E.
The Gift of
WILLIAM H. BUTTS, Ph.D.
A.B. 1878 A.M. 1879
Teacher of Mathematics
1898 to 1922
Assistant Dean, College of Engineering
1908 to 1922
Professor Emeritus
1922
RADCLIFFE
OBSERVATORY
OXFORD.
ક્
ARTES
1817
SCIENTIA
LIBRARY
VERITAS
OF THE
UNIVERSITY OF MICHIGAN
LURIBUS UMUN
TUEBOR
SI-QUÆRIS-PENINSULAM-AMŒNAM
CIRCUMSPICE
*
Entesso
William
1
A
TREATISE
O F
ALGEBRA,
In TWO BOOKS.
+
ན
BOOK
CONTAINING,
I.
The Fundamental Principles of this ART.
Together with
All the Practical Rules of OPERATION.
воо к K II.
CONTAINING,
A great VARIETY of PROBLEMS,
In the most important
BRANCHES of the MATHEMATICS.
Vix quicquam in univerfa Matheft ita difficile aut rduum
occurrere poffe, quò non inoffenfo pede per hanc methodum
penetrare liceat.
SCHOOT. Pref. to DES CARTES.
THE SECOND EDITIO,
LONDON:
RADCLIFFE
UBSENTÍTORY
OXFORD.
Printed for J. NOURSE, in the Strand;
Bookfeller to His MAJESTY,
M DCC LXXX.
•
When Willis. H. B..
}
10.14-17307
825
111
Li
THE
PREFACE.
T
HE fubject of the following bock is Alge-
bra, a Science of univerfal use in the Ma-
thematics. Its business and ufe is to folve
difficult problems, to find out rules and theorems in
any particular branch of science; to discover the pro-
perties of fuch quantities as are concerned in any ſub-
jet we have a mind to confider. It properly follows
theſe two fundamental branches, Arithmetic and
Geometry, but is vastly superior in nature to both,
as it can ſolve questions quite beyond the reach of either,
of them.
THIS is an art truly fublime, and of an unlimited
extent; for if the conditions of a problem be never ſo
complex, and though the quantities concerned are never
So much entangled with one another, yet the Algebraist
can find means to diffolve and feparate them; or if
they be ever fo remote, his art can furniſh him with
methods to bring them together and compare them. It
is true he is often obliged to traverse by many round-
about ways, to get the relation of the quantities con-
cerned; yet by certain rules he can purſue the computa-
tion of his problem through all theſe intricate turnings
and windings; and by his ſkill and fagacity can hunt
it through all these labyrinths, till he arrives fafely
A 2
at
رت
,
iv
PREFACE:
The
*
at the end of the chace, viz. the folution of the pro-
blem.
THE extent of this curious art is fo great that it
bas gained the title of Univerfal Mathematics; and
is called by way of eminence, The Great Art; and bas
been esteemed the very apex of human reafon. It is alfo
called Specious Arithmetic, Univerfal Arithmetic,
The Analytic Art, The Art of Refolution and
Equation; with a view to fome or other of its pro-
perties or operations.
THE nature of this excellent art is fuch, that it may
be applied to any fubject, provided the principles of
that fubject, it is applied to, be understood. Its great
beauty is, that it deals in generals. For whilst other
branches go no farther than their own particular
fubject, and can only find folutions in particular caſes ;
this art finds out general folutions, general rules, ge-
neral theorems, and general methods.
THIS noble Science has alſo this peculiar property,
that it not only investigates rules in all the other parts
of the Mathematics; but by the most fubtle art and in-
vention, it finds out its own rules, models them ac-
cording to any form, and varies them at pleasure, fa
as to answer any end proposed. It would be in vain
to attempt to enumerate all the uses of this admirable
art.
By making use of letters instead of numbers, it has
one great advantage above arithmetic, viz. that in
the feveral operations of arithmetic, the numbers are
loft or fwallowed up, and changed into others but
here they are preferved distinct, visible, and unchang-
ed. By which means general rules are drawn from
particular folutions, to answer all cafes of like nature.
By
1
The PREFACE:
By help of algebraic characters, geometrical de-
monftrations are often rendered more fhort, compendi-
ous and clear. So that by this means we avoid the
tedioufneſs of a long verbal proceſs, which otherwiſe
we should neceffarily be involved in, and which never
fails to darken and obfcure the fubject.
Ir is highly probable the ancients made uſe of ſome
fort of analyfis, whereby they found out their noble
theories. For it is hardly poffible fo many fine theorems
in Geometry, should be groped out or stumbled on, with-
out fome fuch method. But as it was then only in its
infancy, it must have been far fhort of the perfection
we have it in at prefent.
As to the Reader's qualifications, it is abfolutely ne-
ceſſary that he understand Arithmetic and Geometry, as
the keys to all the rest. And it is also neceffary that
be understand the principles of every branch of science,
to which he would apply algebraic calculations; other-
wife it would be in vain to attempt the folution of any
problems therein, by the help of Algebra.
THEN as to the method I have followed, it is this.
I have gathered together the most valuable rules and
precepts, which lie fcattered up and down in all the
best books of Algebra; and what was deficient, I have
Supplied as well as I could. Then I have thrown all
thefe precepts and rules of working, into ſo many pro-
blems; which I have reduced into as fort a compaſs,
and expreſſed in as plain terms as poffible, fo as they
may be clear and intelligible. And the method I have
taken I fuppofe will appear to be very fimple and eaſy,
and will readily be apprehended by fuch people as have
found confufion and difficulty in other methods. 1 be-
lieve I have omitted nothing that is fundamental; and
if
མ་ན་
Vi
PREF Á C E.
The
F
if any thing of lefs moment is paſſed by, it is either
because it is of little ufe, or is supplied by fome other
method or rule. And all the rules and problems arê
in Juch order, that the eafieft appear firft, and lead
on to the harder, which follow in due course after-
wards: thefe make up the first book. And the fecond
book contains the application of Algebra to all forts of
problems, of which there is great variety, and many
of them perfectly new; others that are not jo, have
generally new folutions to them. So I hope I have
delivered both the principles and the practice at large,
and yet have not clogged the Reader with any fuper-
fluity.
W. Emerfon.
9
!
I
TH1.
$
1
THE
CONTENTS.
D
Efinitions
Characters
Notation
Axioms
BOOK I.
The fundamental Principles.
Sect. I. The Operations in Integers
Sect. II. Fractions
Sect. III. Surds
Sect. IV. Of managing Equations
Page
I
3
5
7
31
48
Sect. V. Subftitution, Extermination, &c.
Sect. VI. Infinite Series
· 90
110
Sect. VII. Several fundamental Problems
132
difference is given
Prob. LXV. To find two quantities whofe fum and
198
198
198
Prob. LXVI. To find the leaft common dividend
Prob. LXVII. To find the difference of the fquares,
having the fum and difference given
201
Prob. LXVIII. Two quantities given, to find the ſquare
201
of the fum
wả
203
202
Prob. LXIX. Two quantities given, to find the ſquare
of the difference
Prob. LXX. Given the fum and difference to find the
rectangle
Prob. LXXI. Given the power of a binomial, to
find the difference between the fquare of the ſum of
the odd terms, and the fquare of the fum of the even
Prob. LXXII. To find any root of a binomial furd 204
203
Prob. LXXIII. To explain the properties of o, and
infinity
terms
209
212
Prob. LXXIV. To find the value of a fraction, when
both numerator and denominator are o
Prob. LXXV. To find whole numbers anſwering the
equation axby + c
215
Prob. LXXVI. To find a number that being divided by
given numbers, will leave given remainders
219
Prob.
鳥
​[
t
vili
The
CONTENTS.
Prob. LXXVII. To find the limits of an equation con-
taining feveral unknown quantities
pag. 225
Prob. LXXVIII. To find the limits in two fuch equa-
tions
228
Prob. LXXIX. The investigation of the rule of alli-
gation
231
234
Prob. LXXX. Inveſtigation of the rule of falfe
Prob. LXXXI. Inveftigation of the rule of exchange 236
Prob. LXXXII. To find rational fquares, cubes, &c. 237
Prob. LXXXIII. To find the maxima and minima of
quantities
241
Prob. LXXXIV. To turn numbers into logarithmic
feries
1
244
Prob. LXXXV. To turn logarithms into numerical
feries
247
Prob. LXXXVI. To demonſtrate a propofition fynthe-
tically, from the analytical folution
250
Sect. VIII. The refolution of equations, and extraction
of their roots in numbers
251
Sect. IX. The geometrical conftruction of equations 297
Sect. X. To inveftigate a problem algebraically
316
BOOK II.
The Solution of particular Problems.
Page
Sect. I. Numerical Problems
328
Sect. II. Problems concerning intereſt and annuities 348
Sect. III. Problems in arithmetical and geometrical pro-
greffion
358
Sect. IV. Unlimited problems
368
Sect. V. Problems for finding rational fquares, cubes,
&c.
8
¿
Sect. VI. Geometrical problems
Sect. VII. Problems in Plain Trigonometry
Sect. VIII. Problems in Spherical Trigonometry
Sect. IX. Problems of the Loci
Sect. X. Mechanical problems
Sect. XI. Phyfical Problems
375
385
419
434
450
464
475
489
497
507
ALGEBRA,
Sect. XII. Problems concerning feries
Sect. XIII. Problems concerning exponential quantities
Sect. XIV. Problems of Maxima and Minima
[ 1 ]
ALGEBRA.
I.
AL
DEFINITION S.
LGEBRA is a general method of com-
puting Problems, by help of the letters of
the alphabet, and other characters. It is
of the fame nature as Arithmetic, but more gene-
ral, and therefore it is called Univerſal Arithmetic,
as likewife the Analytic Art. The peculiar practice
of this method is, to affume the quantity fought as
if it was known, and proceeding to work by the
rules of this art, till at laft the quantity fought, or
ſome powers thereof, be found equal to fome given
quantity, and confequently itſelf becomes known.
2. Like quantities, are thoſe that confift of the
fame letters; as a, 44, -3a. Alſo bb, 3bb,
-11bb; alſo 2abc, 15abc, -abc; &c.
3. Unlike quantities, are thofe confifting of dif-
ferent letters, or of the fame letters, differently re-
peated. As a, b, 2c, -3d. Alfo a, 2aa, -5aaa.
4. Given quantities, are thoſe whofe values are
known.
5. Unknown quantities, are thoſe whoſe values
are not known.
6. Simple quantities, are thofe confifting of one
term only; as 5b, 3a²c, 13dcc, &c.
7. Compound quantities, are thofe confifting of fe-
veral terms, as a+b, za−3c, a+2b-3d, &c.
8. Pofitive
B
2
DEFINITIONS.
8. Pofitive quantities, are thofe to be added.
9. Negative quantities, are thofe to be fub
tracted.
10. Like figns, are either all +, or all, (See
the Characters.)
11. Unlike figns are and -.
12. The Coefficient, is the number prefixed to any
letter or letters in any term.
As 3 is the coeffici-
ent of gaa. If no number be prefixed, then r
muſt be understood, as a a fignifies aa.
13. A Binomial quantity, is one confifting of twó
terms, as 2a +36. A Trinomial of 3 terms, as
a + b C. A Quadrinomial of four, &c.
Refidual is a binomial, where one of the quantities
is negative.
A
14. Power of a quantity, is its fquare, cube, bi-
quadrate, &c.
15. An Equation, is the mutual comparing of
one thing with another, by the ſign of equality put
between them.
16. A dependent Equation, is an equation which
may be deduced from fome others.
17. An independent Equation, is one that cannot,
by any means, be produced from the others.
18. Pure Equation, is an equation containing but
one power of the unknown quantity; as a fimple
Equation, a pure Quadratic, a pure Cubic, &c.
19. An affected Equation, is that which contains.
feveral powers of the unknown quantity; and is de-
nominated according to the higheſt power in it; as
an affected Quadratic; an affected Cubic; an affected
fourth Power; &c. Thus a fimple equation con-
tains only the fimple quantity itſelf. A quadra-
tic, a quantity of 2 dimenfions; a cubic, a quan-
tity of 2 dimenfions; a biquadratic, of 4 dimen-
fions; &c.
20. Index or Exponent, is the number ſet over
a letter, fhewing what power it is as a3; here 3
fhews
DEFINITIONS.
3
thews it is the third power; or that a is equiva-
lent to a a a. And thus at is the fame as a aaa;
as the fame as aa aaa; &c. the index always
fhewing how oft the letter is repeated.
21. A Fraction, confifts of two quantities placed
one above another, with a line between them,
a
as; the upper (a) is called the numerator, the
lower (b) the denominator.
22. A Surd, is a quantity that has not a proper
root, as fquare root of a (a), cube root of
bb (3/bb), &c. Roots of compound quantities that
contain other furds, are called Univerfal Surds.
23. A rational quantity, is a quantity that has no
radical fign.
A
+
Characters used in Algebra.
more, to be added, being the fign of ad-
dition. This is called an affirmative
fign. Thus a + b fignifies added
to a.
b
lefs, abating, the fign of fubtraction.
This is alfo called a negative fign.
Thus b fignifies b ſubtracted
from a.
a
Thefe figns always affect the quan-
tity following; and are always to be
interpreted in a contrary fignification.
If+fignifies upward, forward, gain,
increaſe, above, before, addition, &c.
then is to be interpreted down-
ward, backward, lofs, decreaſe, below,
behind, Jubtraction, &c. And if be
underſtood of theſe, then
interpreted of the contrary.
B 2
+
is to be
is dif
'
4
CHARACTERS.
difference; as ab, fignifies the differ-
ence between a and b.
x multiplied by; as a x b, fignifies a mul-
tiplied by b. Likewife a b, fignifies
a multiplied by b. All letters joined
together fignifies a multiplication. For
brevity's fake points are often ufed
inſtead of X, as n
12- 1. N~2
2
N2
nifies n ×
X
2
3
figè
3
divided by, as ab, fignifies a divided.
4
by b. And fignifies the ſame.
b
equal to, as a + b = 2d, fignifies a and
b equal to 2d.
greater than, as ab, is a greater
than b.
Cleffer than, as a □ b, is a leſs than b.
a root, as a, is fquare root of a; 3/a,
cube root of a ; /a, fourth root of a;
&c. It is called a Radical Sign.
2, involved to the
involved to: as
fquare;
3 involved to the cube;
&c.
Еш
extracted: lw 2, fquare root; lu
سا
3,
cube
root; &c.
a+b+c, a line, or vinculum, drawn over feveral
quantities a, b, c, denotes that com-
pound quantity to be confidered jointly
as a fimple quantity.
EXPLANATION.
a a − b b † 3 c d, fignifies bb fubtracted from
aa, and 3 cd added.
a a b b — c d dd, fignifies, that cc-dd is
cd —
fabtracted from a ab b.
aat zab
1
EXPLANATION.
5
aa + 2 aborr-ss, fignifies the difference
between a a + 2 ab and rr SS.
abcc fignifies the product of a and b and c c.
a + bx aa, fignifies the fum of a + b multi-
plied by a a.
a + bxa a, fignifies the product of b into a a
is to be added to a.
2
ea-2 ab, fignifies the fquare of the com-
pound quantity a a-2 ab.
✔bb+cc fignifies the fquare root of bbc c.
√2ab — cc, fignifies the cube root of 2ab—cc.
3
a a
X X
b-
fignifies a a divided by a — b.
a3
vided by xx
a a
fignifies the fquare root of a³ di-
aa.
a³ b² fignifies a aa × bb, or the cube of a mul-
tiplied by the fquare of b.
3ax--xx √ 5ax, fignifies the fquare root of
5 a x multiplied by 3 axxx; and fo of others.
Quantities that have no fign prefixed, muſt be
underſtood to have the fign+, leading quantities
ſeldom have the figns put down, when they are af-
firmative.
If A B and CD be two lines; then A B × CD,
in a geometrical fenfe, fignifies the rectangle made
by the lines A B and C D.
A B
Alfo fignifies the ratio that A B has to
CD.
CD'
NOTATION,
1. In the computation of problems, put the firſt
letters of the alphabet, b, c, d, f, g, h, &c. for
known quantities, and the last letters of the alpha-
bet for unknown ones. Yet fome put vowels for
B 3
unknown
6
}
AXIOM S.
}
unknown quantities, and the reſt of the alphabet
for known ones.
2. For general forms, put the capitals A, B, C,
D, &c. for the general quantities.
3. Or in univerfal forms, let the quantities be
denoted by the Greek capitals, г, ^, Z, ®, A, II, Σ,
T, Þ, Y, N, and indices, coefficients, &c. by the
fmall letters, d, e, n, O, λ, µg V₂ π, T₂ Q.
>
4. In caſe of neceffity, make uſe of any other ſort
of letters, or of any characters, that have names,
as ½, 2, ♂, 0, ?, ☀, D, 8, &, ‡, x, &c.
J
AXIOM S.
1. If equal quantities be added to equal quanti-
ties, the fums will be equal.
2. If equal quantities be taken from equal quan-
tities, the remainders will be equal.
3. If equal quantities be multiplied by equal
quantities, the products will be equal.
4. If equal quantities be divided by equal quan-
ties, the quotients will be equal.
5. The equal powers or roots of equal quantities,
are equal.
6. If to or from equal quantities, unequal ones
be added or fubtracted; the fums or remainders
will be unequal.
7. If equal quantities be multiplied or divided
by unequal quantities; the products or quotients
will be unequal.
8. Quantities feverally equal to a third, are
equal to one another.
$
9. The whole is equal to all the parts taken to-
gether.
10. If a quantity be added, and the fame quan-
tity fubtracted, they deftroy one another, and are
both reduced to nothing.
BOOK
}
Į
1
1
1
[ 7 ]
*
I
воок
І.
The fundamental Principles of Algebra.
SECT. I.
The primary Operations of Algebra in Integers.
PROBLEM I.
.
To add feveral Quantities together.
I RULE.
F the quantities are like and have like figns;
add all the coefficients together, for the coef-
ficient to that quantity, and prefix the fame fign.
Ex. 1.
to + 5 a
to
16 ab
add + 70
add
5ab
to + 4a
add + 5a
3 x
X
add
2 a b
add +
a
5x
Sum + 12 a
23ab Sum + 10 a
— 9 ×
Ex. 2.
to + 135 ab²
202 x x y³
add + 17abb
105 xxy3
+
3abb
17 xxy3
+
abb
324xxy3
Sum + 156 a b b
2 RUL E.
In like quantities with unlike figns; add all the
affirmative coefficients, into one fum; and all the
BA
negar
ADDITIONO N.
B. I.
*
.
negative ones into another; fubtract the leffer fum
from the greater, and to the difference prefix the
fign of the greater, with the proper quantity.
Ex. 3.
1
to + 6 a
16d
add
3 a
+ 3d
3a7b
3 a +86
Sum + 3a
13 d
+ b
¿
S
++1
125 ab
37 a b
ab
+ 99 ab
162 a b
+100ab
1
Sum
62 a b
2 aa
Ex. 4.
Ex: 5:
+ 34x²y
8 x²y.
xzy
+92x²y
- 67x²y
+126x²y
- 76x²у
+ 50 x²y
9 b c d + d d + 2 e
+7aa-20 bcd-dd + 5e
+3aa + 4 b c d
Sum +8aa- 2 5 b c d
3
RULE.
+7e
1
Set down all the unlike quantities with their pro-
per figns.
1
Ex. 6.
+2a
+36
+1+
C
d
Sum 2a + 36 - ç + d
1
f
Ex:
Sect. I.
$
ADDITION:
Ex. 7.
413a a
+1+
4ab
2 a
3 a
bc + 5 a
2 d d + 6 d
Sum + 13 aa→ 4 a b + b c — 2 d d + 6 d.
Ex. 8.
zee+3ef-ƒƒ + 17
·3ee5ef + 2 ƒ ƒ
+ 6ee
II
ef + ƒƒ- 3
Sum + 5ee + 7 eƒ + 2ƒƒ + 3
The reaſon of this rule is evident for like figns;
and in unlike figns, it follows from the nature of
affirmative and negative quantities, that the diffe-
rence ought to be taken, to make up the total.
if a man owes 10%. then 10%. ought to be deduct-
ed from his ftock to find his real worth.
As
Cor. 1. When feveral quantities are to be added to-
gether, it is the fame thing, in whatever order they
are placed.
Thus a + b
c = a − c + b = −c+ a + b
=b+a-c, &c. for all theſe are the fame.
Cor. 2. Hence the fum of any number of affirmative
quantities, is affirmative; and the fum of any number
of negative quantities, is negative.
PROBLEM II.
To fubtract quantities from one another.
RULE
Change the figns of all the quantities to be fub-
tracted; and then add them all together by Prob. I.
and the fum will be the remainder fought.
Ex.
10
B. I.
SUBTRACTION.
}
+ 8a
+3a
Rem. 8 a
or
5 a
from
за
Ex. I.
500
166
56
16b5b
or 21 b
20
6
20+ 6
or 14
Ex. 2.
6a3x+6y - 7
take + 8a+ 4x + 6y + 5
II C
+3¢
II C — 30
Or
140
t
}
Rem.
2 a
-7x+0
I 2
Ex. 3.
from a + b
a + b
a+b
take a
b
Rem. + 2b
24
Ex. 4.
from aa + 2 a b + b b
take +4ab
+bb
Rem a a
2 a b + bb
Ex. 5.
from a a-
· bb
take cc
d d
Rem. a a
bb-cc+dd
Ex. 6.
from
take
24 a
5 a
3aa2a+ c d -d d -ff
Rem. 5a a + 3 a + c d + ab + dd—ff
a b
2 d d
Cor.
}
t
I
Sect. I.
II
MULTIPLICATION.
}
Cor. Hence, To fubira one quantity from ano-
ther, is the fame thing as to add them together, when
ail the figns of the fubtrabend are changed.
a- •b = a + b.
For it is the fame thing to fubtract, as to
add ; and to add, as to fubtract +. For
ſuppoſe a man to owe 10/; becauſe it is a debt it
muſt be writ 10%. therefore if any body would
take away this -10, it is the fame thing as if he
added 10 to his ftock: but before it is diſcharg-
ed, this 10 is the fame, as + 10 deducted out
of his ftock.
PROBLEM III.
To multiply one quantity by another.
RULE.
Multiply every particular term (or fimple quan-
tity) of the multiplier, into every term of the mul-
tiplicand, one after another; fo that the coefficients
be multiplied into the coefficients; and the letters
into the letters, by placing them all together, like
letters in a word. And prefix + to products of
like figns, and to unlike ones.
- to unlike ones. The fum of all
is the product fought.
Ex. 1.
ta
a
+ 3 a
+ b
h
2 b
4 c
+5d
tab
tab
6ab
-20 cd
Ex. 2.
a+b
a+b
Q
a + b
b
a a + a b
+ab+bb
aa +2abbb
catab
a b - b b
a a
b b
Ex.
12
MULTIPLICATION. B. I.
1
1
}
за
Ex. 3.
2 b
5a+4b
15a a 10 a b
+12ab - 8bb
15aa + zab
Ex. 4.
aa+ab - b b
8 bb.
ЪЪ
a
- b
a³ + a ab
a b b
Bab
a b b + b 3
a3
zabb + b³
Ex. 5.
a b - 3 c d + r s
5r-7d
5 rab
15red + 5rrs 7abd +21cdd-
Ex. 6.
3aa2ab+ 5
aa + 2 a b — 3
3 at — 2 ba³ +5 aa
+6ba³ — 4 aabb + 10 ab
-7rsd
gaa, + 6 ab
15
3a4 + 4ba3 — 4 bbaa
-4aa + 16 ab
15
Ex. 7.
a a + b b
CC
d d
c ca a + c c b b
ddaad db b
}
ccaa-ddaa + ccbb - d db b
}
}
t
Ex.
J
{
Sect. I. MULTIPLICATION.
Ex. 8.
a³
a²
463
365
13
25
1268
له
That every term in the multiplicand muſt be
multiplied by every term in the multiplier, is thus
made evident. Let a + b be multiplied by c+d;
it is plain, a+b muſt be taken ſo often as there are
fuppofed units in c and d, that is, as often as there
are units in c, and alfo as oft as there are units in d.
Therefore the product will be a + bxc+a+bxd.
But for the fame reafon a + b x c = ac + bc,
alſo a + bx d = a d + b d.
Whence the pro-
duct will be a c + b c + a dbd; that is, the
fum of all the products of every term multiplied
by every term.
That like figns give +, and unlike figns, in
the product, will appear thus.
Cafe 1. Let + a be multiplied by + b. Then
fince this multiplication fuppofes, that + a is to be
fo often added together as there are units in+b;
and the fum of any number of affirmatives is affir-
mative, therefore the whole fum is affirmative, that
is + ax + b = + ab.
Cafe 2. Leta be multiplied by b. Now
fince this implies that a is to be as often fub-
tracted as there are units in b; and the fum of any
number of negatives, is negative, therefore that
whole fum, is negative, that is, axb =
+
ab.
Cafe 3. Leta be multiplied by + b.
plain here, thata is to be fo often taken as
there are units in b; and the fum of any number
of negatives being negative, therefore the whole
fum is negative; that is, ax + b = −ab.
It is
Otherwife,
}
14
B. I.
MULTIPLICATION.
Otherwife, Let da be multiplied by +b;
then (Cafe 1.) the product will be bd together
with ax+b: but bd is too big, as being
the product of d by b, inftead of da by b
(da being less than d); therefore bd, being
too much; the product- ax + b muſt be ſub-
tracted; that is, the true product will be db-ab;
and confequently ab ax + b.
=
Cafe 4. Leta be multiplied by -b. Here
a is to be fubtracted as often as there are units
in b: but fubtracting negatives is the fame as add-
ing affirmatives (Cor. Prob. 2.); confequently the
product is + ab.
Or thus. Since aao, therefore a- a x
-bo, becauſe o multiplied by any thing pro-
duces o; therefore fince + a − a x − b = 0;
and the first term of the product is a b (Cafe 2);
therefore the laft term of the product muſt be
+ab, to make the ſum o, or a b + ab = 0 ;
that is,
а х
ax. b = +ab.
-
1
b.
Other wife. Let da be multiplied by
Then (Cafe 2.) the product will be bd toge-
ther with ах b; but bd the quantity
to be ſubtracted is too big, being the product of d
by -b, inſtead of d a by — b, (d - a being
lefs than d); therefore the quantity
bd to be
fubtracted being too much, fomething must be
reftored, that is a X-b muſt be added; and
the true product will be bd + ab; and there-
fore+ab — axb.
=
Cor. 1. If feveral quantities are to be multiplied
together; it is the fame thing in whatever order it be
done. Thus abc acb cab = bca, &c. for
all these are equal.
=
Cor. 2. The powers of the fame quantity are
multiplied together, by adding their indices. Thus
a² × a³ = a²+3 = as.
Cor.
1
Sect. 1.
MULTIPLICATION
15
}
Cor. 3. Any odd number of -, multiplied together
produce -; and any even number of -, pro-
duce +.
SCHOLIUM.
In the multiplication of compound quantities, it
is the best way to fet them down in order, accord-
ing to the dimenfions of fome of the quantities.
And in multiplying them, begin at the left hand,
and multiply from the left hand towards the right,
the way we write, which is contrary to the way we
multiply numbers. But this will be moft expedi-
tious, and the feveral producs will by this means
be fo ranged under one another, that like quantities
will fall in the fame places, which is the eafieft way
for adding them up together.
In many cafes, the multiplication of compound
quantities is only to be performed by writing
their fums, each under a vinculum, and putting
the fign (X) of multiplication between. As if
the fquare of a axx was to be multiplied by
ag bh, and that by ac + bd, it may be writ
ten thus, ex- ≈≈ × ag b b xa 6 + bd.
PROBLEM IV.
To divide one quantity by another.
I RULE.
In fimple quantities, which will divide without
a remainder; divide the number by the number,
and put the anſwer in the quotient. Then throw
out all the letters in the dividend which are found
in the divifor, and place the remaining letters in
the quotient. And like figns produce +, and un-
like figns-, in the quotient..
Ex.
16
DIVISION.
B. I.
a a) a ab (b
aab
— aa) a ab (—
aab
Ex. I.
3 ab) 1 5 a a b c d ( 5 a c d
1 5 a a b c d
Ex. 2.
b|-zab)-15 a abcd (5acd
→ 15a a b c d
7
Ex. 3.
- 3 cdd) — 6 cc d d (+ 2 c
-6ccdd
1
Ex. 4.
b³a³d³c
1
+ 6a²b²) — 18 b³a³ d³c (— 3 a b c d³
18 b³a³d³c
O
Ex. 5.
- 5 a²b) 10 a²b b d ( — 2 b d
10 a² b b d
Ex. 6.
9x²y) — 9ײy²b (—y b
9x²x²b
1
!
今
​Ex.
Sect. I.
$7.
DIVISION.
h
-
Ex. 7.
8xx) — 16 x³ († 2 *
16 x³
2
RUL E.
In compound quantities, range the terms of the
divifor and dividend, according to the dimenfions
of fome letter. Then, by Rule 1, divide the first
term of the dividend by the firft term of the divi-
for, placing the refult in the quotient. Multiply
the whole divifor by the quotient, and fubtract it
from the dividend, to which bring down the next
term of the dividend, call this the Dividual.
Divide the firft term of the dividual by the firft
term of the divifor; then multiply and fubtract as
before, and repeat the ſame proceſs till all the quan-
tities be brought down. This is in effect the very
fame rule as is uſed in arithmetic.
$
Ex. 8.
a) ab + ac➡a (b + c 1 the quotient
ab
+ac
+ac
A
a
Ex. 9°
3 caa (a a
2 b 3 c) 2 ba a
−
z baa-3 ca a
C
B
**
18
B. I.
DIVISION.
I
A
"
>
Ex. 10.
a + b) ac + b c + ad + b d (c+d
a c + b c
+ ad + b d
+ ad + b d
Ex. II.
3yy4y+ 12 (y-3
¶y — 4) y³
4) y³ — 3yy
y3
-4y
39 9
+ 12
399
+ 12
Ex. 12.
a + b) a a−bb (a-b
aa+ab
ab-bb
a b - b b
O
Ex. 13.
34—b) 3a³—12aa-baa+10ab—2bb (aa—4a+2b
3a3
-12aa
-12aa
-baa
+10ab
+ 4ab
1
*
+ 6ab-2bb
+6ab-2bb
3 RUL E.
When the divifor does not exactly divide the di-
vidend; place the dividend over the divifor, in
form
Sect. I.
19
DIVISION.
1
&
form of a fraction; throwing out fuch letters, as
are found in all the terms of both the dividend
and divifor.
Ex. 14.
a
x) a
に
​Ex. 15.
ax−xx) ax+xx(ax+xx
ax XX
Ex. 16.
the quotient.
a+x
quote.
a-x
(1+x+~~+x³+x4 + &c. fine fine.
:)
I ---X
+x
X-
XX
+xx
(or 1+x+x+x³+
I
{
ca
ee) aaе
XX-X³
+x3
aae- e3
e3
X3 X4
+x+ &c.
Ex. 17.
(e +
e3
+
a a
es
e3
+
+
aa
es
aa
es
e7
aa
a4
es
04
+
+
C 2
ет
Q+
Rem.
*
e7
a6
·eea+
This
20
B. I.
DIVISION.
This and fuch like examples will be better under,
ftood after the next fection.
Ex. 18.
a³) 2as (243
245
= 20².
26²) 1866
a a
XX
Ex. 19.
1866
62
Ex. 20.
5
a a — x x (a a
aa - X X
5
-9b4.
3
X X
That like figns give +, and unlike figns,
in the quotient, will appear thus. The divifor
multiplied by the quotient mult produce the divi-
dend. Therefore, i. When both are +, the quo-
tient is, becauſe then + x +
1. +)+(+ muſt produce + in the dividend.
2. —)-(+ 2. When they are both the
3. +(quotient is again, becauſe +x-
4. −)+(— must produce. in the dividend.
Again, 3. When the divifor is +
and the dividend, the quotient is, becauſe
- ×+ muſt produce in the dividend.
Laftly, If the divifor is, and the dividend +,
the quotient will be, becauſe X- - produces
in the dividend.
4.
Cor. 1. One power of a quantity, is divided by
another power thereof; by fubtracting the index of
the divifor, from the index of the dividend. Thus,
as
Q3
a
5-3
=a²: And
!6b 4b3 4b-2 4
1263
3
==
3
==
3bb.
Cor,
Sect. I.
Zi
INVOLUTION.
Cor. 2. Hence any power of a quantity may be ta-
ken out of the denominator and put into the numerator,
and the contrary; by changing the fign of the index.
a
262
ab-2
b
= ba³.
Thus
And
2
a-3
Cor. 3. Hence
divided by +, or
+ divided
That is,
a
a
-b
by give the fame quotient, viz. -.
응
​PROBLEM
V.
To involve a quantity to any power.
IRUL E.
Multiply the quantity fo often into itſelf as the
index denotes. And where the root is +, all the
powers are +. And where the root is, all
the odd powers are, and all the even pow-
ers +.
Ex. 1.
a root
a a fquare
a³ cube
04 4th power
&c.
ab root
aabb ſquare
a3b³ cube
&c.
2a3 root
a² root
24 fquare
26 cube
as 4th power
&c.
+ 40° fquare
849 cube
+16a¹² 4th power
&c.
Ex. 2.
3abb root
+ gaab+ fquare
bo
-27 a³ b6 cube,
C 3
&c.
Ex.
22
B. I.
INVOLUTION.
A
f
Ex. 3.
Involve a + b to the cube or 3d powers
fquare
a+b
a + b
aa+ab
+ab+bb
a a +2ab+bb
a + b
å³ + 2 a ab + a b b
+ aab+za b b + b²
cube a³ + 3a ab + 3 abb + b³
2
RULE.
Multiply the index of the quantity, by the index
of the power, and make the figns as in Rule 1.
root a or
or a
ſquare a¹×2 or a²
cube a¹×³ or a³
th
om
power a
Ex. 4.
--- 2 bba; or 25²à
X 2
+ 4 b² x ² a
+46
8 b 2 x 3 ax
IX2
or + 4b+q²
1 X3 or
866a3
2″ × b2m a
773
Ex. 5.
root
a-x
fquare a-x
2 X 2
4
or a—x
-2X3
-2 X 3
6
cube
a-X
or a-x
th
2 X m
2772
m" power a-x
or A-X
:
3 RULE.
?
Sect. I
23
INVOLUTION.
A
3
RULE.
I
In a binomial. The power will confift of 1 term
more than the index of the power. The higheft
power of both is the index of the given power, and
the index of the leading quantity continually de-
creaſes by one in every term, and in the following
quantity, the indices of the terms are 0, 1, 2,
3, 4, &c.
Then for finding the unciæ or coefficients. The
firft is always 1; the fecond, the index of the
power. And in general, if the coefficient of any
term be multiplied by the index of the leading
quantity, and divided by the number of terms to
that place; it gives the coefficient of the next fol-
lowing term.
Laftly, When both terms of the root are +, all
the terms of the power will be +; but if the ſe-
cond term be then all the odd terms will be
+, and all the even terms
Ex. 6.
Involve ate to the 5th power.
The feveral terms without the coefficients will be
a³, ate, a³ee, a²e³,
5X4 10X3
coefficients 1, 5,
2
>
ae4,
es; and the
10X2 5XI
;
3
- 4
LO
5
5,
I.
that is, 1, 5, 10, 10,
And therefore the 5th power is
as + 5ate + 10 a³ее + 10 a²e³ + 5 ae4 + es.
Ex. 7.
Involve a-x to the 4th power.
thepower is a+-4a³x+4×3 a²x²-
a4
2
that is, at 4a³x + 6a²x²
C A
6×2
4XI
´ax³ +
3
4
40x3 + x4.
4 RULE.
24.
B. I.
INVOLUTION.
4 RULE.
مصر
In trinomials, quadrinomials, &c. Let one
letter remain, and put another letter for the reſt
of the quantities; then involve this binomial by
Rule 3; then inſtead of the powers of the affum-
ed letter, find (by Rule 3.) the powers of the
compound quantity it reprefents, which put in its
ftead.
Ex. 8.
Involve a + bx to the third power.
Put e for bx, then the cube of a te is
à ³ + з a α é + з a еe + e³ (Rule 3), that is,
аз засе ез
2
2
3
3
¿³ +zaa × b−x+zexb−x + 5—x. But (Rule 3.)
b—x — bb−2bx + xx, and b—x —b³—3bbx+3bx²
-x. Therefore a+bxa³ + 3aab - zaax +
3abb—6abx+ zaxx+b³—3bbx+ 3bxx-x³
3
Cor. 1. The nth power of a +e, that is,
12
I
a+e =a"+na²-¹e +nx a^ − 2 ee + n x
N I
2
ท
I
2
n-2
ท
11
2
X
an-3e3+nx
X
X
а
n~ 3 an― A e 4 + &c.
3
2
3
4
This rule is proved by involving a+e as far
as you will, for the feveral powers will always agree
with the rule.
Cor. 2. All powers of an affirmative quantity, are
affirmative. And all oad powers of a negative quan-
tity, are negative; and all even powers affirmative.
Cor. 3. The index of the power of any quantity,
is the product of the index of the power, and index of
the quantity.
Cor. 4. The nth power of any product, is equal të
the nth power of each factor, multiplied together.
in
12
a b² ²² = a² x b²".
X
PRO.
3
ป
Sect. I.
25
EVOLUTION.
PROBLEM
VI.
To extract the root of any quantity.
Evolution is juft the reverſe to involution; and
is performed as follows.
I RULE.
For fimple quantities; extract the root of the
coefficient for the numerical part, and divide the
index of the letter or letters, by the index of the
power, gives the index of the root.
Ex. I.
3
The cube root of a³ is a³ or a.
the Square root of 25at is
the Square root of 2a2b
4
5a² or gaa.
6
is a b√//2.
or ab³√2.
9
the cube root of -1256 is—563 or-56³.
2 RULE.
For the fquare root, of a compound quantity
range the terms according to the dimenfions of
fome letter. Then find the root of the firft term
(1 Rule), and fet it in the quotient subtract its
fquare, and bring down the next term, which di-
vide by double the quotient, and let the anſwer
in the quotient. Multiply the divifor and quo-
tient by this laſt quotient, which fubtract from the
dividual. Proceed thus, juſt as in common arith
metic.
Ex.
26
B. I.
EVOLUTION,
Ex. 2.
Extract the fquare root of aa+4ab+4bb—2ax
4bx + xx.
aa+4ab+4bb—2ax−4bx+xx (a+2b—x root
aa
2a+2b) 4ab+466
4ab+4bb
2a+4b-x) -2ax—4bx+xx
-2ax-4bx+xx
Ex. 3.
Extract the Square root of aa-6na+2za+9nn➡
6nz + zz.
aa
+22
-6n a +9nn (a
бnza
•3n2
十万
​aa
+zz
20
·3210
+21
+9nn.
6n6nz
a
+22+zZ.
-6n+gnn
a
+22 6nz
+zz
Ex:
Seat. I.
27.
EVOLUTION.
Ex. 4.
Extract the Square root of aa + **.
XX 2.4
24 893
ㅇ
​·+
16as
&c.
aa+-xx
aa
(a+
XX
za+
0+xx
aa
~4
+xx+
4a6
XX 24
X4
20+
a
8a3
44a
1
X4
дов
+
4aa
8a+ 64a6
20+)
x8
o +
8a+ 64a³
3 RUL E.
In higher powers. Find the root of the firft
member, which place in the quotient: fubtract its
power, the remainder is the refidual. Involve this
root to the next lower power, and multiply it by
the index of the given power, for a divifor by
this divide the firft term of the refidual, the quo-
tient is the next term of the root. Then involve
the whole root as before, and fubtract: and repeat
the operation, till all the terms of the root be had.
Ex. 5.
Extract the cube root of x+6x5—40x³+96x-64.
x+6x5—40x³ +96x-64 (xx+2x—4 root.
246
3x4) 6x5 (+2x
x+6x5 + 12x4+8x³=xx +2x.
3x4) 0 — 12x+(-4
3
·3
x+6x-4x³+96x-64=x+2x-4 ·
1
EN
28
B. I.
EVOLUTION.
1
Ex. 6.
Extract the 4th root of 16aª—96a³b+216aabb
·216ab³ +81b4.
16a4—96a³b+216aabb-216ab³ +81b4 (2a-4b
1624
24a³) 0-96a³b (—4b
двазъ
16a+—96a³b+216aabb-216ab³ +81b4
О
root.
4 RUL E.
The roots of compound quantities, may fome-
times be diſcovered thus. Extract the roots out of
all the fimple powers or terms in it; then connect
theſe roots by the figns + or, as you judge
will beft anfwer. Involve this compound root to
the proper power; then if it be the fame with the
given quantity, you have got the root.
If it only
differs in the figns, change fome of them, till its
power agrees with the given one throughout.
Ex. 7.
To extract the cube root of a³—6a²b+12ab²—8b3;
Here the root of a³ is a, and the root of -8b3
is -2b. Then a-2b is the root, for its cube
is a³—6a²b+12ab²-8b³, as required.
Ex. 8.
t
Extract the 4th root of 16a4-96a³x+216a²x²
216ax³ +81x4.
The roots of 16a4 and 81x4, are 2a and 3x.
Therefore if 2a+3x be made the root and in-
volved,
J
Sect. I.
29
EVOLUTION.
volved, it is 16a++96a³x+216aaxx+216ax³ +
81x4, which differs in the figns, from the quantity
given. Therefore make 2a-3x the root, which
being involved fucceeds; the power being
16a+ — 96a³x + 216aaxx — 216ax³ + 81x+.
5 RUL E.
When the quantity given has not fuch a root as
is required, fet it down in form of a ſurd.
Ex. 9.
Square root of a³, is
3
a³.
Cube root of 15aa, is 15aa.
4th root of 2a5x³, is √2a³x³.
Ex. 10.
The cube root of a³—6a²b+12ebb +8b³, is
3
√ a³ —ba²b+12abb+86³.
Ex. II.
What is the 5th root of as-xs.
the root is 5
@5-x5.
Cor. 1. The fquare root, or any even root, of an
affirmative quantity, may be either + or -.
For the fquare root of a a may be a ora,
for+ax+aaa, and -ax-aaa: alfo the 4th
root of at is +a or —a, for the 4th power of
a is at, as well as of +a.
Cor. 2. Any odd roet of a quantity, will have the
fame fign, as the quantity itself.
}
For the root of +a and -a, will be +a and
-a; for a cubed is a³, and a cubed is
1
Cor.
{
30
B. I
EVOLUTION.
Cor. 3. The Square root, or any even root, of a
negative quantity, is impoffible.
For neither +axta, nor aX-a, can pro-
duce —aa.
Cor. 4. The nth root of a product, is equal to the
nth root of each of the factors, multiplied together:
VAB VA XVB.
√B.
4
}
*
SECT.
T
SECT. II.
Of FRACTIONS.
31
HE operations of algebraic fractions are ex-
actly the fame as thofe of vulgar fractions in
arithmetic; therefore he that has made himfelf
maſter of vulgar fractions, will eafily underſtand
how to manage all forts of algebraic fractions, as
in the following problems.
PROBLEM VII.
To reduce a given quantity to a fraction of a given
denominator.
RULE.
Multiply that quantity by the given denomina-
tor, and under the product write the fame de-
nominator.
Ex. I.
Let a+b have the denominator x.
F
a+bxx
ax + bx
anſwer.
>
*
X
Ex. 2
Let xx-yy have the denominator 1:
xx-yyXI
xx-yy
, anſwer.
I
I
Ex. 3.
Let 2 have the denominator b—c.
a
x b-c
ac
a
b
b
b-c
b.
11
bc anſwer,
Cor.
K
32
B. I.
FRACTIONS.
Cor. The value of a fraction is not altered, by
multiplying both numerator and denominator by the
rab rabd ab
Tame quantity. Thus
rc rcd
=
PROBLEM VIII,
To reduce a mixed number to a fraction.
RULE.
Multiply the integral part by the denomina-
tor of the fraction, and to the product add the
numerator, under which write the common deno-
minator.
Ex. 1.
b
ac-b
be given.
Then is the
C
C
Let a ---
fraction required.
Ex. 2.
aa
ax
Suppoſe a-x+ x
ax-
-xx+aa ax
aa XX
Here
or
is that re-
X
XX
quired.
PROBLEM IX.
To reduce an improper fraction to a whole or mixed
number.
RULE.
Divide the numerator by the denominator, as far
as you can, gives the integral part; and place the
remainder over the denominator for the frac-
tional part.
!
Ex:
Sect. II. FRACTIONS.
33
Ex. I.
aba a
Given
b) ab-ac
1
b
ab
い
​aa + xx
Suppose
A-X
a-x) aa + xx
aa
ax
-44
Ex. 2.
1
a a
b
anſwer.
(a+x+
2XX
1
anfwer.
+ax+xx
+ax-xx
- 2xx
PROBLEM
X.
To find the greatest common divisor, for the terms
of a fraction, or for any two quantities.
RULE.
The quantities being ranged according to the
dimenfions of fome letter; divide the greater
by the leffer, and the laſt divifor by the laſt re-
mainder, and fo on continually till nothing remain;
then the laft divifor is that required. But obfèrve,
firſt to throw out of each divifor, all the fimple di-
viſors, (or others) that will divide it; and then
proceed. The fimple divifors are had by infpection,
Let
cd + dd
aac+aad
Ex. 1.
37
be the fraction propoſed.
cd + dd) aac + aad!
or c + d ) aac + aad (aa
aac + aad
D
There-
{
34
B. I.
FRACTIONS.
1
"
Therefore c+d is the greateſt common divifor.
cd + dd
+d) aac + aad
Ex. 2.
ď.
aa
03 abb
Let
aa + zab + bb
be proposed.
aa +2ab+bb) a³
abb
(a
zaab-zabb remainder.
a³ + zaal Tabb
-2aab—2abb) aa+2ab+bb (
or a + b) aa+zab+bb (a+b
aat ab
+ab+bb
+ab+bb
Therefore a + b is the greateſt common divifor.
Ex. 3.
a4-64
Suppose
be given.
asbbas
24b4) as—bbas (a
a5b4a
rem. —bba³+b^a) a4—b4 (-
or
a a - bb) aª— b4 (aa + bb
a4-bbaa
the common divifor is aa
+bbaa64
+bbaa-b4
O
bbª.
PRO
Sect. II.
35
FRACTIONS.
PROBLEM XI.
To reduce a fraction to its lowest terms.
RUL E.
Find the greateſt common meaſure (Prob. X),
by which divide both numerator and denominator
of the fraction; the quotients will be the numera
tor and denominator of the fraction required.
à
Ex. I.
cd+dd
Let
be propofed.
cd+dd
d
aac+aad
the greateſt common divifor is r+d. Therefore
+d)
the fraction required.
aac+aad
aa
Ex. 2.
a3-abb
Let
be proposed.
aa+2ab+bb
Here a+b is the greatest common diviſor: then
a3abb
a+b) aa+2ab+bb
aaab
the fraction
a + b
fought.
Ex. 3.
a4-64
Suppoſe
to be given.
asbba³
the greatest common divifor is aa—bb; then
Bbb
bb).
a4b4
as_bbas(
aa+bb
a3
the fraction re-
D 2
quired.
PRO-
J
36
B. I.
FRACTIONS.
}
و
PROBLEM
XII.
To reduce fractions of different denominators, to frac-
tions of the fame value, having a common deno-
minator.
I RULE.
Multiply each numerator, into all the other de-
nominators, for a new numerator; then multiply
all the denominators together for a common de-
nominator.
G
Let
2
a + b
Ex. I.
be given.
C
ac ab+bb
theſe become
bc"
bc
Ex. 2.
f
a
Let b'
? J'
they become
£ be propoſed.
مة
adg cbg fb d
bdg bdg bdg
2 RULE.
Divide the denominators by their greateſt com-
mon divifor, then multiply both numerator and
denominator of each fraction, by all the other quo-
tients, which will produce as many new fractions.
Ex. 3.
a
C
d
Suppose
or
2bb' 26
b
26 2 I
2a 2bc 4bd
4bb 465 466
a bc 2bd
•
2bb´ 2tb 216
the fractions required.
Ex.
L
Sect. II. FRACTIONS.
37
Ex. 4.
2aa
Given
zab-zbb
aaab
2ac
a-b
2.6
400€
2aac-zabc
4008
2030
gab--ṛbby, a—.b
2αac 2abc
3aab-5abb4263
2abc
2aac
zabc
,
that is,
PROBLEM
XIII.
To add fractional quantities together.
R U L E.
If the fractions have not a common denomina-
tor, reduce them to one (Prob. XII); then add
the numerators, and under the fum, write the
common denominator.
Ex. i.
Add 2 to 29
}
C
d
bc
reduced and
bd
a
aa
bd
Add 'g
ad + b c
; then
= fum.
bd
Ex. 2.
C
f
together.
reduced
adg bcg
bcg bdf; then
bdgʻ bdg.
bdg bdg
adg+bcg+bdf
bdg
fum.
Ex. 3.
d--b
2b1a+d
Add
to
30
30
ab+2b-48+d
b-za+d.
the fum=
3C
36
D 3
Es
1
38
B. I.
FRACTIONS.
41
}
To a
add b +
a a
6
Ex. 4.
C
ab-bb-caa
fum a+b+
сь
PROBLEM
XIV.
To fubtract one fraction from another.
RULE.
Reduce them to a common denominator; then
fubtract the numerators: and under the difference,
write the common denominator.
Ex. 1.
a+b
C
From
d
Subtract •
d
a+b-c
difference.
d
Ex. 2.
a+b
ab+bb
From
==
d
bd
aad
Subtract 22
bd
v
ab+bb-aad
then
remainder.
bd
Ex. 3.
ab
From
take
26 40
30
5d
reduced
remainder
5ad5bd 6bc-12ac
15cd. 15cd
5ad—5bd—6bc+1200
15cd
Ex.
Sect. II. FRACTIONS.
39
}
Ex. 4:
a a
aac
take b +
From a -
72.
a--b
C
or a
bc
2
or b +
ab-bb
bc
difference a-b +
-aac-ab-bb
bč
PROBLEM XV.
To multiply fractions.
IRUL E.
In fractions, multiply the numerators together
for a new numerator; and multiply the denomi
nators together for a new denominator.
Ex. I.
Multiply 2
a
by /
d
ахс
then
or
b x d
ac
bd = product,
b
Ex. 2.
a+b
Multiply
by
b+c
b
a + b
ab + bb
here
X
>
product.
b + c
bc + co
Ex. 3.
aabb
aa+bb
Multiply
by
bc
b+c
abb
aa ÷bb
a4b4
then
X
product.
bc
b for C
bbc + bcc
D 4
2 RULE.
40
B. I,
FRACTION S.
1
3
2 RULE.
When the numerator of one, and denominator
of the other, can be divided by fome common di-
vifor, take the quotients instead thereof.
C
Ex. 4.
Let multiply
aa
I
reduced X
I 3dd
#1-
9
aabb
3cdd
bb
bb
3dd
product.
}
Ex. 5.
dà
cd-dd
by
1
a+b
reduced
a+b d
ad+bd
X
c-d
I
c-d
>
› product.
!
Multiply
aa+2ab+bb
3 RUL E.
If a fraction is to be multiplied by an integer,
which happens to be the fame with the denomina-
tor; take the numerator for the product.
Ex. 6.
aa-2bb
Multiply
by a-b.
a-b
quotient aa-2bb.
4 RULE.
When a fraction is to be multiplied by an inte-
ger; multiply the numerator by the integer.
Ex. 7.
Multiply
aa+3bb
by xx.
>
3cd
then
aaxx+3bbxx
aa+3bb
or
3.cd
3cd
XX
xx = the prod.
Ex.
Se&t. II.
41
FRACTION S.
Ex. 8.
2 A--2 X
Multiply
by a + x
35
200--- 2. XX
then
36
product.
Ex. 9.
b -
C
Multiply a +
d
b + c
by b
d
product ab +
d
bb_bc — ab+ac __bb—cc
d
dd
Schol. By this rule, a compound fraction may be
reduced to a fimple one.
PROBLEM
XVI.
To divide one fraction by another.
IRUL E.
In fractions, multiply the denominator of the
divifor by the numerator of the dividend, for a new
numerator; alſo multiply the numerator of the di-
vifor into the denominator of the dividend, for a
new denominator.
Ex. 1.
Divide
Ђ
by 2
d
응​)
ad
bc
the quotient.
Ex
42
B. I.
FRACTION S.
3
Ex. 2.
a + c
a+b
Let
divide
a
bi
a
at
a+baa−bb
quotient.
a-b a aatac
2 R
RULE.
If the fractions have a common denominator;
take the numerator of the dividend, for a nume-
rator; and the numerator of the divifor, for the
denominator.
Ex. 3.
Divide
aa-bb
zab-bb
by
a+d
a+d
aa-bb
quotient
zab-bb
Ex. 4.
aa+2ab+bb
a³-abb
Let
divide
c-d
c-d
a³-abb
then
=quotient.
aa+2ab+bb
aa
ab
or
=quotient reduced.
a+b
3 RUL E.
When fractions are to be divided by integers;
multiply the denominators of the fractions, by fuch
integers.
Ex. 5.
a
Ђ
Divide
by d.
C
a
b
quotient is
cd
Ex
Se&t. II. FRACTION S.
43
Ex. 6.
Let ab divide
aa-2bb
then
a=bxq+b
а aa-
-2bb
aa-2bb
aab5
9
quotient.
!
4 RUL E.
When the two numerators, or the two denomi-
nators, can be divided by fome common divifor;
throw out fuch divifor, and proceed by Rule 1.
Ex. 7.
a-b
a abb
Let
divide
cd
c+d
reduced
cd
d) Fd (
a+bacd+bcd
quotient.
c+d
Ex. 8.
aa+ab
Let
divide
ab
aa-zab÷bb
a a3a²b+ abbb3
reduced
a³-aab abb — b3
I
B b
acab
the quotient.
bb
that is, the quotient + a
From hence may be deduced the following co-
rollaries.
Cor. 1. The value of any fractional quantity is not
at all changed, by changing all the figns of both nu-
ab-ac ac ab
merator and denominator. Thus
Cor. 2. The value of any compound fractional
quantity, is equal to the sum of all the particular
fimple
Į
44
B. I.
FRACTION S.
1
Simple fractions, that compofe it. Thus
rx+2cx—11rz
3r-2x
rx
2CX
+
3r-2x 3r2x
IITZ
3r-2x
Cor. 3. If a fraction be multiplied by any given
quantity; it is the fame thing whether the numerator
be multiplied by that quantity, or the denominator di-
dabd dab
dab
vided by it. xd=
dc
dc.
=
C
Cor. 4. The product of two fractions, is equal to
the fraction, that has the product of the numerators
for the numerator; and the product of the denomina-
tors for its denominator.
a
b + x
X
axr-c
=
ar-ac
b+xxx bx+xx
Cor. 5. If a fraction is to be divided by fome
quantity; it is the fame thing whether the numera-
tor be divided by it, or the denominator multiplied.
For
24Z
X
÷r=
2az
rx
And
2ar
X
20
==
X
Cor. 6. If any fort of quantity is to be divided by
a fraction; it is the fame thing, as to multiply the
faid quantity, by the fraction inverted. Thus
γ
S
ab ÷ = = ab × =—= .•
a
And
小
​b
or
r
a
ar
c
X
Ъ
C
b
b c
PROBLEM XVII.
To involve fractional quantities.
RULE.
Involve the numerator into itſelf, for a new
numerator; and the denominator into itſelf for a
new denominator; each as often as the index of
the power.
Ex.
Sect. II. FRACTION S.
45
Ex. 1.
a
Involve
and
I
b
a a
a
root
b
aa
a a
I
fquare
cube
4th power
&c.
a4
b b
a4
аз
1
b3
I
as
Ex. 2.
Let
3bc
-ad
and
zad'
466
>
3bc
ad
root
zed
4bb
fquare
9bbcc
aadd
4aadd
+
1664
cube
276303
a3d3
8a3d3
6466
&c.
be involved.
aa
Involve
bc
a4
Ex. 3.
to the Square, &c.
a + c
2acbc + bbc
aa + 2αc + cc
aε — 3a+bc + 3a³bbcc —
a³ + 3a²c + zac² +
>
the Square.
b³ç³
b3c3
,
cube, &c.
c³.
Ex.
5
t
$
1
46
B. I.
FRACTIONS.
ax
Involve
26
it is
Ex. 4.
to the 4th power.
a4-4a³×бacxx—40x3+x4
1064
4
4
ax
or thus
a-X
ΟΙ
2 b
1664
XVIII.
PROBLEM
To extract the root of a fraction.
RULE.
Extract the proper root of both numerator and
denominacor, if it can be done. If not fet the
radical fign (✔) before one or both of them,
as they happen to be furd.
Ex. r.
What is the Square root of
9a2b4
4dd
завъ
root
2d
Ex. 2.
What the cube root of
the root is
-b
a+b
a³ —— ·3a²b+3ab² —bi
a³ +3a²b-f3ab² -63
Ex. 3.
aabb
The Square root of
>
is Jaabb
or
ab
#
Ex.
Sect. II. FRACTIONS.
47
Ex. 4.
What is the cube root of
-27a3b3
a3b3
·3ab
the root is
√ a³ — b³
Ex. 5.
What is the cube root of
a³+4abd-d³
8a0b3
the root is 3a3+4abd—d³
√ a³+4abd—ď³
2 or
8a6b3
24ab
Ex. 6.
What is the 4th root of
x434
8ax³—8x²yy+yª
the root is 4
or
4
x4—y4
8ax³—8x²yy+34.
√x+—y4
8ax³—8x²y²+y4
Cor. The nth power or root of a fraction, is equal
to the nth power or root of the numerator, divided
by the nth power or root of the denominator.
a
an
= =
And */
xn
a
X
n
SECT.
48
✡
穹
​Su
SECT.
III.
1
Of SUR D S.
URDS are fuch quantities as have not a pro-
per root. Simple Surds are thofe which confiſt
but of one term. Compound Surds are thofe
which confiſt of feveral fimple ones. And Uni-
verfal Surds are thofe confifting of feveral terms
under any radical fign.
Surds are faid to be commenfurable, when they
are as, one number to another; and incommenfura-
ble, when their proportion cannot be expreffed in
numbers.
PROBLEM
XIX.
To defignate or express the roots of quantities by
fractional indices.
I RULE.
Divide the index of the quantity by the number
expreffing the root; the quotient is the index of
the root required.
Ex. I.
Let the quantity a be propofed.
2
3
4
then √a = a³, Va = a³, î/a =*a³, &c,
Ex. 2.
Let 3ab² be propofed.
✔ zab² — zabbi³ = a*b./3. √3ab²
= a*b•3• √3ab² = a³b}\/·3.
√3ab² = a²b¹³/3. √zabb = a¹³√3, &c.
4
I
a*b*
4
as b
4
.
Ex.
Se&t. III.'
49
SURD S.
1
Let as be given.
3
3
Ex. 3.
then a¹ = a¹, Va¹= a* or a, v/a = a¹, &c.
Ex. 4!
Let aa-xx be propoſed.
then ✔aa xx = aa
then✔
then ✔
Vaa—xx
xx = aa
Ex. 5.
Let I be given.
Let
a--b
a
४
abbc³
4 A4
&c.
Ex. 6.
be given.
I
X3
XX
플
​XX &c.
&e.
* ==>看
​abbc³
a-b
až b c
abbc³
Ex. 7.
a3b3c
a a
Let
a
a a
3
63
be propoſed.
a a
a a
&c.
v#=jv#=tv#=q»««
63
63
E
Es.
50
B. I.
9 URD S
}
Let
a+2x
a + 2x
a a -- XX
HIN
Ex. 8.
be proposed.
a+2x
wl.
e+2x1
2
a+2x
=
aa-xx
aa-xx¡
Ι
X
ag- XX
aa-xxl
3
a + 2x3
1
aa
2
RULE.
When any quantity is in the denominator of a frat-
tion; fet it in the numerator, and change the fign of
the index.
Ex. 9.
I
I
Q Z
H
a
= a
A
1
Ex. 10.
Let 1
I
I
I
I
I
a
aa
Q3
a4
as
&c. be given.
then they become a1, a2, a¨³, a¨¹, a¨³, &c,
reſpectively.
Ex. II.
ab
Given
This becomes abx-2y-³.
x²y3
Ex. 12.
I
I
I
Given
29
&c.
1
3 9
aa
XX
aa XX
aa
XX
they become aa-xx
2
aa -XX
A A XX
&c.
Ex.
Sect. III
51
SURD S.
Let
aab
Ex. 13.
aab
a + x a + x
2
they are aab xa + x
aab xa + x³.
aab
a + x
I
2
3
be given.
2
aab × a + x +
In order to explain this; let there be a rank of
powers; as I, a, aa, aaa, aaaa, aaaaa, &c. the
fame will (by Def. 20.) be denoted by 1, a, a,
a³, at, as, &c. Now thefe quantities. a, a², a³,
&c. are in geometrical progeffion, and their in-
dices, in arithmetic pregreffion, as is plain. Now
fince 1, a, a, a, &c are geometrical propor-
tionals, therefore thefe will alfo be geometrical
proportionals,
I, √a, Vaª, √a³‚ √aª, √a³, √aº, √a”,
Va³, &c.
3
that is 1, a, a, Ja³, aa, Ja³, a³, √a¹, aª, &c.
and 1, ✔a, Vaa, a, Vat, Vas, a³, Vo,
Vas, &c.
4
and I, Va, Vaa, Ja³, a, Jas, Ja,
†/a7, aa, &c. and ſo on.
Therefore by the rule of analogy, the indices of
all theſe, are alſo in arithmetic progreffion.
3
3
3
Take any one of theſe ſeries as 1, √a, √ào,
a, Vat, &c. thefe will be equivalent to 1, a
a², a, a³, &c.
Suppole now the feries 1, a, a, a³, &c. con-
tinued backwards, the powers of a will come in-
to the denominator; and the indices, which con-
tinually decreaſe, will then become negative, and
will stand thus:
E 2
Powers
52
B. I
SURD S.
P
Powers
Indices
?
I
a 3 a²
2
I
I,
a, a², a³, at, &c.
3,
2,
1,
therefore a 3, a², a, a, a',
a~3
will reprefent thefe powers; that is,
2, 3, 4, &c.
a², a³, aª, &c.
1,
I
I
-39
a³
Q2
I
I
a-¹, I : a°, &c.
a
In like manner, let the feries 1, a³, a³, a', a³,
&c. be continued backwards; thefe powers, and
their indices will be as follows:
4, ---I,
ત
244
WIN
a¹, as, &c.
I
I
I
I, a,
a
4
I
a
a
03
аз
2
I
I
2
0,
, [,
3
3
3
3
Then,
I
号
​3
a
a
Q
2
2
Q
2
3
4, &c.
3
$
a
a°, a˚‚ª‚à°‚&c
will denote the fame powers; that is,
1
-
I
I
---3-3
2
a
a
a
2.
2
a
a
03
I
a3
@° = 1, &c. And therefore the feries,
季
​=2
I
I
2
3
3
3
I
✔ aaaa ✓ aaa ✔ aa
3
3
Jaa, Vaaa, ✔aaaa, &c.
3
Ι
ร
Vas
3
3
I
?
I
Va³ va²
Vat, &c.
I
I
1
4
2
a
शु
A3
1
3
, I;
I; Va,
√ a
may be expreffed thus,
I
3
3
3
2
3
Va
Or thus,
I
1
1, aï, a³, a³, a‡, &c.
Of
Sect. III.
53
SURDS.
I
or thus, a¯³, a¯¹; a¯¾³, a¯³, aº, a³, a³, a',
4
@*, &c. and the fame is equally true of any of
the other feries.
Cor. 1. The powers of any quantity are a fet of
geometrical proportionals from I; and their indices,
a fet of arithmetic proportionals from ọ.
1, a, a², a³, at,
indices 0, 1, 2, 3, 4,
•1.
increafing.
thus, powers
::
ட்
increafing.
alfo, powers I,
I
1 I
I
decreaf.
,
a a a
a;
a4
indices
3,-4, decreaf
- ÷
0,-1,-2,
Cor. 2. Hence the double, triple, quadruple, &c.
the index of any quantity, is the index of the fquare,
cube, biquadrate, &c. of that quantity.
Cor. 3. Hence also, the index of the product of
any two powers (whole or fracted) of any quantity,
is equal to the fum of the indices of thefe powers. And
therefore to multiply any two powers together, is to
add their indices. Thus a² Xa³=a³, a² Xa³ —a—,
&c.
I
Cor. 3. The index of the quotient of two powers,
dividing one another, is equal to the index of the di-
vidend the index of the divifor; whatever the
indices be. And therefore, to divide by powers, is to
fubtract their indices. Thus = a², and
a²
05
=a-3. Alfo
a
2
a3
= as, &c.
Cor. 4. Any power is taken out of the denomina-
tor, and put into the numerator, by changing the
hign of the index: and the contrary, Thus
I
b
= a
-ba-2
Alfo
Q
I
a² 6-3 --
&c.
• bi
E 3
Cor
54
B. I.
SURDS.
}
Cor. 5. In fractional indices, the numerator fhews
the power, and the denominator the root.
Schol. In all the following problems, it will be
the beſt way to reduce the furds to fractional in-
dices.
PROBLEM XX.
To reduce a rational quantity to the form of a furd.
RUL E.
Multiply the index of the quantity, by the
index of the furd root given; to which fet the
radical fign, or index of the furd.
Ex. 1.
Reduce 6 to the form of √
Here 6¹×2 or 6² = 36,
: 36, and 36 is that
required.
Ex. 2.
3
Reduce a to the form of b.
I
a³, and 3/a3 is the anſwer.
Ex. 3.
Here a¹×3
Reduce a + b to
the form of ✔bc.
Anfw.
a + b²,
or √ aa + zab + bb.
Ex. 4.
a
Reduce
to the form of vão
aa
Anf. is of the form.
bbc
PRO
Sect. III.
55
SURD S.
PROBLEM
XXI.
To reduce quantities of different indexes, to other equal
ones, that fhall have a common index given.
RUL
ULE.
Divide the indexes of the quantities by the gi-
ven index; the quotients will be the new indexes
for theſe quantities. Over thefe quantities with
their new indices, place the index given.
Ex. i.
I
Reduco 12% and 7% to the common index 1.
2
2 4
1/1)/(
I
I
first index.
2
then 12212 and 713
are
I
I
=) ( fecond index (the quantities required.
2
I
6
·3
Ex. 2.
Reduce a and b, to the common index 3
3 I
=) = (6 first index.
3) 2 (2 fec. index.
3
then 41¹, and 62³are the
furds.
XXII.
PROBLEM
To reduce quantities of different indices, to others, equal
to them, that fhall have the least common index.
RULE.
Then
Reduce the indices of the given quantities, to
a common denominator, in the leaſt terms.
involve each quantity to the power of its nume-
rator; and take the root denoted by the co-
mon denominator.
E 4
Fx
56
B. I.
SURDS.
Ex. I.
Reduce b and c to the least common index.
I
I
and
4
6
I
3
are= and
12
3
2
2. Therefore
12
bª =·b™³½³, and c³ =¿¹²².
I
or b* and become band,
bbb¼”
and ccl™½.
2
Ex. 2.
OF
20
2
9
Let b³ and dcl be given.
2
3
6
and are reduced to and
2
9
Therefore band de
or 6313 and del³,
or
Let Va+b, and
I
2
9
9
become band de³,
1 and ddcc.
Ex. 3.
Vaa
aa-xx be propofed.
xx³. The indices
Theſe are a + b² and aa-xx³.
are reduced to 3 and 2/20
5
3
6
6
Therefore the furds
become a+ble and aa-xx
2
or
a³+zaab+3abb+b³, and a-za²xx+x+1%, or
✅a+zaab+3abb+b³, and Va4—2a²xx+xª.
