have "~ ""g-^tt-^ therefore from l = a + (n- l)d i=i + (4-Drf; \ 3d= -j, or d= - 1 "12" HARMONICAL PROGRESSION. Hence the common difference is - yo : therefore the terms are 111 ,121 3"I2 = 4 and 3"I2 = 6' or the arithmetical means are - and -. 4 6 Hence the harmonic means are 4 and 6. EXERCISES. LIV. 1. Define harmonic progression ; insert 4 harmonic means be- tween 2 and 12. 2. Find the arithmetic, geometric, and harmonic means between 2 and 8. 3. Find a third term to 42 and 12. 4. Find a first term to 8 and 20. 5. The sum of three terms is l^" , if the first term is ^, what is the series ? 6. The arithmetical mean between two numbers exceeds the geometric by two, and the geometrical exceeds the harmonical by 1*6. Find the numbers. 7. A h.p. consists of six terms ; the last three terms are 2, 3, and 6, find the first three. 8. Find the fourth term to 6, 8, and 12. 9. Insert three harmonic means between 2 and 3. 10. Find the arithmetic, geometric, and harmonic means between 9 2 and =, and write down three terms of each series. m MATHEMATICAL TABLES. Each candidate at the Examinations of the Board of Education (Secondary Branch) in Practical Mathematics, Applied Mechanics, and Steam is supplied with a copy of Mathematical Tables similar to those here given. TABLE II. USEFUL CONSTANTS. 1 inch = 25 *4 millimetres. 1 gallon = -1604 cubic foot = 10 lbs. of water at 62 F. 1 knot = 6080 feet per hour. Weight of 1 lb. in London = 445, 000 dynes. One pound avoirdupois = 7000 grains = 453*6 grammes. 1 cubic foot of water weighs 62*3 lbs. 1 cubic foot of air at C. and 1 atmosphere, weighs *0807 lb. 1 cubic foot of hydrogen at C. and 1 atmosphere, weighs '00559 lb. 1 foot-pound =1 -3562 x 10 7 ergs. 1 horse-power-hour = 33000 x 60 foot-pounds. 1 electrical unit =1000 watt-hours. t i> i ** * i tt / 774 ft. -lb. = 1 Pah. unit. Joule s equivalent to suit Kegnault s H, isi logoff -i-u _i p f 1 horse-power =33000 foot-pounds per minute = 746 watts. Volts x amperes = watts. 1 atmosphere =14 '7 lbs. per square inch = 21 16 lbs. per square foot = 760 mm. of mercury = 10 6 dynes per square cm. nearly. A column of water 2*3 feet high corresponds to a pressure of 1 lb. per square inch. Absolute temp., t = d C. +273 *7. Regnault's #=606*5+ '305 C. = 1082+ *305 9 F. log 10 p = 6-1007-f~, where log 10 = 3*1812, log 10 <7= 5*0871, p is in pounds per square inch, t is absolute temperature Centigrade, u is the volume in cubic feet per pound of steam. tt = 3*1416. 1 radian = 57 "3 degrees. To convert common into Napierian logarithms, multiply by 2*3026. The base of the Napierian logarithms is e = 2*7 183. The value of g at London = 32* 182 feet per sec. per sec. 312 PRACTICAL MATHEMATICS FOR BEGINNERS. TABLE III. LOGAKITHMS. 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 4 8 12 17 21 25 29 33 37 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 4 8 11 15 19 23 26 30 34 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 3 710 14 17 21 24 28 31 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 3 6 10 13 16 19 23 26 29 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 3 6 9 12 15 18 21 24 27 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 3 6 8 11 14 17 20 22 25 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 3 5 8 11 13 16 18 21 24 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 2 5 7 10 12 15 17 20 22 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 2 5 7 9 12 14 16 19 21 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 2 4 7 9 1113 16 18 20 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 2 4 81113 15 17 19 21 3222 3243 3263 3284 3304 3324 3:545 3365 3385 3404 2 4 6 8 10 12 14 16 18 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 2 4 6 8 10 12 14 15 17 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 2 4 6 7 9 11 13 15 17 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 2 4 5 7 9 11 12 14 16 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 2 3 5 7 910 12 14 15 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 2 3 5 7 8 10 11 13 15 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 2 3 5 6 8 9 11 13 14 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 2 3 6 6 8 9 11 12 14 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 1 3 4 6 7 9 10 12 13 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 1 3 4 6 7 9 10 11 13 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 1 3 4 6 7 8 10 11 12 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 1 3 4 5 7 8 9 1112 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 1 3 4 5 6 8 9 10 12 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 1 3 4 5 6 8 9 10 11 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 1 2 4 5 6 7 9 10 11 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 1 2 4 5 6 7 8 10 11 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 1 2 3 5 6 7 8 9 10 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 1 2 8 5 6 7 8 9 10 39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 1 2 3 4 5 7 8 9 10 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 1 2 3 4 5 6 8 9 10 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 1 2 3 4 5 6 7 8 9 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 1 2 3 4 5 6 7 8 9 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 1 2 3 4 5 6 7 8 9 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 1 2 3 4 5 6 7 8 9 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 1 2 S 4 5 6 7 8 9 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 1 2 3 4 5 6 7 7 8 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 1 2 8 4 5 5 6 7 8 48 6812 6821 6830 6839 6848 6857 6S66 6875 6884 6893 1 2 3 4 4 5 6 7 8 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 1 2 3 4 4 5 6 7 8 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 1 2 8 3 4 5 6 7 8 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 1 2 3 3 4 5 6 7 8 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 1 2 2 3 4 5 6 7 7 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 1 2 2 3 4 5 6 6 7 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 1 2 2 3 4 5 6 6 7 MATHEMATICAL TABLES. 313 TABLE III. LOGARITHMS. 55 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 1 2 2 3 4 5 5 6 7 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 1 2 2 3 4 5 5 6 7 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 1 2 2 3 4 5 5 6 7 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 1 1 2 3 4 4 5 6 7 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 1 1 2 3 4 4 5 6 7 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 1 1 2 3 4 4 5 6 6 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 1 1 2 3 4 4 5 6 6 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 1 1 2 3 3 4 5 6 6 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 1 1 2 S 3 4 5 5 6 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 1 1 2 3 3 4 5 5 6 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 1 1 2 8 3 4 5 5 6 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 1 1 2 3 3 4 5 5 6 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 1 1 2 3 3 4 5 5 6 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 1 1 2 3 3 4 4 5 6 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 1 1 2 2 3 4 4 5 6 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 1 1 2 2 3 4 4 5 6 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 1 1 2 2 3 4 4 5 5 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 1 1 2 2 3 4 4 5 5 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 1 1 2 2 3 4 4 5 5 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 1 1 2 2 3 4 4 5 5 75 8751 8756 8762 8768 8774 8779 8785 87911 8797 8802 1 1 2 2 3 3 4 5 5 76 8808 8814 8820 | 8825 8831 8837 8842 8848 8854 8859 1 1 2 2 3 3 4 5 5 77 8865 8871 8876 ! 