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ATICS
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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 and»that 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. 293- = f = 3-^8= -375; | = 7 4- 8 = -875.
Ex.2. Tijj = '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 0 — — "9" 9 "9 0 0 _ 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. 6£d.=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.
T52 = -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 0
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 0 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 0 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- -066
' T9<y of -138' '022
21. Compute by contracted methods 23 '07 x 0'1354, 2307 -f 1 '354.
22. Compute 4 • 326 x 0 '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 small 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 in 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.D2s'»
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 = — 0^.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 = To5o x To°o =T§o-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. — £5T% — 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 22,
and the fourth power of 2 either as 2 x 2 x 2 x 2, or 24. Similarly
the square, cube, fourth or fifth power of 4 would be written
42, 43, 44, and 45.
This method is also adopted when a number containing
several digits is used. Thus 25742 signifies 2574 x 2574, and
633 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 :
102 = 100, 103= 1000, 106 = 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 106.
Again, 5,830,000 could be written as 5'83 x 106, and 5,830 as
5*83 xlO3.
The squares of all numbers from 1 to 10 are easy to remem-
ber ; they are as follows :
12==1) 42=16, 72 = 49,
22=4, 52 = 25, 82 = 64,
32 = 9, 62 = 36, 92 = 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.
As3002 = 90,000and 4002 = 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) 39TV ; (ii) 40008-0004.
10. 90018 0009. 11. 6877219041. 12. 998001. 13. 42436.
14. 00501264. 15. 18671041. _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 0 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.
0°
10°
20°
30°
40°
50°
60°
70°
80°
0°
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 0 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 0 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 2V
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 0 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 0 . 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
i3
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 0 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 at C and placing it on B.
~^\T£gletre^r °' ^ iS°SCeleS ™S ma7 be effected 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^afto0^
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, 32 + 42 = 25 = 52.
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
, - — - ,P2 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 APXB Fig. 36, the angle formed is
greater than a right angle, and when as at A P2B 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=ED2.
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
PT2 = 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 icngth 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 0 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 — 5°C. 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 a2 ;
b x b x b is called the cube of b and is written 63.
Similarly, the multiplication of any number of the same
letters together may be represented; thus cn 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 a4 are quite
different terms.
Let a = 2, then 4a = 8 ; buta4 = 24 = 16.
The use of signs may be exemplified in the following manner :
Ex. 1. In the expression a2+ b - c,
Let a = 4, 6 = 7, and c = S.
Then a2 + 6-c = 42 + 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
aar2 + 62 ^ when a = 3 h = ^ c = ^ X = Q
bx-az-c
Here axi + 62= 3 x62 + 52 =133,
Also te-a2-c=5x6-32-2= 19,
" 6«-a2-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 + b2cd + c2ad - 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-b2 + c3-d4.
Find the value of
5. a? + 3a26 + 3a62 + 63 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 c2-d?
iJ. 1 1
a-b c-d bd + ac a2 + b2
When a=10, 6 = 3, c = 7 of,
10. 6 + c 11 3c 4&<"
2c -- 36 " 4a + 2~r10a-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 : lx2 - 9y2 + 20xy, - 11a;2 + 10y2 - §xy,
8x2 - 7y2 - 4xy.
First adding together the coefficients of the terms in x2 we get
7-11 + 8 = 4.
In a similar manner the coefficients of y2 are -9+10-7= -6;
and of xy are 20 - 6 - 4 = 10.
Hence, the required sum is Ax2 - 6y2 + 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. ax2 - bx2 + cy2 - ab2c, U ax2 - 3$ bx2 + cy2 + 4ab2c,
5 \ a x2 - 5j bx2 -3cy2- 2a62c, -lax2 + Ibx2 + cy2 - ab2c.
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. ax2-2dx2-2x + 2c-Sf, ax2 + 2dx2-bx + cx-l,
ax2 - dx2 -bx-cx-c+1, -x2 + 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=]32.
9- i«* + ? y4 + £*% " 5 a-'V and far4 + f y4 - J a% - £a%2-
Simplify
10. 4a:2 + 8a^ + 4ty2-9a:2+18a^-92/2.
11. a;6 + 3x* + lx4 + 15a;3 + 10a;2 - 3a;5 - 9a;4 - 21 x3 - 45a;2 - 30a;
+ 2a;4 + 6ar> + 14a;2 + 30a; + 20.
12. Simplify and find the value of x2 + xy + x - x2 + xy x2
+ y2 + 66, when a; =100, y = 50.
13. 4a?-16a;2+162/2 + 4a;+16a;2- I6y2-4y 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 a2 -2b- 2c from 3a2 - 46 + 6c.
Here, after arranging as in addition and changing ga2 _ 4^ . gc
the signs, we proceed as in addition, thus : _ a.2 + 26 + 2c
3a2 - a2 = 2a2 ; 26 - 46 = - 26 ; and, finally, 2a2 -26 + 8c
6c + 2c = 8c.
Hence, the result is 2a2 - 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 3x2 + 5x - 1.
Here we may, as in Ex. 2, write the terms under each other.
Then, after mentally altering the sign of Sx2, 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 _nx,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 Zxl will give 7a2. 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 ABy
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. 9x2-3x + 5 take 6x2 + 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. x2 - 3y2 + Qxz - 3xy from x2 + 4xy - 5xz + z2.
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 x2(l +2y) -y2(l+2x) to give as remainder
2xy{x-y)t>
13. Add together 4a3 + 36c2- 6a26, 5a26 - c3 - 3>.c2, c3-2a3-2a26.
14. From
6x?y + 10x2y2 + 13xy* + I9y4 take 5x*y - 2x2y2 + 3xy3 - 2y4.
15. Subtract
2a4 + 3a36 + 5a262 + 8a63 + 1 164 from 4a4 + 6a36 + 8a262 + 10a63 + 1264.
16. Find the sum of 4a~> - 5ax2 + 6a2x - 5a3, 3a3 + 4ax2 + 2a2x + 6a3,
- 17a3 + I9ax2 - \5a2x + 8a3, I3ax2 - 27a2a + 18a3, 12a3 + 3a2x - 20a3.
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 3a262.
Here the product of the coefficients is 4x3 = 12. Adding the
indices of like letters we have a x a2 = ai+* = a3, and b3 x b2=b3+2=b5.
Hence the product is 12a3&5 :
.-. 4a63x3a2fi2=12a3&5.
Ex. 2. 4a362c5e4 x 6a463c4e3 = 24a76 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 = x2. Write <r + 5
the x2, and after it the product of x and 5 or 5x. x + 6
As the signs are alike, the sign of each of these x2 + 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 x2 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) = a2 + 2ab + b2,
{a-bf, or (a-b){a-b) = a2-2ab + b\
{x -5) (x -6) = x2 -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
14a2c2- 3a26c+ 2ac
- Ua?bc + Sa2b2 - 2ab
+ \4ac -Sab + 2
14a2c2 - 1 la2bc + 1 6ac + 3a2b2 -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=14a2c2.
Next - 3ab xac= - 3a26c,
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 x2 + 5x + 6.
Also the product of x and x + 1 is x2 + x.
Hence x2 + 5x + 6
x2 + x
x4 + 5xs+ 6x2
x3 + 5x2 + 6x
x* + 6xs+llx2 + 6x
EXERCISES. XI.
Multiply
1. x2 - ax + a2 by x2 + ax + a2.
2. 2a3 - a2b + 3ab2 - b3 by 2a3 + a2b + 3ab2 + 63.
3. x6 + x*y - xPy3 + xy5 + y6 by x2-xy + y2.
4. x* + 3x* + 1x2 + \5x + lQhy ic2-3cc + 2.
5. x2 + 4xy + Ay2 by x2 - 4xy + 4y2.
6. x3- 12# -16 by x*-\2x+\Q.
7. a6-3a462+666bya6-2a264 + 66.
8. x4 + x2y2 + yi by x2 - y2.
9. 4a36 - 6a262 - 2a&3 by 2a2 + 3ab - b2.
10. a4-&4 + c4 by a2 -62-c2.
11. l-y2-y3by \-y*-y*.
12. 3x* - x2 - 1 by 2x* - 3x* + 7.
13 x* + §x2 + Sx-§hy x2-2x + 4.
14. 4a2 + 12a& + 962 by 4a2 - 12a6 + 962.
15. 13a2 - 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 2ax2y ;
. IQabxPy
2ax2y
As in Arithmetic the work may be done by cancelling, thus,
70 -r 2 gives 35,
and abxPy -5- ax2y gives bx ;
hence the required quotient is 356a;.
