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Full text of "Practical mathematics for beginners"

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IN MEMORIAM 
FLOR1AN CAJOR1 





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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 <y of -138' '022 

21. Compute by contracted methods 23 '07 x 0'1354, 2307 -f 1 '354. 

22. Compute 4 326 x '003457 and 0*01584 -"-2'104 each to four 
significant figures, leaving out all unnecessary figures in the work. 



CHAPTER II. 

RATIO, PROPORTION, PERCENTAGES. 

Ratio. The relation between two quantities of the same kind 
with respect to their relative magnitude is called Ratio. 

In comparing the relative sizes of two objects it is a matter of 
common experience to refer to one as a multiple two or three 
times, etc., the other ; or a sub-multiple one-half, or one-third, 
etc., the other. This relation between two quantities of the 
same kind in respect of their relative magnitude, and in which 
the comparison may be made without reference to the exact size 
of either, is called Ratio. 

Ratio may be written in three ways ; thus, if one quantity be 
12 units and another 6 units the ratio may be expressed as *, 
12^6, or omitting the line 12 : 6. If there are two quantities 
in the ratio of 12 to 6, then the statement 12 : 6 or ^ indicates 
that the first number is twice the second ; or, the second 
quantity is one-half the first. Again, if two quantities are in 
the ratio 5 : 7, then the first is fy of the second, or the second is 
\ of the first. 

Quantities of the same kind are those which may be expressed 
in terms of the same unit. 

The ratio of 12 things to 6 similar things is definite, and 
indicates that the number of one kind is twice that of the 
other ; but the ratio of 12 tables to 6 chairs conveys no meaning. 
Also it will be obvious that it is impossible to compare a length 
with an area, or an area with a volume, as, for example, the 
ratio of 3 inches to 4 square inches, or 4 square inches to 20 
cubic inches, although in comparing two quantities of the same 



RATIO. 17 



kind we can assert that one is twice, three times, or some 
multiple or sub-multiple of the other, without defining what the 
unit implies. 

In making the comparison the magnitudes may be either 
abstract or concrete numbers, but the ratio between them must 
always be abstract, that is, merely a number. 

Hence, it is necessary, in comparing magnitudes, that the 
quantities be written in terms of a common unit. For example, 
the ratio of 3 tons to 14 lbs., or the ratio of 10 feet to 4 inches 
is obtained by considering that as there are 2240 lbs. in a ton, 
the first named ratio would be 3 x 2240 : 14 ; the second, since 
12 inches make 1 foot, would be 10 x 12 : 4. 

When it is required to divide a number in a given ratio, it 
is only necessary to add together the two terms of the ratio for 
a common denominator, and take each in turn for a numerator. 

Ex. I. Divide 35 in the ratio of 2 : 5. The denominator becomes 

2 + 5, and the required amounts are f of 35 and y of 35 = 10 and 
25 respectively. 

Beginners are often confused when required to divide a given 
number in the proportion of two or more fractions, and begin by 
taking the given fractions, instead of proceeding to reduce them 
to a common denominator. The way to proceed may be shown 
by an example : 

Ex. 2. Divide 70 in the ratio of 3 and ^. This does not mean 

3 and ^ of 70 ; but, as fractions with the same denominators are in 
the same proportion as their numerators, it is necessary to write ^ 
as y 2" and T as tV Then the question is to divide 70 in the ratio 
3:4, and the required amounts are y of 70 = 30, and y of 70 =40. 

Ex. 3. Find the ratio of 1 ft. 3 in. to 6 ft 3 in. 
Here as 1 ft. 3 in. = 15 in. and 6 ft. 3 in. =75 in. the required ratio 
is y f = - ; or, the quantities are in the ratio of 1 to 5. 

Ex. 4. A yard is 36 in. and a metre 39*37 in. Find the ratio of 
the length of a yard to that of a metre. 

The ratio is ^^= '9144. 
39 "37 

Ratios of very small quantities. In finding the ratio of 
one quantity to another, it is only the relative magnitudes of 
the two quantities which are of importance. The quantities 

P.M. b. b 



18 PRACTICAL MATHEMATICS FOR BEGINNERS. 

themselves may be as small as possible, but the ratio of two 

very small quantities may be a comparatively large number. 

Thus Tflta = '001 * s a sma ll quantity, and so is '00001, but the 
*001 
ratio of -001 to '00001 is = 100. Again '0000063 is a very 

small number, and so is '0000081, but the ratio of - is 

'OOOOOoo 

simply fj or f = lf. This very important fact concerning ratio 
is often lost sight of by beginners, and it must be carefully 
noted in making calculations. 

Proportion. The two ratios 2 : 4 and 8 : 16 are obviously 
equal, and their equality is expressed either by 2 : 4 = 8 : 16, 
or by 2 : 4 : : 8 : 16. The former is the better method of the 
two. When as in the given example, the two ratios are equal, 
the four terms are said to be in proportion, hence : 

Four quantities are proportional, when the ratio of the first to the 
second is equal to the ratio of the third to the fourth. That is, 
when the first is the same multiple or sub-multiple of the second, 
which the third is of the fourth, the quantities are proportional. 

We may thus state that the numbers 6, 8, 15, and 20 form a 
proportion. The proportion is written as 6:8 = 15:20, and 
should be read " that the ratio of 6 to 8 is equal to the ratio 
of 15 to 20." 

The first and last terms of a proportion are called the 
extremes, and the second and third terms the means ; in the last 
example 6 and 20 are the extremes, and 8 and 15 are the means 
in the proportion. 

When four quantities are proportional, the product of the 
extremes is equal to the product of the means. 

Thus 6x20 = 8x15, or f = ^oS i n which the proportion is 
written as the equality of two ratios. 

Since the product of two of the terms of a proportion is equal 
to the product of the other two, it follows at once that if three 
terms of a proportion are given, the remaining one can be 
calculated. 

Ex. 1. Find the second term of a proportion in which 34, 12 
and 15 are respectively the 1st, 3rd and 4th terms. 
14 : required term = 12 : 15 ; 

., , 15x14 ... 

.'. required term = r^ = 1 / . 



PROPORTION. 19 



To find the fourth proportional to three given quantities. 
When the first three terms of a proportion are given to obtain the 
fourth we proceed as follows : Multiply the second by the third term 
and divide the product by the first term. 

Ex. 2. Find the fourth proportional to 2*5, 7 '5 and 4*25. 

Fourth proportional = ~7^ = 12 "75. 

Hence 25 : 7*5 = 425: 1275. 

Mean proportional to two given numbers. This may be 
taken to be a particular case of the last problem, in which the 
second and third terms are alike. Thence, we have the rule : 

Multiply the two given numbers tbgether and find the square root 
of the product. This will be the mean proportional required. 

Ex. 3. Find the mean proportional to 10 and 40. 
Here, 10x40 = 400_; 

Mean proportional = \/400 = 20. 
Hence, 20 is the mean proportional required. 
[The rules for square root are explained on p. 27.] 

Unitary method. By the previous methods of simple pro- 
portion, we may proceed to find the remaining one when three 
out of the four terms are known. In practice this plan of pro- 
cedure may often be replaced by a convenient modification of it- 
called the Unitary Method, in which, given the cost, or value, of 
a definite number of articles, or units, we may, by division, find 
the value of one unit, and finally, the value of any number of 
similar units by multiplication. 

The method may be shown by the following simple example : 

Ex. 1. If the cost of 112 articles be 10s., what will be the cost of 
212 at the same rate ? 

Using the three given terms, we may write the following pro- 
portion : 

112 : 10 = 212 : required term, 

10 x 212 
.". required term = p^ =18s. llfd. 

By the unitary method we should proceed as follows : 
If the cost of 112 articles be 10s., then the cost of one article 
at the same rate is tS. D 2 s ' 

therefore the cost of 212 articles is fe by 212s. = l8s. llfd. 



20 PRACTICAL MATHEMATICS FOR BEGINNERS. 



EXERCISES. IV. 

1. If a train travel 215 miles in 10 hrs. 45 min., what distance 
will it travel in 24 hrs. at the same rate ? 

2. In what time will 25 men do a piece of work which 12 men 
can do in 15 days? 

3. Divide 814 among 3 persons in the ratios : f : j . 

4. If the carriage of 8 cwt. for 120 miles be 24s., what weight 
can be carried 32 miles, at the same rate, for 18s. ? 

5. Find a fourth proportional to 45, '8 and '367. 

6. Divide 56 between A, B, C and D in the ratio of the numbers 
3, 5, 7 and 9. 

7. Divide 204 into three parts proportional to the numbers 7, 8, 9. 

8. Find the number that is to 7 in the ratio of 3. Is. 3d. to 
4. 13s. lid. 

9. A sum of 32818 is to be divided among four persons in the 
proportion of the fractions f , f , and f . Find the share of each. 

10. Define ratio. Does it follow from your definition that it 
would be wrong to speak of the ratio of 5 tons to 3 miles, and if so 
how does it follow ? 

11. What should be the price of 194 dozen articles, if 391 such 
articles cost 21. 3s. 7d. ? 

12. If a parish pays 2165. 12s. 6d. for the repairing of 7| miles 
of road, what length of road would 1500 pay for at the same rate? 

13. If the carriage of 3| tons for a distance of 39 miles cost 14s. 7d. , 
what will be the carriage of 20 tons for a distance of 156 miles at 
half the former rate ? 

Percentages. The ratio of two quantities, or the rate of increase 
or diminution of one quantity as compared with another of the same 
kind, is often expressed in the form of a percentage. The word 
" cent " simply denotes a " hundred," hence a percentage is simply 
a fraction with a denominator of a 100. This fact enables a com- 
parison to be made at once, without the preparatory trouble 
of reducing the fractions to like denominators. Examples on 
percentages occur so frequently, and are so varied, that it is 
difficult to select typical illustrations. The following, however, 
may make the matter clear. 

Suppose that two classes, of 20 and 50 students respectively, 
are expected to attend an examination. In the first named, 18 
students, and in the second, 47 students, present themselves. 
Then, we may say that 2 in 20 and 3 in 50 were away from the 



PERCENTAGES. 21 



examination ; but the comparison is most easily made by finding 
the percentage in each case. Thus, in the first case we 

2 
have absent ^r x 100 = 10 per cent. ; 

Q 

in the second case x 100 = 6 per cent. 
50 

These results would be written as 10% and 6%. 

Ex. 1. Suppose the population of a town in 1885 was 15,990, 
and in 1890 was 20,550. The actual increase is 20550 - 15990 = 4560 ; 
but although the actual increase is useful, it is much better to be 
able to state the rate at which the population is increasing for each 
100 of its inhabitants. The increase for each 100 of its population 
is found by simple proportion as follows : 

15990 : 100 :: 4560 : increase required. 

.*. Increase for each 100= r=^ 28*5. 
15990 

Thus, the increase for each 100 of its population, is 28*5. This 
number is called 28 "5 per cent., and is written 28*5%. The rate 
per cent, permits of a ready reference to an increase or diminution 
of any kind. 

Ex. 2. The population of another town in 1885 was 20,400, and 

in 1890 was 24,960. The actual increase (as before) is 4560, but it 

does not follow from this that the two towns are increasing at the 

same rate. In this case the rate of increase is obtained from : 

20400 : 4560 :: 100 : rate of increase ; 

. . . 4560 x 100 00 _ 

.*. Rate of increase = ^. ftA =22 3. 
20400 

From these examples it is clear that the population of the 
latter town is not increasing as fast as the former by 6 per 
hundred, or, as usually written, by 6%. 

In like manner, percentages are often used to compare the 
proportions of lunatics, paupers, criminals, etc., in the population 
of different towns. 

Rate or debt collectors and others in many cases are paid at 
the rate of so much per cent. If a rate collector is paid at the 
rate of 2 per cent., for example, this would mean that for every 
100 collected he is allowed 2 ; for every 50, 1, etc. 

Ex. 3. If in a machine it is found that a quarter of the energy 
expended is wasted in frictional and other resistances, we should say 



22 PRACTICAL MATHEMATICS FOR BEGINNERS. 

that 25 per cent, is wasted, meaning that is useless. This does 
not tell us the actual numerical amount of the loss ; all that we can 
infer is that for every 100 units of work expended on the machine 
25 units disappear. Such a percentage also enables a comparison 
to be made, and is a convenient method of expressing the efficiency 
of machines. If one machine has an efficiency of 75 per cent, and 
another of 80 per cent., we know that the second is 5 per cent, 
more efficient than the first. 

If, in addition, we know that 25 per cent, is the total loss due 
to all resistances, but 10 per cent, of this is due to the resistance 
of a particular part of the mechanism, this gives a percentage of 
a percentage and its numerical value is 

iVo of tw = To 5 o x Too =To-o-o =2-5, or 2| per cent. (%). 

Ex. 4. A reef of quartz contains -0044 per cent, of gold. If the 
quartz produces 5. 12s. per ton, find the weight of a sovereign in 

grains. 

5. 12s. 5 T % 5 6 sovereigns, 

1 ton = 2240 x 7000 grains, 

0044 
and -0044 per cent. - -^ - "000044. 

.-. weight of 5-6 sovereigns = -000044 x 2240 x 7000 

= -44 x 224 x 7. 

. ., . _ -44 x 224 x 7 
.*. weight of 1 sovereign = ^ 

= 123-2 grains. 

Percentages and averages. The data for practical calcula- 
tions are in many cases either the result of measured quantities, 
or experimental observations in each case liable to error. To 
obtain a trustworthy resulct a omparatively large number of 
observations may be taken and the average or mean result 
calculated. Such an average may be obtained by adding all the 
results together and dividing by the number of them. When 
the average is thus arrived at it may generally be accepted as 
the best approximation to the truth. Accepting it as correct, 
then the difference between it and any single value got by 
observation can be ascertained, and the error expressed most 
conveniently as a percentage. 



PERCENTAGES. 23 



EXERCISES. V. 

1. The composition of bronze or gun metal is to be 91 % 

copper and 9 % tin. Find the weight of a cubic foot of the . 

material. Also find the amount of copper and tin required to make 
1000 lbs. of the alloy. (See Table I., p. 123.) 

2. The composition of white or Babbit's metal is to be 4 parts 
copper, 8 antimony and 96 tin. Express these as percentages, find 
the weight of a cubic foot of the alloy, also the amount of each 
material required to make 200 lbs. of the metal. 

3. If the cost of travelling by rail for 42 miles is 5s. 3d. , what is 
the cost of travelling 35 miles at a price per mile 20 per cent, higher? 

4. A collector receives 5 per cent, commission on all debts col- 
lected, and this commission amounts to 4. Find the amount 
collected. 

5. In a class of 80 boys, 12| % failed to pass an examination. 
How many passed ? 

6. If the annual increase in the population of a state is 25 per 
thousand, and the present number of inhabitants is 2,624,000; what 
will the population be in three years' time ? and what was it a year 
ago? 

7. By selling coal at 15s. a ton a merchant lost 12 per cent. 
What would he have gained or lost per cent, if he had raised the 
price to 18s. 9d. per ton ? 

8. A man sold a horse for 36, losing 4 per cent, of the cost " 
price. How much did he pay for the horse ? 

9. A 56-gallon cask is filled with a mixture of beer and water, in 
which there is 84 per cent, of beer. After 8 gallons are drawn off, 
the cask is filled up again with water. What is the percentage of 
beer in the new mixture ? 

10. If a dozen eggs are bought for Is. 8d., for how much must 
they be sold' singly to make a profit of 20 per cent.? 

11. In an examination 60 per cent, of the candidates pass in each 
year. In 5 successive years the numbers examined are 1000, 840, 
900, 1260, 1400 ; what is the average number of candidates per 
annum, and the average number of failures ? 