8882 8887 8893 8899 8904 8910 8915 1 1 2 2 3 3 4 4 5 78 8921 8927 8932 1 8938 8943 8949 8954 8960 8965 8971 1 1 2 2 3 3 4 4 5 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 1 1 2 2 3 3 4 4 5 80 9031 9036 9042 j 9047 9053 9058 9063 9069 9074 9079 1 1 2 2 3 3 4 4 5 81 9085 9090 9096 : 9101 9106 9112 9117 9122 9128 9133 1 1 2 2 3 3 4 4 5 82 9138 9143 9149 ! 9154 9159 9165 9170 9175 9180 9186 1 1 2 2 3 3 4 4 5 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 1 1 2 2 3 3 4 4 5 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 1 1 2 2 3 3 4 4 5 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 1 1 2 2 3 3 4 4 5 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 1 1 2 2 3 3 4 4 5 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 1 1 2 2 3 3 4 4 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 94S9 1 1 2 2 3 3 4 4 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 (J 1 1 2 2 3 3 4 4 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 1 1 2 2 3 3 4 4 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 1 1 2 2 3 3 4 4 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 1 1 2 2 3 3 4 4 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 1 1 o 2 3 3 4 4 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 1 1 2 2 3 3 4 4 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 1 1 2 2 3 3 4 4 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9868 1 1 2 2 3 3 4 4 97 9808 9872 9877 9881 9886 9890 9S94 9899 9903 9908 1) 1 1 2 2 3 3 4 4 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 1 1 2 2 3 3 4 4 99 9956 9961 9965 i 9969 9974 9978 9983 9987 9991 9996 (1 1 1 _> 2 3 3 3 4 314 PRACTICAL MATHEMATICS FOR BEGINNERS. TABLE IV. ANTILOGARITHMS. oo 1 2 3 4 5 6 7 8 9 1 2 3 1 4 5 6 7 8 9 1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 1 1 1 2 2 2 01 1023 1026 1028 1030 1033 1035 1038 1040 1042 1045 1 1 1 I 2 2 2 02 1047 1050 1052 1054 1057 1059 1062 1064 1067 1069 1 1 1 1 2 2 2 03 1072 1074 1076 1079 1081 1084 1086 1089 1091 1094 1 1 1 J 2 2 2 04 1096 1099 1102 1104 1107 1109 1112 1114 1117 1119 1 1 1 1 2 2 2 2 05 1122 1125 1127 1130 1132 1135 1138 1140 1143 1146 1 1 1 1 2 2 2 2 06 1148 1151 1153 1156 1159 1161 1164 1167 1169 1172 1 1 1 1 2 2 2 2 07 1175 1178 1180 1183 1186 1189 1191 1194 1197 1199 1 1 1 1 2 2 2 2 08 1202 1205 1208 1211 1213 1216 1219 1222 1225 1227 1 1 1 1 2 2 2 3 09 1230 1233 1236 1239 1242 1245 1247 1250 1253 1256 1 1 1 1 2 2 2 3 10 1259 1262 1265 1268 1271 1274 1276 1279 1282 1285 1 1 1 1 2 2 2 3 11 1288 1291 1294 1297 1300 1303 1306 1309 1312 1315 1 1 1 2 2 2 2 3 12 1318 1321 1324 1327 1330 1334 1337 1340 1343 1346 1 1 1 2 2 2 2 3 13 1349 1352 1355 1358 1361 1365 1368 1371 1374 1377 1 1 1 2 2 2 3 3 14 1380 1384 1387 1390 1393 1396 1400 1403 1406 1409 1 1 1 2 2 2 3 3 15 1413 1416 1419 1422 1426 1429 1432 1435 1439 1442 ] 1 1 2 2 2 3 3 16 1445 1449 1452 1455 1459 1462 1466 1469 1472 1476 1 1 1 2 2 2 3 3 IT 1479 1483 1486 1489 1493 1496 1500 1503 1507 1510 1 ] 1 2 2 2 3 3 18 1514 1517 1521 1524 1528 1531 1535 1538 1542 1545 1 1 1 2 2 2 3 3 19 1549 1552 1556 1560 1563 1567 1570 1574 1578 1581 1 ] 1 2 2 3 3 3 20 1585 1589 1592 1596 1600 1603 1607 1611 1614 1618 1 ] 1 2 2 3 3 3 21 1622 1626 1629 1633 1637 1641 1644 1648 1652 1656 1 1 2 2 2 3 3 3 22 1660 1663 1667 1671 1675 1679 1683 1687 1690 1694 1 1 2 2 2 3 3 3 23 1698 1702 1706 1710 1714 1718 1722 1726 1730 1734 1 1 2 2 2 3 3 4 24 1738 1742 1746 1750 1754 1758 1762 1766 1770 1774 1 1 2 2 2 3 3 4 25 1778 1782 1786 1791 1795 1799 1803 1807 1811 1816 II 1 1 2 2 2 3 3 4 26 1820 1824 1S28 1832 1837 1841 1845 1849 1854 1858 1 1 2 2 3 3 3 4 27 1862 1866 1871 1875 1879 1884 1888 1892 1897 1901 (1 1 1 2 2 3, 3 3 4 28 1905 1910 1914 1919 1923 1928 1932 1936 1941 1945 ] 1 2 2 3 3 4 4 29 1950 1954 1959 1963 1968 1972 1977 1982 1986 1991 1 1 2 2 3 3 4 4 30 1995 2000 2004 2009 2014 2018 2023 2028 2032 2037 II 1 1 2 2 3 3 4 4 31 2042 2046 2051 2056 2061 2065 2070 2075 2080 2084 1 1 2 2 3 3 4 4 32 2089 2094 2099 2104 2109 2113 2118 2123 2128 2133 (1 1 1 2 2 3 3 4 4 33 2138 2143 2148 2153 2158 2103 2168 2173 2178 2183 II 1 1 2 2 3 3 4 4 34 2188 2193 2198 2203 2208 2213 221S 2223 2228 2234 1 1 2 2 3 3 4 4 5 35 2239 2244 2249 2254 2259 2265 2270 2275 2280 2286 1 1 2 2 3 3 4 4 5 36 2291 2296 2301 2307 2312 2317 2323 2328 2333 2339 1 1 2 j 3 3 4 4 5 37 2344 2350 2355 2360 2366 2371 2377 2382 2388 2393 1 1 2 2 3 3 4 4 5 38 2399 2404 2410 2415 2421 2427 2432 2438 2443 2449 1 1 2 2 3 3 4 4 5 39 2455 2460 2466 2472 2477 2483 2489 2495 2500 2506 1 1 2 2 3 3 4 5 5 40 2512 2518 2523 2529 2535 2541 2547 2553 2559 2564 1 1 2 2 3 4 4 5 5 41 2570 2576 25S2 2588 2594 2600 2606 2612 2618 2624 ] 1 2 2 3 4 4 5 5 42 2630 2636 2642 2649 2655 2661 2667 2673 2679 2685 1 1 2 2 3 4 4 5 6 43 2692 2698 2704 2710 2716 2723 2729 2735 2742 2748 1 1 2 3 3 4 4 5 6 44 2754 2761 2767 2773 2780 2786 2793 2799 2805 2812 1 1 o 3 3 4 4 5 6 45 2818 2S25 2831 2838 2844 2851 2858 2864 2871 2877 1 1 2 3 3 4 5 5 6 46 2884 2891 2897 2904 2911 2917 2924 2931 2938 2944 1 1 2 3 3 4 5 5 6 47 2951 2958 2965 2972 2979 2985 2992 2999 3006 3013 J 1 2 3 3 4 5 5 6 48 3020 3027 3034 3041 3048 3055 3062 3069 3076 3083 1 1 2 3 4 4 5 6 6 49 3090 3097 3105 3112 3119 3126 313313141 3148 3155 1 1 2 3 4 4 5 6 6 MATHEMATICAL TABLES. 315 TABLE IV. ANTILOGAEITHMS. 50 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 3162 3170 3177 3184 3192 3199 3206 3214 3221 3228 1 1 2 3 4 4 5 6 7 51 3236 3243 3251 3258 3266 3273 3281 3289 3296 3304 1 2 2 3 4 5 5 6 7 52 3311 3319 3327 3334 3342 3350 3357 3365 3373 3381 I 2 2 8 4 5 5 6 7 53 3388 3396 3404 3412 3420 3428 3436 3443 3451 3459 1 2 2 8 4 5 6 6 7 54 3467 3475 3483 3491 3499 3508 3516 3524 3532 3540 1 2 2 3 4 5 6 6 7 55 3548 3556 3565 3573 3581 3589 3597 3606 3614 3622 1 2 2 3 4 5 6 7 7 56 3631 3639 3648 3656 3664 3673 3681 3690 3698 3707 1 2 3 3 4 5 6 7 8 57 3715 3724 3733 3741 3750 3758 3767 3776 3784 3793 1 2 3 8 4 5 6 7 8 58 3802 3811 3819 3828 3837 3846 3855 3864 3873 3882 1 2 3 4 4 5 6 7 8 59 3890 3899 3908 3917 3926 3936 3945 3954 3963 3972 1 2 8 4 5 5 6 7 8 60 3981 3990 3999 4009 4018 4027 4036 4046 4055 4064 1 2 3 4 5 6 6 7 8 61 4074 4083 4093 4102 4111 4121 4130 4140 4150 4159 1 2 3 4 5 6 7 8 9 62 4169 4178 4188 4198 4207 4217 4227 4236 4246 4256 1 2 3 4 5 6 7 8 9 63 ! 