Ex. 2. Divide 15a262 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 :
a2 + 2aa + x2 ) a5 + 5a4x + I0a3x2 + IttePx3 +5ax4 + x5 ( a3+ Sa2x+3ax2 + x3
a5 + 2a4x + a3x2
Sa4x + 9a3x2 + lOaPx3
3a4x+ 6a?x2+ Sa2xs
3a?x2+ laW + bax*
3a3a^+ 6a23r* + Sax4
a2x3 + 2axi + x5
a2x3 + 2ax4 + x5
Divide the first, or left-hand, term of the divisor into the dividend.
Thus a2 divided into a5 gives a3 ; write this quantity on the right-
hand side as shown, and put the term a5 under the first term of the
dividend. In a similar manner by multiplying the remaining two
terms 2ax and x2 by a3, and subtracting, we obtain Sa4x + 9a3x2.
Now bring down the next term IQatx3, and proceed as before.
70 PRACTICAL MATHEMATICS FOR BEGINNERS.
EXERCISES. XII.
Divide
1. 4a5-a3 + 4aby 2a2 + 3a + 2.
2. x2 + 4yz - 4y2 -z2 by x-2y + z.
3. a3 + a26 + a2c - abc - b2c - 6c2 by a2 - be.
4. 2x4 + 27xy3-81y4by x + 3y.
_ x3 - Sx2y + 3xy2 - ys , x2-2xy + y2
x3 + ys x2 - xy + y2 '
6. 9a2-462-c2 + 46cby 3a-26 + c.
7. 25a2 - 62- 4c2 + 46c by 5a-6 + 2c.
8. - 207a567c4 by 23a463c2.
9. 4x2y-8xy2-4y2 + 3x2 + 4x+l by a;-2y + l.
10. I2x4 - llx* - 9x* + 13x - 63 by 4rc2 - 3x + 7.
11. 2a3 + 6a26 - 4a2c - 2a62 + 3ac2- 6&3 + 462c + 96c2-6c3 by a + Sb-2c.
12. a5-a4 + a?-a2 + a- 1 by a3- 1.
13. (a2 - 6c)3 + 863c3 by a2 + 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 (a2 + x2) - 7 (a2 - x2) + 6 (a2 - 2a;2)
= 8a* + 8x2 - 7a2 + 7x2 + 6a2 - 12x2
= 7a2 + 3x2.
Ex. 5. Simplify
4x - [{4x - 4y) {4x + Ay) - {4x + (4x + 4y) {4x - 4y)\ + 4y\
Multiplying the terms in the brackets we get
4x - [16ic2 - I6y2 - {4x + 16a;2 - 1 6y2} + 4y]
or 4x - [16a:2 - 16y2 - 4x - lQx2 + 16y2 + 4y\
= 4x-l6x2+ I6y2 + 4x + 1 6x2 - IQy2 - 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
2a2 -6 a2 -26 3a°{b-x)
b-x x + b 25 - x2
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 , , » , 0 , „
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. x2 + y[x-{y + z)]-[x2 + 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) - 662.
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
x2 + 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 = a2 + 2a& + 62 (l)
(a-b)2 = a?-2ab + b2 (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 = x2 + 2xy + y2.
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)(x—y) we obtain x2—y2. Conversely
given x2—y2 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. 402 - 392= (40 + 39) (40 - 39) = 79 x 1 = 79.
Ex. 2. 10002-9982= (1000 + 998) (1000 -998) = 1998x2 = 3996.
Ex. 3. To obtain the factors of (a2 + b2 - c2)2 - Aa2b2. [This may
be written (a2 + o2 - c2)2 - (2a6)2. ]
Using the last rule given above we get
(a2 + b2 - c2 + 2ab) (a2 + b2 - c2 - 2ab),
or {{a + b)2-c2}{(a-b)2-c2}.
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* - y4.
First (x4 - y4) = {x2 + y2) {x2 - y2) .
Also as x?-y2={x+y){x-y),
we can write x4 - y* — {x2 + y2) {x + y) {x - y).
Ex. 5. Multiplying a2 - ab + b2 by a + b, the product is found to be
a3 + b3.
.-. a3 + bs = (a + b) (a2 - ab + b2).
Similarly a3 - b3 = {a - b) (a2 + ab + b2).
The quantities (a + b) (a2 - ab + b2) are the factors of a3 + b3, and
(a - b) {a2 + ab + b2) are the factors of a3 - b3.
Generally an+bn is divisible by a+b when n is an odd number,
1, 3, 5, etc. Thus, in Ex. 5, n is 3.
Also an — bn 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
an - bn is divisible by both (a + b) and (a - b).
Ex. 6. Let n = 6 ; :. an-bn becomes a6 - 66.
We know that a6-b6 = {a3 + b3){a3 -b3), and in Ex. 5 the factors of
(a3 + b3) and (a3 - 63) have been obtained.
Hence the factors of a6 - &6 are
(a + b) (a2 -ab + b2) (a - b) {a2 + ab + b2).
Thus a6 - 66 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 an±bn, and
to find that :
a*»+bn iS divisible by a+b when n is odd.
an— bn „ ,, a— b ,, ,,
an_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)=x2 + 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 x2.
„ 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 x2 + 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 x2, of the two last is 7 ; and the sum of the diagonal products
7#+#=8.z\ Hence, these are the factors of x2-\-8x + 7.
76 PRACTICAL MATHEMATICS FOR BEGINNERS.
Proceeding in like manner the product of the terms in the
last is x2 + 8x + 12.
Similarly (x - 6) (x - 2) = x2 - 8x + 1 2,
(x+6)(x-2)=x2 + 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
x2 + 8#+12 as x2 + 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
x2 + 8x+12 = (x + 2)(x + 6).
Ex. 8. In a similar manner,
x2 - 9x + 20 = x2 - 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 x2 — 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 = 0 ;
.'. x — 5 is a factor.
Hence x2- 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 ;
/. x2 + 6z-55=(a; + ll)(a;-5).
FRACTIONS.
77
EXERCISES. XIV.
Resolve into factorfa
1. x2 -7x + 10.
2. a;2 -#-90.
3. a;2-3a;-4.
4. a?+2a;-15.
5. 27a3 + 863.
6. 8a^-27.
7. a;2 -a: -30.
8. x1 + 12a; -85.
9. x2 - 2xy - xz + 2yz.
10. 3a;2-27y2.
11. x2 + 18a; -175.
12. a;2-3a;2;-2a;y + 62/z.
13. 625a^-^.
14. 10^ + 79a; -8.
15. arJ-13a;2y + 42a;y2.
16. (a2 + 62-c2)2-'
ia2b2.
17. {x-2yf + y3.
18. (i)a2-62 + c2
-d?-2(ac-bd);
(ii) (p2 + q2-^)2-4p2q2;
(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* + x2y2 r-v x2-\-3x-\-2
Ex. 3. Simplify (i) __£, (11) ^^
Here, (i) ^ aV= *2W)
*»Ty \i* + y*)\7*-y*) tf2-y2*
1 \ x2 + 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.
Xs + Sx2 - 4
To find the h.c.f. we proceed as follows :
x3 + Sx2-4)x4 + xs + 2x -±(x-2
-2X* +Gx- 4
-2^-63? + 8
6^ + 62:- 12
= 6{x2 + x-2);
x2 + x-2)xs + 3x2-4(x + 2
x* + x2 -2x
2x2 + 2x-4
2x2 + 2x-4:
Therefore the h. c. f. = x2 + x - 2.
SURDS. 79
Hence a?4 + a?3 + 2a;-4_(a;2 + a;-2)(a;2 + 2) _x* + 2
.x3 + Sx2-4: ~ {x2 + 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
x3-3x2-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) (x3 + 7x2 + 5^-25) or (x - 5) (x + 5) (x2 + 2a? - 5).
Ex. 6. Simplify the following :
1 1
Xs -3a;2 -15a; + 25 Xs + 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){x2 + 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?-x2 3 x? + as 4 4 + 12a; + 9a;2
15a2' ' a + x' ' x2 + 2ax + a2' ' 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; + 6Xa;2 + 5a; + 4' ' a;2 + 3a;-4 * 2a; + 8
11. Express as the difference of two squares 1 + x2 + x*, and thence
factorise the expression.