6
PROBLEM XXIII.
To reduce furds to their meft fimple terms.
RULE.
Divide by the greatefi power contained in it, and
fet the root before the furd containing the remain-
ing quantities.
Elite
Sect. III.
57.
SURDS.
Reduce
Ex. 1.
48 to a fimpler form.
√48 √3 × 16
43 the furd required.
Ex. 2.
Let 64aabc be propofed.
√64aa8a. Then 64aabc8a√/bc.
Reduce a³x
Ex. 3.
Hère ✔aa = a, and the furd becomes a×ax+xx
or avax
Ex. 4.
Let Ja³b-4
a3b-4aabb + 4ab³
be given.
CC
The furd is
aa
4ba4bb
Xab. And
cc
aa-4ba4bb
CC
a-26
C
a
25
becomes
× Vab, or
Vab.
C
C
Therefore the furd
Ø-26
Given
3
✓
270453
8b
Ex. 5.
3
Sa
reduced, becomes 3ab
2.
3
27a3b³ × a
8 x b-
V
a
PRO
1
58
B. I;
SURD S;
PROBLEM XXIV:
To find whether two furds are commenfurable or not.
RUL E.
Reduce them to the leaft common index, and
the quantities to a common denominator, if frac-
tions, except when like terms are commenfurable.
Then divide them by the greateſt common diviſor,
(or by fuch a one as will give one quotient ratio-
nal;) then if both quotients be rational, the furds
are commenfurable; otherwiſe not.
Ex. I.
Let ✓18 and Vo be propofed.
Theſe are 2 x 9 and
2 X 4. Divide by 2,
and the quotients are 9, and √4; that is, 3
and 2; therefore they are commenfurable.
Ex. 2.
Let the furds be✓
50
16
and √12.
72
25
Theſe are 50 and
72. Divide by 2,
5
4
and the quotients are 25 and 36, that is 5
and 6, and the furds become 52 and
4
6
5
5
and are therefore commenfurable, being as
6
to
5
Ex. 3.
Let 48 and 8 be propofed.
Divide by 8, the quotients are 6 and VI QË
1; therefore they are incommeaſurable,
I
Ex.
Sect. III.
51
SURD SI
Let
Here
Ex. 4.
21 and 1² be given.
C
b
61
- C
N
and
C
b
MIN
are reduced to
b
C
I
HIN
2
64
A
and
and theſe to
and
Die
b3
cb3
vide by s
and the quotients are 64
and
cb3
{
that is, bb and cc; therefore the furds are
commenfurable,
Ex. 5.
Suppoſe ✔a++aabb and ✔aabb + 64
Theſe are aa xaa + bb, and
bb xaa + bb.
Therefore dividing by aa + bb, the quotients are
bb, or a and b, and therefore they
aa, and
are commenfurable.
Ex. 6.
andaa be given.
16aa
Let
146
86
That is,
4a
and
3a
Divide the de-
√146
√86°
49
nominators by 2,
за
and
776 √46
commenfurable.
or
then they are reduced to
3a
and are therefore in-
PRO
{
SURDS:
B. 1.
4
PROBLEM
XXV.
To add furd quantities together.
RULE.
Reduce quantities with unlike indexes, to thofe
of like indexes.
!
Alfo reduce fractions to a common denomina-
tor, or elſe to others that have rational denomina-
tors (or numerators).
Then reduce the quantities to the fimpleft terms
(Prob. 23.) This being done; if the furd part
be the fame in all, annex it to the fum of the ra
tional parts, with the fign (x) of multiplication.
If the furd part is not the fame in all, the quan-
tities can only be added by the figns and
Ex. 1.
Add √6 to 2√6.
The fum is I + 2 × √6 or 36.
Ι
Ex. 2.
Add 8 to √50.
√822, and √50 = 5√2, and the fum
=2+5×√2 = 7√2 = √98.
3
3
Ex. 3.
3
Add 500 to 108.
3
3
√500 =√4X 125 = 54. And
108 =
3
3
== √4×27 = 3√4. Therefore the fum
3
3
5 + 3 × √ 4 = 8√/4.
Ex. 4.
Add 48a4b to ✔zaab³.
They are reduced to 4aa3b and ab3b.
And the fum4aa + ab × √3k.
Ex.
'
Sect. III
SURDS,
Ex. 5i
Given ✔4a and
4
aº.
4
= √16aa =
√4a=4a² = 4a² = 16aa* =
4
4
2/aa. And Vaº
a√√aa.
And their fum
4
= a + 2 × √ aa = a + 2 × √.a.
x
Ex. 6.
Add 24 to
to v
25
}
61
-<
Theſe reduced to a common denominator, be-
come ✓12 and
75
that is, 62
2
75
1
75
75
50, or
or ✔
2 X 35
2 X 25
and ✔
75
75
and
2
and 5√ whoſe fum is
75
Or thus,
ex |
11
5
6
6 And their fum
Here
24
√24
√ 4 X 6
25
5
5
2
Alfo
11
11
3
9
H
2
5
+
1
× √6
3
Add
3
4
I I
3
V6
I I
1/6 = = = √1/6 =
= = = √1/23/1
3
15
Ex. 7.
to 16.
3
شد
I
27
Theſe become and ✔
4.
5
3
64-
or
108
3
4
and
62
B. I.
SURD S
3
and 4√
4 X 27
that is
whoſe fum is I +
Vand
4
X
3
4
=
3
I
4
V² = 127
شد
4
108
and √16 — √64
V====
27
108
And the fum=
Or thus,
==X27 =
=
3√
33.
I
4 X 27
3
4
I
4 3
4√
3
4
73
3 4
4
bb
b
b
Ex. 8.
C
3
Add $104.
C
to
Theſe are reduced to
I
I
64 1 1/2/25
and
C4/2
cb31
or
cb3
ve and And their fum is
bc
CC
I
b
bc
bb + cc
I
bb + cc
b
bc
bbc
Ex. 9.
Add √ccddaa-ccddxx√√2 to √d+aa—d+xx√√✅2•
They are reduced to cd
aa-xx√2, and
dd √ aa-xx√2, and the fum is
cd + dd × √ aa—xx/2•
Ex. 10.
To 23/aa-√a³ + √/13.
Add Vaa +2√α-√7.
Sum 33/aa—√a³ + √13+2√/a−√7.
PRO-
Sect. III.
63
SURD S
PROBLEM XXVI.
To fubtract furd quantities.
R ULE
Reduce, as in the laſt rule; then fubtract the
rational quantities, and annex the difference to
the common furd, with the fign (X) of mul-
tiplication.
1. Subtract
EXAMPLE S.
6 from 26, the remainder is
2—1 × √6 = √6.
2. √50 √/8=5√2—2√/2 = 3√2.
3
3
3
3. ✔500—108 = 5√4 −3 √4 = 2√46
4. ✔48a4b ·√3aab³ = 4aa√3b — ab√35
= 4aa―ab × √36.
4
·√4a=a√a—
5. √/a° — √ 4à = √✓ a³ √4å a√ a-
2√/a = a−2× √a.
6. √24 - √
2
72
50
2
= 6√
25
3
75
75
75
2
2
- 5√
75
75
or ✔
24
25
2
24
=
3
25
2
√6
6
3
16 3
7.
11
+1
27
3
3
3
V.
3
4
4
6
3
Or thus,
27
3
4
I
I
===
3
4
4
3
3
I
4
√6.
I
4
3
9
4
3
4
64
4 X 27
8.
64
SURD S..
B. 1.
b 1ā
C
b
8.
1
C
I
2
C
b
bc
9.
cdy aa-xx
X
J
bc
I
cc x 1 |² = b b = cc x = ;
ccddaa-ccddxx
b
bc
✓ d+aa-d+xx =
dd√ aa—xx cd-ddx✓aa-xx.
=
to. From 23/aa- ·√a³ + √13⋅
take
3
√aa + 2√a — V70
3
rem. Vaa√a³ +√13−2√a+√7•
PROBLEM XXVII.
To multiply furds.
1
R UL E.
Surds by furds; if they have not the fame index
already, reduce them to the fame; then multiply
the quantities under the common index.
Ex. 1.
Multiply 5 by √3.
15155
the product 15
Ex. 2.
Multiply2ab by gad
30
26
}
{
8aabd
product =
=
obc
3aad
Ear
Se&t. III.
65
SURD S
Ex. 3.
3
Multiply ✔d by ✔ab.
I
Reduced to do and aabble; the produc
3
a²b²¿°³ =
aabbd³.
Ex. 4.
Product a³ x a² = a²¾³±± — a
Multiply a³ by a³.
X
I
3
+
IL
I 2
а
2
RULE.
A furd by a rational quantity; connect them
with the fign (X) of multiplication; or elfe re-
duce the rational quantity to the form of that furd,
and multiply by Rule 1.
Ex. 5.
√4a
Multiply 4a 3x by 2a.
The product is 2a XV 4a 3x
Or 2a = √4aa, then the produ&t=
1623
12aax.
3 RULE.
1
When rational quantities are annexed to furds.
multiply the rational by the rational, and the furd
by the furd.
Ex. 6.
a
Multiply 26
a
x by c
-d Vax.
axc -
d
a
xXax
The product =
ac ad
26
26
aax
axx
F
66
B. I.
SURD S
a
Multiply ✔ax by
a
Here & Vax =
b
Ex. 7.
!
b—
b- *
3
b
a
xax¹² = 2x4x1.
b-xx
b
And b-x
Vax³
b
aax
b
xx
Therefore
•
bb
ЂХ
b
H
a
de
× ³×³° × into b X X
аахб
ab = ax × aixļi — ab — ax
b
ab
6
ax
X
b
bb
a5x3
b
bb
6
/* the product.
bb
Multiply a+b
by a-√b
Ex. 8.
aa + a√b
d
ad
who
a5x9
bb
-ab-b+d √ b
product aaad — b + d√b
Ex. 9.
Multiply 24- 30√d
by 36-20/d
6ac
-9ac√d
·4ac✓d+bac✓dd
product 6a613ac√d + 6acd
Exi
Sect. III.
SURDS.
#
Multiply
product
a
Ex. 10.
√6−√3
bý √a + √b-√3
Jaa_av b√3
or Vaa
+ a
b
aa − b + √3•
√3−6+√3,
Schol. If impoffible or imaginary roots be muf-
tiplied together, they always produce, other-
wife a real product would be raiſed from impoffi-
ble factors, which is abfurd. Thus,
BD
√~ax√-b=ab, and—ax-√-b=
✓ab, &c. Alfo -ax√ — a = −8₂
and Vax-√—a =+a, &c.
PROBLEM
I
To divide furds.
XXVIII.
RULE.,
In furds of the fame fimple quantity; fubtract
their indices from each other.
Ex. I.
2
Divide a³ by a*.
quotient a3—1
I
=
72
a
S
ΙΣ
Ex. 2.
Divide a " by a
I
I
=
777
quotient a "
m
me
a
F 2
2 RULE.
68
B. I.
SURDS:
1
2 RULE.
If they be different quantities, reduce them to
the fame index, if they are not fo already. Then
divide the quantities under the common index.
Ex. 3.
Divide 15 by √5.
5) 15 (3 the quotient.
Ex. 4.
Divide √3aad, by √2ab
аб
C
30
2ab) Baad (9ad, quotient.
30
6
C
Ex. 5.
Divide aabbd³ by √d.
6
3
√d=√d³. d³) aabbd³ (✅aabb = √ab, quot.
3 RUL E.
If rational quantities are annexed; divide ra-
tional quantities by rational quantities, and furds
by furds.
Divide √16a3
Ex. 6.
12aax by 2a.
✓16
"баз
12aax
4aa
quotient√16a3 — 12aax = √
40
20
3*•
11
Ex.
:
1
Sect. III.
SURD S.
Ex. 7.
Divide
ac ad
26
Vaax
a
axx by
a
X
26
a
x) aax
axx (✔ax,
2) ac = ad (c = d.
26
に
​I
Then cdxax = quotient.
Divide
ab
3 ax³
b
—
b
ax
=
6
6
Ex. 8.
3
by b
3 ax³
x
x asx³ by
б
Vaaxs
b2
b — x ) ab — axx (ax.
b
bb
Then the quotient
b.
*
aax's) art (2
bb
11
6
ax
b
ax
a
a
=
[ *
=
ax.
X3
b
x
—√b)
aa
Ex. 9.
b) aa—ad — b+d√b (a + √b−d
ab
+a√b
+a√b-b
quotient.
Q...
O
ad
F 3
ad + d√b
70
B. I.
SURD S:
Ex. 10.
aa + a√/bc) a³b — abbc (ab — b✅bc
a³b + a²b√bc
a²b✓ bc — abbc
a²b√ bc
✓bc
abbc
Ex. II.
Divide Vaa-b+√3 by
√b−√3) aa−b+√3 (Va+√b=√3
aa—a√ b—√3
+a√b−√3−b+√3
+a√b—√3−b+1/3
{
Divide
Ex. 12.
bbca+✓aab-bc-✓abc
by √bc + √a
√bc+√a)√kbca+√ aab_bc—✅abc(✅ba—//bc
✔bbca+✔aab
-bc - ✓ abc
bc abc
O
4 RULE.
<
↓
When the quantities will not divide, fet them
down in form of a fraction.
Ex.
Se&t. III.
7 *
SURD S.
Ex. 13.
Divide bcd+abb by ✓ab - √4bc
√bcd + √abb
The quotient is
Vab
√ ab — √ 4bc
PROBLEM
XXIX.
To involve furd quantities to any power.
IRUL E.
Multiply the index of the quantity, by the in-
dex of the power to be raiſed.
V2 be cubed.
Ex. 1.
Let
√2
2 = 21.
Then 2
1/2 X 3
that is 23 or
What is the Square
3√bec
34
or 2 is the cube,
8 = the cube of √2.
Ex. 2.
3
of 3bcc.
3×
= 3x bed. Its fquare =
3
3
9 × bccl³ = 9√/bbc4 = 9cv/bbc.
Ex. 3.
What is the cube of ava-x.
.I
a√ a — x = a¹×a-x; cubed it is a³Xax
that is, the cube a√ a 3a²x ± zax² -x³
—
X3
K'N
Exi
F 4
72
B. I.
SURD S.
$
Ex. 4.
What is the 4th power of
a
24
1
a
26
Here ✓2 = 2 x 24.
:
power is
26
a
26
=
2a
And its 4th
2a
a4
24
c-b
1654
c-b
26
2
}
16b4xc-b
464 × cc.
2bc + bb
496
2
RULE.
If quantities are to be involved to a power de-
noted by the index of the furd root; take away
the radical fign.
Ex. 5.
Let ab be Squared.
CC
3
-
Its fquare is
4ab
CC
Ex. 6.
・b³
What is the cube of Va³ — b³ + 3b√abb..
Anfwer, a³-b³ + 3b√abb.
3 RUL E.
Compound furds are involved as integers, obferv
ing the rule of multiplication of furds.
Ex. 7.
Let 3+✓5 be ſquared.
3+√5
3+√5
9+3√5
+3√5+5
the ſquare
14+6√5
Ex.
Sect. III
73
SURD S.
L
Ex. 8.
Let a-b be cubed.
a-√.b
a-√b
aa-a✓b
-a/b+b
aa-2ab+b
the cube
a-√b
a³ —2aa√b+ab
— aa√✓b+zab―-bb
a³—3aa√/b+3ab—b√✅✅b.
PROBLEM
XXX.
To extract any root of a furd.
RUL E.
Divide the index of the quantity or quantities,
by the index of the root to be extracted.
Ex. I.
Extract the Square root of a³.
The root
3
a√ã³.
Ex. 2.
Extract the cube root of ab".
I 2
3
The root is a³b³ = √abb.
Ex. 3.
What is the 4th root of 3aa.
2 4
4
The root is a√3 = a³ √3 = √ a × √3•
Ex.
74
B. I.
SURD S.
}
Ex. 4.
What is the cube root of ✔aa xx.
6
The root is aa- XX
= aa-xx =
aa-xx²
aa-xx.
2 RULE.
When the index of the root to be extracted, is
the fame as the index of the power of that quan-
tity; take away that index, and the quantity itſelf
is the root.
Ex. 5.
What is the Square root of 32a.
Anfw. 3a, the root.
Ex. 6.
What is the cube root of 5x-3xx
Anfw. 5ax-3xx, the root.
3 RUL E.
3
Compound furds are extracted as integers, due
regard being had to the operations of fimple furds.
When no fuch root can be found, prefix the radical
fign.
Ex. 7.
For the fquare root of aa- 4a√b + 46.
aa — 4a√ b + 4b (a — 2 √ b
aa
2a — 2 √ b) 0-4a√/b + 4b
4a√6 +46
Q
Ex
Sect. III.
75
SURD S.
Ex. 8.
Anfw. Vax-
What is the cube root of aa
3
√ az
XX.
√ ax
xx, the root.
PROBLEM XXXI.
To change a binomial furd quantity into another.
RULE.
This reduction is performed by an equal involu-
tion, and evolution. Involve the binomial to the
power denoted by the furd or furds, then fet the
radical fign of the fame root before it.
Ex. I.
To transform 2 + √3 to another.
Its fquare,
4+3+4√3=7+4√3
the fquare root, √7+4√3.
Ex. 2.
Reduce √2+√3 to a univerfal furd.
Its fquare 2+ 3+2 √6=5+2√6
the root
√5+2√6.
Ex. 3.
Let Va-2√x be given to reduce.
The fquare a + 4x-4√/ax.
the root
√a + 4x — 4√ax.
Ex. 4.
3
3
Let
Va
a +
a+b
b be given.
The cube a + 3√aab + 3√abb + b
3
the root
3
3
a + 3√aab + 3√abb + b.
Cor.
76
B. I.
SURD S.
√a + √bl; and in ge
Cor. √a+√b = √ √ a +
I
I
neral a" + b
+Б
12
א
n
I
12
a
+ b
12
PROBLEM
XXXII.
To extract the Square root of a binomial (or refidual)
furd, AB, or A-B; or a trinomial, &c.
I RULE, for binomials.
AABB D. Then A+B =
A-D
Take
A+ D
+ √
2
2
and √A — B = √
A+ D
A
D
2
2
For if VA+D+A
A-D be involved by
2
Prob. 29. it will produce A + √AA
that is A+B, as it ought. And
A-D will alſo produce A-B.
2
Ex. 1.
To extract the root of 7 +✔20.
Here A7, B = √20, and ✔A A — BB
√29 = D.
Then the ſquare root of 7 +20=
2
DD,
A+ D
2
✔AA—BB:
√ √ 7+ √29
+
/7 — √/29
2
2
Ex.
Sect. III.
77
SURD S.
?
Ex. 2.
What is the fquare root of 3-2/2.
=
Here VAA BB =√1=1 D, and
A+ D
2
—
A-D
= 2,
= 1. And
2
√3—2√/2 = √/2−√1=√2−1, the root.
Ex. 3.
To extract the root of 27 + √704.
√AA—BB=√25=D=5. And the
32 +22 that is,
root✓
2
2
√27 + √704 = √16+ √11 = 4+√11:
Ex. 4.
What is the ſquare root of 6 — 2√5.
Here VA ABB √36—20 = D = 4.
=
And ✔A
A+ D
= 5, and
№5,
A-D
= 1,
2
2
And the root √51.
Ex. 5.
Extract the root of √21 + √5:
√AA—BB = √16 D = 4. And
A+ D
2
==
√21+ 4,
2
=D=4.
A-D √21-4
4.
2
√√21 +4
2
And the root 21+ 4 + √√214,
2
2
Ex
78
B. I.
SURDS.
Ex. 6.
Extract the root of a² + 2x√aa
2x√ aa— xx.
Here Aaa, B = 2x√
√AA—BB =√ aa—4a²x²+4x4 = aa
XX.
Then
2xx = D.
A+ D
Then
AD
aa
xx, and
2
-xx,
xx, and
2
the root = x + √ ax
XX.
Ex. 7.
What is the root of 6+√8—√12−√24.
Let A=6+/8, B=√12+/24.
Then
√AA—BB=D=√44 + 12√8—36—2/12×24
A+ D
A-D
= √8.
= 3 + √8,
= 3.
2
2
And the root = √3 +
3+ √8-√3.
3. But
√3 + √8 = 1 + √2, (fee Ex. 2.); there
fore the root=1+ √2 −√3.
2 RULE, for trinomials, &c.
For trinomial, quadrinomial furds, &c. divide
half the product of any two radicals by a third,
gives the fquare of one radical part of the root.
This repeated with different quantities, will give
the ſquares of all the parts of the root, to be con-
nected by and But if any quantity occur
oftener than once; it must be taken but once.
For if x+y+z be any trinomial furd, its fquare
will be x²+y²+z²+2xy+2x≈+2yx; then if half
the product of any two rectangles as 2xyX2xz (or
2x22) be divided by fome third 2yz, the quotient
2x²yz
xx, muſt needs be the fquare of one of
2yZ
the parts; and the like for the reft.
{
Ex
Sect. III.
79
SURDS.
Ex. 8.
To extract the Square root of
6+√/8−√12/24.
Here ✓8X12
I, and
2√24
and
✓ 8× √24 = √4=2,
2√12
√12X√24 = √93. And the root is
12X1/24
2/8
I + √2 −√3.
Ex. 9.
To find the Square root of
12+√32−√48+√80−√24+√40-√60.
Here 32X48 = √24,
24, this produces no-
2/80
thing. Again, 32X48
2√24
5
=
√16 = 4. And
V40x60 =√/25=5; and √24×40 =√4=2;
2√24
and √ 48X24
2√32
260
= √9 = 3; and√32 × 80=√16=4,
240
&c. therefore the parts of the root are 4, 5,
√3, √2, √4, &c. and the root 2+2 — √3
+5; for being fquared it produces the furd
quantity given.
Cor. 1. In binomials, if D be a rational quan-
tity, the root will confift of two furds; and the
parts of each under the radical fign will confift of a
rational quantity (D), and a furd (A).
Cor. 2. If both A and D be rational, the root
will confift either of the two furds, or elfe of a ra-
tional part and a furd; which is the only cafe that
is useful in this extraction.
PRO.
80
B. I.
SURDS:
PROBLEM
XXXIII.
To extract any root (c) of a binomial furd A + B₂
or A B.
RULE.
Let AA-BB=D, take Q fuch, that QD=cò
the leaſt integer power.
the neareſt integer number.
C
Let A+BXQ=r;
Reduce AQ to the fimpleft form p/s.
r + n
r
Let
=t, the nearest integer.
215
Then the root =
tv/s± √tts—n
2C
, if it can be
extracted.
!
Note, is for the binomial A+ B, and
for the refidual A- B.
Ex. I.
What is the cube root of
968 + 25.
Here D = 343 = 7×7×7. QX7³n³, and
3
3
Q=1, n=7. Then VA+B×√Q=√56+
=r=4. A√Q = √968 = 222 p√s, and
=
n
r +
4 + 7
4
√s = √2.
= t = 2. And
21/5 2√2
¿√5=2√2, √tts—n =√8—7—
6
√Q=1:
And the root
2√2+ I =
2√2 + 1, which
fucceeds.
Ex.
Sect. III.
SURD S.
Ex. 2.
Extract the cube root of 68√4374.
Here D=250=5X5X5X2. And 5³×2³=4D
=QD n³, and Q = 4, n = 2 x 5 10. And
X
3
A + B × √ Q = √134×2 = 6 = r.
A√Q = 136√1 = p√s, and √s = 1.
r + =
7323 3
उ
23
=4=t. t√s = 4•
21/5 2
6
And the root
n
12
√16
16- ΤΟ
4-6
6
3
√4
for its cube is 68–276.
t
Ex. 3.
Extract the 5th root of 296 +41√3.
Here D=3, n=3, Q=81, r=5, √s=√/6,
20
ΤΟ
t = 1, t√/s = √6, √its—n = √3, VQ=√1
5
9. And the root to be tried
SCHOLIU M.
√Q=√8$
√6+√3
19
If the quantity be a fraction or has a common
divifor, extract the root of the denominator or of
that common divifor, feparately. They that would
ſee the demonftration of this rule, may confult
Gravefande's or Mac Laurin's Algebra For as it
feldom happens that fuch quantities have a proper
root; it is not worth while fpending any more
time about them,
PRO-
82
B. I.
SURD S.
PROBLEM
XXXIV.
A compound furd being given, confifting of two, three,
or more terms; which are furd Square roots: to
find fuch a multiplier or multipliers, by which mul-
tiplying the given furd; the product will be rational.
RULE.
Change the fign of one of the terms in a bi-
nomial, or trinomial, or the figns of two terms
in a quadrinomial; and by this multiply the gi-
ven furd.
Ex. I.
Let a +√3 be given.
Multiply by a √3
product
Multiply by
product
aa
3.
Ex. 2.
Given √5-√.
✔5
√5+ √x
5-x rational.
Ex. 3.
Let √5+√3-2 be given.
5+√3+√2
Multiply by
5+✓12-10
+√15+3−√6
6+215
+√10+√6—2
product
multiply by
product
6+2√15
60 — 36=24.
Ex.
Se&t. III.
83
SURD S.
Ex. 4.
There is given √a+√b-√c + √d
Multiply by Va+v/b+√c−√d
product
or
a+b−c−d+2✓ab+2√dc
multiply by ƒ+2√ab−2√✓dc
product
or
ƒ+2√ab+2√dc.
ƒƒ+4f√ab+4ab-4dc
8+4f√ ab
88-16ffab
multiply by g-4ƒ√ab
product
In this proceſs f is put for the rational part
+b-c-d; and g for ff+4ab — 4dc.
Cor. A binomial becomes rational after one opera-
tion, a trinomial after two, and a quadrinomial after
three, &c.
PROBLEM XXXV.
A binomial being given, confifting of one or two furds,
whofe index or root is any power of 2; to find
a multiplier or multipliers that hall make it ra-
tional.
RULE.
Multiply it by its correfponding refidual (that
is when one fign is changed); and repeat the fame
operation, as long as there are furds.
Ex. 1.
Let Va-b be given.
Multiply by a + √b
product a-b rational.
G 2
Ex.
84
SURDS.
B. I.
Xy
4
Ex. 2.
4.
Let √5+ √3 be propoſed.
4
Multiply by ✔5
4
√3
I product √5 -√3
multiply by ✔5 + √3
2 product 5 - 3 = 2, rational.
Ex. 3.
8
%
Let there be given √a + √b.
8
Multiply by ✔a - √ b
I product
4
4
a—b
4
multiply by ✔a + √b
2 product ✔a —
mult. by ✔a + √b
3 product
a
b
rational.
Ex. 4.
4
Let a +√b be given.
Multiplier a- √b
I prod. aa-
mult. aa + √b
2 prod. a4 — b
Cor. The number of operations is equal to the power
of 2 in the index.
PROB.
Sect. III.
85
SURD S.
PROBLEM
XXXVI.
Any binomial furd being given, to find a multiplier
which shall produce a rational product.
RUL
E.
If the furds have not the fame index, reduce
them to the fare, (Prob. 21.)
Take the two quantities (throwing away the
radical fign or index); change the fign of one of
them. That done, involve theſe to the next in-
ferior power denoted by the index of the root
(Prob. 5. Rule 3), but leave out the unciæ or co-
efficients then place the common radical ſign be-
fore each quantity, but after its fign. And this
will be your multiplier.
n
Shorter thus,
Binomial A±/B.
✔A”-
I
Multiplier A¹ ± 7/A”−² B + "/A”—3 B²
±²/A”−4B³ + &c.
The upper figns muſt be taken with the upper,
and the lower with the lower; and the feries con-
tinued to ʼn terms.
Ex. 1.
3
3
Let √7 + √3 be given.
3
3
3
Multiplier √7×7−√7×3 + √3×3
3
3
7 + √7X7X3
✓7X7X3 — √7×3×3
3
+ √7×3×3+3
product 7+3=10, rational.
G 3
Ex,
86
B. I.
SURD S.
Ex. 2.
Let a-3/2 be proposed.
Va³—3/2.
3
Multiplier aa + a³/2 + √2X2
product
Ex. 3.
Let a +
Mult. Vaa
3
b be propofed.
Vab + 3bb
product
C
+ b
a3 -- 2.
4
Ex. 4.
Let 5+ √3 be given.
4
reduced 5+/9, given.
4
4
Multiplier ✓53—✓5³×9 + √5×9² — ŵ/93
product 5
-9=-4.
}
Or thus,
4
Surd 9+5.
4
4
4
mult. √ 9³ — √9³×5 + √9×5²
4
√ 53
product 9
-5
Ex. 5.
3
4
=4
Let /a³/b³ be given.
4.
4
3
Multiplier Va+ŵ/a6b³ +ŵ/a³b³ + ÷/bº.
Or
Sect. III.
8.7
SURDS.
Surd
Or thus,
Va³/b³.
mult. cat/a + a‡aab³ + b¾ƒa³bb + bb‡ƒb
3
product a³ b³.
Ex. 6.
3
Let ab be propoſed.
6
a³ — ✓bb
put x = a³, y = bb.
6
reduced to
3
6
Surd ✔x-Vy
6
X
6
6
6
6
4
mult. √x³+√x+y+√x3y²+√x²y³ +√xy++√ys•
product x-y — a³ — bb.
=
PROBLEM
XXXVII.
Afraction being given whofe denominator is a com-
pound furd, to reduce it to another whofe deno-
minator is rational.
RULE.
Find fuch a multiplier (by Prob. 34, 35, or 36),
as will make the denominator rational. By this
multiply both numerator and denominator.
3
Ex. I.
Let √ √ be propoſed.
Here
3 × √5+ √2
3√5 + 3√2
√5−√2 × √5 + √ ²
=√5+ √2.
5 -2 = 3
G 4
11
Ex
$8
B. I.
SURDS.
>
A
Ex. 2.
Let there be given
Multiply both terms by
√42 — √18
becomes V42—
7-3=4.
√6
√7+ √3
7-3, the fraction
Ex. 3.
Suppose
خو
2
3-√2
Multiply by 3+√2, then 3/2+2 is the
fraction required.
9-2-78
Ex. 4.
ab
bbc
Let
be propoſed.
a + √ bc
Multiply by abc, then
is the fraction fought.
aab-2abbc + b² c
aabc
Ex. 5.
3√a +2√//b
Let
be given.
5-√3
Multiply by 5+√3; then
15√a + 10√b + 3√3a + 2√36
25322
Ex. 6.
10
Suppose
3
3
-5
3
3
3
Multiply by √7+√7×5+√5³, and the frac-
3
3
3
tion becomes 10/49 + 10/35 + 10/25
3
51/19+
51/19 + 51/35 +5√25•
19+5√35+5√25.
7-5
=
Exi
Sect. III.
89
SURDS.
Let
Vab
Ex. 7.
4
√5+ √3
4
Multiply by ✓535²·3 +√5·3² −√3³,
4
And the fraction is
4.
4
✔ 125 −√75 +√45 —√27 √✓ab.
5-3=2
339
Or thus,
Multiply the terms of the fraction
Vab
√5+ √ 3
4
4
4
by 5-3, and it becomes
√5−√3
✔ab;
√5−√3
again multiply the terms of the laft fraction by
√5+√3, and it becomes
5—5*3+*3*5*—3* √/ab.
5-3=2
Let
Ex. 8.
8
be the fraction.
√3+ √2+ I
Multiply by 3+√2-1, and the fraction will
be √3 + 84/2 — 8
4√3 + 4√/2
4
=
5+2√6
2+ √6
Again, multiply by
26, and it becomes
4/6
8√/3 −81/2 + 8 + 4√18 + 4√12
6 4 = 2
-
= 4+2 √/18+ 2√/12—2√/6 — 4√3 — 4√2
=4+6√2 + 4√3 → 2√6 — 4√3—4√2
4+21/2 21/6.
SECT.
1
90
1
1
1
SECT. IV.
Several Methods of managing Equations.
A
N Equation is the mutual comparing of two
equal quantities, by the help of this charac-
ter (=); the part on the left hand is called the
firft fide of the equation; that on the right, the
fecond fide. And the fingle quantities are called
terms of the equation.
An equation is either two ranks of quantities
equal to one another, and feparated by this mark
(=); or one rank equal to nothing. And they
are to be confidered either, as the laft conclufion
to which we come in the folution of a problem;
or as the means whereby we come to it. In the
firſt cafe, the equation is compofed of only one
unknown quantity mixed with known ones, and
may be called the final equation. But thoſe of
the laſt fort involve feveral unknown quantities;
and therefore they are to be fo managed and re-
duced, that out of all the reft there may emerge
a new equation, with only one unknown quantity,
which is that we feek. And this is to be made as
fimple as it can, in order to find the value of the
unknown quantity.
A
An equation is named according to the dimen.
fion of the highest power of the unknown quan-
tity in it.
A fimple equation is that which con-
tains only the quantity itself; as ab-c.
quadratic equation, is when the higheſt power is
a fquare, as aa-bad. A cubic equation, when the
d. A
higheſt power is a cube, as a + ba² — cad.
fourth power when the higheſt power is fuch, as
a² — za³ + ad, &c.
PRO
Sect IV. Managing EQUATIONS,
91
PROBLEM
XXXVIII:
To turn proportional quantities into equations;
equations into proportions.
and
1
In the folution of problems, it often happens,
that we have ſeveral quantities in geometrical pro-
portion, which are to be reduced into an equation;
which will be done thus:
RUL
E.
Multiply the extremes together for one fide of
the equation, and the two means for the other
fide; or the fquare of the mean, when there are
but 3 terms.
On the contrary in a given equation, divide
each fide into two factors; and make the two fac-
tors of one fide the two means; and the two fac-
tors of the other fide, the extreams.
Ex. I.
If a:b::c+f: d. Then ad = bc + bf.
Ex. 2.
C
Let a+b:a------
::
d
ˇ
aa
xx :
S
ar + br
ca-cb
Then
Vaa
XX.
S
d
Ex. 3.
If ad be+bf. Then a:b::c+f: d.
Exi
92
B. I.
Managing EQUATIONS.
Ex. 4.1
ar + br
ca
-cb
If
=
S
d
Jaa
XX.
a+b
ca
ch
Then
::
aa
xx: r
S
d
or a−b: a+b::
✔aa
xx.
S
d
Ex. 5.
Let bcbdda-cg.
Then I
1: b:: c + d: da— cg.
or b: √da-cg:: ✓ da — cg: c + d.
PROBLEM XXXIX.
To reduce an equation.
3
When a queſtion is brought to an equation, the
unknown quantities are generally mixed and en-
tangled with the known ones; and therefore the
equation muſt be fo ordered that the unknown quan-
tity may ſtand clear, on the firſt fide of the equa-
tion; and the known ones on the fecond fide.
Which is done thus:
I
RULE.
When any quantity is on both fides the equation,
throw it out of both.
Ex. I.
If 3x+6b = 4c — d + 9b.
Throw out 66. Then 3x 4c — d + 3b.
2
RULE.
1
When the known and unknown quantities are
both on one fide; tranſpoſe any of them to the
contrary fide, and change its fign.
}
Ex
Sect. IV. Managing EQUATIONS.
93
Ex. 2.
If 5x + 3b = rx + bd.
Then 5x rx + bd - 3b.
And 5xx bd3b.
rx =
For to tranſpoſe a quantity with a contrary fign,
is the fame thing as to add it, or elfe to fubtract it
from both fides; therefore the quantities on each
fide, remain ſtill equal, by Axiom 1. and 2.
3 RULE.
If there be fractions in the equation, multiply
both fides by the denominators.
aa
Suppoſe
+c=f=ax
Ex. 3.
dx
b
a
Multiply by b, a a + cb — fb =
bdx
a
multiply by a, a³ + bca - bƒ a = bdx.
This proceſs is plain from Axiom 3.
4 RUL E.
When any quantity is multiplied into both fides
of the equation, or into the higheſt term of the
unknown quantity; divide the whole equation
thereby.
1
Ex. 4.
If 7ba³ + bcaa = bcda.
aa + ca = cd.
ca
Divide by ba,
divide by 7,
aa +
7
cd
7
The truth of this appears by Axiom 4.
5 RULE.
J
94
B. I.
Managing EQUATIONS.
5 RULE.
If the unknown quantity is affected with a furd;
tranſpoſe the reſt of the terms; then involve each
fide according to the index of the furd.
Ex. 5.
If✓ aa—ba + c = d.
✔ea
Then Vaa-bad-c.
fquared, aa- badd2dc + cc.
This proceſs is plain from Axiom 5.
6 RULE.
1
When the fide containing the unknown quantity
is a pure power; or if being adfected, it has a ra
tional root: then extract fuch root on both fides
of the equation.
}
Ex. 6.
if a3b3bbc.
Cube root. a = 3/b³
3/b3-bbc.
Ex. 7.
If
xx+6x + 9 = 20b.
Square root
+ 3 = ±√ 20b.
and
"
x = ±√206 — 3.
Schol. All theſe rules are to be uſed promifcu
ouſly, as one has occafion for them, till the equa-
tion be duly cleared.
PROBLEM
XL:
To explain the nature and origin of adfelled equations.
1. Any adfected equation may be confidered as
made up of as many fimple equations, as the di-
menfion
Sect. IV. Nature of EQUATIONS.
95
menfion of the higheſt power is. Suppofe xa,
xb, and x = c, &c. then x a = 0, x
b=0,
x-co. And if all theſe be multiplie i together,
ax х b x x
then x
=0; that is,
x3
x³ — ax² + abx-abco, a cubic equation,
b + ac
c + bc
whoſe roots are a, b, c.
In like manner, x—a x x -b xxc × x-d=0,
produces a biquadratic equation,
x4 — a x³ + ab x²-abc abcd=0,
-b +ac
-abd
·C
+bc
-acd
-d d+da
- -bcd
x
+db
+dc
whoſe roots are a, b, c, d.
Theſe two equations may be written or de-
noted thus,
2:4
X3
px² + qx
rx + s=0.
7
and
· px³ + qx²
And any fuch e-
quation being found in the folution of a problem;
the business is then to refolve it into its original
compounding fimple equations, and fo to find the
roots a, b, c, &c. For each of theſe ſimple equa-
tions gives one value of x, or one root. And if
any one of theſe values of x be fubftituted in the
equation inſtead of x, all the terms of the equa-
tion will vanish and be
0.
For fince
o. It is plain, when
≈-axx-bx xc, &c.
one of the factors xa is o, the whole pro-
duct will be ≈ 0. And of confequence there are
three roots in the cubic equation, and four in the
biquadratic; and in general there are as many
roots, as is the dimenfion of the higheſt power in
it, and no more.
2. If
96
B. I.
Nature of EQUATIONS.
2. If it happen that the roots a, b, c, &c. be
equal to one another, then xa will be = 0,
or x
4
3
a o, &c. and x a is had by evolu-
tion, fince the given equation is generated by in-
volution.
3. That there are no more roots than theſe is
plain; for if you put any quantity, as f for x,
which is equal to none of the roots a, b, c, &c.
Then fince neither fa, f-b, nor f-c, &c. is
o, their product cannot vaniſh or beo, but
muſt be ſome real product; and therefore ƒ is
no root of the equation.
4. Since the fquare root of a negative quantity
is impoffible; therefore if we have ſuch an equa-
tion as this, xx + aa = 0, or xx =
aa, then
x=±√ aa, which are two impoffible roots of
that equation. So that a quadratic equation has
either two impoffible roots or none. And there-
fore in any equation, there is always an even num-
ber of impoffible roots; fince each quadratic that
goes to the compounding it, muft have either two
or none. Therefore no equation can have an odd
number of impoffible roots. Hence therefore the
number of real roots in a cubic equation, will ei-
ther be one or three; in a biquadratic, four, two,
or none. In a fifth power, 5, 3 or i
1 ; &c.
5. From the foregoing equations it is plain, that
the coefficient of the first term (or that of the high-
eſt power) is 1. The coefficient of the fecond
term, (or next higheft power), is the fum of
all the roots, a, b, c, &c. with their figns chang-
ed. The coefficient of the third term, the fum
of the products of every two of the roots. The
coefficient of the fourth term, the fum of the pro-
ducts of every three of them, with contrary figns,
&c. The odd terms having always the fame fign,
and the even terms a contrary one. And the
abfolute
Sec. IV. _Nature of EQUATIONS.
97
abfolute number is always the product of all the
roots together.
6. Hence it follows, that when the fum of all
the negative roots is equal to the furn of all the
affirmative, the fecond term vanifhes, and the con-
trary. And if all the negative rectangles be equal
to all the affirmative ones, the third term vanishes.
And if all the negative folids be equal to a!! the
affirmative ones, the fourth term vanishes, out of
the equation; and fo forward.
7. But the roots of equations may be either +
or, yet ftill the fame rules hold good. For let
the fign of any of them as c be changed intoc,
that is, let x+c=0; then in the cubic equation
the ſecond term will be-a-bc; that is, the
fum or the roots with a contrary fign; the third
term will be + ab
bc, that is, the
fum of the products of all the roots; and fo
of the reft.
ac
8. Hence alfo in every equation cleared of frac-
tions and furds, each of the roots, each of the rec-
tangles of any two of the roots, each of the folds
under any three of them, each of the products of
any four of the faid roots, &c. are all of them
juft divifors of the laft term or abfolute number.
Therefore when no fuch divifor can be found,
it is evident there is no root, no rectangle of
roots, no folid of roots, &c. but what is furd.
For in the cubic equation, a, b, c, and ab, ac, bc,
are all of them divifors of the laft term abc: and
ſo of higher powers.
-
9. In any equation, change the figns of all the
terms but the fift; then let the coefficients of the
firſt, fecond, third, &c. terms be 1, p, q, ", S,
t, v, &c. refpectively.
H
Then
98
B. I.
Nature of EQUATIONS.
Then obferving the figns, we fhall have
pfum of the roots, a+b+c &c.
+29
pA + 29 = fum of the fquares of the roots
= a² + b² &c.
the fum of their cubes
pB + q A + 3r
= a³ + b³ &c.
pC + q B + r A + 45
drates, &c.
Where A, B, C, &c.
&c. terms.
the fum of the biqua-
are the firſt, ſecond, third,
For +pa + b + c &c.
2
A.
Alfo pA or a + b + c = a² + b²+c²+2ab+
2ac+2cd=B 29. Therefore B-pA+29, &c.
To go through the calculations of the reft would
be tedious, and of little uſe.
But if the
10. In equations of the third and fourth power,
we find, when the roots are all affirmative, the figns
are + and alternately; fo that there are as
many changes of the fines as is the index of the
power, or as the number of roots.
roots are all negative, the figns are all through.
out, there being no changes of the figns. Whence
in theſe cafes, there are as many affirmative roots,
as changes of the figns in all the terms, from +
to, and from to +. And the fame rule
holds in general, that is, there are as many affir-
mative roots in any equation as there are changes
of the figns. But the equation is fuppofed to be
compleat, that is to want no terms, and to have
numeral coefficients. And likewife the number of
negative roots is known thus; as often as two of
the figns, or two of the figns ſtand next
one another, fo often there is a negative root. It
would be needlefs to trouble the reader with the
proof of thefe things; fince it can only be done
in particular cafes, and not in a general way.
And
Sect. IV. Nature of EQUATIONS:
99
And befides when impoffible roots happen to lie
hid in the equation, they cauſe the rule to fail.
+
11. When the roots are all affirmative, the
terms of the equation are alternately and
through the equation; but when the roots are all
negative, the figns are all ; and therefore, as
by changing the figns of the roots, the ſigns of
the alternate terms are changed; fo on the con-
trary, changing the figns of the alternate terms,
changes the figns of all the roots. And this holds
in general, as will be evident by producing two
equations from the fame roots, with contrary figns.
as x a = 0, x
3
12. Since any adfected equation, as x³ — px² +
qx-ro, is made up of fimple equations, fuch
b =o, &c. Therefore if one root
as a be known, the whole equation may be exact-
ly divided by x-a; and fo reduced to a lower
dimenfion. Also when all the roots a, b, c are
found out, then will the continual product of
x—a, x — b, x-c, exactly produce the fame equa-
tion. It is no wonder that an equation has feveral
roots; becauſe in fuch cafes, there are more folu-
tions to a problem than one. So that in one cafe
of it, x isa, in another cafe xb, in a third
x= c, &c. and they are all comprehended in the
general equation. And hence though there be fe-
veral roots in an equation, yet only one of them
will answer one cafe, or the particular queftion
propofed.
12. That any root fubftituted for x in the given
equation, will make the whole equation to vanish,
by deſtroying all the terms, is proved thus. Let
the equation be,
a x² + ab x
abco.
-b
C
+ ac
+bc
H 2
And
100
B. I.
Nature of EQUATIONS.
And let the roots be a, b, c, as before.
Then
fubftitute any one, as a, inſtead of x, and the
equation will become
3
a³ — a³
3 + baa abco,
baa
+ caa
caa
+ a b c
Where the terms manifeftly deftroy one another.
And the fame will happen, by fubftituting b or c,
for x.
13. If the laft term of an equation vanifhes (as
abc, Art 12), then one root will be o; for then
the whole equation may be divided by the un-
known quantity ≈ or x 0. If the two laft terms
vaniſh (abx + acx + bcx, and — abc), then two roots
are o; if the three laft terms vaniſh, then three
roots will be o; &c.
And on the contrary, if one, two, or three roots,
&c. beo, the laft term, the two laft, or the
three laſt terms, &c. will vanifh out of the equa-
tion, and the remaining part of the equation will
contain the reft of the roots. Thus in the equa-
tion, Art. 12. if the roots b, c be = 0; there
remains only ³— a + b + c × x²=0, or x—a=0,
an equation containing the remaining root a.
x
14. And in any power of a binomial, if each
term be multiplied by the index of the unknown
quantity therein; it will thereby be reduced to the
next inferior power. To prove this, we muſt ob-
ferve, that the coefficients of a binomial, are the
very fame, whether you reckon forward from the
beginning, or backward from the end; that is,
the first and laft are the fame; the ſecond and
laft but one; the third and last but two, &c.
For the coefficients of any power of x + b, are the
fame as of b+x. In the quadratic xx+2bx+ bb,
the
Sect. IV. Nature of EQUATIONS.
ΙΟΙ
1
the coefficients are 1, 2, I. In the cubic
x³ + 3׳b + 3xb2 + b³, they are 1, 3, 3, 1. In the
fourth power they are 1, 4, 6, 4, 1. In the fifth
power, 1, 5, 10, 10, 5, 1; and ſo on.
Therefore, let any power of x+b be denoted
I
thus, x" + x²-1b +
4
น n
n.n
92- I
2.3
1. n
2
·
3
n.n
I
Mn-26b +
2
2
x^2-363 &c.
+
2.
x36x-3+
n.n
I
x² En-2 +
2
; n being the index of the power,
and let m be that of the next inferior power, or
I. Now let each term be multiplied by
m = n
the index of x in each term; that is, by n,
n — I,
珏
​2,
&c. and we fhall have
nxn + n. n → [.x²-1b +
•
xπ-2bb +
2
12 n
I. n. 2.12
3
xn-3b3 &c.
+
2.3
N. N — I. n
2
x3b-3+n.n— 1. x²bπ-2 +
2
nxt-o.
I
And dividing all by nx, it becomes
12
AM-! + N — I . **-26 + 1. n
2
..ท 3 x²-4b3 &c.
2 072-366 +
N
I. n
· +
2.3
N
i .n
2
n
x²/"_3+12—1b-2b-1; that
2
is, reſtoring m, xm + mxm-1b + m • ? — ▲
. 232 sm-26b+
2
m.
712
I .m
2.3
2
x²¿m −2 + mxbm−1 +bm, which is manifeftly the th
2 Xxm-3b3 &c. .... + 712 772
I
power of
b.
H 3
15. Alſo
f
102
B. I.
Nature of EQUATIONS.
15. Alfo if the equation refulting from the last
operation be taken, and its feveral terms again
multiplied by the index of x in each term; it will
be reduced to the next power below that, and ſo
on for more operations. And therefore after each
operation one root will be deſtroyed; or ſo many
roots will be deſtroyed as there are operations, and
the reſt will remain.
16. And further: If there be feveral equal roots
of one fort, and alfo ſeveral equal ones of another
fort, in any equation. And if the terms of that
equation be multiplied by the feveral indexes of the
unknown quantity in each term; an equation will
arife wherein one of the equal roots of each fort
will be deſtroyed. And in general, whatever roots
there be in any equation, if the terms be refpec-
tively multiplied by the indexes of the unknown
quantity therein, an equation will come out where.
in one root of every fort will be deftroyed, whether
there be equal roots, or all different. But theſe
things being of little confequence, I fhall not de-
tain the reader any longer about them.
17. As impoffible roots are fuch as are produced
from the ſquare roots of negative quantities: fo
impoffible equations are thofe produced from im-
poffible roots; as this equation a4-4a³ +aa+ica
220, which is produced from theſe two,
aa + 2a + 2 = 0, and aa 6a+ 11 = 0; the for-
mer produced from a + 1 +√ I, and a +1-
√-1; and the latter from a 3 + √— 2, and
a—3—√~2. Thefe fort of equations have roots
that are barely impoffible.
-
Likewife, there are equations that are doubly
impoffible, or impoffible equations of the fecond
degree. And thefe are produced from equations
involving two degrees of impoffibility, as this
a*+4a³+8aa+8+5=0, which is produced from
the
Sec. IV. Nature of EQUATIONS.
103
I
the equations, aa + 2a + 2 + √ — 1 = 0, and
aa+za+2-✓-10. Such as thefe cannot be
reduced into rational quadratics, as the others may.
PROBLEM XLI.
To increase or diminish the roots of an equation, by
any given quantity.
RULE.
For the unknown letter fubftitute a new letter,
the given increment, or the given decre-
ment. And fubftitute the powers thereof, in the
equation, inſtead of the powers of the unknown
letter.
x3
Ex. I.
Let x³ — px² + qx-ro, be given; and let the
roots be leffened by the quantity e.
Suppoſe y=x-
e, or x = y + e. Then
≈³ = y³ + 3ey² + 3e²y + e³
X3
Px²
+qx
r
Dayz
2
2pey
Pe²
+qy + qe
O
=0, which
is the equation required.
Ex. 2.
Increaſe the roots by 4, of this equation
a³ + a²
1
10a + 8 = 0.
+8
Suppoſe a + 4 =
e,
or a = e
4.
Then
a3
+ a²
—
=e³ 12e² + 48e64
ee
8e + 16
10a
+8
£3
10e + 40
+ 8
11ee + 30e =0, the
*
equation required; reduced, e²-11e +30=0, a
quadratic.
H 4
Cor.
104
B. I.
Nature of EQUATIONS.
Cor. 1. The last term of the transformed equation,
is the very fame as the equation given, baving e in
the place of r (in Ex. 1.)
Cor. 2. When the last term vanishes, the number
affumed (-4, Ex. 2.) is one of the roots in the
equation propoſed.
Schol. By this rule, all the roots of an equation
may be made affirmative; by increafing them by a
proper quantity.
PROBLEM
XLII.
To multiply or divide the roots of an equation, by e
given number or quantity.
RULE.
Affume a new letter; and divide or multiply it
by the given number; and fubftitute its powers in
the equation, inftead of the unknown quantity.
Ex. 1.
Multiply by 3, this equation y3-4 y — 145 0.
Suppofey, then fubftituting
for y,
23
we have
4
146
Z
= o, or reduced
27
9
27
3
12Z
146 = 0.
Ex. 2.
Divide by 3, the equation x3
2x + √3 = 0,
Let xy√3, which
3√3, which put for x, we have
3.1³ √3.—— 2y√3 + √3 = 0, or 3y³ — 2y +1=0.
Cor. By this rule, frallions or furds may be taken
out of an equation; by dividing the new letter by the
common denominator; or by multiplying the new letter
by the furd quantity.
PRO-
+
Sect. IV. Nature of EQUATIONS.
105
PROBLEM
XLIII.
To change the roots of an equation into their
reciprocals.
RULE.
In the given equation, inſtead of the root, fub-
ftitute a unit divided by fome other letter.
Example.
Let 3y³ — 2y + 1 = 0, be given.
Put y ==
I, then 3
2
+I=0.
Z
Z3
reduced 3-
or 23
2x² + 23 = 0.
2≈² + 3 = 0.
Schol. By this rule the greateſt root is changed
into the leaft, and the leaft into the greateft, &c.
PROBLEM.
XLIV.
To compleat a deficient equation.
An equation is compleat, when it has all its
terms, or thoſe containing all the powers of the
unknown quantity; and deficient, when any power
is wanting.
RULE.
Increaſe or diminish the roots of the equation,
by fome given quantity (by Prob. 41).
Example.
Suppoſe a³ + 2a 50, deficient.
Let e+1 = &, then
—
= e³ + zee + 3e + I
аз
+ za
-5
=
+ 2e + 2
5
e³ + 3ec + 50—20, compleat.
o,
Schol.
106
B. I.
Tranjmutation of
Schos. An equation may be rendered compleat,
by multiply ng by the fame letter with fome quan-
tity added, as a + 1; but then it raifes the equa-
tion a degree higher.
PROBLEM XLV.
To deprefs an equation to a lower dimenfion; one of its
roots being given.
I RUL E.
Put the equationo, and divide it by the un-
known quantity the root given.
Example.
a—4.
Given a³+a²-10a +8=0, one root a = -4.
8+4=0) a³+a²-10x+8=0 (aa-3a+2=0 the
a3
a³ +4a²
·3a² — 10a
-3a² - 12α
+20+8
+20+8
O
equation req.
2
RUL E.
Put a new letter added to that root, equal to the
unknown quantity; and fubftitute that and its
powers in the equation.
3
Example.
Let a³+a²-10a +8=0, be given, and a=-4.
Put a = e 4. Then
+ @³ = e³
ез
12e² + 48e — 64
+ a²
+
ee
8e of 16
100
10e + 40
+ 8
O
11 11
8
11e² + 30e +
reduced &
Je + 30 = 0
PRO
Sect. IV. EQUATIONS.
107
PROBLEM XLVI.
To find how many roots are affirmative, and bow
many negative, in a given equation.
RULE.
Range the terms of the equation according to
the dimensions of the unknown quantity. And
if the equation is not compleat, make it fo by
Prob. 44.
-
Then obferve how often + follows, or -
follows, that is, how many changes of the
figns there are; and there are fo many affirma-
tive roots in the equation.
Alfo, as often as two like figns ftand together,
fo often there is a negative root.
Ex. 1.
30 = 0;
+
and
Given 4x³-19xx + 49x
Here the figns are
are +
there are three changes; from the first to the fe-
cond, from the third to the fourth, and from the
fourth to the fifth term: therefore there are three
affirmative roots. Alſo, in the fecond and third
terms, two negatives ftand together, and in none
elfe, confequently there is one negative root.
Ex. 2.
Suppoſe x++ 5x³ — 7x²
x4 5223
29x+30= 0.
The figns are ++
+
roots
neg. af. neg. af.
So there are two affirmative, and two negative
roots.
Ex.
108
B. I.
Tranfmutation of
Ex. 3•
Let the equation be a37a+6=0.
This equation being defective is to be compleated.
a3 *
-7a+6= o.
mult. by
a + I
O.
24
*
7a² + ba
*
ча
70 +6.
702
a + 6 =
+ a³
04 +03
So there are two affirmative, and two negative
roots in this laft equation, and one of the negative
roots being-1, (by the multiplication of a+1=0,)
therefore, the given equation contains two affirma-
tive roots, and one negative.
The reaſon of this rule appears from Art. 10.
Prob. 40.
SCHOLIU M.
This rule does not hold good, if there be im-
poffible roots in the equation; except fo far as
thefe impoffible roots may be taken for ambiguous
ones, that is, for either affirmative or negative roots.
As in the equation x³-6x² + 13x-10=0, which
by this rule gives three affirmative roots, but in
reality it has but one root, which is 2, the reft are
imaginary.
There are alfo fome rules whereby to judge how
many impoffible roots are in an equation, but they
are fo very tedious, and of fo little ufe, that I fhall
not trouble the reader with them. See Newton's
Univerſal Arithmetic, p. 197.
PRO-
Sect. IV. EQUATIONS.
109
PROBLEM
XLVII.
To change the affirmative roots into negatives, and the
negatives into affirmatives.
RUL
E.
Place cyphers for the deficient terms, if there be
any; then change the figns of all the even terms,
that is, of the fecond, fourth, fixth, &c. terms of
the equation.
Ex. I.
Given x³ + 8x + 24 = 0.
That is, x³ +o+ 8x + 24 = 0.
transformed ³ — + 8x240.
o
In the given equation x——2, in the transform-
ed equation x = + 2.
Ex. 2.
Suppoſe +x+- 4x³·
19x² + 106x
120 = 0.
transformed +4 + 4x³—19x²-106x-1200.
In the former equation the roots are 2, 3, 4
and 5; and in the latter 5, 2, 3, and
4.
The reaſon of this proceſs is plain from Art. 11.
Prob. 40. and may be demonftrated thus. In the
given equation, we have + x for the root. Now
fuppoſe-x to be a root. Let this be ſubſtituted in the
given equation, and it produces-x3-8x+24=0,
that is, x³+8x 240, as in Exam. 1. And
*4 + 4x³ —— 19x² — 106x—120—o, as in Exam. 2.
For it is plain, all the odd powers of x will now
be negative, which before were affirmative, the reſt
remaining as before. Whence the figns of all the
odd
powers will be changed, according to the rule.
SECT.
110
SECT. V.
Ranging the terms; working by general forms;
fubftitution and reftitution; taking away any
term of an equation; extermination of un-
known quantities; the defignation of quantities
by letters; regiſtering the steps.
PROBLEM XLVIII.
To range the terms of an equation, or difpofe of them
in the best manner for any operation.
TH
RULE.
HIS is done by placing theſe terms foremoſt
that contain the higheſt power of the un-
known quantity; and in the following places,
thoſe of leſs dimenſions; fo that the powers in the
feveral terms may continually decreaſe from the
highest, according to the feries of the natural num-
bers. But in many cafes, the contrary method is
to be followed, and the loweſt power taken firſt.
Ex. I.
Let az³ + 24 — b.x3b4 + ab³o.
0.
Place it thus, z✦ + az³ * * + ab³ = 0.
z4
-b
64
Ex. 2.
Suppoſe x4+ax³ + bx²-bx³+cx=dx—ab³+b4.
ranged x4 + ax³ + bx² + cx + ab³ = 0.
b
-d-b4
PRO.
Sect. V. GENERAL FORMS.
III
PROBLE MXLIX.
To work by a general form.
RULE.
Write down each letter or quantity in the ge-
neral form, and after it (with the fign =), each
letter it reprefents in that particular cafe; which
will give feveral equations.
1
Then caft your eye over the general form, and
obferve the general quantities therein, and look for
them on the first fide of the equations; and what
you find them equal to, on the right hand, write
down, inſtead of them, each one by one, till you
have gone through the general form; and you will
have the folution.
When the quantities are many, it will be the
beſt way to write down the general form firft, and
the particular one under it, each quantity under its
correfpondent; then it will appear by inſpection
what letters to fubftitute.
Ex. I.
To involve aa -xx to the 5th power.
This is to be done by the general form in Cor. 1.
Prob. 5. therefore we have
a = a a
e
XX
n = 5,
n
Whence a + e = aa
xx + 5×
5 - 2
3
2
5
2
× a a × ×+ + 5 ×
a a x − x²+5×5—1x
×
a a xx³ +5X
5
5
-4
xx = aa +
5 ×
a a X
-3
5
I
X
2
5
2
X
5-3
2
3
4
5
2
5 - 3
X
X
X
5-4
2
3
4
5
x10 = aro
X
5a8x² + 10°×4
-x, the power required.
10a4x6 + 5aax³
Ex.
112
B. I.
GENERAL FORMS.
Ex. 2.
I
Extract the fquare root of 28-300.
This is to be done by the form in 1 Rule, Prob. 32°
Here A 28, B=√300, D=V784-30022,
A+D
28 +22
A-D
==
<= 5,
2
2
2
11
✓ 28 — 22=✓ 3. Therefore v/A — B = 5 —√/31
2
the root required.
Ex. 3.
To find a quantity, by which if 2-√6 bè mul-
tiplied, the product will be rational.
This is to be done by Prob. 36.
Heren 5, A2, B6.
5
=
5
And the multiplier
5
√16 + √/8 × 6 + √⁄4 × 36 + √2 × 216 +
S
√1295.
mult. ✔16+✔8×6+√4×36+√2×216+✓1296
by 2-6
5
5
5
5
5
S
5
√32+ √96+ √⁄8 × 36 + √ 4 × 216 + √2592
✔96—√/8×36—√4×216–√2592—✓7776
2-6=
=-4. product.
PROBLEM
L.
To fhorten the work by fubftitution and reftitution.
In any operation, when the quantities grow very
numerous, or very much compounded, it will
make the work very tedious; and therefore it
ought to be made fhorter as follows.
RULE.
Sect. V. SUBSTITUTION, &c. 113
:
RULE.
Affume a new letter to reprefent or ftand for any
number of given quantities; and likewife fome dif
ferent letter to ftand for the coefficient of any power
of the unknown quantity; do fo for as many of the
coefficients as are compounded. Likewife, put let-
ters for the numbers concerned; then work with
theſe inſtead of the or ginal quantities, which will
make the work eafier. And this is called Subſti-
tution.
When the operation is over, each number or
compound quantity muſt be restored again inſtead
of its letter; and this is called Reftitution.
Ex. I.
Let aa + baca + dade.
Putsb-c+d. Then the equation becomes
aa + sa=dc.
Ex. 2.
bx
Let a
2x X
aa
XX
dx² + cx:
Put c-d=p. Then
a
2x aa
xx =
Xx
pxx + cx
bx
multiply by pxxcx.
прих 2px³ + acx
qxx
Then
2cxx × √aa — xx — bx.
Put ap2cq. Then
2px³ + acx × Vaa -xx = bx.
or acx+qxx 2px³ × √ aa — xx — bx.
X
fquared acx+qxx - 2px × aa-xx bbxx, &c.
where the values of p, q, may be reftored.
I
PRO
114
B. I.
SUBSTITUTION, &c.
12
PROBLEM
LI.
To take away the fecond term of an equation.
RULE.
Divide the coefficient of the fecond term by the
index of the higheſt power; annex the quotient,
with its fign changed, to fome new letter, which
ſubſtitute for the root, in the given equation.
Ex. I.
Suppoſe a³ + aa — 10a + 8 = 0.
1
Put e
= a. Then
3
3
a³ = e³
e² +
3 2
+
I
27
I
+ aa =
+ ee
3
9
•10a =
+ 8 =
O
es
*
required.
+
10e +
10
3
+8
1oje + 111, the equation
Ex. 2.
27
Let y4 — Say³ + a+ =0, be given.
8a
Let y=x+ = x + 2a. Then
4
y4 = x++ 8ax³ + 24a²x² + 32a³x + 16a+
8ax³ 48a²x² 96a³x — 64a4
Say3=
04
=24
24a²x²
2
+
a4
64a³x
47a4=0.
Schol. Hence by this and the 43d problem, an
equation may be found, which wants the laft term
but
Sect. V. EXTERMINATIOŃ.
115
but one. For if the fecond term be taken away by
this problem, and the equation transformed by
Prob. 43, you will have the equation required.
PROBLEM LII.
To take away any term out of an equation.
RULE.
Take a new letter for the root, to which add an
unknown quantity; and ſubſtitute this fum and the
powers thereof, into the given equation. Then any
term put equal to nothing, will determine the va
lue of that affumed unknown quantity.
1+ 1
Ex. I.
Suppofe x+3x³ + 3x² — 5×2 = 0.
Put y+ex,
x. Then
**
3x3
+3x²
•5x
= y++ 4y³e + 6yyee + 4ye³ + 84
3y3 gyye 9ye² 3e3
+ 2 =
+ 31y +6ye + 3e²
5y
56
+ 2
=0.
Then, if the ſecond term is to be taken away,
make 49³e—3y³=0, or 4e3; therefore e = 3
મ
Ex. 2.
4
The fame fuppofed, to take away the third term.
Here we fhall have 6y²ee — gy²e + 3jy = 0; re-
duced, 2ee3e + 1 = 0, the refolving of which
quadratic equation gives the value of e. Then y + e
gives the value of x, fo that the third term may
vaniſh.
1.2
Ex
116
B. I.
EXTERMINATION.
Ex. 3.
The fame thing still fuppofed; to take away the
fourth or fifth term.
For the fourth term, 4e39e² + 6e5=0, a
cubic equation whofe root is e; and y+ex,
makes the fourth term vaniſh.
For the fifth term, et3e3 + 3e² — 5e + 2 =0, a
fourth power whofe root is e. Then
+ex, which ſubſtituted in the equation,
makes the laft term vaniſh.
Cor. 1. Hence the third, fourth, fifth, &c. term,
may be taken out of the equation; by refolving a qua-
dratic, cubic, fourth power, &c. equation.
Cor. 2. Hence if the last term of an equation (as
e4 — 3e³ + zee — 5e + 2) beo, then one root (x) is
зез
=0; for then ao, or x will divide the equation.
If two of the laft terms beo, two values of the root
will beo, and so on. But if the last term does not
vanish, there is no root
o.
Schol. After the fame rule any term may be made
equal to any given quantity; by putting the faid
term equal to that quantity.
PROBLEM
LIII.
To exterminate a fingle letter, or a quantity of one di-
menfion, out of feveral equations.
I RULE.
Seek the value of the quantity to be expelled,
in two equations; and put thefe values equal to
one another.
Ex.
Sect. V. EXTERMINATION.
117
12
Ex. 1.
Let a + x = b + y
and 2x+y=36
to exterminate y.
By tranfpofing b, a + x — by, and by tranfpo-
fing 2x, y=3b-2. Therefore a +xb=3b2x,
And by reduction 346a, and z = a
Ex. 2.
2 by = ab}
Let ax
and xy = bb,
46
3
to exterminate y.
黾
​Here ax
ab2by, and y =
ax
ab
26
bb
Alfo y =
ax ab
b b
therefore
and re-
X
26
263
a
ducing xx- -bx =
2 RUL E.
Find, by reduction, the value of one unknown
quantity, in one equation; and ſubſtitute that va-
lue for it, in all the other equations. Proceed thus,
with another unknown quantity, &c.
Ex. 3.
Let a += 2b
and
3axyx d
=
"}
to expell y.
=
By the first equation y 2bax, put this
value in the ſecond equation; then
зах
3ax-xx2b-a-d, that is, 3ax-2bx+ax
+xxd, or 4ax 24x + xx=d.
2ḥx
1 3
Ex.
118
B. I.
EXTERMINATION.
}
Ex. 4.
Suppoſe x+y+z = a }
3y = x + 2x
az = xy
to expunge x and y.
By the first equation, za-X —y,
By the fecond equation, 2y+ 2a — 2x — 23,
By the third, a xa — x — y = xy,
or aa ax ay xy.
The former reduced, 5y = 2a-x.
and fince
ca ax ay = xy.
From theſe to expunge y.
By the former y =
20
5
aa—ax = ay + xy, and y =
20
X
ac -- ax
==
>
5
a + x
X
aa
9
and by the latter
ax
Therefore
a + x
in which equation there is only
one unknown quantity x.
Cor. 1. By each given equation, one unknown quan-
tity may be taken away. And confequently when there
are as many equations as unknown quantities, they may
be all taken away but one.
Cor. 2. If there be more unknown quantities than
equations, there will remain in the last equation more
unknown quantities by 1, than that excefs amounts to.
PROBLEM LIV.
To exterminate an unknown quantity of feveral di-
menfions.
IRUL E.
Find the value of its greateſt power in two equa-
tions; then if they are not the fame, multiply the
leffer
Sect. V.
119
EXTERMINATION.
leffer power, fo that it may become equal to the
greater. Then put theſe values equal to each other,
and there will come out a new equation, with a lefs
power of the unknown quantity. And by repeat-
ing this operation, the quantity will at laft be ta-
ken away.