4266 4276 4285 4295 4305 4315 4325 4335 4345 4355 1 2 3 4 5 6 7 8 9 64 4365 4375 4385 4395 4406 4416 4426 4436 4446 4457 1 2 3 4 5 6 7 8 9 65 4467 4477 4487 4498 4508 4519 4529 4539 4550 4560 1 2 3 4 5 6 7 8 9 66 4571 4581 4592 4003 4613 4624 4634 4645 4656 4667 1 2 3 4 5 6 7 9 10 67 4677 4688 4699 4710 4721 4732 4742 4753 4764 4775 1 2 3 4 5 7 8 9 10 68 4786 4797 4808 4819 4831 4842 4853 4864 4875 4887 1 2 3 4 6 7 8 9 10 69 4898 4909 4920 4932 4943 4955 4966 4977 4989 5000 1 2 3 5 6 7 8 9 10 70 5012 5023 5035 5047 5058 5070 5082 5093 5105 5117 1 2 4 5 6 7 8 911 71 5129 5140 5152 5164 5176 5188 5200 5212 5224 5236 1 2 4 5 6 7 8 10 11 72 5248 5260 5272 5284 5297 5309 5321 5333 5346 5358 1 2 4 5 6 7 9 10 11 73 5370 5383 5395 5408 5420 5433 5445 5458 5470 5483 1 8 4 5 6 8 9 10 11 74 5495 5508 5521 5534 5546 5559 5572 5585 5598 5610 1 3 4 5 6 8 9 10 12 75 5623 5636 5649 5662 5675 5689 5702 5715 5728 5741 1 3 4 5 7 8 9 1012 76 5754 5768 5781 5794 5808 5821 5834 5848 5861 5875 1 3 4 5 7 8 9 11 12 77 5888 5902 5916 5929 5943 5957 5970 5984 5998 6012 1 3 4 5 7 8 10 11 12 78 6026 6039 6053 6067 6081 6095 6109 6124 6138 6152 1 3 4 6 7 8 10 11 13 79 6166 6180 6194 6209 6223 6237 6252 6266 6281 6295 1 3 4(6 7 9 10 11 13 80 6310 6324 6339 6353 6368 6383 6397 6412 6427 6442 1 3 4 6 7 9 10 1213 81 6457 6471 6486 6501 6516 6531 6546 6561 6577 6592 2 3 5 6 8 9 11 12 14 82 6607 6622 6637 6653 6668 6683 6699 6714 6730 6745 2 3 5 6 8 9 1112 14 83 6761 6776 6792 6808 6823 6839 6855 6S71 6887 6902 _> 3 5 6 8 9 11 13 14 84 6918 6934 6950 6966 6982 6998 7015 7031 7047 7063 2 3 5 6 8 10 11 13 15 '85 7079 7096 7112 7129 7145 7161 7178 7194 7211 7228 2 3 5 7 8 10 12 13 15 86 7244 7261 7278 7295 7311 7328 7345 7362 7379 7S96 2 3 B 7 8 10 12 13 15 87 7413 7430 7447 7464 7482 7499 7516 7534 7551 7568 2 3 5 7 9 10 12 14 16 88 7586 7603 7621 7638 7656 7674 7691 7709 7727 7745 2 4 6 7 9 11 12 14 16 89 7762 7780 7798 7816 7834 7852 7870 7889 7907 7925 2 4 5 7 9 11 13 14 16 90 7943 7962 7980 7998 8017 8035 8054 8072 8091 8110 2 4 6 7 9 11 13 15 17 91 8128 8147 8166 8185 8204 8222 8241 8260 8279 8299 2 4 8 9 11 13 15 17 92 8318 8337 8356 8375 8395 8414 8433 8453 8472 8492 2 4 6 8 10 12 14 15 17 93 8511 8531 8551 8570 8.590 8610 8630 8650 8670 8690 2 4 6 8 10 12 14 16 18 94 8710 8730 8750 8770 8790 8810 8831 8851 8872 8892 2 4 C 8 10 12 14 16 18 95 8913 8933 8954 8974 8995 9016 9036 9057 9078 9099 2 4 6 8 10 12 15 17 19 96 9120 9141 9162 9183 9204 9226 9247 9268 9290 9311 2 4 6 8 11 13 15 17 19 97 9333 9354 9376 9397 9419 9441 9462 9484 9506 9528 2 4 7 9 11 13 15*17 20 98 9550 9572 9594 9616 9638 9661 9683 9705 9727 9750 2 4 7 9 11 13 16 18 20- 99 9772 9795 9817 9840 9863 9886 9908 9931 9954 9977 2 5 7 9 11 14 16 18 20 316 PRACTICAL MATHEMATICS FOR BEGINNERS. TABLE V. Ingle. Chords. Sine. Tangent, Cotangent. Cosine. Deg. Radians. 00 1 1-414 1-5708 90 1 0175 017 0175 0175 57-2900 9998 1-402 1-5533 89 2 0349 035 0349 0349 28-6363 9994 1-389 1-5359 88 3 0524 052 0523 0524 19-0811 9986 1-377 1-5184 87 4 0698 070 0698 0699 14-3006 9976 1-364 1-5010 86 5 0873 087 0872 0875 11-4301 9962 1-351 1-4835 85 6 1047 105 1045 1051 9-5144 9945 1-338 1-4661 84 7 1222 122 1219 1228 8-1443 9925 1-325 1-4486 83 8 1396 139 1392 1405 7-1154 9903 1-312 1-4312 82 9 1571 157 1564 1584 6-3138 9877 1-299 1-4137 81 10 1745 174 1736 1763 5-6713 9848 1-286 1-3963 80 11 1920 192 1908 1944 5-1446 9816 1-272 1-3788 79 12 2094 209 2079 2126 4-7046 9781 1-259 1-3614 78 13 2269 226 2250 2309 4-3315 9744 1-245 1-3439 77 14 2443 244 2419 2493 4-0108 9703 1-231 1-3265 76 15 2618 261 2588 2679 3-7321 9659 1-217 1-3090 75 16 2793 278 2756 2867 3-4874 9613 1-204 1-2915 74 17 2967 296 2924 3057 3-2709 9563 1-190 1-2741 73 18 3142 313 3090 3249 3-0777 9511 1-176 1-2566 72 19 3316 330 3256 3443 2-9042 9455 1-161 1-2392 71 20 3491 347 3420 3640 2-7475 9397 1-147 1-2217 70 21 3665 364 3584 3839 2-6051 9336 1133 1-2043 69 22 3840 382 3746 4040 2-4751 9272 1-118 1-1868 68 23 4014 399 3907 4245 2-3559 9205 1-104 1-1694 67 24 4189 416 4067 4452 2-2460 9135 1-089 1-1519 66 25 4363 433 4226 4663 2-1445 9063 1-075 1-1345 65 26 4538 450 4384 4877 2-0503 8988 1-060 1-1170 64 27 4712 467 4540 5095 1-9626 8910 1-045 1-0996 63 28 4887 484 4695 5317 1-8807 8829 1-030 1-0821 62 29 5061 501 4848 5543 1-8040 8746 1-015 1-0647 61 30 5236 518 5000 5774 1-7321 8660 1-000 1-0472 60 31 5411 534 5150 6009 1-6643 8572 985 1-0297 59 32 5585 551 5299 6249 1-6003 8480 970 1-0123 58 33 5760 568 5446 6494 1-5399 8387 954 9948 57 34 5934 585 5592 6745 1-4826 8290 939 9774 56 35 6109 601 5736 7002 1-4281 8192 923 9599 55 36 6283 618 5878 7265 1-3764 8090 908 9425 54 37 6458 635 6018 7536 1-3270 7986 892 9250 53 38 6632 -651 6157 7813 1-2799 7880 877 9076 52 39 6807 668 6293 8098 1-2349 7771 861 8901 51 40 6981 684 6428 8391 1-1918 7660 845 8727 50 41 7156 700 6561 8693 1-1504 7547 829 8552 49 42 7330 717 6691 9004 1-1106 7431 813 8378 48 43 7505 733 6820 9325 1-0724 7314 797 8203 47 44 7679 749 6947 9657 1-0355 7193 781 8029 46 45 7854 765 7071 1-0000 1-0000 7071 765 7854 45 Radians. I Deg. Cosine. Cotangent. Tangent. Sine, Chords. Angle. BOARD OF EDUCATION. ELEMENTARY PRACTICAL MATHEMATICS. 1901. eight questions are to be answered. Two of these should be Nos. I and 2. 1. Compute 30-56-f 4105, 0-03056x0-4105, 4-105^, -04105- 2 *. The answers must be right to three significant figures. Why do we multiply log a by 6 to obtain the logarithm of a b ? 2. Answer only one of the following, (a) or (b) : (a) Find the value of ae~ bt sin (ct + g) if a = 5, 6 = 200, c = 600, g= -0-1745 radian, ='001. (Of course the angle is in radians. ) {b) Find the value of sin A cos B - cos A sin B if A is 65 and B is 34. 3. A tube of copper (0*32 lb. per cubic inch) is 12 feet long and 3 inches inside diameter ; it weighs 100 lb. Find its outer diameter, and the area of its curved outer surface. 4. ABC is a triangle. The angle A is 37, the angle G is 90, and the side AG is 5 "32 inches. Find the other sides, the angle B, and the area of the triangle. 5. An army of 5000 men costs a country 800,000 per annum to maintain it, an army of 10,000 men costs 1,300,000 per annum to maintain it, what is the annual cost of an army of 8000? Take the simplest law which is consistent with the figures given. Use squared paper or not, as you please. 6. In any class of turbine if P is power of the waterfall and H the height of the fall, and n the rate of revolution, then it is known that for any particular class of turbines of all sizes In the list of a particular maker I take a turbine at random for a fall of 6 feet, 100 horse-power, 50 revolutions per minute. By means of this I find I can calculate n for all the other turbines of the list. Find n for a fall of 20 feet and 75 horse-power. 7. At the following draughts in sea water a particular vessel has the following displacements : Draught h feet 15 12 9 6.3 Displacement T tons - 2098 1512 101 S 586 What are the probable displacements when the draughts are 11 i and 13 feet respectively ? 318 PRACTICAL MATHEMATICS FOR BEGINNERS. 8. The three parts (a), marks. (6), (c) must all be answered to get full (a) There are two quantities, a and b. The square of a is to be multiplied by the sum of the squares of a and b ; add 3 ; extract the cube root ; divide by the product of a and the square root of b. Write down this algebraically. (&) Express as the sum of two simpler fractions. (c) A crew which can pull at the rate of six miles an hour finds that it takes twice as long to come up a river as to go down ; at what rate does the river flow ? 9. A number is added to 2*25 times its reciprocal ; for what number is this a minimum ? Use squared paper or the calculus as you please. 10. If y = \x 2 - 3x + 3, show, by taking some values of x and calculating y and plotting on squared paper, the nature of the relationship between x and y. For what values of x is y = ? 11. The keeper of a restaurant finds that when he has G guests a day his total daily profit (the difference between his actual receipts and expenditure including rent, taxes, wages, wear and tear, food and drink) is P pounds, the following numbers being averages obtained by comparison of many days' accounts, what simple law seems to connect P and G ? G P 210 270 320 360 -09 + 1-8 + 4-8 + 64 For what number of guests would he just have no profit ? 12. At the end of a time t seconds it is observed that a body has passed over a distance s feet reckoned from some starting point. If it is known that s '25 + \50t-5t 2 what is the velocity at the time tl Prove the rule that you adopt to be correct. If corresponding values of s and t are plotted on squared paper what indicates the velocity and why ? 