12 168a362c 13 a?+{a + b)ax + bx2 14 a;4 + a;2+l
' 48a26c3' ' tf-bW ' ' x3-!. '
1K a;2-a;-6 33-2a?-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) 9x2-4y
18 l 6y 1 19 3a;3-6a? + a;-2
' x + 3y + x2-9y2 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?-b2 + 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 x2 ; 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=^gt2 ; 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—x2-\-2xy+y2, 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 — ly 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? = b2-—.
ox ax
First remove the fractions by multiplying all through by abx
:. a2 - a?bx = ab3x - b2,
transposing, - a?bx - ab3x = - a2 - b2,
changing sign or multiplying by - 1,
x{a2 + b2)ab = a2 + b2;
a2 + b2 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 , . 3.0 , 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\ a2 + b2
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—\gt2. 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 + b2y = bc
,, (ii) by a, we get abx + a2y — ad
By subtraction, b2y - a2y = bc-ad
or y{b2-a2) =bc-ad;
_bc-ad
" y~b2-a*'
To obtain x we may either substitute for y, or proceed to eliminate
y from (i) and (ii).
Thus multiplying (i) by a, a2x + aby = ac
,, (ii) by b, b2x + aby — bd
Subtracting the upper line from lower, (b2 -a2)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.
0 (#-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 **-* 9a; + 20 2 7
Ex.9. Show that ^ + 5x + Q = ^2 + ^rS-
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 x—y, 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=y2;
,\ 6y = y2 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
xl\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= 0
.\ 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 + qdy
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,
• Pa_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 a1 and the time by t, then s <x t2,
:. s=kt\
64-4 = &x22; or, &=16-1.
Hence, in 5 seconds s=16*l x 52 = 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 nnnr>
Strain = j—. — fl 7$ — = ^ — — = -0002,
original length 12 x 12
''• ^=^^ = 25'000'000 or 2'5 x 10? lbs' Per 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=hOim.
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 202 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=kT-
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~ X2(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 0 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 a5, a7, a9, 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 ab 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 a3 x a2 = (a x ax a) (ax a) = a3+2=a5.
Ex. 1. a3xa5 = a3+5 = a8.
Ex. 2. a*xa3x a*=a2+3+4=a9.
This may be expressed in a more general manner as follows :
am =(a x a x a. . .to m factors)
and an = (axaxa. .. to n factors),
.'. am xan = (ax ax a... to m factors) (ax ax a... to n factors)
= (axaxa...to m+n factors)
= am+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 a5 axaxaxaxa ,„ „
Also -7= = a5~3 = a2.
a6 ax ax a
0. ., , am a x a x a to m factors „ „
Similarly — = , x = a™-n.
an ax ax a to n factors
108 PRACTICAL MATHEMATICS FOR BEGINNERS.
Ex. 3. Explain why the product is a? when a3 is multiplied by a4,
and why the quotient is a when a4 is divided by a3.
a4xa3=(axaxaxa) x (ax ax a) = a4+3=a7.
A1 ataxaxaxa
Also —,= — a.
a6 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.
a2 xa2=a
*+i
=aL=a,
Hence, the meaning to attach to a? is the square root of a ; to
a5 is the cube root of a squared, and to a3 the cube root of a.
Thus
si a can be written as a^,
%/a can be written as d3.
Also
and
sja
va
Again
-r=rxa
a*
¥_„£-¥_„-¥
Also
Similarly
axaxa
axaxa
= „3-3^„0
Generally, since amxan=am+n is true for all values of m and
If n be 0, then
amxa° = am+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 a2 we have
(a?)s = (a xa){ax a) (a x a) = a2x3 = a6,
where the index is the product of the indices 2 and 3.
Ex. 2. Find the value of (2'152)3.
(2-152)3=2-152X3
= 2156=98-72,
or, expressing this rule as a formula, we have
(am)n =amn,
.'. a quantity am 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 = amxbmxcmxdm.
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= l2 x 2? x 32 x 42
= 242=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 an. This may be written in the form v am, which is read
as the nth root of a to the power m.
Ex. 6. 2*= Z/25= 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
4I v/64 8*
Ex. 8. Find the value of 64* + 4l * + 2s 5 + 27**
Here 64* = 8, 41<5 = 4* = 64*=8,
22'5=2^=32* = 5-656,
27^=3.
Hence 64* + 41-5 + 22,5 + 27*=24'656.
Ex. 9. Find to two place's of decimals the value of x2 - 5x* + x~2
when x=5.
110 PRACTICAL MATHEMATICS FOR BEGINNERS.
Here x2- 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 xlO6.
Similarly 6340- 6'34 x 103,
•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
P2 means PxP, but P2 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 P0, Plt P2 ..., etc., may each refer to forces, etc., of
different magnitudes, and in different directions.
Binomial theorem. — We have already found that
(a + b)2 = a2 + 2ab + b2,
and by multiplying again by a + b we obtain
(a + bf = a3 + 3a25 + Zab2 + b3.
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\„ « waM-16 n(n — 1) „ 07„ ,
(a+b)n = an + — — + vx 2 'an-2b2+....
Take n=2, then
and — :; — = 2ab.
Hence (a + b)2 = a2 + ^ + bK(20'1) = a2 + 2ab + <
l l . z
APPROXIMATIONS. 1 1 1
Take ?i = 3 ; here an=a? ; — ^ — =3a?b, etc.
*b
1
, , ,., Sl 3a26 , 3.2a&2 , 3.2.1 0/3
= a3 + 3a26 + 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 63, 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)a2
(l+a)«lsl + _+_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. 3J— = ( 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 + a2.
.'. (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
a2 and a3.
:. 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/45" + 2* + #27 - 9*.
(\i)J'¥ + sl2E + y&.
Simplify
fi (pl>-g)g+gx(qy)«+« am-nam-Snam-6n g _
(a*>)*>~* ' ' an_TOaw~3man"6w'
9. If a glass rod 1 inch long at 0° 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 0° 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 x3 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 107!r, 101**, 10*.
Explain how you would illustrate that 10°= 1.
lfr. Divide (x3ymn)m by (x2ymn)n.
17. Raise {a?b{a?bc)T}^ to the 7th power.
18. Find the mth root of 2™amb2mc2.
/20 {(o»ho»)V r (a\*y
' \Vb" . a/by}™* ' Iw /'
21. (ilX'W^vlJAr1)- (ii)
ax~l + g-lx + 2 ,
asx ^ + a %T-1
22. Find the value of a;2 - f x* + ar1 when a? =3.
23. Va'^'Hv^-1.
24. Find the value of
2-W + 2"33_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 1010sq.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
6m
'' 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 4°C.
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 xl06gm.
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 = 23, 3 is the logarithm of 8
to the base 2. This can also be expressed by writing log28 = 3.
In a similar manner, if the base be 5, then 3 is the logarithm
of 125 to the base 5 ;
.'. log5125 = 3.
Also 64 = 26 = 43 = 82.
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. ;
.-. log264 = 6; log464 = 3; log864 = 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* ;
•'• log10iV=#.
Or by definition, substituting positive numbers for A7,
as 1 = 10°
Also, 10 = 101
Again, 100 = 102
.'. 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 103010=2.
In a similar manner, the logarithm of 200 = 2 x 100 might be
written as 102xl03010,
/. 200 = 102'3010.
Hence we write log 200 = 2 '30 1 0.
Also -0002 = j^tjo o = 2 x 10~4-
Hence -0002 = 10T3010.
:. 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 102.
Hence as before the characteristic is 2.
Similarly, 4700 = 4*7 x 103, '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 log10a& = 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 108357 = 6'85,
.-. 2-784 x 6 -85 = 10-4446 x 108357 = 101'2803.
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 log10|=^-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. 1—3 048 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
0 = 1-811, d='642, 0 = 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 an = (10*)n ;
.'. log10an = nx=n log a.
Ex. 1. Find log a3, also log cfi.
In the first case the index is 3, and, hence log a3 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. :
logaw=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.
.-. -053= -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 2a3 + (62)376 + 2c " 3 76 - d~lin.
When a = -07, b = 3, c = -578, d = 8 '4.
Let x denote the value required, then
x = 2a3 + (62)3'76 + 2c - 3^ _ d ~ ™.
Here the four terms must be separately calculated.
•073 is found to be '000343 .'. 2a3 = -000686.
From Ex. 3. (32)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 32 - (log 33 + Mog 15 + log 82) ;
.-. 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^^^-y6) find ^ yalue of y
x x
Multiplying both sides by x, then
x6 = 7-4x6-7'4y6,
or 7*4i/6 = 6-4a^;
x \7'4'
log^ = l(log6-4-log7-4)
x 0
= h -8062 --8692)
6
=1-9895 ;
/. ^=-9761.
x
Ex. 4. Given 7* = 3X+1 + 2X~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 _ *dTx*!" '"'
log k= log 7 -26 -flog 1-5 =-4207; *rl\zW
:. 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 V3D* or
HP = JcVsD%. Given F= 17, D= 19700, and #P= 13300, find k.