Ex. I.
to
expunge e.
Let aee + be + c =
and fee +ge + b = oS
By tranfpofing and dividing,
ee-
ge+b. Therefore
f
be + c
ני
and
ee=
a
be + c
ge+b.
a
f
And multiplying, bef + cf = age + ab, and by tranf-
pofing, bfe age = ab cf, and dividing,
abcf. And multiplying bye,
e =
bf — ag
-abe + cfe
be + c
-abe+cfe
Whence
ee-
bf — ag
a
bf — ag
e =
bbf
And multiplying alternately,bbfe+bcf — abge — agc
= acfe-aabe. And tranfpofing and dividing,
agc-bef
abgacfaab
Therefore
ab cf
agc
bcf
=
Then multi-
bf-ag
bbf — abg — acf + aab
plying and reducing, bhaa + cgga
+
bbfb = 0.
2cfb
bg fc
bgb +
ccff
2 RULE.
For two quadratic equations.
ax² + bx + c = 0₂
}
to exterminate x.
and fxgx + b = o, S
Here a, b, c, f, g, h, are either given quantities,
1 4
Of
120
B. I.
ÉXTERMINATION.
or compoſed of given quantities, and fome other
unknown quantity y. Thus
make bfag A, bbcg B, and ef
cg=
=D, then AB + DD = 0.
To prove this rule, we have - x² =
8x + b
f
that is,
>
therefore
bx + c
a
ab
which reduced is bf-agxx + cf-ab=0;
Ax+Do. Whence Ax² + Dx = 0;
Dx
bx + c
xx =
which reduced
>
A
a
са
Dx
is x =
In like manner
XX
eD — bA
A
gx + b
bA
which reduced is x =
Whence
f
ƒDgA
CA
hA
And this reduced is
aD — bA
ƒD — gA
cf — abxD+bb —cgxA=o, that is AB+DD=0.
The Newtonian Rule is,
ab × ab — bg—zcf +bf xbb-cg+cxagg+cff=0.
3 RUL E.
For a cubic and a quadratic equation.
ax³ + bx² + cx + d = o,
and fx²+gx + b = o.
Make fab
cb = D, fb — ag=A.
2
Then D-gAxbD-fdg + dff — bao.
For, multiplying the first equation by f, and the
fecond by ax, and fubtracting one from the other,
we have bf-ag Xx² +ƒc — ab xx + fd=0; and
fince fx²+gx+bo, these two equations come un-
fc
fc
der
Sect. V.
121
EXTERMINATION.
der the laſt rule,making abf-ag, b=fc-ab,c=fd.
And Afxfc-ab-gxhf—ag, B=bxfc-ab-fdg,
D=ffd― bxbf ag. Whence by that rule,
fxth h -g x vj ag xbx fc — ab -fdg +
ffd - b x bf — ago, that is, according to the pre-
fent defignation of the letters A, B, C ; ƒD — gÂ×
bD — fdg + ffd — bA” — 0.
The Newtonian Rule is,
abbah — bg — 2cf + bfbx bb — cg — 2 df (=0.
+cb-ig ×agg + cff + dfx sagh+bgg+dff S
4 RULE.
For a quadratic and a fourth power.
ax+ + bx² + cx² + dx + e = 0,
and
fx² + gx+b=0.
Make Abfag, D= cf — ab.
Then df³-gfD+gg-fbx Ax dbff—egff—bbA
+ eƒ³ +gbA —ƒ bD¹² = 0.
For, multiply the first equation by f, and the lat-
ter by axx, their difference will be bf-ag xx3 +
cf — ab × x² + dfx + ef = o. Or Ax³ + Dx² + dfx
+ef=0. And ſince fx² + gx + b = o. Therefore
theſe two equations come under the laft rule; in
which writing A for a, D for b, df for c, ef for
d; and laſtly ƒD-gA inſtead of A, and ffd-bA
for D, you will get the rule, as above.
The Newtonian Rule is,
ab³ Xab-bg-2cf + bfbb× ób —
+ agg + cff × cbb dgh egg-
+
+ dfb × zagh + bgg + dff
+ eff × 2abb + 3bgb — dfg + eff
efgh × bg + zak
cg--2df
zefb
=0.
5 RULE.
}
122
B. I.
EXTERMINATION.
5 RUL E.
For two cubic equations.
ax³ + bx² + cx + d = o,
and ƒƒ³ +gx² + bx + k − 0.
Make Abf ag, C = df — ak, D = cf — ab ;
2
and PCA —aAC —bAD + aDD,
Q = c AC — aCC — dAD,
RdAAbAC + aCD.
Then PQ+ RR = 0.
For, multiply the first equation by f, and the
latter by a, and their difference will be found
bf — ag xx² + fc — ab xx +ƒd — ak = 0; that is,
ab×x+fd
Ax² + Dx + Co. And fince ax + bx + cx + d
= 0; theſe two equations come under the third rule;
in which writing A, D, C for f, g, h, refpectively;
and likewife cA aC for A, and bAaD for D;
the rule will be evident.
The Newtonian Rule is,
I
+ ab- bg
2cfx adbb
achk
cg
zdf
+ bdfb×ak + bb
+ aakk× bk — ak + 2gc + 3df
+ bbfk × bk — 2 dg
+ cdb―ddg cck + 2bdk Xagg + cff
+ zagh + bgg + dff3afk x ddf
+ bcfk × cg + df зак
bb
agk× bbk + 3adh + cdf = 0.
6
RULE.
For a cubic and a fourth power.
ax4 + bx³ + cx² + dx + e='o,
fx² + gx² + bx + k = 0.
and
Make A =fbog, C=fd—ak, D =
cf — ab.
Then
Sect. V. EXTERMINATION.
123
५
Then put
P=C׃D—gA˚— A׃D—gA ×ffe-kA
2
DxƒD — gAxƒC — bA + A xƒC — bA³,
Q=CxƒD—gA×ffe — kA — A×ffe — kĀ¹²
2
-efxfDgA × ƒC — bA,
R = ef × ƒD — g A²-DxƒDgA× ffe
kA
+ Ax ffe — kA x fC-bA.
Then PQ+ RR = 0.
Or thus,
Put EfD-gA, Fffe-kA, G =ƒC—bA,
PCEAF x E+ AG-DEX G,
Q = CEAF x F-feEG,
RfeEDF x E + AFG.
Then PQ+ RR = o, as before.
For, multiplying the firſt equation by ƒ, and the
laft by a, the difference is Ax + Dx² + Cx+ef=0.
And fince fx³ + gx² + bx+ k = 0; it will come un-
der Rule 5, in which write A, D, C, ef, for a,
b, c, d refpectively; and likewiſe ƒDgA,
ffe kA, and ƒC-bA, for A, C, D, refpective-
ly; and the rule will appear.
Ex. 2.
Let xx+5x-3yy=0,
5×
and 3xx-2yx + 4 = 0,
to exterminate x.
By Rule 2, a = 1, b = 5, c =
3yy, ƒ = 3,
8 = 2y, b = 4, and A = 15 + 23, B = 20—6y³,
D =
9yy 4.
Then AB + DD = 15 + 23 × 20 + 6y³ +
2
-9yy 4 = 300+ 40y—90y² — 1 234 + 81у4 +
72y²+16= 0.
Exi
124
B. I.
EXTERMINATION.
Ex. 3.
Suppoſe y³ - xyy — 3x0,
and y²+xyxx + 3 = 0,
to expunge y.
Here by Rule 3, a=1, b=—x, c=o, d=—3x.
and ƒ = 1, g = x, b = − xx + 3.
Ax-x=2x, D=xx
alfo
ƒD—gA
f
-
3.
fD-gA = xx 3 + 2xx = 3xx 3,
bD—fdg — — x++6x²→→9+3x² ——x++9x²—9
dffbA
3x2x³+6x=3x-2x³.
Then 3xx-3X−x49x²—9+3×—2׳1² = 0.
Or,
3x5 +27x427x²+3x4-27x² +27 +9x²
12x4 4x6 = 0.
And reduced 618x445x² + 27 = 0.
x
Let y4
Ex. 4.
3x³y + 3 = 0,
and 2y³ + xy² - 4x³ = 0,
to expunge y.
4x³.
Whence
By Rule 6, a=1, b=o, c=o, d=-3x³, e=3°
f = 2,g=x, b = 0, k =
Then Ax, C=
Exx, F 12
=
2x³, Do.
4x4, G-4x³. And
4x5 Xxx + 4x4 X
3
16x7 = 12x³
6x7
P
2x512X
=12x¹
Q =
2x5 + 12x
12X
4x5 X 12
4%3
22x',
4x4 + 24x5
6x5 X 12 4x4 + 24x5
14496x5 + 24x³,
R=6x++ 4+ X 12 -
4.2454241642
Whence
Sect. V.
125
DESIGNATION.
Whence
PQ+ RR = 12x³-22x7 X 144x96x5 +24x9
8
+54x+16x³)² = 1728x4—4320x8+
2400x¹²—528x16+2916x*—1728x¹²+256x¹ =0.
Reduced, 68x12 — 168x8 + 351x4—432
SCHOLI U M.
=
0.
In the folution of determined problems, you will
often have three or more equations, involving as
many unknown quantities. Then theſe muſt be
exterminated one after another, by degrees, by re-
peating the foregoing rules, till at laft there re-
mains only one unknown quantity contained in one
final equation. But a perfon uſed to theſe forts of
computations, will often find fhorter methods than
by theſe particular rules, but the finding thofe, is
only to be attained by conftant practice.
PROBLEM
LV.
To defignate or denote any affections of literal quanti-
ties, as fums, products, &c.
RULE.
The original quantities being written down; any
affections of them, as fums, differences, products,
quotients, &c. are got by the rules of algebraic
addition, fubtraction, multiplication, divifion, &c.
before laid down.
Ex.
126
B. I.
DESIGNATION.
Ex. 1:
There are two quantities, a the greater, and e the
leffer, to find the fum, difference, product, &c. as
follows.
The fum
..
difference
product
greater divided by the leſs
leffer divided by the greater
fum of their ſquares
difference of their fquares
fum of their fum and diff.
diff. of their ſum and diff.
prod. of the fum and diff.
fquare of the fum
fquare of the difference
fum of the fquares of the
fum and difference
difference of the ſquares of
the fum and diff.
fquare of the product
cube of the greater
cube of the leffer
cube of the fum
cube of the difference
a te
a
a e
a
e
e
a
aa+ee
aa- ee
2a
2e
atexa-e, or aa-et
aa + zae + ee
aa
Qae + ee
2aa + 2ee
4ae
aaee
a3
C3
a³ +3a²e + 3ae² + e³
a³ — 3a²e + 3ae² — e³.
Ex.
Sect. V.
127
DESIGNATION.
Ex. 2.
There are two quantities, whofe fum is b, and the
greater is a; what is the leffer, the difference, &c.
Leffer
difference
2a
Ъ
product
ab-
aa
greater
by the leffer
a
b.
a
2aa+bb-2ba
zba
bb
fum of their ſquares
difference of their fquares
fum of the fum and difference 2a
difference of the fum and dif-
ference
product of the fum and dif-
}20—20
ference 24
zab
bb
4aa
4ab + bb
4aa
fquare of the difference
difference of the fquares of the
fum and difference 4ab —
fquare of the product
!
a²b²
2ba³ + a4
Ex.
128
B. I.
DESIGNATION.
1
Ex. 3.
There are two quantities the greater is a, and
the greater is to the leffer as r to s, what is the
leffer, &c.
The leffer (r: s :: a :)
the fum
difference
product
1
fum of the ſquares
}
A
sa
r
sa
a t
sa
a
*
11
saa
r
ssaa
aa+
rr
ssaa
difference of the fquares
a a
gr
greater divided by the leffer
product of the fum and differ. a a
fum of the fquares of the fum'
and difference
S
ssaa
rr
fum}
2ssaa
zaa+
rr
}
4saa
r
difference of the fquares of the fum
and difference
the fum divided by the greater
the difference divided by the leffer
I
до
I.
Ex
Sect. V.
129
DESIGNATION.
Ex. 4.
The product of two quantities is p, and the leffer
is e, what is the greater, &c.
Greater
ium
difference
0
8
P
e
༤
e
P
+ e.
leffer by the greater
fum of their ſquares
difference of their ſquares
•
fum of the fum and difference
diff, of their fum and diff.
ee
P
PP
+ ee
ee
PP
ee
ee
2P
e
- 2e
ſquare of the fum
fquare of the difference
diff. fquares of the fum and diff.
the fum by the difference
PROBLEM
PP
ee
+ 2p + eë
2p + ee
}
PP
ee
4p
4P
p + ee
p-
ee
LVI
To keep a fhort account of the steps in any operation.
In long and tedious operations, it is neceffary to
fhew, how one ſtep is produced from another, or
K
one
$30
TRACING THE STEPS. B. I.
one equation derived from other foregoing ones;
which to explain in words would take up a great
deal of room. Therefore the method of tracing
the ſeveral ſteps, will be beft done by regiftering
them in the margin.
RULE.
Against every step write the numbers 1, 2, 3,
&c. in order, and fet down, in the margin on the
left hand, the ftep or steps in figures, that each
ftep is produced from; with the figns +X,
c. according to the feveral operations, ufed; by
which means one may fee at one view how any
equation comes, or is produced; and when an ab
folute number is registered, it must be put in a pa-
renthefis (); and if any quantity is added, fub.
tracted, &c. it must be put down.
Let
Example.
112 + e = b.
24
e
= C.
I + 2
1
2 2
I X 2
312a = b + c
ize = b
baa
I 2 6
I lw 2p
4 &
G
2p
3 + 7
4 X 5
=
a te
a
е
C
ee
be
b
C
7 √o +
at e =
ск
4ee = bb
2bc + cc
9/2a + √a+e = b + c + √b
10/2Дaе 2e3bbc bcc
112a + 4 = b + c + 4
3 + (+)
e
b
C
4 (4)
12
2
4
b_
9-√ate 132a = b+c+ √b - √ate
་
3 = 13 I ab + c = b+c+ √b-√ata
&c.
EXPLA
1
Sect. V. TRACING THE STEPS. 131
:
EXPLANATION.
1+2 fignifies that the third ſtep is found by
adding the first and fecond ſteps together. 1-2
fignifies, the fourth ftep is got by fubtracting the
fecond from the firft. Likewife, the fifth ſtep
(1X2) is had by multiplying the firft and fecond:
the fixth ftep, by dividing the first by the fecond:
the feventh, by extracting the ſquare root of the
firft: the eighth (402p) is had by fquaring the
fourth the ninth (37), by adding the third
and ſeventh ſteps: the tenth (4 X 5), by multi-
plying the fourth and fifth fteps: the eleventh
(3+(4)), is had by adding the number 4 to
the third ftep: the twelfth (4(4)), fhews that
it is gained by dividing the fourth ftep by the num•
ber 4 and the thirteenth (9a+e), is had by
fubtracting a te from the ninth the four-
✔a+e
teenth (313) is got by making the third and
thirteenth equations equal; and fo for others.
K 2
SECT.
132
*
A
SECT VI.
Infinite Series.
N infinite feries is formed, either by actually
dividing any fractional quantity having a
compound denominator, or by extracting the root
of a furd; and fuch feries being continued will run
on ad infinitum, in the manner of a decimal frac-
tion. And in many cafes the law of the progref-
fion of the terms will be evident, by obtaining à
few of the foremoft; and confequently may be
continued without actually performing the whole
operation.
7
PROBLEM LVII.
To find the value of a fraction or furd, to be defignated
by an infinite feries.
IRUL E.
Proceed in the fame manner as is taught in
Prob. iv. Rule 2. for divifion; or in Prob. vi.
Rule 2 and 3, continuing on the operation at
pleaſure.
Ex.
+
Sect. VI. INFINITE SERIES.
133
}
Ex. 1.
I.
í
ax
Let
be given.
a
X
XX
x) ax
( x +
a
ax
XX
+ xx
+ xx
a
+
+
*||s &
X3
a
X+
aa
x4
+
aa
Therefore
X3
/ *
+
x4
аз
ཅི|
+
ax
x +
+
aa
६
+
213
aa
+
&c.
24
ི།
+ &c.
}&
24
+
05
&c.
ad infinitum.
Exi
K 3
134
B. I.
INFINITE SERIES.
1
Let the fraction
o
b + x) aa + 0 (
aa +
aax
b
Ex. 2.
be proposed.
aa
b + x
aa
aax aaxx
aax³ aax4
+
+
&c
bb
63
b4
bs
Anſwer.
}
-aax
+o
b
-aax
aaxx
b
+
bb
bb
aaxx
1
aaxx
a²x3
+
+
bb
63
-32x3
&c.
Ђз
Or thus,
aa
ŵ + b) aa + 。 (aª — aab + a²b² — ab
o
X2
a2b3
X4
&c.
a2bz
213
baa
aa +
X
-baa
-aab
aabb
X
XX
+
aab²
XxX
&c.
Ex.
Se&t. VI. INFINITE SERIES.
135
}
Ex. 3°
Suppoſe
I
I + xx
×4
+ x8
I + xx) I + O( I − xx + x + — x6 x³ — &c.
I + xx
XX
XX
x4
+x4
+x++x6
доб
226
Ex. 4.
Let the fraction be
I
I
४
x8
४
+xs &c.
I
2x2
I
X
MN
I
1 + x²-
2
·3%
2x + 7׳ — 13x²
1 + x²---- 3x) 2x²— x² +0 (2x² -— 2x + 7׳
I
S
+ 34 x = &c.
2x² + 2x
6 x ž
- 2x + gx²
2x
2x² + 6x²
+ 7 x 212
3
6x2
3
2
+7x² +7x²
- 21X
K 4
5
13x² + 21x ?
13x² — 13x *
I
+ 31"
NC.
Ex.
136
B. I.
INFINITE SERIES.
Ex. 5.
Extract the Square root of aa + xx.
aa + xx
XX
+
X4
26
+
5x8
24
8a³
a5
12797
XX
aa
za + **) 0 + xx
20+
24
X4
+ xx +
4aa
XX
4
114
С
8a
)
4aa
+ &c. = √ aa + xx
24
208
+
4aa
8a+
= 496
2a +
XX
X4
a
893
&c.) +
210
23
8a3
649°
X6
+ +
8a+
48
&c.
16a6
578
6486
&c.
Here fuch terms are neglected whofe dimenfions
exceed thofe of the laft term
root is to be continued.
50
128a7
و
By the fame way it may be
to which the
By the fame
extracted in this form
a a
a4
xx + aa = x +
2X
8x3
-
&c.
16x5
1
Ex.
Sect. VI. INFINITE SERIES.
137
Ex. 6.
Extract the cube root of 1
x³.
X3
From 1
take
5x9
X3 I
&c.
I
3
9
81
3
-3) - x³
From
take I
I ·
3).
४
3
26
3
From I-x³
take I-
X3 *
+
206
X9
X3
3
I
3
3
9
27
27
M
5x9
+
&c. =
X3
४
xo
= [
27
3
9
3)
5x9
&c.
27
2 RUL E.
Affume a feries with unknown coefficients, to re-
preſent it. Which feries being multiplied, or in-
volved, &c. according as the queftion requires;
the quantities of the fame dimention must be put
equal to each other; from which equations, the
coefficients will be determined.
Ex. 7.
I
Let
be given.
a
X
I
Suppoſe
= A + Bx + Сx² + Dx³ + Ext
a
*
&c. the feries required. Multiply by
Multiply by a -x. Then
I
138
B. I.
INFINITE SERIES.
1 = aA + aBx + aСx² + aDx³ + aEx¹ &c.
Ax Bx2 Cx3 Dx4 &c.
Whence equating the coefficients of the fame pow
ers of x, we have cA1, aB — A = o, aC — B
=0,aD-C=0, aE-Do, &c. Therefore A
I
I
A
B =
1, C=
B
C
D=
ร
I
a
aa
a
Q3
a
a4
D
I
E =
a
feries is
$188
aa
+
218
I
+
a
+
1
a4
&c. by reduction. Therefore the
X
aa
+
+
24
95
+ 2
I
I
&c. or
+
a
a
+ &c.
Let
CC
Ex. 8.
be given.
cc + 2cy-yy
Suppoſe it = A + By + Cy² + Dy³ &c. Mul-
tiply by cc2cy-yy. Then
—
ccccA + cc By + ccCy²+ ccDy³ &c.
+ 2cAy + 2cBy² + 2cCy³
Ay²
By³
3
And equating the homologous terms, ccccA,
ccB + 2cA
ccc2cBA 0,
= :
ccD + 2cC — B = 0, &c. and by reduction,
2 A
2
A
2cB
A = 1, B
C =
C
C
CC
I + 4
5, D =
B — 2cC
-2
10
CC
CC
CC
CC
&c. Whence
≈ al
215
+
5 y²
cc + zcy-yy
C
CC
1
12y3
&c.
63
Ex.
Sect. VI. INFINITE SERIES. 139
What is ✔aa
Ex. 9.
XX.
Let Vaa-xx = A + Bx² + Cx4 + Dx &c.
which being ſquared,
2
aa — xx = A² + 2ABx² + B²x4 + 2ADx &c.
+2ACx4 + 2BCx
Here Aaa, 2AB —— I,
2AD + 2BC= o, &c.
1, BB + 2AC
BB+2AC = 0.
Whence A = a,
BB
I
T
I
B =
C =
2 A
2a
2 A
8a3
BC
I
D =
2
&c. Therefore Vaa-xx
A
Ibas
XX
24
доб
=a
&c.
LVIII:
24 803 16as
PROBLEM
To reduce any binomial furd to an infinite feries, or
to extract any root of a binomial.
RUL
ULE.
This is done by fubftituting the particular let-
ters or quantities, inftead of thefe in the following
general form, duly obierving the figns.
A
B
m
m
m
P + PQ|" = P
D
m
211
371
m
C
+AQ + "=" BQ
E
+ "CQ + " = 3" D Q + &c.
422
2n
Where P is the first term, Q the ſecond term
divided by the first,
m
or root, A B, C, D,
with their figns.
22
the index of the power
&c. the foregoing terms
!
Ex.
I 40
B. I.
INFINITE SERIES:
Ex. 1.
Extract the Square root of rrxx.
Here Prr, Q =
-XX 772
I
There-
диза
12
2
-xx
MM
fore rr-
xx|² = r +
A X
B x
2
rr
3
-xx
-XX
X
6
Сх
DX
&c. =r
A
rr
2rr
XX
+
B + c+
3xx
EXX
D&c. that is, reftor-
4rr
orr
8rr
ing the values of A, B, C, &c. rr
XX -
Xx
204
5x8
&c.
27 8r3 1675
12877
Ex. 2.
дида
r + x
What is the value of
Here
vy
• + x
= rr x r + xl and P = 1, Q==",
9
I, n = 1. Therefore
go
m
n
I, or m
·I
X
x
r + x1
=7-
IAX
IB X
r
X
X
IC X
r
།༞།*
X
B-
› D x = &c.
C &c. And rr x r + xl
xr
I
-A
I
I
=rrx:
g
203
z 2
213 +
+
XX
X3
rz
XX
&c; that is,
x +
добр
r + x
X4
&C.
g3
Ex.
Sect. VI. INFINITE SERIES.
141
Ex. 3.
I
To find the value of Trx-xx
I
I
XX
27x
xx, and P = 27x;
27x
Q =
2rx
xx]
X
27
-플
​by N
XX
m
= 2rxl
플
​MIN
I,
n = 2. Then
I
2
1÷4
X-
3
AX
BX
27
4
27
-동
​-X
-X
1
сх
DX
&c.
=
2r
2r
2rx
+ 11
-A +
X
+ A
4r
I
1/2rx
+
√2rx
X: 1+
&c.
X
4r√2rx
3B +
5x
8r
127
C + 7x D + &c.
16r
D+
X
3xx
I
+
&c. =
32rr2rx
√2rx
3.5x3
anfox
+
+
3.5.7x4
+
4r
3272
4.8.1273
4.8.12.1674
Ex. 4.
What is the cube root of 1-µ³.
Here P1, Q = — x³, m = 1,
=
I
x3.
n = 3. Whence
I
3
3
x ³ 1 ³ =
· = 1 +
3
— A × − ׳ —
X X3
2
— B×—x³ —
5
8
II
9
CX X3
DX — x³
EX-
x³ &c.
12
15
3
X3
==
-A +
5x
B +
2x3
2X
11x³
C+
D+
E
3
3
9
3
15
3
X 3
xo
&c. that is,
3
5x9
X
Į
3
81
10X12
22x15
&c.
243
729
}
{
Ex
142
B. I.
INFINITE SERIES.
What is 3/
aa
Ex. 5.
2
in an infinite feries.
aa + xxl
3
This reduced is a xaa + xxi. Here P = aa,
XX
Q =
II
12
+
D
aa
aal
aa
2
2
110x8
243a93
m = — 2, n = 3. And aa + xxl³
aa
1
2
XX
A
3
동​B
Xx
8
C
XX
aa
9
aa
XX
&c.
I
2XX
==
5x4
a
3 a ³ 35
+
40x6
3
9a53
81973/3/30
I
&c.
I
2x2
I
X:
5x4
a
a
3a3
9a5
40x6
I 10X8
+
8107
&c therefore
a 33
2
a3
243a9
aa
+ xxl
I
2x²
Vaa
X : I
+
5x4
40x6
3a²
+ &c.
9a4
8125
Ex. 6.
What is the value of Vaa xx.
5
t
5
a a
xx = aa
xxl s
Here
Paar.
Xx
Q =
m
>
I, n
5. Therefore aa
xxl³
aa
2 CX
= aa|³ + — -A X
-XX
5
-xx
4
10
AB X
-xx
В
ал
aa
14
-XX
DX
15
20
aa
aa
2XX
+
5aa
5aa
&c. = a ³
³
C + 7xx D &c.
3xx
B +
10aa
===
2
5aa
a3 x :
2
XX
A
XX
2X4
6x6
21x8
I
&c.
5aa
2564
12520
625a8
Ex.
Sect. VI. INFINITE SERIES. 143
}
}
Ex. 7·
4
To reduce a + x x √ a — x to a
x to a feries.
Va — x = a - *|². Where P = 4,
4.
Then a
4
Q =➡,
X-
m = 1, n = 4.
Then a — x² = a + A ×
7
3BX
12
8
a
3x
8a
4
- Bx-/Cx= &c. = a* —
+ B + 7x C &c.
124
a
-X
x
сх
A
a
40
I
X
322
7x3
3
4a+
32at
I I
&c.
128a
Multiply by a + *
I
5
Then
a4x
+
a
32.2
7x3
4 32dt
&c.
128a7
x+2
+ a*x
3x3
4a
fax
32a4
&c.
a + x x √ a = x = a $
4
3a²x
IIX²
aª +
19x3
3
&c.
7
4
32a
128a
4
= √/ax: a +
3x I Ix²
2
19x3
-&c.
4
32a
128aa
Ex.
144
B. I.
INFINITE SERIES.
Ex. 8.
To defignate↓
aa + xx
a a
aa + xx = aa + xxl.
XX
by a feries.
Where P = aa,
m = 1, n = 2, and aa + xxl² = a +
LB 3/C
XX
Q
ea
LA
XX
XX
2
aa
ad
Xx
24
x6
+
&c.
24
16as
XX
&c.
= a +
aa
i
Q
Again, Jaa
-xx
I
aa
XX
xx. Here Paas
XX
n
"
I, n = 2.
2. And aa
HIN
xx
aa
I
I
-xx
11
A X
a
2
3B X
-NX
aa
4
aa
// c x
-XX
I
XX
& C
====
+
+
aa
a
24
3x4
+
5000
8a5
Ibar
&c. Whence ✓
aa + xx
aa
XX
√ aa + xx
XX
X4
205
or
XX
I
XX
+
2 A³
3
✓ca-
X +
244
a
+
25
286
&c. by multiplication.
=a+
+
&ca
2a
баз
16as
3x4
+
546
xx
&c. = 1 +
805
16a7
aa
Ex.
Sect. VI. INFINITE SERIES.
145
Ex. 9:
What is the value of aa
ax
ax + xx
Put
This may be treated as a binomial.
Then aa ax + xx = aa j. And
y = ax xx.
ax =ax × aa—y. Here Paa,
ax + xx
aa
Q =
aa
-y
m
,
I
And aa—yl +
+...
- IC x2
y
aa
aa
او
1, n = 1.
-1D X 2
y
IA X Y
3c. =
c+ D &c.
aa
aa
I
い
​y
1B X
+ A +
24
aa
+
yz
ー​ツ
​aa
y
aa
30 x
B
J:3
$100
aa
-y
&c. =
aa
aa
I
+
aa
y4
of
&c. =
I
aa
+
a4
ax
X
ax
xx15
3
÷
ax
+
f
xxl
4
as
a
98
&c. which
аго
+
(by reftitution)
ax
a
xx
involved and reduced into order will be
I
X
and
+
aa
a3
04
XX 2x3
4
+
ི་
+
+
હું **
as
3x4
&c.
X4
+
&c.
26
Ι
X
+
> હું કહું છું
*
X3
X4
&c.
♦
as
26
|
+
*
X4
X5
&c.
04
as
L
The
ax
aa
aa —ax + xx
[
146
B. I.
INFINITE
SERIES.
The truth of this rule will appear by induction.
For if any of thefe feries be involved according to
the index of the root, it will produce the original
quantity. Thus if r-
XX
x4
27 8r3
&c. be fquar-
ed, it produces rrxx, as in Examp. 1.
2.6
If
&c. be cubed it produces 1-x3, Ex.4.
X3
I
3
9
XX
2x4
If a³ X : I
:
&c. be involved to
1
5aa 25a4
the 5th power, it gives aa-xx, Ex. 5. and the
like of others.
m
m
m
Cor. 1. P+ PQ|"
=P" X: I +
LQ +
n
m
n
m
m
12
m
2n
X
Q² +
X
X
Q³ +
12
2n
n
m
m.
12
2n
}
X
X
X
n
2n
Cor. 2. a + x\"² = a² +
372
m
2n
4n
3″ Q+ &c.
3n
7X
n + I
A+
X
B
a + x
2
a+x
n+ 2
X
n + 3
X
+
X
C +
X
D+&c.
3
a + x
4
a + x
where n is any index;
terms with their figns.
A, B, C, &c. the foregoing
X
ay
For put y =
then x =
and
a +x
I
-y
72
a
a + x =
Therefore a + x1"
a
==
I
y
a" X 1
Here P1, Q = −y, m = n;
=
y". Here P
n = 1 (fee Prob. xlix); then by this problem,
I
او -
-yl
N
Ах
= 1
y
[
n + 2 Cx-y-
3
CX
I
n + 1 B x − y
2
n + 3 D x − y &c.
4
1 + ny A
Sect, VI. INFINITE
147
SERIES.
I + ny A +
n + I B y +
n2 + 2
Cy+
n + 3 Dy
2
-12
&c. and a" X 1—y"
=
4
a" x 1 + ny A +
:
(reftoring the value
3
n+i
n + 2
By +
Cy &c.
2
n
3
of y) a X: 1 +
NX
n + I
A +
X
-B+
a + x
2
a + x
22 + 2
X
X
C+
n + 3
X
X
D &c.
3
a + x
4
a + x
PROBLEM
LIX.
To involve the feries z ×: a + bx + cx² + dx³ + ex4
&c. to any power m, whole or fractional.
RULE.
Subſtitute the particular letters or numbers in
the given feries, inftead of theſe in the following
general form.
Z
m
z" × a + bx + cx² + dx³ + ex+ &c.]™
x into a
mb A
+
X
a
+
2mċ A + m
1 . bB
24
+
Befo
+
3mdA+2m-1 . cВ + m — 2. bC
за
4me A+2m-1.dB+2m-2.cC+m3.bD
4a
203
X4
5mfA+4m-1. eB+3m-2. dC+2m—3. cD
+ m
4.be
50
L 2
+ 6mg A
1
<
148
B. I.
INFINITE SERIES.
+
6mg A+5m—1 .ƒB+4m—2. eC+3m−3 ⋅ dD
ба
+ 2??? 4. cĘ + m−5. bF 26
+
7mbA+6m+1.gB+5m−2.ƒC+4m−3 ⋅ eD
+372
7a
4.dE+2m-5.cF + m
6.bG
x7 &c.
Where A, B, C, D, &c. are the coefficients of
the terms immediately preceding thoſe wherein
they first appear. And the law of progreffion is
evident.
Ex. I.
What is the Square of 1 + x + xx + x³ + x4 +
&c.
d = 1, &c.
c d
:
Here ≈ 1, a = 1, b = 1, ( = 1,
z
And m2, then 1+x+x² + x³ &c |² =
2 A
i +
of
x +
2
2
3
4A + B 6A + 3B + 0 ׳ +
x+ &c.
C
D
E
8A+ 5B + 2C - D
A B
4
= 1 + 2x + 3x² + 4x³ + 5*4 &c.
Ex. 2.
What is the Square root of 1 + x + xx + x³ &c.
Here ≈
z
1
1, a = 1, b = 1, c = 1, d= 1 &c.
and m ==. Whence 1 +∞ + x² + ׳ + ** &c.]²
2
A-B
A+0-C
Ax+
x² of
X3
2
2
3
2 A + B — C
D
3
of
+
x+, &c. = 1+ - - x + 3
8
24
4
+ 5 x³ + 35 x 4 &c.
16
3
128
Ex.
Sect. VI. INFINITE SERIES.
149
Find the cube
Here z = 1, a =
Ex. 3.
of 1 + x + x²
+ x³ &c.
Etc. m = 3.
Then
1, b = 1, c = 1, d = 1, e = 1,
+ x + x² + x³ &c. 1³
3
1 +
3 Ax
6A2B
9A + 5B + C
2
+
x² +
x³ +
2
3
12 A + 8B + 4C
4
1544 + &c.
x+ &c. = 1 + 3x + 6x²+10x³ +
Ex. 4.
What is the value of
2
I
rr zyy +
y4
5
ys
ys
8
472
+
&c.
824
1 бро
I
Here ≈ = 1, * = yy, a = rr, b = — —, c = rr
x
2
I
d
>
ļ =
874
16763
&c. m
I. Then
I
14
rr
2 yy f
yo
&c.
=
422
876
rr
I
A+ B
3
A -
3
2rr
814
+ 31
+
A
x +
2rr
B+ 3C
xx +
4rr
N-
x³ &c.
2rr
3rr
11
I
rr
I
+
x + ox² + ox³ &c. =
I
274
+
rr
I
x
2r+
+
yy'.
Ÿr
274
Ex. 5.
$
To Square the feries y-y³ + y -- y7+ y⁹ &c.
This is equal to yx: 1-y+y+ —y° + y³ &c•
Herey, yy, a = 1, b = — 1, c = 1, d= —I,
ند
I
£ 3
e = 1.2
1
150
SERIES. B. I
INFINITE
e = 1, &c. and m2. Then I—y²+y4 &c.l² =
I-2Ax +
4A B
-6A +3B
x² +
x³ &c. =
2
3
I- 2x + 3x²
4x³ &c. and y — y³ —ys &c.[² =
y² × 1 − 2x + 3x² &c. = yy — 234 + 336 — 438 +
&c.
Ex. 6.
To Square the feries.
1
23
√2rx:07 +
3v 2
5
3.5.vž
+
The feries is 2rvl² ×: 1 + +
2.2.3r 4.2.4.5r2 8.2.4.6.7r3
I
+
&c.
ข
302
+
3
с
( =
d =
5
2
>
896r3
I
บ
2rv|
127
3
A +
+ I
40rr
2
503
896r3
127 160r2
I
&c. Here z = √2rv, a = 1, b =
Iбory
2rul² x : 1 + +
GA x +
&c. m = 2. Then
2
127
&c. = 2ru ×: 1 +
3v² &c.
160r2
I
В
12r
x² +
30
A +
9
B
89брз
3
16orr
3
x³ &c = 2 ru X : 1 +
3
$།&
2x2
57x3
+
+
&c.
45rr
4480r3
Ex. 7.
Find the m power of·
• + c x
за
r+2n
+ dx
n
r+3n
&c.
xx: a + bx" + cx²
=
go
n
x, xx, m = 2,
2n
&c.
r + n
ax + b x
This reduced is
dx3" &c. Here ≈
Then
Sect. VI.
151
INFINITE SERIES,
Then xx: a + bx" + cx²² &c. lx
TM
TIL
x: a
mb
+ Ax² +
2mc A + m
1.bB 212
X +
a
20
3mdA +2m
1. cB + m
2. bc
372
+
за
4me A +3m-1.dB + 2m-2.cС + m −3.ED
bD
X42
&c.
4a
2 RUL E.
Subſtitute each letter in the given feries, inſtead
of the correfpondent one, in the following ge-
neral form.
m
zm xa + bx + cx² + dx³ + ex+ &c.1" = 2" X into
m-I
a"
712
+ mbam-x x + m
2
a71-2 bb 2
777-I
+m.
m
2
+m.
m
+ ma C
2
a
m-3 b3
3
I
2am-2
bc
X3
2
+ mam- d
m
I
2 m 3
+ m
m-4 64
2
3
4
I m
+ m
2
3
m
I
+ m
2
+ ma
2
3am-3 bbc
за
m-2
a
m-x
L 4
e
X
S2bd
+cc
X+
+ 2:3;
152
B. I,
INFINITE SERIES.
m
I
m
2
m- 3 m
4
+ ግንጌ
am-5 bs
2
+m.
m
2
+ m
3
m
m
2
3
+ m
4
5
2
m 3
4
I m
2
· 3
m
I
2
m-4 b³c
4am
39-3 Sbcc
2 +bbd
Scd
2 be
m-2
20m-
+ ma'
m-I
f
17
5
4
m
I
m
2
72
3
+m.
2
3
4
m
1. m
2
m
3
m
+ m
2
3
4
5
m
I
772
2
m
+m.
2
3
m
+ m.
1
N||
+m.
4
m
3
3
132
6
5 712-
a.
.6
Z6
5am-5b4c
m-4
a
2
772-3
a
I
2
a
M-2
6bbcc
{4b3d
3bbe
6bcd
C3
S2bf
+ mam-i g
2ce
dd
&c.
доб
For let y bx + cx² + dx³ &c• p
m
2
q =
3
m
p₁r = " = 3 q, s =
4
m
=m,
I
m,
2
m 4
r, &c. Then
5
a + bx + cx² + dx³\" &c. = a + yl" = am+mam-1 y
+ pam-2 y² + 9am-3 y³ + rom-4
3 = bx + cx² + d∞³ + e~4
yy = bbxx + 2bcx3 + 2bd
+ cc
X4
M4
y4 &c.
&c.
But
+ zře
+2cd
x5, &c.
y3 = b3
Sect. VI. INFINITE SERIES.
153
y³ = b³x³ + 3bbcx4 + 3bbd x5 + &c.
+ 3bcc
34 = 64x4 + 4b³cx² + &c.
jsb5x5
6505
&c.
Then the power a" + ma"-ly + pam-2xy
qam-³y³ &c. becomes
да
ат
m-I
+ ma x: bx + cx² + dx³ + ex4 + fxs &c.
+ pa
+qa
712-2
M-3
M-4
X:
X:
+ ra X:
M-5
- så X:
&c.
bbx² + 2bcx² + 2bdx4 + 2bexs
+ cc +2cd
b³×³ + 3b²cx4+ 3bbdxs
+ 3bcc
b4x4 + 46³cx³
bsxs
Theſe being actually multiplied, and the coef-
ficients of each power of collected; will give
the feveral terms as in the form above.
And the firſt Rule is in effect the fame as this:
For let a + bx + cxx + dx³ &c.l" =
2
A + Bx² +
Cx³ + Dx4 &c.
Then by Rule 1, A,
a", as in Rule
mb A
2d. Alfo B =
= mba
M-I
as in Rule 2d.
a
McA
m
Likewife C
+
a
24
m
7712-2
mbba
>
as in Rule 2d.
bB = mca
7-I
+
2
Again
1
154
B. I.
INFINITE
SERIES.
2m
Again D =
m-i
= mda +
M-2
mca +
m
2
mdA
+
a
2m
I
3
I
за
X mcba
mb ba™-3
I.CB
772-2
+
+
M-I
m
2
ЂС
3
= mda +
× mbcam-2 + m
за
2
b x
m-2
2m
I
3
am~3
+ m
+
3
m
2
63.
2
3
m
I
77-2
m-I
+ m.
zbca
+
2
m
I m
2
= mda
2
fo for the rest.
3
123
¿³am-3, as in Rule 2d, and
In ufing this laft rule, it will
be the eaſieſt way to divide all by the firft term,
that a may be 1.
Ex. 8.
What is the fourth power of 1 + x + x² + x³ &c.
Here z = 1, a = 1, b = 1, c = 1,
I, &c.
d = I,
4
m = 4. Then I + + x² + x³ &c.]+
= 1 + 4bx+ 6b²x² + 4b³x³ + b²x4 &c.
+ 46 +12bc + 12bbc
+ 4d
+ 6cc
+12bd
+4e
= 1 + 4x + 10x² + 20x³ + 35x4 &c.
Ex. 9.
What is the Square of
I
X
+
+
XX
I
I
X
X
X
= 1, &c. m
2.
In this Example, z=
c = 1, d d = 1,
I
X3
I
+ &c.
X4
a = 1, b = 1,
b=1,
Then
Sect. VI. INFINITE SERIES.
155
I
1
Then
+
+
&c. |
I
=
X:
X
XX
I
X3
I
XX
I
I
X
XX
+2c
I
=
X: 1+
+
XX
3 + 26 × — + bb × 1 + 2bc × 1 + 2bd× — &c.
2/8
3
XX
+ 2d
+
4
263
+
X3
+ cc
+ 2e
5
1
&c. ==
X4
XX
2
3
4
+
+
X3
+
+
&c.
X5
200
Ex. 10.
To Square the feries y —y³ +ys —y7 + &c.
Here zy, a = 1, b = 0, c =
e = 1, ƒ = 0, g-1 &c. and m = 2.
=
y — y³ + ys &c.]²
I + ox
-
20x2
+0x5
&c.
28
= y² × :
Ox³ + ccx4
+ 2e
1,
d= o,
d=0,
Whence
2cex &c. = y² X: 1 — 2y² + 334 — 476
=y? — 2y+ + 336-438 &c.
Or thus,
z = y, x = yy, a = 1, b = − 1, c = 1, d= — 1,
&c. and m = 2. Then y-y+ys &c." = yy x.
2
1 + 2bx+bbx² + 2bcx³ + 2bdx4 &c.
+ 20 + 2d + ce
+ 3c
x:
= y² × : 1 — 2y² + 3y4 — 4y6 + 5µ³ &c. = y² —
2y+ + 3y6 — 43·8 +570 &c.
Ex.
156
B. I,
INFINITE SERIES.
Ex. II.
What is the fquare root of rr-zz +
Z8
+
&c.
3156
226
3r2
45% +
70100
Here z = 1, x = zz, a = rr, b = −1, c
=
=3
3r²
-2
I
d =
e =
>
&c. and m
4524
31576.
2
112
2
I
&c. Then rr
I
M
2
I
m
3
4
3
2
4
Z+
N
I
L
ZZ +
&c.
=r+ X
४
3rr
2
r
I
I
ZZ
Z4
x² +
x² &c. =r.
&c.
873
6r3
?r + 24r3
Rather thus,
The quantity reduced is rr x: 1-
ZZ
Z4
+
rr
зна
226
4520
&c. Here z = rr, a = 1, b
I
==
rr
2
d =
2
&c. Whence
324
rr
Zz+
24
45%G
&c.]²
Ι
HIN
2
X
rx : I
3rr
2rr
I
I
x²
x3
82+
подой
x³ &c. =rX : I
X
2rr
I
I
1
+
+
+
+
I
x²
1276
x³ +
247+
72026
&c.
4520
ZZ
ལྟུ¢
zo
f
+ &c.
1.27
1.2.3.473 1.2.3.4.5.6.rs
ENA
Sect. VI. INFINITE SERIES.
15
Ex. 12.
What is the Square root of
I
ZZ
Z4
+
2
Z6
૮૪
+
&c.
8 доб
4rr 674
The quantity reduced is
I
I
X
rr
ZZ
Z+
I
+
&c.
2rr
424
бро
I
I
Where z =
x = ZZ, a = 1,
b =
rr
2rr
I
-I
C =
d =
676
&c. and m
2
3 M-2
5 M-3
7
2
>
3
4
100
&c.
I
And
X.:
ZZ
rr
+
&c.
2
4rr
X
I +
+
322
+
5218
&c.
4rr
3274 12876
1
3
Ext
327°
I
-
+
1275
I
X
+
I IX3
+
&c.
473
3275
38427
SCHOLIU M.
From this problem the powers of a compound
quantity are deduced as follows, which will be
ſerviceable upon particular occafions.
If
58
B. I.
INFINITE SERIES.
If y = A+B+C+D &c. Then
y=A+B+C+D+E+F+G+ H
²=A²+2AB+2AC+2AD+2AE+2AF+2AG+2AH &c.
+ BB +2BC+2BD÷2BE÷2BF + 2BG
+ CC+2CD +2CE +2CF
+DD + 2DE
¿ƒ³—A³+3A²B+3A²C+3A²D+3A²E+3A²F+3AAG &c.
+3ABB+6ABC+6ABD+6ABE+6ABF
+ B³
+3ACC+6ACD+6ACE
+3BBC +3BBD +6BCD
+3BCC +3ADD
+3BBE
+ C³
ƒª—Aª+4A³B+6A²R² + 4AB³ + 6A²C² + 4B³C &c.
+4A³C +12A2BC+12AB2C+12ABC2
+ 4A³D +12A²BD+12AB²D
+ 4A³E +12A CD
+ B4 +12A2BE
+4A³F
y5=A³+5A4B÷10A³В²+10²³ + 5AB÷ + B³ &c.
+ 5A4C +20A3BC +30A2B2C+20AB³C
+ 5A4D +20A³BD+30A2BC²
+10A³CC+30A2B2D
+5A4E +20A³CD
+20A³BE
+5A4F
➡A6+6A5B+15A4B2+20A³B³ + 15A²B+ &c.
+ 6A°C +30A+BC +60A3B2C
+ 6A³D +30A4BD
3
+15A4CC
+ бASE
7=A7+7A°B+21A5B2+35 A+B³ + 35A³B4 &c.
+7A°C +42A5BC +105A4B2C
+ 7A´D + 42A BD
42A³BD
+ 21A5CC
+ 7 A°E
y³—A®+8A?B÷28A°B²+56A³B³ + 70A+B+ &c.
+8A7C +56A 6BC+168A5B2C
+8A'D + 56A6BD
+ 28A6CC
+ 8A7E
y⁹A
Sect. VI. INFINITE SERIES.
159
yº≈A⁹+9A³B+36A7B²+84A6B³ +126AB4 &c.
+ 9A°C +72A7BC+252A6B2C
+ 9A³D + 72A7BD
3
+ 36A7CC
+ 8A8E
¸¹º—A¹º+10A9B +45 A° B²+120A7B³ +210A°B4 &c.
+10A°C +90A8BC +360A7B¹C
&c.
+ 10A⁹D + 90A³BD
+45A°CC
+ 10A9E
In making uſe of any of thefe forms, the
terms of the given feries muſt be ranged in or-
der (Prob. xlviii.), and the whole terms thereof
fubftituted one by one, in the room of the quan
tities A, B, C, D, &c. (Prob. xlix).
Ex. I.
Let a + bx + cx² + dx² + ex4 &c. be cubed.
A+B+C+D + E &c.
= a + bx + cx² + dx³ + ext &c.
that is Aa, B = bx, &c. Then
(y³) A³ + 3A²B + 3A²C &c.
+3AB2
@³ + 3aabx + 3aacx² + zaadx³ + 3aaex+ &c.
+zabbx² + 6abcx³ + 6abdx4
+ b³ x³ +3accx4
+36bcx4
Ex. 2.
What is the fourth power of
2
༣༩
+
P
2cd
+ + +
A + B + C + D
A+
2
X
P
M
2cd
Then
160
B: I:
INFINITE SERIES:
4
Then y*—*—4³ 2+6xxx — — 4x ×
y=x
443
X
12
&c=
X
XX
+ 4x³ x 1
-I2XXX
20
X3
x
2cd
X5
4
8x²+24+4P
1
32 +240 + 8cd
Ex. 3.
5
XX
II
2
&c.
Involve 2x + 3x² — 4x² + 5x2 - 6x &c.
to the 5th power.
5x²
A + B + C + D + E &c.
= 2x² + 3x½ — 4x² + 5x2 — 6x²² &c.
y³ = 32x² + 80xª × 3x² + 80x³ × 9x¹2
+80x + × 4x
7
= y.
= y.
i z
160x X.12x 2
+ 80x± × 5x2
&c.
II
80xª × 6x³½³ &c. = 32x + 240x² + 720x™½
1920x½-480x¹ &c. that is
3
420x¹³½ &c.
js = 32x² + 240x2. -320x² + 720x
Orys = 32x² + 240x²
I S
2
+ 400x
2
320x
LI
&c.
一​+
1920x125
480x15
320x² + 1120x
13
2400x &c. Here I omit all theſe terms,
where I ſee the index of x exceeds 15.
2
Or
A
Sect. VI. INFINITE SERIES.
161
Or thus,
A+B+ C + D + E + F &c,
7
ร
II
2
= 2x²+0+ 3x² —4x² + 5x² — 6x²² &c. then
ys = 32x² + 0 + 80x1
4
7
2
80xª × 4x² + 80xª × 9x
— 160x* × 12x2
4
+ 80x 2 × 5x2
I I
2
X
80x1 × 6x2
— 32x² + 240x² — 320x + 1120x³-2400x
&c.
}
Ex. 4.
If y = 1 + x³ — 2x, what is y³.
A+B+C+D+E+ F &c.
= 1 2x + x³ + 0 + o &c.
I
ys—1—16x+28×4x²-56×8x³ +70×16x4 &c.
+ 8x³-56×2x++ 168 X 4xs
×
+ 28x6
= I
16x + 112x²
+ 8x3
448x³ + 1120x4 &c.
112x4 + 672x5
=116x + 112x²+
8x3
112X4 &c. that is,
+ 28x6
448x3 + 1120x4
3
y81-16x+112x²-440x³- 1008x4 &c. This
is fuppofing x to be very fmall; but when x is
very great, then x muft begin the feries;
Thus,
A + B + C + D + E + F &c.
2x + I + o &c. Then
µ³ = x²4+0—8x21x2x + 8x¹¹×1+28x¹³×4xx
or y³ = x²4 — 16x²² + 8x²¹ + 112x20 &c.
M
PRO-
162
B. I.
INFINITE SERIES.
PROBLEM LX.
To abridge an infinite feries, or denote it in a short
manner for working.
When a feries confifts of terms very much com-
pounded, or having a great many factors; it is
very laborious to reduce them into numbers. And
when feveral factors in any term are contained in the
fucceeding terms; the work may be fhortened, by
making ufe of the preceding term or fome part of
it, inſtead of ſuch factors as are equivalent to it,
in the following terms; as follows.
I RULE.
Put A, B, C, D, &c. for the firſt, ſecond,
third, fourth, &c. terms of the given feries. Then
to get the coefficients thereof, divide every term
by the preceding one, gives the coefficient of
that term. Whence you will have a new feries.
equal to the former, and fhorter defignated.
ZZ
24a
Ex. 1.
A B
C
D
E
23
325
If z+
242
Z3
+
Then x) (=
2a2
:+3.5.729 &c=Ÿ.
+3.527
2.4a4 2.4.626
2.4.6.8a8
B
- coefficient of B =
A
Α'
23 325
322
C
coefficient of C=
2a* 2.4a44a2
B
325
)
3.527
2.4a4) 2.4.6a° | 6a²
Hence the feries
א
x +
ZZ
24A
522
coefficient of D-
010
› &c.
becomes
322
A +
4aa
522
B +
C+
723
D&c. y
6aa
Saa
Ex.
Seft. VI. INFINITE SERIES.
163
Suppoſe 1 +
+
2
Ex. 2.
v2
1.3.5
23
V4
+
+
+
507-9
1.3
3.5.7
05
&c. = y.
79.11
B
Here
A
야
​v D
3
B
5
E
30 F
50
&c.
2
D 9 E
Then the feries is,
II
||
ข
I +
A +
B +
• c +
30
D +
3
5
7
9
50
5° E +
7ལ
72 F &c. = y.
II
13
Ex. 3.
322
52·3
7x+
Let x
&c.
1.2
1.2.3.4
1.2.3.4.5.6
be given.
-5x3
x ) = 3 +127
3.4)
-7x4
1.2.3.4 ) 1.2.3.4.5.6 5.5.6
And the feries is
X
3 A +
3.3.4
2
-3x
-3x²
1.2
1.2
)
-5x3
1.2.3.4
(
5x
3.3.4
(
7x
&c.
|
-3x12
2
)
5303
(
2.3.4\3.4
5x
B +
5.5.6
7x
C &c.
Or thus,
5
"5x3
-7x+
2.3.4) 2.3.4.5.6 (5.6
M 2
&c.
And
7
V
164
B. I.
INFINITE SERIES:
And the ſeries
= x - 3* A
+
૬×
B +
5
2
3.4
Where A, B, C,
with their figns.
४
5.6
C &c.
&c, are the foregoing terms
Ex. 4.
bz³
3
bzs
bz7
Suppose bz
2.3aa
5.2.4a4
7.2.4.6a6
bz9
9.2.4.6.8.as
&c. d.
Then bz)
-bz3 -ZZ
coefficient of B,
2.3aa2.3aa
-bz3
-bzs ZZZ
)
2.3aa / 5.2.4a4 \ 4.5aa
2.3aa) 5.2.4a+(
-625
5.2.4a4 ) 7.2.4.6a6 (6.7aa
•
And the feries is
coefficient of C,
-bz7 522
coef. of D, &c.
ZZ
32
bz
A +
B +
522
C
2.3aa
4.5aa
6.7a2
+
7zz D &c.
8.9aa
2 RUL E.
If there be fome fingle factor or factors, which
are not in all the terms; fet them afide at preſent.
Then put A, B, C, D, &c. for the remaining
terms; and proceed as before. And at laſt re-
ftore theſe ſingle factors into their proper terms.
Ex.
Sect, VI. INFINITE SERIES.
165
$
Ex. 5.
3x²
5x³
7x4
If x
1.2
1.2.3.4
1.2.3.4,5.6
&c. y.
=
9x5
1.2.3.4.5.6.7.8
Here the factors 3, 5, 7, 9, &c. are not in
all the terms, and being left out, the feries is
x²
X3
*
1.2
1.2.3.4
X4
1.2.3.4.5.6
&c.
X
abridged to x-
1.2
X
x
5.6 7.8
&c. and the factors restored, the feries becomes
X
3.4
A + B+ C+ D
X
*
AX 3+
X
X
BX 5 +
1.2
3.4
5.6
Cx7+
50 D x 9 &c. y.
=
7.8
Where A, B, C, &c. are
the ſeveral terms with their proper figns; without
the numbers, 3, 5, 7, &c.
Ex. 6.
If bz-
bz3
bzs
bz7
+
3.2aa
5.2.4a4
7.2.4.690 +
bz9
&c. = y.
9.2.4.6.8a8
Then the factors 3, 5, 7, 9, &c. not being
common to all the terms, are left out, and the
feries is
bz3
bzs
bzi
bz
+
&c.
24a
2.484
2.4.6a
ZZ
ZZ
ZZ
ZZ
Бать
A
B
24A
4aa
C-
баа
D &c.
8aa
M 3
And
166
INFINITE SERIES. B. I.
And restoring the numbers, the feries will then be
ZZ
ZZ
ZZ
A
B
C
D
bz 2aa
400
6aa
8aa
I
3
5
7
9
&c. = y.
Where A, B, C, &c. are the fore-
going numerators, with their proper figns.
Ex. 7.
There is given
32
curtailed, --
x
or ſhortened, x—
+
2 2.4
compleat, x- AX
0x3
bxs
+
5.2.4
X3
x+5
CX7
dx9
+
&c.
XX
XX
A
B
7.2.4.6 9.2.4.6.8
X7
+
2.4.6 2.4.6.8
XX
C
X9
&c.
XX
D &c,
2
4
6
8
XX
a
XX
BX
2
3 4
015
b
XX
с
XX
CX
6
7
8
DX &c.
9.
Where A, B, C, &c. are the foregoing terms,
exclufive of the following quantities.
Cor. 1. If the first term of any transformed feries
be multiplied by any number or quantity; the whole
feries is multiplied thereby. For the first term is vir-
tually contained in all the following terms. This is
made plain by Ex. 4.
Cor. 2. In like manner A, B, C may be made
to stand only for the coefficients, or otherwife, as
any one pleaſes.
PRO.
Sect. VI. INFINITE SERIES.
167
PROBLEM LXI.
To find the finite value of an infinite feries, or what
furd it is involved from.
RULE.
Divide all the terms by the firft; then the first
term will be 1. Then compare three terms of this
feries with three terms of the feries Rule 2, Prob.
lix. each with each, fuppofing a to be 1, and
c, &c. o; which two equations will find the in-
dex, and the fecond term, if it is a binomial. If
this does not fucceed, compare four terms with
four, for a trinomial; or five terms with five, for
a quadrinomial; making do, or eo, &c.
Ex. 1.
Suppose this feries I -
༧*
Compare this with... 1+mbx +m=
y
+
+
&c.
a
aa
Q3
Q$
m
I
bbx².
2
Then mbx =
2, and
and m .
m
I
yy
bbxx
·
*
2
aa
m
I
and dividing the laft by the first,
bxx =
y
2
a
777
I
mbx; therefore
=m, and 2m = m
I,
2
bx =
y
a
}
whence m = -1. Therefore mbx = -
or bx. Whence the index is -1, and
a
the ſecond term of the binomial (if it is one) is
Y. And the binomial 1 + y
ajo
a
that is
a + y
I
or
>
a
I +
y
a
the root required; which fucceeds.
M 4
Ex.
168
B. I.
INFINITE SERIES.
Ex. 2.
Suppoſe a +
xx
x6
+
&c.
24 8a3
1645
XX
X4
Reduced ax: 1+
&c.
200
8a+
I
Rule
1 + mbx+ m.
bbxx.
2
XX
m
I
Here mbx =
and m
24
bbxx
2aa
2
8a4
112
XX
and by divifion,
bx =
Then
2
4aa
xx = mbx × 200 = -
ท
I
2aa
4aa X
bx;
whence
2
I
mm
1, and 2m
I, or m=
the index.
2
I
And mbx =
bx =
2
XX
200
XX
or bx =
the fe-
aa
94
XX
cond term. And the furd is a × 1 +
or
aa
aa + xx√ž.
Radikal
Ex. 3.
Let x/8
5aa
/8
75a4
8x
512x3
7/8-&c.
be given.
4
Rule
ting bx=y.
Here my =
-5aa
8.xx'
m
and m.
I
-75a4
-yy=
2
512x
and
Reduced /8x: 1-
I + my + m. -y', put-
5aa
75a+
&c.
8xx 512x4
m
I
2
Sect. VI. INFINITE SERIES.
169
and by dividing
5aa
Iy 15aa
then y
3
m
y =
;
2
64xx
15aa
T
and
m
4 X m
8xxm 32xx.m I
-4m + 4 = 3m, and 7m 4, whence m =
-5aa
index. Alfo y =
8mxx
5aa
32
XX
7
2
or
4
the
7
35aa
32xx
the ſecond term. And the binomial furd is
4
* 8 × 1
35aa
32xx
+
Ex. 4.
Let the feries
až - y
4a
5y²
+
+ _5y3
+
35%*
96a2
384a 2
&c. be
proposed.
18432at
This example reſolved like the foregoing, gives
3
m
and
y for the fecond term of the
2
ба
Mic
binomial. But až × 1 +
y
alı
does not produce
6a
the given feries. Whence we may conclude it has
not a binomial root.
For a trinomial root; for brevity's fake put
1, z, v for a, bx, cxx in the Rule, Prob. Îix.
which rule then becomes 1 +z+ulm =
m
I
m
2
3
I + m² + m.
zz + m.
2
2
3
m
I
+
of-
Mv
+
m
22V
2
and
170
B. I.
INFINITE SERIES.
플
​2
and a
y
512
5y3
X I
+
aigh
is the
4@
given feries reduced.
equations, mz=
and m.
9&aa 384a³
Then we have theſe three
zz + mv
-y
m
I
m
4a
2
m
m
2
m
23 +m
2
3
2
5yy
96aa
>
533
.22V=
384a³
Divide the third by the firſt, and there comes out
M - I
m I m
2
zz+m
2
3
m
2
zz + 2m
3
ZZ + 2m
the fecond, and we have m.
-5yy
I.V=
;
add this to
96aa
m
I
m
τ
zz +
2
2
m
4112 -
2
X
2
3
1.v=o, or
I.V=0. And fquaring the firft,
yy
MMZZ =
and
yy
=16mmzz.
Alfo
16aa'
aa
5yy
m
I
5
ตบ
m.
ZZ = × 16mmzz
96aa
2
96
913 - I
•
ZZ.
And v =
5
71 - I
MZZ
ZZ =
2
6
2
2m + 3
6
712
I
ZZ.
Therefore
4m
2
ZZ of
2
3
122
I
2m - I.V
• 2m
- 1.22 + 2m
3
2m + 3
6
2m−2+2m+3
22=0; or
6
×2M —1=0,
4m + I
that is
6
X 2m
2 113
Z
V
4am
I
4
O,
Which
=0, and 4m + 1 x
equation has two roots.
I
I
and m =
If m =
then
>
2
2
4
y
and v =
2112 ++3
ZZ
6
Sect. VI. INFINITE SERIES.
171
X
6
aa
5yy
12aa
3
فولو
And the furd root is a X
ax
-1
y
5yy
I + +
which involved produces
a
12aa
four terms of the feries but not the laft.
I
And if m=
Then z =
7
世​,
2
4am
24
2m
and v =
3
yy
yy
ZZ =
Х
And
6
4aa 6aa
46
I
then the furd is a X: I
y
yy
+.
2
which
2a 6aa
involved, produces all the
terms of the given fee
ries; and therefore is the root required.
PROBLEM
LXII.
To revert an infinite feries; or to find the root of fuch
a feries.
IRUL E.
If the ſeries confifts of all the powers of z, as
Az + Bz² + Cz³ + Dz4 + Ez³ &c. y; then
ſubſtitute the values of the coefficients, A, B, C, D,
&c. into the following form, for the root.
+
+
+
B
2 BB — AC
z= y
·y² +
y³
A³
As
A7
5 ABC — A¹D — 5B³
14B4-21 ABC + 6A BD + 3A°C² — A³E
y+
A⁹
-42Bs84ABC-28A2B2D28A2B2C
A"
+ 7A³BE + 7A³DC A4F
ys
yo &c. This rule
often diverges when y is great. See another Rule,
Prob. XCIII.
For
172
B. I.
INFINITE SERIES.
For put zay + by²+cy3 + dy4 &c. Then
ZZ
aay² + 2aby³ + bby+ &c.
+2ac
a³y³ + 3a²bу4 &c.
a4y4 &c.
Whence
Aay + Aby² + Acy + Ady4 &c.
Ba²y² + 2Baby³ + Bbby &c.
23
24
&c.
Az
+ B2² =
+ Cz³
+ Dz₁ =
&c.
+ 2 Bac
Ca³y³ + 3Ca²by* &c
Da4y4 &c.
and a =
Then making the homologous powers equal, Aay=y,
I
-B
A'
A・ And Ab + Ba² = 0, or b =
Α'
2BB-AC
Likewife Ac + 2Bab + Ca³=0, and c
In like manner Ad+Bbb+2Bac+3Ca²b+Da+=0,
5ABC — A³D — 5B³
whence d =
; and fo on.
As
A? 7
1
Ex. 1.
Suppoſe xxx + x³ — x4 + x³ &c. =y, to find
the value of x in terms of y.
Here z=x, A=1, B=-1, C=1, D=-1, &c.
B=—1, D=—1,
I
2 — ]
Whence x =+²+ -33 +
I
I
&c. = y + y²+y³ + y4 + ys &c.
XX
Ex. 2.
-5+1+54
I
Let x = x + x + x² + + + +
z
2
find x in a fries of ≈.
3
4
Here, yz, A = 1, B =
y*
&c. to
5
I
I
C=
2
3
D=
Sect. VI. INFINITE SERIES.
173
I
I
D=
E =
&c. and x =
I
z +
4
I
2
I
I
I
I
I
5 X
5 X
+
2
I
3
4
+100
23 +
24 &c.
I
I
I
I
I
x² +
24 +
zs &c.
2
24
120
2:3
22
1
+
2.3
Z
A
2
ZZ
I. 2
that is, x = x
where A, B, C,
with their figns.
&c. are the foregoing terms,
24
25
+
&c.
2.3.4
2.3.4.5
Z
Z
Z
B-
В
C
-
D &c.
3
4
5
&c. c, to find a.
Ex. 3.
aa
04
аб
Suppoſe r
+
21
24r3
720rs
as
+
4032017
Put rcv. Then
aa
a4
25
+
27
a8
2473 720rs
40320r7
&
&c. = v. Here aa, y = v,
I
A =
1/12 B =
-I
2r
2423
&c.
Whence
C = _L
720rs'
I
-I
D =
4032017
I
I
-I
3
24r
ġa = 2rv
vv t
I
28826 1440r6
I
v³ &c.
823
32rs
4
I
= 2ro +
vv +
3
45r
jj³ f
35r2
v4 &c. And
V
extracting the root, a√2rv X: 1 +
+
127
302
503
+
16orr 896r³
+ &c.
2 RULE.
174
B. I.
INFINITE SERIES,
2 RULE.
If the ſeries confifts of the odd powers of 2, as
Az + B2³ + Cz³ + Dz' &c. y. Subftitute the
values of the coefficients A, B, C, &c. into the
following form; which will give the root.
2 =
B
A A+Y³ +
A4
8ABCA'D — 12B³
3BB
AC
A'
·ys +
A 10
دو
}
55B4-55AB²C +10A2BD + 5A¹C²—A³E
+
&c.
A¹
13
For put zay + by³ + cys + dyr &c.
|| ||
Then z³
=
a³y³ + 3a²bу³ + 3a²cy¹
25
+ zabb
asys + 5a+bу7
a7y7
ورو
Aay + Abys+ Acys + Ady &c.
+ Ba³y³ + 3Ba³bys + 3Ba²cy"
Z7
27 =
&c.
And Az
+ Bz³ =
+ Czs=
+ D2 =
+ 3Babb
+ Dały,
+ Casys + 5Ca+by7
Then equating the coefficients of like terms;
Aa≈1, Ab + Ba³ = 0, Ac + 3Ba²b + Cas = 0,
Ad+3Ba²c+3Babb+5€a+b+Da¹o, &c. whence
I
a =
b
==
A
c =
3BR-
AC
A¹
"
Ba3
d =
A
B
:
Likewife
A4°
8 ABC — A¹D→ 12B³
1.
&c.
A 10
Ex.
Sect. VI. INFINITE SERIES.
175
Ex. 4.
Q3
as
Let a
07
+
2.3dd
2.3.4.5d+
+&c.
2.3.4.5.6.7d6
y; to find a.
Hereza, y = y, A = 1,
= 1, B = —
I
2.3da'
I
C=
2•3•4•5d4
>
D =
I
I
y
2.3.4.5.6,7do
Whence a =ý + 2.3dd33 +
I
> &c.
I
3.1d+
2.3.4.5d4
I
I
xys +
1
+
2.3.3.5d°
2.3.4.5.6.72°
+
Xyl
2.3.3do
&c. = y +
I
3
yз +
2.3dd
2.4.5d+
ys +
3.5
}?
2.4.6.7.d's
+ &c.