12. A farmer purchased 120 lambs at 30s. a head in the autumn. 
During the winter 12 died, but he sold the rest in May at 45s. each. 
What was his gain per cent.? 

113. If 3500 baskets are purchased at lfd. each and sold at 2^d. - 
apiece, what will be the total gain and the gain per cent.? 
14. For what amount should goods worth 1,900 be insured at 
5 per cent. , so that, in case of total loss, the premium and the value 
of the goods may be recovered ? 



CHAPTER III. 

POWERS AND ROOTS. 

Squares and cubes. Powers. When a number is multi- 
plied by itself the result is called the square or the second power 
of the number. Thus, the square of 3, or 3 x 3, is 9 ; and the 
square, or second power, of 4 is 4 x 4, that is, 16. 

When three numbers of the same value, are multiplied to- 
gether the result is called the cube or third power of the number ; 
thus, the cube of 3 is 3 x 3 x 3, that is, 27. The cube of 4 is 
4x4x4, or 64. 

In the same manner, the result of multiplying four numbers 
of the same value together is called the fourth power of the 
number. Hence, the fourth power of 4 is 4x4x4x4, that 
is 256. 

By multiplying five of the same numbers together we should 
obtain the fifth power of the number. The sixth, seventh, or 
any other power, of a number is determined in a similar 
manner. 

Index. A convenient method of indicating the power of 
a number is by means of a small figure placed near the top and 
on the right-hand side of a figure ; thus the square or second 
power of 2 may be written 2x2, but more conveniently as 2 2 , 
and the fourth power of 2 either as 2 x 2 x 2 x 2, or 2 4 . Similarly 
the square, cube, fourth or fifth power of 4 would be written 
4 2 , 4 3 , 4 4 , and 4 5 . 

This method is also adopted when a number containing 
several digits is used. Thus 2574 2 signifies 2574 x 2574, and 
63 3 means 63x63x63. The small figure used to show how 



INVOLUTION. 25 

many times a given number is supposed to be written is called 
the index or exponent of the number. 

Involution. The process by which the powers of a number are 
obtained is called Involution. The number itself is called the first 
power or the root, and the products are called the powers of the 
number. 

The powers of 10 itself are easily remembered, and are as 
follows : 

10 2 = 100, 103= 1000, 10 6 = 1,000,000, etc. 

This method of indicating large numbers is very convenient 
in physical science, in which such numbers as 5 or 10 millions, 
etc., are of frequent occurrence ; for, in place of writing 
5,000,000, for instance, we may write it more shortly, as 5 x 10 6 . 

Again, 5,830,000 could be written as 5'83 x 10 6 , and 5,830 as 
5*83 xlO 3 . 

The squares of all numbers from 1 to 10 are easy to remem- 
ber ; they are as follows : 

12==1) 4 2 =16, 7 2 = 49, 

2 2 =4, 5 2 = 25, 8 2 = 64, 

3 2 = 9, 6 2 = 36, 9 2 = 81. 

The squares of all numbers from 10 to 20 might with 
advantage also be written down. 

When a number consists of three or more figures its square or 
any higher power can be obtained by multiplication, or in many 
cases better by logarithms, which will be described later. 

Evolution. The process which is the reverse of Involution 
is that of extracting, or finding, the roots of any given 
numbers. 

The root of a number is a number which, multiplied by itself a 
certain number of times, will produce the number. 

Thus, the square root of a given number is that number which, 
when multiplied by itself, is equal to the given number. 

The root of a given number may be denoted by the symbol 
*J placed before it, with a small figure indicating the nature 
of the root placed in the angle. 

In this manner the cube root of 27 is denoted by \^27, the 
fourth root of 64 by \/64, and so on. 

The square root would be denoted by V9,Jbut the 2 is usually 
omitted, and it is written more simply as V9. 



26 PRACTICAL MATHEMATICS FOR BEGINNERS. 

Another, and for many purposes a better method, is to in- 
dicate the root by a fraction placed as an index, and referred to 
as a fractional index ; thus, for example, the square root of 9 
is written 9* and is read as nine to the power one-half. Simi- 
larly, the cube root of 27 is written as 27"*, meaning 27 to the 
power one-third. 

Square root. Method I. To extract the square root of any 
quantity in cases where it is not possible to ascertain such root 
by inspection, we have to adopt a rule. The following example 
will illustrate the method of extracting a square root. 

Ex. 1. Method /.Find the square root of 155236. 

155236(300 + 90 + 4 
90000 



(2 x 300 ) + 90 = 690 ) 65236 
62100 

2x390 + 4 = 784)3136 
3136 

The process is as follows : 

Divide the given number into periods of two figures each, by 
putting a point over the unit's figure, another on the figure 2 which 
is in the second place to the left of the 6, and a third also on the 5, 
as shown. The given number consists of six figures ; the required 
square root contains three. 

As300 2 = 90,000and 400 2 = 160,000, the required square root lies 
between 300 and 400 ; hence, we put 300 to the right of the given 
number, and subtract its square 90,000 from the number the quare 
root of which is required ; this gives a remainder of 65236. 

Put twice 300 to the left of the remainder 65236 ; this 600 divides 
into 65236 a little over 90 times ; place 90 with the 300 in the place 
occupied by the square root and add 90 to 2 x 300, and thus obtain 
690 ; this result multiplied by 90 gives a product of 62100 ; subtract 
this product from 65236, and the remainder 3136 is obtained. 

Next, set down to the left of the remainder 3136, 2x390 = 780; 
this will divide into 3136, 4 times. Place 4 with the answer as shown. 

Add 4 to 780, obtaining 784 ; multiply by 4 and obtain 3136 ; 
this subtracted from 3136 leaves no remainder ; or 394 is the square 
root required. 

Ex. 1. Method II. The ordinary practical method is as follows: 
Point as before, and find the largest number the square of which 



SQUARE ROOT. 27 



is less than 15 ; 3 is such a number. Set the figure 3 to the right 
of the given number and its square 9 under the 
first pair of figures 15; subtract 9 from 15, 15523$ ( 394 

obtaining a remainder 6. 

Bring down the next two figures, making the 69 ) 652 
number 652. *>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 <J'd or 1732... 

in order to obtain the numerical value of the fraction ; but it is 
better, and simpler, to multiply both numerator and denominator 

by \/3. This gives -j= ;=.= : and in this form, knowing 

that v3 = l "732..., it is only necessary to move the decimal 
point two places to the right and divide by 3. 

Square root of a vulgar fraction. In finding the square 
root of such a fraction, it is necessary to obtain the square root 
of numerator and denominator. 

Ex. 7- Find the square root of 2f-. 

Herev^|= N /=f. 

In a similar manner the square root of 20^ is 4'5. 

When the denominator is not a perfect square, we may proceed 
in some cases to first multiply both the numerator and deno- 
minator by the number which will make the denominator a 
perfect square. Or, we may multiply both numerator and 
denominator by the denominator. 

Thus, to find the square root of J, we might find the square 
root of 3 and of 8 and divide one long number by another ; 
but it is better to multiply thus 

y/3^ V3x\/8 = \/2i_ 4-898 

J$ 8 8 ~ 8 ' 

when we only require to find one root instead of two ; or, convert 
the given fraction to a decimal fraction, and find the root in the 
usual manner. 

Ex. 8. Find the square root of J. 

Here if the numerator and denominator be multiplied by 2, the 

fraction becomes and its square root is ~-, which leaves only 



one root to be extracted. 



30 PRACTICAL MATHEMATICS FOR BEGINNERS. 

Contracted method. In practical calculations the square 
root of any quantity is never required to more than a few 
significant figures, and when more than half the required 
number of digits have been found, the remainder may be found 
by contracted division. 

Ex. 9. Obtain the square root of 13 to five places of decimals. 

13(3-60555 
Here, proceeding as in the preceding ex- _9 

amples, the square root of 13 3*605 is ob- 66) 400 

tained together with a remainder 3975. The 396 

remaining figures of the square root may now 7205 ) 40000 

be obtained by contracted division (see p. 36025 

13), viz., by dividing 3975 by 7205, giving 7205 ) 3975 ( 55 

55, which is placed with the number already 3602 

obtained. 373 

Hence the required root is 3-60555. 360 

18 
Cube root. The arithmetical method of ascertaining the 
cube root of a number in all except the simplest cases is too 
tedious and unwieldy to be of any practical use. Indeed, it is 
not worth the time necessary to learn it, and it will be better to 
leave a consideration of cube roots until the student is familiar 
with the use of logarithms or the slide rule, by the help of 
which cube roots can be found easily and readily in any case. 

EXERCISES. VI. 

Find the square root of : 
1. 37249. 2. 4-9284. 3. 1006009. 4. 18671041. 5. 122-1025. 

6. 65 and 50 in each case to four decimal places, also of 8 to 
six decimal places. 

7. 3263-8369. 8. 450643*69. 9. (i) 39 T V ; (ii) 40008-0004. 
10. 90018 0009. 11. 6877219041. 12. 998001. 13. 42436. 
14. 00501264. 1 5. 1867104 1. _16. 1085*0436. 

17. Add together s/5'Sl 11*81 16, n/20J and \/9. 

18. Divide the square root of *04 by v5|. 

19. Find the value of 

(i) lWf, (ii) lW|, (iii) lW&, (iv) lW 
in each case to two significant figures. 



CHAPTER IV. 

PLANE GEOMETRY. 

Use of instruments. Graphic methods are applicable to 
the majority of the problems which a practical man is called 
upon to solve. By means of a few mathematical instruments 
results may often be obtained which could only be arrived at 




Pig. 1. Two forms of protractors. 

by mathematical methods after the solution of many difficulties. 
Even in problems in which sufficiently accurate results are not 
obtainable by graphical methods mathematical instruments may 
be used with advantage to check roughly the conclusions reached 
by calculation. To take a simple example ; if the lengths of the 
three sides of a triangle are given, then, by means of a simple 



32 PRACTICAL MATHEMATICS FOR BEGINNERS. 

formula, the area can be obtained to any degree of accuracy 
necessary. But it is also advisable to draw the triangle to a 
fairly large scale, for, since the area is one-half the product of 
the base and the perpendicular drawn from the base to the 
opposite vertex, the length of this perpendicular and the base 
can be measured and the product obtained, then half the 
product gives the area, and furnishes a ready method of checking 
the calculated result. 

Accuracy and neatness are absolutely necessary in graphic 
work of any description. Distances should be measured and 
circles drawn as accurately as the instruments will permit. The 
instruments should be of fairly good quality ; the following are 
necessary, but others may be added if it is thought desirable. 

(a) Pair of pencil compasses, (b) pair of dividers, (c) protractor 
The last may be rectangular, as shown at BC(Fig. 1), or, better, 





Fig. 2. (ii.) A 45 set square. 



Fig. 2. (i) A 60 set square. 



a semi-circular one about 6" diameter, (d) A 60 set square 
about 9" long, (e) 45 set square about 6" long, (i), (ii) (Fig. 2), 
(f) a good boxwood scale (Fig. 10), (g) drawing board, imperial 
size (30" x 22"), or if preferred half imperial, and (h) a tee-square 
to suit the board. 

The best point for the compass-lead is the flat or chisel point, 
and the lead used should be of medium hardness, for if it is too 
soft the point requires constant sharpening, and it is difficult to 
draw a good firm circle or arc ; if it is too hard scratches are 
made and the surface of the paper is spoilt ; when closed the 
point of the compass and lead should be on the same level. 



MEASUREMENT OF ANGLES. 



33 



Pencils. Two pencils are requisite ; one an H., H.H., or 
H.H.H. sharpened to a flat chisel point (Fig. 3) ; the other an 
H.B., sharpened to a fine round point. 
The chisel-point should be sharpened so 
that when looked at from the end of the 
pencil the edge is invisible. The edge is 
made and maintained by using a small 
rectangular slip of fine glass paper. 

Measurement of angles. If a straight 
line OA (Fig. 4) initially coincident with 
a fixed line 00, rotate about a centre 0, 
and in the opposite direction to that in 
which the hands of a clock turn, then the 
number of degrees in the angle OOA is 
the numerical measure of the angle. 

Draw a circle of any convenient radius, 
and divide its circumference into 360 
equal parts, then if two consecutive 
divisions be joined to the centre, the lines 
so drawn contain a length of arc equal to 
3^oth part of the circumference of the 
circle, and the angle between them is 
known as an angle of one degree. If, 

in the circle, two radii are drawn perpendicular to each 
other, they enclose a quarter of the circle, and hence a right 
angle consists of 90 degrees, written 
90. Each degree is divided into 60 
equal parts, or minutes ; and each 
minute is again subdivided into 60 
equal parts called seconds. 

Abbreviations are used for these 
denominations. 52 14' 20"*5 denotes 
52 degrees 14 minutes 20'5 seconds. 

Fractional parts of a degree may be expressed either as 
decimals of a degree, or in minutes, seconds, and decimal parts 
of a second. 

Thus, an angle may be expressed as, say, 54 -563, or, 





Pig. 4. 



an angle may be expressed as, say, 
multiplying the decimal part by 60, we get 33*78 minutes ; 
again, multiplying 78 by 60 we get 46*8 seconds. Therefore 
p.m. b. c 



34 PRACTICAL MATHEMATICS FOR BEGINNERS. 



the given angle may be written either as 54*563 or 54 33' 

46"*8. 

The length of the lines forming the two sides of the angle 

have no connection with the magnitude of the angle. Hence 

with centre and any con- 
venient radius describe a 
circle CEDE as in Fig. 5 ; 
the line OA may be assumed 
to be a movable radius of 
the circle free to move about 
a centre 0. When at A if an 
arc one-sixth of the circum- 
ference has been described, 
then the angle CO A is an 
angle of 60 degrees, written 
as 60. When coincident 
with OB the angle traced 
out will be an angle of 90. 
When in a position OA' the 

angle CO A' is greater than a light angle and is called an obtuse 

angle. The angle CO A is less than 90 and is called an acute angle. 
Comparison of the magnitudes of angles. A comparison 

of the magnitudes of two angles ABC and DEF (Fig. 6) may 




Fig. 5. Measurement of angles. 





Fig. 



a E 

Comparison of the magnitudes of two angles. 



be made by placing the angle DEF on the angle ABC, so that 
the point E exactly falls upon the point B, and the line BE 
coincides with the line A B. Then if the line EF falls on the 
line BC the angles are said to be equal. The angle DEF is 
less than the angle ABC if EF falls within BC, as shown by 
the dotted line BC. It is larger if it falls outside BC. 



USE OF PROTRACTOR. 35 

This method of superposition is readily performed in the 
following way : Draw from centres B and E two equal arcs AC 
and DF, so that DE and EF in the one case are equal to AB 
and BC, respectively, in the other. If the point A be joined to 
the point C, and D to F, then, if AC is equal to DF, the angles 
ABC and DEF are obviously equal. Or, using a piece of 
tracing paper, make a tracing of DEF, and by placing the 
tracing on ABC, the comparison is readily made. 

When, as in the angle DEF, there is only one angle at E it is 
usually written simply as the angle E. 

To copy a given angle ABC. This is obviously only a 
modification of the preceding construction. Thus, with centre 
B (Fig. 6) and any radius describe an arc cutting the two sides 
of the triangle in points A and C. With centre E, and same 
radius, describe an 
arc cutting ED in D, 
mark off a length 
DF equal to AC, join 
E to F DEF is 
the required angle. 

Ex. 1. Set out by 
a protractor an angle 
of 53. 