13. The three rectangular co-ordinates of a point P are 2*5, 3'1 and 4. Find (1) the length of the line joining P with the origin, (2) the cosines of the angles which OP makes with the three axes, and (3) the sum of the squares of the three cosines. ELEMENTARY PRACTICAL MATHEMATICS. 1902. 1. Compute by contracted methods without using logarithms 23 07x01354, 2307 -=-1354. Compute 2307 065 and 2307" 1 " 5 using logarithms. The answers to consist of four significant figures. Why do we add logarithms to obtain the logarithm of a product ? 2. Answer only one of the following (a) or (6) : (a) If w=lU{p 1 (l+\ogr)-r{p 8 +\0)} and if ^ = 100, p 3 =\7 ; find w for the four values of r, 1, 2, 3, 4. Tabulate your answers. \b) If c is 20 feet, D = Q feet, d = S feet, find 6 in radians if . D + d sind== -2c- Now calculate L the length of a belt, if L ^ D+d il +e+ tL}- 3. The three parts (a), (6), and (c) must all be answered to get full marks. (a) Let x be multiplied by the square of y, and subtracted from the cube of z, the cube root of the whole is taken and is then squared. This is divided by the sum of x, y, and z. Write all this down alge- braically. x 13 (b) Express , as the sum of two simpler fractions. x 2 - 2x - 15 (c) The sum of two numbers is 76, and their difference is equal to one-third of the greater, find them. 4. What is the idea on which compound interest is calculated ? Explain, as if to a beginner, how it is that A = p ( 1+ mT where P is the money lent, and A is what it amounts to in n years at r per cent, per annum. If A is 130, and P is 100, and n is 7 '5, find r 320 PRACTICAL MATHEMATICS FOR BEGINNERS. 5. Suppose s the distance in feet passed through by a body in the time of t seconds is s= 10t 2 . Find s when t is 2, find s when t is 2*01, and also when t is 2*001. What is the average speed in each of the two short intervals of time after t = 21 When the interval of time is made shorter and shorter, what does the average speed approxi- mate to ? 6. If z = ax-by' d x*. If 2=1*32 when x=l and y = 2 ; and if 2 = 858 when #=4 and y\ ; find a and 6. Then find 2 when x=2 and y=0. 7. A prism has a cross- section of 50*32 square inches. There is a section making an angle of 20 with the cross-section : what is its area ? Prove the rule that you use. 8. In a triangle ABG, AD is the perpendicular on BG ; AB is 3*25 feet ; the angle B is 55. Find the length of AD. If BG is 4*67 feet, what is the area of the triangle? Find also BD and DG and A G. Your answers must be right to three significant figures. 9. It is known that the weight of coal in tons consumed per hour in a certain vessel is 0'3 + - 001^ where v is the speed in knots (or nautical miles per hour). For a voyage of 1000 nautical miles tabulate the time in hours and the total coal consumption for various values of v. If the wages, interest on cost of vessel, etc. , are repre- sented by the value of 1 ton of coal per hour, tabulate for each value of v the total cost, stating it in the value of tons of coal, and plot on squared paper. About what value of v gives greatest economy ? 10. An examiner has given marks to papers ; the highest number of marks is 185, the lowest 42. He desires to change all his marks according to a linear law converting the highest number of marks into 250 and the lowest into 100 ; show how he may do this, and state the converted marks for papers already marked 60, 100, 150. Use squared paper, or mere algebra, as you please. 11. A is the horizontal sectional area of a vessel in square feet at the water level, h being the vertical draught in feet. A h 14,850 14,400 13,780 13,150 23 6 20-35 17-1 146 Plot on squared paper and read off and tabulate A for values of h, 23, 20, 16. If the vessel changes in draught from 20*5 to 19 "5, what is the diminution of its displacement in cubic feet ? 12. Find a value of x \yhich satisfies the equation a; 2 -51og 10 #-2'531=0. 13. If cc = a(0-sin0) and y = a(l-cos0), and if a = 5; taking various values of

) where is always 10, find x when 6 has the values 30, 40, 50, 60, and plot the values of x and of 6 on squared paper. About what value of 6 seems to give the largest value of x ? (6) At speeds greater than the velocity of sound, the air resistance to the motion of a projectile of the usual shape of weight w lb., diameter d inches, is such that when the speed diminishes from v 1 feet per second to v, if t is the time in seconds and s is the space passed over in feet, * = 7,000 d*\v v 1 ) i s = 7,000|logA If v 1 is 2,000, find s and t when v= 1,500 for a projectile of 12 lb. whose diameter is 3 inches. (c) Find the value of *-4-***+Ji(i-*) if t x = 458, t 3 = 373, l x = 796 - -695 t v P.m b x 322 PRACTICAL MATHEMATICS FOR BEGINNERS. 3. The four parts (a), (6), (c), and (d) must all be answered to get full marks. (a) Write down algebraically : Add twice the square root of the cube of x to the product of y squared and the cube root of z. Divide by the sum of x and the square root of y. Add four and extract the square root of the whole. (b) " x^-Sx-4 as the sum of two simpler fractions. (c) Find two numbers such that if four times the first be added to two and a half times the second the sum is 17*3, and if three times the second be subtracted from twice the first the difference is 1 *2. {d) In a triangle ABC, G being a right angle, AB is 14*85 inches, AG is 8*32 inches. Compute the angle A in degrees, using your tables. 4. The following are the areas of cross section of a body at right angles to its straight axis : A in square inches - 250 292 310 273 215 180 135 120 x inches from one end 22 41 70 84 102 130 145 Plot A and x on section at x = 50 ? volume ? squared paper. What is the probable cross What is the average cross section and the whole 5. The following table records the heights in inches of a girl A (born January, 1890) and a boy B (born May, 1894). Plot these records. The intervals of time may be taken as exactly four months. Year 1900. 1901. 1902. 1903. Month Sept. Jan. May. Sept. Jan. May. Sept. Jan. A 54 8 55 6 56-6 58-0 59 2 60-2 60'9 61*3 B 48-3 49'0 49-8 50-6 51-5 52-3 53 1 53-9 Find in inches per year the average rates of growth of A and B during the given period. At about what age was the growth of A most rapid ? State this rate ; divide it by her average rate. 6. In any such question as Question 5, where points on a curvt have coordinates like h (height) and t (time), show exactly how it that the slope of a curve at a point represents there the rate growth of h as t increases. EXAMINATION PAPER. 323 7. Find accurately to three significant figures a value of x which satisfies the equation 2a; 2 -101og 10 a?-3-25=0. 8. Answer only one of the following {a) or (6) : (a) A cast-iron flywheel rim (0*26 lb. per cubic inch) weighs 13,700 lbs. The rim is of rectangular section, thickness radially x, size the other way 1 '6x. The inside radius of the rim is 14a;. Find the actual sizes. (6) The electrical resistance of copper wire is proportional to its length divided by its cross section. Show that the resistance of a pound of wire of circular section all in one length is inversely pro- portional to the fourth power of the diameter of the wire. 9. It is thought that the following observed quantities, in which there are probably errors of observation, follow a law like y = ae bx . Test if this is so, and find the most probable values of a and b. X 230 310 4 00 4-92 5 91 7 20 y 33 39 1 50 3 67 2 85-6 125 10. Plot 3y = 4 Sx + 0-9 Plot y =2-24-0 -7x. Find the point where they cross. What angle does each of them make with the axis of x ? At what angle do they meet ? 