Ex. 5.
Suppoſe y +
уз
335
3.5y
+
2.3dd
2.4.5d4
+
&c. = a, io find y.
2.4.6.7do
+
Here≈y, ya, A = 1, B =
I
C=_3
2.3dd
2.4.5d+
D-
3.5
2.4.6.7d6
&c. Then y = A
I
a³ +
2.3dd
1
3
xas +
5
I
2.2.3d+
2.4.5d+
X
2.5
2.4.2.7
2.3.3
a7
I
&c. = a
3
I
+i
as
2.3dd
2.3.4.5d+
I
a7 + &c.
2.3.4.5.6.7d
€
Exì
176
B. I.
INFINITE SERIES:
!
Ex. 6.
Given bz
bx³
bzs
bz7
bz9
6ag
4044
d
11
&c. =d, to find z.
Dividing by b, z-
n. Then
= n.
Then yn, A = 1, B =
33бa
3456a8
Z3
25
27
6ea 40a4
336a6
&c.
I
?
6aa
C =
I
I
D =
40a4
336a6
&c. then will
n3
I
I
z = n +
+
+
x ns +
6aa
8
I
12
+
+
6.40 336 6.6.6
1244
ከ7
40a4
X + &c. = n +
N3
26
бай
13
463
+
ns +
n² + &c.
12044
840
3 RULE.
m
When the feries confifts of any powers of z de-
noted by m and n, as Az" + Bzm+n+ Czm+2n
+ Dzm+3″ + Ezm+4″ &c. =y. Then ſubſti-
tute the values of the coefficients, A B, C, &c,
into this form, for the root or value of z.
Put v =
नांद
Then
1+ 1
22
+
I
B
V
m
མ
mA
m + I + 2n.BB
2mm AA
- 2 MAC
1+218
抛
​"
2mm
Sect. VI.
177
INFINITE SERIES.
2mm +9mn + 9nn` + 3m + 6n + 1
B³
1+372
m
6m³ A³
3
m + 3n+ I
+
BC
mm A²
Ꭰ
MA
&c.
1+n
i+an
m
m
+ co
ข
=
1+ 332
i do
m
+ &c.
1
For put z vm + bo
Then dividing the given feries by A, we have
B
C
m
z" +
m + n
m + 2n
y
+
&c. =
V.
A
A
A
AK OKAK
א
Z
m
Whence by involution,
= v + mbv
m + n
m + n
zm+zn
312
=
11
PA
C
A
X
m+3 = &c,
m+ 2n
in
772
+
mcv
&c.
111
I
+m.
bb
2
m + n
m + zn
=vi
m
m
V
+m+n.bv
&c.
m+ 2n
m
V
&c.
Then equating the coefficients, mb +
And mc + m
-B
b=
MA
= o, and c =
B
A
=0, and
O,
m
I
bB C
bb +m+n.
2
A
+
A
&c.
2m² A²
N
Note,
m + 1 + 2n.BB — 2mAC
178
B. I
INFINITE SERIES.
Note, in all theſe rules, I have only purfued
theſe ſeries to a few terms; to have gone farther
would have taken up too much room: but the
method is vifible.
Ex. 7.
Suppoſe 1xx + 3׳ + 1x4 + 3x5 &c. =y.
Here zx, v = 2y,
工
​A, B = 1, C = 1/2
D, &c. and m 2, n = 1.
m2,
Whence
x = vž
I
I
vž
5BB — 4AC
v² +
vž &c. = v
***IN
3
I
V + 2-18
3
3
+
270
8AA
I
v² &c.
Ex. 8.
24
аб
Let x
+
+ &c. =y=v.
2X
6x3
24x5
aa
Here ≈ = x, m = 1, n = — 2, A = 1, B =
2
c = /%
04
аб
D =
&c. and x = y +
y}
aa
24
2
-2BB
2 AC
aa
+
y³ &c. = y +
5a4
2A2
2y
1233
586
+
&c.
83.5
12/
I
I
Ex. 9.
I
16
X
8
Let x" — — — — — — *-**
2
&c. z, to find x.
In this Ex. z=x, v = 2, m ==
I
=
I
16
-
2
22 == 1, A=1,
1
16 E
E =
5
B = 1, C, D = 128
2
&c.
Whence
Sect. VI. INFINITE SERIES.
179
Whence x
-2
Z
-4
2BB+C -6
+
Z
t
2
I
14
14B³ + 14BC + 2D × 2-8 &c. that is
I
x =
ZZ
1
I
I
+
&c.
Z4
Z6
28
Cor. 1. If you would find any power of y; find
y in a ſeries of z, and then involve that feries to the
power required, or elſe put s = y; then find s (yr)
from juch a jeries as this,
m
r
As + Bs
by the last rule.
m+n
m+2n
r
+ Cs
z
&c. =y,
Cor. 2. The reverted feries is of the fame form as
the given feries; for otherwife they are not convertible
into one another.
PROBLEM.
LXIII.
To extract the root of a feries containing all the powers
of two letters.
I RU
RULE.
If the feries confifts of all the fingle powers of
z and y, as az+bz² + cz³ +dz+ &c. =gy+by² +jy3+
ky4 &c. fubftitute the values of the coefficients
in the following form, for the root.
b- bA² j — 2b AB — cA³
-26AB-CA³
y² +
a
—
2bAC-3cA2B dA4
A
y4
y³
7—2bBC bAD-3cAB'—3c A¹C-4dA³B
༧
z = &y +
a
a
k— bB²
+
+
-eAs
ys
m-2bBD-bC²-2bAE-cB³-6cABC-
+
a
a
3 c A²D—6d A²B²-4d A³C—5€ A4B —ƒA6
N 2
ys &c.
Where
180
B. I.
INFINITE SERIES.
Where A, B, C, &c. are the coefficients of the
firſt, ſecond, third, &c. terms.
Let ≈ Ay + By² + Cy³ + Dy &c. Then
azaAyaBy² + aCy³ + aDy4 &c.
+bA²ð² +2bABy³ + bBBy4
+ 26AC
+ bz² =
3
+ cz³-
+dxt =
&c.
+ CA³y³ + 3cA2By+
+ dA+y+
=gy + by² + jy³ + ky4 &c,
And equating the coefficients, aA = g, and A =
2
Alfo aВ+bA² = b, and B =
b-bAA
a
g
~
a
Alfo
aC + 2bAB + cA³ =j, and C —j—2bAB—cA³
a
Again aD+bB+2bAC+3cA2B+dA+k, and
k— bВ² — 2bAC 3cA2BdA4
D=
Co
&c.
Let x
1
2
+
216
Ex. 1.
X4
&c. =
24
I
اد
2
Y po
I
I
— y² + y² + y² &c. to find x.
3
Here
z
d=-
I
I
— —
5
x, y = y, a = 1, b = —
24&c. and g =
k =
= —, &c.
>
2
る​=
3
I
I
j
4
The
}
2
Sect. VI. INFINITE SERIES.
181
Then
HIM
II
I
+
x =
+
2
**
2
1
3
II
I
8
y+ y² +
24
y² +
100
+
4
48
48
y³ &c. or
I
II
1381
y³ +
yu &c.
24
2880
Ex. 2.
Suppoſe z +
2:3 325
+ +
527
6dd 40d4 112d6
&c.ny+
ny3
6dd
+
of 3nys
5
5ny7
40d+
+
112d6
&c. to find z.
Whence
22
n
I
6dd
Comparing this
with the rule, and we have
d=0,e= 3 ƒ=0,&c.
I
a = 1, b = 0, c =
6dd
n
→ k = 0,1 =
6dd
g = n, b = 0, j =
z= ·y + oy² +
40d4
зи
40d4 m = 0, &c.
>
A³
6dd
xys + Oy4
+
3n
AAC
3As
40d+
2dd
40d+
Xys &c. where B, D, &c.
N n3
=o; that is z=ny +
3n
n3
+
40d4
12d4
120ďt
9n
Ion³ + n³
120d+
225
y³ +
× y³ &c. =ny +
n I
y³ = ny + 1/2 x =
= 6dd
I
X
6dd
n
n³
-y³ +
6dd
nn
n
n³
·№3 +
6dd
9
nn
X
ys &c.
20dd
I
Or z = ny +
nn
2.3dd
where A, B, &c. are the foregoing terms.
N 3
yy A+
9
nn
4.5dd
yyB &c.
2 RULE.
3
182
B. I.
INFINITE SERIES,
2 RUL E.
In two feries confifting of the powers and pro-
ducts of z and y; as
az+bz²+cz³+dz4 &c. +fy+gzy+bz²y+jz³y &c.
+ly²+my²z+ny²z² &c. +py³+qy³z &c. +sy4 &c.
= 0.
Then fubftitute the values of the coefficients, into
the following form;
f
y
a
l+gA+b A²
a
·y²
2bAB +¢A³ + p + gB + mA + hA²
-y³
a
2bAC+bBB+3cA²B+dA++s+gC+mB+qA
а
+ 2bAB + nA² +jA³
y+ &c.
Where A, B, C, &c. are the coefficients of the
firft, fecond, third, &c. terms.
For put z Ay + By² + Cy³ + Dy4 &c.
az
+ bzz
+ cz³
+ fy
+ ly²
+
py³
+gyz
+ my²z
11
|| || ||
Then
aAy + aBy² + eСy³ &c. 1
bA²y²+2bA Bу³
3
3
cA³y³
fy
+ Zy²
o.
+ pys
+gAy² + gBy³
+ mAy³
+ byz²
&c.
+ bA²y³
Then equating the coefficients, aA +ƒ=0, aB+
¿A²+1+gA=ò, &c. whence A =— B =
bA²+1+gA
C=
a
+bA²
&c.
f
a
2bAB+cA³+p+gB+mA
a
Ex.
Sect. VI. INFINITE SERIES.
183
Ex. 3.
I
I
I
Suppoſe 2y + ——y³ +
12
+ x²y² = 0; to find y.
2
Here zy, y = x, a = 2, c =
xx + xy - 3 xyy
8
I
I
,
f =
2
I 2
n=1; b, d, j, m,
P, q, s = 0.
3
g= 1,
b =
1 =
,
8
4
Therefore
I
2
72
A³ +B-3AA
8
y
4
2
2
I
x4 &c. =
+ x
4
I
212
4
J = = = = = = = + A
A¹B+C¬AB÷A²
173
768
2
43
x³ + x4 &c.
384
PROBLEM
LXIV.
To extract the root of an adfected equation, by a feries.
I RULE.
t
+D++'
If the equation confifts of terms which contain
the powers of x and y; and you want the value
of y, in a ſeries of x. Make the equation = o,
and affume an indetermined feries for the root,
asy Ax" + Bx"+" + Сx"++ Dxn+ &c. where-
in the indices n + r, n + s, &c. continually increaſe
if x be very fmall; but they decreaſe if x be
great; the first is an afcending feries, and the
latter a defcending one. By this means the feries
will converge; every following term growing ftill
lefs, till they vanish or become of no moment.
For y and its powers in the given equation, fub-
ftitute the first term Ax" and its powers. Then
to determine n, put the two leaft indices equal to
N 4
each
184
B. I.
INFINITE SERIES.
each other, for an afcending feries; or the two
greateft, for a defcending one.
And if it ap-
pears not at first fight, which is the two leaft, or
two greatest; it will be known, by comparing
every two of the indexes.
Then to determine r, s, t, &c. fubftitute its va-
lue for n, in all theſe indices, and having taken
the leaft for an afcending feries, or the greateſt for
a defcending one; fubtract it from each of the
reft. Then take theſe remainders, and add them
to themſelves and to one another, all poffible ways;
and theſe remainders, and the fums refulting,
taken in order, will be the values of r, s, t, &c.
which will be affirmative, in an afcending feries;
but negative in a deſcending one. Then put theſe
values in the feries, Ax² + Bx”+” + Cx²+2r &c.
n
Then to find the coefficients A, B, C, D, &c.
fubftitute the laft feries for the powers of
y, in
the equation; and put the coefficient of each pow-
er of x, fucceffively o; and A, B, C, &c. will
be gradually found from thefe equations.
Let a*x²-a^xy + x
Ex. I.
= ay³, to find y.
By reduction a4x² — a¹xy + x6 — ays = 0.
ayo. Put
y = Ax” + Bx²+r + Cx²+s + Dxn+ &c. ſub-
ftitute Axn for y, in the equation, and we have
aˆx² — aªÂƒ²+¹ + x³ — @ŧ×5” — o. Then equa-
ting the indices, 212, for the leaft, or 526
for the greateſt indices.
4
12
a A sxs n
For an afcending feries.
#
Here +2, and
1.
Then the indices
2, n† 1, 6, 51, become 2, 2, 6, 5. Subtract 2
from
Sect. VI.
18.5
INFINITE SERIES.
from the reſt, and you have 3, 4; out of which
is compoſed this feries 3, 4, 6, 7, 8, 9, 10, &c.
for the values of r, s, t, &c. whence the form of
the feries will be y≈ Ax + Bx² + Cx5 + Dx² + Ex³
&c. This feries fubftituted for y in the given
equation, will be as follows:
a²x² = a²x²
+x6 =
-a4xy =
xo
a4 Ax²
a4Bx5
a4Cx6
a4Dx8 &c.
*
*
+
208
-ays
O
|| | ||
a Asxs
·5a A4Вx³ &c.
O
Then equating the homologous terms a4-a4A=0,
and A1. Alſo—a4В'—a A³ = 0, i
Alfo-a4B-aAso, and B =
I
аз
5
Again, 24C + 1 = 0, and C =
-
a4D5aA4Bo.
Likewife
D = +
quired is
2.
Whence
-As
I
a3
a4°
&c. Then the feries or root re-
X4
y = x
+
a3
218.
X5
+
5x7
&c.
For a defcending feries.
Here 576, the greateſt indices, and n = 1,
and fubftituting this value of n, the indices 2,
n + 1, 6, 5n become 2, 23, 6, 6, and the re-
mainders
4
3 +/-, -4; and r, s, t, &c. will be
4,
7, -8, &c. and the feries
3
Ax's
+ B✩ 23
becomes y = Ax
2.
+ Cx-23
+ Dx-6323
&c. which fubftituted in the given equation, will
be
186
B. I.
INFINITE SERIES.
a4x²
*
*
+ 2343²
a4xy
04 Ax
-13
a4 Bx
+xo = + x6
&c.
— @µ³ — — aA³ x6 — 5aA+Bx² - 5aA+Cx² — 5aAª¹Dx˜¯
ay5 =
= 0.
*
10ª A³B²
&c.
Then equating the coefficients of like terms,
I-a Aso, and A =
5aA4B=0, and B = -
3
I
aš
33 3
Likewife-a & A --
ša³¹³, alſo e4—5@A¹C=0,
5
and Ca³. Alſo — aªB —
=
= o, and D =
root is
-
a4B -5aA&D — 10a A³B²
73 &c.
2250
Whence the
1 풍
​3
a
a
a73
y
+
&c.
3
2
a
5x
25x
5x23
If you put n + 16, the indices will be 2, 6,
6, 25; but 6 is neither the greateſt nor the leaft,
therefore this fucceeds not.
2;
If you put 52 = 2, the indices will be 2, 12, 6,
but here alſo 2 is neither the greateſt nor the
leaft. Therefore this will not fucceed.
If we put n + 1 = 52, the indices will be 2, 14,
6, 14; and 1 being the leaft, this will do for
an afcending feries; and the form of it will be
+ Dx²² &c.
y =
Ax
Ax² + Bx + Cx
I
2
Ex. 2.
Let a3x + 2x3 =
Putting Ax for y, the equation becomes
a³x ax³- @³ Ax"
+
a
a³y — y4 = 0, be propofed.
n
A 4x4 = 0.
0. Then put n = I
for
Sect VI.
187
INFINITE SERIES.
for the leaft indexes, then the indexes become
1, 3, 1, 4; and the differences 2, 3; and r, s,
t, &c. 2, 3, 4, 5, 6, &c. and the feries
Ax+Bx³ + Cx4 + Dxs &c. y. Whence
a³× = a³x
x
+ax³ =
+ ax³
— a³y — — a³Ax—a³Bx³—a³Cx+—a³Dx5 —— a³ Ex˜
=
&c.
-
•y4
a4x4
-4a³Bx6
&c.
Then equating the coefficients, a³A a³, and
Ar. In like manner B = C=
I
aa
&c, and the root is
D0, E =
4
Q5
y=x+ X3
I
I
4
аз
4x6 &c.
as
I
a3
3
aa
Otherwise for a defcending feries:
3
Put 34, then n, and the indices are
1, 3, 2, 3; and the differences -2, -2, and
", s, t, &c. = 2,
3
4
y= Ax+ Bx
Whence
a³x
*
2,
4,
41, &c. and
- 11/
+ C x = 1 ½
+ Dx-3
&c.
+ a³x
*
*
a³ Ax
+ax³ = ax³
a³y =
−y+ —— A¹µ³ — 4A³Bx — 4A³Cx*
4A³Dx
6A2B2
Then by equating the coefficients, A4 = a, and
I
A = a; alfo B =
D =
12 11
2호
​a
a
C =
4
I
&c.
32
327
за
And
188
B. I.
INFINITE SERIES.
And
y = at x =
+
a
21
4
зада
a
2-13
4
x-34 &c.
32
Ex. 3.
Suppose y³ + aay+axy
x³-2a30, to find y.
Put Ax, for y, and the equation becomes
A³x3+ aa Ax² + aAx²+¹ — X3 20³x° 0.
For an afcending feries.
Put the leaft indices no, and the indices be-
come 0, 0, 1, 3, 0; and the differences 1, 3;
I,
and r, s, t, &c. = 1, 2, 3, 4, 5, &c. and the
feries y A+ Bx + Cx² + Dx³ &c.
Then
y3 = A³ +3A2Bx + 3AB²x² + 3A²Dx³ &c²
÷ 3A²C + B³
+6ABC
+ a²y = a²A + aaBx + aaСx² + aaDx³
+ aAx + aBx²
+axy =
X3
2a3
203
+ aСx³
Xx3
3
Then equating the coefficients, A³+aaA-2a³=0,
and extracting the root, Aa; alfo 3A²B+aaВ+,
aAo, and B =
I
4
In like manner C =
I
640
and D =
131
51244
"
&c. Whence ya —
| 8
4
XX
64a
131x3
I
+
&c.
512aa
E
Sect. VI. INFINITE SERIES. 189
Ex. 4.
Let y³ + y²+y — x³ = 0, to find y in a defcend-
ing feries.
3
Putting Ax" for y, the equation becomes
A³×³¹² + A²y²² + Ay²
for the greateſt indices.
indexes are 3, 2, 1, 3;
-2; and the feries — 1,
I
x³o.
O. Put 32 = 3,
3n
Then n = 1, and all the
and the differences - I,
2, 3, 4, &c. and
2
y = Ax + B + Сx˜¹ + Dx˜2 &c. Then
y³ | A³x³ +3A2Bx² + 3A²Cx + 3A¹D
+ y²
+ "
x²
x3
+ 3ABB +6ABC > &c.
+ B₁ S
+ A²x²+2AB+ BB } &c.
+ 2AC
+ Ax.. + B
&c.
Then equating the coefficients, A³ 1, and A = 1.
I
2
Likewife B-
C =
D = 1, &c. and
3
9
therefore y =
y = x
I
2
7
+
&c.
3
دو
81xx
2 RULE.
n
ntr
n+zr
Affume y = Ax” + Bx + Cx + Dx
n+ 3r
its value into
&c. and having found 7, and put
the indices, as in Rule 1; fet them down in or-
der, and fubtract each of them from the next
greater; and you will have a ſeries of differences.
Then find the greateſt number, which will mea-
fure all theſe differences; and this is the value of r,
which must be affirmative in an afcending feries,
or when x is fmall; and negative, in a deſcending
one,
1
B. I.
190
INFINITE SERIES.
one, when x is great. Then the values of n and
r muſt be fubftituted in the affumed feries.
The proceſs muſt then go on as in Rule 1; and
if there be any fuperfluous terms, which will be
known by ſome of the coefficients A, B, C com-
ing out o; theſe terms must be thrown out of
the feries, and the operation begun anew.
Ex. 5.
Let y³ — axy + x³o, be given.
ኪ
0,
Put Ax for y, and the equation becomes
312
=
A³x
aAx +x³ = o. Let n + 1 = 3,
= 3, and
n = 2; and the indices are 6, 3, 3; that is, 3, 6.
Then 6-3 3, then r = 3; and the leaft indices.
being compared, the feries will be an aſcending one,
which is this yAx²+ Bxs + Cx8 + D &c.
Dx¹¹
which ſubſtituted in the given equation will be as
follows:
y3
3
+ A³x + 3A²Bx⁹
axy — a Ax³ — aBx6
3
+ x³ + Ix³
Then aA = 1, and A =
aCx?
+ 3A²Cx¹² &c.
+3ABB
aDx12
+
C = 3
3x8
a7
I
B =
I
a
205
y =
+
a
Q4
D= &c. Whence y
12
210
12xII
+
&c.
1
B7
Ex. 6.
Let ys — by²+9bx².
x3 = 0.
Subſtitute Ax* for y and the equation is
´A³× 5” — bA²x² + 9bx² — x³ = 0.
Put 2n = 2;
whence
Sect. VI.
INFINITE SERIES. 191
}
whence n = 1, and the indices are 5, 2, 2, 3;
and the differences 1, 2. Whence r 1.
= 1. There-
fore yAx+ Bx² + Cx³ + Dx &c. Then
315
3
+ Asxs
by² —b A²x² - 2bABx32bACx+-2bADX³
+ 9ỏx² 1 + 9bx²
3
X3
-bBB
26BC
2
=
Here bА² 9b; and A=3: alfo B =—
I
6b
I
8 I
I
C =
D=
Whence
216bb
26
388853
x3
81
I
y = 3x
+
X X4
66
2166b
26
3888b³
&c.
3
RULE.
If the equation determining A, be an adfected
equation, which has feveral equal roots or values
of A, then you muſt divide the leaft remainder,
found by Rule 1, by the number of equal roots,
one of which you take for A; and take this quo-
tient for another remainder. Or elfe divide r
found by Rule 2, by that number, and make uſe
of the quotient, inſtead of r.
Ex. 7.
14
Let y⁹ — xy³ + 2x³y² — x³y — x¹4 = 0, to find y'.
y9
for y, and the equation becomes
Put A
32+1
212 +2
Ax⁹
A³x
3
+2A2x
n+3
Ax
x140.
Let 3n+1=2n+2; whence n = 1, then the in-
dices are 9, 4, 4, 4, 14. But the fum of the
coefficients for the leaft index 4, is A³ + 2A²
A=0, or A² 2A +1 = o, which equation has
two
192
B. I.
INFINITE SERIES.
J
two equal roots AI, and A1. Now the
difference of the indexes will be 5, 10; there-
fore divide 5 by 2, gives, and we have 5
5, 10 for the differences. Therefore r, s, t, &c.
will be , 5, 7—, 10, &c. Or (Rule 2) r — 5 ;
5
2
therefore =
2
the ſeries will be
I
2
5
94
2
=
2
to be taken for r; whence
y = Ax+Bx³² + Cx³ + Dx³
ولو
муз
6/1/1
8/1/125
&c. Then
+ Ax⁹
A3x4 3A Bx · 3A²Сx⁹
6/1/1
3AB2
+2x²y² +2A²x² + 4ABx² + 4ACx⁹
- x³y
X¹4
Ax4
+ 2BB
Cx⁹
&c.
3
Hence - A³ + 2A² Ao, and A = 1;
4B-4B = 0, and B may be taken at pleaſure.
Suppoſe B-1, then 1-3C-3+4C+2—C=0,
or 4C=4C, and C may be taken at pleaſure. Let
C= 1; then y=x
x6 &c.
=
Or thus; In the fecond equation, 4B 4B;
which concludes nothing; alfo 1-3C-3BB +
4C+2BB-C=0; that is, 1-BB=0, and B = 1
-I, &c.
or
Ex. 8.
Let A4y²-2a4xy + a*x² + x4y² = ¯`
Put Ax" for y, and the indices become 21,
n + 1, 2, 2n + 4. Let 2n = 2, or n= 1, and the
indices
Sect. VI.
193
INFINITE SERIES.
indices are 2, 2, 2, 6; and the difference 4. The
2 n
2 2n
equation is a¹A²x²” — 2 aª Ax²+¹+ a4x² - A² µ²² +
or a4 A²x²-2a4Аx² +a4x²- A²x60, where the
coefficient of the firft term is A2A + 1 = 0,
+I=
which has two equal roots A = 1. Therefore di-
vide the difference 4 by 2, and the quotient 2 is
r or the common difference; whence the feries is
y = Ax+ Bx³ + Cx5 + Dx7 &c. o. Then
a¹Â²x²+2aªABx² +2a¹AСx+2a¹ADx³
a4yy
+a+BB
=
+2a4BC
&c.
-2a+xy-2a+Ax²—2a+Bx4—2aªCx6—2aªDx$
+a+xx
x4y²
+ a4x²
A²x6 2 ABx8
Hence A² 2A+10, and A1; again,
BB, and B may be taken at pleafure. Sup-
poſe B = 1. Again, oC + a¹B² — i = 0, or
I
aa
CC=1-10, and C may be taken at pleaſure,
Let C = 1. Then oD = 2AB — 2a4BC = 0, and
a+
I
D may be taken at pleafure. Let D= &c.
X3
Then y =
x
+
a a
X5
X7
+ + + &c.
a+
a6
Or rather thus, when A is determined to be 1,
the firft and third lines vanish; whence a4BB =
I
2
A² = 1, and B =
;
alſo 2a4BC = 2AB, and
aa
1
C-
Q4
&c.
4
RULE.
If the quantity forming the feries (x) be nearly
equal to fome given quantity, put a new letter +
that quantity for it, and fubftitute it in the equa-
a
tion;
194
B. 1.
INFINITE SERIES.
tion; then find the root in an afcending feries of
the new letter, Or if the quantity (x) be very
great, and the feries for y is to afcend by x'es.
Take fome quantity nearly equal to x, and fub-
ftitute the fum of that and a new letter for x.
Ex. 9.
2
Let yaay-x30, where x =
a, nearly.
3
2
8
4
Put
a v=x. Then x³
a3
aav +
3
27
3
8
a 3 +
4aav
·aav—2av²+03
27
3
zav²-v³, then y³+aay—
o. Let y
0, 1, 2, 3.
Av", then the indices are 3n, n,
Let no, then the indices become
0, 0, 0, 1, 2, 3; and y = A + Bv + Cv² + Dv³
&c.
1 +
Then
A³ + зA²Вv + 3A¹Cv² &c.
a a A + aa Bv
33
+ day
8
8
03
a3
27
27
4
4
+
a²v
+ + +
+ av
3
3
·2av²
+23
t
+
3ABB
+ aaСv²
3
Then A³ + 2
4
B =
3
26
C:
2002
+ &c.
I
8
aaA
a³o, let A=r. Alfo
3
27
; and 3A2C+3ABB+aaC=2a. Whençe
53 r
3rraa
&C.
"
1
Ex.
Sect. VI. INFINITE SERIES: 195
Ex. 10.
Let y4-x²y²+xy²+2y²-2y+1=0, where x=2
very near.
Let x=2+%, which fubftituted for x, there
arifes y — z¹² — 3zy² — 2y + 1 = 0.
4
ay
Let y Az", then the equation is A44
—A²±²²+² — 3A²x²+¹ — 2Az” + 1 = 0. Let
n = o, and the indices are 0, 2, 1, 0, 0; and
the differences 1, 2, 3, 4, &c. whence y = a +
Bx + Cz² + Dz³ &c.
Then
n
y4
A+
+ 4A³B≈
+ 4A³Cz² &c.
+6A2BB
111+
z'y²
A Azz
2
3zyz
3A2z
-6AB≈2
2y
+ I
2A
2 Bz
•2Cz²
+1
Here A4 2A + 1 = 0, and A1; alfo
4B — 2B = 3, and B = 3; and 4C + 6BB—1—
2
6B2C = 0, and C —— 1, &c. and y = 1 +
3
2
א
7
zz &c.
4
4
Cor. 1. In all these cafes of extracting roots, the
feries must be made to converge, or else they are of no
ufe. For in a converging feries, the terms grow con-
tinually less and less, and fo approach nearer and
nearer to the true root, till the difference is as small
as you will. But a diverging feries always runs far-
ther from the root, and therefore gives a falfe value
thereof.
Cor. 2. If y be denoted by a ſeries of x afcend-
ing; the leffer x is, the fafter the feries converges.
O 2
And
196
INFINITE SERIES:
B. I.
So
And in a ſeries of x defcending; the greater x is,
the fafter likewiſe it converges. Therefore we are
to contrive the feries, that we may have the least
quantity in the numerators, or the greatest in the de-
nominators.
Cor. 3. If the equation for finding the first term A,
be an adfected equation; as many roots or different
values of A, as that equation has, ſo many different
feries will arife. For the first term A being differ-
ent in each, the coefficients B, C, D, depending
thereon, will also be different. Likewife, if two
roots are equal, thecond term will vanish, and the
coefficient B will be found in the third, which will
be a quadratic equation. And if there be three equal
values of A, the fecond and third equations vanish,
and the fourth contains a cubic equation of B, &c.
Cor 4. An equation will also admit of feveral
different feries for the roots, according to the different
values affumed for n. Alfo there are other equations
that are impoſſible, and will admit of no roots.
Cor. 5. When the first equation, or that for de-
termining A, bas feveral equal roots; then the va
lues of r, s, t, &c. must be divided by that number.
Or, which is the fame thing, the indices of x (r, s, t,)
found by Rule 1, must have others interpofed between
them, according to the number of equal roots.
As for
two equal roots, the feries Ax" + Bx"+" + Cx"+s
&c. must be reduced to this, Ax² + Bx"+er inf
플
​ints
Cx²+ + Dx²+½s + Ex²+s &c. If this be not
aone, the fecond term B will be infinite, and all the
following ones.
Cor. 6. If the ſeries A+B+C××+D+E+F
Xz² + G + H + ! + K × ≈³ &c.
!+ 23
o, z being
an indetermined quantity; then whatever value is put
upon z, it will be Ao, B+C=0, D+E+F=0,
G+H+I+K=0, &c.
For
<
Sect. VI.
SERIES. 197
INFINITE
For this being a general equation, where z may
be of any value; therefore put zo, and then will
Ao, and B+Cx≈+ D + E + F × z² &¿. =0,
divide by z, then B+C+D+E+F×≈&c, =0.
Again, put zo, and then B + C = 0; whence
D+E+FX≈ + G + H + I + K × ≈² &c. = 0•
Divide again by z, and D+E+F+G+H+1+K
xzo. Again, put z = 0, then D + E + F = 0,
and G+H+1+K×%=0, and G + H+I+K
= 0, &c.
•
Reversion of feries, and the extracting the roots
of all infinite feries, depends upon this. For the
coefficients of the feveral powers of the indeter-
mined quantity, must be puto, or elfe the whole
equation cannot vanifh, as it ought to do. And
this being done, the feveral affumed coefficients A,
B, C are determined as in the problems above.
SCHOLIU M.
1
in the feries agile + byl to + cyμ +24
To find y
&c. = fx" + gx'
y = Ax" + Bx'
π + S
n + r
+ bx
π+28 &c.
n+s
+ Cx &c.
Affume
&c.
Then by ſub-
aA
+ bAμ + nx
un ‡ v n
+ &c.
ftitution we get,
π
= ƒx" + gx
Whence making the leaſt
indices equal, n; then n =
un
πν
ferences will be ę; &c.
μ
greatest common divifor of
form of the feries will be
T
TV
π
and the dif.
,
شا
Then find 7 the
.and
g; and the
"0"
3
πT
+9
y = Ax + Bx
سم
+ Cx!
TO
+29
+39
+DxM
&c. in which the coefficients will be determined as
before.
O 3
*
SECT.
198
1
1
SECT. VII.
Some general and fundamental Problems, useful
and neceffary in algebraical calculations.
PROBLEM
LXV.
The fum and difference of two quantities being given;
to find the quantities.
then
L
ETs
d
a
e
the fum
the difference
greater quantity
the leffer.
a+es, by the problem.
and a e = d.
then 20s+d by addition
and zesd by fubtracting.
Whence a
or
s+d and e=
2
S
d
2
a=s+d, and es- d.
Cor: 1. Half the fum added to half the difference
of two quantities, is equal to the greater.
Cor. 2. Half the difference of two quantities, ta-
ken from half the fum, gives the leſſer quantity.
PROBLEM LXVI.
To find out the least common dividend, or the leaft
quantity, that can be divided by several given
quantities.
RULE.
Refolve each of the quantities into all the fim-
ple divifors contained therein, by firft dividing by
the
Sect. VII. FUNDAMENTAL PROBLEMS. 199
*
the leaſt, and then by the next, and fo on, till
they are all exhaufted; and collect thefe divifors
together for each quantity. Then if there be any
divifors in the fecond quantity which is not in the
first, multiply the first by fuch divifors. Likewife,
if there be any diviſors in the third quantity which
is not in this laft, multiply it thereby, or put them
into that quantity. Likewife fuch divifors as are in
the fourth quantity and are not in this laft, muſt
be put into it, and ſo on. And lastly, all theſe di-
viſors, in this last quantity must be multiplied to-
gether for the leaft common dividend.
Or shorter thus,
Divide the product of any two of the quantities
by their greateſt common divifor, (found by Prob.
x. Sect. II.) take this quotient and a third quan-
tity, and divide their product, by their greateſt
common divifor. Take this quotient and another
quantity, and proceed as before; and fo on to the
laft quantity. And the last quotient will be the
leaft common dividend.
Ex. I.
What is the greatest common dividend of a'bc, and
2ab2d.
The divifors of abc are a, a, b, c
of 2abbd are 2, a, b, b, d.
Here 2, b, d are in the laft but not in the firft;
therefore a xaxbxcx2bd, or 2aabbed is the leaft
common dividend.
Or thus,
The greatest common meafure is ab, then the
product is 2a³b³cd.
leaft dividend.
ab)za b³cd(zaabbed the
0 4
Ex.
200
B. I.
FUNDAMENTAL
t
Ex. 2.
Let abcd and ac + bd be propofed.
Theſe have no divifors but 1.
Therefore
ab + cd × ac + bd or aabc + abbd + accd + bcdd, is
the dividend required.
Ex. 3.
Let 3a²b, a³ +a'b, and aa —
bb, be given.
The greatest common divifor of 3ab and
a³ + a³b is aa. Then
aa) 3a5b+3a4bb (3a³b + zaabb.
·bb
Then the greateſt common divifor of 3a3b + zaabb
and aabb, is a+b; then 3a3b + 3aabb × aa—
divided by a + b is
a+b) 3a5b+3a4bb — 3a³b³ — 3a²b4 (3a4b—3a²b³
the leaſt common dividend.
Ex. 4.
Let the given quantities be a b₁, aa + ab,
at + a²b², and a + b.
a+b.
Thefe quantities refolved into their divifors are
aa+bb×a+bxa−b, axa+b, a×a×aa+bb, and
Now becauſe there is one factor a in the
fecond which is not in the firft, put it in the firſt,
which becomes na + bb × a + bx a ~ b xa, the
leaft dividend for the first two quantities.
Likewife, there is a, one factor in the third,
which is not in this laft; let it be inferted, and
it becomes aa + b b x a + b xa-bx aa, the leaſt
dividend for three quantities.
Laftly, Since a+b the last given quantity is
in the laft dividend; it will be the dividend for
all four; that is, aa + bb xa + bxa-bx aa, or
во
Sect. VII.
201
PROBLEM S.
q6a4bb is the leaft common dividend for the
four given quantities.
SCHOLI U M.
All the fimple divifors of a quantity, are found
the fame way, as in Prob. 6, 7. Chap. iv. B. II.
Arithmetic.
PROBLEM
LXVII.
The fum and difference of two quantities being given;
to find the difference of their ſquares.
Let s fum, d= difference, A = greater quan-
s+d
tity, E the leffer. Then A = and E
(Prob. lxv).
Whence
AA =
2
ss + 2sd + dd
2
s-d
2
4
SS-
and EE -
2sd + dd
4
and AA-EE =
4sd
= sd.
4
Cor. The product of the fum and difference of
two quantities, is equal to the difference of their
Squares.
7
PROBLEM
LXVIII.
Two quantities being given to find the square of the
Sum.
Let a be the greater quantity, e the leffer ;
then the fum is a +e; and a +e being fquared
is aa + zae + ec.
Cor. 1. Hence the fquare of the fum of two quan-
tities is equal to the fum of the fquares of the quanti-
ties, increafed by double their product.
Cor.
202
B. I.
FUNDAMENTAL
Cor. 2. The fquare of the sum of any number of
quantities, a + b + c &c. is equal to the fum of all the
Squares, together with twice the sum of all the pro-
ducts of every two.
For by this Prob. a+b+cl² =a+bl² + 2 ×a+b
Xc+cc; that is aa+2ab+bb+2ac+2bc+cc, and
fo for more quantities.
Schol. By the fame way, theorems may be
found for the cube of the fum of two or more
quantities.
PROBLEM
LXIX.
Two quantities being given to find the Square of their
difference.
Let a be the greater, e the leffer; then the
difference is ae, which being fquared, pro-
duces aa
2ae + ee.
Cor. Hence the Square of the difference of two
quantities, is equal to the fum of their Squares abating
twice their product.
Schol. By the fame method a rule may be found
for the cube of the difference of two quantities.
PROBLEM LXX:
The fum and difference of two quantities being given;
to find their rectangle.
Lets
fum, d difference, A the
difference, A the greater, E
the leffer. Then A+ Es, and A-E-d; and
adding theſe equations 2A =s+d; and ſubtract-
ing, 2Es-d. Then 2A X2E or 4AEs+d
SS dd
× 5-d=ss-dd, and AE =
4
>
Cor. The fquare of the fum, leſs the fquare of the
difference of two quantities, is equal to four times their
rectangle.
PRO.
Sect. VII.
203
PROBLEM S.
PROBLEM LXXI.
Given the nth power of the binomial a +b; to find
the difference between the Square of the fum of
the odd terms, and the fquare of the fum of the
even terms.
The nth power of a + b, that is a + bl" = an +
nan-1b + n.
ท
n
I n 2
a²² bb +n.
2
2
3
an-3b3
&c. Put A, B, C, D, E, &c. for the firſt, ſe-
cond, third, fourth, &c. terms. Then A+C+E
&c. fum of the odd terms; and B + D + F&c.
fum of the even terms. But A + C + E &cl
B + D + FĽ² = A+B+C+D+E &c. X
A-B+C-D+E &c. -
a" + na"-1b + n.
2
N
L
a"-2 bb + &c. X
2
n
7- I
Q
na
b + n
a7-2 bb
&c.
2
= a + bl²² × a
12
612
=aa
bbl".
Cor. 1. Hence aa
66172
ស
N.NI
N—1. 22—2 11-3
a² +
a^2-2 bb +12.
an-4b
2
2
3
4
&c.
n
I 12
2
n- I
12-X
no
b + n
Q2-3b3+n.
2
بن
2
ท
2 n
3 n
4
а
272-5 b5 &c.
&c. 1²
3
4
5
Cor. 2. aa
Cor. 3. aa
6613
bbl² = aa+bbi²
bb\³ = a³ ÷ 3abbl² — zaab + b³1².
+
·zabl
Cor,
206
FUNDAMENTAL
}
B. I.
22
√A+B+ √A - B
whence x =
12
บ
2
n
VA+B-VA-B
2
Therefore xv will be had, at leaft in decimals,
+1
Cor. Hence, VA±B =
n
n
√A + B −√A — B
2
VA
-
12
22
✔A+B+✔A−B
Ex. 1.
2
Extract the Square root of 11+6√/2.
Here A11, B = 6/2, AA — BB = 49 = D;
XX •7
and D7. Therefore x²+
✓D
x² = 11, or
XX
2xx-7=11, and xx 9, or 3. Likewiſe
√3 = +√9=7=+√2, and x +√v=3+√2
the root.
Or thus,
A+B=11+6√/2=19.484, and A—B=2.516.
√19.484 + √2.516
Whence =
2
4.4141.586
3, and v =
4.414
- 1.586
2
2
1.414√2, and x+v=3+2, the root.
Ex. 2.
Suppoſe 37-203 be given, to extract the
Square root.
Here A=37, B=20√3, AA—BB=169=D,
and ✔D=+13. Therefore 2xx-13=37, and
xx 25, or x 5.
Alfo
}
Sect. VII,
207
PROBLEM S.
Alſo √y = √25 — 13 = √12, and
vy
*+√y=5−√12. Or the root may be 12-5,
putting ✓D—13.
Or thus,
√71.64+ √2.36
20
ט
2
✅/2.36—√71.64
2
10.00
= 5.
2
6.92
=
3.46
2
=-2√3, and x + v=5—2√3 the root.
Ex. 3.
Let 7-5√2 be given to find the cube root.
Here A=7, B=5√2, A²-B¹=-=D, and
D³ = — I.
2
2
xx I
Then x3 + 3X
x³ = 7, or 4x³ + 3* =7,
XX
and the root of this equation is x = 1.
Alfo vy
√xx
* + √y = 1−√2.
xx + 1 = −√2, and
7
Or thus,
3
VA
A + B :
3
3
VA-B
V-
Then x
3
B =
14.07 = 2.414, and
-.07 = -414;
-4142.414
= 1.
I
2
-.414
2.414
And v =
1.414=-√2
2
for here B is negative; therefore x+v=I√//2.
Ex. 4.
What is the cube root of 25 +✔968.
Here A 25, B√968, AA BB -
3/D
343 D, and D7.
Then ³ + 3x³ + 21x = 25, and x = 1.
And √y = √8, and the root, I + √8.
A
Ex.
208
B. I.
FUNDAMENTAL
ī
Ex. 5.
+
Extract the cube root of — 10+√243.
Here A²- B 100+ 243 = 343 D, and
D7. Therefore x3 + 3x xxx-710, or
4x3 21*= 10, and the root is x 2, whence
= =
x³
=
vor√y=√ 47 =√−3, and x + √y=2+
✓ -3, as required.
ΙΟ
In like manner the cube root of 10√243
is 2√3.
Ex. 6.
Extract the 5th root of 843-589 √2.
Here AABB = 16807, and D' = 7:
7.
And 16x5-140x³+245x=843, and the root is
x=3; and √/y =—√9—7——√2, and
x + √y = 3−√2 the root required.
Ex. 7.
What is the 7th root of 568 +3284/3.
Here A-B2128 D, and D 2.
128D, and/D=
1136.112, A-B-112, and
Then A+B
X
√1136.112 — V112
2
2.732 —.732
= 1.
I.
2
√1136.112 + √~.112
And v =
2.732 +.732
2
= the root.
2
=1.732=√3, and x+v=1+√3
SCHOLIU M.
In the former method, if "/D is not rational,
neither member of the root will be rational, and
هم
"
in
Sect. VII.
209
PROBLEMS.
in the fecond, if neither the fum nor difference
n
n
of A+B and AB, is rational; neither
member of the root will be fo: and in thefe cafes
the rules are of no ufe, Logarithms will be ufe-
ful here in finding thefe roots, being exact enough
in finding whether any of the quantities be ratio-
nal or not. When none of thefe quantities are
rational, multiply the given equation by fome
number, till
12
n
12
But
D, or √A+B±√A−B, comes
out rational; then extract the root as before.
remember to divide the values of x, v, at last,
by the root of that number. Thus 22 + √486
has not fuch a cube root; but multiply by 2,
and then 44+ √ 1944
will have a cube root,
2
for the numerator.
>
PROBLEM
LXXIII.
To explain the feveral properties of (0) nothing, and
infinity.
It is plain, nothing added to, or fubtracted
from, any quantity, makes it neither bigger nor
lefs.
Likewife, if any quantity is multiplied by o,
that is, taken no times at all; the product will
be nothing.
b
Let =q; that is, let the quotient, of b di-
a
vided by a, be q. Then if b remains the fame, it is
plain the lefs a is, the greater the quotient q will
be. Let a be indefinitely fmall beyond all bounds,
then q will be indefinitely great beyond all
bounds. Therefore when a is nothing, the quo-
tient a will be infinite. Whence
P
Alfo
*
210
B. I
FUNDAMENTAL
b
infinity, therefore nothing
Alfo fince
X infinity.
Let there be feveral geometrical proportionals,
x, x², x³, x4, xs, &c. If this feries be continued
3
backwards, it will be x, 1,
I
I
X XX
that is, x',
x°, x-I, x-², the indices continually decreafing
by 1.
Then its plain x is equal to 1, whatever
be; for it may ftand univerfally for any thing.
Therefore o° is I.
Let x be an indefinitely fmall quantity, beyond
all conception; then in the feries x, x², x³, &c.
each term will be indefinitely greater than the fol-
lowing one. And when is o, then in the fe-
ries 1, 0, 0', o*, &c.
O
nothing, by what goes
I
is infinite, and o is
before.
Therefore the
mean o° is a finite quantity. Suppoſeb, whence
I
IXO
1×0 = bb, that is bb = = 1, and ɓ = 1,
whence it is plain again, that
Let
O
(b) 0° = 1.
a
be an infinite
+ I
a
a
- I
or its equal
quantity, then by actually dividing,
+e+, and
Therefore
I
a
a
I+1
=a+
ca + a
a
a a—a—a+
- I+I
a
+a+a+a &c. =
a
II
I- I
a &c. that is, an infinite quantity is neither
increaſed nor decreaſed by finite quantities.
Cor. 1. If o multiply any finite quantity, the pro-
duct will be nothing.
Cor.
Sect. VII.
211
PROBLEM S.
Cor. 2. If o multiply an infinite quantity, the
product is a finite quantity. Or a finite quantity
is a mean proportional between nothing and infinity.
For o infinity
b.
Cor. 3. If a finite quantity is divided by o, the
b
quotient is infinite ( inf.).
O
Cor. 4. If o be divided by o, the quotient is a
finite quantity of ſome fort.
For (Co. 1.) ¿ ×0 = 0, and therefore, a
Sinite quantity, or nothing.
Cor. 5. Hence allo 0° 1, or the infinitely fmall
power, of an infinitely small quantity, is infinitely
near 1.
A
Cor. 6. Adding or fubtracting any finite quantities
to or from an infinite quantity, makes no alteration.
Cor. 7. Therefore in any equation, where are fome
quantities infinitely lefs than others; they may be thrown
out of the equation.
Cor. 8. An infinite quantity may be confidered ei-
ther as affirmative or negative.
h
b
For infinity =
or
+○
SCHOLIU M.
There is fomething extremely fubtle, and hard
to conceive, in the doctrine of infinites and nothings.
Yet although the objects themſelves are beyond our
comprehenfion; yet we cannot refift the force of
demonftration, concerning their powers, proper-
ties, and effects; which properties, under fuch
and ſuch conditions, I think, I have truly explain-
ed in this propofition. Any metaphyfical notions,
that go beyond thefe mathematical operations, are
P 2
not
1
212
B. 1:
FUNDAMENTAL
2
not the bufines of a mathematician. But thus
much may be obferved, that o, in a mathemati-
cal fute, never fignifies abfolure nothing; but
always nothing in relation to the object under con-
fideration. For illuftration thereof, fuppofe we
are confidering the area contained between the baſe
of a parallelogram and a line drawn parallel to the
bafe. As this line draws nearer the bafe, the area
diminiſhes; till at last, when the line coincides
with the baſe, the area becomes nothing. So the
area here degenerates into a line; which is no.
thing, or no part of the area. But it is a line
ftill, and may be compared with other lines.
PROBLEM
LXXIV.
To find the value of a fraction, when the numerator
and denominator, is each of them nothing.
IRUL E.
Confider, from the nature of the queftion pro-
pofed, what quantities are infinitely greater than
others, when they are all taken infinitely fmall.
Then throw out of the equation, all thofe terms
that are infinitely lefs than others; retaining only
thoſe that are infinitely greater than the reft; by
which expunge one of the unknown quantities, and
the value of the fraction will be known.
Ex. I.
Let x³ +y³ = axy, and y infinitely greater than x,
when they vanish; to find the value ofÿ, when x
and y are = 0.
x
Here ³ is infinitely lefs than axy or y³, whence
yзaxy, or yy ax.
Then y
X
value of the fraction propofed.
ax
= =a, the
X
Ex.
Sect. VII.
213
PROBLEMS.
t
Ex. 2.
If 2ax + xxyy, what is the value of
x and yo, and y infinitely greater than x.
X
when
yy
Here reject xx being infinitely leſs than the reft;
then yy 2ax, and
X
yy
I
=
2a
Ex. 3.
What is the value of 1, when 2ay + yy = rx ;
y, being 0.
x
=
Here yy is infinitely less than 2ay. Whence
2ay = rx, and
y
↑
응​.
X
24
2 RUL E.
Obferve what the unknown quantity is equal to,
when the numerator, &c. vanishes; put the un-
known quantity that value e, where e is
fuppofed infinitely fmall. Which being fubftitu-
ted for that unknown quantity, and the roots of
all furds, extracted to a fufficient number of places
of e; at laſt you will have fome terms in both the
numerator and denominator, which will determine
the value of the fraction.
What is the value of
Ex. 4.
XX when x = a.
avax
a Vax
Put x=a+e, then expunging x;
P 3
}
a√ax
XX
a
Vax
214
B. I.
FUNDAMENTAL
a × aa + ael½
a + · el²
a
a
aa + Lae &c.
aa
2ae &c.
e
a aa + ael
a ×: a + ½ e &c.
že &c.
aa
2ae &c.
- ae
-że
зае
3a, the value of the fraction.
What is the value of
x = a.
Ex. 5.
2a3x
X4
aaax
4
a
Vax³
>
when
Let the fraction =y, and put = a—e, then
✰
3
2a³ X a
e
a
el*
y =
But
4.
3
а
a xa-el
√2a³ Xa
ax a
2a³ × a—e—a-
a—el
a+ + 2a³el½ = aa + ae
je = aa
1+
=
√ 2a+—2a³e—a++4a³e
&c.
3
Alſo ava³ — aae
4
ae &c.
And
a. x a
e] =
a+ — 3a³el == = a
3
e &c.
Whence
aa + ae &c.
aa +}ae &c.
4-ae
16a
} =
3
a
a + 2e &c.
3
9
Ex. 6.
av
Let
a²² 4a³ + 4x³
ax
aa
=y, what is its
√zaa + 2ax
*
a
Let a
e = x.
value when x = a.
And expunging *,
3
3
a
4a³ + 4x a
е
-aa + ae
aa
=y. But
2
zaa + 2 x a·
el
X
a
av
Sect. VII.
215
PROBLEMS.
3
a² 4a³+4Xa—el³=ax8a³—12aae+12aee³ 2aa¬
ae+ee &c. And
2aa+2xx=4aa-4ae +2eel½
ce
20
-et
&c. Whence
40
20a
ae + Lee &c.
>
aa + ae
+ Zee
y: =
ee
24
e + &c.
44
ee
za te
+
46
2aee
= 20°
ee
Here, if I had gone no farther, than the firſt
power of e, it is evident by infpection, that all the
terms would have vanished; by which nothing
could have been concluded.
SCHOLIU M.
If e remains at laft in the numerator, the value
of the fraction is o, and if e remains in the de,
nominator, the fraction is infinite. But if all the
terms vaniſh out of both numerator and denomi-
nator, the feries must then be carried to more
places, to have a ſolution.
PROBLEM
LXXV,
To find two whole numbers x, y; in the equation
axby+c, being in its leaft terms: a, b, c, bea
ing given numbers.
ULE.
Let wh. ftand for the words a whole number.
Reduce the equation, then x =
by + c
a
= wh. By
an abridged fraction, I mean the fraction refult-
ing by throwing all whole numbers out of it, till
the terms in the numerator be less than the deno-
minator.
P 4
216
B. I
FUNDAMENTAL
れ
​by + c be a-
a
minator. Thus let the fraction
bridged to
dy+f. Then to find y.
a
The method confifts in leffening the coefficient
of y continually, till at last it becomes 1.
this is done by fubtracting dy+f
a
And
or fome multi-
ple of it, from y, or any multiple of it, which
comes very near it; that is, from ay, 2ay 3ay
a
a
a
&c. or this from it. And the refulting fraction
abridged, or its neareft multiple, is in like man-
ner to be fubtracted from the neareft foregoing
fraction; or from any wh. which is nearer; or
this from that. And theſe wh. may ay
be
&c. or
2ay
a a
ay ±a ay ± 2a 2ay±34 &c. or any you
a
2
a
a
can find, which has the nearest coefficient to y.
By this means the coefficient of Y is continually
leffened, till at laſt we have " +3
=whp; then
a
will yap-g: where p may be any whole num-
ber taken at pleafure. And y being known, x
will be found from the given equation.
You muſt obſerve in this whole proceſs, to keep
the fame denominator a, throughout.
For whole numbers fubtracted from one ano-
ther, will always leave whole numbers. And
whole numbers multiplied by whole numbers, will
always produce whole numbers. And upon thefe
principles the rule is founded.
گر
Ex.
Sect. VII.
217
PROBLEM S.
Ex. I.
Let 19x 14y-II, to find x, y in whole num-
bers.
By reduction x =
14y
I I
=
wh,
Alfo
Alfo 19
·y
19
19
wh. Then by fubtraction,
19y
[ 4y — I I
19
19
5y + 11
19
20y+44
19
19
=
197
wh. Subtract
=
= wh. And multiplying by 4,
20y+6
19
+2=wh. And 20y+6
; and
Whence y 19p-6. Let p =
firmative value of y, and y = 13.
wh. = P.
Whence x = 9.
19
y+6
19
1, for the leaft af-
Or thus,
19% + II
5x + II
y
x +
=wh. Then
14
14
5x + II
=wh. And multiplying by 3,
15x +33
14.
14-
14x + 28
=wh. But
=wh. And fubtracting,
14
x+5=wb.
14
x
wh.p. And ≈
to have the leaſt; and x = 9, and y = 13.
Ex. 2.
Suppoſe 3x=8y-- 16, query x, y.
x
14p-5. Let p=1,
*
бу 16
Here x =
= 2y―5+
3
2y I
3
= wh.
2y
And
= wh.
And multiplying by 2,
3
4Y
2
= wh. But 3y
wh. And their dif
3
3
ference
14.
218
B. I
FUNDAMENTAL
ま
​ference
2
3
wh. p. Whence y = 3 + 2;
and taking p = 0, y = 2.
Ex. 3.
Let 24x13y+ 16.
3P
Whence xo.
Here x =
13y+16
wb. multiply by 11, and
24
1437 + 176
- wh. But
6 × 243 + 7 X 24
or
24
1441 +168
24
24
y
8
wb. From which ſubtract the for-
mer, and =wh.p. And then y=24p+&,
24
and putting po, y = 8, and x = 5.
Ex. 4.
Let 14x4y+7.
Then x4y+7
=
wh. And multiplying by
14
28y +49
28y+7
7,
or
+3=
+ 3 = wh. And
14
14
283+7
28y
= wh. But
wh. Therefore their
14
14
difference 7 wh. which is abfurd; for an even
14
number cannot divide an odd number, nor a
greater number a leffer. See Cor. 2. Prop. VIII.
B. II. Arithmetic.
}
Ex. 5.
16y.
Let 27x1600
1600 — 16y
Here x =
wh. abridged
7-163
27
27
16y
wb.or
•7
27y
wh, Subtract it from
and
27
27
Se&t. VII.
219
PROBLEM S.
and 11y +7 — wb. multiply by 2, and 229 +14
27
27-
wh. Subtract it from 27y+27, and 5y+13
27
27
wh. multiply by 2, and
10y+26 = wh. ſub-
27
19-wh. = p, and
27
27
tract it from 11y +7, and
y=px27+19, and if p = 0, y = 19, and x = 48.
Cor. 1. All the values of y are bad, by continual-
ly adding the coefficient of x; as y, y + a, y + 24,
y+ 3a, &c. And all the values of x are bad, by
continually adding the coefficient of y; as x, x + b,
x + 2b, &c; or by fubtracting them, for negative
numbers, and both are in arithmetical progreffion.
Cor. 2. When the process brings out an odd num-
ber divided by an even number, or a leffer number di-
vided by a greater, which ſhould be a whole number;
the question is impoffible.
Cor. 3. If it be required to find y a whole num
ber, ſo that the fraction
by + c
a
may also be a whole
number. You must proceed the very fame way, by
y ±
abridging the fraction to 8, and then find
a
y=aPg, where P is any whole number, taken at
pleaſure.
PROBLEM
LXXVI.
To find fuch a whole number x, that being divided
by the given numbers a, b, c, &c. fhall leave the
given remainders f, g, h, &c.
RULE.
Since the fractions
x-ƒ x-g x-b &c. are
whole
220
B. I.
FUNDAMENTAL
whole numbers; put the first f
x-f =P wh.
Then xaP+ƒ.
aPf. Put this value of x in the ſe-
cond fraction; then
b
== wh. Then
(Cor. 2. laft Prob.) find PbQ+m, where
Qwh. then will x = abQ+am + f. Put this
abQ+am+f—b
value of x in the third fraction; then
C
wh. Then, as before, find QcR+n; and
put this inftead of Q in the laft value of x; then
this value of x must be put into the fourth frac-
tion; and proceed the fame way through all the
fractions. This is the method of proceeding; but
numbers muſt be uſed all along inftead of the
fmall letters. And the leaſt wh. number R may
be taken at pleaſure.
Ex. 1.
To find a number which divided by 3, 5, 7, and 2;
will leave the remainders 2, 4, 6, 0, respectively.
5
4 X 6
>
7
X 2
X
Let the number be x, then
3
X
and
are whole numbers.
Let *
2
= P,
2
X
and
3P + 2; then
4
5
3
3P+24
5
3P — 2
wh. fubtract it from
5P
2P + 2
and
5
5
5
-wh. Subtract this from
wh. Q, and P≈ 5Q+4, and ≈ =15Q+14.
X 6
Again
7
wb. and Q+i
15Q+8
= wb.
7
7
=wh.=R, and Q-7R—1, and x=105R I.
3P
2
P - 4
; then
5
5
Laftly,
I
Sec. VII.
221
PROBLEM S.
X
Lattly
105R-1
= wb. and R!=wh.
2
2
2
=S, and R2S+1. Whence x 210S+104,
the number fought; and putting So, the leaſt
value of x is 104.
Ex. 2.
To find a whole number, which being divided by
16, 17, 18, 19, 20; will leave 6, 7, 8, 9, 10,
remainders.
Let x number. Then
X
7, x
8
16
17
18
X 9 X
ΙΟ
>
are whole numbers. Put
X 6
19.
20
16
=P, then ≈≈ 16P + 6.
x =
Then
X-7 $6P I
wh. And thence
17
17
P+I = wh. =Q, and P = 17Q — 1, and
17
x=272Q
IO.
x-8
Alfo
18
=R, and Q9R, whence x = 2448R
18
2.72Q— 1.8
18
2.Q
= wb.
wh. and =wb.
10.
Again
-X 9
2448 R
19
= wh. and
19
19
-3R
= wb.
wh.or
3R
18R
wh. and
= wh. whence
19
19
19
R
wh.
S, and R = 19S. Then
19
* = 46512 S
10.
X
IO
Laftly
465128 - 20
wh. and
20
20
12S
=
Whence
20
= =
wh. T, and S 5T.
x=232550T-10.
And if T = 1, then the
leaft value of x=232550.
Ex.
222
B. I.
FUNDAMENTAL
Ex. 3.
To find a number (x), which being divided by 3
7, 14, 20; there fhall remain 1, 3, 7, 14.
X
I X
Here
3
x
14
7,
are whole
3
7
14
20
numbers. Let
=
P, and x =
P, and x 3P + 1.
3
Then
x-3
3P-2
wh. and
6P — 4
= wh.
7
7
7
P+4
Whence
7
x=21Q-11.
II.
Alfo
x - 7
2IQ — 18
wh. Q, and P = 7Q—4, and
= wb. and 70-4
14
14
14
=wh, and
14Q — 8
8
wh. Whence = wh.
8
14
14
which is abfurd.
Hence the queftion is impoffible for the three
firft fuppofitions; but will hold good for two of
them in which cafe x
leaſt value of x is 10.
2
21Q-11, where the
RULE.
When two divifors and their remainders are
given; then find two fixed multipliers M, N:
fuch, that dividing them,
leaves o, and leaves I remaining.
M
M
a
b
N
N
and
leaves 1, and
leaves o remaining.
a
b
Then divide Mg + Nf,
ab
x, the number fought.
and the remainder is
Likewiſe
Sect. VII.
223
PROBLEM S.
Likewife for three divifors and remainders;
find three fixed multipliers M, N, P; fuch, that
by dividing them,
M
M
leaves I, and
leaves o, remaining.
a
bc
N
leaves I,
leaves o, remaining.
ac
P
P
leaves I,
leaves o, remaining.
C
ab
Then dividing
Mf+Ng+Ph the remainder
abc
is, required; and the like for more quantities.
To prove the truth of this. Since (Cafe 1)
N
M
as alſo
leave o, by divifion; therefore
a
Ъ
Mg, and
Nf
leave o.
a
b
M
N
And fince
as alſo
leave 1. Therefore
b
a
M-I
N
and
b
a
and
Nf
a
a
leave o. Therefore
Therefore Mg-g
الله
leave o; that is, Mg leaves g,
b
Mg + Nf
leaves f. Therefore
and Mg + Nf
b
leaves g+o.
a
b
and
leaves o + f,
But fince Mg + Nf may exceed ab, and there-
fore is not the leaft number; therefore divide by
ab, and the remainder is the leaft number re-
quired. And the fame way, Cafe 2, or any
other, is proved.
Ex.
224
B. I.
FUNDAMENTAL
Ex. 4.
Having the cycle of the dominical letter f, and
cycle of the moon g; to find the year
nyfian period.
11
Let be the year fought. Then
of the Dio-
!
x - f
and
28
are whole numbers. Here a 28, and
wh. P, and M 28P. Alfo
; and multiplying by 2,
x-g
19
M
M—I
—
—wh.
28
19
28P
I
56P — 2
19
19
=wh. Alfo
57P
wh. Therefore
P+ 2
!
=whi
19
19
=Q, and P=19Q-2. Whence M-28x19Q-2
=532Q-56, and if Q = 1; then M
N N
Then = =wh.=P, and N = 19P. Alſo
476.
Ъ
19
N I
=
28
19P - 1
28
wh, multiply by 3; then
57P — 3
wh. and
56P
wh. therefore
P - 3
28
28
28
=wh.=Q, and P=28Q+3. Whence N=28×
19Q + 57, and if Q = o, N = 57.
Therefore remainder of 476g + 57f, which
x
532
ferves in general, for any numbers, ƒ, g•
Let f 10, g= 12; then x = 430.
Ex. 5.
2
Having the cycle of the Sunday letter f, the golden
number g, and indillion h; to find the year of the
Julian period.
Here a
=
28, b = 19, c=15, ab=532, ac=420,
bc 285, and abc = 7980.
Then
Sect. vii.
225
PROBLEM S.
M
Then
285
M
28
I 285P
wh. P, and M 285P. Alfo
1
28
wh. This at laft gives
'
P 17 the leaft; and then M 4845-
wb.P, and N = 420P. Alſo
N
Again,
420
N-I
420P — I
2 P — I
= wh, and
wb.which
19
19.
19
P
Laftly,
will give P10, and N4200.
wh.Q, and P=532Q. Alfo
532
P-
532Q — I
wh. and
= wh. at
15
15
15
laſt Q13, and P = 6916. Whence the re-
mainder of 4845f + 4200g + 6916b
7980
is = x.
Let f 0, or 28, g=1, b = 2; then x2072.
=
Cor. When the operation brings out a leffer num-
ber divided by a greater, instead of a whole number
the problem is impoffible.
PROBLEM
LXXVII.
;
An equation being given, containing feveral unknown
quantities; to find their limits.
When an equation contains feveral unknown
quantities, the values of all of them, except one,
may be taken at pleafure; and when their values
are affigned, and numbers put for them in the
equation, that fingle quantity may alſo be found,
by reducing the equation. And fuch equations
will admit of an infinite number of folutions, if
we admit of fractional and negative numbers.
But fince thefe folutions are moft ufeful where af-
firmative quantities are concerned; and more ufe-
ful ftill, when only affirmative whole numbers are
admitted;
226
B. I.
FUNDAMENTAL
admitted; therefore I propofe to confider only
theſe two cafes, and particularly the laft: becaufe
in that cafe fuch an equation will have a deter-
mined number of folutions. And therefore it is
neceffary to know the limits of the unknown
quantities; left we go about to feek their values
beyond theſe limits.
RULE.
Tranſpoſe the negative quantities to the con-
trary fide; that all the terms may be affirmative.
Then to find the limits of any one, put all the
refto, or ſuppoſe them to vaniſh; and from
hence find the value of that quantity, which will
be one limit thereof. And to know which limit
it is, conceive the other quantities to increaſe and
have fome certain value; then if by this, the
value (of the unknown quantity under confidera-
tion) increaſes; it is the leaft limit you found
if it decreaſes, it is the greatest limit. And in cafe
you find no leait limit, then o is its leaft limit.
This procefs relates to fractional quantities.
But if you only defire whole numbers; put
for each of the other quantities, which is the leaft
value they can have; then from the refulting
equation, find your unknown quantity and its li-
mit, as before directed.
Proceed the fame way with all the unknown
quantities.
Let 30+ 5e = 28,
Let eo, then 3a
Ex. I.
to find the limits of a, c.
28
28, and a = =9
3
Now let e be fome real quantity; it is
I
9;
3
plain the
greater e is, the lefs a muft be; therefore 9 is
the greater limit.
Whence a 93.
For
Sect. VII.
1
PROBLEM S.
227
For e let a = 0, then 5e = 28, and e=
28
5
53. But if a increaſes, e decreaſes; therefore
53 is the greater limit, and e53, and the leffer
limit of both a and e, is ó. All this including
fractions.
For whole numbers.
Let e 1, then 3a 28
= =
=73, the leffer limit, and a 73.
23
523, and a =
3
Again, let a
1, then 5e
25, and e=
25
25
5
5.
=5, the greater limit, and eor
Ex. 2.
Let 3a5e 28, to find the limits of a, e in
whole numbers.
Then 3a 28 +5e; let è = 1, then 3a 33,
33 =11; but when e increaſes a in-
3
and a =
creaſes; therefore 11 is the leffer limit, and a =
or Il.
Let a
=
1, 285e3, and e will be negative,
which we exclude. But whilft a increaſes e in-
therefore o is the leaft limit of e, or
eo and it has no greatest limit.
creaſes;
Ex. 3.
Let 3x+5y+ 82 = 10003, to find the limits in
whole numbers.
Suppofe y=1%, Then 3=10003-13, and
9990
3
z
=3330. And fince x decreaſes, whilft
y and increaſe; therefore 3330 is the greater
limit, and x or 3330.
Q &
Again,
228
B. I.
FUNDAMENTAL
Again, let x 2 = 1; then y 9992, and
9992 1998; and y decreaſes whilft x, z
5
increaſe: whence 1998.
J
Lastly, for z; let xy=1, then 8z10003
8
89995, and z = 9951249. But ≈ de-
creafes, whilft x, y increafe; therefore ≈ 12493.
Ex. 4.
Let 13x-59 + 8% = 10, to find the limits of y.
Sz
I
Here 13x + S≈ = 10 + 5y; let x = 1 = 2;
then gy+10=21, and 5y = 11, and y2; and
whilft x and z increafe, y increafes; therefore 2 is
the leaft limit, and y C 2.
Note, the limits of x and z cannot be found
till the value of y be affigned.
PROBLEM
LXXVIII.
Two equations being given, containing three or more
unknown quanities; to determine their limits
RULE.