At the point P p* $t 

( Fig. 7 ) place the mark Fig. 7. To set out an angle by means of a protractor. 
* shown on the pro- 
tractor in Fig. 1 coincident with P, and the edge of the protractor 
BC with the line PM. 

Make a mark opposite the division indicating 53 on the pro- 
tractor. Remove the protractor and join the mark to P. An 
angle MPN containing 53 will have been made with the given 
line PM. 

How to use a protractor to measure an angle. When 
used to measure a given angle, the edge BC of the protractor is 
placed coincident with one of the lines forming the angle. The 
mark * on the protractor is made to coincide with the vertex of 
the angle, and the point where the other line crosses the 
division on the scale is noted ; this shows, in degrees, the angle 
required. 




36 PRACTICAL MATHEMATICS FOR BEGINNERS. 



Scale of chords. The protractors in general use, even when 
expensive instruments, are often inaccurate. It is not, more- 
over, an easy matter to read off an angle nearer than half a 
degree, and for many purposes this is not sufficiently accurate. 
A scale of chords is consequently often used. Such a scale is 
shown below. 

Thus, to set out an angle of 53 '7. From the table of chords 
the chord of 53 is found to be '892, difference for '7 to be 
added is 'Oil .-. chord of 53'7 = -903. 

With centre P (Fig. 7), radius 1 or 10 units, describe the 
arc MN. With centre M, and a distance *903 if the radius is 1, 
or 9*03 if radius is 10, cut the arc at N; join P to N. Then 
MPNis an angle of 53 7. 

Conversely given an angle MPN (Fig. 7), with a radius I (or 
10), describe arc MN, measure MN and refer to the scale of 
chords. 

CHORDS OF ANGLES. 




10 
20 

30 
40 
50 

60 
70 
80 





1 2 3 


4 5 6 


r 8 r 


Difference to be addtd. 


T-2-3 


4 5 6 


7 8 


000 
174 
347 

518 
684 
845 

1-000 
1-147 
1-286 


017 -035 -052 
191 "209 -226 
364 -382 -399 

534 -551 -568 
700 -717 -733 
861 -867 -892 

1-015 1-030 1-045 
1-161 1-175 1-190 
1-209 1-312 1-325 


070 '087 -105 
243 -261 -278 
416 -433 -450 

585 -601 -618 
749 -765 '781 
908 -923 -939 

1-060 1-075 1-089 
1-203 1-218 1-231 
1-338 1-351 1-364 


122 -140 157 
296 -313 -330 
467 '484 -501 

635 -651 -667 
797 -813 "829 
954 -970 '985 

1-104 1118 1-133 
1-245 1-259 1-272 
1-377 1-389 1-402 


2 3 5 
2 3 5 
2 3 5 

2 3 5 
2 3 5 
2 3 5 

13 4 
1 3 4 
13 4 


7 9 10 
7 9 10 
7 9 10 

7 8 10 
6 8 10 
6 8 9 

6 7 9 
6 7 8 
5 6 8 


12 14 
12 14 
12 14 

12 13 
11 13 
11 12 

10 12 

10 11 

9 10 



To bisect a given angle EBF. With centre B (Fig. 8), 
and any convenient radius, draw an arc of a circle, cutting 
EB and BF in the points A and C. With A and G as centres 
and any equal radii, draw arcs intersecting at D. Join D to Z?, 
then BD bisects the given angle, EBF. 

In this construction it is desirable to make the distances BA 
and BO as large as convenient, and also to arrange that the two 



SCALES AND THEIR USES. 



87 




Fig. 8. To bisect a given angle. 



arcs cutting each other shall cross as nearly as possible at right 

angles, for the point of intersection is then easily seen. If the 

given lines do not 

meet, then we may 

either produce them 

until they meet, or 

draw CB and AB two 

lines meeting at B 

parallel to and at 

equal distances from 

the two given lines. 

Angles of 60 and 
30. To set out OE 
at an angle of 60 to OG (Fig. 9). With as centre and any 
radius, draw an arc cutting OC at B. From B as centre with 
the same radius draw another arc cutting the former at E. 
Join to E. Then BOE is an angle of 60. Bisecting the 
angle we obtain an angle 
of 30, by again bisecting 
an angle of 15, and so 
on. 

The angles referred to 
may be set out also by 
using the 60 set-square, 
or by the protractor. 

Scales and their use. 
The majority of problems 
considered are supposed 
to be solved by the process 
known as drawing to 
scale. In making a drawing of any large object, such as a build- 
ing, it would be inconvenient, if not impossible, to make it 
as large as the object itself. In other words, to draw it full size 
is out of the question, but if a drawing were made in which 
every foot length of the building were represented on the draw- 
ing by a length of half an inch, the drawing would be said to be 
drawn to a scale of J inch to a foot, or to a scale of 2 V 

By means of a suitable scale any required dimension could be 
obtained as readily as if the drawing were made full size. In a 




Fig. 9. To set out an angle of 60 



38 PRACTICAL MATHEMATICS FOR BEGINNERS. 







ft 






a" 




E 




o 


f 










fit 


'- 








E 











> 


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 <r + 5 

the x 2 , and after it the product of x and 5 or 5x. x + 6 

As the signs are alike, the sign of each of these x 2 + 5x 
products is + . Next multiplying by the second 6a; + 30 

term 6 we get 6a; + 30 ; the term Qx is placed a; 2 + 1 la; + 30 
immediately below the corresponding term 5x, and ^ 
the term 30 on the extreme right-hand side; finally, 
add the terms together to obtain the product. By arranging the 
terms one under the other, and multiplying, the result can always 
be obtained. But this is not enough ; in such a simple expression 
the student should be able to at once write down the product by 
inspection. 

This is effected by noting that the first term x 2 of the product is 
obtained by multiplying together the two first terms in the given 
expressions ; the last term is the product of the two second terms 
6 and 5, and the middle term is the product of x and the sum of 
the two second terms. 

.'. (a; + 5)(a; + 6)=a; 2 + lla; + 30. 

In a similar manner, 

(a + 6) 2 , or (a + b){a + b) = a 2 + 2ab + b 2 , 
{a-bf, or (a-b){a-b) = a 2 -2ab + b\ 
{x -5) (x -6) = x 2 -11* + 30, 
(a;-5)(a; + 6) = a; 2 + a;-30, 
(a: + 5) (a; -6) = a; 2 -a; -30. 

When the product of two expressions containing more than two 
terms is required, it is usually convenient to arrange the terms 
one under the other, and to proceed as in the following example : 

Ex. 4. Multiply together I4ac - 3ab + 2 and ac - ab + 1, 
Uac-3ab + 2 
ac- ab + 1 



14a 2 c 2 - 3a 2 6c+ 2ac 

- Ua?bc + Sa 2 b 2 - 2ab 

+ \4ac -Sab + 2 

14a 2 c 2 - 1 la 2 bc + 1 6ac + 3a 2 b 2 -5ab + 2 



68 PRACTICAL MATHEMATICS FOR BEGINNERS. 

Proceeding as in Ex. 3 we multiply each term of the top line by 
ac, by -ab, and finally by 1, thus we have 
14ac xac=14a 2 c 2 . 

Next - 3ab xac= - 3a 2 6c, 

and finally 2 x ac =2ac. 

By multiplying by the second term -ab the second line is 
obtained. After writing down the third line, the terms are added 
and the product is thus found. 

Continued Product. When several quantities are multiplied 
together the product obtained is called the continued product of 
the quantities. 

Ex. 1. The continued product of 3b, 7c, and 2a is 42abc. 

Ex. 2. Obtain the continued product of a; + 2, x + 3, x, and x + l. 
The product of x + 2 and x + 3 is x 2 + 5x + 6. 
Also the product of x and x + 1 is x 2 + x. 
Hence x 2 + 5x + 6 

x 2 + x 



x 4 + 5x s + 6x 2 

x 3 + 5x 2 + 6x 

x* + 6x s +llx 2 + 6x 



EXERCISES. XI. 

Multiply 

1. x 2 - ax + a 2 by x 2 + ax + a 2 . 

2. 2a 3 - a 2 b + 3ab 2 - b 3 by 2a 3 + a 2 b + 3ab 2 + 6 3 . 

3. x 6 + x*y - xPy 3 + xy 5 + y 6 by x 2 -xy + y 2 . 

4. x* + 3x* + 1x 2 + \5x + lQhy ic 2 -3cc + 2. 

5. x 2 + 4xy + Ay 2 by x 2 - 4xy + 4y 2 . 

6. x 3 - 12# -16 by x*-\2x+\Q. 

7. a 6 -3a 4 6 2 +66 6 bya 6 -2a 2 6 4 + 6 6 . 

8. x 4 + x 2 y 2 + y i by x 2 - y 2 . 

9. 4a 3 6 - 6a 2 6 2 - 2a& 3 by 2a 2 + 3ab - b 2 . 

10. a 4 -& 4 + c 4 by a 2 -6 2 -c 2 . 

11. l-y 2 -y 3 by \-y*-y*. 

12. 3x* - x 2 - 1 by 2x* - 3x* + 7. 
13 x* + x 2 + Sx-hy x 2 -2x + 4. 

14. 4a 2 + 12a& + 96 2 by 4a 2 - 12a6 + 96 2 . 

15. 13a 2 - 17a - 45 by -a-3. 



ALGEBRAICAL DIVISION. 69 

Division. In Algebra, as in Arithmetic, the terms divisor, 
dividend, and quotient are used. From a given dividend and 
divisor, we can by the process of division proceed to find the 
quotient of two or more algebraical expressions. When the 
divisor is exactly contained in the dividend, then the product of 
the divisor and the quotient is equal to the dividend. When 
the divisor is not exactly contained in the dividend, and there is 
a remainder, the remainder must be added to the product of the 
quotient and the divisor in order to give the dividend. 

Ex. 1. Divide lOabxPy by 2ax 2 y ; 
. IQabxPy 
2ax 2 y 
As in Arithmetic the work may be done by cancelling, thus, 
70 -r 2 gives 35, 
and abxPy -5- ax 2 y gives bx ; 

hence the required quotient is 356a;. 

Ex. 2. Divide 15a 2 6 2 by - 5a ; 

- 5a 

When the dividend and divisor both consist of several terms, 
we arrange both dividend and divisor according to the powers 
of the same letter, beginning with the highest. The following 
example worked out in full will show the method adopted : 

a 2 + 2aa + x 2 ) a 5 + 5a 4 x + I0a 3 x 2 + IttePx 3 +5ax 4 + x 5 ( a 3 + Sa 2 x+3ax 2 + x 3 
a 5 + 2a 4 x + a 3 x 2 

Sa 4 x + 9a 3 x 2 + lOaPx 3 
3a 4 x+ 6a?x 2 + Sa 2 x s 

3a?x 2 + laW + bax* 
3a 3 a^+ 6a 2 3r* + Sax 4 

a 2 x 3 + 2ax i + x 5 
a 2 x 3 + 2ax 4 + x 5 

Divide the first, or left-hand, term of the divisor into the dividend. 
Thus a 2 divided into a 5 gives a 3 ; write this quantity on the right- 
hand side as shown, and put the term a 5 under the first term of the 
dividend. In a similar manner by multiplying the remaining two 
terms 2ax and x 2 by a 3 , and subtracting, we obtain Sa 4 x + 9a 3 x 2 . 
Now bring down the next term IQatx 3 , and proceed as before. 



70 PRACTICAL MATHEMATICS FOR BEGINNERS. 



EXERCISES. XII. 
Divide 

1. 4a 5 -a 3 + 4aby 2a 2 + 3a + 2. 

2. x 2 + 4yz - 4y 2 -z 2 by x-2y + z. 

3. a 3 + a 2 6 + a 2 c - abc - b 2 c - 6c 2 by a 2 - be. 

4. 2x 4 + 27xy 3 -81y 4 by x + 3y. 

_ x 3 - Sx 2 y + 3xy 2 - y s , x 2 -2xy + y 2 
x 3 + y s x 2 - xy + y 2 ' 

6. 9a 2 -46 2 -c 2 + 46cby 3a-26 + c. 

7. 25a 2 - 6 2 - 4c 2 + 46c by 5a-6 + 2c. 

8. - 207a 5 6 7 c 4 by 23a 4 6 3 c 2 . 

9. 4x 2 y-8xy 2 -4y 2 + 3x 2 + 4x+l by a;-2y + l. 

10. I2x 4 - llx* - 9x* + 13x - 63 by 4rc 2 - 3x + 7. 

11. 2a 3 + 6a 2 6 - 4a 2 c - 2a6 2 + 3ac 2 - 6& 3 + 46 2 c + 96c 2 -6c 3 by a + Sb-2c. 

12. a 5 -a 4 + a?-a 2 + a- 1 by a 3 - 1. 

13. (a 2 - 6c) 3 + 86 3 c 3 by a 2 + 6c. 

14. (i) Express in algebraical symbols : The difference of the 
squares of two numbers is exactly divisible by the sum of the 
numbers. 

(ii) The sum of the cubes of three numbers diminished by three 
times their product is exactly divisible by the sum of the numbers. 

Use of Brackets. In Algebra it is frequently necessary to 
group parts of an expression, and tbe use of brackets for this 
purpose is very important. There are several forms of brackets 
in general use ; for example, ( ), { }, [ ]. Sometimes a line is 
placed over two numbers, and such a line has the same meaning 
as enclosing in brackets. Thus, if a quantity b + c has to be 
multiplied by d+f the terms may be written as b + exd+f, or 
as (b + c)(d+f). In this way the use of brackets gives a short 
method of indicating multiplication. The use of the different 
forms of brackets can be shown by the following examples : 

Ex. 1. 3a -(46 -7c). 

Here, the brackets indicate that the expression 46 - 7c is to be 
subtracted from 3a. We have already found (p. 63) that in the 
process of subtraction we change the sign of each term and 
then add ; hence it is obvious that the result obtained will be 
the same whether we first subtract 7c from 46 and afterwards 
subtract the remainder from 3a, or first add 7c to 3a and subtract 
46 from the sum. 



USE OF BRACKETS. 71 

If a positive sign occur before a bracket the signs of all the 
terras remain unaltered when the brackets are removed ; if a 
minus sign is placed before the bracket the signs of each term 
inside the brackets must be changed when the brackets are 
removed. 

Ex. 2. 3a + (4b-7c + 3d) = 3a + 4b-7c + 3d. 
3a-(4b-7c + 3d) = 3a-4b + 7c-3d. 

The other forms of brackets which are used are [ ] and { }. In 
each case they denote that whatever is in one pair of them is to 
be regarded as one quantity to be added, subtracted, multiplied, 
or divided, as a whole, in the manner which the signs and quan- 
tities outside the brackets indicate. 

Ex. 3. Express the product of 2a + 3b and 4c + 5d. 
The quantities may be written as (2a + 36) (4c + 5d). 

Further, to indicate that 3/ is to be subtracted from the 
product and the result multiplied by 7e, we use another pair of 
brackets, thus, 7e{(2a + 3b)(4c + 5d) 3f\ ; and to express that 
when 3x is subtracted from the last obtained product the whole 
must be multiplied by 8, we have to use another bracket, thus, 
8 [7e{ (2a + 36) (4c + bd) - 3/} - &e]. 

In removing the brackets the work may commence either 
from the inside pair, removing one form of bracket at each 
operation until the outside pair is reached. Or, the work may 
with advantage commence at the outside pair, repeating until 
the inside pair is reached. Moreover, to prevent mistakes, it is 
advisable only to remove one pair of brackets at each step. 