11. A firm is satisfied from its past experience and study that its expenditure per week in pounds is 120 + 3-2a; + -^ + 0-0lC, a; + 5 where x is the number of horses employed by the firm, and C is the usual turnover. If C is 2,150 pounds, find for various values of x what is the weekly expenditure, and plot on squared paper to find the number of horses which will cause the expenditure to be a minimum. 12. Assuming the earth to be a sphere, if its circumference is 360 x 60 nautical miles, what is the circumference of the parallel of latitude 56 ? What is the length there of a degree of longitude ? If a small map is to be drawn in this latitude, with north and south and east and west distances to the same scale, and if a degree of latitude (which is of course 60 miles) is shown as 10 inches, what distance will represent a degree of longitude ? 13. At a certain place where all the months of the year are assumed to be of the same length (30*44 days each), at the same 324 PRACTICAL MATHEMATICS FOR BEGINNERS. time in each month the length of the day (interval from sunrise to sunset in hours) was measured, as in this table. Nov. Dec. Jan. Feb. Mar. April. May. June. July. 8-35 7-78 8 35 9-87 12 14-11 15 65 16-22 15 65 What is the average increase of the length of the day (state in decimals of an hour per day) from the shortest day which is 7*78 hours to the longest which is 16 "22 hours? When is the increase of the day most rapid, and what is it ? 14. At an electricity works, where new plant has been judiciously added, if W is the annual works cost in millions of pence, and T is the annual total cost, and U the number of millions of electrical units sold, the following results have been found : u W T 0-3 0-47 0-78 1-2 103 1-64 2-3 1-70 2-73 3-4 2-32 3-77 Find approximately the rule connecting T and W with U. Also find the probable values of W and T when U becomes 5, if there is the same judicious management. PRACTICAL MATHEMATICS. 1904. STAGE I. Answer questions Nos. 1, 2 and 3 and five others. 1. The three parts (a), (b) and (c) must be answered to get full marks. (a) Compute by contracted methods to four significant figures only, and without using logarithms, 3*405 x 9123 and 3*405-r 9'123. (6) Compute, using logarithms, V2*354x 1*607 and (32-15) 1 " 52 . (c) Write down the values of sin 23, tan 53, log 10 153*4, log e 153*4. 2. Both (a) and (6) must be answered to get full marks. (a) If F=EIir* + 4J?, If /=&*-*- 12, If E=S x 10 7 , 7T = 3'142, 1 = 62, b = 2, t = 0% find F. (b) Two men measure a rectangular box ; one finds its length, breadth, and depth in inches to be 5 32, 4*15, 3 29. The other finds them to be 5*35, 4 '17, 3 33. Calculate the volume in each case ; what is the mean of the two, what is the percentage difference of either from the mean ? 3. All of these (a), (b) and (c), must be answered to get full marks. (a) Write down algebraically : Square a, divide by the square of b, add 1, extract the square root, multiply by w, divide by the square of n. (/>) The ages of a man and his wife added together amount to 72*36 years ; fifteen years ago the man's age was 2*3 times that of his wife ; what are their ages now ? (c) ABO is a triangle, C being a right angle. The side AB is 15*34 inches, the side BO is 10*15 inches. What is the length of AC? Express the angles A and B in degrees. What is the area of the triangle in square inches ? If this is the shape of a piece of sheet brass 0*13 inch thick, and if brass weighs 0*3 lb. per cubic inch, what is its weight ? 4. If y = 3# 2 -201og 10 a*-7-077, find the values of y when a: is 1 *5, 2, 2*3. Plot the values of y and x on squared paper, and draw the probable curve in which these points lie. State approximately what value of x would cause y to 326 PRACTICAL MATHEMATICS FOR BEGINNERS. 5. It has been found that if P is the horse power wasted in air friction when a disc d feet diameter is revolving at n revolutions per minute, P=cd 5 V 5 . If P is 0*1 when tZ = 4 and ?i = 500, find the constant c. Now find P when d is 9 and n is 400. 6. There is a district in which the surface of the ground may be regarded as a sloping plane ; its actual area is 3 '246 square miles ; it is shown on the map as an area of 2*875 square miles ; at what angle is it inclined to the horizontal ? There is a straight line 20 17 feet long which makes an angle of 52 with the horizontal plane ; what is the length of its projection on the horizontal plane ? 7. A British man or woman of age x years may on the average expect to live for an additional y years. Age x. Expected further Life y. Man. Woman. 70 60 50 40 30 8*27 1314 18-93 25 30 3210 8-95 14-24 20-68 27 46 34-41 Plot a curve for men and one for women, and find the expectations of life for a man and for a woman aged 54 years. 8. The following tests were made upon a condensing- steam- turbine-electric-generator. There are probably some errors of observation, as the measurement of the steam is troublesome. The figures are given just as they were published in a newspaper. Output in Kilowatts K, - - - 1,190 995 745 498 247 Weight W\b. of steam consumed per hour, 23,120 20,040 16,630 12,560 8,320 4,065 EXAMINATION PAPER. 327 Plot on squared paper. Find if there is a simple approximate law connecting K and W, but do not state it algebraically. What are the probable values of K when W is 22,000 and when W is 6,000? 9. If y= 2x +, * X for various values of x, calculate y ; plot on squared paper ; state approximately the value of x which causes y to be of its smallest value. 10. A series of soundings taken across a river channel is given by the following table, x feet being distance from one shore and y feet the corresponding depth. Draw the section. Find its area. * 1 10 16 23 30 38 43 50 55 60 70 75 80 y 5 10 13 14 15 16 14 12 8 6 4 3 11. The value of a ruby is said to be proportional to the \\ power of its weight. If one ruby is exactly of the same shape as another, but of 2*20 times its linear dimensions, of how many times the value is it? [Note that the weights of similar things made of the same stuff are as the cubes of their linear dimensions.] 12. x and t are the distance in miles and the time in hours of a train from a railway station. Plot on squared paper. State how the curve shows where the speed is greatest and where it is least. What is the average speed in miles per hour during the whole time tabulated ? t 05 10 15 2 25 3 35 40 45 5 X 25 1-00 3 05 5-00 5-85 610 6*10 6 35 7*00 7 65 PRACTICAL MATHEMATICS. 1905. STAGE I. Answer questions Nos. 1, 2 and 3, and five others. 1. The three parts (a), (&) and (c) must all be answered to get full marks. (a) Compute by contracted methods to four significant figures only, and without using logarithms, 12*39 x 5*024 and 5 024-f 12*39. (b) Compute, using logarithms, #2*607 and 26-07 1 ' 13 . (c) Write down the values of cos 35, tan 52, sin -1 0*4226, log 10 14*36, log e 14*36. [Note. sin -1 ?i means the angle whose sine is n.] 2. The three parts (a), (6) and (c) must all be answered to get full marks. (a) If x=a(- sin MH =-~> where a* =20, a = 18, 7=169, * = 13*3, tt = 3*14. Use the tables provided. 5. Construct an equilateral triangle whose area shall be 3 times that of a given equilateral triangle, explaining every step in your work. 6. Give some method of finding the formula for the area of a circle whose radius is r. What is the circumference of a circle whose area is 1 acre? (tt = 3*1416). 