Here 13300 = k x 173 x 19700§ ;
.-. log *= log 13300- 3 log 17 - 1 log 19700 = 3-5697 ;
.-. k= -003713.
Hence the equation becomes HP= -003713 V3D%.
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 log102 or more simply as log 2, but
the hyperbolic logarithm of 2 is written as loge 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=ax = by or b = a& J
.-. ~ = \og b and *-=* *.
y &<l x logab
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
loge 10 =2 -30258509 = 2 -3026 approx.,
and 2«509= '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 e2-3026=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 = 101'29 = e2-3026xl'29 = e2'9703
or log1019'5 = l-29, log e 19 "5 =2 '9703.'
Ex. 2. Find log e 3 and log e 8 '43.
.-. loge 3 = -4771 x 2-3026 = 1 -0986,
log 8 43 ='9258;
.-. loge8-43= -9258 x 2-3026 = 2-1317.
Ex. 3. Find log 13 to base 20.
Here log 13 = 1 -1139, also log 20= 1 3010.
•'• log2ol3=^=-8562.
Find the value of
1. -03571 x -2568.
Divide
EXERCISES. XXVI.
o 8352 x 3-69 0 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_ 80x33-62x16
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)4x -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 loga6 + log6a = 1.
24. If a = 10'4, 6 = 2-38, #=-2064, and y=-09SQ, determine the
values of ab and ah~%(a? - y2f.
25. Find the value of p^kr'1 -r~k) -p3, when p1 = 80,p3 = 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^x2, 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-8602 + 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) 0-252-19. (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-5x 9300x1 -05 xM\*
~-\ 240 )'
M 11 *6 x -4785
33- * = n^r
MISCELLANEOUS EXAMPLES. XXVII.
Calculate
1. #23-51. 2. 6*78234. 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 + loge r ) -*- r.
5. If 2>m10646 = 479 find p when w is 12*12, and find u when p is 60*4.
Compute
6. 1-683365. 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 d0 to d1 from the equation :
10. Find the ratio of a to 6 when a?= — -.
11. Find the ratio of y to x from the equation xs= —
12. Find the ratio of x to y from y4=°'8(x5 + y5\
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 0 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 103 be multiplied by 104 the
result is 103*4 or 107.
From the definition of a loga-
rithm,
2 is the logarithm of 100,
since 102=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 jpjjj
146 PRACTICAL MATHEMATICS FOR BEGINNERS.
Also since 103010 = 2 ; .*. log 2 = '3010.
Hence '7781 is the log of 6 = 103010+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 0 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 =-r1 — ~=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
a2 = 62-c2 = 102-52=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
DC2 = AC2-AD* = 4-l=3;
:. dc=sJz.
D<0_JI
AC~ 2 '
AD_\
AC~2;
60c
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*angTeIdTianglIight* AB aild BC (FiS« 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=JAB2 + BC2 = 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 Je'4C=(180o-30°j = 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
0° 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 0 ; the value is indefinitely great when
the denominator is 0 : 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.
0°
30°
45°
60°
90°
120°
135°
150°
180°
sin
0
1
2
]
v/2
2
1
v/3
2
1
\/2
1
2
0
cos
1
n/S
2
1
s/2
1
2
0
1
2
1
s/2
2
-1
tan
0
1
v/3
1
v/3
QO
-s/S
-1
1
0
From the figures already drawn and also from these tabulated
values, it will be seen that as an angle increases from 0° to 90°,
the sine of the angle increases from
0 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 0
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) sin2A + cos2A = l.
• a BG A AB
For sin.4 = -j~; cos.4=-j^.
™ /• ,Na,/ am BG2 AB2 BC2 + 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 = cos2J.
(ii) sec2A=l+tan2A.
A AC
secA=AB'
aecA-AB2- AB2 -1+^^-1 + tan 4.
Aiso cosec2A=l+cot2A.
_, tA AC2 BC2 + AB2 AB* •-- 3.
For cosecM = ^r2 = — ^ ~ = 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 #=sin1-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_1f and 0=tan-1f 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 :
sin235° + cos235° = 1 ; ^|£ = tan 35°.
cos 35
sec235° = 1 + tan235° ; cosec235° = 1 + cot235°.
2. If sin 6 = f , find cos 6 and tan 6.
3. The cosine of an angle is y2^, 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) cot2^+sinM.
(iv) Show that cosecM - cosM =cot2A + ain2A.
6. If A =90°, £ = 60°, (7=30°, Z) = 45°, show that
sin B cos C+sin G cos B = smA,
and that cos2D - sin2Z> = cos A .
7. The tangent of an angle is 0 675 ; 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 0° 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
/. #tan3°15'=-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° tan2 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 ebt sin {at) when
6 = -0-7, t = l% and a =3*927.
13. Find the numerical value of e^(a2 - 62) tan 0, where
c = 25*2, a = 90, 6 = 49*6, 0 = sin-1(-528). (0 is less than 90°.)
14. Find to the nearest integer the value of the expression
700s/n{|Si„-M0-1426) + §l.}
when the given angle is positive and less than ^.
15. If ?? = 11 *78, q = 5'61, 0 = 4712 radians, calculate the value of
p° sin 0 (p2 + q2)~% ;
also the value of sjpq .
16. In the following formula, a = 25, 6 = 8*432, c = 0*345, 0=0*4226
radians ; find the value of
aii57 b~ £ * 0 (c3 + a loge b . tan 0).
17. sin 162° tan2 140° -r Vsec 105°.
18. Find the value of a * (a2 - 62)^ -*- sin 0 loge £ «
6
If a is 9*632, 6 = 2*087, 0 is 0*384 radians.
168 PRACTICAL MATHEMATICS FOR BEGINNERS.
/
19. Find the value of ah+ \/c2 + 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
.: JBC=45tan60o=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
*
*»
0
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
0
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
y—ax+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
/. 0 = 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
0 -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.
a6
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
0 -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='007xr.
0
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 pIa 8is
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 (xy 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 0
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
0
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 0
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 T2 is given by L = aT2 + 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
p1 = Pitch of Rivets \
(single riveted), - /
2-06
2-25
2-3
2-37
2-40
2-63
p2 = 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-, j7^-, and ^-.
(iv) Plot d and px and d and p2 and obtain the constants in the
relations px = d + b ; p2 = 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=CVn, 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 100'5 x 10°"25 = 3*162 x 1-778 = 5-623 ;
.-. 10075 = 5-623.
In a similar manner we obtain 10¥ = 1*336, 10T?r= 1*1548, and
10^ = 1*0746;
.*. 10^ x 10^ = 10^ = 3*162 x 1*0746=3*398.
Also 10* = 1000* = 31*62,
and 101* = 10^ =56*23.
10° = 1 ; .*. log 1=0.
101 = 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 0° 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 0° 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 x2 - Ax -f 3 the factors {x - 1) (x - 3)
can be obtained by substitution; these values x=l, x—3 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 x2 - 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
0
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=0t 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 x3 - Sx - 1 = 0. Let
y=x3-Sx- 1.
As before, put x=0, 1, 2, etc., and calculate corresponding values
of y as follows :
■
X
y
-2
-3
-1
1
0
-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 .r2 - 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 Xs- 6x + 4=0. ■,
Write the given equation in the form of two equations.
y=x3(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
0
i
2
3
4
5
y
0
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=axM, y=ae6*,
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 = axn becomes
y=0'25x2x
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
/
?
0
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,
0
1
2
3
4
5
Corresponding \
values of y, /
0
•25
1
2 25
4
6-25
As y is 0 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 = axM becomes when a = l, y=xn. Giving
various values 2, 3, ^, J, 1, etc., to the index n then functions of
the form y=x3, 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 = x3,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 = axn
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-
0 > 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 pvn = 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'lex 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 0
34-2
41 0
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. x2- 5 -45a; + 7 '181 = 0. 8. 0-24x2-4'37«- 8'97 = 0.
9. 2 3a;2 -6'72a;- 13-6 = 0. 10. x3-7x2+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. 2'42ar5-3-151ogea--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
pu11 — 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) xs+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
0
•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. xs-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 Qx (Fig. 92), its co-
ordinates (5*4, 1*88). The increments 8y and 8x become 0-38
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 = x2 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±
0
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 0
4t
Fig. 93.— Graph of y=x2.
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 = x2 (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 = x2 + 2x5x + {5x)2 (ii)
Subtracting (i) from (ii),
.-. dy = 2x5x+{8x)2.
Dividing both sides by 8x,
8v
jL = 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 = x3 + 3x2dx + Sx{dx)2 + {dx)3.