Ex. 4. Simplify 

8 (a 2 + x 2 ) - 7 (a 2 - x 2 ) + 6 (a 2 - 2a; 2 ) 
= 8a* + 8x 2 - 7a 2 + 7x 2 + 6a 2 - 12x 2 
= 7a 2 + 3x 2 . 

Ex. 5. Simplify 

4x - [{4x - 4y) {4x + Ay) - {4x + (4x + 4y) {4x - 4y)\ + 4y\ 
Multiplying the terms in the brackets we get 

4x - [16ic 2 - I6y 2 - {4x + 16a; 2 - 1 6y 2 } + 4y] 
or 4x - [16a: 2 - 16y 2 - 4x - lQx 2 + 16y 2 + 4y\ 

= 4x-l6x 2 + I6y 2 + 4x + 1 6x 2 - IQy 2 - 4y 
= 8x- 4y. 



72 PRACTICAL MATHEMATICS FOR BEGINNERS. 

Ex. 6. Simplify 

3a- [a + b- {a + b + c-{a + b + c + d)Y] t 
3a - [a + b - {a + b + c - a -b - c - d)~\ 
=3a-[a + b + d] = 2a-b-d. 

EXERCISES. XIII. 

Simplify 

1. (tt.-y+i^yfezjg, and find the value when 

a+b b+c a+c 

a = 5, 6 = 3, c=l. 

2. When a = % 6 = 5, x = 4, of 

2a 2 -6 a 2 -26 3a{b-x) 



b-x x + b 25 - x 2 

3. a-[26 + 3c + {2a-46-(a-26 + 4c)}]. 

4. 6.r + y - [3x + 2y - { 7x - 2z - {Qy - 18z) }], and subtract the result 
from 13a: - 7 (a; - y) - i6 (z + a;). 

5. Remove the brackets from the expression 

4(a + 26)-[3a-4{a-(26-3a)}]. 

6. Find the value of 

a + b + c abc , , , , 
j r- when a = f, b= -*, c = #, a = 4. 

a - b- c a * " 

Simplify the following, find the value in each case when 
x=l,y=-$,z=-2. 

7. 3a;-4y-(2a;-3y + z)-(5a; + 2y-3z). 

8. x 2 + y[x-{y + z)]-[x 2 + 5xy-y{x + z)~\. 

9. lla; + 2y-[4a;-{7y-(8a? + 9y-3z)}]. 

Simplify 

10. 12c-[{36-(26-a)}-4a + {2c-36-a-26}]. 

l+gZ-a-a? 
1 -a 

12. (a + 6 + c) 2 -(a-6 + c) 2 + (a + 6-c) 2 -(-a + 6 + c) 2 . 

13. Simplify 7 (a - b) (6 - 2a) - (2a - 6) (6 - 7a) - 66 2 . 

14. From 9 (a - 6) 2 take the sum of (3a - 6) 2 and (a - 36) 2 . 



CHAPTER VII. 

FACTORS. FRACTIONS. SURDS. 

FACTORS. 

A knowledge of factors must be obtained before the student can 
hope to deal successfully with algebraic expressions and their 
simplification. 

Factors. When an algebraic expression is the product of two 
or more quantities, each of these quantities is called a factor of it. 
Thus, if x + b be multiplied by # + 6, the product is 
x 2 + 11^ + 30, 
and the two quantities x + 5 and x +6 are said to be the factors of 
^ 2 + ll^ + 30. 
The determination of the factors of a given expression, or, as 
it is called, the resolution of the expression into its factors, may 
be regarded as the inverse process of multiplication. 

The following results, easily obtained by multiplication, occur 
so frequently, and are of such importance, that they should 
be carefully remembered : 

(a + b)(a + b) or (a + 6) 2 = a 2 + 2a& + 6 2 (l) 

(a-b) 2 = a?-2ab + b 2 (2) 

The results are equally true when any other letters are used 
instead of a and b. We can write with equal correctness 
(x + y ) 2 = x 2 + 2xy + y 2 . 

Or, The square of the sum of two quantities is equal to the sum 
of the squares of the quantities increased by twice their product. 

Similarly, The square of the difference of two quantities is equal 
to the sum of the squares of the quantities diminished by twice their 
product. 



74 PRACTICAL MATHEMATICS FOR BEGINNERS. 

By multiplying (x+y)(xy) we obtain x 2 y 2 . Conversely 
given x 2 y 2 we can at once write down the factors as x+y and 
x-y. 

The first of these relations may be expressed as : The product 
of the sum and the difference of two numbers is equal to the difference 
of their squares. 

Ex. 1. 40 2 - 39 2 = (40 + 39) (40 - 39) = 79 x 1 = 79. 

Ex. 2. 10002-998 2 = (1000 + 998) (1000 -998) = 1998x2 = 3996. 

Ex. 3. To obtain the factors of (a 2 + b 2 - c 2 ) 2 - Aa 2 b 2 . [This may 
be written (a 2 + o 2 - c 2 ) 2 - (2a6) 2 . ] 

Using the last rule given above we get 

(a 2 + b 2 - c 2 + 2ab) (a 2 + b 2 - c 2 - 2ab), 
or {{a + b) 2 -c 2 }{(a-b) 2 -c 2 }. 

Again using the rule we get 

{a + b + c) {a + b - c) {a - b + c) {a - b - c). 

Ex. 4. To obtain the factors of x* - y 4 . 
First (x 4 - y 4 ) = {x 2 + y 2 ) {x 2 - y 2 ) . 

Also as x?-y 2 ={x+y){x-y), 

we can write x 4 - y* {x 2 + y 2 ) {x + y) {x - y). 

Ex. 5. Multiplying a 2 - ab + b 2 by a + b, the product is found to be 
a 3 + b 3 . 
.-. a 3 + b s = (a + b) (a 2 - ab + b 2 ). 
Similarly a 3 - b 3 = {a - b) (a 2 + ab + b 2 ). 

The quantities (a + b) (a 2 - ab + b 2 ) are the factors of a 3 + b 3 , and 
(a - b) {a 2 + ab + b 2 ) are the factors of a 3 - b 3 . 

Generally a n +b n is divisible by a+b when n is an odd number, 
1, 3, 5, etc. Thus, in Ex. 5, n is 3. 

Also a n b n is divisible by a- b when n is an odd number 
The case of n = 3 is shown, and by actual division, assuming 
n to be any odd number, the rule can be further verified. 

When n is an even number, 2, 4, etc., it will be found that 
a n - b n is divisible by both (a + b) and (a - b). 

Ex. 6. Let n = 6 ; :. a n -b n becomes a 6 - 6 6 . 

We know that a 6 -b 6 = {a 3 + b 3 ){a 3 -b 3 ), and in Ex. 5 the factors of 
(a 3 + b 3 ) and (a 3 - 6 3 ) have been obtained. 
Hence the factors of a 6 - & 6 are 

(a + b) (a 2 -ab + b 2 ) (a - b) {a 2 + ab + b 2 ). 
Thus a 6 - 6 6 is divisible by both a + b and a - b. 



FACTORS. 75 



When the preceding simple examples are clearly made out it 
is advisable to consider the more general expression a n b n , and 
to find that : 

a*+b n i S divisible by a+b when n is odd. 

a n b n ,, a b ,, ,, 

a n_bn }> 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 



~/<K-0 



CHAPTER IX. 

SIMULTANEOUS EQUATIONS AND PROBLEMS 
INVOLVING THEM. 



Simultaneous equations. If an equation contains two un- 
known quantities denoted by x and y, then by giving definite 
values to one of the unknown quantities, a corresponding series 
of values can be obtained for the other. 

Ex. 1. Solve 3#-5y = 6. 

This means that we require to find two numbers such that five 
times the second subtracted from three times the first number will 
give 6. 

By transposition, Sx = 5y + Q ; and giving values 1, 2, 3, etc., to y, 
we may obtain a corresponding series of values of x. 

If, y=l, then 3x = ll ; /. x = Q 

y=2, then3a=16; .'. x = J . 

Proceeding in this manner, a table of values can be arranged as 
follows : 



X 








7 







y 


1 


2 


3 


4 







Thus, for any assigned value of y a corresponding value of x 
can be obtained. 

In a similar manner if values are assigned to x, corresponding 
\alues of y can be found. 

If, now, we have a second equation 4x+3y 37, then as before, 
by giving any assigned value to either x or y, a corresponding 
value of the other unknown is obtained, and a table of corre- 
sponding values of x and y can be tabulated as in the preceding 



SIMULTANEOUS EQUATIONS. 91 

case. Comparing the two sets of values so obtained it will be 
found that only one pair of values of x and y will satisfy both 
equations at once, or the two simultaneous values are x= 7, y = 3. 

Equations such as 3x by = 6, 

4# + 3y = 37, 
which are satisfied by the same values of the unknown quantities, 
are called simultaneous equations. 

To find two unknown quantities, we must have two distinct and 
possible equations. 

Ex. 2. 4x + 3y = 37, and 12# + 9y=lll. 

These form two equations, but they are not distinct, as the second 
can be obtained from the first by multiplying by 3. 

To solve simultaneous equations, we require as many distinct 
and independent equations as there are unknowns to be found, 
i.e. if two unknowns have to be determined, two distinct 
equations are required ; if three unknowns, three equations, 
and so on. 

If only one equation connecting two unknown quantities is 
given, although the value of each of the unknowns cannot be 
determined, it is still possible to obtain the ratio of the 
quantities. 

Ex. 3. If 5x ~^ = 4, find the ratio of x to y. 
3x-2y 

:, 5x-4y = 4{3x-2y) = 12# - 8y. 

/. 4y = 7x, 

or - = -. 

V 7 

Elimination. When in the data of a problem the given 
equations are not only distinct, but are sufficient in number, it is 
possible from such data to obtain others, in which one or more 
of the unknown quantities do not occur. The process by which 
this is effected is called elimination. At the outset it is convenient, 
in a few simple cases, to show some of the methods which may be 
adopted in dealing with simultaneous equations containing two 
or more unknown quantities. 

Solution of simultaneous equations. In the solution of a 
simultaneous equation containing two unknown quantities, there 
are two general methods by which their values may be obtained. 
The first is by multiplication or division, which processes are 



92 PRACTICAL MATHEMATICS FOR BEGINNERS. 

used to make the coefficients of one of the unknowns the same in 
the two equations. Then, by addition, or subtraction, we can 
eliminate one unknown quantity. This leaves an equation con- 
taining only one unknown, the value of which can be found in 
the usual manner. 

The other method is to find the value of one unknown in 
terms of the other unknown in one of the equations, and then 
to substitute the value so found in the other equation. 

Ex. 4. 3#-5y= 6, (i) 

4x + Sy = 37 (ii) 

To apply the first method, multiply (i) by 3 and (ii) by 5. This 

will make the terms in y the same in both equations, and as these 

have opposite signs their sum is zero. 

.'. 9a;-15y= 18 

20^ + 15^^185 

By addition, 29a; =203 

203 - 

*=29" = '- 

Substitute this value in (i) ; 

.-. 21-5y = 6; 

or 5y = 21-6=15; /. y = S. 

On substituting these values of x and y in the given equations the 

equations are satisfied. Thus, substituting the values in (i), we get 

3x7-15 = 6. The values obtained should always be substituted in 

this manner to ensure accuracy. 

By the second method : 

From (i) 3a;=6 + 5y; 

6 + 5y 
or X =~^L; 

. . 24 + 20y 
.. to- 3 



Substitute this value in (ii) ; 

24 + 20y 



+ 3y=37. 



3 

Multiply both sides of the equation by 3 ; 

or 24 + 20*/ + 9y = lll. 

Hence 29y=lll-24= 87; 

87 Q 
- ^ = 29 = 3 - 
Having found the value of y, then by substitution in (i) or (ii), 
the value of x is readily obtained. 



SIMULTANEOUS EQUATIONS. 93 

Miscellaneous examples. As the solution of simultaneous 
equations is of the utmost importance, a few miscellaneous 
examples are worked here. 

Ex. 5. 6x+3y=33,) (i) 

13x-4y = 19.J (ii) 

Multiplying (i) by 4, we get 24#+12y=132 

(ii) by 3, we get 39a; -12y = 57 

By addition 63a: =189 

189 o 
* = -63- = 3 ' 
and by substitution in (i), y = 5. 

If the known quantities are represented by the letters, a, 
b, c, d, the solution is effected in the same manner. 

Ex. 6. Solve ax + by = c, (i) 

bx + ay = d, (ii) 

Multiplying (i) by b, we get abx + b 2 y = bc 
,, (ii) by a, we get abx + a 2 y ad 

By subtraction, b 2 y - a 2 y = bc-ad 

or y{b 2 -a 2 ) =bc-ad; 

_bc-ad 
" y ~b 2 -a*' 
To obtain x we may either substitute for y, or proceed to eliminate 
y from (i) and (ii). 

Thus multiplying (i) by a, a 2 x + aby = ac 

,, (ii) by b, b 2 x + aby bd 

Subtracting the upper line from lower, (b 2 -a 2 )x = bd- ac 

bd-ac 



From the preceding examples the student will have seen that 
in solving two simultaneous equations, the object is to determine 
from the two given equations a value of one of the unknowns. 
Using the value so obtained we proceed to find the other. The 
methods which may with advantage be employed in solving 
equations quickly can only be seen by practice. 

Simultaneous equations of more than two unknowns. 
The general methods previously explained may usually be 
employed. The following methods are also made use of when 
more than two unknowns have to be found. 



94 PRACTICAL MATHEMATICS FOR BEGINNERS. 

Ex. 7. Solve x+ y + z = 53 (i) 

x + 2y + 3z = 105 (ii) 

a; + 3y + 4z=134 (iii) 

Subtract (i) from (ii), .*. y + 2z = 52 (iv) 

,, (ii) from (iii), y + z= 29 (v) 

By subtracting (v) from (iv), z= 23 

Substitute this value for z in (v), 

.'. y + 23 = 29; 
or y = 29-23 = 6. 

Again substituting for y and z in (i), 

a; + 6 + 23 = 53; 

.'. a: = 53- 29=24. 
Hence the values are x- 

y 



Partial fractions. When the denominators of two or more 

fractions are alike we can proceed tc add, subtract, or compare 

the fractions ; in like manner the converse operation would be to 

replace a given fraction by two or more simpler fractions as in 

the following example : 

3x - 16 
Ex. 8. Express 2 '_ Q as the sum of two simpler fractions. 

Here the given fraction is 7 ^- 77. 

(#-3)(a;-4) 

We may express this as the sum of two simpler fractions 

A B 
--\ 7, and we require to find the numerical values of the 

numerators A and B. 

1 3a; -16 A B 

Then clearing of fractions we have 

3x-16 = A(x-4) + B(x-3) = {A+B)x-(4A+3B). 
The coefficients of the terms in x are 3 on the left hand side, and 
A+B on the right. 
Equating coefficients : 

A+B= 3 (i) 

4.4+3^=16 (ii) 

Multiplying (i) by 3 and subtracting from (ii) we find 
A =7, and B= -4. 

Hence *>-*\ 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 



<j- 



13. 



11 11,11 -ia x v \ 

- + - = a, - + - = &, - + - = c. 14. - + ^- = m.\ 
v y y z z x p q \ 

x y [ 
Q P ) 



15. If 7 (x - y) = 3{x + i/), find the ratio of x to y. 

16. t^=*-7. 1 . ^ + ?^ 

40 I b a 



a 6 

18. {a+p)x+{b-q)y = n.} 
{b-q)x + (a+p)y = n.j 

Problems leading to simultaneous equations. It will be 
found that some practice is necessary before even the data 
of a simple question can be expressed in algebraic symbols,. 