7. The tangent of one acute angle is 7, and the sine of another is 0*7 ; find graphically the cosine of the difference between the angles, explaining the constructions and measurements which you make. Check your result by measuring the difference of the angles with your protractor and finding its cosine from the tables. EXAMINATION PAPER. 333 8. Solve completely a right-angled triangle in which a=68 07, ^4=39. Show that A, the area of the triangle, may be found by the formula log 2A = 2 log a + log cot A. 9. Prove the formula . tf + ct-a? COsA = 26c - and use it to find to the nearest degree the largest angle of a triangle of which the sides measure 3, 4, and 6 inches. Construct a triangle of this shape with its longest side equal to 2*4 inches; measure its angles with your protractor, and check by adding the results. 10. Taking rectangular axes, plot off the points ( - 1, 2) and (3, 4), and draw the line represented by 2x-y-3=0. Find the co-ordinates of the point on the given line which is equidistant from the given points. 11. Find the equation of the circle which passes through the points ( - 1, 2), (3, 4), and has its centre on the line 2x-y-3=0. Give a diagram. Prove that the tangents to this circle which cut the axis of x at 45 are represented by x-y-l 2^5 = 0. ANSWERS. Exercises I., p. 3. 1. 124 971046. 2. 26-010801. 3. 706-42724. 4. 38-732229. 5. 280-68054. 6. 1290-657788. 7. 332-72973. 8. 32-04147. 9. 107 060597. 10. 472-979307. 11. 100-610704. 12. 98-0246457. 13. 1-15S55. 14. 11-200568. 15. 05444. 16. 101-68787. Exercises II., p. 11. 1. 0-34118. 2. 014955. 3. 501-8551. 4. 0312034. 5. 6248501. 6. 2074272. 7. 756-872. 8. 5-329956. 9. 5-20163. 10. 2-824575. 11. -1481883. 12. 4-41063. 13. 3349313. 14. 183-6587. 15. 049265. 16. 10-84589. 17. 1-5581. 18. 1-15421. 19. 73-93787. 20. 6-955714. 21. 2-114. 22. 0560682. 23. 2-332714. 24. 014056093. 25. 189. 26. 3472. 27. 8-304 pence. 28. 33-7708hrs. 29. 37-072 lbs. 30. 4621 -32 ft. 31. 8s. lid. 32. 75, 33. 325. 34. 759-725. Exercises III., p. 14. 1. 00198. 2. 02665. 3. 575. 4. 30-16. 5. 470. 6. 012. 7. '0645. 8. 296000. 9. 00545. 10. 0125. 11. -00892. 12. 01733. 13. 846; rem r , 0047 ft. 14 . 6-453. 15. 34118, -01733. 16. 29 7. 17. 17404. 18. 1217 6. 19. 53 05. 20. 563-54. 21. (i) 3-123, (ii) 1704. 22. (i) '01495, (ii) -007529. Exercises IV., p. 20. 1. 485 miles. 2. 7^. 3. 224 ; 240, 350. 4. 22 cwt. 2 qrs. 5. -6525. 6. 7, 11. 13s. 4d., 16. 6s. 8d., 21. 7. 59^, 68, 76f. 8. 5. 9. 7173. 6s. Sd., 8070, 8608, 8966. 13s. 4d. 11. 126. 2s. Od. 12. 5yy miles. 13. 8. 6s. Sd. Exercises V., p. 23. 1. 543-9 lbs., 923-5 lbs. copper, 76*5 lbs. tin. 2. 37 %, 7-4 %, 88-9% ; 462 lbs., 8-8 lbs., 13'4 lbs., 177 '8 lbs. 3. 5s. Sd. 4. 80. 5. 70. 6. 2,825,761, 2,560,000. ANSWERS. 7. Gained 11 6 per cent. 8. 37. 10s. 9. 72 percentage of beer. 10. 2a\ 11. 1080 candidates, 432 failures. 12. 35 per cent. 13. 10. 18s. 9d.; 32* per cent. 14. 2000. Exercises VI. , p. 30. 1. 193. 2. 222. 3. 1003. 4. 4321. 5. 11 '05. 6. 8 0623,7 0711, 2-828428. 7.57 13. 8.671'3. 9. 6 '25 ; 20002. 10. 300-03. 11. 82929. 12. 9 99. 13. 206. 14. '0708. 15. 4321. 16. 32-94. 17. 237 96. 18. ft. 19. (i) -73, (ii) -85, (iii) -9, (iv) 1-12. Exercises VII., p. 55. 1. 8-66". 2. 28 ft. 3. 33-11 ft. 4. 4*9 ft., 9*8 sq. ft. 5. 5-3 miles. 9. 6 91. 11. 104 '5, 46 "5, 29, 1 -38". 13. 151, 16. 14. 4-16. 15. 48 8'. 16. 1-115, 109, 34. 17. 10,6. 18. 29 56'. 19. 34 8', 4114 sq. ft. 20. (i) 23 69; (ii) 1147 sq. ft. ; (iii) 22, 126. 21. 2-6624 ft.; 6*217 sq. ft.; 1864; 2'806; 3'868. 22. 36 2', 53 56', 90 2'. 23. 9196. 24. 252, 1'92. Exercises VIII., p. 61. 1. 20. 2. 0. 3. 0. 4. 3. 5. 27. 6. 1*058. 7. 172800. 8. 4022. 9. 0. 10. 2. 11. lj. 12. -2. 13. 1. 14. 3. Exercises IX., p. 63. 1. la + lb + ic. 2. 3ax 2 -3bx 2 . 3. 16m- lira. 4. 5a + 76-6c; 1. 5. 2x + 3y. 6. 36. 7. Sax 2 - x 2 - dx 2 - 2x + bx -/. 8. 8a + 26 + 4c; 2. 9. x^ + y^-xh/ 2 . 10. 26xy - 5x 2 - 5y 2 . 11. x*-2\x 2 + 2Q. 12. 2xy + x-x 2 + y 2 + Q6; 2666. 13. 8x -4y; 8. Exercises X., p. 65. 1. 3a + 6 -c. 2. 3a -106. 3. 3a; 2 -8a-+8. 4. 106. 5. ax + cy. 6. ax -ex -ay- cy. 7. 3m -n- 2p. 8. xy - xz. 9. a + 36-c + 3rf + 4e. 10. 3y 2 + 7xy- llxz + z 2 . 11. a + 26 + 3c + 4d; 2a+26 + 2c + 2d. 12. 2c-a-6 + a*; x 2 -y 2 . 13. 2a 3 -3a 2 6. 14. x*y+ 12x 2 y 2 +10xy 3 + 21y*. 15. 2a 4 + 3a 3 6 + 3a 2 6 2 + 2a6 3 + 6 4 . 16. 2ar* + 31a.r 2 - 31a 2 * + 7a 3 . 17. 44a6 + 33aaj + 24cy + 43ez. Exercises XI., p. 68. 1. a^ + a^ + a 4 . 2. 4a 6 + 1 la 4 6 2 + 7a 2 6 4 - 6 6 . 3. x* + xy + y*. 4. a 6 -21^+20. 5. x*- S^ + Uy*. 6. x 6 - 24x* + lUx 2 - 256. 336 PRACTICAL MATHEMATICS FOR BEGINNERS. 7. a 12 - 3a 10 6 2 - 2a 8 6 4 + 13a 6 6 6 - 3a 4 6 8 - 12a 2 6 10 + 66 12 . 8. xP-y*. 9. 8a 5 6-26a 3 6 3 + 2a6 5 . 10. a 6 - a 2 6 4 - a 4 6 2 + aV + 6 6 - a 4 c 2 - 6 2 c 4 + 6 4 c 2 - c 6 . 11. l-y 2 -y 3 -y 4 -y 5 + y 6 +2y 7 + y 8 . 12. 6X 8 - lla^ + 22ar*-4a^-7. 13. ar 5 * 43^ + 48a; -32. 14. 16a 4 -72a 2 6 2 +816 4 . 15. - 13a 3 - 22a 2 + 96a +135. Exercises XII., p. 70. 1. 2a 3 -3a 2 + 2a. 2. x + 2y-z. 3. a + b + c. 4. 2^-6x 2 y + 18rcy 2 -27y 3 . 5. ^*. 6. 3a + 26-c. x + y 7. 5a + 6-2c. 8. -9a6 4 c 2 . 9. 4xy + 2y + Sx + 1. 10. 3a- 5 -2a -9. 11. 2a 2 -26 2 + 3c 2 . 12. a 2 -a+l. 13. <*-Wbc + lW. 14. (i) *Z* (ii) * z + *>*+ I*- ^ K+&- f : Exercises XIX., p. 98. 1. $. 2. 16, 2. 3. T 7 7 . 4. 72. 5. j%. 6. 13, 10. 7. Jf . 8. 9, 15. 9. l| hrs. 10 Horse costs 25 ; cow 18. 11. f. 12. ^4 27, 22. 13. 2:5. 14. 10 gallons. 15. i*-v?=2f8. 16. 90 : 89. 17. 1250000 ; 128048. 15*. Id. Exercises XX., p. 105. 1. A by 1. 5s. OeZ. 2. f . 3. 5^ hours. 4. 16 days. 5. 25 lbs. per sq. in. 6. 15. 7. 4*5. 8. 15. 9. 18. 10. 35 days. 11. 1000. 12. 9. 14. 15012. P.M.B. Y ' 338 PRACTICAL MATHEMATICS FOR BEGINNERS. Exercises XXI., p. 112. 1. a-rb 2 + c 3 -Sahh. 2. (i) 21*656; (ii) 17 656. 3. A 4. a~h^. 5. 2a. 6. a^ p+x K 7. a 12 < m ~ n >. 8. x 4 < m -4:X zm + n + 6x* m +* n -4:X m+in + x 4 ' n . 9. '000024. 10. -0015 per cent. 11. 1-001, 992, '00162%. 12. '0008%. 13. 42172, 2-3713. 14. 1.006, 999. 15. 1-2432, 1-6548,2-3758. 16. * y- 1 2 17. a^&W 18. 2"*ab 2 c m . 19. 6*. 20 21. (i) a- 1 ^, (ii) a V"- 1 + cfix~% + a~*x$ + ofM . 22. 5-44. 23. ar 1 ^. 24. 12-127. Exercises XXII., p. 119, 1. -929, 8-361. 2. 1011 '68. 3. 836113. 4. -2019. 5. -645137. 6. 10. 7. latter, '033. Exercises XXIII., p. 124. 1. 4-167. 2. 48tol. 3. 97-25 lbs., 145 9 lbs. 4. 36-4 c.c. 7*5. 5. *729 inches. 6. 29-92 in., 339 ft., 14 7 lbs., 2116, 1034. 7. 4-15kilog. 8. 719-6. 9. 92 9 tons. 10. 1'4. 11. '72. Exercises XXIV., p. 130. 1. -0007736. 2. -00001573. 3. -07502. 4. 34-67. 5. -2025. 6. 2 583, 000744. 7. -01374. 8. 4 08. 9. 78*77. 10. (i) -1097; (ii) 973-6; (iii) '09761; (iv) '00007381. 11. (i) 157'8; (ii) 416-8. Exercises XXV., p. 132. 1. 4-799. 2. 00025, 250000. 3. -000009687. 5. 165000. 6. (i) 1262, (ii) '8042. 7. -0006398. 8. '02665. 9. -06039. 10. (i) 50*67; (ii) '0004511. 11. (i) 1-285, (ii) 33-29, (iii) 53'32. 12. (i) 3*468; (ii) 346 8. 13. (i) -6797, (ii) 67-97. 14. -624. 15. -1394,2-283. 16. -01496, '00753. 17. 7 446, '01254. Exercises XXVI., p. 140. 1. -00917. 2. 1-078. 3. 0001404. 4. 1-4779,1-6797. 5. 1-6796. 6. 0*5611. 7. 11999. 8. 1*3865. ANSWERS. 9. -003176. 10. -6869. 11. T'4577, 3-0701. 12. 2095. 13. 3546. 14. I'565xl0 7 . 15. 2311. 16. 9'2xl0. 17. -4055, -6931, "9163, 1-0986, 1*2528, 13863, 1-6041, 16094. 18. 5 435. 19. 6-575. 20. 1'948. 21. 4409. 22. 2928. 24. 263 3, -2353. 25. -17 75. 26. 2682. 27. 567. 28. a=-571, 6 = 26*63, 671'2. 29. 5-228, 1-222, '4956, '2563, '1665. 30. 04801; (ii) '2869; (iii) '2291. 