Having pitched upon the quantity you would
limit; expunge one of the other quantities, and
you will have one limiting equation. Then ex-
punge another of them, and this gives another
limiting equation. By thefe two equations find
the limits of the quantity pitched on feparately, by
the last problem.
But note, In any limiting equation, all the other
unknown quantities therein, (being put on the
fame fide of the equation, with the abfolute num-
ber,) must have the fame fign: otherwife, (if they
have different figns) they cannot limit the quantity
propofed, till the value of fome of the reft be
known.
If
F
1
Sect. VII.
229
PROBLEMS.
If there be more equations, the procefs is the
fame with any of them,
Let a + e + y = 56.
Ex. I.
and 32a + 20e + 16y=1232; to limit a.
Multiply the first equation by 20, produces
20a + 20e + 20y=1120.
Subtract this from the
fecond, and you have 120-43112: whence
(Prob. lxxvii.) e ≤ 93.
2
Multiply the first equation by 16, gives
16a+ 16е+ 16y 896. Subtract it from the fe-
cond, and 162 + 4e = 336; whence a 203.
In like manner, to limit y, multiply the firſt
equation by 32, and 320+32e+32y=1792. Sub-
tract the fecond from it, and 12e+16y=560 This
gives y 34. And the equation 124-4y=112,
gives yo.
To limit e; the equation 16a + 4e336, gives
e80. And the equation 12e + 16y=560, gives
e453 But here is no leffer limit for e; there-
fore eo, and 45.
Ex. 2.
Let 3x - y + 24 = 202
and 12x + by + 5 = 150°
5u
1ft x 5 is, 15% 5y + IOU = 100
2d X 2 is, 24x + 12y + 104 = 300
difference
9x+179
This equation gives x 20, and y
=
= 200
4
I7
Ift x 6 is, 18x6y+izu 120; add this to
the ſecond: then 30x+174
whence x 83, and u
1ft x 4 is 12x-47+ Su 80.
fecond,
=
101312 = 70.
270,
1477•
Subtract from the
And y 7%, and uco.
There is no leaft limit for x; therefore C. O,
and 83.
Q 3
Ex.
*
;
I
230
B. I.
FUNDAMENTAL
Ex. 3.
1 × (6)⋅ |
3
>
Let a + e + y + u
100.
and 16a+ 10e + 8y + бu = 1200.
u
1 a + e + y + z = 100
216a + 10e + 8y + 6u = 1200
36a + 6e + 6y +
410a + 4e + 2y
600—4—2
6u = 600
= 600
594
4
4÷
5
Į X (10)
5a
10
10
IO
=59 at moſt
6a5910.
2
8 tr.
7100 + 10e + 10y + 10 = 1000
86a zy 41 200
96a = 200 + 2y + 4ų
9 ÷ (6) |10|a =
206
6
I
34 at leaſt, and a□34;
fo a is between 341 and 593. Then
for the other quantities.
Sa + 8e8y + Su= 800.
16u = 1600.
I X (8) I
Į X (16) |12|16a + 16e + 16y +
2
N.
II
13
Sa + 2e
au = 400.
น
This equation will limit u but not e.
Here uo.
6e + 8y + 10u = 400.
400-14_386
12
2
14
14,
15 u
38%, or 38%.
IO
10
600-4e-Ica
586
4,
Isy=
= 293, or
2
2
}
y293; but, fince the limits of a,
are known; y may be determined
more exactly; thus
595-10 × 35 246
y =
y123.
2
2
123, or
14
Sect. VII.
231
PROBLEM S.
14
17 Again y=
400-16 384
= 48, or
8
8
yor 48. But there are all
greater limits of y, and there wants
the leffer limit; therefore yo, and
48.
400-18 382
14
18e =
6
6
= 633, or e=
633, But the leaſt limit of a can-
not be found; therefore take eo.
SCHOLI U M.
When three numbers are fought by two equa-
tions; all the values of each of them, in whole
numbers, make three feries of arithmetical pro-
greffion, taken within the limits of thefe numbers.
And if four or more numbers are fought, the va-
lue of each is to be found in feveral arithmetical
progreffions. But yet the values of any three will
be in arithmetic progreffion, when the values of all
the reft are affigned, as before for three numbers.
For in the cafe of three numbers, and two equa-
tions; any one of the three may be expunged; and
then you will have but one equation, and two un-
known quantities; which brings it under Prob. lxxv.
But by Cor. 1. of that problem, theſe two remain-
ing quantities are contained in two feries of arith-
metical progreffion. And as any of the three may
be expunged; therefore any two of them will con-
ftitute two feries of arithmetical progreffion.
PROBLEM LXXIX.
The prices of feveral ingredients being given, to find
the quantities thereof; fo that the mixture may be
jold at a given price.
Suppofe four fimples A, B, C, D, are to be
mixed; and their prices to be as follows:
Q 4
Mean
232
B. I.
FUNDAMENTAL
i
Mean price = m
Price of A = m + a
of Bm + b
of C m
of D = m
C
d.
And let the quantities to be taken of A, B, C,
D, be x, y, z, v, refpectively. Place them in
order, thus:
prices quantities
1
m + a
m + b
x
y
m
m
C
m
d
א ←
Z
Then by the nature of the queſtion; if each
quantity be multiplied by its price, the fum of the
products will be equal to the fum of all the quan-
tities multiplied by the mean price; that is,
d X v
m + a x x + m
+m+ bxy + m
=x+v+ y + zxm.
> CX Z
Let m+a X x + m²
xoxoxm
And m + bxy + mcxz = y + zxm.
That is,
mx + ax + mu
my + by + mz
by the former, ax
dv = mx + mv
cz = my + mz.
dvo, or ax = dv.
by the latter, by - cz
cz=0, or by = cz.
o,
Now fince x and y may be taken at pleaſure.
Therefore put xd, and y = c. Then will v = a,
and b. Whence the quantities will be ranged
thus:
z
713
Sect. VII.
233
PROBLEM S.
m
m + a
m + b
d
m
C
b
112 d
a
which gives this
RULE.
Couple every greater rate with one leffer than
the mean price (m+a and m-d; alfo m + b and
m c); then take the difference between each rate
and the mean rate, and place it alternately, that
is, against the quantity it is coupled with; do the
fame with all the rates, (thus place a againſt
m-d, b againft mc, c againſt
c, c againſt m + b, d againſt
m + a); then if none of the quantities of A, B,
C, D, be given. Then d, c, b, a will be the
quantities of each to be taken for the mixture.
But if any one quantity be given; then all the
quantities d, c, b, a muſt be increaſed or de-
creaſed in proportion. Or if the fum of the quan-
tities be given, then other quantities muſt be taken
in proportion, ſo that d+c+b+a may be to
the fum given, as any of the differences d, c,
&c. to the refpective quantity required. And this
is the common rule of Alligation Alternate.
Again,
Since ax = dv, and by = cz.
y = no;
Take xmd, and
then v v=ma, and ≈=nb. Then putting
md, nc, nb, ma, for x, y, z, v refpectively; and
the cafe will ftand thus:
m + a
m + d
md
no
772
711
C
ub
172 d ma. which gives this
RULE.
Having coupled the rates as before directed,
and taken the differences. Then inftead of any
couple
a
234
B. I.
FUNDAMENTAL
couple of the differences, you may take any equi-
multiples thereof; and place them alternately.
And theſe (or other quantities proportional to
them), will be the quantities required. And this
is the Rule of Alligation improved.
PROBLEM LXXX.
If the numbers A and B be produced from a and b,
by any fimilar operation; to find the number from
which N is produced, by the like operation. Sup-
pofing the differences of the numbers A, B, N, to
be as the differences of a, b, and the unknown
number.
a
b
Let z be the number fought,
Z and put the differences NA=r,
ABNN-Bs. Then by the queſtion,
↑ (N — A) : s (N − B) : : ≈ — a:
z-b. Then rz-rb=sz-sa. And by tranfpo-
fition, rz-sz-rb-sa, and z =
rb. sa
; or if s
до
S
be negative (or B greater than N), then z =
rb + sa
r+s
2
the number fought.
Cor. 1. Hence is derived the practice of the double
Rule of Falfe. For if both
N, or both greater; then z
A and B be leffer than
rb sa
=
But if only
r
S
one as В be greater than N, then s is negative,
and z =
rb + sa
r + s
That is, if each fuppofed number be multiplied by
the error of the other, and the difference of the pro-
ducts be divided by the difference of the errors, when
the errors are like; or the fum of the products di-
vided by the fum of the errors, when the errors are
unlike; the quotient gives the number fought.
Cor.
Sect. VII.
23
PROBLEM S.
Cor. 2. Hence alfo is derived another method of
avorking the Rule of Approximation, or Rule of Falfe,
which is this.
Multiply the difference of the fuppofed numbers, by
the least error, and divide the product, by the dif
ference of the errors, if like; or by the fum if unlike.
The quotient is the correction of the number belonging
to the leaft error.
Then this correction is to be added or fubtracted,
according as that number was too little or too great.
For lets be the leaft error, being the error of
, and q = the correction; then if A, B be lefs
g
rb - ṣa
than N, 6+q=≈, and q=z—
rb sarb + sb
-
gue
S
b =
b =
g
- S
è
a
S.
↑
S
But if B is greater than N, then b-qz, and
rb + sa rb + sb — rb- sa
q=b―z=b —
r+s
rts
b
a
S.
g s
S.CHOLIU M.
Since it has been fhewn, that the number fought
will come out exactly, by this rule, when the er-
rors are exactly proportional to the differences of
the ſuppoſed numbers from the true one. There-
fore it follows, that when the errors are nearly pro-
portional to thefe differences, that the anfwer will
come out nearly true. And theſe proportions will
be the nearer to an equality, the nearer theſe
fuppofed numbers are taken to the true number.
And therefore in all queftions where this rule is
applied, every operation will bring us
the true anfwer, if we always take the neareſt
numbers, (where the errors are leaft) for new fup-
pofitions. And thus repeating the operation, one
nearer
may
236
B. I.
FUNDAMENTAL
may continually approximate to the true number,
within any degree of exactnefs required; let the
particular queftion be of what nature it will.
Upon this rule alfo is founded the rule of finding
proportional parts.
PROBLEM
LXXXI.
Suppoſe A, B, C, D, &c. to be feveral forts of
goods; and m, n, p, q, &c. given numbers;
and the values of theſe goods are
mA = nB
pB = qC
~C
SD
tD = vE
To find what quantity of the last fort is equal to a given
quantity of the first: and the reverſe.
Let z times the laft be y times the first, that
is, let
zE ≈yA.
Multiply all theſe equations together; the firſt
fide by the firſt, and the fecond by the fecond.
Then we have
mAxpBxrCxtDxzE=nBxqCxsDxvExуA. Then
nqsvy. Then if the quantity of the laſt
mprtz = nqsvy.
ngon
=
fort be required, ≈ngsty. But if the quantity
mprt
of the first fort be fought; y =
mprtz
Whence
ngsv
this
RULE.
Place the terms in two columns, fo that there
may not be two terms of a fort in either column.
Then multiply the numbers in the leffer column
for a divifor; and the numbers in the greater co-
lumn (with the odd term) for a dividend. The
quotient
Sect. VII.
237 S
PROBLEM S.
quotient is the quantity of that fort which ftands
fingle in the two columns. And this is the Rule
of Exchange in arithmetic.
PROBLEM LXXXII.
To inveſtigate numbers for rational ſquares, cubes, &c•
Problems of this fort are often capable of an
infinite number of anfwers; and yet none of the
quantities can be affumed at pleaſure, but muſt
be inveſtigated as follows.
RULE.
Put one or more letters to denote the root of
the ſquare, cube, &c. Which letters muſt be ſo
affumed, that when the equation is involved, ei-
ther the given number, or the higheft power of
the unknown quantity, may be on both fides of
the equation, and confequently vanishes out of it.
And then if the unknown quantity be but of one
dimenfion, the problem is folved, by reducing
the equation. But if the unknown quantity is
ftill a fquare or higher power; you muſt farther
affume other new letters, to denote the root, and
proceed as before; till you get the unknown
quantity of one dimenfion; and from this un-
known quantity all the reft are to be determined.
For the whole art is, fo to denote the root of the
given power, that the unknown quantity may be
reduced to one dimenfion.
But no general rule of proceeding can be given
to fuit all cafes; and therefore the folution will
often be left to the fagacity of the analyſt, in con-
triving fuch a defignation of letters as is proper
for the purpoſe.
J
Ex
238
B. I.
FUNDAMENTAL
Ex. 1.
To find two fuch numbers, ſo that the fum of their
fquares is a fquare.
Let x, y, z be the roots of the fquares, fo that
xx+yỳ=zz. Affume z=y+r, then xx+yy=ZŻ
=yy+2ry+rr, and xx=2ry+rr, and 2ry-xx-rr,
where y the unknown quantity is of one dimen
fion, which reduced gives y=
XX
2r
rr
; and
XX
дада
xx + rr
y+r=
y + r
+r=
%. Therefore
27
27
XX
за да
xx + rr
the numbers are x,
and
"
where x
27
27
and r denote any numbers taken at pleaſure.
But if the answer is required in whole numbers,
then 2rx, xx - rr, xx + rr will denote the roots of
the fquares, where the fum of the two firft is equal
to the laſt fquare.
Cor. The three fides of a right-angled triangle will
only be commenfurable, when xx + rr denotes the hypo-
thenuſe, and xx-rr, and 2rx the two fides; x, r
being any numbers taken at pleasure, fo as x is
greater than r.
Ex. 2.
To find two numbers, the fum of whofe fquares is
equal to the fum of two given Squares.
Let x, y be the roots; aa, bb the given fquares.
Affume xa-v, j = vz-b. Then xx+y=aa
+bb = aa— zav + vv +vvzz-2bvz+bb; and
vv + vvzz = 2ev+2buz, and v+vzz=2a+2bz;
2a +2bz
and v =
ZZ + I
Where is any number ta-
ken at pleaſure. Then x=
AZZ
2bz
a
and
2
ZZ + 1
2
2 ax + bzz — b
y =
22 + 1
Or
Sect. VII
239
PROBLEM S.
Or thus,
Let xav, then aa-zav+vv+yyaa+bb; and
Yy—2av+vvbb. Put y=vz-b; then vvzz-
2bzv+bb-zav+vv=bb, and vvzz+vv=2bzv+
2bz+2a
zav, or vzz + v = 2bz + 2a, and v =
as before.
Ex. 3.
ZZ + I
To find two numbers, fuch that when either of them
is added to the ſquare of the other, the fum will be a
Square number.
24
Let the numbers be *, y; then xx+y= 0›
and yy+x=0, Let xx + y = r — x = rr — 2rx
+xx; then y = rr2rx, and 2rxrry, whence
z = rr — y
2r
Again, affume yy+x or yy +
gro
Y
= y + v =
27
gy+2yv + vv. Then
rr — y
= 2yvvv, whence
Friy = 4ryv + 2rvv, and 4rvy+y=rr2rvv ;
2r
ry 2rvv
whence y =
400+ i
2rro + vv
And x =
4rv + I
where r, v may be taken at pleaſure, provided
*
be greater than 2vv.
Otherwife,
зада
・y
y
Since x =
= 17 — ——, and yy+w or yy—
27
2r
2 + r = 0, put yy - 2+vr=y−2+m=
2r
yy-
I
Then r =
1бr r
3
and r³, which is
a cube
240
B. I
FUNDAMENTAL
I
a cube number. And therefore will answer the
queſtion; and we have r = 1; whence x = —y,
and y may be any thing less than
Ex. 4.
I
2
I
4.
To find two numbers in a given ratio, ſo that either
of them added to the fquare of the fum, may make a
Square.
t
Let the ratio of the two numbers be as b to c,
and put b+c=d, and let the numbers be bx and
cx. Then the fquare of the fum is bx + cxl² =
ddxx. Therefore ddxx+bx=0, and ddxx+cx=0.
Put ddxx+bx dx—vľ²=ddxx—2dxv+vv; then
bx=vv-2dxv, or bx+2dxvvv, and x =
Then ddxx+cx or ddx + c x x =
ขบ
b+2dv
ddvv+bc+2cdv
b + zdv
VV
VV
X
=0, but
b + 2dv
= therefore
b+2dvl²
ddvv-2dvz
ddvv + bc + 2cdv□ (See Cor. 27. II. Arithm.);
affume ddvv + bc + 2cdv=dv—zľª — ddvv
+ zz; then 2cdv + 2dzv = 2Z bc; and
bc
ZZ
V
Where zz muſt be greater than
2cd + 2dz
ZZ
zz bc2
bc, and expunging v, x =
4ddz × b + x × c + z
SCHOLIU M.
It appears from theſe operations, that when a
quantity, which is to be a fquare by the problem,
is not an algebraic fquare; we muſt make it ſo,
by affuming fome new quantities to compleat it.
Then thefe fquares being compared, an equation
is had for determining the unknown quantity. And
in
Sect. VII.
241'
PROBLEMS.
in working, one may multiply or divide by any
quantity which is a fquare, and what is left will
be a ſquare, in a more fimple form. The like
for other powers.
PROBLEM. LXXXIII.
To determine the maximum or minimum of a quantity
proposed.
When a quantity is required to be the greateſt
or leaft poffible, it is called a maximum or mini-
mum. And at the time it becomes fuch, it is at
a ftand, and at that moment neither increaſes nor
decreaſes. Therefore to compute it.
RULE.
Calculate the value of the maximum or mini-
mum two different ways, which is done by in-
creafing the unknown quantity therein, by an ex-
ceeding fmall part; then thefe values are to be
put equal to one another. The fame muſt be
done, if there be feveral variable quantities. But
go no farther than the firft power of the ſmall
added part. Or,
If the maximum or minimum confifts of two
parts; compute the exceeding ſmall increment of
one, and the decrement of the other; and put them
equal to one another.
Ex. 1.
What fraction is that whofe fquare exceeds its cube
the greatest poffible.
3
3
Let x be the fraction, then x²-x³-max. Take
e an exceeding fmall part to be added to x, then
you will alſo have x + el-x+el³
x+el = max. that is,
xx + 2xe 223 3x²e max.
max. Whence x²
2xe X3 3x²e, and 2xe
2xe, and 3x=2, or x 3.
R
2
Whence x-x³-xx
3*²e=0, or 3*²e
Or
242
B. i.
FUNDAMENTAL
Since
x²
Or thus,
max. let e be the finall increaſe
of x, then 2xe is the increment of xx, and
is the decrement of x3; therefore 2xe
1x20
3x²e, and
*
3
as before.
Ex. 2.
To divide a given quantity into two parts, that
one of the parts multiplied by the cube of the other
part; the product may be a maximum.
Let a be the quantity, and one part, and '
ax the other part, and e a ſmall additional part
tỏ x. Then x³ X a x or ax³
3
x4 = max: =
ax³ + 3ax²e—x44x³e. Then 3axe 4x³e, and
xa, for one part, and a-xa, the other
part.
Ex. 3.
To find a³ — a²x + x³ a minimum, x being un-
known.
Putxe for x. Then a3a²x + x³ min.
= =
a³——a²x—a²e+x+3x²e, and -a²e+3x²e=0, and
3xxaa; whence x = a. Then a³-a2x+x³
3
J
@³—a³ √ {} + {{a³ √ ÷ a³×1-√, the minimum.
//
2
3
3
Ex. 4.
baaxaaxx-6x3
x 4
Let
-a+x be a maximum.
baa + x³
This reduced to a common denominator is
2bbaax+aaxx-bx³-ba³-ax³
max. Put x + e
baa + x3
for y
Then
Sect. VII.
243
PROBLEMS.
2baax + aaxx
bx3
ba³
0x3
Then
baa + x³
2baax + 2baae + aaxx+2aaxe—bx³ — 3bx²e—ba³ —
ax³ захте
;
baa + x³ + 3xxe
Then multiplying alternately,
zbaax+aaxx-bx³-bas-ax³ x baa+x³ + 2xxe-
: 2baax+2baae+aaxx+2aaxe-bx³-3bx²e—ba³ —
2
ax²
3ax¹e: × baa + x³.
what is common on both ſides,
2baax + aaxx
bx3
And throwing out
baz
ax³ × 3xxe = baa + x³
36x5
× 2baae+2aaxe—-3bxxe-zaxxe. That is (dividing
4
by e), 6baax³ + zaax+ —
2bba4 + 2ba+x
—
3ba3xx 3axs =
3bbaaxx-3ba xx+2baax³+2aax4
36x5-3ax. Reduced, 4baax³ +aax+ =2bba+ +
2ba+x-3bbaaxx; or dividing by a, and tranfpofing,
*4 + 4bx³ + 3bbxx — 2baax — 2bbaa = 0.
Ex. 5.
Suppoſe y³-3yyx + 3yxx = nyx — nxx.
and x
4
y = max.
Zy
x
Suppole the maximumm. Then my
This fubftituted in the firft equation, and reduced,
gives y³+3m²y=uyy-mmn. And y+12mmy-
nyy + 4mìnn = o. Where m is a fixt quantity. Put
ye for y; then y³ + 12m³y-nyy + 4m²n =0=
y³ + 3y²e + 12m³y + 12m²e — ny²-2nye+4m²n=0,
and 3y²e + 12m²e 2nyeo, whence 3y+12m²-
From this equa-
tion, and y+12m³ynyy + 4m no, the quan-
tities y and m will eafily be determined.
2ny=0, or 2ny
33y = 12mm,
Ex. 6.
Through a given point P within the angle BAC, Fig.
to draw a right line BPC, making the area of the 1.
triangle BAC, the least poffible.
Draw AP, and bPc extremely near BPC; then
the area ABP + ACP
minimum,
R 2-
In the very
fmall
244
B. Í.
FUNDAMENTAL
Fig. fmall triangles BPb, and CPc, the vertical angles
I. at P are equal, and BPbP, as alſo CPCP,
extream near. Therefore the areas BPb, and CPc,
are to one another as BP to CP (Geom. 19. II).
But CPC is the increment of the area APC; and
BPb is the decrement of the area APB. There-
2.
2.
fore BPb CPc, or BP CP2; therefore
=
=
BPCP. Whence if PD be drawn parallel to
CA, then DB = DA.
Ex. 7.
To find the greatest triangle inscribed in a circle
ACBD.
Draw the diameter AB, and CD perpendicu-
lar thereto; alfo draw AC, AD. Let AB = d,
AE¤×, EC≈y: then triangle ACD = xÿ=max.
x,
or xxyy = max. but yy dx *** therefore
dx³-x+= max. dx³ + 3dx²e-x-4x³e (putting
xe for x), and 3dx'e 4x³e, or 4x 3d, whence
+
x = 3d.
x4 = 3
SCHOLI U M.
When any quantity is a maximum or minimum,
its root, or its fquare, or its cube, &c. will like-
wife be a maximum or minimum. Alfo when any
quantity is a maximum or minimum, any given
quantity may be added to it, or fubtracted from
it, and it will still be a maximum, or minimum.
Likewiſe it may be multiplied or divided by any
given quantity, and ftill remain a maximum or
minimum.
PROBLEM
LXXXIV.
A number or quantity being given; to find its loga-
rithm by a feries, or to turn numbers into logarithms.
X
Let be the quantity given; M = 1, for
y
Neper's logarithms, or M =,434294482, for the
common
Sect. VII.
245
PROBLEMS.
common logarithms. And let x-y=v, x+y=z.
Then the logarithm of will be denoted theſe
y
feveral ways following, deduced from the nature
of logarithms.
X
1. Log: =MX:
y
X
2. Log: Mx:
y
४ c
บ
V 22 V} 74 95
+
+
y . 2y² 3y³ 434 535
Or,
+
v2
23
23 24
5x5
&c.
+
3203
+
25
+
4x4
+ &c.
Or,
+
+
323
525
3. Log:2MX:
y
212.
V7
+ + &c.
727
Cor. 1. If v be far less than 1. Then
ขะ
VV 23
Log: 1Mx:~—
+
+ &c.
2
3 4
5
I..
For then
This is plain by putting y = 1.
* = I +V, and
X
I + v.
Y
ข
Cor. 2. Log: y+v=lógy,+Mx: --
y
73
74
+
&c.
2y
3y3 4y4
or log:y+v=lag:y, +Mx: དུ
or log:y+v=log:y, +2Mx;
For log: ≈ or y+v=logy×
+ +
V
222
V3
+
Z
V
323
213
V4
+
&c.
4204
vi
&c.
727
X
+ +
323525
=log: y+log:
R 3
Cor
246
B. I.
FUNDAMENTAL
logarithm of n,
Cor. 3. If I
and 1+ slogarithm of n + v. Then the ad-
ditional part of the logarithm, that is,
S=MX:
or s = Mx :
or s = 2MX:
V
་
73
24
+
n
272 3n3
&c.
424
ข
22
V3
+
+
&c.
n + v
2.1+01²
3.n +01³
3
ข
Vi
+
+
&c.
2n+v
3.2n+v³
S
5.2n+v
For fince 1+s=log: n+v, and
25
log: n; there-
n + v
foreslog: n+v-log: n= log:
n + v
that is, slog:
n
n
And by this prop. (writing a
for y, n+v for x, and 2n+v for z); s or log:
will come out as above.
Cor. 4. If x be far less than a, then
n+v
n
Log: a + bx + cx² + dx³ &c. = log : a + M × :
bx + cxx + dx³- &c.
bx + ccxx &c.l
2
a
2aa
3
bx &c.
+
&c.
3a3
a
and log:
-MX:
a
bx
CXX
dx³ &c.
bx + cxx + dx³ &c.
3
bx + cxx &c [2.
+
:
a
of
bx &c. 13
303
&c.
200
and
Sect. VII.
247
PROBLEM S.
a + bx + cxx + dx³ &c.
and log:
!
a
bx
2M X ::
CXX dx³ &c.
bx + cxx + dx³ &c.
bx + cxx &c |
3
+
a
303
bx+ &c.15
+
&c:
5a5
and
The firſt cafe appears from Cafe 1, Cor. 2.
writing a + bx + cxx &c. for x, a for
bx + cxx &c. for v.
ولا
The ſecond appears from Cafe 2. of this prop.
writing a for x, abx cxx &c. for y, and
bx + cxx &c. for v.
The third appears from Cafe 3 of the prop.
writing a + bx + cxx &c. for x, abx
for y, 2bx+2cxx &c. for v, and 2a for z.
SCHOLIU M.
cxx &c.
2 Mv
The log: y+v = log : y : +
very near,
2y+v
when
is very fmall, which is only the firft term
of the feries, Cafe 3. Cor. 2.
PROBLEM
LXXXV.
A logarithm being given; to find the quantity belong-
ing to it, or its number, by a feries. Or to turn
logarithms into numbers.
$
บ
Lets be the logarithm given, n + its num-
ber, and let / be the logarithm of the number n.
I
Put m = 2.302585093
2.302585093 = √, for the common lo-
M
garithms, or m = 1, for Neper's logarithms. Then
by Cor. 3. laft Prob. s = MX:
V
n
23
+ &c.
2n² 3223
R 4
and
248
B. I.
FUNDAMENTAL
1
and orms =+ &c. Then by re-
M
n
verfion of ſeries (Prob. Ixii ),
23
3n3
ข
ms/2
ms +
n
2
ms [³
3
ms 4
+
+
&c. Then
2.3
2.3.4
ms!²
msl³
1. v=nx: ms +
m2.5
ms
2.n+v=nX:1+ms+ + +
+
&c. Whence
2
2.3
ms|3 ms|4
&c. and
2
2.3 2.3.4
n+ v
msl²
ms/3
3.
ள்
= I + ms +
+
&c.
n
2
2.3
That is,
Number of + s = number of / × : 1 + ms +
ms 2
ms 3
ms! 4
+
+
&c.
2
2.3
2.3.4
Cor. 1. If n =
I + v or
1, and l = 0; then
number of
= I + ms +
ms 2
ms13
ms14
+
+
&c.
2
2.3
2.3.4
l
-
Cor. 2. If I log: n, and l+s=log: n+v; then
the additional part of the number, that is,
4
v=nX : ms +
ms²
+
ms/3
msl¹
+
&c.
2
2.3
2.3.4
Cor. 3. If L be the log:
N° 1+mxL +
of the number N, then
3
mxL[²
+
mx Ll mx LI
MxLİ
+
4
&c.
2
2.3
2.3.4
For
f
Sect. VII.
249
PROBLEM S.
For, by Cor. 1. I + v (numb. of s log.) =
ms 2
ms13
I + ms +
+
&c. Where I
2
may
2.3
There-
repreſent any number, and s its logarithm.
v = N, and sL; then
fore let
mLl²
2
N (numb. of L log.) = 1 + mL + +
2
&c. therefore by the nature of logarithms,
N* (numb. of xL log.) = 1 + mxL +
mx L
3
&c.
mL 3
2.3
mxL|z
2
2.3
Cor. 4. If yn+ v, x = r + e, l=log: n. Then
e
3* or n + vl =n'×: 1+ mel +
mell
mell³
3
+
2
2.3
&c. X: I +
r te
rte r+e-
VV
X
+
X
X
I
ท
I
2
nn
rte
r+e-
r te
2 203
+
X
X
X
&c.
I
2
3
223
=
te
For n+vr+en" +ex I +
but by Cor. 3. ne = 1 + mel +
v
irte
= nr x ne X
n
irte
1+
n
mellz
mell³
irte
v
+
&c. and i +
=i+
r te
X
2
2.3
n
I
17
r+e
r+e
I
ขา
+
X
X
&c.
I
2
nn
Cor. 5. If I
=
log: n; then
x:
For here = 0.
v
Note = n² X : I + mel +
mell²
mell³
3
+
&c.
2
2.3
Cor.
250
FUNDAMENTAL, &c. B. I;
13
Cor. 6. If v, è be exceeding fmall, then
rv
n + v² + e = n² × : 1 + mel +
nearly, being on
n
ly the first power of e and v.
Cor. 7. If n = number of the logarithm a; then
the number of the logarithm a + bx + cxx + dx³
1
=nx into I + m × bx + cx² + dx³ &c. +
m3.
&c.
m²
2
X
n24
bx + cxx &c.¹² +
× bx + cxx &c.1³ +
3
X
263
2.3.4
bx + &c.1+ + &c.
This follows from this problem, putting a,
and s bx + cxx. &c.
PROBLEM
LXXXVI.
A problem being refolved analytically, to demonftrate
it fynthetically.
RUL E.
When a problem has been folved algebraically,
the demonftration of it is to be deduced from the
ſteps of the algebraic procefs; by going backward
from the end of it to the beginning; obferving
how each step is formed from the foregoing, and
forming your procefs accordingly.
SECT.
.1
251
SECT.
VIII.
The Refolution of Equations; and the extraction
of their roots in numbers.
PROBLEM LXXXVII.
To find the limits of the roots of an equation.
W!
HEN an equation is propofed to have its
root extracted, it is proper to find the li-
mits of the roots; left we lofe our time in feeking
the roots beyond thefe limits.
RULE.
Reduce the equation, that the higheſt term may
have I for its coefficient; then fquare the coeffi-
cient of the fecond term, from which ſubtract twice
the coefficient of the third term, then the fquare
root thereof is greater than the greateft root of
the equation. But the equation fhould be clear
of impoffible roots.
For that quantity is the fum of the fquares of
the roots, by Prob. xl. Art. 9. and that fum, is
greater than the fquare of any one root.
Or thus,
Subſtitute ſeveral numbers fucceffively for the
unknown quantity; till at laft you find two num-
bers which give, one a pofitive, and the other a
negative refult. Then the root is between theſe
numbers.
There are other rules among the writers of Al-
gebra, which come nearer; but then they are more
laborious.
Ex.
252
B. I.
RESOLUTION of
Let x+3x-5%
Ex. I.
200.
=
Then 3 × 32x-5=9+10 19.
and 194.3, &c. Therefore 4, 3 is greater
than any of the roots.
Ex. 2.
Suppoſe xxx-5=0.
If x 2, then the refult is 2—5—3.
=
If x 3, the refult is 6-5 +1.
=
=
Therefore the root is between 2 and — 3.
PROBLEM LXXXVIII.
To refolve a quadratic equation, and extract its root
in numbers.
I comprehend all equations under the name of
quadratics, in which are two terms involving the
unknown quantity; and where the index of one is
double to that of the other. As in theſe,
aa + ba = d
a4 + ba² = d
a+ba³d, &c.
where b, d, may repreſent any numbers, affirma-
tive or negative.
Every quadratic equation has two roots, though
perhaps only one of them will answer the queftion
propoſed, And to find theſe roots the equation
propoſed muſt be first reduced, by dividing all,
by the coefficient of the higheſt term; and then
tranfpofing the known quantity to the contrary
fide. Which done, the equation will appear thus,
aa+bad. Now add to both fides bb the ſquare
of half the coefficient of a, and we have
ɛa + ba + 1 bb bbd, where the firft fide is a
compleat
}
Sect. VIII.
EQUATIONS.
253
compleat ſquare; therefore extract the ſquare root,
and a +16=±√bb+d, tranfpoſe b, then
a = − b ± √ ÷ bb+d. So a becomes known, be-
ing either equal to 1b + √ 1bb + d, or to
Whence this
b
1
— 16 - ✓ 1bb + d.
I
RULE.
The equation being cleared, compleat the fquare
by adding to both fides the ſquare of half the co-
efficient of the ſecond term. Then extract the root
of both fides, which may be either+ or -;
then tranſpoſe the known quantity.
Note, If the abfolute number is negative, and
greater than the ſquare of the coefficient; the
equation is impoffible.
4
If aa + bad,
Then a√ ±bb + d — 2 b.
And the root extracted in numbers gives a; but
ifbb is leffer than d, and d negative; it is im-
poffible.
If aa+5a=68.
Ex. I.
Then a =±√68+6=-=±√74.25 — 2.5 3
74.25 (8.6168 &c.
64
166) 1025
+6/996
17212900
+1/1721
17226) 117900
+6 103356
17232) 1454400
+ 8.6168
2.5
+ 6.1158 = a
11.1168 = a
Ex.
254
·B. I.
RESOLUTION of
Į
Ex. 2.
Let aa
6a= 27•
9+27=3±√36.
Then a 3+
=
that is, a = 3 + 6 = 9.
or
a 3- 6 = -3.
Ex. 3.
Suppoſe aa - ·236a1155.
Then a
118±√1182—1155;
that is, a = 118 + 113 = 231
or a = 118-113 = 5.
2 RULE.
When you have large numbers to deal with
it is better to proceed thus. Clear the equation,
And if aa + bad,
d
then a =
the form.
ว
b ta
d
To find the firſt quotient figure, take, when
b is far greater than a; or take √d, when a is
far greater than b; or take when a and b
d
26
are nearly equal; thus it will eafily be found by
a few trials. Or in general, take the firft figure
fuch, that when it is multiplied by the fum of it.
felf and b, it will produce the first figure or fi-
gures of d, or the next lefs: this is all the dif
ficulty. Then multiply and fubtract as ufual, the
remainder is the refolvend.
Then to continue the divifion; you muſt find
a new divifor for each quotient figure, thus. Add
the last quotient figure to the laft divifor (duly
obferving their places), for a new divifor; fee
how
}
Sec. VIII.
EQUATION S. 255
how oft this is contained in the refolvend, fet the
anſwer in the quotient, and alſo add it to the di-
vifor; then multiply the whole divifor by that quo-
tient figure; and fubtract the product, for a new
refolvend. But when any of the figns are nega-
tive, the proper quantities are to be fubtracted, in-
ftead of being added. This work is always to be
repeated for each quotient figure.
When any quotient figure is fo great that the
product exceeds the refolvend, place a leſs figure
in the quotient.
When you have got more than half your in-
tended number of figures in the quotient, you
may continue the divifion without adding the new
quotient figures to the divifor.
Obferve, each quotient figure is to be added
twice to the divifor; once before multiplication,
and once after; juft as in extracting the fquare
root, and for the fame reaſon. For this method
extracts the fquare root, when bo.
When one root is had, the other is found, by
adding this to the coefficient b; for the fum,
changing its fign, is the other root.
This rule is the foundation of the method for
extracting the roots of adfected equations.
品
​Ex. 4.
Let aa + 32a = 4644.
then
4644
a
32+ a
4600
Suppofe
= 100 too great for a.
32
4644 60, which is alfo too great for a. Take
4600
64
=7, too great. Take a 50.
32
256
B. I.
RESOLUTION of
32
+ 50·
82) 4644 (50
+ 54 410.
136) 544 (4
544
·| 54 = a
Ex. 5.
Let aa + 35a=28349994
aa+35a
28349994
a
35 a
5000 nearly.
Here a√28 &c.
+35
5000
5035) 28349994 (5307 ≈ ₫
5300/ 25175
10335
31749
307 31005
10642)
74494
74494
Ex. 6.
==
Suppose aa—5307a —— 184520.
184520 184520
then a =
5307+a
5307-a
184
Here a=
= 30 nearly.
5
5307
Se&t. VIII.
257
EQUATIONS.
5307) 184520 (35 = a.
-30 15831.
5277) 26210
-35
26210
5242)
Ex. 7.
Let aa +463a = 26698
26698
a =
463+a'
463) 26698 (51.855342 a
+50 2565.
513) 1048
+51
564
564)
484.00000
+1.8 45264
565.8) 31.3600
+.85 283325
566.65) 30275
+.5 28335
566.70
1940
1700
240
226
14
II
3
Scholium. If x++ bx²=d.
Put a xx, then
aa + bad; and find a as above. Then x = √ɑ,
S
by
!
2-58
RESOLUTION of B. I.
by extracting the root. And the fame for higher
equations.
To prove the truth of this rule. Let x+y+z &c.
be the true value of a; the firft figure, y the
fecond, and z the third, &c.
aa + bad, the value of d will be
Then fince
bxx+y+z+x+y+zl³, whence x+y+≈ &c. or
{
d
@=
b+ a
3
=
bx + by + b≈ + x + y +
b + x + y + z
The operation.
2.
bx+by+bx+xx+2xy(x+y+z &c.
b
+) bx+by+
+yy + 2xz
+zz+2yz
+x/
I divifor b+x) bx + xx
by+bz+yy+zxy refolvend
+zz+2xz
+x+y
2 divif. b + 2x+y)
+y+z
+232
by+2xy+yy
3 divif. b + 2x + 2y +z) bz+zz+2xz refolvend
+2yz
bz+2xz+2yz+zz
Here bx+by &c. being divided by b gives
in the quotient, and x added to b, gives b+x for
the divifor, and bxxx for the product, which
fubtracted, leaves the refolvend by + bz + 2xy
+yy &c.
Then in order to get the fecond figure y, the
refolvend by + 2xy + yy &c. is to be divided by
b+2x+y. Therefore x + y is to be added to the
Therefore+y
laſt diviſor b + x, to get the new divifor b+2x+y.
This divifor multiplied by y, gives by + 2xy + yy,
which
Sect. VIII.
EQUATION S.
259
which fubtracted, leaves bz + 2xz + zyz + zz for
the refolvend.
Then to get the third figure z, it is plain, the
refolvend bz + 2xz + 27% + zz muſt be divided by
the divifor b + 2x + 2y +z, but this new divifor is
b+2x+y+x+z, that is, it is the old divifor with
*+≈ added. Then this divifor multiplied by 2,
z
ż,
and ſubtracted, o remains. Therefore the root is
rightly extracted, and the rule true.
As I am upon this fubject, I fhall alfo fhew
the truth of the rule for extracting the fquare root
in Arithmetic, which is the cafe here, when b=0.
Let x + y + zl be the fquare, that is,
1 div. x) xx+2xy +yy+2x+2yz+zz (x+y+≈
+ x) xx
2xy+yy+2xx+2yz+zz
2x)
+y
2xy+yy
2 div. 2x+y)
+y+z
3 div. 2x + 2y + z)
+2xz+2y%+zz
+2x+2yx+zz
Here being the root of the first term, its
fquare fubtracted, leaves the refolvend 2xy+yy &c.
Then to find y, the refolvend muſt be divided by
2x+y. That is, to the old divifor x, add x+y
for a new divifor 2x+y; this multiplied by y, and
fubtracted, leaves the refolvend 2xz +2yz + zz.
Again to find z, the refolvend is to be divided
by 2x + 2y + z that o remain; that is, to the
old divifor 2x + y add y +≈, the fum is the new
divifor 2x+2y+z, which multiplied by z, is
equal to the refolvend, ſo that o remains; and the
root is x + y + zo
S 2
PRO-
266
B. I.
RESOLUTION of
PROBLEM
LXXXIX.
To extract the root of a cubic equation.
iRU
RULE.
Take away the fecond term of the equation,
(by Prob. li) which then will be in this form,
a³ + bad.
Then fubftitute numbers in either of the following
forts, and extract the roots, by which means a
will be found.
avid+ √÷dd+z7b³
3
I
3
¿d+✓ dd+b³.
or a = √1d+✓ dd +376 +
27
27
Note, When b is negative, and 63 greater
than dd, the equation is impoffible.
Ex. I.
and — 4
Let x³
6x
ود = 60
Here b-6, d-9, and dd + b³
√/124 = 31/1
3
+ 34 = √/ — 1 — — I.
3을
​2
27
1. Therefore
a =
I
1 — 2 — — 3•
I
Or
a
4 z
— 32 = √// — 8 =
3
- I — 2
2, whence
-3, as before.
Èx.
Be&. WHI.
261
EQUATION S.
Ex. 2.
Let a³ + 6a — 20.
Here b6, d=20, and √dd+,6³
3
27
Ando+ 1981+√3 (Prob. Ixxii.) and
ご
​108.
4
3
V10108 = 1−√3.
√108 I
Whence
2
a = 1 + √ 3+1 −√3 = ?.
Ex. 3.
Let a³ 15a = 4.
Here b =
15, d = 4, and ✔áá + ½², b³
دن
27
121 =
3
And √2+11-1 = 2 + √ - 1.
3
2.
√2-11-1 = 2 — √ — 1.
I.
And
Whence a 2+ √−1 +2 -√-1 = 4.
=
Ex. 4.
Suppoſe a³ + 24a587914.
Heré b = 24, d=587194. ✔dd + 1763 =
293957.000878.
8
3
And Vid + 29 &c. 83.7731. And
83.77
=.0958; therefore
=
a = 83.7731.0958 83 6773.
SCHOLIU M.
It fometimes happens that the root may be
found, though the negative quantity be great-
I
271³
er than dd; and that is when the furd cubic root
can be extracted. For then the irrational parts,
in different parts of the equation, will deftroy one
another, and vanifh; as in Ex. 3.
S 3
To
$
262
B. I.
RESOLUTION of
To prove the truth of this rule.
Put
r = √da+ 2,b³, s=Vid+r. Then a=s-
b
bb
b³
bb
and as³-bs+
and babs -
>
35
35
2753
35
63
2b3
therefore a³ + ba = s³
id + r
27
2753
ed + r
Add + dr + gq —
id + r
I
27
63
=
(reftoring rr)
Add + dr + Idd + 2,
1b3
I f³
ъз
ždd + dr
27
27
=d;
½ d + r
id + r
that is, a³ + bad, according to the first part of
the rule.
And the fecond part is proved, by fhewing that
b
35
Vidr. It is plain id + rx id-r
+r
— \dd—rr —— b³, therefore d—r—
=
b3
› and
27532
be proved.
3
27
i d + r
1
d — r =
b
35
Which was to
Some of the cafes of cubic equations may alſo
be refolved trigonometrically by the table of fines.
As fuppofe the equation x3pxq, to be gi
ven. By Prop. 24, 25. Trigonometry, if rra-
dius, y fine of an arch; then gy
thrice the arch. And by Prop. 26. if x
arch, then
4x3
3
rr
4
4y3
rr
- S. of
cofine of an
x = cofine of thrice that arch.
Theſe equations reduced give y³
ry
3
rry
4
4
X fine of thrice the arch. And x³-3rrx+rr X
cofine of thrice the arch. Or putting y for ei-
ther the fine or cofine of the arch, C for the
fine or cofine of thrice the arch; then
y3.
Sect. VIII.
263
EQUATIONS.
y3 — 4rry = ± — C, the fign+being for colines,
93
and
-
for fines.
4
Then, if the given equation x3px q is to
+
be refolved; it must be compared with the fore-
going, and all the parts made fimilar in both.
Therefore let the equation x3pxq, be de-
noted thus, x³-2RR≈≈±‡RRS, S being the fine
or cofine of thrice the arch. Therefore RR = p,
and R = √p. Alſo 9 = RRS=1pS, and
S = 39. Whence by proportion R (√): S (39)
S=
P
::r: C =
3rq
4
P√ AP
4
P
the cofine or fine of an arch.
Of which, y is the cofine or fine of the third part.
Then y being found, it will be r: y :: R (√‡p)
as required. Hence this
:*=
3
2 RULE.
Take away the ſecond term (by Prob. li.) if it
have any; and the equation will be reduced to
this form,
x³- px = ± 9.
Then take 39
X3
3rq
the cofine of an arch (if
PV P
Find y = cofine
J√ +P
до
it be+q), or the fine (if).
or fine of that arch; then
quired.
3
4
=* re-
And this laft arch may be either that we found,
or, that + 120°, or the fame + 240°. By which
means you will have three roots or values of y.
But note, when
3rg
4
P√ + P
is 8 greater than I, the
There-
queftion is impoffible by this rule.
S 4
264
RESOLUTION of B. I.
Therefore this rule fupplies the defect of the
first rule, which only folves equations that have
but one root real, and two impoffible ones: while
this rule folves fuch as have three roots real.
Ex. 5.
Let as 91a =
-
330,
Here xa, p=91, q=330, and √p=11.015,
and if r = 1,
3rq
4
PV
P√ 3 P
.987655 fine of 81° very
near; and the third part is 27, or 147, or 267;
whoſe fines are, y = .45399, or .54467, or
-.99863; thefe multiplied by 11.015 produce
5.0004, and 5.9991, and
the three roots are 5, 6, and
Ex. 6.
Suppoſe x³- 19* = 30.
10.9998; therefore
- II.
4
Here p 19, 9 = 30, and ✔√†p = 5.03323 ;
and 3rg
= .94112 cofine of 19° 45', and
PV + P
the third part is 6° 35', or 126° 35', or 246° 35',
whoſe cofine is y.99340, or .59599, or
-.39741; which multiplied by 5.03323, pro-
duce 4.99998, and 2.99974, and 2.00024.
So the three roots are theſe, 5,
3, and
PROBLEM XC.
2.
To refolve a biquadratic equation, by diffolving it
into two quadratics.
Take away the fecond term (by Prob. li.), and
let the refulting equation be x + qx² + rx + s=0.
Suppoſe it to be generated by the two quadratics,
x + ex +ƒ= o, and xx-ex+go. Thefe being
multiplied
Sect. VIII.
265
EQUATION S.
multiplied together produce x*+fx²+egx+fg=0.
+g -ef
ee
Comparing the terms of this with the firft equa-
tion, we have f+g—ee=q, eg—ef=r, and fg=s;
whence g +ƒ = q + ee, and g—ƒ == ; and confe-
quently g =
(fg =)
q+ee + ==
2
99+2gee+e4-
4
f=
e
9+ee-=-=
e
and f=
e
And
2
ee
s. And by reduction,
gy
eε +29e4 + qgee — rr = 0. Put yee, and then
45
y³ +2qy²+qqy―rro. A cubic equation; whence
45
the following
RULE.
To refolve the biquadratic equation x4+qx²+ra
+so. Take the cubic equation y³+2qyy+qqyrr
45
o; out of which take away the ſecond term (by
Prob. li.); and find the root by the laft problem,
or otherwiſe; and from thence find y.
take e√y, and f
q + ee e
==
and
2
Then
q + ee + e
g
2
Laftly, find the roots of thefe two quadratic e-
quations, xx+ex+fo, and xx-ex + g = 0.
And theſe will be the four roots, of the biquadratic
*4 + qx² + rx + s = o.
Example.
266
B. I.
RESOLUTION of
Example.
Let x4-25x² + 60x360.
From this you have the cubic equation y³----
50y²+769336000.
co. Take away the fecond
term, by writing v +
193
1150
v
3
27
50 for y.
3
And we have
And by Rule 2.
2. Prob.
I
laſt, v = 8.3333 &c. =8—, whence y = 8 1 - 3
++
Il w/
60
50
11
25, and e
e = 5; therefore f
f =
3
60
·25+25—
5
—6, and g
g=
2
2
―25+25+
5
+6. Whence xx+5x-6=o, and xx-5x+6
o; and the roots of the former equation are i
and -6; and of the latter, 3 and 2. Therefore
the four roots of the biquadratic, x425x+60x
360, are 1, 2, 3, and 6.
And the fame roots will be found, by making
ufe of the other values of v, which
-
are
2
and
>
23
3
3
Schol. But this and fuch like rules are of little
value; for there is far more labour here in get-
ting the roots than by the method of converging
feries, which is to follow.
PROBLEM
XCI.
To extract the root of any pure power in numbers.
Let G be the number given to be extracted;
m the root required, r the neareft root, and e
the
Sect. VIII.
2.67
EQUATIONS.
the remaining part of it; then rel" G, that
is (Cor. 1. Prob. v.) rm + mrm-s e + m.
m.
I m 2
gm-2 ee + m
2
3
m
I
ym-3 e³ &c. = G, and
rejecting e¹ and the higher powers, as very ſmall;
we have mym-se + m.
and
this
m.
m
I
go42 −2 el = G —— gesti,
2
mrm-x
G — gr ???
e tee =
Hence
m
I
m
I
g-11-2
m.
gr M-2
2
2
Let G
↑
IRUL E.
abfolute number.
the neareſt root you can find.
the true root.
the index of the root.
27
re
rte:
m
b
M - I
G- gottle
D=
m
1
m.
rm-2
2
Then beee D, or e=
D
b + e
nearly.
Which equation is to be refolved by Prob. lxxxviii.
When e is had, then re is to be taken for a new
value of r, and the operation repeated, perhaps of-
tener than once. This rule generally triples the
number of figures.
But if the third power of e be takea in, then
mrm-se + m
m
I
M I m
2
rm-2 ee + m.
2
2
3
rm
D, there-
fore
gr M−3 e3 = G — and fince beee
268
B. I.
RESOLUTION of
fore ee-D-be, and e³ De-bee, and m.
m
I
2
m. 2
m
m
2
rm-3 e3 m.
gM = 3 × De-
bee,
3
2
3
m
I
111
M2
whence mrm-1 e + m.
ym-3X De-bee Gym, and rre +
pm-2 ee + m.
2
2
—
3
2
992-2
Det
3
M-1
2
M-I }} -- 2
ree-
bee=
G-gm
=rF,
2
3
Mrm-3
Gg772
by fubftitution, (putting F=
D, then rre +
трт-г
); expunge b and
712
I n
2
G- ·7-722
e t
2
3
712
I
m.
z Mne 2
2
MI
M—Į M— 2
2r
ree
X
eerF; that is,
2
2
3
172-I
m
2
m
I
712
2
gre +
× Fe +
ree
reerF,
3
2
3
m
2
m + I
Fxe+
reerF,
3
6
*
:
that is, rr +
whence this
Let G
2 RULE.
abfolute number.
r = neareſt root you can find.
rte = true root.
g
m = index of the root.
G-
y m
mrm-2
F =
6r +
2m-4. F
6F
Then
e tee
nearly.
m - 1
m + I
Which
}
Sect. VIII:
269
EQUATIONS.
Which is to be folved as Prob. lxxxviii, and re-
peated with new r, if there be occafion. This
rule commonly quintuples the number of figures
in the root, true; at each operation
The root of any number may alſo be extracted
by Prob. lviii. after this manner.
3 RULE.
Let P+Pq, be the number given to be extracted.
P, the greatest power contained in it.
Pq, the remainder; and
q, the quotient arifing by dividing the re-
mainder by the greateft power.
n, the index of the root.
2/ P + Pq = P ^ +
די
1
Then
I
n
Aq
Bq
ท
2n
3n
Cq
Dq &c.
Where A, B, C,
3n
4n
2n
&c. are the preceding terms. In this rule, when
two or three figures are got, put them equal to
I
>
P " and begin the operation anew; and the ſeries
will then converge exceeding faft; and fo much
fafter as q is lefs.
Cor. Hence it follows, that
I
VP + Pq = √P + =/ Aq —
2
7
-Dq — — Eq &c.
10
I
I
— Bq — 3 Cg —
4
for the ſquare root.
2
VP + Pq = VP + − Aq — — Bq ·
V
8
Dq
12
I I
15
√ P + Pq = √P +
3
6
-
5 Cq -
9
Eq &c. for the cube root.
4
7
Aq — 3 Bq — — Cq ·
8
12
I I Dq &c. for the biquadrate root.
16
V
}
270
B. I.
RESOLUTION of
I
4
9
V P + Pq = VP + — Aq − 1 Bq - 2 Cq-
14
20
5
10
Dq &c for the fifth root.
2
15
P + Pq|³ = P³ + — Aq —
6
— Bq
Bq — — Cq -
3
1Dq - &c. for the
7
fquare; and fo on.
Ex. 1.
What is the cube root of 2.
9
cube root of the
Here G2, r=1, m = 3,
by Rule 1, 6}
b =
D=
3
I.
1+
=.3333; and e=
+2)
.3333 (.26e
24
1.2 ) .0933
+26 876
D
andre1.26
b+e
1.46
57
L
Again, for a fecond operation;
Let new r 1.26; then Gg3.000376,
and m."―r=3r=3.78, and D= 000376
2
3.78
.000099471, and becauſe is negative here,
.000099471
ļ =
1.26-
1
1.2600000
Sect. VIII.
271
EQUATIONS.
1.2600000) .0000994710 (-.000078950106
-7
881951
1.2599300
11275900
-78
10078816
1.2598520
11970840
-89
11338587
1.259843
632253
629921
2332
1259
73
до I .26000
e=.000078950106
√2 = 1.259921049894
Ex. 2.
Extract the 5th root of 2327834559873*
Firſt point every fifth figure thus
2327834559873,
Then for brevity's fake, take only the firft pe-
riod, as an integer, that is 232. Then proceed-
ing by Rule 2, we fhall find 2 the root of the
greateſt power contained therein; and thence,
r = 2, and
and
Whence
232 G
32 деть
200 = G―rm
40
200
40
mrm-2
= 5 = F.
Therefore
!
1
B. I.
272
RESOLUTION of
2
Therefore 4е + ee = 5, or e=
41+
4.5) 5.00 (.92 = e.
4.5+ e
5
+9 4.86
5.4) 1400
whence r = 2.92;
+92 1264
6.32) 136
Suppoſe again new r = 290.
Then r³ 24389000.
G = 2327834559873
r52051114900000
G-rs 276719659873
=
F 2269.217 whence
297.825e+ee = 2269.217
2269.217
ore=
297.825 + e
297.825) 2269.217 (7.43375
+7 2133.775
304.825) 1354420
+7.4 1248900
312.225)
105520
+ 43°
93796
312.6515
11724
+ 33
9380
312.68
2344
2188
156
Whence re=297.4337, which may be ta
ken for new r, and the operation repeated, if
there be occafion.
Ex.
Sect. VIII.
273
EQUATIONS.
Ex. 3.
What is the 7th root of 100000.
The neareſt root of 100000 is 5, whence by
Rule 3d,
P + Pq = 100000
Р
= 78125
Pq
21875.
whence q
9
.28 &c.
And VP+Pq = VP + — Aq
Aq-Bq-
6
14
13 Cq — 20 Dq - 27 Eq &c.
28
21
That is,
7
35
VP = +5.000
+ Aq=
-3B q
- 2 232 C q
- =
5
- Dq =
&c.
.200
024.
+.004
ΠΟΙ
5.204.0255.179
But becauſe this converges flow, take 5.179 for
the root, and involve it to the 7th power, and
we have
P+ Pq = 100000.
P
Pq
9
===
11
99435.8652873094
64.1347126906
=.0006417587
5.1790000000 = A
Then VP =
+÷Aq = +.0004748097 = B
7
-3 Bq
1306 C
2
~ 1 3 C q =
+
I
D
5.1794748098
-1306
5.1794746792 = √1000CO.
T
S.bol.
1
274
B. I.
RESOLUTION of
Schol. i. If the root is required for only a few
places of figures; the eafieft way by far, is to ex-
tract it by the help of logarithms.
Schol. 2. From the foregoing process, the rule
for extracting the cube root in arithmetic, may be
demonftrated.
3
;
Let a + be the root, a the firſt figure, e the
fecond. Then the cube is a³ + za¹e + zað² + e³
then a the greateſt cube contained in it, being
fubtracted; there remains 3ate + 3ae, fetting afide
e as being very fmall. Divide this remainder by
3, and we have a'e - aee, from which to find e,
this remainder or refolvend muſt be divided by
aa + ae. That is, the refolvend muſt be divided
by aa, the fquare of the root, and then to the
divifor, there must be added de, the product of
the root by the quotient figure; and the whole
will be the true divifor for finding e. But as e¹
was left out of the account; the root got this way
will deviate from the true root; and therefore you
muft, after a few figures are had, begin the ope-
ration again, with the new root which you have
already got.
PROBLEM XCII.
To extract the root of any adfelled equation, in
numbers.
Preparation.
=
Suppoſe Ax+Bx²+Cx³+Dx44-Exs &c. N.
Put rex, r being the firft figure of the root;
and to find r, put 1, 10, 100 fucceffively for x;
and the neareſt value of thefe being found, try the
intermediate numbers 5, 50, &c. then expung
ing *, we have
Ar + Ac
Rect. VIII.
275
EQUATIONS.
1
Ar + Ae
Br²+2 Bre + Bee
Cr³ +3Cr²e +3Cree +Ce³
=Ax~
=Bxx
=Cx)=N.
N,
Dr++4Dr³e+6Dre²+4Dre³ +De+ Dx4
&c.
&c.
Sum P+ae + bee + ce³ + de4 &c. = N.
And ae+bee+ces+de+ &c.
f
N-P=f.
Then fince e= nearly, we ſhall have
f
a
+ be x = = f. Or ae + bf e=f. From whence
ae +
a
we ſhall have this
a
IRUL E.
f
nearly.
a+b x 1
f
a
Or, if more exactnefs be required, we may
bring in ee; then ae + beef, whence this
2
RULE.
b
Ge+ a = /
f
b
ce³ +de+ &c.
or e
a
te
a + be
Ъ
nearly, to be wrought by LXXXVIII. Rule 2.
Or if e³ be taken in for more exactneſs; pro-
ceed thus, bee =ƒ—ae, and ce³ =
ae bee + ce² = ae + bee +
+
cfe-caee whence
of
ca
b
eef, whence
T 2
3 RULE.
246
B. I.
RESOLUTION of
3 RULE.
e+
of
f
e + ec =
>
b
ca
very near; to be wrought
ca
b.
b
b
as Prob. lxxxviii. Rule 2.
In any of theſe rules the operation muſt be re-
peated after a few figures are had, by taking a
new value of r, and proceeding as before.
Ex. I.
Let 120x³ +3657** — 38059x
38059* = 8007115.
By a few trials, you will find x to be greater
than 30, and lefs than 40.
Therefore fuppofe
30, and 30+ex the root fought, which be-
ing involved, and taking the leaſt powers firſt, as
in the rule, we have
-1141770-38059e
+3291300+219420e+ 3657ee
+3240000 + 324000e + 10800еe + 120e³
Which being added,
=8007115.
5389530+505361e+14457ee+120e38007115
505361e+14457ee+120e=2617585.
and
or
ae +
bee+ce³ =f.
Then to ſhorten the work, divide by 1000, and
then 505 +14ee &c. 2617, and by Rule 1,
e=2617
2617
505
5.18; or rather e =
2617
505 +14X5.18
4.53.
Whence re or 34.5 for a
577
x
new operation. Which being involved, beginning
at the higheſt power first, we have
4927635 + 42849e + 12420ee)
+ 4352744 + 252333e + 3657ee8007115.
1313035 38059e
That
Seal. VIII. EQUATION S.
277
That is,
79673434 + 257123e + 16077ee = 8007115.
257123e + 16077ee = 39771.25
or
whence
15.9932e+ee = 2.473799
and by rule 2d, e =
2-473799
15.9932 + e
15 9932) 2.473799 (.1532 = e
1.60932.
+.l
16.0932) 864479
r = 34.5
+15
812160
e =
16.2432)
52319
.1532
r+e=34.6532 = x.
+.53
48888
16 29612)
3431
+3
3260
16.30
171
Ex. 2.
Let 24 3ײ — 75% = 10000.
Here by a few trials z will be found very near
10. Therefore let r 10, and rez.
24
322
=
= 10000 + 4000e + 600ee &c.
300
+750+
+75% = +750 +
60e
75e
Being added,
3ee
Then
}
= 10000.
104504015e +597ee10000.
or
or
40150 +597ee4.50
6.725e+ee == 0.753769
therefore e is negative, and by Rule 2.
-0.753769.
e
6.725+6
I 3
6.725
5
278
B. I.
RESOLUTION of
:
6.725) -0.753769 (-0.114 e
I
6625)
-6625••
9126
I I
6515
6.515)
26119 -
14
26004
115
↑ 10.000
e
.114
r+e=9.886z.
ર
6 501
Again, put r 9.886, and r+ez; then
9551.738507135 + 3864.753593824e
-293-198988
+741.450
+586.397976ee
-zee
59316e
+ 75€
= 10000; that is by addi-
And e=
tion, 9999.989519135 + 3880.437593824€ +583.
397976ee10000; and tranipofing,
3880.437593824e + 583.397976ee010480865.
.010480865
3880.437
=.00000270095, nearly.
1
=
Then re 9.88500270095 %.
=
Ex. 3.
Suppoſe 7ys + 2100y3
8000y² = 3850000000.
By a few trials y will be found between 50 and
60; therefore put r = 50, and rey; then ex-
punge y. Or rather thus: Since the numbers are
Jarge, transform the equation (by Prob. xlii.), by
I
F
putting *==-y, or y = 10x, which done we have
700000x5+2100000x³-800000x² = 385000cooo,
or 7x + 21x³- 8x38500. Then to extract the
root of this, put r5, and re or 5 + e = x;
and x being expunged, we have
3
21875
Sect. VIII.
279
EQUATION S.
21875 + 21875e + 8750еe + 1750e³
+2625 + 1575e + 315ce + 21es >= 38500.
200
80e
8ee
That is,
2430022370e +9057ee + 1729e³ 38500.
23370e +9057ee1729e³
and
of
ca
b
14200.
Then by Rule 3,2711, 4461, whence
5.675e+ee = 3.0900.
5.675) 3.0900 (.5004 = €
+ 5
30875
6.175
25
+ 5
6.675
Again, put r5.5, and repeating the operation,
35229.90625+ 32027.187e+11646.25ee'
+3493.875
242.
+ 1905.75 et 345.5 ee
88. 548.5 ees
¿
That is when added,
= 38500.
38481.78125 + 33844.9376 + 11984.75ee =
38500.
And
33844937e + 11984.75ee = 18.21875.
Then
18 21875.0005383 = £e nearly.
33845
a
Then bx=6,451, and (Rule 1.)
a
18.21875
33844 937 +6.451
exactly.
= .000538198 = e, more
Then
r+e= 5.500538198, and y=10x-
55.00538198.
T 4.
The
}
280
B. I.
RESOLUTION
of
The root may alſo be extracted as follows. Hav.
ing got aebe² + ce³ + def, as before direct-
ed; let v be the firſt figure of the value of e, s the
fecond. Then putting v+s for e; a × v +st
bxv+s¹² +cxv+sl³ &c. =ƒ; that is, av+as+
bv² + 2bus &c. + cv³ + 3cv²s &c. = f.
f. And
as + 2bvs + 3cv's &c. fav bv² — cv³ &c.
Whence
T
S=
ƒ— ev — bv² — cv³ &c.
av
a +260 + 3cv² &c.
4 RUL E.
----
Whence this
Having any equation given, proceed as in the
other rules, till you get ae+bee+ce³+de+ &c. =ƒ.
Then find by repeated trials, the firſt figure v, of
the value of e, fo that vxa + bv + cv² + dv³ &c.
may be nearly f; and take that product from
f to find the remainder.
Then to find the next figure or figures; divide
this remainder by a+b+3cvv +4dv³ &c, the
quotient is the faid figure, which must be added
to, for a new value of v. Then repeat the ope-
ration with new v, viz. take v xa + bv + cv²+dv³
&c. from ƒ, and divide the remainder by a + 2bv
+ 3cv² &c. and add the quotient to laft v; and
fo on.
And note, after the divifor once takes place, each
new quotient may be continued to near as many
gures, as all the preceding ones. Alfo in the
divifor, you need not continue the parts of the di-
viſor 2bv, 3bv? &c. any farther in decimals, than
to anſwer the number of figures, you would have
true in the root.
General
Sect. VIII.
281
EQUATIONS.
General form.
ย
f
a
= 2
f
f
; or v
nearer; or v
2=
at bo
?
a + bv + cvV
fav
bv²
nearer ftill; &c. then, next figure
cv³ &c.
a + 2bv + 3cvv &c.
Ex. 4.
Let 23- 17%² + 54% = 350.
Here ≈ is greater than 10, and lefs than 20.
Letr 10, r+ez; then
3
1000 + 300e + 30e² + e³
1700 340e
+ 540 + 54e
17ee
e3
ww
= 350.
or 160 + 14e + 13e² + e² = 350.
that is
14e + 13eee³ = 510.
To find e, try 1, 2, 3, &c. and you will find
e very near 5, but fomething lefs. Therefore
take v = 5, and vxa+bu+cv² = 5×14+65+25
520, and 510-52010, then a +260 +
219, and
=
3082
3cv²
10
"
= .045
219
5.000
.045
e = 4.955
Let new v4.95; then a + bv + cv² × v =
509.119875, and 510-509.119875
=
Allo a + 2bv + 3cv² = 216.2075.
0.880125.
Whence
0.880125 = .00407, and e= 4.95407;
216.2075
whence
282
B. I.
RESOLUTION of
whence z 14.95497. Or, if you pleaſe, put
ย 4.95407 for a new operation.
Ex. 5.
Let 2x* 4 -16x³ +40x² 30x=-1.
By a few trials, it appears that x is between i
and 2.
Therefore put r = 1, r + e
putri, r+ex.
expunging x,
Then
2 + 8e +
12ee + 8e3 + 2e4'
16 48e
48ee - 16e³
+40 + 80e + 40ee
30 - 306
The fum is
or
a
4 + 10e + 4ee
10e + 4ee
b
8e³ + 2e4 =
8e³ + 2e4 = 3
I m
- I
C
d
1
1.
Here we have e- 38
3
e =
10+ 4e
3
II.2
ΙΟ
= .26 = v.
f
3, or more exactly
་ས
Then for the next figures of the root,
vxa+bv+cv²+dv³ = 2.73893, and 3 -2.73893
=.26106. Alſo a +2bv + 3cv²+4dv³ 10.598,
and
.26106
= .0246
10.598
v=.26
thene = .2846
Take new v.2846, then vXa+bv+cv²+dv³
2.99869539, and 3--2.99869539.00130461.
Alſo a +260 + 3cv² + 4.dv³ = 10.51728.
Then
.00130461
Sect. VIII.
283
EQUATIONS.
.00130461
10.5173
ད
= .00012404
.2846 = v
.284; 2404. l,
whence re or x 1.28472404.
=
The roots of equations may alſo be extracted by
help of the Rule of Falſe Poſition in Arithmetic,
as follows.
5 RUL E.
In fuch equations as contain furds, exponential
quantities, &c. make two fuppofitions in numbers,
for the root, as near as you can get them. Then
each of theſe being put in the equation instead of
the root, you must mark the errors (that is, the
exceſs or defect) arifing from each of them.
Then multiply the difference of the fuppofed
numbers by the leaft error, and divide the pro-
duct, by the difference of the errors, if they are
like, (that is, both exceffes or both defects); or by
the fum, if unlike. Then ..