96 PRACTICAL MATHEMATICS FOR BEGINNERS. 

and it is necessary to remember that in all cases there must be 
as many independent equations as there are unknowns to be 
determined. Thus if a simultaneous equation contains two 
unknown quantities, then two independent equations are requi- 
site ; if only one equation is given the ratio of one unknown to 
the other can alone be determined. 

If x and y denote two numbers, then the sum of the two is 
x+y, the difference of them is xy, the product xy, etc. 

Ex. 1. Find two numbers the sum of which is W> 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 <x t 2 , 

:. s=kt\ 

64-4 = &x2 2 ; or, &=16-1. 
Hence, in 5 seconds s=16*l x 5 2 = 4025 feet. 

Stress and Strain. Stress is directly proportional to strain. 

Hence stress oc strain, or = constant. This is known as 
strain 

Hooke's Law ; the word stress denoting the force per unit area 

or the ratio of load to area, and strain the ratio of alteration of 

length to original length. The constant is called the modulus 

of. elasticity for the substance and is usually denoted by the 

letter J?. 

Ex. 2. The area of cross-section of a bar of metal is 2 sq. in., and 
when a load of 10,000 lbs. is applied the alteration in the length of 
the bar is '0288", find E. The length of the bar is 12 feet. 

Here Stress = - = ^ = 5,000 lbs. per sq. in. 
area 2 r ^ 

o, . alteration in length '0288 nnn r> 

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<x &Rt, or H=hO i m. 



104 PRACTICAL MATHEMATICS FOR BEGINNERS. 

When H is 777600, G is 10, R is 18 ohms, and = 30 minutes, find 
H when G = 20, R = 60, t = 60. 

Here 777600 = &x lO^x 18 x 30 x 60, 

^ 777600 _ 

" 100x18x30x60" 
Hence #= -24 x 20 2 x 60 x 60, 

.'. H= 345600. 
Inverse proportion and variation. One quantity is said to 
vary inversely as another when the product of the two quantities is 
always constant. Or, 

xy = k; 
t 

y 

Hence, as one quantity increases the other decreases in the same 
ratio ; or, one quantity varies as the reciprocal of the other. 

The reciprocal of a quantity is unity divided by the quantity ; 

thus, the reciprocal of y is -. 

y 

For a given quantity of gas the force exerted varies inversely as 
the volume ichen the temperature remains constant. This relation 
is known as Boyle's Law for a gas. 

Denoting the pressure and corresponding volume of a gas by 

p and v, the law gives p x -, 

or pv = constant = Jc (i) 

Ex. 4. When the pressure of a gas is 60 lbs. per sq. in. the 
volume is 2 cub. ft. If the gas expands according to Boyle's Law, 
find the pressure when the volume is 3 cub. ft. 

From (i) we have 60 x 2 = k ; .'. h= 120. 

If the volume change to 8 cub. ft. , then the pressure is given by 

p = - = -jr- = 40 lbs. per sq. m. 

The load W that a bar, or beam, of length I, breadth b, and 
depth d will carry, varies directly as the breadth, as the square 
of the depth and inversely as the length, or 

r* T . 

Ex. 5. A bar of fir 10 in. long, 1 in. broad, and 1 in. deep will 
carry a load of 540 lbs.; find the depth of a bar of fir similarly 



PROPORTION. 105 



loaded to carry a load of f ton when the breadth is 2 in. and the 
length 5 feet. 



w = k T- 

To find k we have 



540= *** ** ; .% =5400. 



Hence fx2240: 



10 

5400x2x<ff 
60 



3x2240x60 , AK . 

J= 4x5400x2 ; <* =305m - 
Simple and compound proportion. By introducing alge- 
braical symbols questions which involve arithmetical difficulties 
are readily solved, as in the following example : 

Ex. 6. If 40 men working 9 hours a day can build a wall 50 ft. 
long in 16 days, find how many men will be required to build a 
similar wall 25 ft. long in 20 days working 8 hours a day. 

Here the number of men required will vary directly as the length 
of wall to be built, and inversely as the number of days and the 
number of hours per day. 

Denoting by m, I, d, and h, the number of men, length of wall, 
number of days, and number of hours per day. 

Then the statement is, m oc -jr, 
dh 

or m=k. -jt (i) 

an, 

To find the value of k it is only necessary to substitute the given 

values for m, I, d, and h . 

._ , 50 .40x16x9 

.-. 4Q=k . a , or k- R . 

16 x 9 5 

To find m the number of men required substitute the known 

values for the right-hand side. Then 

4x16x9 25 

m = 5~ X 2(bT8 

= 18. 

EXERCISES. XX. 

1. The rents of an estate should be divided between A and B in 
the proportion 5:3; 470, however, is paid to A, and 280 to B. 
Which has been overpaid, and by how much ? 

2. If the ratio of 2x + y to Qx - y equals the ratio of 2 to 3, what 
is the ratio of x to y ? 



106 PRACTICAL MATHEMATICS FOR BEGINNERS. 

3. If an express train, travelling at the rate of 55 miles an hour, 
can accomplish a journey in 3^ hours, how long will it take a slow 
train to travel two -thirds of the distance, its rate being to that of 
the express train as 4 to 9 ? 

4. A'a rate of working is to i?'s as 4 to 3, and Z?'s is to C"s as 2 to 
1. How long will it take G to do what A would do in 6 days ? 

5. A gas is expanding according to the law pv = const., if when 
> = 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 <j> = 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 &<l x log a b 

Hence, if the logarithm of any number in the system in which the 



140 PRACTICAL MATHEMATICS FOR BEGINNERS. 

base is a be multiplied by -, we obtain the logarithm of the 

number in the system in which the base is b. 

The common logarithms, or, as they are usually called simply 
logarithms, have been calculated from the Napierian logarithms. 
Let I and L be the logarithms of the same number in the common 
and Napierian systems respectively, then 

log e 10 =2 -30258509 = 2 -3026 approx., 
and 2509= '43429448= 43429 approx. 

Hence, the common logarithm of a number may be obtained by 
multiplying the Napierian logarithm of the same number by 
43429.... 

To convert common into Napierian logarithms multiply by 2 3026 
instead of the more accurate number 2 30258509. 

The preceding rules will be best understood by a careful study of 
a few examples. 

Ex. 1. Log 10 to base e is 2*3026. 

.-. loge 10=2-3026, 
or e 2-3026 =10. 

From this relation any number which is a power of 10 may be 
expressed as a power of e. Thus in Table III. log 19*5= 1 "29. 

.'. 19-5 = 10 1 ' 29 = e 2-3026xl ' 29 = e 2 ' 9703 
or log 10 19'5 = l-29, log e 19 "5 =2 '9703.' 

Ex. 2. Find log e 3 and log e 8 '43. 

.-. log e 3 = -4771 x 2-3026 = 1 -0986, 
log 8 43 ='9258; 
.-. log e 8-43= -9258 x 2-3026 = 2-1317. 

Ex. 3. Find log 13 to base 20. 

Here log 13 = 1 -1139, also log 20= 1 3010. 

' log 2 ol3=^=-8562. 



Find the value of 
1. -03571 x -2568. 
Divide 



EXERCISES. XXVI. 

o 8352 x 3-69 1 -265 x -01628 



(30 -57) 3 2-283x64-28 

4. (i) 5-3010 by 9. (ii) 4-4771 by 11. 



EXERCISES. 141 



Find the logarithms of 

6 - (f )* * VSr 

* 3 l& 



9. (1500) ,( i. 
W3 

10. Find the tenth root of -0234. 

11. Find log 4/ -00054 x 3 6 and log ^TJn^j: 

12. Find the value of \/2543 x 1726. 

13. Why do we divide the logarithm of a number by 3 to obtain 
the logarithm of its cube root. Find the 'cube root of 44*6. 

y 14. Find the value of E from the equation 

F _ 80x33-62x 16 

1-5 x ( -375) 2 x -7854 x -0166* 
15. Find the fourth proportional 4/8*37, *84, and #-05432. 
z 16. Find the value of N from the equation 

7M2x(l-25) 2 
1406 x(-25) 4 x -022' 
17. Find the hyperbolic logarithms of 15, 2, 2*5, 3, 3*5, 4, 4*5, 5. 
Find the value of x from 



(ir 



100. 19. (l-05)*=(8-25)*- 

20. Find the eleventh root of (39 -2) 2 . 

21. Find the seventh root of '00324. 

22. Find a mean proportional to V4756 and ( -0078) 7 . 

23. Prove that log a 6 + log 6 a = 1. 

24. If a = 10'4, 6 = 2-38, #=-2064, and y=-09SQ, determine the 
values of a b and ah~%(a? - y 2 f. 

25. Find the value of p^kr' 1 -r~ k ) -p 3 , when p 1 = 80,p 3 = 3, k= -8, 

r=vm - . S^-Ai' 

ys 26. If s varies as i* and s is 20*4 when t is 1'36, find t when 
' .s = 32-09. U^x t 

27. If Q varies as H* and Q is 7 "26 when H is 1 -5, find Q when H 
is 1-359. 

2 

28. Having given the equation y ax + bz^x 2 , find the values of a 
and b, if y = 49'5 when x=l, and z = 8 and y = 356 when a-=l*5 and 
z=20. What is the value of y when x = 2 and z = 20 ? ? .. . 

>* 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<mB. 

Involution with the slide rule. On inspection, the numbers 
on the upper scale A are seen to be the squares of the numbers 
on the lower scale D. 

To obtain the square of a fractional number some difficulty 
would be experienced in noting the coincidence of divisions on 
A and D, separated as they are by the slide ; in this case we can 
make use of the runner, thus : 

Set the runner to coincide with the given number on the scale 
/), and, by its means, read off the square of the number on scale 



SLIDE RULE. 149 



A. In this manner, having obtained the square, the fourth, or 
any even power, can be obtained. 

Square root by the slide rule. In extracting square roots the 
process for involution is reversed. 

'Hie number, the root of which is required, is found on the 
scale A , and its root is, as before, found directly below. The 
runner enables the coincidence of the two divisions denoting the 
number and its root to be readily obtained. 

As shown on p. 225, the area of a circle is 3*1416 xr, or 
*7854cP, where r is the radius and d the diameter of the given 
circle. Conversely, if the area of a circle is given, the diameter 
can be obtained from : 

diameter =V^. 

Ex. 1. Find the area of a circle 3" diameter. 

The mark on the runner is set to the 3 on the lower scale ; 
the upper mark over the top scale registers 9. Then moving the 
slide to the right until its 1 coincides with the mark, we have 
coincident with *7854 (which is marked on the scale) the required 
area 7 '06 square inches. 

Ex. 2. Find the area of a circle 2*5" diameter. 

The square of 2*5 is seen to be 6 '25, and multiplied by '7854 the 
area is 4*9 square inches. 

To obtain the cube of a number with a slide rule. First 
method. Bring the right-hand 1 of scale C to the given number 
on D. Then over the same number on the scale B read off the 
required cube on A. 

Second method. The slide may be inverted, keeping the same 
face upwards. The scale B will now move along scale D. Put 
in coincidence on the scales B and D the two marks indicating 
the number the cube of which is required, then opposite the 
right-hand 1 on the slide the cube required will be found on 
scale A. 

Cube root with the slide rale. First method. Set the 
given cube on scale A and coincident with it on scale B any 
rough approximation to the root, move the slide to the right or 
left until on B coincident with the cube on A, the same number 
is simultaneously found on D opposite the right or left hand 
index of C. 



150 PRACTICAL MATHEMATICS FOR BEGINNERS. 

Second method. The inverted slide is used. This is placed 
with the right-hand 1 of the slide coincident with the number 
ths cube root of which is required. Now find what number on 
the scale D coincides with the same number on the inverted scale 
of B ; this number is the cube root required. 

Ex. 1. Find the cube root of 64. 

Move the slide from right to left, and it will be found that 4 on 
scale B coincides with 64 on scale A, simultaneously with 4 on D 
and 1 on G. Hence 4 is the cube root required. 

Invert the slide, keeping the same face uppermost. Set 1 on G 
inverted, to 64 on scale A, the division on B which coincides with 2) 
is 4. Hence 4 is the cube root. 

When the product or quotient of several numbers is wanted 
the runner may be used with advantage to record the results of 
the partial operations. 

Ex. 2. Find the value of 3 ' 4x2 ' 8 . 
1-7 x -3 

We may use either scales A and B or C and D. 

Using G and D move the slide until the graduation 1 on scale G 
coincides with 3*4 on scale Z>. 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<V3=173...ft. 

Ex. 2. From the top of a tower, the height 
of which is 100x^3 ft., the angle of depression 
of an object on a straight level road on a 
line with the base of the tower is found to 
be 60. Find the distance of the object from' 
the tower. 

In this case, draw GD a horizontal line 
(Fig. 72) through G, the point of observa- 
tion, making GB 100 \/3 ft. on any scale, 
and BA at right angles to GB. Then the 
point at which a line GA, drawn at an 
angle of 60 to the horizontal line DC, meets BA gives the distance 
BA required =100 ft. 




wo * 

Fig. 72. 



EXERCISES XXXI. 

1. ABG is a triangle with a right angle at G, GB is 30 feet long, 
and BA C is 20. If GB be produced to a point P, such that PAG is 
55, find the length of GP. 

2. The angular height of a tower is observed from two points A 
and B 1000 feet apart in the same horizontal line as the base of the 
tower. If the angle at A is 20 and at B 55, find the height of the 
tower. 

3. The angle of elevation of the top of a steeple is 30. If I walk 
50 yards nearer, the angle of elevation becomes 60. What is the 
height of the steeple ? 

4. The elevation of a tower from a point A due N. of it is 
observed to be 45, and from a point B due E. of it to be 30. If 
AB = 240 feet, find the height of the tower. 

5. If from a point at the foot of the mountain, at which the 
elevation of the observatory on the top of Ben Nevis is 60, a man 
walks 1900 feet up a slope of 30, and then finds that the elevation 
of the observatory is 75, show that the height of Ben Nevis is nearly 
4500 feet. 

6. The angle of elevation of a balloon from a station due south of 
it is 60 ; and from another station due west of the former, and 
distant a mile from it, the angle of elevation is 45. Find the height 
of the balloon. 



170 PRACTICAL MATHEMATICS FOR BEGINNERS. 

7. Two ships leave the harbour together, one sailing N.E. at the 
rate of 7| miles an hour, and the other sailing north at the rate of 
10 miles an hour ; find the shortest distance between the ships an 
hour and a half after starting. 

8. A and B are two points on one bank of a straight river and C 
a point on the opposite bank ; the angle BAG is 30, the angle 
ABC is 60, and the distance AB is 400 feet ; find the breadth of the 
river. 

9. From the top of a cliff 1000 feet high, the angles of depression 
of two. ships at sea are observed to be 45 and 30 respectively : if 
the line joining the ships points directly to the foot of the cliff, find 
the distance between the ships. 

10. Two knots on a plumb line at heights of 7 feet and 2 feet 
above the floor cast shadows at distances of 1 1 *4 feet and 1 "65 feet 
respectively from the point where the line produced meets the 
floor. Find the height of the source of light above the floor. 

11. From a ship at sea the angle subtended by two forts A and B 
is 35. The ship sails 4*26 miles towards A and the angle is then 
51 ; find the distance of B from the ship at the second point of 
observation. 

12. A tower stands at the foot of an inclined plane whose inclina- 
tion to the horizon is 9; a line is measured up the incline from the 
foot of the tower of 100 feet in length. At the upper extremity of 
this line the tower subtends an angle of 54. Find the height of 
the tower. 