Subtracting y = x3,
dy = 3x2dx + 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 <& = 3x2.
dx
The slope of the curve at the point x =25 is
3x2-52 = 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 = axn then ~=naxn~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)xsc
dx
Ex. 14. y=4x3 + 3x2 + 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~2X6X -2*'
206 PRACTICAL MATHEMATICS FOR BEGINNERS.
Ex. 16. y
i xG-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 = axn 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
0
•l
•2
•3
•4
•5
•6
•7
•8
•9
1-0
y
0
•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=ae6x, 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 = 0 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.
0
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 = 0 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— 0 0 = 1 "785.
3x8
Proceeding as in preceding problems,
If y=aebx, -^- = abebx
dx
dx
= -53 x -26x2 -7 18 78 =-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 Zx3- I5x2 + 24# + 25 = 0 ; 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
0
•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
0
0 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 = 2ar5-15x2 + 24# + 25, ^| = 6x2 - 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 x2-5x + 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; - x2
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 - x2, 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- 12}=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 - l2}
= 16(M-1) = 9;
9
/. average velocity during "25 second = ^ = 36.
If the interval be from 1 to 1*01,
space-16(1012-l2)
= 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 = 16tf2... (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 :
*
0
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'lt2 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^ + It2, 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 = 20t2 : 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
0
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 + 4t2.
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 = 0 and x= 10.
Solve the equations :
8. 0 35a:2 -5 '23a; -7 "86 = 0. 9. 2-065- 048* = 0 '826.
10. Find, correctly to three significant figures, a value of x
which will satisfy this equation :
9ar* - 41x08 + 0 -5e2* - 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=x2-5'45x + 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=0-53eQ™*.
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: = 0 to a; = 8.
i -1
16. Given (i) y = axn, (ii) y = ax5 + bx~* + ex* + dx qy
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 + ex2,
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. respectivelye
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 53p, 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.
a2 + b2>c2, a2 + b2 = c\ a2 + b2<c2.
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 Per 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) = irr2.
If d is the diameter of a circle, then as d—2r the formula 7tt2
becomes — d2.
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 = irr2.
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 0 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 „ dr2
area of sector = —irr8 - ^-Trr2 = — . Fia 10i._Area 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 52 ;
area of sector = -£foir x 152=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
XyX252:
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 _ 0
~~ &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(R2- r2) = 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,
A3 2 _
area==2^ + 3cA>
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 0
= 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 0
.-. 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 hu A2, h3...h7 ; then, if s denotes the common distance or,
space between the ordinates, we have
Area of GFFD^{h1 + h7-^4(h2+hi+h6)-h2(h3+h5)}
= 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
0
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
'
0
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 0 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.
0
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
0
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 = irr2 ; .\ volume — irr2 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 + 27rr2
= 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 = -(Z2 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 0 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 = ttR2 - irr2 = tt(B? ~ r2),
And volume = tt(jR? - r-%
;•; V='7854(B2-d2)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 (182- 102)
= •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., 1—10 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)
= i7rr2xA,
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
wThich 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=i1Tr2h ; 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 = 672x ^=3525*66 sq. in.
Volume of cone = ^(3525 '66 x 30) = 35256 '6 cub. in.
Slant height = V33 '52 + 302 = 44 '98.
.'. Surface = J ■> x 67 x 44*98) = 4733 -85 sq. in.
EXERCISES. XLIL
1. From the two formulae V=^irr2h 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?rr2 (i)
Volume of a sphere -firr3 (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 ttt2, 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 27rr2.
Hence the area of the surface of the sphere is 2 x 27r2 = 47ir2.
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 = 7rr2,
Fig. 125.— Sphere and its and height of cylinder = 2r ;
circumscribing cylinder. . volume of cylinder = 2^ ;
two-thirds of 27rr3 is Jttt3, 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 = ^ d3
= -5236d3 (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 52 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 r2 and rx respectively, then the
volume of the material forming the sphere would be
i^3-^3, or £-7r(r23 - 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(93-43) = -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 27rCDy 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 = 4ir2Rr (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 = 7rr2.
Mean length = 2ttR.
Volume = irr2 x 2irR.
;. V=2ir2Rr2 (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 4ir2Rr '~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 = 4ir2 x 41 x 5 sq. in.
tt2x 20x41 ,. _„0 '
= yta scl- ft. =56 '2 sq. ft.
Volume. — Area of cross-section = -k x 52.
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) = 52 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 s2 x a and volume
s3 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 22. As there are
6 similar faces the surface is 6 x 22=24 sq. in.
In a similar manner the surface of the second cube is 6x 42=96
sq. in.
Thus the surface of the second is 4 times that of the first,
The volume of the first cube is 23=8.
The volume of the second cube is 43=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 83 times that of model ;
.-. weight = 512 x 100 = 51200 lbs.
Area of surface is 82 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.
0
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
0
s
10
Lr
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
0
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.
0
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.
0
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.
0
13
23
33
47
62
78
91
104
120
Values of A.
0
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.
0
5
10
17
21
25
29
33
38
42
A.
0
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.
0
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 plt
266 PRACTICAL MATHEMATICS FOR BEGINNERS.
z
yPz
j / 2
\ ' ^*
* "
«---/j -*/?
Fig. 137.
their point of intersection, to the origin 0. Then 0pk is the pro-
jection of OP on the plane xy, its length is 2*5".
In a similar manner the projections, 0p2 on the plane yz and 0pz on
the plane xz are obtained ; Op2 = 2'5" and Op3 = 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 0px is
the base and the perpen-
dicular pxP the height of
P above the plane of xy,
or simply the z co-ordinate
of the point. Hence draw
pxP perpendicular to 0p1
and equal to 2". Join 0
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 0 as centre and length of OP = 3 2"
as radius. Describe a circle cutting the lines passing through pv
p2, and 2h at P\> A> an(^ l\ respectively. Join 0 to Px, P2,
and P3, 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 0 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 0 cos <£ (i)
y = 0G= Op cos pOC= Op sin cf> ;
:. y=OPsin #sin</>, (ii)
and z = OP cos 0 (iii)
Ex. 3. Let 0P= 100, 0 = 25°, 0 = 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.
OP2 = r2 = Op2 + pP2 also Op2 = OB2 + Bp2;
.-. r2=OB2 + Bp2+pP2
= x2 + y2 + z2 = 9 + 16 + 25 = 50;
.-. r=\/50 = 7'071.
From (iii) z=rcos 0 = 7*071 cos 0;
.'. 008 0 = ^1^= -7071; /. 0 = 45°.
From (i) 3 = r sin 0 cos 0,
CO8»=^L=7-071x-7071; •'• 0 = 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~x2-\-y2-\-z2. 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
cos2a + cos2^ + cos2(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 cos2a + cos2/3 + cos20 = l or its equivalent
l2 + 7n2 + n2 = 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=\/22+l'52 + 22=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 + cos245° + cos20 = 1,
J + l + cos20 = l,
or cos^^l -|=J;
.'. cos 0=£ and 0 = 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 pq = ^ - 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 r2 = x2 + y2 + 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
=\/22+l-52 + 2-32=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'.
0 = 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 0 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 0 can be
obtained.
Ex. 11. Let r=20, 0 = 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, 0 = 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 0 of the room where these
three mutually perpendicular planes meet. If a string were
stretched from 0 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 0 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", 0 = 32°, 0 = 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, 0 = 65°, 0 = 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 0 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,
v2==(t)2r2 = 227r2n2r2.
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 0 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 0 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 0 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/A2 + B2=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 0 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 0 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 0 = 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 0° 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 0 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 0 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 0 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=\/29-162 + 69-142
= 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 0 (where 0 denotes
the inclination to the axis of z) can be calculated from the relation
cos2a + cos2/3 + cos20 = 1.
Thus, for vector A , we have
cos20 = 1 - cos2a - cos2/3 = 1 - i - t = T 5
;. cos0=l and 0 = 60°.
Similarly for B,
cos20 = l-(-866)2-(-1736)2='23; .\ 0 = 61° 21'.
And for G, cos20 = l - J- J = J ; .'. 0 = 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 0 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 = 20cos61°21' = 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; .-. 0 = 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 cara
then the work done is
A B cos d (i).
As A is 300, £ = 100, and 0 = 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 0° = 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° = 0 ;
.'. 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, - ' -
0°
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,
0°
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 '730o + 1 -482° + 2-6i57° •
Find also B and /3 from the equation
£/3 = 3-73o°-l-482<> + 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 a2 + 2ab + b2. Hence,
when any expression of this form is given, its square root can
be seen at once and written down, e.g. sjx2 + 2xy+y2=x+y.
We may, from this example, proceed to derive a general rule
for the extraction of the square root.
a2 + 2ab + b2(a + b
a2
2a + b)2ab + b2
2ab + b2
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 + b2 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 4x2 + 24xy + 36y2.