31. 20*78. 32. 02147. 33. 2 885. Miscellaneous Exercises XXVII., p. 142. 1. 2-865. 2. 88-1. 3. 1658. 4. (i) -894 ; (ii) -891. 5. 33-64, 6-995. 6. 6 -686. 7. 2892. 8. 1588. 9. 1-443. 10. 1015. 11. 9035. 12. 7578. 13. 98-51. 14. (i) 1503, a =243 '9, 6 = -26-6, y = 671'86. Exercises XXVIII., p. 153. 1. 13-75 ft. 2. The former. 3. 5 '73. 4. |, 6 22'. 5. 108. 6. 17 '19. 7. -3708. 8. 2-36, 135. 9. 1 foot. Exercises XXIX., p. 162. 43 ft 9 9 99 . 2 njE o o 5- I> ^T> V- 7. 2-38 in. 8. 2'64 in., 5-014 sq. in. 9. 2-843, 1-991. 10. S6-52', 96 sq. ft. Exercises XXX., p. 167. 1. 1-4603, -5371. 2. -8102. 3. 7903. 4. 1-8492, -6141. 5. -419. 6. -6362. 7. 2057, '4429. 8. -0887. 9. -248. 10. -8461. 11. 1-15, 19918, 2-2216, 22216, 1 9918, 1-6261, 2-2. 12. - -4317. 13. 30140. 14. 336. 15. -1526, 1088. 16. 2007. 17.- -1387. 18. 7-718. 19. -02076. Exercises XXXI., p. 169. 1. 117-7. 2. 488-5. 3. 43-3. 4. 120. 6. 1-225 miles. 7. 10 '62 miles. 8. 173 2 ft. 9. 732*1. 10. 12-13 ft. 11. 8 869 miles. 12. 151 '5 ft. 13. 8768 yds. 14. 3960. 15. 1034 ft. 16. 3'18 miles per hour. Exercises XXXII., p. 185. 1. #=-0427i? + 4-4 ; 100 lbs. 2. (i) E= -118i?+ 1 -84, F= -0736i?+ 1 -83 ; (ii) E= -042/?+ -35, F=2R + 25; (iii) ^=-118/? + 1-75, ^=-077^ + 1-75. P.M.B. Y2 4 3, -1376. 10. 1, 2, 4. 2-22. 14. w = l*08. 1-2953. 17. 225. 20. -2,5-898, -3-898. 340 PRACTICAL MATHEMATICS FOR BEGINNERS. 3. (i)rc = 2-02 logiV-4'14; (ii) rc=2*33 logi*/-4*79; (iii) w=2*32 log ^-4*47. 4. 795-8 lbs. per hour. 5. B= -0208^ + 6-3, '84%. 6. ilf =l*42 + 4*66iV. 7. (i) a=*041, b= 173; (ii) a=119, 6 = 45*7, error 2*6%. 8. L = 1 -49 7' 2 + -537. 9. a = 8 '8, b = - 14. 10. a = 2500, 6 = 26, JT=2500 + 26P, W =4320, TT~P=76, 51, 61*7. 11. (i)d=*75*+*48"; (ii)d=l-2V; (iii) '67, *79, '9. 12. 1 -7d + -23 in. , A = *6d 2 , A = ( '593d 2 - -3) sq. in. 13. C=*344, w = l-79. Exercises XXXIII., p. 198. 1. 38d.,59d. 2. -39 2, 1-47", 2-25". 3. 13'77. 5. 22-1, 38-2, 63-3, 27*8, 0*5 million per annum. 6. 4" is 68 -3s., 5" is 91 65s. ; 63^s. 7.2-23,3-22. 8.20 06,-1-86. 9. 11. 218. 12. 211. 13. 15. (i) -3594; (ii) 1*4435. 16. 18. -3, -1, 4. 19. -4, -13, 17. Exercises XXXIV., p. 214. 1. 5 + 4'2, 26. 2. =255 when t is 5, aver, vel., 82*007, actual 82. 3. 81-8, 8002, 80*0002, actual speed 80. 4. 3*15, when r=0*5. 5. Aver, force = 3535 lbs. ; work = 3535x70 =247450 ft. lbs. 6. 104, 10-004, 10-0004, 10. 7. aver. val. = 1924. 8. 16-32, or -1-376. 9. 5491. 10. 235. 11. 8 and 4. 12. 2*23, 3 22, aver, value 0-57. 13. Rate of increase 2'014, aver, value = 10*08. 14. 0-1558, 1*902. 15. 0*30056, 1*785. 16. {\)nax n ~ l ' y (ii) 5a^ + ibx~ s +pcxP~ l 17. 210, 210. 19. 5, 15. Exercises XXXV., p. 219. 1. 10*87 sq. ft, 2. 13| ft. 3. 488*87 ft. 6. 60 ft. 7. 600*3 sq. ft. 8. 8 and 6. 9. 21*82 sq. ft. 10. 480. 11. 4. 7s. 6cZ. 12. 1 ft. 6 in. 13. 2376*9. Exercises XXXVI., p. 222. 1. 6*186 sq. ft. 2. 84 sq. ft. 3. 210 sq. in. 5. 60 sq. yds. 6. 2390. 7. 3000 sq. ft. 8. 150, 200, 250, 45,000 sq. yds. 9. 270 sq. ft. 10. *538 sq. ft. 11. 15 ac. 12. 1*155 miles, *2421 sq. miles. ANSWERS. 341 Exercises XXXVII., p. 223. 1. 5, -75, 1-5, 3-75, 5*499, 40 75. 2. 5 498, 7*854, 14-92, 25 13, 95 82, 212 058. 3. 22 ft. 7 434 in. 4. 4967. 5. 26400, 6 365 ft. 6. 63 65, 58 "76. 7. 5f miles. 8. 180. 9. 1-91 ft., 2-228 ft. 10. 5712 ft. Exercises XXXVIII., p. 227. 1. 64 in. 2. 3820 sq. in. 3. 4854 sq. in. 4. (i)-944;(ii)-004;(iii)-02;(iv)-2. 5. (i) "003218 ;(ii) -00933; (iii) 8*553. 6. 2. 18s. 10*9d. 7. 140*3 sq. ft. 8. 11385*3 sq. ft. 9. 13-36 sq. in. 10. 7 '658 sq. in. 11. 43 '5 in. 12. 82-47 sq. ft. 13. 488 '9 sq. ft., 64 ft. 14. *982 sq. ft. 15. 524-8 sq. in. 16. 1472 sq. ft. 17. 56 ft. 8 in. 18. 65-2 ft. 19. 196 ft. 20. 102*09 sq. ft. 21. 2065 03 sq. ft. Exercises XXXIX., p. 239. 1. J x5 2 = 19*635, Simpson's 19*45, error 0*8%. 2. 12797 cub. ft. 3. 2720 lbs. 4. 80*2 sq. yds. 5. 236 lbs. 6. 49988 sq. in., 320 4 in. 7. 58"2, 58*92. 8. 375*2 sq. ft. 9. 674*08 sq. ft. 10. 2853*9 sq. ft. 11. 2794*7 sq. ft. 12. 923 3 sq. ft., 15*39 ft. Exercises XL., p. 244. 1. 2l ft., 13s. lid. 2. 64 cub. ft., 398*72. 3. 13 cub. ft., 81, 810 lbs. 4. 6*191. 5. 3000 kilos. 6. 2359. 7. 11*51 cub. ft., 9*245 cub. ft. 8. 3-984 cub. ft., 230-9 lbs. 9. 151*4 lbs. 10. 3 -578 tons. 11. 33-48 tons. 12. 9*048 in. 13. 3*329 feet. 14. 5-556 cub. ft., 10*44 cub. ft. 15. 1500 kilos. Exercises XLL, p. 248. 1. (i) 2*82, 106*4 sq. in. ; (ii) 3'538", 66*68 sq. in. ; (iii) 502*62 cub. in., 251*3 sq. in. 2. 16 ft. 3. 8 ft. 4. -06186 in., 2*64 in. 5. 23*58 cub. in., 66*16 sq. in. 6. 2222 lbs. 7. 238*8 sq. ft.; 4352 lbs. 8. 39 5 in. 9. 6*443 in. 10. 9563 yds. Exercises XLIL, p. 250. 1. (i) 4*887"; (ii) 5*3"; (iii) 301*6 cub. in., 188*6 sq. in. 2 10*47 ft. 3. 50*264 cub. in., 13*068 lbs. 4. 47*13 cub. ft., 54-95 sq. ft. 5. 7". 342 PRACTICAL MATHEMATICS FOR BEGINNERS. 6. 33 cub. ft. 7. 132 cub. ft. 8. 92-02 sq. in., 92 '5 sq. in., -26%. The third figure is only approximately correct and hence seven of the ten figures are unnecessary. Exercises XLIIL, p. 253. 1. 8-555", 452-4 sq. in. 2. 655 8 lbs. 3. 2267 lbs. 4. (i) 7-442"; (ii) 3-385". 5. 11 -62". 6. 4-083". 7. -79 lbs. 8. 2-33 tons. 9. 16 ft. 3 in. 10. 1". 11. 40-62 sq. in. Exercises XLIV., p. 256. 1. 631-7sq. in., 6317 cub. in., 176*9 lbs. 2. 7843 lbs. 3. 3|". 4.9-8". 5. 128-03 cub. ft. 6. 6636 cub. in., 1725 36 lbs. 7. 3327 lbs. 8. (i) 118-4 cub. in., 2369 sq. in. ; (ii) 1", 5-065"; (iii) -9187. 9. External radius 5 -2", Internal radius 3". 10. 702-6 cub. in., 182-7 lbs. Exercises XLV., p. 259. 1. 91-6 cub. ft., 92-6 cub. ft. 2. 104 sq. in., 14980 cub. in. 3. 28-9 cub. ft. 4. 54 '86 cub. ft. 5. 133 cub. ft. 6. 3405-7 cub. yds. 7. 40421 cub. ft, 8. 2544966 cub. ft. 9. 100-7 cub. ft. 10. 792000 cub. ft. 11. 3*69 ft., 73 4. Exercises XL VI., p. 273. 1. 7-07. 2. 105 3. 14-15, 15-36, 17*2, 1822. 4. 44-2", 48 17', 22 36', 32 52'. 5. 3-55", 22 -7, 40 -5. 6. 2'5", 2'24'\ 1-8", 2'69". 7. 3-283, cos a ='4568, cos j3= '7004, cos0 = '5483. 8. a* = 3-624,y = 9*959, 2=1696. 9. 59'7, 30*3. 10. (i) 7*071 ; (ii) cos a= -4242, cos 0= '5657, cos 0='7O71. 11. x- 1 -747, y = 2-083, 2=1-268. 12. 849*6 miles per hour. 13. 2439 miles, 15320, 42-55. 14. 69*1 miles. 15. a; = 1-293, y = 1-477, 2=2-298. 16. (i) 5-643, (ii) -4429, -5493, '7087. 1*0001. Exercises XL VII., p. 276. 1. 4*188 radians ; 10*47 ft. per sec. 2. 2*2 radians, 13*2 ft. per sec. 3. 2 62 radians. 4. 4*4; 61*58. 5. 1 radian; 38*2. 6. 9-425; 56 55 ft. per sec. 7. 12 -56; 21*99. ANSWERS. 343 Exercises XLVIIL, p. 288. 1. 145-5, 20 N. of E. 2. 16, 44 -5. 3. 6 '75 knots, 21 S. of E. 4. (i) 3-08, 42-5 N. of E. ; (ii) -94, 35 4 W. of S. ; (iii) 3*79, 8'3 E. of N. 5. 2035, 7'8 W. of S. ; 577, 25 E. of N. ; 6*5, ll-5 W. of S. .4.5=2-472, A. C=2-863. 6. (i) 50-6, 26 ; (ii) 425, -6 '7. 7. 31 3, 52 50'. 8. 6000 ft. -lbs. per sec. ; (ii) 2652 ft. -lbs. per sec. ; (iii) ; (iv) - 1060. 9. (i) 25-07, 44 26'; (ii) 25-07, 44 26'; (iii) 23'68, 2 46'; (iv) 23'68, 2 46'. 10. 7-36 miles per hour, 28 5 W. of N. 11. = 60, 51 46', 60, A+B + G=IU'8, a = SS 42', /3 = 101 6', = 53 36'. 12. ^4=4-368, a = 7640', 5=16, 0=67 24'. Exercises XLIX., p. 292. 1. 3x*-7x-2. 2. 2a 2 + 3*-5. 3. x 2 + 2ax-a?. 4. ^-11^+ 17. 5. 5^-6 *> -y i5 - jf WA- 16. (l+V2)\/(2 + 2\/2j. 17. A /5. 18. 4, i 2, | \ 2 4 5 19. | | -2, -| 20. ^?- 6 , + 21. x=4, 2; y =2, 4. 22. a; = 6i 3; y=-2|, I. 23. x=6S, 4; y= -5-4, 3. 24. a+b, a-b; x 2 -2{a 2 + b 2 )x+{a 2 -b 2 ) 2 =0. 25. 4-2426,-14-142, =0. Exercises LI., p. 300. . 2. (i) 10. (ii) *. 4. 6000 sq. yds. 5. 60 miles. 6. 10, 15. a 1. (i) 9-5 ft. (ii) 32. 2. (i) 10. (ii) I. 3. 