13. From two stations A and B on shore 3742 yards apart a ship 
G is observed at sea. The angles BAG, ABG are simultaneously 
observed to be 73 and 82 respectively. Find the distance of A 
from the ship. 

14. Three vertical posts are placed at intervals of one mile along 
a straight canal, each rising to the same height above the surface of 
the water. The visual line joining the tops of the two extreme 
posts cuts the middle post at a point 8 inches below the top ; find 
to the nearest mile the radius of the earth. 

15. The altitude of a certain rock is observed to be 47, and after 
walking 1000 feet towards the rock, up a slope inclined at an angle 
of 32 to the horizon, the observer finds that the altitude is 77. 
Find the vertical height of the rock above the first point of observa- 
tion. 

16. A balloon is ascending uniformly, and when it is one mile 
high the angle of depression of an object on the ground is found to 
be 35 20', 20 minutes later the angle of depression is found to be 
55 40' ; find the rate of ascent of the balloon. 



CHAPTER XVII. 



USE OF SQUARED PAPER. EQUATION OF A LINE. 

Use of squared paper. Two quantities, the results of a 
number of observations or experiments, which, are so related 
that any alteration in one produces a corresponding change in 
the other, can be best represented by a graphic method, in which 
it is possible to ascertain 



by inspection the relation 
that one variable quantity 
bears to the other. 

For this purpose squared 
paper or paper having 
equidistant vertical and 
horizontal lines ^", J", 
1 mm., etc., apart is em- 
ployed ; these cover the 
surface of the paper with 
little squares (Fig. 73). 

Commencing near the 
lowest left-hand corner of 
.the paper, one of the lower horizontal lines may be taken as the 
axis of x and a vertical line near the left edge of the paper as 
the axis of y. 

The two lines ox and oy at right angles are called axes. The 
horizontal line ox is called the axis of abscissae ; oy is the axis 
of ordinates 

One or more sides of the squares, measured along the line ox, is 
taken as the unit of measurement in one set of the observations ; 
and, in a similar way, one or more sides of the squares bordered by 
the line oy is taken as the unit of the other set of observations. 



x 



Fig. 73. Squared paper. 



172 PRACTICAL MATHEMATICS FOR BEGINNERS. 

Then for any pair of observations two points are obtained, 
one on ox and the other on oy. 

If a vertical line were drawn upon the squared paper, from 
the point on ox, and a horizontal line from the point on oy, the 
two lines would intersect. It is not necessary to draw such 
lines ; they are furnished by the lines on the squared paper 
By dealing with a number of pairs of observations a series of 
points may be obtained upon the squared paper ; these are 
usually marked with a small cross or circle. 

In this manner a sufficient number of points separated by 
short distances may be obtained, and the points joined by a 
straight or curved continuous line called a graph or a locus. 
From such a line intermediate values may be read off. 

When the plotted points are obtained by calculation from a 
given formula, the curve is made to pass through each point ; 
but when the points denote experimental numbers, and there- 
fore include errors of observation, any attempt to draw a 
continuous line through the points would give a curve consisting 
of a series of angles or sharp bends. As such an irregular curve 
would be for many purposes practically useless it is better to 
use a piece of thin wood ; this when bent may be used to draw 
a fairly even curve lying evenly among the points, probably 
passing through a few of the plotted points, above some, and 
below others. The curve also furnishes a check on the plotted 
values and gives a fair idea not only as to the numbers which 
are in error, but also in each case their probable amount. 

In the case of experimental results, the points can often 
be arranged to lie as nearly as possible on a straight line, 
and the line which most nearly agrees with the results may be 
obtained by using a piece of black thread. This thread is 
stretched and placed on the paper, moved about as required 
until a good average position lying most evenly among the 
points is obtained. 

A better plan is to use a strip of celluloid or tracing paper on 
which a line has been drawn, the transparency of the strip 
allowing the points underneath it to be clearly seen. When the 
position of the line is determined, two points are marked at its 
extremities, and the line is inserted by using a straight edge or 
the edge of a set square. 



USE OF SQUARED PAPER. 



173 



The word curve is often used to denote any line, whether 
straight or curved, representing the relation between two 
quantities. 

Ex. 1. The following table gives the number of centimetres in a 
given number of inches. Plot these on squared paper and find from 
the curves the number of centimetres in 1^ inches, and the number 
of inches in 8 centimetres. 



Inches. 


1 


2 


24 


3* 


Centimetres. 


2-54 


5-08 


6-4 


8-8 



Use the vertical axis oy to denote inches and the horizontal axis 
ox to denote centimetres (Fig. 74). Read off 2*54 on the horizontal 
and 1 on the vertical axes ; these two values receive various names, 



5 


r 


















































































































































































































































































































































































































































































4 
















































































































































































































































































































































































i 


r' 


































































































3 






































































































































































" 


















































































































to 
































































. 





























































































































a . 






























































































E 
















































r 














































"~" 




















































































































































































































































































































a 




























































































* 




































































































































































































































































































































































* 





































































































i 






























' 


\ 

e 


n 


ti 


m 


hi 


re 


s 






t 




















i 










i 





Pig. 74. Relation between centimetres and inches. 



with all of which it is important to become acquainted. Thus, the 
two values of the point a may be called the x-coordinate and the 
y-coordinate of point a ; or simply the point (x, y) ; this when 
the given values are substituted becomes the point (2*54, 1). 

The next point b is the point (5*08, 2), i.e. its abscissa is 5*08, and 
its ordinate 2. 

The two remaining points c and d are obtained in like manner, 
and finally a fine line is drawn through the plotted points. 



174 PRACTICAL MATHEMATICS FOR BEGINNERS. 

In a similar manner the relation between square inches and square 
centimetres, or pounds and kilograms, may be obtained. 

Interpolation. When a series of values have been plotted 
on squared paper, and a line drawn connecting the plotted 
points, any intermediate value can be read off by what is called 
interpolation. Thus, at the point, the ordinate of which is 
l inches, we read off 3*8 centimetres. Similarly, we find 
3*15 inches correspond to 8 centimetres. 

Positive and negative coordinates. To denote negative 
quantities conventional methods similar to those already 

referred to in measuring angles 
(p. 159) are adopted. All distances 
measured above the axis of x are 
positive, and all distances below 
negative ; distances on the right of 
the line oy are positive, and those 
on the left negative. This may also 
be expressed by the statement that 
all abscissae measured to the right 
of the origin are positive, all to the 
left negative. All ordinates above 
the axis of x are positive, and all 
below are negative. 

Ex. 2. In an experiment on a spiral spring the following values 
for loads and corresponding extensions were observed : 
Original length of spring 17*2 centimetres. 

(a) Plot on squared paper the given values of extension and load 
and find the law connecting them ; (b) also find E the modulus of 
elasticity. 



36 



























/ 


/ 










/ 


/ 










^ 


/ 












/ 












/ 












/ 












/ 















5 10 15 20 2-5 30 
7%x tensions. 

Fig. 75. 



Load on spring 
in pounds. 


2 


4 


6 ' 


8 


10 


12 


14 


Extension in 
centimetres. 


60 


96 


1-42 


1-88 


2-35 


2-82 


3-28 



We can use the vertical axis for loads and the horizontal axis for 
extensions ; as the coordinates of the first point are 0'6 and 2, mark 
off the first point as in Fig. 75. 

The next point ( *96, 4) is marked in like manner, proceeding with 



PLOTTING A LINE. 



175 



the remaining observations a series of plotted points are obtained. 
These are observed numbers, and therefore contain errors of observa- 
tion ; consequently the line is made to lie evenly among the points 
instead of being drawn through them. 

To obtain E ; at one point we find a load of 10*8 lbs. produces an 
extension of 2*5 cm. 

Hence, an extension of 1 cm. would require -, 

2*5 

.*. to double the length, i.e. to increase the length of the spring by 

17 '2 cm. requires 



17-2x10-8. 



2 5 



; /. #=74-13 lbs. 



Plotting a line. When the relation between two variables 
is given in the form of an equation, or formula, then assuming 
values for one, corresponding values of the other variable can 
be found and the line plotted. 

Ex. 3. Let y = x + 2 be a given equation. 
When x = 0, y = 2, ; x=l, y = 3; x = 2, y 4=. 

Corresponding values of x and y may be tabulated in two columns 
thus : 



Values of x 





1 


2 


3 


4 


5 


6 


Values of y 


2 


3 


4 


5 


6 


7 


8 



When x 0, y = 2; there is no distance to measure off on the axis 
of x, and asy = 2we measure 2 units upwards, and, as in Fig. 76, this 
gives one point in the line. When x = 2, y = 4; at the intersection 
of the lines through 2 and 4, make a cross or dot. Proceeding in 
this manner, any number of points lying on the line are obtained, 
and the points joined by a fine line, which will be found to be a 
straight line. The line plotted may be written y = ax + b, where 
a=l, 6 = 2. 

Conversely, assuming that the straight line represents a series 
of plotted results of two variables such as E and R. To find the 
law connecting the two it is only necessary to substitute in the 
equation E=aR + b, the simultaneous known values of two 
points in the line. This will give two equations from which 
a and b can be determined as in the previous example. 



176 PRACTICAL MATHEMATICS FOR BEGINNERS. 



Let the values along oy represent values of E, and those along 
ox values of R ; :. from the line (Fig. 76), 
when E is 3, R is 1, 
when E is 8, R is 6. 
Substituting these values in the equation : 

.-. 3 = axl+6 (i) 

8=ax6 + 6 (ii) 

Subtracting 5 = 5a ,*. a \. 

Hence from (i) 6 = 3-1=2. 

Substituting these values for a and 6, the required equation is 

E=R + 2. 
It will be noticed that the value of the term b gives the point 
in the axis of y from which the line is drawn. By altering the 

value of 6, the term a re- 
maining constant a series of 
parallel lines are obtained. 
Thus, let 6=0, then equation 
(i) becomes y=x. /. when 
#=0, y=0, and the result 
obtained by plotting values 
of x and corresponding values 
of y is a line parallel to 
the preceding, but passing 
through the origin. 

Again, let b = 2. 
when x = 0, y% and we 
obtain a line parallel to the 
preceding, intersecting the axis of y at a distance - 2, or 2 units 
below the origin, the equation is now y=x-% The three lines 
are shown in Fig. 76. 

When the term b remains unchanged, but the magnitude of 
a is altered, then when the values of a and b are plotted a series 
of lines are obtained, all drawn from the same point, but each 
at a different inclination to the axis of x, or better, the slope of 
each line is different from that of the rest. 

Equation of a line. When as in Ex. 1, the relation between 
two variable quantities can be represented by a straight line, 



/ v\ Mill 



12 3 4 5 6 7 8 
Fig. 76. Plotting Lines. 



the equation of the line is of the form 

yax+b.. 



0) 



PLOTTING A LINE. 177 

where a and 6 are- constants. Then, if in (i) simultaneous values 
of x and y are inserted, the values of a and 6 can be found. 

Thus, in Fig. 74, when y = \, # = 254 ; 
also when y=% #=5*08. 

Substituting these values in (i) we obtain 

2 = ax 5-08 + 6 (ii) 

l=ax2-54 + 6 (iii) 

By subtraction 



1* 


= ax2'54 


a = 


=4=- 39 - 



Also substituting this value for a in (ii) or (iii) we find 6 = 0. 

Hence the equation of the line is y = '39^, and the line passes 
through the origin. 

Again, in Ex. 2 the equation of the line as before is of the 
form y = ax + b (i), if simultaneous values of x and y be inserted 
for two points the values of a and 6 can be found. 

Referring to Fig. 75 we see that when #=1, y=4; when 
j?=3, y = 13. 

Substituting these values in (i) we obtain 

13 = 3a + 6 .....(ii) 

4= ct + 6 (iii) 

By subtraction 9 = 2a 

/. a = 4*5. 

Also substituting this value for a in (ii) or (iii) we find 
6= -'5. 

Hence the required equation, or straight line graph as it is 
called, is y = 4*5 #- '5 (iv). 

At the point where the line cuts the axis of y the value of x 

is 0. Substituting this value of x in (iv) we get y - *5, i.e. the 

line intersects the axis of y at a point *5 below the origin. The 

point where the line cuts the axis of x is in like manner 

obtained by making y=0. 

'5 
/. = 4*5^- *5 or #= . ='111. 
4-5 

Line passing through two points. From the preceding 
example it will be obvious that we can readily find the equation 
to a straight line passing through two given points, and if 
necessary from the equation proceed to plot the line. 
P.M. b. M 



178 PRACTICAL MATHEMATICS FOR BEGINNERS. 

Ex. 1. Find the equation of the straight line passing through 
the points (1, 4), (3, 13), 

The equation of a straight line is y=ax+b (i) 

Substituting the coordinates of the first point ; .'. 4 = a x 1 + b . . . . (ii) 

,, ,, ,, second,, 13 = ax3 + & ....(iii) 



Subtracting (ii) from (iii) 9 = 2a 

from either (ii) or (iii) we find b= - '5 or a = 4 5. 

Hence the required equation is y \'hx- 5. 

The line can now be plotted, and is shown in Fig. 75. 

As the line passing through two given points can be obtained, two 
such lines will give at their point of intersection simultaneous values 
of x and y, and this may be made to give the solution of a simultane- 
ous equation. 

Simultaneous equations. Two general methods of solving 
simultaneous equations have been described on p. 91; another, 
which may be called a graphical method, is furnished by using 
squared paper. The method applied to the solution of a simul- 
taneous equation containing two unknown quantities, consists in 
plotting the two lines given by the two equations. "When this 
is done the point of intersection of the two lines is obtained. 
This is a point common to both lines, and as the co-ordinates of 
the point obviously satisfy both equations, it is the solution 
required. 

Ex. 1. Solve the simultaneous equations 

(i)3ar+4y=18, (ii) x-2y = 2. 

v rx 18-3# ..... 

From (l) y = - (in) 

From (ii) y = 2x-l. (iv) 

To plot the lines it is sufficient to obtain two points in each and 
join the points by straight lines. 

In (iii), when x = 0, y = 4*5 and when x = 5, y='75, the first gives 
point a (Fig. 77), the second point b. Join a and b and the line 
corresponding to Eq. (iii) is obtained. 

In Eq. (iv), when x=l, y\ and when x = 3, y = 5, the first gives 
point c, the second gives d. Join c and d and the two lines are 
seen to cross at /. This point of intersection is a point common to 
both lines, its coordinates are seen to be x = 2, y = 3, and these 
values satisfy the given simultaneous equations. 



SIMULTANEOUS EQUATIONS. 



179 



In the preceding examples it has been possible to draw a 
straight line through the plotted points, and to obtain an 
equation connecting the two variables. When, however, the 
plotted points lie on a curve, it would be difficult, if not im- 
possible, to express the relation between the two variables by 
means of a law or an equation. In such cases a straight line 
may often be obtained by plotting instead of one of the 
variables, quantities derivable from it, such as the logarithms, the 
reciprocals, or the squares, etc., of the given numbers. 



7 


/ 




t 




71 




r 


is;-:: : : j 




4 :"5;::::::::7: 




N v 




s 7 




51 ' 




=S 7 




* > 




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<W> 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 = <X)x s c 



dx 
Ex. 14. y=4x 3 + 3x 2 + 2x + 3. 

( ^- = \2x* + x + 2. 
dx 

As the differential of a constant is zero the constant 3 connected 

to the variables by the sign + disappears during differentiation. 