Here 2x is clearly the root of the 4*2 + 24^ + 3%2 ( 2* + 6y
first term 4x2. 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 + 36yi
and subtract its square from the given 2Axy + 36?/2
expression, Bring down the other
two terms 24xy + S6y2. 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 + 36y2.
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* + 4x2y2 - \2xh2 + y*- 6yh2 + 9z4.
4a:4 + 4x2y2 - 1 2xh2 + y*- Qy2z2 + 9z4 ( 2x2 + y2-3z2
4x*
4x2y2 - \2xH2 + y*- Qyh2 + 9z4
4x2 + y2 ) 4a;y jV
4a;2 + 2y2 - 3z2 ) - \2x\2 - Qy2z2 + 9z4
-12x2z2-6y2z2 + 9z4
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+2X 8X +16 128 +
292 PRACTICAL MATHEMATICS FOR BEGINNERS.
Viol =n/(!oo+I) = 1(>V(1+iuo)
= 10(1+200~80000+ 16000000 ~ 12800000000 + etc7
= 10-04988....
EXERCISES. XLIX.
Find the square root of
1. 9a4 - 42a3 + 37a2 + 28a + 4. 2. 4a4 + 12a3 - 1 la2 -30a + 25.
3. a4 + 4axs + 2a2a2 - 4a3a + a4.
4. a6 - 22a;4 + 34a3 + 121a2 - 347a + 289.
5. 25a8 -60a6 -34^ + 84a2 + 49. 6. a4 - 2a3 + 9a2 -8a + 16.
7. 25a4 - 30a3 + 49a2 -24a +16. 8. a4 + 8a3 -26a2- 168a + 441.
9. 16a4 -4005 + 89a2 -80a + 64. 10. f^L + ^-S.
y2 a2
11. 9a4 - 30a3y + 31a2y2 - lOay3 + y4.
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
5a2 -8a +13 and 4a2 + 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 a2 -9 = 0.
We have *2 = 9 = 32; .\ 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
x2 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 x2 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 x2 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 x2 + 4x - 21 = 0.
We have #2 + 4x=21 (i)
Add the square of one-half the coefficient of x to each side ;
.-. a2 + 4# + (2)2=21+4=25;
i.e. (a + 2)2 = 52;
.'. a? + 2=±5; (ii)
/. x= -2±5 = 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 x2 + 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 = .r2 + 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
ax1 + 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/ 4a2
2a
0
a
— 4ac
Aac
2a
•(i)
The following important cases occur. If b2 is greater than
4ac i.e. b2 > 4ac, there are two values, or roots, satisfying the
given equation. If b2 = 4ac the two roots are equal ; each is
-— . If b2<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
0
1
2 j 3
y
2
0
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
0
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 x1 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 Qfa-x\
\x^rs) ~S\x^b)
15.
If we write y for —
x-
Hence
x
the equation becomes
y2-8y + (4)2:
.'. y=4±l
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 yf the equation could be written as
(H)2-8(H)+<*>2=-^=i>
4±1 = 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 x2 + y = 8, 3x + 2y = rJ.
x2 + y=8 (i)
3x + 2y = 7 (ii)
Multiply (i) by 2 and subtract (ii) from it ,
.*. 2x2 + 2y=lQ
3x + 2y~ 7
2x2-3#= 9
X 2X+\4j ~2+16_16'
q q
* = 7±7 = 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) x2 + xy = 84; (ii) xy + y2 = Q0.
Adding (ii) to (i) we get
x2 + 2xy + y2=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* - 17x2+ 16 = 0.
The equation may be written
(x4-8x'2+16)-9x2=0, or (#2-4)2- (3#)2 = 0 ;
:. {x2 + 3x - 4) {x2 - 3x - 4) = 0.
Hence x2 + 3x-4 = 0, (i)
or x2-3x-4:=:0 (ii)
From (i), x2 + 3x-4: = {x + 4)(x-l) ;
.-. x= -4, or 1.
EQUATIONS REDUCIBLE TO QUADRATICS. 297
From (ii), x2-3x-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) = 0 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
ax2 + bx + c=0.
We have ax2 + bx + c = a(x - a)(x - /5) ;
\ ax2 + bx + c = a(x2 — ax- /3x + a/3)
= a(x2-(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 x2 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) = x2-5x + 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 x2 + 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 x2 x + 1=0.
EXERCISES. L.
Solve the equations :
1. x2- 60 = 80 -Ax. 2. x2 + S2x =320.
3. ?^-x + ~ = 0. 4. 2x2-4x-6 = 0.
5. 3#2-84 = 9x. 6. x2+'402x='\63.
7. 6x2-13x + 6 = 0. 8. x2-{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. x3-2x2-3# + 4 = 0. 14. (x2-4a: + 3)2-8(:r2-4a; + 3) = 0.
15. l+2xs/(T^x2j=9x2. 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. xs + y* = 72, xy{x + y)=A8. 22. *2-4y2 = 8, 2{x + y) = 7.
23. 2x2-3y2 = 5, 3x + y=15.
24. (i) Find the roots of the equation x2-2ax + (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 x2 + 7 W2 = 60. Form the quad-
ratic equation having roots a and -.
a
26. If a and (3 are the roots of the equation ax2 + 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-s2.
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;
.-. x2 + 239a; - 240 = 247*,
or x2- 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; '
/. *=-2±3=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 ax2 + 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 9th 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, ar2, 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 ar2.
Thus, if I denote the last term and n the number of terms then
we have Z = arn_1 (i)
Let S denote the sum of n terms then
S = a + ar + ar2+... arn~2 + ar71'1 (ii)
Multiplying every term by r
Sr = ar + ar2 + ar3 + ... ar^ + ar" (iii)
(Subtract ii) from (iii).
.*. rS-S=arn-a,
or £(r-l) = a(rw-l).
0 a(rn-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 ar2 where a denotes the first term and r
the common ratio ;
.-. 5r2=20 or r = 2.
From l^ar71'1
we get by substitution 1=5^ = 5 x 27
= 640.
a a(r"-l) 5(28-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 ar2= 20 and ar7=640;
ar7_640.
•'' ar2- 20 '
or ^=32; .-. r=2,
and a=— - = 5.
4
r-1
_5(2»-l)
" 2-1 J
. 2M_1=20475 = 4095
5
or 2W=4096 = 212;
/. 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, rn~1 = -; or arn=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 rn decreases as n
increases. Thus when r is ^, r2=Y^, ^3 = j0,5o"? e^c') wnen **
is indefinitely great, rn is zero, and (v) becomes
S~r (vi)
1—7*
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
+ 102
6
:+103 +
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, ar2 + ...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 = arn~1
we obtain rn~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 '
r5 = 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 6th 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\ c—a-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 0° C. and 1 atmosphere, weighs *0807 lb.
1 cubic foot of hydrogen at 0° C. and 1 atmosphere, weighs '00559 lb.
1 foot-pound =1 -3562 x 107 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 = 106 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 0° C. = 1082+ *305 9° F.
log10p = 6-1007-f~,
where log10£ = 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.