The quotient is the correction of the number
belonging to the leaft error; and is to be added
if that number was too little; or fubtracted, if
too great. This gives the root nearer than before.
In like manner try this root, and the neareſt of
the former, or elfe take a new fuppofed number;
then find their errors, and proceed as before, and
you will get a root ftill nearer. And thus by re-
peating the operation, you may continually ap-
proximate, as near as you will, to the true root.
Ex.
284
RESOLUTION of B. I.
Ex. 6.
:
**
Suppose x
100, to find x.
By the nature of logarithms x x log: x = log.
100 = 2.
=
Herex, by a few trials, will be found greater
than 3, and lefs than 4. Suppofe 3; then
l:x=.5440680, and xl:x=1.9042380, which ſhould
2
—.0957620 = 1 Er. too
be equal to
x =
l:x
little.
Again, ſuppoſe 3.6, then x .5563025,
and xl:x=2.0026890
we have
2
F
+.00268902 Er. too great. Hence
I num. 3.5
2 num. 3.6
diff.
I er. -.095762
2 er. +.002689
fum .09845!
O. I
Then
0.1X.002689
= .00273 cor.
=
.09845
2 num. 3.60000
correct.
-.00273
3.59727 = x;
x
Again, ſuppoſe = 3.597, then l:x=.5559404,
and xl:x = 1.9997176, which fubtracted from 2,
gives -.0002824 the error, too little.
Whence
2 num. 3.600,
2 er..0026890
3 num. 3.597,
3 er.
.0002824
diff.
.003,
fum .0029714
Then
Sect. VIII.
285
EQUATIONS.
Then .003 X.0002824
.0029714
3 num. 3.597
=.000285 the cor.
cor. + .000285
x = 3.597285 as required.
Ex. 7.
If s be the fine of an arch z, rad. = 1, and
4sz = 5, to find s and z.
By divifion sz
1.25. The length of 1 degree
is = .01745329 &c. By a few trials, we may
find z between 70 and 80 degrees. Suppoſe
z = 70 deg. then .01745 X 70 1.2215; alfo S. 70
.939s, and sz=1.1469, and 1.251.1469
=.1031 the first error, too little.
Again, fuppofe z 75 deg. then .01745 X 75
×
= 1.3087, and s.965, and sz= 1.2642; and
1.2642-1.250142 the fecond error, too much.
Hence
=
.1031
2 er. +.0142
1 num. 70
2 num. 75
I er.
5.
fum .1173
5X.0142
.0710
Then
.60 the cor.
.1173
.117
2 num. 75.0
cor. .6
2 = 74.4
Again, let z
74.474° : 24°, s.9631626;
=
then .01745329 X 74.4 = 1.298524, and
1.2985245 = 1.2506895, from which fubtract 1.25,
then .0006895 = 3d error, too much.
2 aum.
}
298
B. I.
RESOLUTION of
2 num. 75
3 num. 74.4
diff. .6
2 er..0143937
3 er. +.0006895
diff.
.0133105
Then .6X.0006895.0310.
.01331
3 num. 74.400
cor. .031
and 74.369 74° 22′ 8″
z =
= :
and s.9630372.
SCHOLI U M.
There are alfo other ways of extracting the roots
of equations, though not much different from ſome
of the foregoing ones, particularly a method of
Sir I. Newton's, which is like the procefs ufed in
the ſecond method foregoing; the principal dif-
ference being, that he every where takes a new
letter, where we find a new value of e.
Alfo furd or tranfcendental equations, may be
refolved by reducing fome of the quantities to in-
finite feries; proceeding by the rules of Sect. VI.
In equations, where the terms involve a great
many factors, which makes it tedious to multiply
them together; it will be a fhorter way to add
the logarithms of the feveral factors together; and
then find the number belonging, which will be
the numeral coefficient of that term. And thus all
the coefficients of the particular terms may be
found.
We may note, that though the third rule con-
verges fafter than the reft; yet, as there is fo
much trouble in finding the coefficients, and di-
vifors, it will be found not fo expeditious as the
fecond, or even the firft.
In
Sect. VIII.
287
EQUATIONS.
In making ufe of the fecond rule, after half the
number of places are found for the value of e; it
will be needleſs to form new divifors; for the reſt
of the figures will be as truly found by plain di-
vifion. For what is added to the divifor, in places
fo far back, does not at all affect the quotient.
The root may alſo be extracted as in the follow.
ing problem, and the coefficients a, b, c, &c.
found as there directed; which is a compendious
method, when the equation confifts of many terms.
PROBLEM
XCIII.
To extract the root of the infinite feries Az + Bx² +
Cz³ + Dz² + Ezs &c. = N, in numbers; fuppo-
fing this feries to converge fast enough.
Preparation.
Taker as near the root 2, as you can find it;
and let rez, and z being expunged, we have
Ar + Ae
Br2 Bre +
+ Cr³ + 3 Cre +
+ Dr4 + 4Dr³e +
Be²
3Cre² + Ce³
6Dr²e² + 4Dre³ + De4
+ Ers +5Erte + 10Er³e² + 10Er²e³ + 5Ere4 +-
Ee' &c.
the fum
3
P + ae + be² + ce³ + det + ges=N
and ae + be² + ce³ + de4 + ges = N —P=ƒ.
Whence this
RUL
2,
E.
Take r very near ≈, and let r+e, then fub-
ftitute the powers of re for thofe of z, till you
get P+ae+be+ces &c. N, and ae + bee &c.
N-Pf, which equation is to be refolved by
Prob. lxxxviii; or elfe the equation ae+be+ce³
+ det
=N
288
RESOLUTION of B. I.
+de+ &c. =ƒ, is to be refolved by fome of the
rules in the last problem, and the operation re-
peated if there be occafion.
And here the coefficients a, b, c, d, &c. are
moſt eaſily had from the terms, which compoſe
the value of P; for we have P Ar + Br² +
Cr³ + Dr4 &c. Whence
a =
z =
Ъ
c :
Ar + 2 Br² + 3Cr³ + 4Dr4 &c.
Br² + 3Cr³ + 6Dr4 + 10Ers
rr
Cr³+4Dr4+ioErs &c.
6
дов
8
5
and fo on; where
the numbers in a, are 1, 2, 3, 4, &c; in b, 1,
3, 2X3, 2X5, 3× 5, 3×7, 4×7, 4×9, &c. in
c=1, 4, sp, q, Ir, is, 2t &c. where p, q, r, s,
t, &c. are the foregoing terms. And in finding
a, b, c, &c. you must go through all the terms,
till they grow very ſmall, and at laft vaniſh. But
you need not find above two or three of theſe co-
efficients a, b, c, &c. and each fucceeding one
may conſiſt of fewer places of figures.
Example.
I
I
I
Let z
x² +
Z3
24 +
ZS
2
2.3
23.4
2.3.4.5
2
&c. =
Here by feveral trials z is found nearly 3;
therefore put r, and r+ez. Then PA+
Br² + Cr³ &c. that is,
+=
Sect. VIII.
289
EQUATIONS.
r = .333333
6172
I
дов
2
1
дов
34
720
+ㅎ ​3
+ Tors
ΤΖΟ
then a
•339539
-.056071
P=.283468
.333333
+ 18516
170-
+
716523. Alfo
b =
I
3
.IIIIIO
2056
055555+.018516
3084 +
30
L
340
12
11
.055555
514
2
-.056071
•352019
-113178
X3=
-.058669
+.018856
× 9
I
-.3583. Hence
.283468+.71652e-.3583ee ==.285714
.71652e-.3583ee.002246
1.9998e
and
and
e =
ee = .00627
.00627
1.999 — e
1.999) .006270 (.00314 = e
5988
3
1.996
282
199
83
79
4
U
Then
1
290
B. I.
RESOLUTION of
Then
r
.33333
e = .00314
Z
z = .33647
or putr.33647 for another operation.
SCHOLIU M.
If the ſeries breaks off, then it is no matter whe-
ther it converges or not.
And in that caſe it co-
incides with the last problem, and may be folved
by any of the rules therein.
And if e be very ſmall, the equation ae + bee +
ce³ &c=f, may be expeditiously folved by Prob.
lxii. Rule 1, in which you need only uſe the three
first terms; which will be fhorter than taking new
". But that rule cannot fo conveniently be ap
plied to the given feries, becauſe it does not con-
verge fo faft as this.
PROBLEM
XCIV.
To extract the root in numbers of the infinite feries
Az + Bz³ + Cz³ + Dz7 &c. =N; fuppofing it
to converge faft.
Preparation.
Taker as near the root as it can be found by
trials, and put rez, and expunging z, we
fhall have,
Ar + Ae
÷Br³+3Br²e+ 3Bree + Be³
+Cr³+5Cr4e+10Cr³ee+10Cr²e³ + 5Cre4&c.
+Dr²+7Dr°c + 21Drsee + 35Dr4e3 +35Dr³e4=N.
+Er9+9Er³e +36Еriee +84Er6e3+126Erset
the fum
P + ae + bee + ce³ +de4 &c. - N.
and
Sect. VIII. EQUATION S¿
29f
and ae + bee + ce³ + de¹ &c. =N—P=ƒ;
Whence this
RULE.
Affume by trials very near z, and r+e=2;
then fubftitute the powers of r+e for 2, in the
given feries, till you get P+ae+be²+ce³ &c.=N,
And ae+be+ct³ &c. N→P=ƒ.
3
Where P Ar + Br³ + Crs + Dr² + Ero &c;
Ar + 3Br³ + 5Cr³ + 7Dr7 &c.
a =
а
b
C
==
&c.
*
3Br³ +10Сrs + 21 Dr7 &c.
rr
Br³ + 10Crs + 35Dr &c.
z3
Where the numbers of a, are 1, 3, 5, 7 &c. of
b; 3, 2X5, 3X7, 4X9, 5X11, 6X13, &c. And
each feries is to be continued till the terms become
very ſmall and vaniſh; which will happen in a lit-
tle time, becauſe the given feries converges. The
terms of a, b, c, are easily had from the terms of
P, as above, without much labour; then having
got ae + bee + ce³ &c. =ƒ, in numbers; find the
root e, by Prob. lxxxviii. or by fome of the rules
in Prob. xcii.
Example.
Let ŷ + // y³
уз
+
3
ys +
4.5
3.5
7.2.4.6
37
źryy
3
3.5.7 ·y³ &c. =
9.2.4.6.8 y⁹ &c. =
.698132, to find ý.
The feries abridged will be y +
U 2
1
f
Qxy
5
37
-1
292
RESOLUTION of B. I.
+ Ryy &c. =.698132; Q, R, S, being the
&
7
numerators. By a few trials, y is nearly.6, put
r = .6; then
= .600000
Q.1080co
= .606000 = Ar
3) Q =
36000 = Br3
R
29160
5) R
S
8748
7) S
T
2756
9) T
5832 = Crs
-
1249 Dr?
306 &c.
V
893
11) V
81
W
295
13) W
23
X
98
15) X
6
Y
33
17) Y
2
Z
II
19) Z
I
.643500
P.
Then a
1.2500, the fum of the first column di-
vided by r. b.585; whence
.643500 +1.250e +.585ee = .698132
and 1.250 .585ee = .054632
+
.0545
and e=
= .043 nearly.
1.25
and e=
.054632
.054632
1.2500+.585X.043
1.2500 + .0251
.054632
11
1.275
add r
= .04284 more exactly.
r = .60000
.64284 = y.
or take new r = .6428 for another operation.
PRO-
Se&. VIII.
293
EQUATIONS.
PROBLEM XCV.
To extract the roots of two given equations, containing
two unknown quantities x,y; though never ſo com-
pounded.
RULE.
By feveral trials find two near values of x and
y, viz. r and s, and put r+ex, and s + v = y.
And instead of the powers of x and y, put in
thofe of r+e, and s+v. Then involve all furds
by the bynomial theoremi (Prob. lviii.), alfo re-
duce logarithmic quantities to feries
feries (Prob.
lxxxiv, lxxxv.), and the like for all compound
quantities; fo that at laft the equations may con-
fift only of ſimple terms. And in doing this, re-
ject all powers of e and y above the firft, and
?
alfo their products.
Then you will have two fimple equations of e
and v, which being refolved, will give their ya-
lues; and from hence x and y will be known.
Then put new r and s for theſe values of x and
3, and repeat the operation, which may be done
as often as you pleate, till you get the roots as
near as you have a mind. And the fame form
may ſtand and ferve for all theſe operations.
Ex. I.
2xy
Suppoſe √yy—xx +
20 = b
yx + 2x
and
log:x + √xx + yy
+= 0.096 = c.
y
S
Let a = 2. And by fome trials we find x near
4, and y near 13; then put r = 4, 13, and
by involution, and putting re for x, and
for y, we have
U 3
v
294
B. I..
RESOLUTION of
SS
rr + 250 2rel + 2rs + 2rv + 256 ×
ss + 25v + ar + ael² = b and
2
I
log: r + e + rr + 2re + ss + 250 1플
​2
= c.
s + v
SV re
but ss-
rr + 250 arel = √ss
rr +
√ss-rr
I
and ss+ar+25v+ael² =
ss+ar
25v+ae
2Xss+arl
re+ sv
allo rr + ss + 2rẹ + 2svl² =✔rr + ss +
3
√rr+ss
Put dd-ss-rr, f=ss+ar, gg=ss+rr. Then we
d
sure
I 25v+ae
have d+
+2rs+zro+2se ×
25
r
ars
xe +
f
d
f3
18
f
+
21
f
2f3
arss
f3
=b, that is, (1)
xv=b―d-
275
f
f
¿
log: r+e+rr + ss +?re + 2solž
Again,
re + sv
log:r+e+g+
5+v
g and log:r+e+8+
>
l =
=s+vxc. Put tr+g, log: t,
C
re+su
g
m. =.4342945, then (Prob. lxxxiv) / +
mte + msv =cs + cv, which reduced, is (2)
tg
stemsv
cgt
-
tg X cs-l. Then numbers being
fubftituted in thefe two equations, give
(1) 1.588€ + 1.075 = 0.190
(2) 7.6430 17.33260 + 0.59536;
And
Sect. VIII.
295
EQUATION S.
And theſe equations
equations being refolved, give
e=.0743, and v——.0671; whence r+e or
= 3.926; and s+vor y = 12.933.
Or for another operation, put = 3.926 and
$12933, and finding new values for d, f, g, t,
and, you will have two equations, which will
give e and v more exactly.
Ex. 2.
Let xlog:yb 8.7679114.
and y+log: xc = 3.4760046.
By a few trials, we find x nearly
3=21.
8, and
2. Putr 8. and r+e=x; alſo s = 2½, and
s+v=y. Alfo M.4342945; then we have
r+e+lis+v=b=r+e+lis+
and s+v+l:r+e=c=s+to+lir +
Thefe equations reduced become
Μυ
(Pr. 84.)
Me
S
Μυ
e +
b.
1:s.
S
Me
v +
l:r.
p
And put into numbers are
e .17370 = .3700.
and v+.0543e = .0730.
Which equations being refolved give e.3608,
and v = .0535; whence
r = 8.0000
+e=.3608
x = 8.3608
U 4
s = 2.5
+ v = .0535
y = 2.5535
Again,
296
RESOLUTION, &c. B. I.
Again, put r 8.3608, and s = 2.5535, for a-
nother operation; whence will be found
.170078,
M
y
M
=.051944; and b-r-l: S = —
0000245, c—slr=.0002568, and from thence
will arife theſe two equations,
e +.170078v = - .0000245,
and v+.051944e =
.0002568.
Which being refolved, give e-.0000688, and
v = .0002604; therefore
g
+e=
8.3608000
8890000'
* 8.3607312
s = 2.5535000
+= .0002604
y = 2.5537604.
?
SECT.
→
297
SECT. IX.
The Geometrical Conftruction of Equations.
HE conftruction of equations, is the drawing
Trights liners or curves, after fuch a manner,
rights lines or curves, after fuch a manner,
as by their interfections, to give the roots of the
equation propofed. This method is ufed for a-
voiding the tedioufnefs of computation; and is
exact enough for finding two or three of the firſt
figures of the root, but not more. For where
great exactness is required, we are not to truft to
a conftruction by lines; but make a computation
in numbers, to find the root.
In geometrical conftructions, the fimpleft is al-
ways to be made uſe of, or that by which we can
come the ſhorteſt way, to the roots of the equation
propoſed.
But fince the extraction of roots by converging
feries, is now brought to fo great perfection; geo-
metrical conſtructions are almoft laid afide. There-
fore I intend to trouble the reader only with the
fhorteft methods of conftructing equations as far as
the fourth power. When we come to higher pow-
ers, there is fo much trouble and difficulty in draw-
ing the lines proper for them, that their interfec-
tions cannot be depended on; and one may fooner
extract the root in numbers.
PROBLEM
XCVI.
To conftruct a fimple equation.
RULE.
1. When there are feveral fimple quantities, con-
nected by the figns + and -
From a certain
point,
298
B. I.
CONSTRUCTION of
है
Fig.
3.
point, draw a right line, from which point fet
all affirmative quantities one way, one ad.
joining to another; from the lat point, fet
all the negative quantities the contrary way,
adjoining to each other as before. Where the
laft ends, the distance from thence, to the firſt
point, gives the fum (or difference) of all;
which is affirmative or negative, according as
it lies on the affirmative or negative fide of
the first point affumed.
2. When you have the fquare root of two
quantities, find a mean proportional between
them, by Prob. 16. B. VIII. Geometry.
3. To reduce two compound quantities to
the fame defignation. By Prob. 15. B. VIII.
Geometry, find one or more proportionals
thus; fay, as the first letter of the first quan.
tity, to the first in the fecond, fo the fecond
in the fecond to that fourth proportional. A-
gain, as the fecond letter in the firft quantity
to the third letter in the fecond; fo the fourth
proportional laſt found, to another fourth pro-
portional. Proceed thus till all the letters in
one quantity be exhauſted.
'Note, when any term is of too low a di-
menfion, make 1 to be one of the factors, as
oft as it is wanted. And when you have fe
veral fimple quantities, add them into one, by
Art. 1.
4. For many compound quantities, reduce
them all to the fame defignation by Art. 3.
Ex. I.
Suppoſe a + b C = X.
=
Draw the line DAB, and from the fixt point
A, fet off ABa, and make BC b, both
forward; laftly, make CEC, backward.
Then AE = x.
+
Ex.
Se&t. IX.
299
EQUATIONS.
Ex. 2.
Fig.
Let ax = bc, to find x.
Make AB (a) : AC (b) :: AD (c) : AE 4.
=x, (by Prob. 15. Geom.)
Ex. 3.
Suppoſe zabcx=5defg.
Make as 2a
5d::e: m (Pr. 15. Geom.)
and b : f : : m : n
and C g::n:
:n: p
then 2abc: 5dfg: :e: p
and 2abcp = 5defg = 2abcx
or px,
Ex. 4.
Let abx-f√bc x x = ddvac-bd.
×
am,
By Prob. 15. Geom. make a : b::d: m;
then bdam; and ac-bd =
✓ ac
make c-mn, then vac bd = √a an.
Find a mean proportional between a and
n, and q a mean between b and c, (Prob. 16.
Geom.) then the given equation becomes
abx — fqx — ddp.
Reduce theſe three terms to the fame defig-
nation, thus @ :f::q:r, whence fq = ar, in
like manner ddas; then the equation is
abx-arx-asp, or bx-rx sp. Put b-rt,
then tx sp, and t:s::p:x required.
Ex. 5.
Let 2abcdd-eefgb+3kllmn=4qrstz-5noplz,
to find z.
Reduce all the quantities to the fame de-
fignation, then
497s%
300
of B. I.
CONSTRUCTION
Fig.
4.
4qrsx = 5nopl
4qrstv zabcdd
4qrstw = eefgh
=
4qrsty=3kllmn.
then the equation becomes
4qrstv—4 grstw+4qrsty=4qrstz—4qrsxz.
that is, tv-tw + ty = tz - XZ
Put v-w+y=A, t-x B, then At B,
or B: A: : t : z.
PROB. XCVII.
To construct a quadratic equation.
IRUL E.
If it is a pure quadratic; reduce the quan-
tities concerned therein to the fame defigna-
tion (Prob, xcvi. Art. 3.) by which means
furds will be denoted by fimple quantities, and
at laſt you will get all the known quantities
equal to a known fquare, whofe fide is the
root.
Suppoſe yy = ab
Ex. I.
cdd
+ d√ aa—bc.
b
Make b c d
::
m, (Prob. 15. Geom.)
cdd
bmd
= md. Alfo
b
b
then cd bm, and
make a bcn, then be an; whence
:
yy = ab — 'md + d√ aa—an.
Let p be a mean between a and a-n,
(Prob. 16. Geom.) then aa-an-p. whence
yy = ab — md + dp.
=
Let dab: q, then ab dq.
then yy = dq
dm + dp.
Laftly, put q-m+pr, and finds a mean
between d and r; then yydrss, and ys.
2 RULE.
Sect. IX.
301
EQUATIONS:
2 RULE.
Fig.
In adfected quadratics, reduced to this form 5.
aa+bann. Draw a right line AD, then take
any point C; and make CBb, towards the
right hand if + b, or towards the left, if - b.
Erect the perpendicular BF n. From the
center C through F, deſcribe the circle AFD,
to cut AD. Then (BD, BA) the diſtances of
B, from the interfections A, D, are the two
roots, the affirmative to the right hand, the
negative to the left of B.
Ex. 2.
Let aa + 3a = 10.
Draw the line AD, make CB = 1½ on the
right; find a mean proportional between I
and 10, fet it in the line BF, perpendicular
to AB, with the radius CF defcribe the circle
AFD; then a BD+2, and a BA=-5,
the two roots required.
Ex. 3.
Suppoſe aa— 3a = 10.
за
5.
Draw the line AD, make CB (on the left 6.
of C) 1, find a mean proportional between
1 and 10; at B erect the perpendicular BF,
and make BF the mean; with the radius.
circle AFD; to cut AD in A
BD=+5, and a=BA=—2,
CF defcribe the
and D; then a
the roots required.
3 RULE.
In fuch quadratic equations as may be re-
duced to this form, aa+ba-nn. From any
point C as a center, in the right line BD, with
radius 1b, deſcribe the circle BFD, erect a per-
pendicular at D on the right, if it be + b, or
on
302
B. I.
CONSTRUCTION of
*
7.
Fig. on the left at B, if it is-b; whofe length is
BA=n. Through A draw AFG parallel to
BD, to interfect the circle in F and G; then
AF and AG are the two roots of the equa-
tion; which are affirmative, if they lie to
wards the right hand from A; or negative,
if on the left.
Note, if the parallel does not cut the circle,
or touch it, the equation is impoffible.
=
Ex. 4.
Suppoſe aa+ja —— 10.
With the radius 31, and center C, defcribe
8. the circle BFD. At the end of the diameter
D, on the right, raife the perpendicular DA,
a mean between 1 and 1o. Through A draw
AFG parallel to the diameter BD, to cut the
circle in F and G; and AF, AG, being on
the left from A, are two negative roots :
a = AF== 2, and a = AG =
- 5.
7.
Ex. 5.
7a=
Let aa-7a
10.
With the radius CB = 3½, and center C;
defcribe the circle BFD; at the end B, of the
diameter BD on the left, raiſe the perpendi-
cular BA, equal to the mean between 1 and
10. Through A, draw AFG parallel to the
diameter BD, to cut the circle in F and G;
then AF, AG, lying on the right hand from
A, are the two affirmative roots; and a=AF
2, and a AG = 5.
4 RULE.
When the unknown quantity is higher than
the ſquare, and the index in one term double
to that in the other; it may be brought to fome
of the foregoing forms, whofe higheſt term is
a fquare.
Sect. IX.
303
EQUATION S.
a fquare. Affume an unknown quantity, whofe Fig.
rectangle with ſome given quantity, is equal to
the fquare of the unknown quantity propoſed;
for this fubftitute that rectangle; and you will
have an equation as required.
Ex. 6.
Let Z4 bzz = n.
Affume dx=zz, then by ſubſtitution, ddxx
b
n
-bdxn, and xx-
X
x =
Let d: b: :
d
dd'
::
Ip; then bdp; alfo make dn: 1:q,
and d: 1 ::q: r, then' dd : n::1:r, and
And the leaft equation becomes
ddr = n.
***
dp
d
ddr
X = that is, xx-px=r, which is
dd
to be conſtructed by fome of the former rules.
To demonftrate thefe rules. Let aa + ba
= nn. Here we have CB¦b, BF = n, and 5.
if BDa, then CD or CF = a + b, and
+žb,
CF CB²+BF2, that is, a+b² =÷bb+nn.
CF²=CB²+BF²,
2
But if BA-a, then CA or CF-a-b,
and —a—;b)²=bb+nn. In both cafes aa +
bann.
Again, if aa-bann, we have as before
CB=1b, BF=n, and if BD=a, then CD or 6.
CFab, and ab² = 4bb + n.
=
But if AB-a, then AC or FC a
+b, and
aa
bann.
a+b)²=nn + bb, in both cafes
Again, if aa-bann; here BCb,
BA=n, and AG BD-AF, therefore if 7•
a, then AG ba, and AG ×
AF
AF
=
=
AB, that is, ba×a = nn.
But
304
CONSTRUCTION of B. I.
Fig.
8.
9.
But if AG=a, then AF-BD-AG-b-as
and AFX AG AB', or baxa = nn. In
both cafes al -ba
nn.
Lattly, when aa + ba—nn, then BC=1b,
BA=n, as before; then if AF-a, then
AG=BD-AF=b+a, and AFX AG=AD',
ora x b + a= nn.
But if AG——a, then AFBD—AG=b+a,
and AFX AG = AD², or b + a × — a = nn.
In both cafes aa + ba = nn.
PROB.
-------
XCVIII.
To construct cubic and biquadratic equations:
To conftruct a cubic equation, that has all
its roots real, by a circle. Let the radius
OB = R, fine EF = s, GH the fine of the
arch GB or 3BE. Then by trigonometry,
35.
453
RR
=GH. Draw CD parallel to AB,
and put SF = c, ES = x, GH = 6, then +x
4
ول
Es, whence 3 Xc+x-
RR
Xc+x³b, this
reduced gives x³+3cx²+3ccx + c³
= 0.
2/ RR + 46R R
3
CRR
Suppoſe this cubic equation be given,
x³+px²+qx+r=o. Comparing this with the
former, and equating the coefficients, we have
p=3c, and cp. Alfo q=3cc — & RR={pp
-RR, whence Rpp-39, and rc²
+RRRR; whence b =
Hence arifes the following
gr -pq
PP-39
+ 23 p.
I RULE.
Se&t. IX.
305
EQUATIONS.
IRUL E.
Fig.
Having the equation x3+p+q+r=0. 10.
given;
I.
2.
3.
4.
With the radius pp
the circle BGAK,
PP 32, defcribe
Draw the diameter AB, and CD parallel
3
to it, at the diſtance of p; above it, if
it bep, but below it, if
Draw alſo ZG parallel to AB, at the di-
ftance gr - pq
2
P.
+3p, above it, if it is af
PP-39
firmative; or below it, if negative. Let
it cut the circle in G.
Take the arch BPBG; and make
PQ = QK KP.
5. From the points P, Q, K, let fall perpen-
diculars, upon the line CD, which will be
the roots of the equation; the affirmative
above the line, and the negative below it.
SCHOLI U M.
If 34 be greater than pp, the equation is
impoffible; for in this cafe the equation has
two impoffible roots.
Allo if p = 0, then the radius of the circle
po,
OB = 3√ — 39; and CD coincides with AB;
and the distance of ZG from AB is
sr
And if q is affirmative, the equation is im-
poffible. Thefe conftructions are extreamly
cafy.
EN. 1.
Let x9m². 22X 120 0.
Here the radius OB = 3√pp ---
81+66=8.0829, and p3,
et CD, above AB.
X
39
the diſtance
And
I IF
306
B. I.
CONSTRUCTION of
Fig. And 9rp q + 3 p -1080+198
PP-39
उ
2 =
81+66
+ 3×6
i1. =—6+6=o, the distance of GZ from AB;
therefore ZG coincides with AB; and the
arch BG and alfo its third part is o, and P
falls on B; and making PQ=QK=KP, and
letting fall perpendiculars on CD, we fhall have
PS3, QT +4, and KT-10, the
three roots required.
C3
:
13.
14.
I
Ex. 2.
Suppose x317x²+82x-1200.
The radius OB√289246
P
3p =
and
4.37.
5.66, the diftance of CD below AB,
gr pq
PP 39
2
+ 3 p =
-1080+139+
209246
24
3
7.302 11.3334.031, the diftance of
GZ below AB. Take BP the third part of
BG, and making PQ = QK=KP; and mea-
juring the perpendiculars upon CD, we have
PS = +4, QT+10, and KV+3, the
roots of the equation.
Ex. 3.
Let y³ — 13y+12= 0.
In this example po, therefore CD coin-
cides with AB; and radius OB=3√—39=
3√394.18; and
3
3r
q
-36
===
= 2.77 the
-13
diftance of ZG above AB. Take arch
BP arch BG, and make PQ=QK=KP;
and let fall perpendiculars on AB, then
PS= +1, QT +3, and KV4, the
=
three roots required.
Cubic equations may alfo be conftrued by
a cubic parabola and a right line. Let FVAC
be a cubic parabola, whofe latus rectum is r.
Draw
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H.
Sect. IX.
307
EQUATIONS.
Draw VE the tangent at the vertex, perpen- Fig.
dicular to the axis VS, and BI parallel to it, 14.
and SBC perpendicular to AB.
Then put VHb, VDc, VI or SB=n,
BC=a, Vox, then SC=n+a, and by
the property of the parabola VS = SC³, or
x = =n+a. By fimilar triangles, VH (b):
3
: VD (c) :: SH (x—b) : SC =
cx-bc
cx-cb
b
; and
BC or a= -n, and bacx — cb — nb,
b
Ъ
whence cx-ba cb + nb, or x— — a = b +
nb
7
=
C
that is (expunging x) n³ + 3n'a + 3na² +
C
h
¿ ³
C
nb
a = b + which reduced is
C
a² + + 3na² + 3nna + 22² = 0.
b
a
b
nb
C
Let a³ + pa² + ga + r = O, be
any cubic equation. By comparing them,
and equating the like terms; we have 3np,
and np. Alfo
b
=3nq = s PP -9.
b
3nn
= 9, and
C
C
b n
Again, nb
I
27
=r, or p³ -b — p
C
X = PP
-q=r, whence bpqP' -r.
1
2
3
3
b
b
And fince
C
pg 2 p³ — 3r
PP - 39
= 3 PP - 9, c =
C
3 PP - q
Whence we have the fol-
lowing construction, by this
X 2
2 RULE.
308
CONSTRUCTION of B. I.
1
Fig.
14.
1.
2
RUL E.
3
Given the equation a³-+pa+qa+r=0.
With the parameter 1, and the axis VS,
deſcribe the cubic parabola FVAC, draw
the diameter RAB, diftant p from the
axis VS, to the right hand, if affirma-
tive; and draw the tangent at the vertex
IVD.
2. In the axis VS take VH = {}
downwards, if affirmative.
I
2
{ 9 — √ √ P³ —
p q ~
3. In the tangent IVD, take VD
=
tive.
pq — ~ p³ — 3r
PP - 39
>
VH
= PP - i
q
to the left, if affirma-
4. Through the points D, H, draw the right
line FDHC, to cut the parabola. From
all the points of interfection, let fall per-
pendiculars on the diameter AB, which
will be the roots of the equation; thofe
on the right hand of AB affirmative;
thefe on the left, negative.
5.
When any of the aforefaid quantities are
negative, they must be laid the contrary
way to what is directed above (Art. 1,
2, 3).
SCHOLI U M.
If the fecond term be wanting, po, and
AB coincides with VS; and then VH-1',
and VD =
g
9
・
If the numbers given in the equation, be
too great for your parabola; the equation is
eafily changed into another with lefs numbers,
by Prob. xlii.
Ex.
Sect. IX.
309
EQUATIONS,
Ex. 4.
Let the equation be a³ + 1.8a².51250 —
1.05 = 0.
g
P
2
Z
The parabola being defcribed, we have
VI=6, the diftance of IB from the axis
VS, on the right; and VHpq — √ √ p² — r
.3105, taken downwards from the vertex V.
VH
195, on the left from V.
And VD=
SPP-9
Through D, H draw the right line FDHC,
cutting the parabola in F, G, C; from which
points letting fall perpendiculars on AB, we
have FR1.75, GL.8, and CB=
+75,, the three roots of the equation.
Let x3
3
Ex. 5.
1x+4=0.
Fig.
Here po, and therefore AB coincides
with VS; then make VH1, up-
3
3
ward; and VD = ———, to the right hand.
9
7
Then through H, D draw the line FDC, to
cut the parabola, in F, G, C; from which
letting fall perpendiculars on VS, we have
FR = — 11, GL = +1, and CB+1, the
three roots required.
14:
15%
Cubic and biquadratic equations may alſo
be conftructed by the common parabola. Let 16.
FVAC be a parabola, VS its axis, AB a dia:
merer parallel to it. EA, and SBC two or.
dinates perpendicular to VS. Draw alfo HD
perpendicular, and HQ parallel to VS;
draw HC.
X 3
Put
#
9
1
310
Fig.
ร
1
CONSTRUCTION of
B. I.
Put EA or SB = c, AD =d, DH=g,
16. HC =ƒ, and BC-a. Then QC=g+a, la-
tus rectum of the figure
1.
Then by the property of the figure AB
FBX BC2ca + aa, and DB or HQ=2ca+
aa-d, and HC²=HQ²+QC², that is, ff=
a4+ 4ca³ + 4ccaa+dd—4cad—2daa+gg+28
aa; which reduced is,
@4 + 4ca³ + 4cca² + 2ga + gg = o.
+
2d 4dc + dd
-f
Let a¹ + pa³ + qa² + ra + s = o, be any
biquadratic equation; compare this term by
term with the other; to find the values of
the quantities c, d, f, g. Then we have
4cp, and c = p.
Again, 4cc-2d+1=9, and 2d=4cc+1—
+ pp + 1 — 9.
I
9 = 4PP + 19, and d =
Again, 2g4cdr, and g =
= pd + r
2
A
2
4cd + r
2
Laftly, gg+dd-ffs, and ff=gg+dd—s.
From hence arifes the following construction.
3
RUL E.
Having the equation,
or
a4 + pa³ + qa² + ra + s = o.
aš + pa² + qa +r=0.
1. Defcribe a parabola FVAC, whofe pa-
rameter is I, and axis VS. Draw the diame-
ter AB at the diftance of p from the axis,
on the right hand, if p is affirmative. Then,
for the central rule.
2. From
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Pl. II. pazo
NIV
OF
CH.
Sect. IX.
311
EQUATIONS.
2. From A, the top of the diameter, take Fig,
AD =
ipp + 1
2
downwards, if affirmative.
2
3. From D in the perpendicular DH, take
DH =
mative.
p× AD + r
2
towards the left, if affir-
4. But when any of theſe quantities are ne-
gative, fet them the contrary way.
5.
Ha
2
From the center H with the radius
VAD² + DH' —- s == VHs defcribe a
circle which will cut the parabola in feverai
points as C.
6. From the points of interfection, let fall
perpendiculars upon the diameter AB, and
thefe will be the roots of the equation; thefe
(BC) on the right fide of AB, are affirmative
roots; and on the left fide, negative. And
there are always as many real roots, as there
are points of interfection; and the reft are
impoffible.
SCHOLIU M.
If the ſecond term be wanting; then po,
and the diameter AB coincides with VS. Then
alfo AD =
I
9
and DH = žr.
2
In cubics s is wanting, and then the ra
dius HC becomes = HA.
If the numbers or coefficients be too large
for your parabola, you must transform the
equation, into another to fuit your parabola,
by Prob. xlii. and then conftruct it; and laft-
ly, reſtore the true roots.
16.
Ex.
X 4
312
B. 1.
CONSTRUCTION of
Fig.
17.
3
Ex. 6.
Suppose y³ +20y² 500у- 6000 = 0.
The numbers being too large, put x=y,
του,
ory IOX; then the equation becomes
1000x200cx-5000x60000, that is,
x³+2x² — ~60, where the numbers are
fmall,
The parabola FVC being defcribed, make
EA, on the right, and draw AB pa-
rallel to the axis VS.
Make AD=
- pp + 1 - 9
2 + 5
11
32,
2
2
downwards.
Draw DH perpendicular to
AD, and make DH =
pxAD + r
2X3-6
2
18.
2
, to the left. From the center H, with
radius HA, defcribe a circle, cutting the pa-
rabola in R, A, C, F; from which letting
fall perpendiculars on AB; we have RA-1,
BC: +2, FB 3, the roots of the equa-
tion x³-2x²—5—6—0, and multiplying by
10, we have 10, + 20, and 30, for the
roots of the given equation y +20у²-500y
6000 = 0.
Ex. 7.
Let x4 - 1.75x³- 4.625 +4.875% +
6.75 = 0.
Defcribe the parabola FVC, and draw the
axis VS, and make EA p.44, to the
left, and draw AB parallel to the axis, make
10p+1-9
3.19 downwards. Draw
AD=
2
DH perpendicular to AB, and make DH
p X AD + r
.36, to the right.
}
From
Se&t. IX:
313
EQUATIONS.
From the
center H,
with the radius Fig.
√AD² + DH' -6.752, defcribe a circle 18.
cutting the parabola in F, R, G, C; from
which drawing perpendiculars to AB, we
have RO 1, FB — — 1½, GI=2, CP=
2.25, the roots required.
=
Ex. 8.
Given the equation x1.5x + 5x² — 9× —
6=0.
Deſcribe the parabola RVC to the axis VS, 19.
and make EA=4p=—.375, to the left, and
draw AB parallel to VS. Take AD =
± pp + 19
2
1.72 upwards. Then (per-
pendicular to AD) take DH =
+
pXAD+r
2
3,21, to the right. With the center H,
and radius 4.39 (= √AD² — DH² + 6), de-
fcribe a circle, to interfect the parabola; from
the points of interfection, letting fall perpen-
diculars on AB, gives the roots, RO.5,
and CB = + 2. The other two roots are im-
poffible, which is known from this, that the
circle interfects the parabola in no more
points than theſe two.
Let x4
Ex. 9.
567x² + .S06x +3.8640.
Here po, therefore deſcribe the parabola 20.
FAC, whofe axis is AS; and make AD =
2
9
= 3.335, downwards; and DH==
403, perpendicular to AD, to the left. With
radius AD+DH² -3.864=2.72, defcribe
a circle cutting the parabola, in R, G, C, F;
and
314
CONSTRUCTION of B. I.
3
}
f
[
Fig. and the perpendiculars from thefe points up-
20. on AS, give RE--.8, GI=+1, CB=+
2. 1, and Fs-2. 3, the roots of the equa-
tion.
21.
SCHOLIU M.
Geometrical equations may be constructed
by lines as well as by numbers. For proper lines
may be found for the coefficients, by proceeding
according to Prob. xcvi; and fo the whole
may be done geometrically..
Quadratic equations, whofe general form is
a² + pa + q = o, may alfo be conftructed by
the laft rule; and then r and s will be = 0;
but the method of conftructing by the circle,
is eaſier.
4 RULE.
Any cubic or b'quadratic equation + + px³
+ qx² + rx + so, may be conſtructed me.
chanically thus:
1
1. Upon a plain fmooth wall, draw a hori
zontal line AB, and CD perpendicular to it,
and take CP = p, to the left hand if p is af-
firmative. Hang a thread and plummet EPF
to any point E, in the perpendicular EP;
make a knot in the thread at n, and tie the
other ends fo to the fixt point E, that Pn may
be = ½. Then with a pin or the point of a
compaſs, move the thread EF fideways toward
CD, till the knot n falls in the point C;
mark the point D in the line CD, where the
pin is, when that happens.
2. From D take DG=PP+I—9
downwards, if affirmative.
pendicular GH, take GH=
2
$
(=d),
And in the per-
dp+r
2
to the left,
if
>
1
:
¡
1
-
D
R
19
V
E
B
+
***
Fig. 17.
R
E
H
02
S
B
R
I
A
E
18
P
DH
C
}
D
S
F.
B
A
R
E
H
I
F
S
H
D
B
1
S
20
Pl. III. pa. 314
UNIC
OF
Sect. IX.
315
EQUATIONS.
:
if it is affirmative. But if any of thefe quan- Fig.
tities be negative, they must be taken the con- 21.
trary way, to what is directed above.
3. Then with the radius. or diſtance
✔HD's, and one foot of the compaffes in
H, move the other foot along with the
thread, round in a circle, and the weight F
will afcend and defcend, as the thread EF
moves laterally. Obferve always, when the
knot n falls in the line AB, and mark all
thefe points, Q, N, O, R. Then the di-
ftances of theſe points from C, are the roots
of the equation; the affirmative on the right,
the negative on the left hand of C; thus RC
is an affirmative root, and QC, NC, OC,
negative ones.
It is plain, this rule is founded on the laft.
For the moving point of the compafs is al-
ways in the curve of a parabola, when the
point n is in the line AB. To prove which,
fuppofe the parabola ADB, to be deſcribed,
whofe focus is E. Then by the property of
the figure, EL + LR EP + ½ parameter =
EPPn or EnED + DC. Therefore the
circle cuts the parabola in L; and the diftance
of L. from DC, that is RC is one root of the
equation; and the like for the reft.
SECT.
316
SECT X.
Rules and Directions for the investigation and
Jolution of Problems.
A
PROBLEM XCIX.
To find if a question be truly limited.
Queſtion is faid to be truly limited, when it
admits but of one folution; or at moft, of
as many as is the index of the higheſt power of the
unknown quantity in the final equation. And whe-
ther a queſtion be limited or not, may be known
from the equations, by this
RULE.
When the number of unknown quantities, is juft
as many as the number of given equations, not
depending on one another; then the question is
truly limited.
But when the number of unknown quantities
exceeds the number of equations given; then the
queftion is unlimited, and capable of innumerable
anſwers.
And when the number of unknown quantities
is less than the number of given equations; then
the queſtion is abfurd and impoſſible, except theſe
equations be dependent upon one another; in which
cafe the dependent ones may be ftruck out. **
Equations are faid to be dependent on one ano-
ther, when they may be formed or derived from
one another, by any operations, with the help
of the known axioms.
For
Sect. X.
317
RULES for folving PROBLEMS.
For by Cor. 1, 2. Prob. liii, one unknown quan-
tity may be taken away by each equation; fo that
at laft there will remain but one equation, and one
unknown quantity in it; and therefore it is truly
limited.
But if there were more unknown quantities than
equations, there will remain more unknown quan-
tities than one, in the laft equation. And then the
queftion is not limited; for all of them, but one,
may be taken at pleafure; and this is the reafon of
queftions being unlimited.
Laftly, if there be more equations than unknown
quantities, then at last there will remain one un-
known quantity for feveral equations; and then
the queſtion is more than limited; and will there-
fore be impoffible. For the unknown quantity be-
ing exterminated, there will be an equation con-
fifting of all known quantities; which must be con-
tradictory to one another, except they were fome
way or other depending on one another, fo as to
make an equality.
SCHOLIU M.
As a problem is truly limited, when the number
of independent equations, is equal to the number
of unknown quantities; fo likewiſe a problem is
truly limited, though there be never ſo many equa-
tions, provided all, above that number, are de.
pending upon theſe, and derived from them. This
is plain from any algebraic procefs; for in the
ration, all the fucceeding equations are derived
from thefe, first given; and all equations fo de-
rived, make no alterations in the limitation of the
problem.
ope-
A problem may be impoffible and more than li
mited, though the number of equations be less than
the number of unknown quantities; and that is
when the equations are contradictory.
As
318
B. I:
RULES for folving
As if a + e + 2y = b,
And 2a + 2e + 4y = c; a, e, y being unknown
quantities, and b, c, known ones. Now if it hap-
pen that c = 2b, the problem is unlimited; but if
is not 2b, then the problem is impoffible.
And therefore in general, problems are abfurd,
when the equations given are derived from abfurd
equations, or may be reduced to fuch: even though
the number of equations be equal to or less than
the number of unknown quantities.
The equations given in a problem, ought to be
independent, otherwife they will either be confe-
quential, or contradictory to one another. In the
firft cafe, you will at laſt find fome quantity equal
to itſelf. And in the ſecond cafe you will arrive at
fome abfurdity, where a greater quantity is equal
to a leſs. And it often happens, that at the end
of an operation, the equations given, are found to
be either dependent or inconfiftent with one ano-
ther; which at firſt, could not ſo eaſily be diſcovered.
PROBLEM
C.
To investigate an algebraic Problem.
RULE S.
1. When a queftion is propofed to be refolved
algebraically; the first thing to be done, is to con-
fider the nature and circumftances thereof, to find
out what is given therein, and what required. And
the nature and tenor thereof being clearly under-
ſtood; reject every condition or circumftance, which
has no neceffary connection or relation with the
thing enquired after. Then give names to all the
quantities concerned in the calculation, whether gi-
ven or fought; that is, for the feveral numbers or
quantities, or at leaſt for the principal of them,
put fo many different letters, as directed in the
notation.
Sect. X.
319
PROBLEMS.
notation; taking care to make the fame letter ftand
invariably for the fame thing, throughout the whole
operation.
And in general problems, it will be convenient
to make choice of fuch letters or fymbols, as may
fome way reprefent to the mind, the things de-
figned by them; as r for radius, s for fine, for
I
latus rectum, v for velocity, t for time, &c.
And if there be never fo many quantities of dif-
ferent forts, we may reprefent them by any num-
bers we like; or even all of them by 1, which is
the moſt fimple notation. Thus we may call any
degree of motion 1, any degree of velocity, and
we may put 1 for any quantity of space, time,
matter, &c. But then we must take care to re-
prefent other quantities of the fame fort, by pro-
portional numbers.
We can alfo meafure any kind of quantity by
any other kind of quantity, by taking parts or de-
grees of one fort, proportional to the parts or de-
grees of the other. Thus, quantities of force may
be meaſured by right lines proportional to them;
bodies or quantities of matter by their weights;
velocities by the ſpaces defcribed in equal times;
and all forts of quantities or things by numbers.
2. But as that folution of a queſtion is reckoned
the more artificial, the fewer unknown quantities
are affumed at firft. Therefore when the princi-
pal quantities are denoted by letters; fome of the
quantities, that may be eafily derived from the
reft, are left without a name. As when the whole
is given and a part, the other part is eafily had
from thence, or the parts being given, you may
find the whole. Alfo when two fides of a right-
angled triangle are denoted in algebraic terms, the
third fide is had from thefe, by addition or ſub-
traction of fquares. Likewife three terms of a
proportion being given, the fourth term is eaſily
derived
320
B. I.
RULES for folving
derived from thefe three; and in all fuch like
cafes, where the values of fome are eaſily derived
from the reft. And by, this means there will be
fewer terms to exterminate.
3. After the defignation of the quantities, by
letters, as aforefaid; we muſt next abftract it
from words, and tranflate it out of the Engliſh
into the Algebraic language: that is, we must de-
note all the conditions of it, by fo many algebraic
cquations; and this is called ftating the question. In
order to this, we muſt fuppofe the thing done
which was required; and then, without making
any difference between the known and unknown
quantities, affume any of them, known or un-
known, to begin your computation from; taking
fuch as you think will bring out the fimpleft equa-
tion, or give the eafieft folution. And it is beft to
affume that quantity to begin with firft, which is
eafieft found or brought to an equation.
And
therefore it is often more convenient, not to begin
with that which is directly required, but with fome
other, from which the quantity required may be
eafily had.
From thefe firft affumed quantities, you must
proceed in a ſynthetic method to find other quan-
tities wanted, and from thefe to find others, &c.
according as the nature of the queſtion directs, till
you get what equations you want.
To this pur
poſe, you must attend ftrictly to the nature, de-
fign, and meaning of the queftion, and fearch in-
to all the circumftances of it, and examine into the
particular relations of the quantities to one another;
To that from thence you may get a proper number
of equations. But fometimes thefe equations can-
not be had from the words of the queftion; but
depend upon the hidden properties of the quanti-
ties concerned therein; and then the equations are
to be deduced from them, by a proper chain of
reafoning,
Sect. X.
321
PROBLEM S
$
10
reaſoning, according to the nature of the fubject
under confideration. Thus, in numerical que-
ftions, we muit proceed by the properties of num-
bers: in geometrical problems, by the elements
of geometry in mechanical problems, by the
principles of mechanics: in trigonometrical pro-
biems, by the rules of trigonometry: in philofo-
phical problems, by the laws of motion; and fo
of other fufjects. And here great care muſt be
taken that your equations do not depend upon
one another; and that there be as many as there
are unknown quantities, otherwife the queftion will
not be limited.
>
4. Having got a proper number of equations,
our buſineſs is now, to exterminate them one by
one, as faft as we can, till there only remains one
unknown quantity, in one final equation: then the
problem is faid to be brought to a folurion. And
by thele equations, you muſt exterminate thefe
quantities first, that are most eafily exterminated;
that is, the fimpleft firft, and fo on; till the quan-
tity that remains at laft, may give the fimpleft
equation poffible; cr more fimple than if any
other of the unknown quantities remained in the
final equation. And in all your procefs, great
care must be taken, to keep to a juft equality;
which will certainly be, if you obferve all along,
to work according to thefe juit rules or axioms, ac
the beginning of this book.
5. As to the chufing fit terms or quantities to
begin the calculation with; it fometimes happens
that there is fuch a relation of two terms of the
queſtion, when compared with the reft, that in
making use of either of them, they will bring out
equations exactly alike; or that both, if they are
made ufe of together, fhall bring out the fame final
equation, as to form. Then it will be the belt way
to make ufe of neither of these terms; but instead
thereof,
1
{
+
322
B. I.
RULES for folving
thereof, to chufe fome third, which has a like rela-
tion to both. As fuppofe the half ſum or half diffe-
rence, or perhaps a mean proportional;, or any other
quantity related to both indifferently, and without
a like.
6. The proper defignation of the terms will often
much abridge the operation. As if two numbers
are fought, whofe fum or difference (7) is given,
it will be convenient to take in+a, and in-a,
for the numbers.
Alfo when feveral numbers are fought in arith-
metical progreffion, where the common difference
(d) is given; we may properly put a➡d, a, a + d₂
d,
for the numbers, when there are three or a-11d,
13 id, a + id, a +rid, for the numbers, when
four are required; and ſo on.
Again, if feveral numbers are fought in geome.
trical progreffion; put aa, ae, ee, for three numbers:
and a³, a²e, aе², e³, for four numbers: and a*, a³e,
a²e², ae¹, e¹, for five numbers; and fo on. Or de-
note them by fuch other feries, as will give them
all, with the feweft letters.
7. Sometimes problems will run up into very high
equations, where the unknown quantities cannot
be expunged without great difficulty. Therefore
in fuch a caſe, if you can ſubſtitute new letters for the
fums, or products, or powers, &c. of fome of the
old quantities; and then expunge all theſe old
ones, and get a proper number of equations; you
may often find the value of thefe new quantities, by
eafy and low equations; from whence the old quan-
tities may be more eafily determined. And you muſt
find theſe new quantities by trials, fuch, that when
they are ſubſtituted, they may render the equations
eafier. See Prob. xxiv, xxv. B. II.
Likewiſe in any operation, when you have a mul
titude of unknown quantities, for the coefficient of
any power of the unknown quantity; put a fingle
letter
Sect. X.
323
PROBLEM S.
letter for them all, which will much abridge the
operation.
8. In geometrical problems, there is often more
labour and ſkill required, than in numerical ones.
In theſe you must first draw a figure, according to
what the queſtion requires to be done. And then it
is often requifite to produce right lines; or to draw
lines parallel or perpendicular to other lines; and to
certain points, or through certain points; or to make
fimilar triangles, and fuch like; all preparatory for
the folution of the problem: always endeavouring
to refolve the ſcheme into fimilar triangles, or right-
angled ones, or given ones. Then affume fuch a line,
&c. for your unknown quantity, as you judge will
bring out the fimpleft equation. For you may begin
your computation with any quantity, known or un-
known: which done, you muft proceed fynthetical-
ly to find the reft. In general, let theſe quantities be
denoted by letters, that lie neareft the given parts of
the figure, and by means of which other parts ad-
joining may be eafily had, without furds. In trian-
gles draw a perpendicular from the end of a given
fide, and oppofite to a given angle. Such prepara-
tions as theſe being made, juſt as you find neceffary
for the method of folution you intend to try; pur-
fue your computation according to the nature and
property of lines, and the conditions given in the
queition, proceeding from the quantities affumed,
to other quantities, as the relation of the lines di-
rect; till you get two values for one and the fame
quantity, or find one quantity denoted two different
ways, by which you will get an equation. The ge-
neral principles for carrying on the computation are
fuch as thefe, the addition or fubtraction of lines,
to find the fum or difference. The proportionality of
lines (arifing from fimilar triangles), where three
terms being given to find a fourth. The addition or
fubtraction of fquares in right-angled triangles, where
Y 2
324
B. I.
RULES for folving
two fides being given, the third may be found.
Likewife the doctrine of proportion will be of fre-
quent ufe.
Befides we must make ufe of fuch pro-
pofitions in geometry as are fuitable to the purpoſe;
fuch as B. I. prop. 1, 2, 4, 8, 10, 11. B. II. 2, 3, 10,
13, 14, 15, 18, 21, 24, 25, 26. B. III. 1, 6, 7, 17,
20. B. IV. 9, 12, 13, 14, 16, 17, 20, 27, 31; and
fome in the following books, as occafion requires.
By help of theſe principles, and a chain of right
reaſoning, we ſhall obtain as many equations as un-
known quantities, which being had, we muſt change
our method, and exterminate the fuperfluous quan-
tities, and find the root of the final equation.
g. If the method you go upon at firſt, for the
folution of the problem, proceeds but badly, as
running into high equations and furds. You muſt
draw freſh ſchemes, and begin your computation
anew, till you have hit on a method as elegant as
poffible. For the principal art, of refolving pro-
blems, is to frame the pofitions with fuch judg-
ment, that the folution may end in as fimple an e-
quation as poffible. For fome methods will pro-
duce more intricate equations and folutions, than
others. But the fkill of finding out the moſt fim-
ple and eaſy ways of refolution is not to be pre-
fcribed by any rules, but is only attained by con-
ftant practice and experience.
10. If you have any doubt what quantity to take
for the quantity fought, fo as to bring out the ſim-
pleft equation. Suppofe you have got a final equa-
tion with x, take fome other quantity y, which you
fufpect may be as fimple, feek an equation between
* and y; then if y be of as high a power as x;
the final equation, if y were uſed, would be as high
as is the final equation with x.
Or, Having got an equation between x and y;
fubftitute for x its value in terms of y, in the faid
final equation with x; and you will find what pow-
er
Sect. X.
325
PROBLEMS.
er y will arife to, without forming the procefs anew
for y. But if the equation between x and y be not
a fimple equation; it will often be as well to begin
the process anew for
ry.
Or, If there be feveral quantities, and you do not
know which will bring out the fimpleft équation.
Put letters for them all, and get as many equations.
Then by expunging fuch as are inoft eafly ex-
punged; you will, for the moft part, get the moſt
fimple equation.
11. Lastly, when the final equation is obtained,
extract its root by Sect. VIII, and you have the
anſwer in numbers.
Note, The numbers given in a queftion, cannot
always be taken at pleaſure, but must often be ſub-
ject to one or more determinations or reftrictions,
which for the moft part are difcoverable by the theo-
rem refulting from the refolution of the queftion.
12. When you have an equation containing the
quantity fought; and the equation is alfo effected
with a fecond unknown quantity, which you want
to get rid of; the extermination of which runs you
to a very high power. Now if it happens that this
fecond unknown quantity, is but in a few of the
terms, which are but ſmall in refpect of the reft.
Then if you can nearly gueſs at its value, you may
retain it in the equation, putting that value for it,
which will make little difference in the equation, a-
mong fo many quantities, if you mifs its value a
little. Then the root of the equation being extract-
ed will give the other unknown quantity very near,
And this being had, the fecond unknown quantity
will then be found more exactly, and may be fubfti-
tuted for it again, and the operation repeated, &c.
And one may often guefs nearly at the value of
this ſecond quantity, from the conditions of the pro-
blem; eſpecially if it be a geometrical one, from the
conftruction of the figure.
Y 3
Theſe
1
326
RULES for folving, &c. B. I.
}
Theſe forts of equations may alſo be refolved by
the Rule of Falſe Pofition, as directed in Pr. xcii.
Rule 5; as alſo by Prob. xcv.
13. When you want to compute a problem for
fome practical uſe in common life, but by purfu-
ing it in its mathematical rigour, you fall upon
fome irrefolvable equations or intricate furds or fe-
ries. Then you may often refolve it on very fim-
ple principles, by neglecting fuch quantities or fuch
conditions, as ferve only to embarraſs the problem,
but make little or no difference for the purpoſe you
want it. In fuch caſe, neglect fuch quantities or
fuch conditions, as are of little moment; and in-
ftead of fuch quantities as make the calculation dif-
ficult, take others nearly equal to them, which will
make the operations more fimple, or as fimple as
poffible. Or fome of the leaſt moment may be en-
tirely left out. And thus one may come at an eaſy
folution of the problem.
Theſe are the general rules of working; all which
will be made clear, by the examples in the follow-
ing Book II.
BOOK
BOOK II.
The Solution of Problems.
327
A
Problem is a queftion propoſed to be re-
folved; and the Solution of a problem, is
the finding fuch numbers, lines, &c. as
will fulfil the conditions of the question.
Of problems thefe are determined, that have a
determinate number of anfwers: and indetermined,
which have innumerable answers.
Problems are of feveral kinds, as numerical,
geometrical, trigonometrical, philofophical, me-
chanical, &c.
A problem of one, two, three, &c. dimenfions,
is that which has one, two, three, &c. folutions
or anſwers.
We have hitherto been laying down fuch rules,
as are neceffary for the inveſtigation and ſolution of
problems. The reader must take particular care,
to make himſelf well acquainted with thefe rules,
and keep them in mind, fo that he may have
them ready for ufe, upon all occafions; for with-
out them no problem can be folved. But as pre-
cepts are but of little ufe without examples, and
generally reach no farther than mere fpeculation;
I fhall therefore, in the next place, apply them to
practice, and that in the folution of a great variety
of problems, in the most material branches of the
mathematics; which I fhall now begin with di-
rectly.
I 4
SECT.
328
B. II.
SECT. I.
Numerical Problems.
PROB. I.
There are two numbers whofe fum is 25,
portion of one to the other is as 2
are the numbers?
Suppoſe {
per queft.
3 X
1a
I
2e
greater.
e = leffer.
3e:a:: 2:3
42a = 3e
4
(2)
5a
per queft.
6 a +
6 X (2)
÷
7 (5)
5,
Suppofe
per queſt.
2 X
3 tranf.
4 (5)
зе
c = e + 32e25
e
7ze + 3e = 5e = 50
50
and the pro-
to 3, what
8 e == 10, the leffer.
9 a = 2e = 15, the greater.
او
Otherwife thus,
a greater, sfum, 25; then
$ a = leffer.
22: 3 :: S
3:
320 = 35
45a= 35
a: a
за
5α = 35 = 15 = 15, the greater num
a
ber.
6sa= 10, the leffer.
PROB.
Sect. I.
329
NUMERICAL PROBLEMS.
PROB.
II.
A man having a certain number of pence, gives to
A of them, to B, to C, and to D, and
then had 3 remaining. How many bad he at
first?
Let
per queſt.
2,
3 tranf.
Ia number of pence he had.
I
2a +÷a+ša + ½ a = a - 3
თა
24
4
a = a -3.
J
4
a = 3
4 X
I
5 a = 72
PROB. III.
A man hired a labourer, on condition, that for every
day he wrought, he should have 12d. and for
every day be idled, he should forfeit 8d.
390 days, neither of them was in debt.
the number of work days and play days.
Let
per quest.
2 tranfp.
3 ÷ (20)
1,
=
After
To find
I a = the working days, b = 390;'
then ba the play days.
212a = 390-ax83120-8a,
3 20a 3120,
320a
2 12a
3120
4 α = =156, the work days.
20
56a234, the playing days.
PROB.
330
B. II.
NUMERICAL
PROB
IV.
Some young men and maids had a reckoning of 37
fhillings; and every man was to pay 3 Shillings,
and every maid 2; now if there had been as many
men as maids, and maids as men, the reckoning
would have come to 4 fhillings lefs. What is the
number of each?
Suppoſe amen, e = maids, b = 37, c = 4.
23a + 2e = b
32a+3e = b = c
4 9a + 6e = 3b
perqueſt{
2X (3)
54a + 6e = 2b
20
3 X (2)
4-5
6 5a
= b + 2c
b+2c
6 ÷ (5) |
7|a =
=9, the men.
5
2 -
за
82eb
за
b
÷
8 (2)
ge
за
=5, the women.
2
PROB.
V.
A man being asked what a clock it was? answered,
that it was between 8 and 9; and that the hour
and minute hands were both together.
Let
x= time required, b = 8, c = 12,
d = 1.
2 fince the two hands divide the
hour, and the whole circumfe-
in the fame proportion,
rence
:: d: x-b₂
therefore
c
2 X
3 tranf.
3 cx
4 cx
cb = dx
dx = cb
cb
96
4+
5x=
h.
m. fec.
= 8 43 38 T
2
PROB.
Sect. I.
331
PROBLEM S.
PROB.
VI.
ठ
of the
the
ठ
I
Ğ
A man gives the first beggar he meets with,
pence be bad and 4d. more: to the fecond
remaining pence and 8d. more: to the third the
remaining pence and 12 d. more, and fo on, in-
creafing 4d. every time, till at last. he had nothing
left, and then all the beggars had equal shares.
Query, the number of pence and beggars.
Suppoſe
per queſt.
I
2
3 ÷ (6)
a pence he had at firſt.
a
2 +4= pence given to the firſt
3
6
동
​363
w/n win olu
beggar.
a-4 remainder.
a
4+ (8) 5 a
+110, +/0
11
the remainder.'
+8=pence given to fe-
cond beggar.
+ + 8, per queſt.
7 6a+ 144 = 5ª — 24+ 288
36
2=5
이응​+4=
5
a
36
吉​+
6×(36)
5a
7 tranf.
2,
120
8, 9
IO
5
24
8 a 120
=
920+ 4 = 24 = ſhare of each.
number of beggars.
PROB.
332
B. II.
NUMERICAL
PROB.
VII.
3
the
other two
There are three numbers, the first with the other
two makes 14, the fecond with
makes 8; the third with the other two makes
8. What are the numbers?
Let
उ
a, e, y be the numbers.
e + y
2a+
= 14
3
perqueft
a ty
3e +
=8
= 8
1
4
ate
4y+
=8
= 8
5
2 X (3
3 X (4)
5-6
5 × (5)
4 X (5)
8 •9
7X (4)
10 X (3)
5 3a + e + y = 42
6 4e + a + y = 32
4e+a+y
7 20 3e = 10
815a + 5e + 5y = 216
9a + e +5y = 40
10 14a + 4e 170
118a
12e = 40
12 42a12e = 510
11 + 12 13 50a= 550
13÷(5) 14 a = II
53e2a IQ
tranf. 15 3e
15 ÷ (3) 16 e
20
IO
4
3
:
5 tranfp. 17y=42— 3a — e = 5•
PROB. VIII.
Having given the sum of two numbers 8, and the
difference of their ſquares, 16; to find the num-
bers.
Let
1x leffer number, a = 8, b = 16.
* greater number.
20
I 2
Seat. I.
PROBLE M S.
333
1 & 2
2 &
2 ~
3 x fquare of the leffer.
aaaxxx fquare of the greater
5aa2ax = b per queſt.
62ax = aa
4.
3
5
tranf
b
a a
b
624
71 x =
=3, the leffer.
24
2,
8a-x=5, the greater.
PROB. IX.
There are three numbers, the fum of the first and
Second is 9, of the first and third 10, of the fe-
cond and third 13. What are the numbers?
Let
perquest
2
४ ४
1x, y, z, be the numbers, a 95
b = 10, c = 13;
2x + y = a
3x + x = b
C
X
X
x + b − x = 6
4+2
5y=a
حب
6z=b
4, 5, 6
7 tranf.
7a
82x = a + b
a+b.
C
8 (2)
a + b -
C
9x=
3
2
51
Io y = a
10
x = 6
6,
Z
11/z = b −x=7
PROB.
X.
Two travellers A and B, 365 miles dijtance, fet out
at the fame time. A travels 10 miles an hour;
How far does each travel before they meet?
B 8.
Suppofe
A travels x miles, then B travels
360 — x.
by propor. 2x: 360-x:: 10: 8
2 X
38x3600 · IOX
3 tranf
}
334
B. II.
NUMERICAL
3 tranf.
4 18x=3600
3600
4 ÷ (18)
5 x =
200, A's journey.
18
6
I,
=
XI.
360-x
-x=160 B's journey.
PROB.
If three agents A, B, C, can produce the effects a,
b, c, in the times e, f, g, respectively. In what
time will they all jointly produce the effect d?
Let
I
time fought.
2 e: x : : a:
ax
A's effect in the
time x.
by pro-
portion
bx
3f: x
:b:
B's effect in time x.
f
CX
48 : x :: c :
g
ax bx
CX
2, 3, 4,
5
+
+
e
f
g
d
5 reduced 6 x =
C's effect in time x.
= di
time required.
a
b
C
+
+
e
g
PROB. XII.
A woman can buy apples at 10 a penny, and pears
at 25 for 2 pence; if he lay out 9 pence for
100 apples and pears together. How many of
each must she have?
Let
by pro-
portion
1a
apples, then 100 a = pears.
a
2 10: Id. : : @ : —,
apples.
a
IO
price of the
3 25 2d.:: 100 a:
price of the pears.
200
24
25
per
Sect. 1.
335
PROBLEMS.
per queſt. 4 +
a
200
2a
= 9-1-1.
50
4x150) 5 5ª + 400
55a+ 400-40 = 475
5 tranfp. ง 75, the apples.
7100
100 - a
a=25 the pears.
PROB. XIII.
A vintner would mix wine at rod. the quart, with
another fort at 5d; to make a 100 quarts to be
How much of each muſt be take?
fold at 7 d.
Let
by pro-
portion
per qu. {
a = quarts of 10 penny, e
quarts
of 6 penny, b10, c = 6, m =
100, ƒ = 7.
21: baba, value of a quarts.
31: c :e: ce, value of e quarts.
4.ba+ce = mf.
5a+e=m
6e = m
7 ba+cm
а
ca = mf
cm
ca = fm
5-
4, 6
7 tr.
a
8 ba
8
9α=
fm
cm
b
C
bm
fm
6,
Io e =
b - c
PROB.
XIV.
A faltor exchanged 6 French crowns and 2 dollars
for 45 billings; and 9 French crowns and 5
dollars for 76 Shillings. What is the value of a
French crown and a dollar?
x
Suppoſe = French crowns, y dollars,
| 1 x
a=6, b=2, d=9, e=5, c=45,
f = 76.
per
"3
336
NUMERICAL
B. II.
per qu.
2 X
3 Xb
2 ax + by = c
3 dx + ey=f
4 eax + eby = ec
5 bdx + ety = bf
6 aex
bdx ecbf
=
I
= 61.
a + by = c
4-5
ec
bf
6 ÷
* -
7
ae
bd
ec
bf
2,
8
ae bd
ec
bf
8 tr.
9 by = c
a
ae
bd
ae
bd
af
cd
I
10 y =
= 44•
ae
bd
bfa-bdc
PROB.
XV.
To divide a number b, into four parts; ſo that the
first being increafed with d, the fecond diminiſhed
by d, the third multiplied by d, and the fourth di
vided by d; may be all equal.
Let
Il a, e, u, y, be the four parts.