Ex. 15. y = 3x* 



dx~2 X6X -2*' 



206 PRACTICAL MATHEMATICS FOR BEGINNERS. 



Ex. 16. y 



i x G-i) x -l 



dx 



x 
4*2 



Ex. 17. y = 2x ~*. 



dx 



-fx^-*" 1 ), 



The process of finding the differential coefficient of a given expres- 
sion, ft. e. the value of ~ is of the utmost importance, and the opera- 
tions involved in many cases consist of simple algebraic processes 
which may be easily carried out as in the preceding and in the 
following examples. 

Ex. 18. Graph the curve y^O'la;-' 25 . Find the slope of the 
curve at the point # = 0'4. 

The equation y = 0'lx~' 25 is obtained from the general equation 
y = ax n by writing 0*1 for a and - '25 for n. 

To plot the curve we may assume values 0, '1, '2, etc., for x, and 
find corresponding values of y. 

Thus when x = 0, y = 0. 

When x=% y = 0l x0'3-" 25 . 

.-. logy = log -1 - 25 log 0-3. =1-0000 -}(T -4771). 
.-. y=1861. 

In a similar manner other values of y can be obtained and 
tabulated as follows : 



X 





l 


2 


3 


4 


5 


6 


7 


8 


9 


1-0 


y 





1778 


1495 


1351 


1257 


1190 


1136 


1093 


1057 


1027 


1 



To find the slope of the curve at the point 0*4 we may find the 
value of y when x= "42 ; 

.-. &c='42--4='02; 

1 



' y=r. 



1243 ; 



C42r 

= 1243- -1257=- -0014. 



Slope of curve = -- - 



0014 
02 



07. 



COMPOUND INTEREST LAW. 207 

This gives approximately the slope of the curve at the point 
.r='4. The approximation becoming closer and closer to the 
actual value as the increments are diminished, and, when they 
become indefinitely small, the slope is that of the tangent at the 

point or -?-. 
ax 



If y='lx 



-:r, 



then %= - -25 x -l^" 25 " J )= -025a;- 1 - 25 or -025aT*. 

ax 

From this we can obtain the value of the tangent at any point by 
substituting the value of x. Thus when x '4, we get 

The numerical value of '025 (4) _4 is readily obtained by logs. 
Thus log -025 - i log -4 = 2-3979 - 1 '5026 

= 2-8953; 

/. -025 (-4)"*= -07857. 

Compound interest law. The curve y=ae 6x , where, a, b, 
and x may have all sorts of values, is known as the compound 
interest law; e is the base of the Napierian logarithms = 2*718. 
When definite numerical values are assigned to a and b, the 
curve can be plotted. 

Let a = '53, b = '26, then the equation becomes y = '53e' 26x 

Ex. 19. Plot the curve y= -53e- 2ti *. 

Calculate the average value of y from x = to x = 8. Also deter- 
mine the slope of the curve at the point where x = 3. 

Assuming values 0, 1, 2, 3, etc., for x corresponding values of y 
can be obtained. Thus when x =0, 

y=-53e='53. 
When x = 2, y = 53e -26 >< 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 ; a<b, that a 
is less than b) : 

a>b, a = b, a<b; a + b>c, a + b = c, a + b < c. 
a 2 + b 2 >c 2 , a 2 + b 2 = c\ a 2 + b 2 <c 2 . 

Circumference of a circle. The number of times that the 
length of the circumference of a circle contains the length of 
the diameter of a circle cannot be expressed exactly, but it is 
very nearly 3 - 14159265. The number 3*1416 is used for con- 
venience, and is sufficiently exact for nearly all purposes. This 
number is denoted by the Greek letter it. 

An approximate value of 7r, sufficiently exact for all practical 
purposes, and very convenient when four -figure logarithms are 
used, is 3| or 3'142. Thus, tt = 3*1 41 59265=^ within 2V P er cent - 

That the length of the circumference of a circle is ire or 2-nr, 
where d is the length of the diameter and r the radius, may be 



CIRCUMFERENCE OF CIRCLE. 223 

shown in several ways. Two simple experimental methods 
will be sufficient in this place. 

1. By rolling a disc of metal or wood on any convenient scale. 
Make a mark on the circumference of the disc. Put the mark 
coincident with a scale division. Slowly roll the disc along the 
scale until the mark is again coincident with the scale, and note 
carefully the distance in scale divisions moved through. Then 
by applying the scale to the disc obtain the diameter. 

Simple division will then show that the length of the circum- 
ference is 3^ times that of the diameter. 

2. Or, wrap a 'piece of thin paper round the disc, and mark, 
by two points, the line along which the edges overlap ; unroll 
the paper, and its length when measured will be found to be 3^ 
times that of the diameter. 

To obtain a good average value the mean of several readings 
should be taken. 

EXERCISES. XXXVII. 

1. Find the diameters of circles, the circumferences in inches 
being 157, 23562, 4712, 1178, 17 28, 128 '02. 

2. Find the circumferences of circles, the diameters of which are 
1-75, 2-5, 4-75, 8, 30 5, 67 '5. 

3. The minute hand of a clock is 6 ft. long. What distance will its 
extremity move over in 36 minutes ? 

4. A carriage wheel is 2 ft. 7 in. diameter. How many turns 
does it make in a distance of 7 miles 1332 yards ? 

5. The circumference of a wheel is 20 ft. How many turns will 
it make in rolling over 100 miles ? Find the diameter of the wheel. 

6. A rope is wrapped on a roller 1 ft. diameter. How many coils 
will be required to reach to the bottom of a well 200 ft. deep? 
What number of coils will be required if the rope is 1 inch thick ? 

7. The wheel of a locomotive 5 ft. in diameter made 10,000 
revolutions in a distance of 24 miles. What distance was lost due to 
the slipping of the wheels ? 

8. How many revolutions per minute would a wheel 56 in. 
diameter have to make in order to travel at 30 miles an hour ? 

9. The circumferences of two wheels differ by a foot, and one turns 
as often in going 6 furlongs as the other in going 7 furlongs. Find 
the diameter of each wheel. 

10. The hind and front wheels of a carriage have circumferences 
14 and 16 ft. respectively. How far has the carriage advanced 
when the smaller wheel has made 51 revolutions more than the 
larger one ? 



224 PRACTICAL MATHEMATICS FOR BEGINNERS. 

Area of a circle. If a regular polygon be inscribed in a 
circle, a series of triangles are formed by joining the angular 
points of the polygon to the centre of the circle. 

The area of each little triangle is one-half the product of its 
base and the perpendicular let fall from the centre of the circle 
on the base of the triangle. 

The length of the base may be denoted by a ; the length of 
the perpendicular by p ; and the radius of the circle by r. The 
area of the polygon will be ^p(a +a+...) or %p2a. The 




F/g 100. Area of a circle. 



symbol 2, which denotes " the sum of," is very convenient, and 
the expression \ip2a simply means the product of j>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 


<r> 




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 




<o 






























1 















s 




10 






L r 




20 




23 






30 







Fig. 130. 
Fig. 130. Dividing the base into eight equal parts we obtain nine 



VOLUMES BY DISPLACEMENT. 



259 



equidistant ordinates, the values of which can be read off. These 
are shown in Fig. 130. The common distance between the ordinates 
is 4 ft. 

Area=*{20 + 7-5 + 4(16-5 + 12-05 + 94 + 7-6) + 2(14 + 10-5 + 8-4)} 
= 367-3 cub. ft. 



Practical methods of finding volumes and weights. In 

many cases a quick method of finding the volume (or weight) of 

a body is required. For example, if a casting has to be made 

from a wooden pattern, the weight of metal in the casting may 

be found approximately by multiplying the 

weight of the wooden pattern by the ratio of 

the weight of unit volume of the metal to unit 

volume of the wood. There are, however, many 

sources of error in such a calculation. Nails, 

screws, etc., which are used in the construction 

of the pattern have a different density to the 

wood. 

Another method is to obtain a volume of 
water equal to that of the given body. When 
the body is of small size it may be placed in a 
graduated cylinder (Fig. 131), and the height 
A before and B after immersion noted, then 
from the difference of the readings the volume 
is at once found. When the body is of com- 
paratively large size it may be placed in a bath 
of water and the amount of water displaced by 
the body obtained, from this the volume is ob- 
tained. The volume of the clearance space in 
a steam or gas engine cylinder may be deter- 
mined by noting the quantity of water required to fill it. 




EXERCISES. XLV. 

1. Find the cubical contents of a body 30 ft. long, the cross- 
sectional areas at intervals of 5 ft. being respectively 7 5, 5 08, 3*54, 
2-52, 1-86, 1-34, 0-92 sq. ft. Find also what the volume would be if 
only the areas of the two ends and the middle were given. 

2. Values of A the area of the cross-section of a body at dis- 



260 PRACTICAL MATHEMATICS FOR BEGINNERS. 



tances x from one end are given in the following table. Find the 
average value of A and the volume of the body. 



A 
Square inches. 


53 




75 


84 


94 5 


123 
62 


139 

78 


134 

97 


106 
114 


76 
128 


45 
144 


X 

Inches. 


9 


22 


41 



3. A body like the trunk of a tree, 13 feet long, its axis being 
straight, has the following cross-sectional areas of A square inches 
at the following distances, x inches from its end. Find its volume, 
using squared paper. 

The following table gives the value of A for each value of x : 



X. 





20 


35 


56 


72 


95 


110 


140 156 


A. 

I 


405 


380 


362 


340 


325 


304 


287 


260 


252 



4. The length of a tree is 16 ft., its mean girth at five equidistant 
places is 9*43, 7*92, 6*15, 4'74, and 3 16 ft. respectively. Find the 
volume. 

5. The areas of the cross-sections of a tree 30 ft. long are as 
follows : 



Distance from one 
end in feet. 





2-3 


45 


7 


122 


18 


24 


30 


Area of cross-section 
in square feet. 


7 


6-3 


5-8 


5-2 


4-8 


4-0 


3-8 


29 



Find the volume. 

6. In excavating a canal the areas of the transverse sections are 
in square feet 687 '6, 8222, 735"8, 809'5, 509*5, the common distance 
between the sections 30 ft. Find the volume in cubic yards. 

7. The transverse sections of an embankment are trapeziums, the 
distance between each section is 25 ft. ; the perpendicular distances 
between the parallel sides and the lengths of the parallel sides are 
given in the following table. Find the volume of the embankment. 



Parallel sides 
in feet. 


22 
46 


219 
46-9 


21-6 
47 6 


21-6 
50 2 


21-8 
524 


21-6 
55-2 


22 

62 


Perpendicular 
distance. 


6 


6-3 


6-6 


7-2 


7'8 


8-4 


10 



EXERCISES. 



261 



8. The height in feet of the atmospheric surface of the water in 
a reservoir above the lowest point of the bottom is h ; A is the area 
of the surface in square feet. 

When the reservoir is filled to various heights the areas are 
measured and found to be : 



Values of h. 





13 


23 


33 


47 


62 


78 


91 


104 


120 


Values of A. 





21000 


27500 


33600 


9200 


44700 


50400 


54700 


60800 


69300 



How many cubic feet of water leave the reservoir when h alters 
from 113 to 65? 

9. A pond with irregular sides when filled to the following heights 
above a datum level has the surface of the water of the following 
areas : at datum level the area is 3 '16 sq. f t ; 4 ft. above datum, 
4-74 sq. ft.; 8 ft., 6'15 sq. ft. ; 12 ft., 7 '92 sq. ft. ; 16 ft., 943 sq. ft. 
What is the volume of water in the pond above datum level ? 

10. Find the cubical contents of a reservoir 42 feet deep, the 
sectional areas A (sq. ft. ) at heights h (ft. ) above the bottom being 
as follows : 



h. 





5 


10 


17 


21 


25 


29 


33 


38 


42 


A. 





2100 


8200 


13100 


15500 


19500 


25400 


32400 


47100 


52000 



11. A log of timber, 20 feet long, has the following cross- sections 
at the given distances from one end. Find the average cross-section 
and the volume in cubic feet. 



Distance from one end 
in feet. 





2-6 


5 


7-4 


10 


12 


15 


17-6 


20 


Area in square feet. 


5-0 


4-3 


4-0 


3-8 


3-46 


3-5 


3-26 


31 


3-0 



CHAPTER XXII. 



POSITION OF A POINT OR LINE IN SPACE. 



Lines. Lines may be straight or curved, or straight in one 
part of their length and curved in another. 

Straight line. A straight line may be defined for practical 
purposes as the shortest distance between two points ; or as that 
line which lies evenly between its 
extreme points. 

Planes. A plane is a surface such 
that the straight line joining any 
two points on it lies wholly in that 
surface. 

Perhaps a clear notion of what this 

definition implies may be obtained by 

using a flat sheet of paper, as in Fig. 

132. If any two points, A and B, on 

the surface of the paper be selected, it 

will be seen that the line joining them 

lies in the surface. Now bend or 

crease the paper between the points, 

as along CD. The surface no longer remains in one plane, and 

the shortest, or straight line, joining the two points A and B, 

does not lie in the surface. 

The intersection of two planes is a straight line, because the 
straight line joining any two points in their line of intersection 
must lie in both planes. 

Projections of a line. The projection of a line A B on a plane 
MN (Fig. 133) is obtained as follows : 

From A and B let fall perpendiculars (as shown by the dotted 




Fig. 132. A plane surface. 



CO-ORDINATE PLANES. 



263 



lines) on the plane MJV. The line joining the points where 
these dotted lines meet the plane is the projection iequired. 

The angle between a line and plane, or the inclination of a line to 
a plane, is the angle between the line and its projection on the plane. 
Thus, if BA produced meets the plane JVM (Fig. 133), the inclina- 
tion of the line to the plane is the angle between the line and its 
projection on the plane. 

B 




Fig. 133. Angle between a line and a plane. 

Three co-ordinate planes of projections. A point or points 
in space may be represented by means of the projections on 
three intersecting, or co-ordinate planes, as they are called ; 
these projections determine the distances of the point from the 
three planes, and hence 
the position of the point 
is known. The planes 
are usually mutually at 
right angles to each 
other, such as the corner 
of a cube, or roughly, 
the corner of a room. 

The floor may re- 
present the horizontal 
plane, sometimes spoken 
of as the plane xy, one 
vertical wall the plane 

























































































/K 















































































































































































xz, and the other vertical 



Fig. 134. Model of the three co-ordinate planes 
of projection. 



wall at right angles to xz the plane zy. 

A model to illustrate this may consist of a piece of flat board 
(Fig. 134) and two other pieces mutually at right angles to 



264 PRACTICAL MATHEMATICS FOR BEGINNERS. 



each other. It is advisable to have the latter two boards 
hinged. This enables the two sides to be rotated until all three 
planes lie in one plane. If the three pieces of wood are painted 
black they may be ruled into squares, or squared paper may be 

fastened on them. By means of 
hat pins many problems can be 
effectively illustrated. 

If preferred, a model can be 

easily made from drawing paper 

or cardboard. Draw a square of 

9 or 10 inches side (Fig. 135). 

Along two of its sides mark off 

distances of 4" and 6" and letter 

as shown. Cut through one of 

the lines OZ, and fold the paper 

so that the two points marked Z 

coincide. 

To fix the position of a point in space, imagine such a point P ; 

from P let fall a perpendicular on the horizontal plane and 

meeting it in p (Fig. 136). pP is the distance of the point P 

from the plane xy, or is the z co-ordinate of P. In a similar 



Fig. 135. 




t 



Fig. 136. 



manner a perpendicular let fall on the plane yz, meeting it in p\ 
will give the distance from the plane yz or the x co-ordinate of 
the point. And the distance Pp" the y co-ordinate of the point 
is the distance of the point from the plane zx. 