0
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
0
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
0
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
0
1 1
2
2 3
3 4 4
88
9445
9450
9455
9460
9465
9469
9474
9479
9484
94S9
0
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
0
1 1
2
2 3
3 4 4
91
9590
9595
9600
9605
9609
9614
9619
9624
9628
9633
0
1 1
2
2 3
3 4 4
92
9638
9643
9647
9652
9657
9661
9666
9671
9675
9680
0
1 1
2
2 3
3 4 4
93
9685
9689
9694
9699
9703
9708
9713
9717
9722
9727
0
1 1
o
2 3
3 4 4
94
9731
9736
9741
9745
9750
9754
9759
9763
9768
9773
0
1 1
2
2 3
3 4 4
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
0
1 1
2
2 3
3 4 4
96
9823
9827
9832
9836
9841
9845
9850
9854
9859
9868
0
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
0
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
0
1
2
3
4
5
6
7
8
9
1
0
2
0
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
0
0
1
1
1
I
2
2 2
•02
1047
1050
1052
1054
1057
1059
1062
1064
1067
1069
0
0
1
1
1
1
2
2 2
•03
1072
1074
1076
1079
1081
1084
1086
1089
1091
1094
0
0
1
1
1
J
2
2 2
•04
1096
1099
1102
1104
1107
1109
1112
1114
1117
1119
0
1
1
1
1
2
2
2 2
05
1122
1125
1127
1130
1132
1135
1138
1140
1143
1146
0
1
1
1
1
2
2
2 2
•06
1148
1151
1153
1156
1159
1161
1164
1167
1169
1172
0
1
1
1
1
2
2
2 2
•07
1175
1178
1180
1183
1186
1189
1191
1194
1197
1199
0
1
1
1
1
2
2
2 2
•08
1202
1205
1208
1211
1213
1216
1219
1222
1225
1227
0
1
1
1
1
2 2
2 3
•09
1230
1233
1236
1239
1242
1245
1247
1250
1253
1256
0
1
1
1
1
2
2
2 3
10
1259
1262
1265
1268
1271
1274
1276
1279
1282
1285
0
1
1
1
1
2
2
2 3
•11
1288
1291
1294
1297
1300
1303
1306
1309
1312
1315
0
1
1
1
2
2
2
2 3
•12
1318
1321
1324
1327
1330
1334
1337
1340
1343
1346
0
1
1
1
2
2
2
2 3
•13
1349
1352
1355
1358
1361
1365
1368
1371
1374
1377
0
1
1
1
2
2
2
3 3
•14
1380
1384
1387
1390
1393
1396
1400
1403
1406
1409
0
1
1
1
2
2
2
3 3
15
1413
1416
1419
1422
1426
1429
1432
1435
1439
1442
0
]
1
1
2
2
2
3 3
•16
1445
1449
1452
1455
1459
1462
1466
1469
1472
1476
0
1
1
1
2
2
2
3 3
•IT
1479
1483
1486
1489
1493
1496
1500
1503
1507
1510
0
1
]
1
2
2
2
3 3
•18
1514
1517
1521
1524
1528
1531
1535
1538
1542
1545
0
1
1
1
2
2
2
3 3
•19
1549
1552
1556
1560
1563
1567
1570
1574
1578
1581
0
1
]
1
2
2
3
3 3
•20
1585
1589
1592
1596
1600
1603
1607
1611
1614
1618
0
1
]
1
2
2
3
3 3
•21
1622
1626
1629
1633
1637
1641
1644
1648
1652
1656
0
1
1
2
2
2
3
3 3
•22
1660
1663
1667
1671
1675
1679
1683
1687
1690
1694
0
1
1
2
2
2
3
3 3
•23
1698
1702
1706
1710
1714
1718
1722
1726
1730
1734
0
1
1
2
2
2
3
3 4
•24
1738
1742
1746
1750
1754
1758
1762
1766
1770
1774
0
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
0
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
0
]
1
2
2
3
3
4 4
•29
1950
1954
1959
1963
1968
1972
1977
1982
1986
1991
0
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
0
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
0
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
0
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.
0°
0
0
0
0
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 ab ?
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 = \x2 - 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 = 0 ?
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-5t2 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 0 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 2307065 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{p1 (l+\ogr)-r{p8+\0)} and if ^ = 100, p3=\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+dil+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.
x2 - 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= 10t2. 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'dx*.
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 + 0-001^ 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;2-51og10#-2'531=0.
13. If cc = a(0-sin0) and y = a(l-cos0), and if a = 5; taking
various values of <p between 0 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, 372-4243, 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 0 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 v1
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 v1)i
s = 7,000|logA
If v1 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 tx = 458, t3 = 373, lx = 796 - 0 -695 tv
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
0 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;2-101og10a?-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 = aebx.
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 0
39 1
50 3
67 2
85-6
125 0
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°, log10153*4, loge153*4.
2. Both (a) and (6) must be answered to get full marks.
(a) If F=EIir* + 4J?,
If /=&*£-*- 12,
If E=S x 107, 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#2-201og10a*-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=cd5V5.
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
0
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
0
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
0
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
0
•05
•10
•15
•2
•25
•3
•35
•40
•45
•5
X
0
•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 26-071'13.
(c) Write down the values of cos 35°, tan 52°, sin-1 0*4226,
log1014*36, loge14*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 = 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 = p2v2113 and if v2/v1 be called r.
If p2 = 6, find r if ^ = 150.
o
5. If y = - + 51og10a?-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
0
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 rf5r?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 + bs3, 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, 0 = 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-|tf2)2
to its simplest form ; and show that when x is a positive proper
fraction the value of the expression is between 0 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 x2 and of 3a; +1. By means of them find
approximate values for the roots of x2 - 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 an 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 logx 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, 4ir2I
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; remr,
•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. 3ax2-3bx2. 3. 16m- lira. 4. 5a + 76-6c; 1.
5. 2x + 3y. 6. 36. 7. Sax2 - x2 - dx2 - 2x + bx -/.
8. 8a + 26 + 4c; 2§. 9. x^ + y^-xh/2. 10. 26xy - 5x2 - 5y2.
11. x*-2\x2 + 2Q. 12. 2xy + x-x2 + y2 + 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. 3y2 + 7xy- llxz + z2.
11. a + 26 + 3c + 4d; 2a+26 + 2c + 2d. 12. 2c-a-6 + a*; x2-y2.
13. 2a3-3a26. 14. x*y+ 12x2y2+10xy3 + 21y*.
15. 2a4 + 3a36 + 3a262 + 2a63 + 64. 16. 2ar* + 31a.r2 - 31a2* + 7a3.
17. 44a6 + 33aaj + 24cy + 43ez.
Exercises XI., p. 68.
1. a^ + a^ + a4. 2. 4a6 + 1 la462 + 7a264 - 66. 3. x* + xy + y*.
4. a6 -21^+20. 5. x*- S^ + Uy*. 6. x6 - 24x* + lUx2 - 256.
336 PRACTICAL MATHEMATICS FOR BEGINNERS.
7. a12 - 3a1062 - 2a864 + 13a666 - 3a468 - 12a2610 + 6612.
8. xP-y*. 9. 8a56-26a363 + 2a65.
10. a6 - a264 - a462 + aV + 66 - a4c2 - 62c4 + 64c2 - c6.
11. l-y2-y3-y4-y5 + y6+2y7 + y8. 12. 6X8- lla^ + 22ar*-4a^-7.
13. ar5* 43^ + 48a; -32. 14. 16a4-72a262+8164.
15. - 13a3 - 22a2 + 96a +135.
Exercises XII., p. 70.
1. 2a3-3a2 + 2a. 2. x + 2y-z. 3. a + b + c.
4. 2^-6x2y + 18rcy2-27y3. 5. ^*. 6. 3a + 26-c.
x + y
7. 5a + 6-2c. 8. -9a64c2. 9. 4xy + 2y + Sx + 1.
10. 3a-5 -2a -9. 11. 2a2-262 + 3c2. 12. a2-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. -6a6-62-a2.
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)(9a2-6a& + 462).
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. (25x2 + y2)(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 + 7y2).
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+a2
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^-y2'
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
x2 + 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]3i 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. 7«f, -ffl; "• * >• I*- ^K+&-f:
Exercises XIX., p. 98.
1. $. 2. 16, 2. 3. T77. 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-rb2 + c3-Sahh. 2. (i) 21*656; (ii) 17 656. 3. A
4. a~h^. 5. 2a. 6. a^p+xK 7. a12<m~n>.
8. x4<m-4:Xzm+n + 6x*m+*n-4:Xm+in + x4'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"*ab2cm. 19. 6*. 20
21. (i) a-1^, (ii) a V"-1 + cfix~% + a~*x$ + ofM .
22. 5-44. 23. ar1^. 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'565xl07. 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 = *6d2, A = ( '593d2 - -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. {\)naxn~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 x52= 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) 0 ; (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. 0 = 60°, 51° 46', 60°, A+B + G=IU'8, a = SS° 42', /3 = 101° 6',
0 = 53° 36'.
12. ^4=4-368, a = 76°40', 5=16, 0=67° 24'.
Exercises XLIX., p. 292.
1. 3x*-7x-2. 2. 2a2 + 3*-5. 3. x2 + 2ax-a?.
4. ^-11^+ 17. 5. 5^-6<c2-7. 6. a2-a + 4.
7. 5x2-Sx + 4:. 8. x* + 4x-21. 9. 4x2-5x+8.
10. —-%!-. ll. 3x2-5xy + y2. 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, Tl. 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; x2-2{a2 + b2)x+{a2-b2)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. n2-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$, 2|.
- 13 o 36. o 13 9. 9 o 9. o 36 9
10. T, d, ^, 2, T, _, 2, 3, -, 2, _, ^
EXAMINATION PAPER, 1901.
1. 7446; 0-01254; 5-68; 1546. 2. (a) 1691; (6)0515.
3. d=3'43"; 1552 sq. in. 4. 4 '009 ; 6*662; 53°; 10 66 sq. in.
5. £1100000. 6. 260. 7. 1350; 1700.