2a + e + u
3 a + d = e
4 a + d = dy
5a+d=
И
d
per qu.
3+ d
4 ÷ d
a + d
7y=
d
6e=a+2d
+ y = b,
d
9 reduced 10 a =
5 x d
8uad + dd
2, 6, 7, 8, 9 a + a + 2d +
bdd32ddd
at d
+ ad + dd = b
d
dd + 2d + I
Sect. I.
337
PROBLEMS.
bd + d³ + 2dd + d
6
II e
e=
dd + 2d + I
b
7
12 y =
dd + 2d + I
bdd
8
13 น =
u
dd + 2d ÷ 1
PROB.
XVI.
A merchant bought (a) bushels of wheat, (b) bushels
of barley, and (c) bushels of oats for (m) pounds.
Afterwards he bought (d) bushels of wheat, (e) buſh-
els of barley, and (f) bushels of oats, for (n)
pounds.
And after that, (g) bushels of wheat, (b) bushels of
barley, and (k) bushels of oats for (p) pounds,
each fort at one price. What was each per bushel?
Let
per qu.
2X fk
3 X ck
4 x cf
5
6
x, y, z, be the prices of the wheat,
barley, and cats.
2 ax + by + cz = m
3
5
dx + ey + fx = n
4 gx + by + kz = p.
afkx + bfky + cfkz = fkm
6 ckdx + ckey + cfkz
7\ cfgx + cfby + cfkz = cfp
8 afkx
& 7 9 ckdx
ckn
ckdx+bfky-ckey-fkm-ckn.
cfgx+ckey-cfby ckn-cfp.
fubftit. 8, 10 Ax + By C
9,11 Fx + Gy = H.
V
1
C - Ax
10 reduc. 12 y =
B
H-Fx
11 reduc. 13y =
G
C- Ax
H-Fx
=
12 13 14
B
G
14 X 15 GCGAx
BH-BFx
Z
15 tr.
338
NUMERICAL
B. II.
15 tr. [16] BFxGAx BH-GC
—
BH-GC
17 x =
BF
GA
by refti-
tution
10, II.
18 x =
8
is_cfk
19] x
cfk
bcfkkn + ccfkep + cffkhm
· cffbpk — cfkkem — cefkbn
bcfkkd + ccfkge + acffkh
-bcffgkacfkke - ccfkhd
bkn —— bfp + cep―kem+fbm—chn
bkd-bfgeeg-aek+afb-cdb
being had, y may be found by ftep 12th; and
then z, by reducing the equation in ftep 2d.
PROB. XVII.
If the number of oxen a, eat up the pasture b, in the
time c; and the oxen d eat up as good a pafture e,
in the time f; and the grafs grows uniformly. To
find how many oxen will eat up the like pasture` g,
in the time b.
State it thus:
Oxen, Weeks. Acres.
a
C
b
d
f
e
b
g
Let
by pro-
portion
y
Il y
21 X
y
number of oxen fought.
grafs upon an acre at firſt.
3grafs growing upon an acre, in
a week afterwards.
I
41 grafs which an ox eats in a
week.
5 bx, ex, gx = grafs on the paſtures
b, e, g.
61 cbz, fez, bgz = graſs
grafs grown
grown af
wards on the paſtures b, e, g,
in
the times c, f, b.
7 ac, df, by grafs eaten by the
oxen, a, d, y, in the times c, f, b.
per
Sect. I.
339
PROBLEMS.
per qu.
I
8 x ef
9 × cb
13 ÷
14 x =
8 ac = bx + cbz
9 df = ex+fez
10 yb=gx + bgz
11 acef = efbx ✈ efcbz
12 cbdfcbex + chefz
+
11-1213 acef- cbdfefbx cbcw.
acef―cbdf
-
efb.
есь
9X8
10 Xe
15gdf = gex+gfez
16 eyb = egx + ebgz
16-15 17 eybgdf
44
ebgz-gfex
17 ÷
182 =
eyb-gdf
r
ebg-efg
ac
8-b 19 = x + cz
19, 14, 18/20
Put
20, 21
Ъ b
ac
b
acef-cbdf
==
ceyb-cgdf
+
efb ecb ebg-efg
21 p = ƒ—c, r = b—f, r + p = b-co
acef — cbdf ceyb cgdf b.
22 ac =
pe
+
reg
22 Xpreg 23 acpregacefrg-cbdfrg +pceybb
23 tr.
24,
pcdfgb
24 peybb apreg-aefrg + bdfrg+
=
pdfgb.
—
25 pebby = areg × p−ƒ + bdfg × r+p.
21, 2526 pebby — — aregc + bdfgxr+p.
26, 21, 27 pebby = aceg × f − b + bdfg × b
f—b - c
aceg xf-b+bdfg xb-c
27
28 y
beb xf-c
Z 2
PROB.
340
B. II.
NUMERICAL
PROB.
XVII.
To divide ten thousand into two fuch parts, that when
each of them is divided by the other, the fum of the
quotients, may
Let
per qu.
be
5.
c
Ila, e be the parts, b = 100ca, 5
< 2a+e=b
a
e
+
e
2
2 +
a
42
4 e
b
a.
3 X ae
5eebb zba + aå
6 cae = aa + ee
4, 5, 6
7 cab
caa = 240
zba + bb
7 ±
82 + c.aa
2b+bc.a = bbi
-bb
8
9:aa
ba =
2+ c
9 reduc.
bb
bb
b
+
4,
11e=1726,732
4
c + z
=8273,268
2
XVIII.
PROB.
A general would range his army in a ſquare battle, but
finds he has 284 foldiers to spare; but if he in-
creafes the fide of the fquare with one man, he
wants 25 to fill up the fquare. How many fol
diers had be?
fide of the ſquare.
2aa284 = number of men
Let
I
a
per qu.
2 = 3
4 tran
52
3 a + 1²
25 number of men
4aa284 = aa + 2a + I
52a = 308
-
25
a
6α = I
154
7aa284 = 24000, the number of
men.
PROB.
Sect. I.
341
PROBLEM S.
PROB. XIX.
Y
Several perfons dining at an inn, the reckoning came
to175 billings, but two of them flinking away, the
reft bad 10 fhillings a-piece more to pay. Query,
the number of perfons?
Put
number of perfons at firft.
- each man's fhot.
a
175
2
a
per qu.
175
31 a
2 X
+10= 175°
a
3 X
4175a +10aa
350-
20a=175a
4 tranf
5 10a
20a= 350
}
5 ÷
6 aa
2a = 35
6 refolved
7
a = 7
2
8
175
25, the club.
a
PROB. XX.
There is a number, confifting of three digits, whoſe
product is 315, and the fum of the first and left is
double the fecond; and that number with 396
added makes a number confifting of the fame digits
but inverted.
Suppofe 1 a, e, y the digits.
per qu.
4 ±
2a + y = 2e
3 aey = 315
ве у
y
e a
4 1000 + 10e + y + 396—100y+102
+a,
5 99ª + 396 = 99y
5÷(99) 6a+ 4 = ÿ
2
a
72e
6 = 7
a = y
8a+ 4 = 20
B
Z3
8,
342
B. II.
NUMERICAL
8,
19a = e
2
6,
10 yea
3, 9, 1011
11 refolv. 12
9, 10 |13
aey = exe-2 Xe + 2 = e³—4e=315
e = 7
=
a = 5, y=9.
PROB.
To find the value of a, when
XXI.
a³-Vaa=4.962a.
3
2
per queft.
2
a
2
3
= ba
Put I b = 4962
2,
affume
3 a
426
9
6
4
ठ
a = ba
a, then
4
5a x, and a = x²
6x9 + x4 = bxo
4,
3, 4, 5
6x+
7 tranf.
7x5
I = bxx
81x5
bxx I
8 refolv.
9 x
1.74256
4,
10 a = 27.998
Given {
2 -
I — e3
PROB. XXII.
=
1 a³e³ + a²e³ = 234000 b.
2/ ae + ae² + ae³ = 1860 = c.
to find a, e.
3a=
C
e + ee + e³
4a³ + a² =
3, 4
5
C3
le + ee + e³]
3
b
3
e
+
CC
e + ee + e³l"
11
b
5 Xe
Sect. I.
343
PROBLEM S.
cce
C3
5 Xe3
6
+
e + e + eel³
6 X
reduced
i te + cel²
= b
76x1+e+ eel³=ccex I +e+ee + c²
8/ be + 3bes + 6be4 + 76e³ + 6be²
зве
+ 360 +8}
+6}=
c²e
= c³
2
C²e³
3 c²e²
PROB. XXIII.
There is a cask of rum, out of which was taken 41
gallons, and filled up with water, and the fame rc-
peated three times more. At last there was found
"by the proof, to remain 25.2935 gallons of rum in
it. What was the content of the cafk?
Let
1| 6=41, c=25.2935, a cafk's con-
tent, then ab firſt remainder.
2 And fince the quantity of liquor is
as the ſpace it poffeffes; therefore
a: a -b (1 rem.) : a b:
a-b12
(2 rem.) ::
abla
a-bl³
a
a
aa
(3 rem.)-::
a-613
a—b 4
(4 rem.)
aa
03
a
614
= C.
a?
3 G
4 tr.
4a4
per queft. 3
5 refol.
a44ba³ + 6bba² — 4b³ i + 64 = ca³
5a4
3
4ba³ + бbba²
C
6 a 124.84 gallons.
·4b³a==b4
24
PROB.
344
B. II,
NUMERICAL
1
PROB.
XXIV.
Given
1
x³ + x²y + y²x + y² = dxy!
2
I,
2,
Put {
3, 5, 6
x6 + x+y² + y4x² +y6 = bbxxyy.
3 x² + y² × x + y = dxy
4x++y4 × x² + y² = bbxxyy.
5 xy = a
6 x + y = e
eezaxe
7 ee
da = x² + y² Xx + y
×x+y
da
7÷e
8 ee
2α =
= xx+yy
e
ddaa
4, 8
9
2aa = x4+y4
ee
9 X
» | ེས་
ddaa
da
10
20a X
= x++y4 X
e
ee
e
IO X
Il dd
x² + y² = bbxxyy = bbaa.
2ee × da = bbe³
bbe³
11-
12 a
d3
2 dee
2bbe+
db2e3
8, 12 13 e3
ds - 2dee
d³
2dee
13 X
14 d³e³
2 des
2bbe+ = dbbe³
14÷
15 23
bb
2
ee +
2dee2bbe dbb
a
e =
dd- bb
which gives e,
ſtep 12, and x
found from itep
and then a by
and y may be
5 and 6.
2
a quadratic
PROB.
Sect. 1:
345
PROBLEM S.
Given
that is,
I, 2
Put
PROB. XXV.
1/ ase — bda+ + 2ba³ee+bbae³-2dbba²e
b³ de² = do
2 a*ee2bda³e + bbdda² + ba²e³ —
2dbbae² + bbd + b³d²e — bªd”.
3 aa + bel² xae — bd — do.
4 ae- bdl xaa + be = b+d²,
5 aa+bex
bd = y
7 x²y = do
6 ae
3, 5, 6
4, 5, 6
8 y²x = b+dd.
7 X 8
9 x³y³=b4d8.
3 سا و
ساو
10xy = bddbdd
d6
d³
710, 11 x =
3
bddbdd
bb bbd
8÷÷ 10 12 y
b+dd
bb
bdd
3
√bbd
bdd
5- be
6, +
14 a
I3| aa = x =
y+bd
X be
e
13, 14
15 aa =
yy + 2bdy + bbdd
=*
- be
ee
15 red. 16 xee-be³ = yy + 2bdy + bbdd, a cu-
bic equation which gives e, whence
a is known by ſtep 14.
PRO B. XXVI.
To find four numbers x, y, z, v, having the produ&
of every three given.
Suppofe
xyz = b
2jzv=
3 zvx = d
=f
41 vxy = f
I X 2 X 3 X 4
345
B. II.
NUMERICAL
1X2X3X4 | 5] x³y³z³v³ = bcdf
5 lw 3
3
6 xyzv = bcdf
Vbcdf
6÷2
7x=
C
3/bcdf
6÷3
8y =
d
Vbcdf
6÷÷4
9
22
f
3/bcdf
6÷1
[0 V =
b
XXVII.
PROB.
To find 3 numbers, x, y, z, having the product of
each and the fum of the other two, given.
1 x x y + z = b
per qu.
2 y x x + % = c
3 xxx+y=d
1+2+3
4 ÷ (2)
5 xy+xz+yz=
5
4 2xy + 2xx + 2yz = b+c+d
6 ÿz = s
jz
b+c+d
=s, by fubft.
2
5-2
7x2 = S
XZ
5-3
8
xy=s
d
|
6x7x8 9 xxyyzz = s—bx S-cX S
x
-d
2 سا و
Io
10 xyz= S
bxs.
c X s - d
106
SX IS
S-
[ 1 x =
S
b
-bxs-
d
107 12y =
S
C
108 13 x = ~
bxs-
P
PROB
Sect. I.
347
PROBLEM S.
Fig.
PRO B. XXVIII.
To find any polygonal or figurate number.
A figurate or poligonal number is the fum 226
of a ſeries of numbers in arithmetical pro-
greffion from 1. And theſe are fo called,
becauſe they denote the number of points,
which fill a regular poligon, placed at equal
diſtances, on lines drawn parallel and equidi-
ftant, to the fides of the figure. The fol-
lowing table fhews the arithmetic proportio-
nals, and the poligonal numbers formed from
them. The numbers of the arithmetical fe
ries fhew what number of points are placed
on the ſeveral parallel lines of the poligon;
and the poligonal numbers, fhew the whole
number of the points contained in the figure.
Rank.
Arithm. propor-
tionals.
2
3
4
Poligonal num-
bers
Names.
I I, I, I, I, I, I 1, 2, 3, 4, 5, 6 laterals.
I, 2, 3, 4, 5, 61, 3, 6, 10,15,21 triangul.
I, 3, 5, 7, 9,11 1, 4, 9,16,25,36 quadran.
I, 4, 7,10,13,16 1, 5, 12,22,35,51 pentang.
1, 5, 9,13,17,21 1, 6,15,28,45,66 hexang.
6 I, 6,11,16,21,26 1, 7, 18,34,55,81 heptang.
7,13,19.25,311, 8,21,40,65,961 Octang.
rany rank, x = poligonal num.
ber fought.
7 I.
Let
I T
n = place of x; then r-com-
mon diff. of the arithm. feries.
2+2 —¡×?—I the nth term in
the arithmetic progreffion.
arith. prop
Pr. 6.
ib. Pr. 7. 3
n
X r
2 ÷ 22
n I X r
2
=
x n = nth term'
in the poligonal numbers.
2 + n
I X r
I, 3,
4 20
X 12.
2
348
B. II.
SECT. II.
Of Intereft and Annuities.
PROB. XXIX.
The principal, time, and rate of intereft being given ;
to find the amount, or money due at the end of that
time; at fimple intereſt.
Let
2
by pro.
portion
3
principal, t=time, r = rate of
intereft of 1. for a certain
time, as a year, &c. s
&c. s = fum
of all the arrears.
1:r::p: rp, the intereft of p
for a year.
Irpt: prt, the intereſt for
the time t.
whole arrear at the end
4p+ prt = whole arrear
of the time t.
5p+ prts, the arrear fought.
I, 4
S
Cor. 1. Hence p =
when s, r, t, are
rt + I
1
given.
Cor. 2. t =
when s, p, r, are given.
pr
S
Cor. 3. r =
Р
when s, p, t, are given.
pt
PRO B.
Sect. II. INTEREST and ANNUITIES.
349
་
PROB. XXX.
The annuity, time and rate of intereft being given;
to find the arrear, at the end of that time, at
fimple interest.
a
Put
I a
2
o
=
annuity or yearly rent; t =
time of forbearance; r = inte-
reſt of 11. for a year, &c. s =
whole arrear.
intereft due at one year's end.
3 ra = intereſt at 2 year's end.
intereſt at 3 year's end.
intereft for 4 years.
by pro-
portion
4 2ra
53ra
₺ —
6 t1.ra= intereft for t years.
I
7 ta rents due at the end of t years.
2,3,4,5,6, 80+1+2+3... to IX into
7
ra + ta = $.
arith.prop
txt - I
0+1+2+3... t − 1
I =
Prop. 7.
2
txt-
I
8, 9, ΙΟ
rata = $.
2
t
I.r++ 2
IO, II
ta=s.
2
25
Cor. 1. a =
t
I.r + 2 x t
2
25
Cor. 2. t =
2
2
js
+
ar
2r
2r
25
ta
Cor. 3. r =
I - IX ta
PROB.
350
B. II:
INTEREST and
PROB.
XXXI.
Let
To find the preſent worth of an annuity, to continue à
given time, at a given rate of fimple intereft.
P = pretent worth, a
time, r
annuity, t=
intereft of 1
Prob. 29.
2 p + prt =S
t
Prob. 30.
I. r + 2
3
ta=s
2
t − 1 . r + 2
2 = 3
4 p + prt =
ta
2
r + I
t-1.7+2
2
4 ÷
5 P =
ta =
ta.
2rt + 2
}
rt + I
rt + I
Cor. 1. @ =
a
X
P.
t -
I
t
r + I
2
2
20
20
Cor. 2. tt +
Ixt=
शु
whence t may
ra
be found.
2ta
20
Cor. 3. r =
>
2pt-1.axt
PROB. XXXII.
The principal, time, and rate of intereft being given ;
to find the amount at the end of that time, at
compound interest.
Let
p = principal, t = time, r = inte-
reft of 1. R= I +r the amount
of 17, and its intereft. sfum
of money due at the end of that
time,
per
!
Sect. II.
351
ANNUITIES.
per queft.
by pro.
portion
21 +r or R = money due at 1 year's
end.
31: R::R: RR = money due at
2 year's end.
3
4 IR :: RR : R³ = money due at
3 year's end.
5 R money due at t year's end.
6 I:
I : Rt :
::p : pR = the amount of
p for the time t.
7 PR' = s.
1, 6
S
Cor. 1. p:
=
R
S
log: s-
Cor. 2. R' =
log: P.
or t
>
log: R
S
Cor. 3. R
or log: R =
log: s- log:p
古
​PROB.
XXXIII.
The annuity, time, and rate of intereft, being given;
to find the arrears due at the end of that time, at
compound interest.
Let
Ia
annuity, or yearly rent, t
time of forbearance, r = intereft
of 1. for a year, &c. RI+",
s=fum of all the arrears.
2 a = money due at 1 year's end.
32a+ra =a+Ra≈ arrear at 2 years
end.
4a+aR+aRR
arrear in three years.
by Prob.
5 a + aR + aR² + aR³
32.
years.
6 a + aR + @R² + øƑ³.
aR
arrear for
4
. to
arrear for t years.
geom.
352
B. II.
INTEREST and
geom.
propor.
prop. 26
=
2
3
1 + R + R² + R³... to R-1
Rx R¹-1
R-I
R-
I
I
r
Rt
I
6, 7
8
а
a = money due at the end
of t years.
R-
f
'1, 8,
a=s.
go
rs
Cor. 1. α =
R
- I
rs
Cor. 2. R' ==+ 1, or i =
S
a
Cor. 3. - R— R' =
a
found, and then r.
rs
log: +1
a
log: R
S
a
whence R may be
PROB.
XXXIV.
To find the prefent worth of an annuity, to continue a
given time, at a given rate of compound intereft.
Let
1pprefent worth, a = the annuity,
t = the time, r = intereft of 1 l
R=1+”.
Prob. 32. 2 PR¢ = s.
Rt. I
Prob. 33
3
a = s.
gro
R₁ — I
2 = 3
4 PR, =
4 ÷ R
5P =
R' — I
TR
2
G
11
I
I
R
121
g
I
a.
Cor.
ļ
Sect. II.
ANNUITIES.
353
Cor. 1. a=
pr
ร
I
I
R
a
or t=
a
pr
?
log:a-log:a-pr
log:R
a
a
Cor. 2. R' =
Cor. 3. R— R² — - Reti
P.
and r will be found.
PROB. XXXV.
whence R
To find the value of an annuity to continue for ever;
at a given rate of compound intercit.
p = preſent worth, a annuity,
r = intereſt of 17. R =ıtr.
P =
2 P
RtI
TR
a
Let
Pr. 34.
ſtep. 5.
Prob. 73.
cor. 7.
2, 3,
4 p =
(TRE
=) /
3 but fince t is infinite, R is infi-
nitely greater than 1, whence
Rt
1 = R.
R
a
a
Cor. 1. apr.
Cór. 2. r =
a
1
P
PROB.
XXXVI.
At what rate of intereft will 1001. amount to 2001.
in 9 years, at compound intereft.
rate of R = 1 + r,t=93
I
Let
r
39
Prob, 32, 2100R
4
=200. per queſt.
A' a
$
L
2 →
354
B. II.
INTEREST and
2-
3 & 4
سا
4 lw 39
5
6X 100
39
3 RF
2
4 R39 = 16
39
=
5 R16 1.0737 by logarithms
6 R R=1==.0737:
77.37 rate of intereft per cent.
PROB. XXXVII.
If a principal x be put out at compound intereft, for
x years, at x per cent. to find the time x, in
which it will gain x.
Prob. 32. 1 PR₁ = 5
X
per queft. 2px, r=
x
R=I+
t = x;
1.GO
100
2X.
I, 2
3*XI+
X
100
= 28.
Xx
3 X
41 +
= 2.
IOO
nature of
X
logs.
5XXI +
100
x M.3010300
Prob. 84.
X
XX.
X3
cor. I. 6xX:
+
100
20000
3000000
&c. =
.30103
:
M
3
xx
X4
6,
7
+
&c
100
20000 3000000
=,693147
by reverf. 8 x 8.49824 years.
PROB
Sect. II.
355
ANNUITIES.
PROB. XXXVIII.
Given the rate per cent. for a year (51.), to find
what the amount of any fum (1001.), will be at
the year's end, at compound intereſt; ſuppoſing it to
arife from the principal and intereſt due every day,&c.
rintereft of 1 l. for a year.
2n=365 the parts of a year.
Let
intereft for 1 day.
r
3
n
3,
4I+
n
172
money due at the year's
money due at one day's end.
Prob. 32. 51 +
12
end.
6nx log: 1+= log:
log: amount
72
for a year .0215694.
amount for
amount for a year.
amount of 100%.
by logs
6,
71.0509
6X100
8 105.09
|n
r
or 5,
I +
rr +
ท
2nn
n.n- I. n
2
r³ &c, the amount
=1+r+
2.3ni
for a year.
If the intereft is fuppofed to gain
intereſt every moment, by be-
coming part of the principal;
then n is infinite, and
r
12
1 + 2 |=1+r+ ²² +
n
за з
74
+
2 2.3
2.3.4
&c. the amount at the year's
end. But this feries is the num-
ber belonging to the hyperbolic
logarithm r, whence
A a 2
The
1
856
B. II
INTEREST and
10,
The number belonging to the lo-
logarithm .43429448r amount
of 1. for a year = 1.0513;
and for 100 = 105.13, to gain
intereft continually.
Schol. If the intereſt for a day be required, fo
that it may amount to 1 + at the year's end, at
compound intereft; then the amount at day's
end, will be Vi +r; which is ſomething less than
1 +
12
12
PROB. XXXIX.
1
A man puts out a sum of money at 6 per cent. to
Continue 40 years; and then both principal and in-
tereſt is to ſink. What is that per cent. to conti-
nue for ever?
The queſtion amounts to this; if 100 7. be
paid for an annuity of 61. a year for 40 years,
what is that per cent ?
Put
a = 6, p = 100, t = 40, r=rate of
1. R=1+r.
Prob. 34
cor. 2. 21t=
log:a
log:a-pr
log:R
2 X
3 Log: R =
log: a
log : a-pr
t
1:6 1:6
40
10Or
Suppofe 4 R = 1.05; then
r = .05,
and
L:R.019454; whence R =
1.046 which is too little.
5 R = 1.053, then r .053, and
L:R.023324, and R = 1.055,
too big.
Suppofe 5 R
=
Then
Sect. II.
357
ANNUITIES.
Then by Rule 5, Prob. xcii. B. I. you will find
R1.052, and the rate = 5.2 per cent. which may
be repeated for more exactnefs.
PROB. XL.
If 2001. be due 3 years hence; and 801. 5 years
bence; in what time must both be paid together, at
5 per cent.
It the time.
Let
Prob. 32.
200
2
cor. I.
1.05|'
=172.76, the prefent worth
of 2007.
80
ib.
3
1.05
5
62.68, the prefent worth
2+3
Prob. 32.
5
of Sol.
4 235.44, the whole prefent worth.
log: 280- log: 235 44
=3.5527
cor. 2.
log: 1.05
years.
XLI.
PROB.
What must I pay for an annuity of 791. to begin 6
years bence, and then to continue for 21 years, at
5 per cent ?
Let
1/a = 70, t =
I
I
R
Prob. 34. 2
if
21, R 1.05, x = 6..
prefent worth of the
annuity 7 years hences.
I
I
Prob. 32 3
R
a
cor. i.
R₂
*R*
Ri-I
*R+ a prefent
worth of s, 7 years hence, =
669 704 7. the preſent worth of the
annuity in reverfion,
A a 3
SECT
>
358
· B. II.
"
SECT. III.
Arithmetical and geometrical Progreſſion.
PROB. XLII.
A traveller feis out and goes 9 miles a day; 3 days
after, another follows him, who travels the first day
2 miles, the fecond 3, the third 4, and ſo on.
what time will be overtake the first?
Ix
X
per queſt. 2 2 x
days the laſt travelled.
1+2 = his laft day's travel.
arith. pro.
greffion.
X
I + 4
3
2
In
xx his whole journey.
5 4 ≈ + 3 = days the first travelled.
per qu. {
5
3=5 6
x + 3×9= firft man's journey.
xx + 3x
2
=9x+27
6 reduced 7 xx15x=54.
7 extr.
8 x = 18.
PROB. XLIII.
There are three numbers in arithmetic progreffion, the
Square of the first together with the product of the
other two is 16; and the Square of the mean to-
gether with the product of the extreams is 17.
What are the numbers?
Put
1 a—e, a, ate for the numbers,
b = ;6, c = 17.
22aa
ae tee = b
per qu.
32aa
لمقدونية
ee
C
2+3
Sect. III.
359
ARITHMETICAL PROGRESSION.
3 tran.
6xaa
52
16a4
2 + 3
4 tran.
4 4aaae = b+cs by fubſt.
4aa — s
5 ae = 4aa
6 ee = zaa
7 aaee2a4
aaеe
C
caa
8sa² + ss
7 = 8
9 1684.
8sa² + ss = 2a4
cac
9
reduced 10
142¹
85
c.aa + ss = 0
10 ext.
11
aa = 9, a = 3.
4aa S
5 ÷ a
12 e=
I.
a
1,
13 and the numbers are 2, 3, 4°
PROB. XLIV.
There are four numbers in arithmetical progreffion,
whofe common difference is 2, and product 3465.
Let
1 262, or b=1, p = 3465; a — 3b,
a —b, a + b, a + 3b, the num-
bers fought.
per queft. 2 aa- ·9bb × aa-
2 X
3 extr.
I,
3 a
bb = P
a+ — 10bba² +9b4 = p.
4aa64, a 8.
a=8.
the numbers are 5, 7, 9, 11
PROB. XLV.
:
To find five numbers in arithmetic progreffion, whose
fum, and product are given.
Put
la
2e, a e, a, a te, a + 2e for
the numbers, bfum = 25,
P = product = 2520.
25a-
→3e +3e5a = b.
= m by ſubſt.
b
per qu.
a
3α =
5
a X aa
4ee xaa
ee = p.
A a 4
3,4
360
B. II.
GEOMETRICAL
3, 4,
5 m x mm
4ee X mm
ee = p.
5 X
6 m x m²
5mmee+4e4 = p
6÷m
74e+
5mmee + m4 =
P
ทะ
7 extr.
8 ee = I, e = 1.
Į
91 and the numbers are 3, 4, 5, 6, 7.
PROB, XLVI.
To find three numbers in geometrical progreffion, where
the fum is 20, and the fum of their Jquares 140.
Let
per qu.
3-y
5 & 2
6, 2
7-yy
3, 8
9 ÷
2 X 4
6-11
12 h 2
1x, y, z be the numbers, b = 20,
c = 140.
2 xz = yy
31 x + y + z = b
4 xx+yy + zz = c
5x+x=by
1
6 xx + 2xz + zz = bb — 2by+yy
7xx + zz + zyy — bb — 2by + yy.
8 xx + zz + yy bb
bb — 2by = c.
9bb
10 y =
1 4x2 = 4yy
=
2by
bb
C
= 6.
6/1/2
zb
2xz + zz = bb
-
2by-3yy
$3
z - Z
2 by
ЗУУ
2 by
་
3yy
X
=
2
12 XX
♡
5+13 14
5 - 13
८२
♡ + √ bo
13 ½ + √ 13 1.
2
by —√bb
2 by ·
3yy
2
I
I
Ca
PROB.
Sect. III.
361
PROGRESSION.
PROB. XLVII.
To find four numbers in geometrical progreffion, whose
fum is 15, and the fum of their Squares 85.
Let
per qu.
22
1v, x, y, z be the numbers, b = 15,
c = 85.
2 v + x + y + z = b.
3 v² + x² + y² + z² = c.
4v+x+y+2]² = x+y)²+v+≈1²+
2 x x + y x v + z = xx + 2xy +
yy + vv + 2VZ + ZZ + 2 x x + y.
xv+x= bb.
5c+2xy+2vZ +2 × x+J ×v+z=bb
3, 4
by propor. 6
Put
2, 7
VZ
vz = xy.
7a=x+y, e=xy=vz by proportion,
8 y + z = b — A.
5, 7, 89c + 4e + 2a x b — a = bb.
-
But
XX
y
z =
yy
=2, by the nature of
proportion.
X
yy
+ x + y + 2/2 = b
:
XX
2, 10
I
y.
X
x3 + y³
+№3
7, II
= b
xy
7 & 3
[ 3
13
13 tran. 14 x³ + y²
3
x + yl³ = a³ =x³ +3x²y + 3xy² + y³
a
x³ + y³
23
14,751
3xy Xx + y
Q3
за
xy
за
34.
xy
e
Q3
12, 15
16 î +
e
16 × e
17 tran.
1723
2ae
3a=b.
be
18 be + 21e = g³
18 -
362
B. II.
GEOMETRICAL
a3
18÷
19 e =
b+2a
4a3
9, 19 20/c+
+ zab
zaa = bb
b+2a
bb
20 reduc. 21 baa + ca
21 extr. 22 a =˝6,
23e8.
24y =
e
b
2
19
7
,
x+
a = x +
*
0183
24 X
25 ax =
xx + e
25 reduc 26 xx
26 extr. 27 x = 2.
10, 24 28y = 4, v = 1, ≈ 8.
ax +e=0.
=
PROB.
XLVIII.
To find four numbers in geometrical progreffion, fuck
that the difference of the means is 100, and the
difference of the extreams 620.
Let
per queſt
But
a, e, u, y be the numbers, b = 100,
c = 620.
2y=a
a—c, u = e
b.
3 au ee, ayeu, by the nature of
progreffion.
abee, aa acee eb.
e - b
Elaa-ac=
e.
ex
cee
3, 2
4 ae
ee
4 -
÷
5a=
4, 5
ee- eb.
е
b
3
6 x
ce x e
b
e
bi
7 tran.
835
c.ee + cb
·3bbie — b³.
==
b3
8 red.
9 ee
be =
C
36°
ез
9 extr.
2, 5
ice = 125
11a = 625, y = 5, v = 2 .
а
1
PROB.
Sect. III.
363
PROGRESSION.
PROB.
XLIX,
The Sum of four quantities in geometrical progreſſion
being given, and the fum of the Squares of the
means, to find the quantities.
Let
1 a³, a²e, ae², e3 be the quantities,
b = ſum of all, c = ſum of the
fquares of the means.
? ?² + a²e + ae² + e³ = b.
aªe² + a²e+ = c.
+ a² + a²e + ae² + e³ = a² + e²xa+e=
aˆe" + a²e+ X are + ae²
a³é³
= b.
Put
5y = a²e + ae²,
6,
52
( yy = ate² + 2a³e³ + a²e4.
~ =
аз ез
су
yyc
2
b
3, 4, 5
!vy — //c
8 red.
9 byy
7,
IC
ae =
2cy = bc
3/yy
3/3 = c = d
2
5,
y
y
late=
=
ae
d
10- a 12 e =
a
d
y
II, 12 13
a +
=
a
14 aa
d
ya+d=o, whence a, e, and
all the numbers are known.
PROB.
354
B. II,
GEOMETRICAL
PROB. L.
There is given the sum of the fquares of the extreams
(b), of four quantities in geometrical progreffion;
and the fum of the means, or their Squares (c); to
find the quantities.
Let
per qu. {
2+3
5 × a4e+
Put
4-7
3, 6, 7, 8
3, 7
10 to 21
lw
t
a³, a²e, aе², e³ be the quantities;
2 a6 + eε = b.
6
3 a4e² + a²e+ = c.
4
51
a6 + a4e² + a²e4 + e² = b + c = d.
a² + e² × a² + e² = aa + eel²
6a" + a¹e² × a²e+ + e° = a+e² + a²e4l²
; y = a + a²ee
8 a²e4 + eε = d y
yxd—y = cc,
·α a⁰ + zatеe + aae4 = y + c
a³ + aee = √y + c = p.
2a4ee + 2a²c+ + e = c + d—y
3, 4, 7,
12 lw 2
13
11 + 13 14
are + e³ = √ c + dy = q
a3 + a²e + ac² +ep+q. Whence
the numbers will be found as in
the laſt problem.
Or thus,
XX
Let
vy
, x, y,
y be the quantities.
y
x+
y4
+
:b
NY
XX
2
3 x + y = Eg
per qu.
3.
X
4y = c
2 x yуxx
5
+
X
x² + y = bxxyy
4, 5
6
2
=
bxx xc → *1
1
6
Sect. III.
365
PROGRESSION.
6
có
3
7/x6 + cε — 6c5x + 15c4xx — 20c³µ³ +
15c²x46cxs+x6bccx²- 2b cx 3
6cxs + 15ccx4.
+ bx4
reduced 82x6
20c3x3
·b
+2bc
+ 15c+xx
6c5x
+ co=0.
· bcc
PROB. LI.
Given the sum of the extreams (b) of five quantities
in geometrical progreffion, and the fum of the three
means (c), to find the quantities.
Let
per qu.
.༨
Put
2, 4, 5
6 reduced
2±5
8w2
a², a³e, a¹è², ae³, et be the quan-
tities.
204 +64 = b.
3
<a³e + a²e² + ae³ = c.
4 a²e² × a++2a²e²+e+ = a²e² ×aa+ee\²
= a³е + ae³|².
5y = aaee.
6 y × b + 23 =
2xy +by+2cy
yy
cc.
8 at ± zaace + eª = b ± 2y
9 aa + ee =√b + 21
+23
10 aa—ee = √b-2y Whence a, e,
and all the quantities are eafily
found.
PROB.
366
B. II.
GEOMETRICAL
PROB.´´ LII.
Of five quantities in geometrical progreffion, there is
given the fum of the extreams (b), and the fum of
the fquares of the three means (c); to find the
proportionals.
Let
per qu.
If a4, a³e, a²e², ae³, e4, be the quan-
tities.
2a4 + et = b.
3 a6ε² + a¹e+ + a²e° = c.
4 a² e² X at + e4 = a°ee & a²eĢ
5 yaaee
56yxb
yx b = c =
c — a4e4 — c — yy.
7 yy + by = c.
Put
2, 3, 4,
6 + yy
2, 5
8 lw 2
10
8 a4 ± 2a²ee + e4 = 6 ± 23
9 ca + ee = √b + 23.
√b+2y.
b
aa—ee = √6—2y.
whence all the reft are found.
PROB. LIII.
There are four quantities in geometrical proportion
difcreet, whofe fum is b, fum of the Squares cy
and Jum of their cubes d; to find the numbers.
Let
per qu.
Put {
Ix, ex, y, ey, be the numbers.
2x + ex + y + ey = b
3x² + e²x² + y² + e²y² = c
4x³ +´e³µ³ + y³ + e³y³ = d
3
x + y = v, x² + y²=z.
61 +e=s, I + eet, 1 + e³ = u
52&c. 7 xy=
VV Z
2Z
ยบ
x² + y³ =
P
2
2
2, 5, 6, 8 sv = b
3, 5, 6
Sect. III.
367
PROGRESSION.
3, 5, 6 | 91 tz = c
4, 6, 710
32
VV
vu = d.
2
h
8, 9 IV=
Z
414
3bcu
b3u
10, 11
12
= d
2st
253
reduced
6, 13,
1
13 3b cs¹³u — b³tu = 2s³td.
14 and reſtoring the values of s, t, u;
then 3bc XI + e³ x 1 + el
b³ x I + e³ x 1 + ee =
63
x I + ee, a 5th power.
2dx
2
2d XI + el³
And e being known all the reft
are eaſily found.
#
###
#
SECT.
368
B. II.
:
SECT IV.
Unlimited Problems.
PROB. LIV.
ミ
​How many old guineas at 21 s. 6d and piftoles at
L
21s,
17s. will pay 1001; and how many ways can
it be done ?
Let
a=guineas, epiftoles; 21 s. 6 d.=
43 fix-pences, 17 s. 34 fix-pences,
and 100 l. 4000 fix pences.
=
per queft. 2 43a + 34e4000
2 3
334e4000
4 e 11613.
5.=34
43a.
3 ÷
34
4000
34e
2-
5 a
= wb.
43
5 abridg. 6
I
34e
= wh.
43
43e
+67
Je + I
= wh.
43
43
7 × (5)
8
45e + 5
wb.
43
43e
8
9
2e + 5 = wb.
43
43
8e+20=wb.
43
9X (4) 10
7
10
II
e — 19 - wh. = P•
43
II X (43) 12 e = 43+ 1919, 62, 105, the pi-
ftoles.
5
13 a =
78, 44, 10, the gui
neas; being three anfwers in
whole numbers.
PROB.
Sect. IV. UNLIMITED PROBLEMS.
369
PROB. LV.
What number is that, which being divided by 2, 3,
4, 5, 6, 7, 8, 9, 10, 11, 12, there will remain
1; but divided by 13, then o will remain.
Put {
It is plain 5 X 7 × 8 × 9 × II, or
27720 is divifible by any of
theſe numbers.
2a föme whole number.
3 13e
per queft. 4 27720a += 13e
the number fought.
277200 + I
4 →
51e=
=wh.
13
4a + i
5
abrid. 6
wb.
6X(3) 7
13
12a + 3
= wh.
13
a
3
(1)-7
8
wh. p..
8 X
5,
13
( a = 13p + 3 = 3, 16, &c."
10 e = 6397.
10X (13)1113e83161 the number fought.
PROB.
LVI.
A man bought 20 birds for 20 pence; geeſe at 4d.
quails at half pennies, and larks at farthings. How
many did he get of each ?
Let
per qu.
=
1 a = geefe, e quails, y larks.
2afety = 20.
340+10+ y
34a + e + y = 20.
2 tr.
4 y 20
a
е
ate
3, 4
5 r ed.
54a + e +5 -
20.
4.
615a+e=60
B b
6,
1
370
B. II.
UNLIMITED
6,
2 tran.
7a4.
8 e 20
a
y
a + y
3, 8
9 red.
940+10+起
​10 70 - y = 20.
1070
I BC 2.
20
2
10
7, 11
12a = 3
2, 12
13 ety = 17.
3, 12
14
že + y = 8.
14 X (2)
15
e + y = 16
13 - 15
16
17 = 1, y = 2.
13,
1710 = 15.
PROB. LVII.
A, B, C, and their wives P, Q, R, went to the
market to buy bogs. Each man and woman bought
as many bogs as they gave fhillings for each bog.
A bought 23 hogs more than Q; and B bought
II more than P. Also each man laid out 3 gui-
neas more than his wife. Which 2 persons were
man and wife.
Let
per qu.
IX
Ix = hogs fome man bought,
xywife's hogs.
2 the money for the man'sxx,
3xx-2xyyy wife's money.
XX
XX
4 xxxx--- 2xy +yy + 63.
5 2xy = 63+yy
4 tr.
63+yy
5 ÷
6x=
= wb.
2y
63-yy
6-y
7 x -y
zy
But
8 In this cafe y must be an odd
number, 1, 3, 5, 7, &c. but
it cannot be 5.
= wh.
8,
Sect. IV.
371
PROBLEMS.
8,
If
9
y = 1, x - y = 31, x = 32
y = 3, x - y = 9, x12,
y = 7, x - y = 1, x = 8
per queft. 10 A has 32 hogs, and Q 9,
Alfo B has 12,
11 Whence B and Q
and PI,
C
C and P
are man and
wife.
A and R
PROB.
LVIII.
+
To find e, y in whole numbers, fo that yyee 22e
= 184.
1) To deprefs the equation, put y=x+e.
2 yy-ee+22e
xx+2xe+22e=184.
184 XX 92-xx
2 →
3e=
22 + 2x
therefore x
2 →
4,
II + x
= web.
muſt be an even
number. Therefore
4 rather make 2e =
184 XX
b.
II + x
5 reduce the equation as low as it
can, then 20
63
II + x
63
I 1 + x
711 + x = P
6
5,
Put
63
6, 7
8
=
Р
7, 8
8
= wb.
}
wb.
x + II +
wb. Therefore take pany
divifor of 63, that is 1, 3,
7, 9, 21, 63.
=
9 then x 10,-8,-4,-2,10, 52
IO
63
P
=63,
63, 21, 9, 7, 3,
5, 9 II 2e = 84, 40, 24, 20,
I
1
4,
4, -40.
II (2) 12 e = 42, 20, 12, 10,
1, 131y = 32, 12, 8, 8, 12,
B b 2
2,
20.
32.
And
372
B. II.
UNLIMITED
And any pair of theſe will folve the problem, which
are all the poffible anſwers in whole numbers.
PROB. LIX.
A vintner bas wine at 24 d. 22 d. and 18 d. per
gallon of which he would mix 30 gallons, to be
fold at 20 d. How much must be take of each?
Let
;
per qu.{
2X(24)
2 X (22)
4-3.
6 ÷ (2)
7,
5-3
1a, e, y be the quantities of each.
2a tety =30,
3/ 24a + 22e + 18y = 600.
424a + 24e + 24 = 720.
5 22a +228 + 22y=660.
62e+6y=120.
7e + 3y = 60.e 60 - 33.
8
20.
92a +4y=60. a 2y 30%
10 4y = 60 + 2a.
—
9+24
10,
11y. 15.
8, 10,
12 y
16, 17, 18, 19.
9,
13 a =
2, 4, 6,
6, 8.
クラ
​14 e
=
12, 9, 6, 3.
PROB. LX.
To find the value of e, y, u, x, z in whole numbers,
in the two given equations following.
Given {
1 X (4)
IX (9)
+
iety J u + x + z = 60.
23e +43 + 5× + 7x + 9z = 440.
3 4e + 4y + 42 + 4x + 42 = 240
4 ge+gy + gu + 9x + 92 = 540
+ u + 3x + 52 = 200.eLa.
6 6e + 5 + 4μ + 2x = 100.e148,
5
2 3
4 2
Suppofe 7
e
e = 10.
8 y + u + x + z = 60 e 50
=
2
·34
Sect. IV. PROBLEM S.
373
1
23e94y+5u+7x+92=440-3e=410.
8 X (5) 10 5y+5u+5x+52=250.
8 × (9) 11 9y+9u+9x+9%=450.
+2x+4x=160. y Co.
9-
10 12 y
11-91353+4+2%
Suppoſe 14 y = 4.
= 40.y 6%.
8-y 15 u + x + x = 50y=.46.
943 16
15 X (5)
15 X 718
17
54 +7x+92 = 410
54 + 5% + 5% = 239.
74 +7x+72=322
4y
43 = 394.
16
17 19
16 - 18 20
21
=
Suppofe 21
2x + 4 = 164.40.
+22= 72.2 or 37.
2 = 40.
15222 u + x = 46—2—6.
1692 23 5u+7x= 394 -92 = 34
22 X (5) 24
24/25
23-
22
5μ+5x=30
2x = 4.x=2
x [26] u = 6- x = 4.
-x
=
And one anſwer is got, viz. e 10, y = 4, u = 4,
x=2, 240. for 10 + 4 + 4 + 2 + 4060.
and
3×10+4×4+5×4+7×2+9× 40 = 440.
PRO B. LXI.
To find a perfect number, or one which is equal to the
Jum of all its aliquot parts.
Suppofey" x a perfect number.
=
2 then y 4-y²... to y², +x+
I
xyxy... to xy-
all the aliquot parts.
fum of
per queft. 3y" x=1+y...y" + xx I + y... J¹²-s
Bb 3
1
1
I
Cor.
2
+
?
1
I
374
B. II.
UNLIMITED, &c.
Cor. 3.
Pr. 26.
4 I + y... y²
y22 =
y² 4. 1
I
y
I
propor.
5+ y
yn
I
J22-1
-I
y
I
I
n
3, 4, 5
6
y
I
yn x
+
X
y-
y - I
y-
6 x tr.
7y
1.y x
yn
•
I x = yn+1
I
I
7 ÷
8
yn + I
I
x =
y
I. yn
yr + I
9 But that x may be a whole num-
1.y" =yn
ber,
y
I. yn
y" + 1 = 1,
9 tr.
10 y
10-
II y
II tr.
12y = 2.
I,
12
8, 13
I = I
132"x a perfect number.
I
142" X 2"+1= perfect number,
where 2"+i muſt be a prime,
as appears by ſtep 2.
15 If n is an odd number greater than
1, then 2"+1
I will be a come
pofite number.
SECT.
SECT. V.
Rational Squares, Cubes, &c.
375
PROB. LXII.
To find two fquare numbers, whofe difference is given.
Let xx and yy be the numbers,
difference,
a
z + v
Z
V
Put
2
x,
= }
2
2
ZZ + 22v + v v
22
تي
3
XX
4
ZZ
ZZV + VV
2 ✪ 2
4
=yy
4
3 4. 5 zv
I, 5
If a
27.
If a = 20.
61 ZV
7
ZVa
XX yv
Take v at pleaſure, then z = a
whence x and y are known.
And
8 If a is a whole number, and x and
y are defired in whole numbers ;
take any two factors that produce
a, fo they be both even or both
odd numbers, if poffible.
therefore a must be either an
odd number greater than 1, or a
number diviſible by 4, to have
x and y in whole numbers.
Take v=1, z=27, or v=3, 2=9.
V=2, 2=10,
B⋅ b 4
PROB.
}
376
B. II.
RATIONAL
PROB.
LXIII.
To divide a given fquare into two other Squares.
Let
the fquares required, bb =
aa, ee
the given fquare,
2a = SV
affume
3e = rv
b
2
2
4 aa = SSVV
3 & 2
5 eerrUV
2rbv + bb
4 + 5
per queft.
6
aa + ee = rr + ss vv
2rbv + bb
7 rr + ss.vv —
2rbv + bb
bb
7 tr.
8
rr + ss.vv = arbu
2rb
8
9v=
rr + ss
2rbs
2, 9
10 a =
rr + ss
gr
SS
32 9
Ile=
b
rr + ss
LXIV.
PROB.
To find a Square number (aa), which multiplied by a
given number (n), and a given ſquare (bb) added
to it; the fum may be a fquare.
Let 1 naa bb = yy.
affume
→
2 va usb = y
3 v²aa 2bva + bb = yy
4/ v²aa
5vvaa
2
2
1=3
4 tr.
6vv
7a=
5÷a
VV
2bva + bb
naa = 2bva
n. a = 2bv
26v
n
at pleaſure.
naa + bb
where may be taken
PROB.
Sect. V. SQUARE S, &c.
377
PROB. LXV.
To find two Square numbers (aa, ee), that their pro-
duct added to a given number (d), may be a fquare.
Let 1 aaee + dyy
affume
2
2 de v = y
zaev + vv = yy
2
3 aaee
I
3
4 aaee + d = aaee
2aev + vv
4 tr.
52aev VV
= -
d
VV d
6c
where a and v may
2av
be taken at pleaſure.
PROB. LXVI.
To find three fuch numbers x, y, z; ſo that yy = xZ;
and x+y, and z+y, may be two fquares.
Affume {
I-y
2 y
3 X 4
5 X
6 tr.
1 x + y = aa
2x + y = ee
2x+y=
3x=aa
42 = ee
ee - y
y
xz = aa—y xee-yyy per queſt.
6yya²e²
aayeeyyy.
+ee.y
7 a²e² = aa + ee.y
8
y =
aaee
aa + ee
where a, e, may be
taken at pleaſure.
3, 8
aaee
a4
X
aa
=
aa + ee
aa + ee
4, 8
aaee
e4
10 Zee
z = ee
=
aa + ee
aa + ee
PROB,
378
B. II.
RATIONAL
PRO B.
LXVII.
To find a number, from which two given numbers
(a, b) being feverally fubtracted; the remainders
Jhall be two fquares.
Let ·
per qu.
1x the number fought.
21 X
a
yy
3
X
b = zz
22
1
2+a
3, 4
affume
61 v
v - Y
y=z
6 & 2
5 = 7
7100
8 a
บบ
b + yy
8 tr.
9 2vy
4 x = a+yÿ
a + yy — b = ZZ
2vy+yy = ZZ
vv + b - a
VV
vv + b
vv + b
a
9-2010 Y =
271
vv + b
4, 10
[ 1 x = a +
40v
2vy+yy
PROB.
LXVIII.
To find three numbers (x, y, z), whoſe ſum ſhall be a
Square, and alſo the ſum of any two to be a square.
Suppofe
4 X (2)
1+2+3
5 = 6
Put
7, 8
9 tr.
зада
1 x + y = page
2x+x=ss
V
tt
3y+z = it
4x + y + z = v v
52vv
52VV
2x + 2y + 22
2Z
62x + 2y + 22 = rr + ss + tt
7/200rr + ss + tt
v,
ss = p-v₁ t = q = v
9 2vv = rr + pp 2pv + vv
+99290 + vv
|10| 2pv + 2qv = rr + pp + 99
2
I I
Sect. V.
379
SQUARE S, &c.
t
II V
ལ=
qq
rr + PP + 99, where r, p, q are
2p+29
taken at pleaſure; whence s and t
are known (ſtep 8).
tt
4-3
12 X VV
บบ
4-2
13y = vv
SS
4-I
[14] 2 = VV
дода
1
PROB.
LXIX.
To find three Squares in arithmetic proportion.
Suppofe
1+3
affume {
I XX y = vv
2 XX
3xx + y = 22
4 2xx vv + zz
50 = S
6 x = t
:
X
25x + xx
7 2xx = 5S -
+ tt 2tx + xx
25 + 21.x = ss + tt
4, 5, 6
7 tr.
8
ss + tt
I
8÷
X
{
25 +2t
5, 9
ΙΟ
10 v =
X
11
6, 9
ss + 2st
tt
25 +20
2st+tt SS
1 x = t
Z
x =
25 + 26
PROB.
LXX.
To find two numbers (x, y) fo that xy +x, and
xy+y, may be fquares.
per qu.
I-X
xy + x = vV
2xy + ya fquare.
VV
3y+1=
X
1
3-1
380
B. II.
RATIONAL
3
4y=
VV X
*
VV
X
2, 4
5
X
5,
6 red.
Let
7, 8
9 red.
XX
× ≈ + I = D
6 to effect this, let
VV
· then
טיט
४
+1 be the fide,
X
=
+ I
7 vv = xx+2x=fquare.
8 r x = fide,
9rr
− 2rx + xx xx + 2x
10 x=
rr
2r + 2
4, 6, 1011y=1+
rr
2 + 2
ken at pleaſure.
where r may be ta
PROB. LXXI.
To find two numbers, whofe fum and difference, ſhall
Let
per qu. {
2
3. 4
Put
be two Iquares.
1 a, e, be the numbers.
2a+e = yy
=
3a-e a fquare.
yy · e
5yy2e
a fquare.
r-y= root of it.
4 a
6 r
5, 6
7 tr.
8 → 27
rr + ze
9y=
· —
7 rr2ry+yyyy 20
8 2ry = rr + 2e
where r, e may be ta
4,
27
"
ken at pleaſure; then
10 a = yy · C.
PROB.
Sect. V.
381
SQUARES, &c.
PRO B. LXXII.
To find three numbers (a, e, y), that the fum of their
Squares may be a fquare.
per queft. 1 aa + ee + j'y = vv
2v=d+y
3 v² = dd + 2dy + yy
4 aa + ee + yy = dd + 2dy + yy
affume
2
2
I = 3
aa + ee
4 =
5y=
2d
dd
where a, e, and d
are taken at pleaſure.
PRO B. LXXIII.
To divide a number into two parts, fo that the fum of
the fquares may be a ſquare.
Let
s
per qu.
2 & 2
4
a
the number; a, e, the parts.
2s=a+e
3 aa + ee = vv.
4 aa + zae + ee = ss.
5aa +ee = ss
61e = s a
2ae
91 ss
2
3, 6, 57 aa + ee = ss
affume
7, 8
2as + 2aa = vv
8 rams = Ve
2as + 2aa = rraa
arsa + ss
9 reduc. 10 a =
X 25
rr
2
r
2
6, 10 [e=
Xrs.
rr
2
PROB.
भ
1
382
B. II.
RATIONAL
PROB.
LXXIV.
To find two numbers in the ratio of b to c, ſo that
either of them added to the Square of the other,
Shall make two Squares.
Let
per qu. {
Put
1 ba, ca, be the numbers.
2 bbaa + ca
a fquare.
3 ccaa + ba = a ſquare.
4bbaa + ca — ba —v² — bbaa
+ vv
52bva + ca = vv.
VV
zbva
4 tr.
5 ÷
6a=
2bv + c
3, 6
C C V V
VV
7 ccaa + ba=
h
+bx
200 to
7,
8 ccaa + ba =
x
8÷0
Put
9, 10
2
260+c
ccvv + 2bbv + bc ×
VV = a fquare.
260 of cl
9 ccvv + 2bbv be a fquare
+ bc =
10 2 = CV
r.
I 1 ccvv + 2bbv + bc = ccvv.
= ZZ.
2 cur+rr
rr
bc
II reduc.
12V =
2bb + 2cr
PROB.
LXXV.
To find a number, to which adding a given cube num-
ber, the fum fhall be a cube; and ſubtracting ano-
ther cube number, the remainder shall be a cube.
Let
per qu.
Il be the number; b³, c³ the two
cubes.
2 x + b² = a cube
31x c³ = a cube
affum
Sect. V.
383
SQUARE S, &છે.
.
affume
Ђз
4,
5x = 3c² α +
c4 x + b³ = 6 +
сб
bb
3674+ aa + 18 a².
bo
3
36 aa + a².
bo
3
а
CC
a
= b³ + 3c² a +
304
b3
affume
6 x
3
C3
c3 = a
cl³ = a³
3a²c +
3
засс
6,
7x
a3
5 = 7
8
a³
3
C³
3a²c + 3acc
зас + засс = засс +
3a²c+3acc
3C4
aa +
b3
сб
a3
03.
8 →
3c4
co
9a
3c=
63 +
a.
bo
9 reduced ca=
b3+c³
3cb3
bu.
•Co
х × 3cb3 =
b³
PROB. LXXVI.
To divide the fum of two given cubes into two other
Let
cubes.
1 x³, y³ be the cubes fought; b³, c³
the given cubes.
x = b + v
3y = c
2
affume
bb
v
cc
2 & 3
4 ׳ = b³
+ 3bbv + 3bv² + vs
3
33
5 y³ = c³ — 3 bbv +
354
56
100
V3
со
4+ 5 6 x³ + y³ = b³+c³+3bπ² +
354
v² + v³
63
b5
сб
v³ = b³+c³, per queft.
6 tr.
384
B. II.
RATION-AL, &t.
6 tr.
3bv² +
384
63
ラメニ
​v2
bo
8] 3bc6 + 3b4c³ — b°v — c°v.
73 ~3.
b³ + c³
8
gv=
х
66
× 3bc3 =
звез
63 + c3
PROB.
LXXVII.
To find three fuch cube numbers, whofe fum may be both
a fquare and a cube number.
Let
a, e, y be their roots; x -the fum
of their cubes.
per queft. 2 a³ + e³ + y³ = x6
2
y3
affume
3e3+ y³ = x6 — a³
X4
aa
5 xx
บ
a
= e
v = y
6] a³ = a cube
4 @ 3
+
5
3
6, 7, 8,
аб
9 X
VV
Suppofe
or fuppofe
со
2012
Q6
8+x6
G
X12
25
IC XI² V
ဒ
4,6
23
3x8
a³
v²+3x4v-a³
=46
per
qu.
3x4v+3x²v2 23
або
3xs
a3.
vv + 3x²v² 23 O.
aбv=x³
x²a³ × 3a³.
× 3a3x² =
IA
v=
х
XIZ
・a'
6
3a3xx
x² + a³·
where
x and a may be taken at pleaſure;
then being known, e and y are
known by step 4 and 5.
x = 1, a = 1/2, then v, whence
2
3
e=3, y=, and the numbers are
I
ठ
8.
1/3 + 1/25 +234 = 1,
ठ
276
x=2, a=1, then u, and e=127,
y = 24; and the numbers
85
2048383
I+
+
274625
15252992 = 64.
274625
SECT
SECT. VI.
Geometrical Problems.
HA
AVING hitherto in all the foregoing fec- Fig.
tions, kept an account of the whole proceſs
by regiſtering the feveral fteps at length in the mar-
gin; fo that the reader may fee at once how each
ſtep is derived from the reft; and by this means
become acquainted with the manner of proceeding,
in any operation. It may be prefumed, that by this
time, he will be able to fee the connection of the
feveral parts of the proceſs in any folution, without
fuch a formal explanation. Therefore, for brevity's
fake, in what follows, I fhall not tie myſelf to this
method, but generally write down the process after
a fhorter way, without notifying all thefe particu-
lars; and content myfelf with mentioning only fuch
deductions as are leſs obvious.
PROB.
LXXVIII.
In the triangle CAD; there are given AC, AD; 23.
and the lines CE, DB, drawn to the given points
E, B; to find the point of interfection F.
Put AB=r, CB=⇒m, AE=p, AD=d, CE=ƒ,
DBg; and the line fought CF = a. Draw EI
I to DB.
By the fimilar triangles CBF, CIE (Geom. II.
BI, and d:p::r:
12,) a:f-a::m:
pr
mf. ma
a
= AI. Then AI + BI = AB, that is,
pro
d
Cc
of
}
385
386
B. II.
GEOMETRICAL
Fig. mf ma
23.
+
a
=r, and multiplying by da, pra +
dmf-dma-dar, and dra+dma-pra=dmf, whence
+
dinf
dr dm-pr
PROB. LXXIX.
24. To divide a triangle ABC in a given ratio, by a line
drawn through a given point P.
Through P draw ED parallel to BA, and put
AB=b, AC=d, BC=ƒ, BE=g, EP÷p, BF=x,
and the ratio as m to n, and m + n = s.
8+x
px
By fimilar triangles, g +x:p::*: = BI;
then (Geom. II. 19.) BIXBF: BAXBC::m: m+n,
that is,
pxx
8 + x
: bfms; then bmf
spxx
& + x
2
and spxx
= bmfx + bmfg: by which equation x is found.
PROB. LXXX.
25. To divide a triangle into two equal parts, by a line of
a given length.
Let BD be perpendicular to AC, KH the given
line, and HL parallel to BD. Put AC-a, BC=b,
HKc, CD=d, CK=x; then (Geom. II. 19)
AC × BC = 2KC × HC, or ab = 2x x CH, and
að
CH=
2X
ab
b:d:
"
and by fimilar triangles (Geom. II. 13.)
abd
2X 2bx
=CL; and KL-KC-CL=>
abd zbxx-abd
But (Geom. II. 21. cor. 5.)
26x
26x
Sa
HK-
Sect. VI.
387
PROBLEMS.
2
HK-KL CH' CL2, or
=
4bbccxx—4bbx4.
4 b b c c x x -4bb x4 — a²b²d²+4ab2dx²
4bbxx
or 4bbx4 -4bbccx²
found.
Fig.
a² b4a²b² d² 25.
4bbx
-4adbb = -ab4, whence will be
PROB. LXXXI.
To find the inacceffible distance AB, by help of the 26.
triangle ACD; CAB being one right line.
Through B draw BEF, and draw EG parallel to
CD. Put AC-a, AD=b, CD=c, AE=d, CF=ƒ,
and ABx. Then, by fimilar triangles, AD (b) :
CD (c) :. : AE (d) : EG = and AD (¿): CA (a)
:: AE (d) : AG =
ad
cd
Then GB=
ad + bx
And by the fimilar tri-
t
b
angles BGE, BCF; CF
(c)
:
GB (ad
ad + bx
E).
cda+cdx
b
(ƒ): CB (a + x) : : EG
adf+bfx
Therefore
bfx-cdx-cda-fda, and x =
PROB. LXXXII.
b
c-f
da
bf-cd
If the line EFB be drawn from the angle E, per- 27:
pendicular to the diagonal AD of a right-angled
parallelogram, and BF, FD are given. To find
the fides of the paralellogram.
Let AF, EF=y, BF=¿, DF c.
angles AFB, AFE, and DFE are fimilar.
Cc 2
The tri-
There-
I
fore
1
}
!
388
B. II.
GEOMETRICAL
1
XX
Fig.
XX
fore b:*
* : *=FE=y, and b:x::y or
: C.
27.
X3
b
Whence = 3
bc, and bbc, and x = Vbbc.
b
x³-
Then AEVx+yy, and ED =√cc + yy.
PROB. LXXXIII.
28. To defcribe a square in the given triangle ATE.
Draw TC perpendicular to AE, and let BFGD
be the fquare. Put AE=b, AC=c, CE=d, TC=p,
BF or BD = x. AB = y. Then
:
The triangles ABF, ACT are fimilar, and
yxcp, whence cxpy. Alfo the triangles
EDG, ECT are fimilar, and ED-b-x-y, whence
b-x-y :x::d: p, and dx=pb-px—py=pb.
d:p,
pb
px
· CX.
Whence x =
pb
d + c + p
6+ P
PROB. LXXXIV.
29. Six equal circles of 2 inches diameter are infcribed in
an equilateral triangle, touching one another and the
fides of the triangle. To find the fide of the triangle.
Draw AF perpendicular to BC, and from the
centers O, S, draw OD, SE perpendicular to AB,
and let DOr, AB = x.
The triangles ABF, ADO, ESB are fimilar, and
(Geom. II. 39. cor.) AF AB. Then BF
rx/23/
(2x):
(1×) : AF (×√√√3 : : DO (r) : AD =
I
2x
= 2rv
EB, and DE 4r, whence AB or x 4r + 4r
3
2 + √3
× 2r = 4 + 2/3.
I
PROB.
Sect. VI.
389
PROBLEMS.
PROB.
LXXXV.
Fig.
There are two circles BDA and BFC touching in B, 30.
and if DE be perpendicular to BA at the center E;
then there is given AC and DF; to find the dia-
meters.
Let radius BE=a, DF b, CA=d; then FE
a-b, EC-a-d, then FE-BEXEC (Geom.
IV. 17), that is, aa—2ab+bb—aa—ad, and 2ba-
bb
da bb, and a =
whence BC
26
d
>
LXXXVI.
za-d.
PROB.
In the triangle ABC, there are given the three perpen- 31.
diculars, from the angles upon the oppofite fides; to
find the fides.
Let AQ=a, CP = b, BR = c, and AB = y.
Then twice the area by ACX CB xa,
= CCB
by
by
whence AC = and CB. And (Geom.
bbyy
II. 21.)
CC
C
2
a
-bb-AP²; and (Geom. II. 23. cor.)
ACAB CB²
bbyy
bbyy
+yy -
CC
aa
AP =
2AB
zy
bby
I
bbyy
bby
I
bby v
+ У
; whence
+ y
200
2
24a
200
2
244
bbyy
CC
10
bb. That is, aabby + aaccy - bbccy =
2aac bbyy bbcc; put aabb+aacc-bbccd, then
dy=2aac√bbyy
aabcc
bbcc; and by reduction,
y =
✓ a + bbc c — AddⓇ
4
Cc 3
PROB
390
B. II.
GEOMETRICAL
Fig.
PRO B.
LXXXVII.
32 In the triangle ABC, there is given the rectangle of
the fides; the rectangle of the fegments of the baſe,
made by a perpendicular; and the area: to find the
rest.
Let the area = b, AD × DC=c, ABXBC=d,
and BD=≈, 2y=difference of the fegments AD,DC.
26
b
Then = AC, and +y=DC,
Z
Z
b
y=
Z
bb
DA. Whence
h
-'c, and 22+
+ulx
Z
b
e
bb
ZZ
ZZ
=d; and fquaring all the quantities,
-c for yy, and v for zz +
and putting
and then + 2by xv- 2by
χυ
Z
Z
4bbyy
ZZ
+cc+4bb-2czz-
2bb
ZZ
= dd = VV
(reftoring the values of v and y) + +
464
24
ZZ
+
or z4
Z4
ZZ
4cbb 464 4bbc
2czz + 4bb + ccdd. Whence z is known, and
!
y =
bb
ZZ
will be found.
c, and then AD, DC, and AB, BC
PROB. LXXXVIII,
33. In the right-angled triangle ABD, there is given the
perpendicular on the hypothenufe; and the radius
of the infcribed circle to find the fides.
Put the perpendicular BQ = p, radius CR=r,
AD=a, AB=e, BDy. Then (Geom. II. 21.)
aa =
I
4
*
}
1
44
•
Fig. 21.
H
G
N
}
?
22
3
0:
R
B
72.
D
P
B
E
E
I
23
B
24
}
B
F
*
کا
1
G
B
F
26
A B*
E
the
.
}
H
25
K
D
L
C
F
G
27
28
D
A
в с
D
S
E
30
R
WOA
29
E
B
31
B
Pl. IV. pa.39
ม
NIV
Sect. VI.
391
PROBLEM S
B
aaee+yy. And (II. 20. cor. 2.) pa
ey. But Fig.
AD or AF+FD=AR+DI, and AD + 2CR = 33.
AB+BD, that is, a+are+y; whence aa+2pa
=ee + 2ey+yy=e+y!" =a+2rl²=aa+4ra+4rr,
therefore 2pa-4ra 4rr, and a=
=
2rr
pir
Alfo
ca-2pa — ee — 2eyyy—e—yl², and ey
aa 2pa. Whence e ==
✔aa
==
a + 2r+ √aa-2pa
2
and y =
a+2r√aa-2pa
2
PROB.
LXXXIX.
There is an ifoceles triangle, in which two circles are 34
infcribed, touching one another and the fides of the
triangle; their diameters are 8 and 12: to find
the fides of the triangle.
From the centers D, F, draw DG, FH to
BC, and FO || to CB . . draw CFDA.
Put DG=r, FHS, DO=r-s=c, FD=r+
s=b. Then FObb-ccd, and CB = a.
The triangles DFE and BCA are fimilar, whence
da
b: d::a:
= AC, and
rb
b r: = CD,
b
C
rb
da
bbr
then r +
= AC =
whence a =
+
C
cd
by
360
√96
C c 4
PRQ
392
B. II:
GEOMETRICAL
Fig.
PROB.
XC.
135. There is given AD, and CD the radius of the femi-
circle CEG, to find the radius of a circle infcribed
between AC, the tangent AE, and the circle CE.
Draw from the center O, the line OI perpendi-
cular to AC; through. O, draw AOF bifecting
the angle DAE, and put radius DE =r, AD = &,
Ola, then AE- ✓dd-rr = b.