POSITIONS OF POINTS. 265 






The three projections of a point on three intersecting planes 
definitely determine the distance of a point from these planes. 

In the example we have assumed the z co-ordinate to be 
above the plane of xy, but the method applies equally to 
distances below the plane. 

Hence, a point or object above or below the earth's surface 
could be specified if two intersecting vertical planes, such as two 
walls meeting at right angles, were to be found in the neigh- 
bourhood. The distances from the two walls, together with the 
remaining or z co-ordinate, would completely define the position 
of the point. Stores of buried treasure may in this manner be 
located. A person unable to carry away treasure might select a 
place in which two convenient intersecting walls are to be found 
in the neighbourhood. If deposited at some depth below the 
surface the treasure could be recovered at any future time, 
provided that the respective distances from the two walls and 
the depth below the surface were known. 

It will be found that the problems dealing with the pro- 
jections of a point, line, or plane, may be solved either by 
graphic methods, using a fairly accurate scale and protractor, 
or by calculation. One method should be used as a check on the 
other. 

Ex. 1. Given the x, y, and z co-ordinates of a point as 2", 1'5", 
and 2" respectively. Draw the three projections of the line OP 
on the three planes xy, yz, and zx, and in each case measure the 
length of the projection. Find the distance of P from the origin 0, 
and the angles made by the line OP with the three axes. 

Let P (Fig. 136) be the given point and O the origin of co-ordinates. 
Join OP. 

The projection on the axis of x is the line OB ; on the axis of y is 
the line OG ; and on the axis of z is the line OD. 
OB='Z', 00 =T5", and OD = 2T. 

Graphic Construction. The relations of the lines and angles 
can be seen from the pictorial view (Fig. 136). To measure the 
lengths of the lines and the magnitudes of the angles, proceed 
as follows : 

Draw the three axes intersecting at O (Fig. 137) and letter as 
shown. Set off along the axis of z a distance = 2", along the axis of 
a distance = 1 *5". Draw lines parallel to the axes, and join p lt 



266 PRACTICAL MATHEMATICS FOR BEGINNERS. 



z 


yPz 


j / 2 


\ ' ^* 


* " 


---/j -*/? 



Fig. 137. 



their point of intersection, to the origin 0. Then 0p k is the pro- 
jection of OP on the plane xy, its length is 2*5". 

In a similar manner the projections, 0p 2 on the plane yz and 0p z on 
the plane xz are obtained ; Op 2 = 2'5" and Op 3 = 2'83". 

The distance of P from 
the origin, or the length of 
the line OP, is the hypot- 
enuse of a right-angled 
triangle, of which 0p x is 
the base and the perpen- 
dicular p x P the height of 
P above the plane of xy, 
or simply the z co-ordinate 
of the point. Hence draw 
p x P perpendicular to 0p 1 
and equal to 2". Join 
to P ; OP is the distance 
required = 3 '2". 

To obtain the angles with 
the three axes it is necessary to rdbat the line into the three planes. 
This is easily effected by using as centre and length of OP = 3 2" 
as radius. Describe a circle cutting the lines passing through p v 
p 2 , and 2h at P\> 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 <f) y 
but Op = OP sin 0; 

:. x = OP sin cos < (i) 

y = 0G= Op cos pOC= Op sin cf> ; 

:. 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 <f> 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 <p between and, say 1 5, calculate x and y and 
plot this part of the curve. 



PRACTICAL MATHEMATICS. 1903. 
STAGE I. 

Only eight questions to be answered. Three of these must be 
Nos. 1, 2 and 3. 

1. Compute by contracted methods to four significant figures 
only, and without using logarithms, 

8-102x35-14, 254-3 -r 0-09027. 

Compute, using logarithms, 

y/Wp2l, s/Wm, 372-4 243 , 0-3724-243. 

What is the theory underlying the use of logarithms in helping 
us to multiply, divide, and raise a number to any power ? 

2. Answer only one of the following (a), (b), or (c) : 

(a) If # = tan0^tan {d + <f>) 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(<f>- sin <p) and y = a(l -cos0), find 

x and y when a is 10 and = 0*5061 radian. 

(6) In a piece of coal there was found to be 11*30 lb. of carbon, 
0*92 lb. of hydrogen, 0*84 lb. of oxygen, 0*56 lb. of nitrogen, 0*71 lb. 
of ash. There being nothing else, state the percentage composition 
of the coal. 

(c) A brass tube, 8 feet long, has an outside diameter 3 inches, 
inside 2*8 inches. What is the volume of the brass in cubic inches ? 
If a cubic inch of brass weighs 0*3 lb., what is the weight of the 
tube? 

3. The four parts (a), (6), (c) and (d) must all be answered to get 
full marks. 

(a) Write down algebraically : Three times the square of x, 
multiplied by the square root of y ; from this subtract a times the 
Napierean logarithm of x ; again, subtract b times the sine of ex ; 
divide the result by the sum of the cube of x and the square of y. 

as the sum of two simpler fractions. 

(c) Some men agree to pay equally for the use of a boat, and each 
pays 15 pence. If there had been two more men in the party, each 
would have paid 10 pence. How many men were there, and how 
much was the hire of the boat ? 



EXAMINATION PAPER. 



329 



{d) The altitude of a tower observed from a point distant 150 
feet horizontally from its foot is 26 ; find its height. 

4. li p^ 1 ' 13 = p 2 v 2 113 and if v 2 /v 1 be called r. 
If p 2 = 6, find r if ^ = 150. 

o 

5. If y = - + 51og 10 a?-2*70, find the values of y when x has the 

X 

values 2, 2 -5, 3. 

Plot the values of y and x on squared paper, and draw the pro- 
bable curve in which these points lie. State approximately what 
value of x woald cause y to be 0. 

6. x and t are the distance in miles and the time in hours of a 
train from a railway station. Plot on squared paper. Describe 
why it is that the slope of the curve shows the speed ; where is the 
speed greatest and where is it least ? 



X 





012 


0-50 


1*52 


2-50 


2-92 


3*05 


317 


3-50 


3-82 


415 


t 


0*00 


005 


o-io 


0-15 


020 


0-25 


0-30 


0-35 


0-40 


045 


0-50 



7. A vessel is shaped like the frustum of a cone, the circular base 
is 10 inches diameter, the top is 5 inches diameter, the vertical axial 
height is 8 inches. By drawing, find the axial height to the 
imaginary vertex of the cone. If x is the height of the surface of 
a liquid from the bottom, plot a curve, to any scales you please, 
showing for any value of x the area of the horizonal section there. 
Three points of the curve will be enough to find. 

8. A circle is 3 inches diameter, its centre is 4 inches from a 
line in its plane. The circle revolves about the line as an axis and 
so generates a ring. Find the volume of the ring, also its surface 



9. If u "is usefulness of flywheels, u cc rf 5 r? 2 , if d is the linear size 
(say diameter) and n the speed. We assume all flywheels to be 
similar in shape. I wish to have the usefulness one hundred times 
as great, the speed being trebled, what is the ratio of the new 
diameter to the old one ? 

10. The total cost G of a ship per hour (including interest and 
depreciation on capital, wages, coal, etc.) is C=a + bs 3 , where s is 
the speed in knots (or nautical miles per hour). 

When s is 10, G is found to be 5 '20 pounds. 
When s is 15, G is found to be 7 '375 pounds. 
Calculate a and b. What is G when s is 12 ? 



330 PRACTICAL MATHEMATICS FOR BEGINNERS. 



How many hours are spent in a passage of 3,000 nautical miles at 
a speed of 12 knots, and what is the total cost of the passage ? 

11. A feed pump of variable stroke driven by an electromotor at 
constant speed ; the following experimental results were obtained : 



Electrical Horse 
Power. 


Power given to 
Water. 


3 12 

4 5 
7'5 

1074 


119 
2-21 
4-26 
6-44 



Plot on squared paper, and state the probable electrical power 
when the power given to the water was 5. 

12. Mr. Scott Russell found that at the following speeds of a 
canal-boat the tow-rope pull was as follows : 



Speed in miles per hour, 


619 


7*57 


8-52 


9-04 


Tow-rope pull in pounds, 


250 


500 


400 


280 



What was the probable pull when the speed was 8 miles per hour ? 
There was reason to believe that the pull was at its maximum at 8 
miles per hour, because this was the natural speed of a long wave 
in that canal. 



UNIVERSITY OF LONDON. 



MATRICULATION EXAMINATION. 

September, 1902. 

ARITHMETIC AND ALGEBRA. 

1. An iron bar is 117 centimetres long and its cross-section is a 
square of which the side measures 9 millimetres. Find its weight 
to the nearest gram, supposing the iron to weigh 7 '6 grams per 
cubic centimetre. 

2. The average of a certain set of p numbers is a, and that of 
another set of q numbers is b ; find an expression for the average 
of all the numbers taken together. 

The population of two towns are 107,509 and 189,160 ; their birth- 
rates per thousand are 27*9 and 25*7. Find to the same degree of 
exactness the birth-rate for the two towns taken together. 

3. From the equation 

find I in terms of the other quantities, and calculate its value to 
three significant figures when 

t=l, = 32-18, tt=31416. 

4. A quantity m is altered in the ratio of a to 6 and the result is 
then changed in the ratio of c to d ; write down an expression for 
the final result. 

A manufacturer reduces the price of his goods by 2| per cent. ; 
what percentage increase in sales after the reduction will produce 
an increase of 1 per cent, in gross receipts ? 

5. State in words the meaning of the formula 

m {a + b) ma -f mb 
and prove it when m, a, b denote positive whole numbers. 

6. Bring the expression 

(l+a0-(l-t4tf-|tf 2 ) 2 
to its simplest form ; and show that when x is a positive proper 
fraction the value of the expression is between and a^-j-8. 



332 PRACTICAL MATHEMATICS FOR BEGINNERS. 

7. Factorise 2a* 2 -a*-l, and find the values of x which make it 
equal to 0. 

8. Draw the graphs of x 2 and of 3a; +1. By means of them find 
approximate values for the roots of x 2 - Sx - 1 = 0. 

Calculate the roots of this equation to three significant figures. 

9. The nth term of a series is 3w-l, whatever whole number 
n may be ; prove that it is an arithmetic progression, and that the 
sum of the first In terms is n{6n + 1). Check this result by giving a 
particular value to n. 

10. The area of a rectangular plot of land is 6,000 square feet and 
the diagonal of it measures 130 feet ; find the length and breadth 
of the plot. 



MATHEMATICS (MORE ADVANCED). 

1. What is the meaning of a n when n is a positive whole number? 

Find meanings for a? and a" 3 , stating clearly the assumption 
which you make. 

Find the values of 128" 7" and log x 2. 

2. Why is the logarithm of *5 written as 1*69897 and not as 
- -30103 ? 

The logarithmic sine of an angle is 9*87314. Make use of the 
given tables to find the angle. May your result be regarded as 
correct to the nearest minute ? 

3. Find the 10th term of the expansion of (2a - 3&) 15 . 

Employ the binomial theorem to find the value of (1*012) 5 to three 
places of decimals. 

4. Find M and H from the following data *. 

M e^tana , r , 4ir 2 I 

h=-^t> 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 + *>*+<?- 3abc 

a + b a+b+c 

Exercises XIII., p. 72. 

1. 3. 2. 5. 3. c. 4. 10a?-7y+16z; -20a;+14y-32z. 

5. 17a. 6. Iff. 7. -4a;-3y + 2z; -6j. 

8. -3xy-yZ; lj. 9. -x + 3z; -7. 10. 2(5c + a). 

11. 1+ai 12. 8a6. 13. 126 (a -6). 14. -6a6-6 2 -a 2 . 

Exercises XIV., p. 77. 

1. (x-2)(x-5). 2. (a;- 10) (a; + 9). 3. (x-4){x+l), 

4. (a; + 5) (a; -3). 5. (3a + 26)(9a 2 -6a& + 46 2 ). 

6. (2a;-3)(4a^ + 6a; + 9). 7. (a? -6) (a; + 5). 8. (a;+17)(a;-5). 

9. {x-2y){x-z). 10. 3(a;-3y)(a; + 3y). 11. {x + 25) (x - 7). 

12. (a;-2y)(a;-3z). 13. (25x 2 + y 2 )(5x-y)(5x + y). 

14. (10a;- 1) (x + 8). 15. x {x - 6y ) {x - ly). 

16. (a + 6 + c)(a + 6-c)(a-6+c)(a-6-c). 17. (x - y) (a; 2 - 5xy + 7y 2 ). 
18. (i) (a + 6-c-d)(a-6-c + a*); 

(ii)(^ + g+r)(p + a-r)(^-g + r)(p-a-r); (iii) (1 -n*)(l -m). 

Exercises XV., p. 80. 



1. = < 2. a - a;. 


^ a^-aar+a 2 


4 2 + 3a; 


3a 


a + a; 


l+5aT 


5 4a^-4a;+l ' x-1 


7 4ary 


8 x* + a* 

X 


4a?-3a;-l *" a; + l" 


* a^-y 2 ' 


9. 1. 10. J?-. 11. 


(a^+l) 2 -^, (a; 2 +l- 


t-a;)(a^+l-a;). 



ANSWERS. 337 



a + X 14. x + l 


15. 1. 


1 18 2 


19 3*2 + 1 


3x- 2y " x-Zy 


x 2 + 2a;-3" 


1 


22 2 


(* + l)(a: a + l) 


Z2, a-b' 


Exercises XVI., p. 85. 




3. 2. 4. 8. 5. 9. 


6. 24. 7. 8. 


10. 2. 11 3. 12. 2. 


13. 23. 14. 43. 


17. a(b-a). 18. ab. 


19. b. 20. L+* 
2 



20. 4^,. 21. 



1. 2. 2. f. 

8. 4. 9. 6. 

15. -5. 16. 6. 

Exercises XVII., p. 88. 

1. 54; 21. 2. 16; 9. 3. 420. 

4. 27] 3 i past 2 ; at 3, and at 32^ min. past 3. 5. 9 oz. , 12 oz. , 16 oz. 

6. A is 54, B 12. 7. A is 37, B is 27, C is 47. 

8. 75. 9. 32, 48, 480. 10. ll yards. 11. 6000, 5000, 3000. 

12. A 400, B 160, G 140. 13. 30 hours. 14. 25, 24. 
15. 10, 15. 16. 15. 17. 120. 18. 3 miles. 19. 19 : 16. 

Exercises XVIII., p. 95. 

1. 5, 6. 2. 5, 4. 3. 30, 20. 4. f , . 5. $ , 2j. 6. 3, J. 

7. f, f. 8. , 2. 9. 7, 2. 10. 3, -4. 11. 4, 5. 12. 16, 35. 

2 2 2 

13. a:= y , v= r z = i 

a-b + c ' a + b-c b-a + c 

1 pq{qm+pn) pq{pm-qn) 
p*+q* ' p*+q* 

15. f. 16. 7f, -ffl; " * > 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<c 2 -7. 6. a 2 -a + 4. 

7. 5x 2 -Sx + 4:. 8. x* + 4x-21. 9. 4x 2 -5x+8. 

10. -%!-. ll. 3x 2 -5xy + y 2 . 14. 3a; 2 - 16a; + 5. 

y x y 

Exercises L., p. 298. 
1. 10, -14. 2. 8, -40. 3. 10,2. 4. 3, -1. 

* 7 ' ~ 4 - 6 ' 4' "250- 7 ' 2' 3* 8 - "' b ' 

9. 2, T l. 10. 4, -3B. ll- |, ~| 12. 8, -1. 

M - * ^ " * 3 > *> -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. 



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