8. (a) s ^ ; (b) - = ; (c) 2 miles per hour.
ax^fb a*-4 x-S
2*25
9. Hint. Let x be the number, then x H =y;
., ^=l-?^=0ora;=v/^25 = l-5.
dx x2
10. x=3± V3 . =4-732 ; 1-268. 11. 230. 12. t>= 150 - 10*.
13. (1)5-643; (2) cos a ='4429; a = 63° "7 ,- cos/S= -5493; 0=56°-7 ;
cos 0 = -7087 ;0=44°-9; (3) 1.0001.
ANSWERS. 345
EXAMINATION PAPER, 1902.
1. 3123, 1704, 1722, 0198.
2. (a) 14407, 16604, 18557, 18815 ; (6) 55 ft.
a
3. (a) fr3-^2)*; (6) 2 _ 1 (c) 45.6 30.4
v ' x + y + z x + S x-5
4. r=3'5. 5. 40*1 ft. per sec, 40 "01 ft. per sec, 40 ft. per sec
6. a=2*2, 6=0-11, z = 4'4. 7. 53'56 sq. in.
8. ^Z>=2'6624 ft., 6-2167 sq. ft., £Z) = 1*864, DC=2-806,
AC =3S68.
9. Value of v is about 9.
10. Converted marks are : 118*9, 160*8, and 213*3.
11. 13550, 14350, 14740. 12. *=2012.
EXAMINATION PAPER, 1903.
1. 284*7, 2817, 3*339, 193, 1768000, 11*03.
2. (a) 40°; (b) t = 1*5 sec, s=2685ft.; (c) 96.
3. (a) |2a?T + ^8T + 4\¥; (b) r^ + ^n ; («) 3*229, 1*753;
{d) 55° 55'.
4. 304 sq. in., 232*2 sq. in., 33669 cub. in.
5. Average rates, ^4=2*8, 5=2*4; ^'s=ll$, 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, 0 072.
14. T=0'95U+0525W=0'6U+0'28; JF=3*28.
(b) i?=— g- x ^j-, where k is a constant.
MATRICULATION EXAMINATION, 1902.
1. 720-2 grams. 2. H±& . 26*8. 3. |£ ; 0*815.
4. ^ ; 3*59 per cent. 6. g*3-^*4- 7. (2a*+l)(a*- 1) ; 1, -1
8. 3*302,-0*302. 10. 120 ft., 50 ft.
1. 0*125, 1. 3. 5005 (2a)6 (36)9; 1*059.
4. M =221*4, #=0*1704. 6. 246*6 yds.
8. 5 = 51°, 6 = 84-05, c = 108*l; A = 2861. 9. 117° 19'.
INDEX.
Acceleration, 213.
Addition, 2, 62, 143.
Algebra, 57-113.
Algebraical sum, 59.
Amsler's planimeter, 238.
Angular measurement, 33, 275 ;
velocity, 275.
Approximations, 111.
Area, British measures of, 117 ;
measurement of, 216 ; of plane
figures, 216-240.
Arithmetical progression, 301
Arithmetical mean, 302.
Averages, 22.
Average velocity, 210.
Binomial theorem, 110.
Boyle's law, 104, 107.
Brackets, use of, 70.
British measures of length, 114.
British measures of area, 117.
Chords, scale of, 36.
Circle, area of, 224 ; circum-
ference of, 222; segment of,
226.
Coefficient, 60.
Common ratio, 301, 304.
Compound interest law, 207.
Cone, 242, 249.
Continued product, 4, 68.
Contracted methods, 8, 9, 13.
Construction of an angle, 37, 161
of a scale, 40.
Co-ordinate planes of projection.
263.
Cross-section, 247.
Cube root, 30, 135, 149.
Curve, slope of, 200, 204.
Cylinder, 241, 245.
Decimal fractions, 2-12.
Density, 122.
Diagonal scale, 41.
Differentiation, simple, 205.
Direction-cosines of a line, 267.
Division, 12, 13, 69, 130, 147 ; of
a line, 39.
Elimination, 91.
Evolution, 25, 43, 135.
Explanation of symbols, 57.
Exponent, 60.
Equations, 191 ; of a line, 176 ;
cubic, 82, 193; simple, 82;
simultaneous, 177 ; quadratic
292; reducible to quadratics,
296 ; simultaneous quadratics,
295.
Equatorial plane, 269.
Factors, 73, 76 ; highest common,
78 ; constant, 301.
Fourth proportional, 19.
Fourth root, 43.
Fractions, 1, 6, 77.
Fractional index, 26 ; equations,
84.
Functions of angles, 161.
Geometrical progression, 301. 304;
mean, 306.
INDEX.
347
Harmonical progression, 308 ;
mean, 308.
Hatchet planimeter, 237.
Highest common factor, 78.
Height and distances, 168.
Hooke's law, 103.
Hollow sphere, 253.
Hollow cylinder, 247.
Hyperbolic curve, 196.
Identity, 83.
Increase, rate of, 184, 201.
Index, 24, 60 ; rules, 107 ; frac-
tional, 26.
Indices, 107.
Inverse, proportion, 104; ratio, 162.
Involution, 25, 107, 133, 148.
Irregular figures, 229, 257.
Interpolation, 174.
Italian method of division, 12.
Latitude and longitude, 269.
Law, Boyle's, 104 ; Hooke's, 103 ;
of a machine, 98 ; compound
interest, 207.
Least common multiple, 79.
Line, plotting, 175 ; slope of , 183;
through two points, 177, 271 ;
equation of, 176.
Logarithms, 125-140.
Machine, law of a, 98.
Maxima and minima, 208, 214.
Measurement of angles, 33, 151 ;
of area, 118, 216 ; of length, 114.
Mean proportional, 19, 42, 102.
Mean, arithmetical, 302 ; geo-
metrical, 306 ; harmonical, 308.
Mid-ordinate rule, 230.
Multiple, least common, 79.
Multiplication, 3, 9, 42, 67, 129,
147.
Parallels of latitude, 269.
Parallelogram, area of, 218.
Partial fractions, 94.
Percentages, 20.
Perry, Prof., 190.
Proportion, 18, 41, 100, 102, 104,
105.
Planimeter, 236; Amsler's, 238;
.hatchet, 237.
Plotting, functions, 194 ; line, 175.
Polar co-ordinates, 273.
Principle of Archimedes, 122.
Prism, volume and surface of, 242.
Progessions, arithmetical, 301 ;
geometrical, 301, 304 ; harmoni-
cal, 308.
Pyramid, 241.
Quadratic equations, 292 ; simul-
taneous, 295 ; relation between
coefficient and roots, 297 ;
problems leading to, 298.
Rate of increase, 184, 203.
Ratio, 16,100; inverse, 162; of
small quantities, 17.
Regular solids, 241.
Resolution of vectors, 281.
Retardation, 214.
Root of a number, 25, 30, 43.
Rule, index, 187 ; mid-ordinate,
231 ; of signs, 66 ; Simpson's,
232 ; slide, 143.
Scalar quantities, 277.
Sector of a circle, 225, 226.
Series, 301.
Significant figures, 5.
Similar figures, 49, 50, 259.
Simple equations, 82.
Simpson's rule, 232.
Simultaneous equations, 90, 177.
Simple differentiation, 205.
Slide rule, 143, 165.
Slope of a curve, 200 ; of a line, 183.
Small angles, 162.
Specific gravity, 122.
Sphere, 242, 251, 253.
Surds, 28, 79.
Surface, of cone, 249 ; cylinder,
245 ; prism, 243 ; solid ring,
255 ; sphere, 251.
Solid ring, 255.
Suffixes, 110.
Symbolical expression, 57, 81.
Squared paper, 171.
Square root, 26, 29, 290.
348 PRACTICAL MATHEMATICS FOR BEGINNERS.
Tee square, 32.
Term, 60.
Theorem, binomial, 110.
Triangles, area of, 220 ; con-
struction of, 53 ; similar, 49.
Units, of area, 116 ; of length,
114 ; of volume, 119 ; of weight,
121.
Unitary method, 19.
Use, of brackets, 70 ; of instru-
ments, 31 ; of squared paper,
171 ; of tables, 173.
Value of recurring decimals, 306.
Variation, 102.
Vectors, 277 ; addition and sub-
traction of, 278 ; multiplication
of, 286.
Velocity, average, 210 ; rate of
change of, 201, 213.
Volume, of a cone, 249 ; acyclinder,
245 ; hollow cylinder, 247 ; prism,
243 ; solid ring, 255 ; sphere,
251 ; units of, 119.
Weight, unit of, 120.
Glasgow: printed at the university press bv Robert maclehose and co. ltd.
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