8 in., 18 in. Si 344 PRACTICAL MATHEMATICS FOR BEGINNERS. 7. 4 miles per hour. 8. 900 for 8 months, 600 for 10 months ; rate 6 per cent. 10. 4, 6, 480. 11. 5 or |. Exercises LIL, p. 303. 1. 52|. 2. 52. 3. 4890. 4. 80. 5. -120. 6. -99|. 7. -133. 8. 4864. 9. (ll-2). 10. 25. 11. 10. 12. 630. 13. 369^. 14. 5. 15. 5, 7, 9, 9, 7, 5. 16. n 2 -n + l. 17. 25. 18. w(w + 1) . 19. n*. m Exercises MIL, p. 307. 1. 5 327. 2. -3 6. 3. 7 556. 4. 765, -255. 5. (i) -34-18. (ii) -103. 6. (i) -12*65 (ii) -21 57. 8. 27,3. 9. 57'6. 10. (i)j| (ii)* (iii) _?ll(x/3-V2). 11. 3, 6, 9.... 12. 2. 13. a. p. - jL . 15. 2, 6, 18. Exercises LIV., p. 309. 1. 2f, 3, 4, 6. 2. 5, 4, 3-2. 3. 7. 4. 5. 5. \ \ | 6. 4, 16. 7. 1, g, | 8. 24. 9. %, 2$, 2|. - 13 o 36. o 13 9. 9 o 9. o 36 9 10. T , d, ^, 2, T , _, 2, 3, -, 2, _, ^ EXAMINATION PAPER, 1901. 1. 7446; 0-01254; 5-68; 1546. 2. (a) 1691; (6)0515. 3. d=3'43"; 1552 sq. in. 4. 4 '009 ; 6*662; 53; 10 66 sq. in. 5. 1100000. 6. 260. 7. 1350; 1700. 8. (a) s ^ ; (b) - = ; (c) 2 miles per hour. ax^fb a*-4 x-S 2*25 9. Hint. Let x be the number, then x H =y; ., ^=l-?^ = 0or a ;=v/^25 = l-5. dx x 2 10. x=3 V3 . =4-732 ; 1-268. 11. 230. 12. t>= 150 - 10*. 13. (1)5-643; (2) cos a ='4429; a = 63 "7 ,- cos/S= -5493; 0=56-7 ; cos = -7087 ;0=44-9; (3) 1.0001. ANSWERS. 345 EXAMINATION PAPER, 1902. 1. 3123, 1704, 1722, 0198. 2. (a) 14407, 16604, 18557, 18815 ; (6) 55 ft. a 3. (a) fr 3 -^ 2 )* ; (6) 2 _ 1 (c) 45 . 6 30 . 4 v ' x + y + z x + S x-5 4. r=3'5. 5. 40*1 ft. per sec, 40 "01 ft. per sec, 40 ft. per sec 6. a=2*2, 6=0-11, z = 4'4. 7. 53'56 sq. in. 8. ^Z>=2'6624 ft., 6-2167 sq. ft., Z) = 1*864, DC=2-806, AC =3S68. 9. Value of v is about 9. 10. Converted marks are : 118*9, 160*8, and 213*3. 11. 13550, 14350, 14740. 12. *=2012. EXAMINATION PAPER, 1903. 1. 284*7, 2817, 3*339, 193, 1768000, 11*03. 2. (a) 40; (b) t = 1*5 sec, s=2685ft.; (c) 96. 3 . (a) | 2a?T + ^ 8T + 4\ ; (b) r^ + ^n ; () 3*229, 1*753; {d) 55 55'. 4. 304 sq. in., 232*2 sq. in., 33669 cub. in. 5. Average rates, ^4=2*8, 5=2*4; ^'s=ll$, 4*1, 1*5. 7. 1*645. 8. (-a) Thickness radially, 7*124 in. ; thickness the other way, 11 '4 in. ; inside radius, 99 *7 in. Wc I 9. y=\W*. 10. 0*84, 1*65, 58, 145, 87. 11. 21 horses. 12. 12080, 33 55, 5*592 in. 13. 0*046, March, 072. 14. T=0'95U+0525W=0'6U+0'28; JF=3*28. (b) i?= g- x ^j-, where k is a constant. MATRICULATION EXAMINATION, 1902. 1. 720-2 grams. 2. H& . 26*8. 3. | ; 0*815. 4. ^ ; 3*59 per cent. 6. g* 3 -^* 4 - 7. (2a*+l)(a*- 1) ; 1, -1 8. 3*302,-0*302. 10. 120 ft., 50 ft. 1. 0*125, 1. 3. 5005 (2a) 6 (36) 9 ; 1*059. 4. M =221*4, #=0*1704. 6. 246*6 yds. 8. 5 = 51, 6 = 84-05, c = 108*l; A = 2861. 9. 117 19'. INDEX. Acceleration, 213. Addition, 2, 62, 143. Algebra, 57-113. Algebraical sum, 59. Amsler's planimeter, 238. Angular measurement, 33, 275 ; velocity, 275. Approximations, 111. Area, British measures of, 117 ; measurement of, 216 ; of plane figures, 216-240. Arithmetical progression, 301 Arithmetical mean, 302. Averages, 22. Average velocity, 210. Binomial theorem, 110. Boyle's law, 104, 107. Brackets, use of, 70. British measures of length, 114. British measures of area, 117. Chords, scale of, 36. Circle, area of, 224 ; circum- ference of, 222; segment of, 226. Coefficient, 60. Common ratio, 301, 304. Compound interest law, 207. Cone, 242, 249. Continued product, 4, 68. Contracted methods, 8, 9, 13. Construction of an angle, 37, 161 of a scale, 40. Co-ordinate planes of projection. 263. Cross-section, 247. Cube root, 30, 135, 149. Curve, slope of, 200, 204. Cylinder, 241, 245. Decimal fractions, 2-12. Density, 122. Diagonal scale, 41. Differentiation, simple, 205. Direction-cosines of a line, 267. Division, 12, 13, 69, 130, 147 ; of a line, 39. Elimination, 91. Evolution, 25, 43, 135. Explanation of symbols, 57. Exponent, 60. Equations, 191 ; of a line, 176 ; cubic, 82, 193; simple, 82; simultaneous, 177 ; quadratic 292; reducible to quadratics, 296 ; simultaneous quadratics, 295. Equatorial plane, 269. Factors, 73, 76 ; highest common, 78 ; constant, 301. Fourth proportional, 19. Fourth root, 43. Fractions, 1, 6, 77. Fractional index, 26 ; equations, 84. Functions of angles, 161. Geometrical progression, 301. 304; mean, 306. INDEX. 347 Harmonical progression, 308 ; mean, 308. Hatchet planimeter, 237. Highest common factor, 78. Height and distances, 168. Hooke's law, 103. Hollow sphere, 253. Hollow cylinder, 247. Hyperbolic curve, 196. Identity, 83. Increase, rate of, 184, 201. Index, 24, 60 ; rules, 107 ; frac- tional, 26. Indices, 107. Inverse, proportion, 104; ratio, 162. Involution, 25, 107, 133, 148. Irregular figures, 229, 257. Interpolation, 174. Italian method of division, 12. Latitude and longitude, 269. Law, Boyle's, 104 ; Hooke's, 103 ; of a machine, 98 ; compound interest, 207. Least common multiple, 79. Line, plotting, 175 ; slope of , 183; through two points, 177, 271 ; equation of, 176. Logarithms, 125-140. Machine, law of a, 98. Maxima and minima, 208, 214. Measurement of angles, 33, 151 ; of area, 118, 216 ; of length, 114. Mean proportional, 19, 42, 102. Mean, arithmetical, 302 ; geo- metrical, 306 ; harmonical, 308. Mid-ordinate rule, 230. Multiple, least common, 79. Multiplication, 3, 9, 42, 67, 129, 147. Parallels of latitude, 269. Parallelogram, area of, 218. Partial fractions, 94. Percentages, 20. Perry, Prof., 190. Proportion, 18, 41, 100, 102, 104, 105. Planimeter, 236; Amsler's, 238; .hatchet, 237. Plotting, functions, 194 ; line, 175. Polar co-ordinates, 273. Principle of Archimedes, 122. Prism, volume and surface of, 242. Progessions, arithmetical, 301 ; geometrical, 301, 304 ; harmoni- cal, 308. Pyramid, 241. Quadratic equations, 292 ; simul- taneous, 295 ; relation between coefficient and roots, 297 ; problems leading to, 298. Rate of increase, 184, 203. Ratio, 16,100; inverse, 162; of small quantities, 17. Regular solids, 241. Resolution of vectors, 281. Retardation, 214. Root of a number, 25, 30, 43. Rule, index, 187 ; mid-ordinate, 231 ; of signs, 66 ; Simpson's, 232 ; slide, 143. Scalar quantities, 277. Sector of a circle, 225, 226. Series, 301. Significant figures, 5. Similar figures, 49, 50, 259. Simple equations, 82. Simpson's rule, 232. Simultaneous equations, 90, 177. Simple differentiation, 205. Slide rule, 143, 165. Slope of a curve, 200 ; of a line, 183. Small angles, 162. Specific gravity, 122. Sphere, 242, 251, 253. Surds, 28, 79. Surface, of cone, 249 ; cylinder, 245 ; prism, 243 ; solid ring, 255 ; sphere, 251. Solid ring, 255. Suffixes, 110. Symbolical expression, 57, 81. Squared paper, 171. Square root, 26, 29, 290. 348 PRACTICAL MATHEMATICS FOR BEGINNERS. Tee square, 32. Term, 60. Theorem, binomial, 110. Triangles, area of, 220 ; con- struction of, 53 ; similar, 49. Units, of area, 116 ; of length, 114 ; of volume, 119 ; of weight, 121. Unitary method, 19. Use, of brackets, 70 ; of instru- ments, 31 ; of squared paper, 171 ; of tables, 173. Value of recurring decimals, 306. Variation, 102. Vectors, 277 ; addition and sub- traction of, 278 ; multiplication of, 286. Velocity, average, 210 ; rate of change of, 201, 213. Volume, of a cone, 249 ; acyclinder, 245 ; hollow cylinder, 247 ; prism, 243 ; solid ring, 255 ; sphere, 251 ; units of, 119. Weight, unit of, 120. Glasgow: printed at the university press bv Robert maclehose and co. ltd. ^1 / >> ~ UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. fifiC 3 1947 MAR 31973 7 24 73 -ft AM 5 LD 21-100m-9,'47(A5702sl6)476 I, I - C^ ^ r 911250 c i THE UNIVERSITY OF CALIFORNIA LIBRARY "% "^ 1 (ipift) 1 - ^*^> = -