Then (Geom. II. 25.) AD AE:: DF: EF,
and AD+AE: AE :: DE: EF; that is, d + b:
= EF, DO = a + r, and DI
b:
br
::7: d+
d + b
DO²—UI²=√rr+ara, and AI=d—√rr+2ra,
and by the fimilar triangles AEF and AOI,
dbr
d-√rrera: a::b:
br
b+ď
Then ba
b+
b + d
br
Vir+zra. Put b+d=c, and reducing,
b + d
ccaa
2r3a
zdrc
24
ddrr.
XCI.
1:
PROB.
36. Through a given point B,to draw the right line BDC,
So that the part DC comprehended between the two
lines AC, AH, equidistant from B, may be of a
given length.
Produce CA to E, and compleat the rhombus
EABH; make the angle CDF CAF, and let
CD=a, AE or AH, BA=d, AÇ=x, AF=y.
The triangles CAD, CEB are fimilar, therefore
CA (~): CD (a) :: AE (b): DB =
ab
X
Since
FDC FAC, therefore their fupplements
FDB =
Sect. VI.
393
PROBLEM S.
FDB CAB, and fo the triangles BAC and BDF Fig.
are fimilar, whence BA (d): AC (x): : DB 36.
(ab) : DF
X
ab
d
+
But the triangles FAD and FDB are fimilar;
for <PDF FAD, (for BAD BAE FAC,
add DAC, then CAB FAD, that is, FDB =
FAD;) and is common; therefore AF (y):
DF (44)
d
:: DF:
ced is ddyy + d³y
gabb
ddy
FB d+y, which redu-
aabb; whence y will be had.
DAF and BAC are alto fi-
ab
Again, the triangles
ab
milår, and CB= a +
then DF (42) : AF
: AF (y)
X
abx aby
:: CB: CA (x); whence
+ay, re
d.
X
Then
duced abxx-adyx abdy, whence x is had.
bx
CE (b+x):EB (3)::AC(x):AD = b + x
PROB. XCII.
Through a given point B, to draw the right line BDC, 36.
So that the part DC, included between the lines AC,
AH, may be given.
Through B. draw BH, BE parallel to EAC and
AH, and put CD=a, AE=b, AH≈c, AB=d,
PHf, BP being perpendicular to AH, and
AC = x.
The triangles CAD and CEB are fimilar, and
CE (≈ + b) : EB (c) :: CA (*) : AD =
CX
b + x
CX
bc
and DHC-
And (Geom.II.
b + x
b + x
ccb b
2 bif
22.) BD =
bb +
and
>
bb+¿6x+xx + b + x
b:
394
B. II.
GEOMETRICAL
Fig.
36.
b:
bb +
ccbb
2bcf
a;
2bcfxx
+
b + x
:: X
bb + 2bx + xx + b + x
whence bbaabbxx +
ccbbxx
bb + 2bx + xx
which reduced is x4+2bx³ +bbx²—2baax—bbaa—….
2cf + 2cf
+
b aa
XCIII.
PROB.
37. The difference of the height of two hills being given,
and their distance; to find their heights.
Let BA, ED be the hills, put radius r = CR =
698000, DE-BA=b=119, AB=a, BE=c=63.
Then CB=r+a, CE=r+b+a. Then (Geom.
II. 21.) BR = √2ra + aa, RE =
bb + 2br + 2ra + 2ba + aa, whence BE =
✔zra + aa +√bb+2br+2ra+2ba+aa=c,
and
√bb+2br+2ra+2ba+aa=c−√2re+aa, and by
fquaring, bb+2br+2ra+2ba+aa=cc+2ra+aa➡
2c√2ra+aa, and 2c√2ra+aa = cc—bb—2rb—
2ba=dd—2ba (by fubftitution); and when ſquared
8ccra +4ccaad44ddba + 4bbaa, and when re-
duced, aa+2ral cc-bb-4rb rrbb
+65
=
+
and
4
CC ᏏᏏ ?
bb
a = 164,69.
PRO B. XCIV.
38. Three lines drawn from the three angles of a triangle
to the middle of the oppofite fides, being given; to
find the fides.
Put AD = b≈ 18, E = c = 24, BF =d=30,
CBx, AB=y, AC = z.
Then (Geom. II. 28.) yy+zz=2bb+xx, yy +
xx = 2dd + 1zz, zz + xx=2cc+yy; and adding
thele
Sect. VI.
395
PROBLEM S.
theſe three equations, 2xx+2yy+2≈≈≈2bb+2cc Fig.
I
I
I
+2dd + 1/2 xx + 1/2 yy + =—=zz, and xx+yy+zz=
2
2
2
4
3
bb 38.
4
+
3
3
4 cc++ dd, from this fubtract the firſt equa-
tions, then xx=dd-
3
+1
2
bb+
3
3
I
CC- xx, or 9xx
2
=8cc8dd-4bb,9yy—8bb8dd—4cc, 9zz=86b+
Scc-4dd, whence x=34,176; y = 28,844; ≈≈20.
PROB. XCV.
ABC is an equilateral triangle, O a point in it equi- 39:
diftant from A, B, C. If the fides, and the line
BO be all produced till they cut the line PD in D,
E, R, P; then there is given DE, ER, RP ; to
find the fide of the triangle ABC, and the area.
Draw EF, EG parallel to BP, BR; and put
DE=a=304; ER=b=121.6; RP=c=159,6;
and DRd, DPs, CL or AL=x, CG or FG=y.
Then (Geo. II. 39. cor.) BL=x√3, EG=yv 3.
The triangles DEF and DPA are fimilar, whence
a: zy :S: = AP, and PB = 2x +
25y
a
2sy
a
Since <PBR=<EBR, therefore (Geo. II. 25.)
2x + 25y: 2x+2y: :c : b, and 2bx+ 2bcy = 2cx +
·ba
a
a
i
2cy, whence cx - bx=
bsy
ca-
cy, and y
ya
x
a
bs-ac
fx
>
by ſubſtitution;
Again DE (a): EG (√√3) :: DR (d) :
=RL = dfx
PB
2x +
ag
✓3, and RBX√3+
2sfx
ag
2
dy
√3
dfx ✓3, and
ag
2fx
g
But
and BE 2x+2y=2x+
396
B. II.
GEOMETRICAL
Fig. But (Geom. II. 26.) BR'+PR-RE=PBXEB,
+bc=4xxx1+
3xxX1+ =
39. that is, 3** X 1 +
df
ag
f
sf
XI+
g
ag
And by reduction, 1 +
I
+
+
g
4saff - 3ddff
aagg
4f 45f 6df
ag
x into xx bc. Whence x 78.4,
y=40, and the area ABC
10646.16.
PROB. XCVI.
40. In the triangle ABC, there is given the baſe, and dif-
ference of the fides, and the area: to find the triangle.
f
Let the area =ƒ= 796; difference of the fides
CA, CB=b=10; bafe AB-d=50; perpendicu-
lar CD =
2f
=p = 35.84, and AD = a.
P
Then
AC = √aa+pp, and CB=√d-al+pp; there-
fore by the queftion Vaa + pp + b =
✔aa PP
✔del—2da + aa+pp, which fquared is aa+bb+pp
+2b√aa+pp=dd—2da+aa+pp, and 2b√aa+pp
=dd-bb-2da, and fquaring both fides 4bbaa +
4bbpp=d++b+2ddbb-4d³a+4bbda + 4ddaa. Which
dd
bbpp
reduced is aa— da =
dd bb
4
bb
whence
a = 16.739, AC = 36, BC = 46, BD = 33.261.
PROB. XCVII.
41. There is given the fide of a rhombus, and the fide of its
inferibed fquare; to find the area.
Let AB BD=d=4%, CO=CE=s=3, BC=x.
4-w, and AC = d +x.
Then DC
d
EAR
The
Sect. VI.
397
PROBLEMS.
The triangles ACE and CDO are fimilar, and Fig.
d + x
d- X
d+x:s::d-x:
s=DO. And (Geom. II. 41.
1
d-
24
2x,) ss--ss X
d +x l²
2
=-x; that is 2ssxdd+xx
dx\
-dd-xx; reduced, x4-- ddxx + d+
255 2ssdd
5
0.
Whence x=√ss + dd ±√ss + 4dd = √, and
8
AC=5, AE=4, DO=24, DQ=54, area=731,
QA = 7•
PROB. XCVIII.
Given the four fides of a trapezium, infcribed in a cir- 42.
cle; to find the diagonals, and diameter of the circle.
Let AB=a, BC=b, CD=c, AD=d, BE=x,
the triangles ABE, and CED are fimilar; for
<ABE = ECD (Geom. IV. 12. cor. 2.); and the
angles at E are vertical; therefore AB (a) : BE
(x) : : DC (c) : CE =
CX
a
; alſo the triangles AED
and BEC are fimilar,
and
BC (b) : CE (*
a
dcx
AD (d): DE =
And BC (6) BE (*) ::
ab
dx
dcs
AD (d): AE =
Then BD = x +
b
ab
and AC= +
dx CN
b
Then (Geom. IV. 32.) AC
a
dxx
ddcxx CXX
xBD=ABXCD+ADXBC, or
+
+
b
abb
a
decxx
+
acbd, whence x is had, and then AC
baa
and BD are known.
Then
C
398
B. II.
GEOMETRICAL
1
Fig.
42.
Then ſuppoſe a perpendicular from A upon BD,
then (Geom, II. cor. 23.) the diſtance of the per-
pendicular from D is =
AD² + DB²
AB
=f.
2BD
p; and
And ✔AD² —ff
ff the perpendicular
(Geom. IV. 28.) p: AD:: AB: diameter of the
AD × AB
circumfcribing circle =
P
XCIX.
PROB.
43. The three femicircles HFG, HEI, and GOI touch
one another in H, G, and I; to draw a fourth
circle FOE to touch all the reft.
From the centers A, B, C draw the lines ADE,
BD, and CD; and DP perpendicular to AC, and
let AG = a, BE or BI=6, CGc, and DE=x.
Then AD=a+x, BD=b-x, CD=c+x, AC=
a+c, BC = b—c.
In the triangle ADC (Geom. II. 22. cor.)
PC =
c+al² +c+xl² —a+xl²
20+ 20
gle BDC, PC =
Ъ
b=d² +c+x1²
and in the trian-
26
26
b
Whence
cc + zca + aa
+ cc + 2cx + xx
aa
22x
c+ a
That is,
XX
bb2bc + co
cc
+ cc + 26x + xx
bb + 2bx XX
វ
2Cc+200-2 CX --- 2 ax
c + a
2CC i 2bc + 2cx + 2bx
;
And
Sect. VI.
PROBLEM S.
And multiplying alternately,
2bcc+2bca+2bcx-2abx-2c3—2bcc+2ccx+2bcx
2c3 — 2 acc-2ccx+2cax
+2cca-2bca+2cax
399
Fig.
43.
+2bax
And
}
4abx+4ccx = 4bcc + 4bca
4bcc + 4bca — 463
4cca,
Whence x =
a + c x b - - c c.
ab + cc
PROB. C.
In the triangle ACB, there is given the fides AC, 44:
CB; and the length and breadth of the infcribed
rectangular parallelogram DEHF; to find the reft.
Draw CP perpendicular to AB, and let CA = b;
CB=c, DE or GP=p, DF-a, CP=z, AB=y;
and let p b + c ₁ q = b — Co
= cs g
The triangles CDF and CAB are fimilar, and
zy::z-p: a; whence za zy-py, and zy-
zapy; therefore z= py
y- a
fegments of the baſe. Therefore AP=
Again, (Geom. II. 24.) y: p::q:
pq
:9
= diff.
y
I
y +23
ра
2
2g
I
pal
2y
But (Geom. II. 21.) bb = zz + 2y +
I
I
ppyy
Z
+ 2 yy + 2 pg + PPqq.
4
Which equation
4yy reduced is
y6—2ays+aay++8abby³+PPqqy²— 2ap¹q'y + aappqq
= 0.
+ 4pp-2apq
+2aapq
+2pq
4aabb
-466
PROB.
}
3
$
400
B. II.
GEOMETRICAL
Fig.
45.
PROB.
CI,
In the right-angled triangle AVP, KD is drawn pè-
rallel to the bafe, and there is given the bafe AP,
and the fegments VD, AK; to find the rest.
Let AK b=200, AP=c=400, VD=d=260,
and DPa; then (Geom. II. 12.) a: b::d:
bd
bd
a
=VK, and VA=b+
b
Xa+d. But
a
a
bb
aa
2
2
AV²-VP² AP², or xa+d² a+di² = cc₁₂
or bb aa xa+daacc, that is,
at + 2da³ + ccan
=
+ dd
bb
2bbddabbdd.
And a 141,727, and a + d or VP = 401.727.
PROB CII.
¿
46. In the figure CAFD; CA, BP, DF are perpendi
cular to AF; and the fides of the triangle being
produced, there is given HA, AC, CB, and BD,
DF, FT; to find the fides of the triangle HBT.
Let HA=n, AC=p, CB=f, BD=d, DF=c,
FTb, and TH = 4, TP≈≈, PH≈v, BP=x,
BT = j, e = HB.
y,
The triangles TBP and TCA are fimilar, and
y: xf+y: p, and py=fx+yx, or py-xy=fx;
whence J'=
fx
P-X
The triangles HBP and HDF
are fimilar, and e:x::e+d c, or ce ex + xd,
dic, =
!
dx
and ce-xe=dx, whence e=
Likewife z:*::
}
€
a + 33
Sect. VI.
PROBLEM S.
401
a+n Fig.
P
a+n: p, and pza+nxx, and z = x. 46.
Likewiſe v:x::a+b:c, and cv=a+bפ, and
a+b
2=
X.
a+n a+b
But v+z=ɑ=
P
x + x, whence pca =
pbx; therefore a =
рс
C
cax+cnx+pax+pbx, and pca-cxa-pxa=cnx+
cnx + pbx
CX Px°
fx
px
Again, TCƒ+
p-x P-x"
-pp = n + a = n +
pcn−ncx―npx+ncx+pbx
rb―n, and sc+p); and by fquaring,
pf
==
and (Geom. II.
21.) PPH
✓
P=x12
pc - cx
ppff. p4 + 2p³x-ppxx
PP-2px + xx
ppccnn+2ppcnrx+rrppxx
РРСС 2 pcsx + ssxX
which reduced is
+ ppss—2p³rr
-2p³ss
-ap³cs
+ p4ss
pprrx4+2ppcnrx³ + ppccnnx²-2p³ccnnx+p+ccnn=0
-4p³cnr +2p+cnr +p°cc
+p+rr -2cpss -ccp off
+2cp³ffs
-Ppffss -2ccps
cnx + pbx
pc-cx- px
pcn + rpx
(putting
PC - sx
+ 4p+cs
+p+cc
PROB.
CIII.
Given the fides and area of a trapezium; to find the 47.
diagonal.
Draw the perpendiculars BE, DF upon the dia-
gonal AC; and put AB a=4, BC=b=6, CD
=c=7, DA=d=5, and ƒthe area, and AC=y.
Dd
Then
402
B. II.
GEOMETRICAL
Fig.
47.
}
yy
Then (Geom. II. 23. cor.) CE + bb — aa
and BE =
=
2y
√ bb_yy+bb — aa
2
✔4bbyy
2y
4bbyy — y² — 2yy × bb — aa — bb — aa
y+
4yy
−y4+yy × 2bb+2aa — bb_aal²
ner DF =
4yy
−y4 +yy × 2cc + 2dd
Put 2aa+2bb × yy
and 2cc +2dd × yy
I
Then y× BE + DF
2
4yy
In like man-
CC
ddj²
2
bb-aaľ = pyy
cc-dal = ryy
2
99
19 = v.
SS = %.
= ƒ, and 2y × BE + DF
=4ƒ, that is, √v—y4+√ z—y4=4f, and by ſquar
ing v + 2—2y4 + 2√
-V
。、
-V
•vz _2y++y³ = 16ff, and
v+z
v z = 2y+ + js = 8 ff— +y4, and by ſquar-
-V
-Z
2
ing, vzy4+y³ = 64ƒ4 — 8ff × v +%+
64₤4
+ 16f—v—z× y++ys. And vz =
0 +21²
2VZ
v+x+
དུལ
+
VV
+ +
4
4
ZZ
~+21
M
4
64f4-8ff x
+16ffy4 = 64ƒ4 — 8 ff x v + x +
v+%+
+ +16ffy4, or 16ffy4 — 8ffxv+%
4
2VZ + ZZ
4
+64f40; that is, 64ff4-
32 f×v+z+ v − 2)² + 256ƒ4 = o, and reftor
the values of v and z, we ſhall have
ing
"
64f
Sect. VI.
403
PROBLEMS.
64ff
Fig.
+p;
}
+2p=2r × ss-qq
y4
}
+55-99
47.
-32 ff
++r
+32/ff×55 +99 (
+25664
=0.
and y = 7.68.
PROB. CIV.
In the right-angled triangle DCF, there is given
DC+CF, and BA parallel to the bafe, and DA;
to find the rest.
Let DC + CFs, BA=b, DA=d, CF=a;
then CDs-a, DF=✓DC_CF²=√55—250.
The triangles CAB and CFD are fimilar, whence
SS- 2sa : a :: b:
ab
SS - 25a
= CA; and
DA' — DF = √dd—ss + 2sa AF; whence
ba
√552502
=
+ √dd ss + 2sa = a, and multi.
plying bv ss
$5254, and tranfpofing,
✓ ddss-2ddsa-s++4s³a—4ssaa ssaa-25a³
ba; and ſquaring,
— 4ssaa + 453a + ddss
2ddsa st
=
25a³ + ssaa √4bbssa+
+bbaa
tranfpofing again,
8bbsas; and
✓ Abbssa+8bbses —— 25a³ + 55saa — 453 a + st
=
+bb
+2dds- ddss
--ca³ + faa+ga +b, by
ſubſtitution, and by fquaring, 4bbssa+ 8bbsas
- 2cfas + ffa4 — 2cha³ +gga² + 2gha +bh.
-2cg + 2ƒg +2fb
=ccao
D d 2
PROB.
48.
$
>
Į
1
404
B. II
GEOMETRICAL
Fig.
PROB.
CV.
49. There are given the three fides of the triangle ABC,
and the angles A, and B, are bifected by the lines
AD, BE; to find the length of one as AD, and
alfo the distance AF to the point of interfection F.
Put AB=a, BC=b, AC=c, and AD=x, AF=y.
Then (Geom. II. 25.) AB: AC :: BD: DC, and
AB+AC: AB:: BD+DC: BD; that is, a+c:
a::b:
BD; likewife a+c:cb:
ab
a+ c
=
bc
=CD. But (Geom. II. 26.) AD'+BDC=BAC
a+c
abb c
bb
that is, xx+
= ac, and xx = ac
X
a+d²
2
a+d
a+cl² — bb
a+c+ bxa+c
a c = ac X
х
= ac X х
a + c²
whence x=
−b
CAD.
a + c ²
√ acx a + c + bxa + c
a + c
Again, AB BD: AD: : AB: AF, that is,
ab
@ +
a + c
y =
ac xa + c + b × a + c
at c
bxeto
ab
a√ acxa+c+b xa + c =
a + c xa + a + c
√ acxa + c + b x a + c
b
11
b
::a:
b
that is,
a+i+b
a c xa + c b
y = √
AF.
a+c+b
PROB
1
الله
#
Fig 32
A
D
}.
A
R
34
33.
C
I
F
D
··A
O
D
H
A
E
35.
F
G
B
F
E
A
R
B
E
36.
D
I C
D
Ꮐ
37.
B
H
P
+
38.
E
F
A
39.
GF
B
B
D
Ꮲ .
R
E
D
B
41.
0
42.
E
B
F
E
C
44
45
F
K
i
40.
E
43.
F
Ħ
D
A B
GPC I
46.
B
E
PHB
A
P
A
H
P
T
47.
48.
F
E
1
D.
B
F
49
E
F
D
}
P1.V. pa. 401
Sect. VI.
405
PROBLEM S.
Fig.
CVI.
PROB.
The diameters of three circles being given, which are 50.
defcribed from the angular points of a triangle, as
centers, whoſe three fides are given; to find the ra-
dius of a fourth circle to touch all the three.
Let ABC be the given triangle, D the center of
the circle required; on AB let fall the perpendi-
culars DE, CK, and draw DF perpendicular to
AC. And put AB = b, AC = c, CB=d, and
AO=r, BR = s, CT = t; and AK=g, KC=h;
and AE x, AF=y, OD a. In the triangle
ADB, (Geom. II. 23.) aa+2as+ss=aa+2ar+rr
+bb--2bx. Whence 2bxrr+bb—ss—2as+zar,
rr+bb-ss-2as+zar
and x =
26
=
And in the tri-
angle ADC, aa+2at+tt=aa+2ar+rr+cc—20),
and 2cy rr + cc tt 2ta + zra, and
y =
rr + cc. tt 2ta + zra
20
The triangles ACK, AFG are fimilar, and
g:c::y:2=AG, then x-GE. Alſo the
су
g
&
triangles DGE (AGF) and ACK are fimilar;
whence b: g
су
: X- : Vaa+2ar+rr—xx =DE.
g
Whence baa + zar + rrxxgx-cy.
Put rrbb-ss, f=25—2r, mrr+cc―tt,
n=2t2r, p = lg bm, q=bn-fg. Then x=
l — fa
26
•
and baazar frr-
-
2b
lg - fga
2b
p+qa. Which ſquared is bhaa+2rbba
m-na
2
2b
D d 3
+ ribb
*
2
}
406
B. II.
GEOMETRICAL
Fig.
+ ribb
bbx
Il — 2lfa + ffaa
pp+2pqa+ggaa
50.
466
and reduced
466
4bbbbaa8bbbkra + 4bbbbrr = 0.
2dfbb bbll
= fub = 2pq
- PP
PRO B. CVII.
1
51. To find the point D, from which three lines DA, DB,
DC drawn to the three given points A, B, C;
fhall have a given ratio.
Draw AC, and DFG, BE perpendicular to it.
Draw BG to EF; and put AE = a, AC = b,
|
EB c; and AF=x, FD=y. Then CF-b-x;
EB=c;
FExa, and let DA, DB, DC, be as 1, r
and s.
Then (Geom. II. 21.) AD² = xx+yy; BD² =
¢+yl² + x—al² =cc+2cy+yy+xx−2xa+aa; and
CD2bb2bx+xx + yy.
2
But by the queſtion 1: rr:: DA: DB² = rr
XDA', and I: ss:: DA²: DC²=ssxDA²; that is,
cc + 2cy + yy + xx — 2xa + aarrxx+rryy, and
bb — 2bx + xx+yyssxx+ssyy, and putting m
rr 1, p = cc+aa, f=ss−1, we ſhall have theſe
(b) (c)
2cy + mxx = 0.
(0)
+ 2ax
two equations, myy
P
(f)
(g)
(b)
and ty
+fxx = 0.
+2bx
bb
Then
Sect. VI.
407
PROBLEMS.
Then to expunge y (by Prob. liv. rule 2.) we Fig.
have
52.
A == 2cf, B =
D = mfxx + 2 afx
mfxx 2mbx
2cfxx - 4bcx + 2cbb.
pf
mbb.
Whence AB + DD = o, that is,
4ccffxx + 8bccfx — 4bbccf
+4aaff-4apff +ppff
4ambf 4ambb + 2mbbpf
+ 4mmbb + 4mbpf + mmb+
+ 4mmb³
PROB.
= 0.
CVIII.
In a triangle, there is given a perpendicular, the dif- 53.
ference of the fides, and the difference of the feg-
ments of the bafe; to find the fides.
Let the perpendicular CD-a, CB-CA-c, and
BD-DA, DA. Then CA =√aa + xx,
and CBaa+xx+c, and AB=2x+b. Then
(Geom. II. 24.), b + 2x : 2√aa+xx + c : : c : b ;
whence bb+2bx = 2caa+xx + ce, and 2bx +
bb-cc=2√ aa+xx, and fquared is 4bbxx+4bxx
cc + bb cc² = 4aa + 4xx × CC 3
bb
reduced
4bbxx + 4b³x=4aacc.
$
-400
4bcc
b4
+2bbcc
Roca C4
Dd 4
PROB
+3
408
B. II.
GEOMETRICAL
Fig.
PROB. CIX.
54. There is given the perpendicular in a triangle, and the
two differences between the leaft fide, and the other
two; to find the fides.
Let the perpendicular AD-a, BC-BA-b, AC
-AB-c. AB=x; then BC=b+x, AC=c+x, and
BD = √xx—aa, and DC = b + x
-√xx—aa;
and (Geom. II. 24.) BC (b+x) : AC+AB (c+2x)
:: ACAB
AB (c) : DC — DB (b+x—2√xx—a 2);
whence b+x) 2b+2x × √xx—aa = cc + zcx;
and 26 + 2x X
XX
aa = bb + 2bx + xx
CC
Put bbccd, 2b-2c=f; then
×
аа
20X
4bb + 8bx+4xx Xxx-aa = xx+fx+dl', which
multiplied and reduced is
3x+ + 8bx³ + 4bbx²
8baax
4bbaao.
2f 4aa
2df
dd
二
​·f
2d
PROB. CX.
55. Having all the fides of a right-angled triangle ACB;
to find either ſegment of the bafe AD, the perpendi-
cular CD, the area, and the radius of the infcribed
circle, &c.
"
a+b+c;
1. Let AC=a, CB=c, AB=b, z= 2
then (Geom. II. 20. cor. 1.) aab x AD, and
AD =
a a
b
fore AD =
But aa = bb-cc=b+cxbc; there-
b + c x b
b
C
that is,
the
t
Sect. VI.
409
PROBLEM S:
the fegment AD =
aa b + c x b = c
Fig.
b
b
55.
2. For the perpendicular CD.
CD2=AC2-AD-aa-
04
aaxbb-aa
and
bb
bb
a
b
a
CD =
b
CD = = √ bb = aa = = √ b + axb—a; or
ac
b
√cc = ==. Therefore the perpendicular
ac
a
a
CD=
√bb
b
bbaa
aa =
b+axb—a.
b
b
AB X CD
2
a√ b+axb-a
2
3. For the area.
area; that is, the area =
ac
And fince aa+cc=bb, add 2ac,
a+d²
then aa+zac+cc or a+c" = bb+2ac, and 2ac
ac
a+cl²- bb, and
or the area =
a+c²-bb
2
4
a+c+6×a+cab
4
=xxx-b. And fince a+c
z—b.
+a-c2aa+2cc=2bb, therefore a+c-bbbb
bb
a
d. Therefore the area
a
cl²
=
4
b + a
- cx by
a + c
=z-axz—c.
Hence the
4
ac
a+c+bxa+c-b
area =
2
2
4
b. + a cxb-a + c
=xxx−b = z
4
4 For
$
410
B. II.
GEOMETRICAL
Fig.
55.
4. For the radius of the infcribed circle.
The area
circle (Geom. IV. 30. cor.) =
a+b+c
a+b+c
× radius of the inſcribed
2
xr, putting
2
a+c-b
=
2
ac
"
that is, the ra-
a + b + c
r for the radius; then r
z-b; or the radius=
א
2 area
a + b + c
dius of the infcribed circle is =
Z
b.
ac
a+c-b
22
2
5. For the circumfcribing circle.
The radius of the circumfcribing circle = b
(Geom. IV. 14.)
6. For the tangents.
I
2
The tangent, or the diſtance from A to the point
of contact of the infcribed circle =
z-c, (Geom. IV. 30).
a+b C
2
And the diftance from the right angle C is
Er the radius.
7. For the distance of the center.
The diſtance from Arr +
√a+c= b² +
2
a+bd
4
2
afico
2
!!
a+b-
2
a+b-
(by Art. 4. and 6) =
amb
a+di²
adi
+
An
4
4
(putting
Sect. VI.
PROBLEM 411
T
S.
(putting d=b—c)=√2aa+2dd
11
aa+bb -2bc+cc
2
bxb-c; that is,
waa+b-
za+b—c] Fig.
55.
2
4
2bb-2bc
✓
= √ bb——bc
2
the diſtance of A from the center of the in-
fcribed circle is
=
= √ bx b = co
PRO B. CXL.
Having the fides of an oblique triangle ABC; to find 56,
the perpendicular CD, the fegments of the bafe, and
the area.
a+b+c
Let AC, AB=b, CB=c,
= 2,
2
a+c=s, a c=d.
>
1. For the Segments.
AB AC + CB:: AC-CB
(Geom. II. 24.); that is, b: s : : d :
I
AD-DB. Then —b +
dif. fegments
ds
b
ds
bb + ds
or
2
26
= great-
bb-ds
er fegment AD, and
BD the leffer feg-
2b
ment.
2. For the perpendicular.
2
bb + as
CD aa-
26
=
4bbaa-b4-2bbds-ddss
4tb
but 2a=s+d, and 4a0=s+d, and 4bbaa—bs+bd,
therefore
CD=
412
B. II.
GEOMETRICAL
Fig. CD=
✓bbss + 2bbsd + bbdd—b4—2bbds—ddss
56.
4bb
ss x b
ss x bb-dd
bb x bb-dd
bb-ddx ssbb
=
4bb
466
26
√ b+dxb-d xs+b.xs-b
+c+b, s — b
:—b=a+c—b, and b+d=a+b—c,
b―d = b+ca, therefore
a + b + c xa +c—b× a+b — c × b + c
But s+ba
CD=
26
a + b + c
b + c
a
But s
Z
a =
2
2
a + b —
a + c
b
Z
- b
b =
Therefore
2
2
2Z X 2.Z
CD =
a X 2.2
2b
C X 2.Z
b
CD =
2√zxz—a.z—biz-c
that is, the perpendicular
a+b+cxa+b-cx a−b+c xb+c=a
26
b + d x b - d xs + bx s
dxs bxs-
b
2b
2 V ZXZ
Z XZ — a X Z
hxx-
C
b
3. For the area.
I
Since the area is = ABX CD (Geom. II.
2
10. cor. 2.), and CD was found by the laſt ar-
ticle, let CD p; fince AD =
aa + bb
CC
26
therefore
Sect. VI.
413
PROBLEM S.
therefore CD =
aa
aa + bb
26
- ccp²
Fig.
11
56.
Jaat
2aabb +2aacc + 2bbcc
a4 - 64
C4 ; therefore
4bb
I
pb =
2
we have the area of the triangle ACB =
21
I
4
11
✔zaabb + 2aacc + 2bbcc 04 br— c+
4
a+b+c
a + b + c xa+b―cxb+c — axa + c—b
b + d x b — d xs + bxs—b =
ZXZ-
-axz
PRO B. CXII.
Having the fides of an oblique triangle; to find the 57:
radius of the infcribed circle, &c.
1. In the triangle ABC, bifect the two angles
A, B, by the lines AF, BE to interfect in O the
center of the infcribed circle. From O, C, let fall
the perpendiculars OD, CP, upon the bafe AB,
And put AB=b, AC=a, CB=c, AP=d, PB=ƒ,
CPP, and DO=x, DP =y; then AD=d-y,
BD =ƒ +y.
Then (Geom. II. 25.) CA: AP :: CS: SP,
and CA + AP: AP :: CP: SP, that is, a+d:
d::p:
pd
a+d
=SP. Likewiſe c+f:f::p:
pf
c+f
= LP. The triangle APS, ADO are fimilar, and
d:
pd
:: d
dy:
a+d
pd-py
a + d
triangles
BPL
and BDO
pf
f: ::fty:
pf + py
c+f
c+f
DO=x. Alſo the
are finilar, and
pd-py; then multi-
a+d
plying
414
B. II.
GEOMETRICAL
Fig. plying, apf+dpf+apy+dpy=cpd+fpd-cpy-fpy,
57. and tranfpofing, apy + dpy +fpy+cpy=cpd-apf;
that is, becaufe d+f=b, apy+bpy+cpycpd-apf,
and y =
cd-af
a+b+c
pcd-pat.
a+dxa+b+c
pba + pbd
a+dxa + b + c
Whence x = pd-py
pd
==
a + d
a + d
pda+pdb+pdc-pdc+paf
a + d x a+b+c
pb
And fince p may
a + b + c
be had various ways, from the laft problem; there-
fore we fhall have the radius of the infcribed circle.
Ja-
→c + bxa+c=bxb+c-a
b+nXbnXs
-nxs—b
==
a + b + c
x—axx—bxx—c
.bp
a+b+c 2
=
I
2
Where z =
s+b
a+b+c
Z
s = a + c, n = a
· C.
2
11
2. For the tangent AD.
We have AD d — y = d —
cd-af
=
a + b + c
ad+bd+axb-d
a + b + c
aa + bb — cc
ad+bd+cd-cd+af
a + b + c
11
=
2
a + b/2
+612
ba + bå
a+b+c
, therefore
CC
2 xa+b+c
a+b-c
2
; that is,
a+b-c
a + d
b
= Zl.
2
But (Geom. II. 23.) bd =
AD =
zab + aa + bb
- CC
2 x a + b + c
a + b + c x a + b
2 xa+b+c
the tangent AD =
C
a + b + c
3. For
Sect. VI.
PROBLEMS.
3. For the central distance.
415
Fig.
57.
AO AD+DO²=;
-·a + dj²
2
PP
bb +
24
a+b+c
aa+zad+dd+aa-dd 2aa +2ad
a+b+c
a+b+c
2
bb=
2abb
bb=
a + b + c = b b
bb=
a+b+d²
Xa+d. But d
aa+bb-cc
26
;
therefore AO² =
zabb
zab+aa+bb-cc
2
X
= ab x
a+bl²-cc
a+b+c
26
a+b+c²
a+b+cxa+b-c
a+b-c
Xab
× ab=
Z-C
a + b + cl²
a+b+c
Xab
× ab =
ab.
Z
That is, the diſtance of the center from the angle
A is
af b
хаб
Xab =
a+b+c
PROB.
CXIII.
א
Xab.
Having the fides of a triangle to find the radius of the
circumfcribing circle.
Let ABC be the triangle, draw the diameter
CF of the circumfcribing circle, and let CP be
perpendicular to AB. Put AC≈a, AB≈b, CB≈c,
a+b+c
CP=P, z=
s=a+c, d=a-6, CH or
HF = R.
2
Then (Geom. IV. 28), p : a : ; c : 2R, and
ac
R =
2P
Now fince we have the value of p va-
rious ways by problem cxi, we fhall have the value
of R fo many ways. Hence R the radius of the
circumfcribing circle =
ac
2 D
58%
416
B. II.
GEOMETRICAL
Fig.
58.
11
abc
a + b + c × a + b − c × a−b+cxb+c—a
abc
acb
√ b + d x b
d x s + bxs - b
||
=
b.z
C
||
4√ z.z —a.z
acb
4 X area
Cor. Hence r the radius of the infcribed circle:
to R the radius of the circumfcribed circle ::
As Z a x z − b x ≈ — 0
C:
to Labc.
PROB.
CXIV.
59. Given the bafe of a triangle, and the diameters of the
infcribed and circumfcribed circles; to find the fides.
Let QRW be the triangle, QDWB the circum-
fcribing circle, DB (perpendicular to QW) its
diameter. Draw BR, which will bifect the angle
R. Let QS bifect the angle Q, then S is the
center of the infcribed circle. Through S draw
ASV parallel to QW. Then AP is the radius of
the infcribed circle. Draw BV, BW.
Let BD=a, QW=b, AP=c, BP=v, BR=x,
then av
1bb, and v±ª±√ aa—bb
vv=1bb,
2
Let
BW = d = √vv+bbav. BA=v+c=s,
X
BV = p =✔BA × BD.
The triangles BRD, BPT and BAS are fimi-
lar, whence x: a :: v :
x:::
av
= BT
BT; and x: a
X
S: =BS. But (Geom. IV. 17. cor.) as = PP,
as
26
and
Sect. VI.
417
PROBLEM S.
dd
and av dd; therefore BT = and BSPP Fig.
>
x 95.
Then RS=x-, and TSPP - dd But
x
X
(Geom. II. 25.) TS: SR:: TQ : QR; and the
triangles TQR and BWR are fimilar (Geom. IV.
12. cor. 2.), and TQ: QR:: BW: BR; whence
pp-dd.
TS: SR:: BW: BR, that is, PP-dd: xx-pp ::
X
*
d:x, whence ppx-ddxdxx- dpp, and dxx +
ddx = dpp + ppx, or d + xxdx = d + xxpp, and
dx=pp, whence x=
pp.
Then BR (x): DR
ལ/
aa
XX
(aa-xx): BP (v): PT =
whence QT, TW are known. Then BW (d):
BR (x) :: QT : QR :: and TW: WR, the two
fides of the triangle.
PROB.
CXV.
There is given the bafe of a triangle, the line that bi- 60.
fects the vertical angle, and the diameter of the cir-
cumfcribing circle; to find the fides.
Let AB = b, EO or OF ≈r, CD=d, HD=x,
FD=y.
I
Then AD =
I
b+x, DB =
2
2.
b.
— b— x, FH
1
1
yy xx.
CDF, or
4
And (Geom. IV. 20. cor. 2.) ADB
bb-xx-dy. The triangles FDH, FEC are fi-
milar, and yyy-xx:: 2r: y+d, and yy+dy=
2 + √ yy — xx = 2 r√ wydybb. Which ſquared is,
E e
y + +
ག
418
GEOMETRICAL, &c. B. II.
Fig y4+2dy³ ddyy = 4rryy +4rrdy —rrbb, and re-
+
60. duced y4+ 2dy3+ ddyy - 4rrdy+rrbb=0. Then
x = √ bb — dy.
4rr
Alfo BFyy — xx + bb, and the triangles
ADF, CDB are fimilar, and AD (b+x) : AF
or BF (Vyy -xx+bb) :: CD (d): CB =
xx + ÷bb
· d√yy
zb + x
Alfo the triangles ADC, BDF are fimilar, and
BD (b-x): BF (√yyxx + bb) :: CD (d) :
CA =
dvy
-xx + 1bb
Ib- X
#
#
###
#
SECT
SE C T. VII.
SECT.
Problems in Plain Trigonometry.
413
Fig.
PROB.
CXVI.
In the triangle ABC, there is given the angle B, the 61.
fide AB; and the sum of the fides BC, AC; to
find the fides.
L
=
ET ABd, BC + ACb, fine Bs,
cof., AC = x, then CB bx.
By plain Trigonometry rad. (1): AB (d) ::
S.PAB or cof. B (c): PB cd.
Then (Geo. II. 22.) xx=dd+bb—2bx+xx+2cdx
b-x,
x, reduced 2bx + 2cdx = bb + dd+2bcd, and
bb + dd + 2bcd
SA
2b+ 2cd
PROB. CXVII.
In the triangle ACB, there is given the two fegments 62.
AD, DB, made by the perpendicular, and the angle
ACB; to find the rest.
Make DE DB, and draw CE, then put
BDb, ADd, CB or CE, S.ACBs,
S.ACE; then AE d—b, AB d + b.
By Trigonometry, (in the triangle ACB) AB
=
(b + d): S.ACB (s) : : CB (y) :
sy
b + d
=
=S.CAB.
Alfo (in the triangle ACE), CE (y): S.CAE
sy
~——) :: AE (d—b): S.ACE (x), and x =
+d,
E e 2
d-b
S.
db b
Then
420
B. II
PLAIN TRIG.
Fig. Then
62.
ACB-ACE
= ACD, and
2
2
ACE + ACB
=BCD. Then S.ACD : AD : : rad : AC. And
S.BCD: BD :: rad: CB.
PROB. CXVIII.
63. In the triangle ABC there is given AB, and the angle
B, and the ratio of AC to BC, to find the fides.
Let fall AD on BC (produced); and put
ABb, AC a, the ratio of AC to CB as I to
r, then CB = ra, and cof. ACB = c. Then rad,
(1): AC (a) :: S.DAC (c) : ca DC. Then
(Geom. II. 22.) bb = aa + rraa + 2craa, whence
bb
b
aa =
?
and a =
rr + I + 2cr
√rr
1
rr + 1 + 2cr
CXIX.
64.
65.
+
PROB.
In the triangle CAB, there is given two fides and
the included angle; to find the area.
+
Let CA, AB and the angle A be given; draw
CF perpendicular to AB, and let AB b, AC=d,
S. AS. Then in the triangle ACF, rad. (1)
<
: AC (d): S.A (s): sd CF.
sdb
area, or
2
Then
CF × AB
2
area; that is, half the rectangle
of the fides multiplied by the fine of the included
angle, gives the area.
PROB.
CXX.
Given all the fides of a trapezium, and two oppofiie
angles; to find the area.
Let the angles B, D be given, and through the
other two angles A, C, draw the diagonal AC.
Let
1
1
1
1
A
Fig.50
D
0
K
E
R
T
F
C
B
52
C
A
53
D
A
B
D
55
A.
D
F
58
B
C
56
D
DR
A
P
A
B
E
F
Ꮐ
51
B
54
57
V
H
S
A
W
PT
H
60
B
+
P
59
61
62
B
P
B
E
D
B
63
F
E
E
D
A
D
P B
B
น
B
2
Pl.VI. pa.420
UN
Sect. VII.
421
PROBLEM S.
Let AB=b, BC=c, CD=d, DA=f, S. <B=p, Fig.
S.D=q. Then by the laft problem, the area of 65.
bcp, and the triangle CDA
the triangle CBA = 2
dfq
bcp + dfq
; therefore the trapezium
2
PROB. CXXI.
In the triangle WNE, there is given the fegment SE, 66.
the angle WNE, and the ratio of NE to NW;
to find the fides.
Let WP be perpendicular to EN, and ſuppoſe
NE to WN as I to p, S.Ns, cof. N = ‹,
SE=b, NE=x, then WN≈px. In the triangle
WNP, rad. (1): WN (px) :: S.PWN (c) :
PN pex, and by the fimilar triangles ENS,
EWP, ES: EN :: EP: EW, or b: x ::
xx + pcxx = EW. But (Geom. II. 22.)
x + pcx
b
x4 + 2pcx4 + ppccx4
bb
:
= xx + ppxx + 2pcxx, or
1 + 2pc + ppcc xxx = bb × 1 +PP+2pc, and x=
吉
​VI + PP + 2pc.
! + P
Or thus,
EWN+E
Let tang. of
then WN+NE
2
tpx tx
(px + x): WN-NE (px-x): : t :
p. I
1+1
t = tang. diff. of the angles
Whence the angles W, E are known.
cof. E: & :: rad: x, required.
Ee 3
px + x
W and E.
Then as
PROB
422
B. II.
PLAIN TRIG.
Fig.
PROB. CXXII.
67. In the right-angled triangle ABC, there is given, the
Sum of AB and BC, the angle CDB; likewife
<ACD=<DCE are given; to find CB, &c.
Let S.ACBs, S.CAB=6, CB≈≈, AB + BC
=6, and BA = b x. Then s:c: :b x: x,
and sxcbcx, whence x =
Then CB,
cb
S+ C
BA are known. Let m, n be the tangents of BCE,
BCD; then 1:~:: 1: BE :: n : BD.
PROB. CXXIII.
In the triangle ADC, there is given AB, BC; and
the angles ADB, BDC; to find AD, DC.
Let S.ADB=s, S.BDC=t, S.ADC=p, cotang.
ADC=q, AB=b, BC = c, AC=d, AD=x.
By plain Trigonometry, b: s::*: =S.ABD
&
or CBD.
And t: c
b
SX CSX
tb
+ CD (x + CSX)
A+C
(9):
2
tb
SX
b
CD. Then AD
: AD—CD (x — CSX)
tbq-csq
tb + cs
:: tan.
A-C
= tang.
Then <SA
2
and Care known; then p: d :: S.C : AD ::
S.A: CD.
PROB. CXXIV.
69. In the triangle ABC, there is given AB, the angle C,
and CD, which is drawn to the middle of AB; to
find the fides.
Draw AF perpendicular to CB, and put AD or
DB=b, CD=d, S.<C=s, cof. Ç=c, AC=x, BC=y,
Then
Se&t. VII.
423
PROBLEM S.
€
Then (Geom. II. 28.) xx+yy=2bb+2dd. By Fig.
Trigonometry I :x :: s: sx = AF, and 1 :x:: 69.
c: cx CF. Then BF-y-cx, and ssxx+yy—
2cyx+ccxx=4bb, fubtract this from the firft equa-
tion, then xx-SSXX+2cy xccxx-2dd-2bb, that
is (becauſe ss + cc = 1.) 2cyx 2dd2bb, and
2dd-2bb
2yx=
C
; therefore yy + 2yx + xx
yy+2yx xx=2bb
2dd-2bb
and yx =
+ 2dd +
C
2ad-2bb
2bb + 2dd +
=m;
m; in
in like manner
C
2dd-2bb
y-x=
2bb + 2dd
= 1. Then
C
m + n
M-N
y =
,
2
2
70.
PRO B. CXXV.
Given the angles of altitude BCA, BDA, the hori-
zontal angle BCD, and the line of ſtation CD; to
find the height AB.
b
47: 30;
BDA = 40: 12; d = S.BCD = 87 5; CD =ƒ:
Let cotang. BCA
cotang.
=f=
283.274 feet; AB = x.
~
=
BC,
db
in the triangle BCD, cx: d:: bx:
=S.BDC
C
Then in the triangle ABC, 1 : : : b: bx
and in the triangle ABD, 1:x::c: cx=BD; and
50 39; whence DBC 42 16; let n
then nfd: BD cx, and ncx
fd
= 355.458.
S.DBC;
fd, and
310
}
E e 4
PROB.
424
B. II.
PLAIN TRIG.
Fig.
PROB. CXXVI.
71. Given the ſum of the fides of a triangle, and all the
angles feverally; to find the fides.
Let S.As, S B≈n, S.C=t, AB+BC÷CA≈b,
ACx. Then by Trigonometry, nxs:
tx
= CB, and ʼn :x::t: =AB.
SK
n
SX
n
+
tx
n
b, whence x =
n
nb
n + s + t
PROB. CXXVII.
And x +
72. In the right-angled triangle VAB, there is given the
perpendicular AB, the fegment VC, and the angle
VAC; to find CB.
Let AB = b, VCc, tang. VAC=t, BC=a.
Then by plain trigonometry, b: 1 :: a :
a
b
=
ta
a
tang. BAC, and (trig. viii.) 1 -
: I:
b
bt + a
:
tang. BAV.
Whence 1: b ::
bt + a
b. ta
bto
bbt + ba
:a+c=
b-ta
b - ta
and multiplying, ba + bc -
taa-tcabbt+ba, reduced aa + ca =
PROB. CXXVIII.
bc
bb.
t.
73. In the right-angled triangle ABC, BE EC, and
< ABD = CBD; and there is given BD and
<CAE; to find the fides.
Draw DF parallel to CB; then in the triangle
DFB, the angles at B, D are 45°, and BD being
given,
Sect. VII.
425
PROBLEM.S.
given, DF and FB its equal, are given; and fince Fig.
CE=EG, therefore DG=GF. Let DG or GF=b, 73.
S.DAGS, AF, then AG=bb+xx, AD
=√4bb+xx. And by plain Trig. AG(/bb+xx)
: rad. (1) AF (x):
:
(√66 + xx)
S. G
に
​bb
X
√bb + xx
=S.G. Alfo
: AD (√4bb + xx) :: S.DAG
(s): DG (b); whence
4
bx
√bb + xx
reduced ✰* + 5bbxx + 46ª = 0.
= s√4bb+xx,
bbxx
SS
PROB. CXXIX.
From the point B, to draw the lines BC, BD, BA, 74.
So that CD, DA, and the angles CBD, DBA,
may be given.
Draw CF perpendicular to AB, and put CD=b,
DA = c, CAs, and S.CBDƒ, S.DBA =d,
S.CBA = m. cof. CBA =n, and CB = x.
Then by plain Trigonometry 1:x::n: nx=BF,
=S.D, and d:c::
fx.cfx
:
b ba
But in the triangle CBA, (Geom. II. 23).
and b:f::x:
fx
b
BA.
ss = xx +
ccffxx
cfx
2nx X
which reduced is
b b d d
bd'
bds
√l bid + ccf — zbdjne
PROB.
426
B. II.
PLAIN TRIG.
Fig.
PROB. CXXX.
2
75. In an oblique triangle there is given, the baſe, and
perpendicular, and angle oppofite to the bafe; to
find the fides.
Draw AD perpendicular to CB; and put
AB=b, S.ACB=s, cof. ACB=c, perp. CF=1,
AC, CB=y.
In the triangle ACD, 1:::s: sw = AD, and
I:~::c:cx=CD. The triangles ABD and CBF
are fimilar, and y: b::p : sx = AD, whence
pbsxy. In the triangle ABC (Geom. II. 23.),
bb = xx+yy2cxy; but xy= and y =
therefore bb = xx + ppbb
pb
S
pb
;
SA
op: which reduced
2cpb
$5
S
bb pp
is w+
bbxx +
SS
2bcp
S
Otherwife,
Let AC+CB=x, AC-CB=y, the reft as be-
pb
fore; then AD=sxx+y, CD=cXx+y₂ =xx-yy.
S
Then in the triangle ABC, bb=x+yl² + x—-y|
2cxxx-yy; that is, 2x+2yy-2cxx+2cyy—bb,
X XX
pb
and putting --- for yy, we have 2xx+2xx-
$
apb
2cpb
2pb
2 CXX2CXN
=bb, or 4xx
S
$
S
2cpb
bb, whence x =
±bb +
x bp = 1,
S
25
51212
and y=465+
bp. Then AC=
25
717-12
BC =
2
Or
Sect. VII.
427
PROBLEM S.
Or thus,
Fig.
Let f cofine of the fum of the angles A, B; 75
v = cof. of their difference; the reſt as before.
Then (Trig. II. io.) v-f;s::p : 1b, and v-f=
2ps
225, and v = +f. Then the angles A and B
b
b
will be known, and confequently their oppofite fides.
PROB. CXXXI.
Given all the fides of a triangle, to find the center of 76.
the circumfcribed circle.
=
On the middle of AB, AC, erect the perpendi-
culars DO, FO, the point of interfection O, is the
center of the circumfcribing circle. From O draw
OI, OG, parallel to AC, AB; and put AB b,
ACd, S.CA=s, col. A = 6, AI = x, 10=y.
The <CGO CAB = OID; and in the right-
angled triangles OID, OGF, it will be y::
ccy = ID, and 1: y::s: sy = OD.
fo 1 : x::c: cx GF, and
I
FO. Then b-cy, and y=d-cx, and
x = 1½
cy = 1b — x = 1 cd-ccx, and x — ccx = bcd,
су
:
OD. Al-
:.
:x::S: SX =
1
d-cb
b-cd
b-cd
or ssx
whence x =
2
>
and y =
2
255
255
d-cb
b-cd
Therefore DO =
and FO =
Like
25
25
✓ bb + dd
2bcd
wife AO =
25
CXXXII.
PROB.
In the given triangle ABC, the anglès AOC, COB,
BOA, about the point O, are given; to find the
distances AO, BÓ, CO.
Produce CO to D, and BO to E.
ABb, ACd, CB=f, S.As, cof.
And let
A = c,
S.B = 4,
428
B. II.
PLAIN TRIG
Fig S.B=q, cof. B=n, S.AOE=g, cof. AOE=m,
S.AODb, S.DOB = p, AO=x.
77.
Then by Trigonometry, AB (b): S.O (g) ::
gx
AO (x) : &* = S.ABO, and
=
b
ggxx
I
= cof.
bb
ABO =y; and (Trig. I. 6. cor.) 83* + my = col.
b
= S.OAB. Alſo
mgx
OAB; and (I. 6.) gy — S.OAB.
b
(Trig. I. 6.) 588* + smy — cgy + cmgx = SCAO;
b
and (I. 6.) qy- gnx =S.CBO.
qy
=CO, and
And by Trigono-
b
metry, p:f::gy-31
gnxfqy
gnfx
b
P
bp
dsmy - degy
+
+
b
b
dcmgx
二
​fay
gnfx
and multiplying, sggdpx +
bb
P
bp
b:d::S.CAO: CO = sggdx
bb
bpdsmy-bpdegy+pdcmgx=bbfqy-gnfbx; and tranf-
pofing, bbfqy + bpdcgy—hpdsmy=sggdx+pdcmgx +
gnfbx, and y =
sggd+pdcmg+gnfb
rx
* = by fub.
bbtq+hpdcg—bpdsm t
88xx
ſtitution, that is, √√1_ggxx
I
rx
ggxx
and I
"
bb
t
bb
ggttxx = bbxx, reduced
rrxx
=
and bbtt
tt
bt
א
*
✔bb +88tt
Or thus,
78.
=
Make the angle BAF fupplement of BOC
and <ABF fup. AOC; through A, B, F de-
fcribe the circle AOBF, to interfect CF in O, the
point required.
Calculation
Sect. VII.
429
PROBLEM S.
Calculation. In the triangle ABF, all the angles Fig.
are given, and the fide AB, to find AF.
In the triangle CAF; CA, AF, and <CAF
are given; to find <ACF.
In the triangle ACO, there is given AC, and all
the angles; to find AO, CO.
PROB. CXXXIII.
78.
In the right-angled triangle ABC, there is given BA, 79.
and angle CBD; alſo AT=TD, and <ABT=
CBT; to find AC.
Let BA695.23, tang. DBC=t=T. 15°,
AT or DT-a, then AD=2a. By trigonometry,
a
b:a:: 1: tang.TBA, and b : 2ª:: 1 :
b
20
b
tang. DBA. Then (Trig. I. 2. Schol.) the tang.
1
zab
2TBA or tang. ABC =
Alfo (Trig. I. 8.)
bb-aa
2ta
2a
bt + 20
: I : : / +
b
6 6
tang.
2ta
2ab
bb-aa
ABD + DBC = tang. ABC. Whence
reduced 2a³-3btaa=b³t, and a=65.13,
bt+za
,
b-2ta
zab
and
xb AC.
bb-aa
PROB. CXXXIV.
Upon a horizontal plane, there ftands a tall pine-tree
leaning towards the fouth A man standing on the
north fide of it 50 yards from the foot, finds the
tree to fubtend an angle of 39 deg. Afterwards go-
ing direfly weft 73 yards, it fubtends an angle of
46°. What is the tree's length?
Let AC be the tree; E, F, the two ftations. 89.
Draw AB perpendicular to the horizon, and AD
perpen-
430
B. II.
PLAIN TRIG.
Fig. perpendicular to FC produced; draw BD, AC. The
30. triangles ABD, ABC, ADC, BDC, CEF, DAF,
BAE, are all right-angled. Put EF=6, CE=d,
CF=c, CD=y, tang. AEB=s, tang AFC=t.
By Trigonometry, It::c+y:c+yxt=AD,
whence ACV yy + tt × c + y. The triangles
FCE, BCD are ſimilar, and d:c::y:2 BC,
су
by
and d: y :: b :
= BD,
Then AB
11
d
2
ccyy
bbyy
yy + ttx c+yl
dd
tt × c + y²
dd
And in the
triangle ABE, d +
су
: I ::
d
bbyy
dds + csy
tt × c + y²
: s, whence
dd
d
bbyy
tt × c+y|²
; and fquared dess+2cddssy+
dd
ecssyy = dd × ttcc+2ttcy+ttyy — bbyy, and reduced
ccssyy + 2cddssy = ddttcc.
ddtt
+ zb
2cddit
d+ss
PROB. CXXXV.
1. Given two altitudes and two azimuths of a cloud in
motion; to find the point of the wind.
Let A be the firft, B the fecond place of the
cloud, O the place of obſervation, ABC the plane
of the cloud's motion; AB its line of direction.
Let AD, BE, CO be perpendicular to the hori-
zon, then DEO is equal and parallel to ABC,
and MDE is the path of the cloud on the earth.
Let OM be the meridian,
Put p=tang. AOD, q=tang. BOE, t=cotang.
DOE, OD = x. In the triangle AOD, 1:*::
p: px = ADBE; and in the triangle BOE,
9:
+
२
{
}
1
1
3
*
i
A
Fig. 64
65
66
MAA
F
67
}
B
B
68
W
C
69.
A A A
D E
B
B
70
B
D
D
73
A
71
C
D
E
74
B
C
B
75
:
A
F
B
F
76
B
F
78
A A 6
B
72
F
· PL.VII. pa. 430
B
Sect. VII.
431
PROBLEMS.
q: px:: 1 px -
II. 6.) Px
:
Fig.
=OE. In the triangle DOE, (Tiig. 81.
9
px
+ x
w::t:
p-
t = tang.
9
q
p + q
and the fum and difference of
2
ODE-OED
2
ODE, OED being had; ODE, and OED will
be known.
In the triangle ODM, there is given ODM and
DOM, therefore OMD is known, which is the way
of the cloud or of the wind.
PRO B.
CXXXVI.
On a clear day, the wind standing N. N. E. I obferved
a fmall cloud W. by S. whofe altitude was 41°, and
whilst the shadow of the cloud moved over 1230 yards
upon a horizontal plane, the cloud itself moved
through an angle of 9° 37′ as I obferved it with
an inftrument. What was the cloud's height?
Let E be the place of obfervation, CA the tract 83:
of the cloud, FG its projection upon the horizon;
AF, CG, BE being perp. to the horizon. Let BD be
perp. to ACD, and AK to ECK.
Let AC or GF=d=1230, S.DCB=c=5 points,
cof. DCB=n, S.GEC=b=41°, cof. GEC=s, tang.
AEC=1=9° 37; CG=x. By Trigonometry, in
the triangle CGE, b:x:: 1:
SX
XX
b
CE, and bi x
S: =CB, and in the triangle BCD, 1 :
SX
b
b
CSX
SX
nsx
=DB, and 1 :
: : 12 :
CD. Then DE
b
b
b
√xx+
ftitution.
CCSSXX
bb
X
√bb+cess = 1/P, by fub-
The
紙
​432
B. II.
PLAIN TRIG.
Fig.
82. x
The triangles CED and CAK are fimilar, and
(CE) : 2* (DE) : : (AC) d : pd = AK; and —
px
b
b
(CE) : 5nx (CD) : : d (AC): snd=CK,´ And in the
b
triangle AKE, snd+ *.(EK): 1 (rad.):: pd (AK)
tx
: † (tang. AEK), whence sndt+=td, and tx=
83.
bpd-sndth, and x =
bpd-sndth
bdp
sndb.
t
t
PROB.
CXXXVII.
In the triangle ACB, there is given AC, CB the
Segment BD, and the angle ACB; to find the
reft.
Draw AF perpendicular to CB; and put
'AC=a, DB=b, CB=d, S.ACD=s, cof. ACD=6,
and S.CDB=x.
bx
In the triangle CDB, d:x:: b:
S.DCB, and
ď
in the triangle CAD, ≈ : a :: s: =AD, then AB
as
=b+
as bx + as
But (Trig.
I. 5. cor. 1.)
Xx
sbx
dd
d
א
z=
I
X
c√1_bbxx
bbxx); and in the triangle ACF, 1:a
dd
cof. ACB=cz- (putting
sbx
d
bsx
absx
:: CZ ---
: cza
=CF. But (Geom. II.
d
d
bx + asl²
absx
23.)
aa + dd - 2d × caz
and
XX
d
multi-
Seat. VII.
PROBLEM S
433
multiplying by xx, bbxx + 2absx + aass = aaxx + Fig.
ddxx2dcaxxz +2absx³, and tranfpofing
2dcaxxż=2absx³+aaxx—2absx—àass, and reſtoring
+ dd
bb
%, and fubftituting for the known quantities,
2dacxx
bbxx
dd
= px³ + qx² -rx-t, and
fquaring, 4ddaaccx4 — 4aabbccx6 =
p²x² + 2pqx5 — 2prx4 2ptx³ •2qtx
- 2gr + rr
+ qq
+artx + tt.
83.
}
Ff
j
SECT
434
B. IL
Fig.
SECT. VIII.
Problems in Spherical Trigonometry.
84.
PROB. CXXXVIII.
Given the latitude of the place and the fun's longitude;
to find the afcenfional difference.
L
e
:
ET b S. greateſt declination, S. fun's
longitude AD, x = S. declination BD, ✰ =
tang. latitude PCH.
Then in the r<fpherical triangle ADB, rad.
(1): S.AD (c) :: S.A (b): bc = S.BD, and (Trig.
I. 1. fchol.) tang. BD =
bc
✓ 1-bbc c
; and in the tri-
angle CBD, rad. (1): co-tang. C (t) :: tang. BD
bc
bct
x =
bb66
(V)
:) : S.CB (~), and ≈
PROB.
CXXXIX,
84. Given the fun's declination, and the fum of the latitude
and amplitude; to find each of them.
Let a fine of half the ſum of CD and PH, b=
the cofine, d=S. declination, s=S. fum of CD and
PH, x=S. half their difference, y=S. whole diffe-
rence. Suppoſe PH greater than CD. Then
(Trig. I. 6. fchol.) b√1-xx-ax cof. PH; and
a√1-xx-bx = S.CD. But in the triangle CBD,
col. C (b√1−xx — 4x) : S,BD (d) : : rad. (1):
---
S.CD
Sect. VIII.
SPHERICAL TRIG. PROBLEMS.
435
S.CD (a√1-xx
aax√1-xx
bx.
or ab— x√1-xx = d
Whence ab × 1-xx
bbx√ 1-xx + abxx =d;
(becauſe aa + bb = 1),
But (Trig. 1. 2. fchol.) 2x VI-xxy, and 2ab
=S. Whence s-y 2d, and y = 5 ---
PROB. CXL.
-2d.j
Fig.
84.
Given the fun's altitude at fix, and also when west; to 85.
find the latitude.
R is the fun's place when weſt, and O at fix
a clock.
Lett
S.RC the altitude weft, s=S.OI
the altitude at fix, x=S. latitude.
In the triangle CIO, S.Ç (≈) : S.OI (s) : : rad.
S
(1) : — = S.CO the declination. In the triangle
X
DCR, S.C (~) : S.DR (~~) : : rad.
S
:: rad. (1): S.CR
(t), and tx= or xx = and x =
PROB.
CXLI.
AEC, BCD are two triangles right-angled at E and 86.
D, and ſtanding on the great circle ECD; alſo
AC=CB, and EC, CD, and the angle ACB are
given; to find the angles and fides.
=
Let atang. DC, b≈tang. CE, s=S. half the
fum of the angles BCD, ACE; c cofine, x =
fine, y cofine of half the difference. Then
sx+cy cof, leffer ACE, and cy-sx cof. greater
cy—sx
BCD. And in the triangle ACE, cy + sx : 1 ::
b
cy of sx
1 : : b :
tang. AC. And in the triangle BCD,
Ff2
Cy
436
B. II.
SPHERICAL TRIG.
Fig.
a
86.
су SX : I :: 0 :
tang. CB. Whence
cy-sx
a
=
b
cy-sx cj+5x
and acy+asxbcy-bsx, and asx
x
+bsx-bey-acy, and
y
=
bc-ac
= tang, half the
bs + as
difference of the angles BCD, ACE; whence the
angles themſelves are had.
PRO B.
CXLII.
87. Given the fun's amplitude, and altitude at fix; to find
the latitude, and declination.
Let P be the pole, Z the zenith, CB the ampli-
tude, AP the altitude at fix.
Let S.CB=b, p=S.AP, x=S. lat. PO, y=S.
twice the latitude. In the triangle CBD, rad. (1)
: S.C (√ï-xx) : : S.CB (b) : b√1−xx = S.BD
or AC. And in the triangle CAP, rad. (1):
S.AC (b√1−xx) :: S.C (*): S.AP (p); there-
fore bx√1-xx = p, and 2xV1−xx =
20
but y = 2x√1-xx; whence y =
20
b
; then x will
be known, and -xx, the declination.
PROB.
CXLIII.
88. Given two altitudes and two azimuths of the fun; to
find the latitude.
Let Z be the zenith, P the pole; S, O two
places of the fun. Let s, f-fine and cofine of
ZS; p, q=fine and cofine of ZO; m=cof. PZS,
Then
n=col. PZO ; x, y=ſine and cofine of PH.
(Trig. II. 38.) cof. SP = sym + fx, and cof.
OP=
Sect. VIII.
437
PROBLEM S.
OP = pyn+qx. Whence sym+fx=pyn+qx, and Fig.
88.
fx — qx = pyn — sym; therefore
x
pn-sm
y
j-q
tang. PH the latitude.
PROB. CXLIV.
Given the latitude of the place; and the fun's altitude 89:
is equal to his azimuth from the fouth, and equal to
the bour from noon; to find any of them.
Z is the zenith, P the pole, ZP is given, and
<ZPO=AZO=DO. Let s S.ZP, c=cof. ZP.
y = S.ZPO = S.AZO.
y : √i-yy : : * : √1-yy
OP. But (Trig. III. 38.)
Then 1-yy
√1-yy = S.ZO,
S.OP, and y = cof.
sv1-yy × √1-yy +
cy=y, or s syy + cy=y, and syy+y=s, or
yy+y=1.
PROB.
- cy
CXLV.
There are two places, whofe latitudes are the comple-
ments of each other to 90°, and the fun's declina-
tion being given, he riſes an hour ſooner in one place
than the other to find the latitudes.
1
Lett tang. declination, & tang. afcenfional 87.
difference, xtang. one latitude, then tang.
I
X
the other latitude. In the triangle CDB, rad (1) :
cotang. DCB (x) :: T.DB (t): tx S.DC the
afcenfional difference in the firft latitude, and I :
I
-:::
*
the afcenfional difference in the
other latitude. But (Trig. I. 9.) 1 + #: 1 ::
Ft
FI 3
tx
438
SPHERICAL TRIG. B. II.
Fig,
87.
t
t
tx 1 : b, and tx —
=b+btt, txx-tbx+
X
X
bttx, and xx
btx = 1.
b
t
*
PROB.
CXLVI.
The ftile of an horizontal dial being turned down, fell
upon the hour line of 8; query, the latitude it was
made for.
Lett
tang, of 4 hours or 60°, anfwering to
8 o'clock, x=S. latitude; then
x
V. 1-XX
tang.
latitude hour angle of 8, by the queftion.
Whence by the known proportion of dialling,
१०.
X
I : x :: t :
.
and tx =
or
I-XX
I-XX
I
WI
1-XX
44
= cof. lat.
1-xx = 1, and ✔
I-XX
PROB.
CXLVII.
To find in what latitude, an erect fouth declining dial
may be made, so that the declination of the plane,
the distance of the fubftile from the meridian, and the .
ftile's height, are all equal.
Let ABC be the right-angled fpherical triangle,
in which are found all the requifites; viz, AB =
co-lat. <A ≈ co-declination, <B = plane's dif
longitude, CB ftile's height, AC fubftile's di-
ftance from the meridian,
=
Let x S.AB, y S.BC. Then (by the pro-
perties of right triangles) : tang. BC: cotang. A
S.AC, or S.AC tang. BC x cotang. A. But
by
Se&t. VIII
439
PROBLEM S.
X
by the queſtion, BC AC comp. A. Therefore Fig.
tang. BC or cotang. A =
y (S.AC) =
y
90.
Whence
I-yy
y
(tang. BC x
1
1-yy
L-yy
cotang. A) =
yy
>
and 1-yyy, or gy+y=1,
1
yy
whence y =
√5-1
2
1
cof. AC::
cof. AC x
1-yy × √ 1—yy
y:
And fince AC
1-xxy.
Again, in the fame triangle, I
cof. BC: cof. AB, whence cof. AB
cof. BC; that is, 1-xx
= Iyy, therefore
=BC, therefore <B<A. Hence all thefe five
are equal; 1. Plain's declination. 2. Diſtance of
the fubftile from the meridian. 3. Stile's height.
4. Latitude of the place. 5. Comp. of the plane's
diff. longitude; and each of them twice the
fine of 18°.618034; whence the latitude and
declination 38° : 10′½•
PRO B.
CXLVIII.
Given the fun's meridian altitude, and alfo his altitude
at two; to find the latitude.
=
Then (Trig. I. 6.
= cof. PO, ay—bx
And in the triangle
Let Z be the zenith, P the pole; B, O two
places of the fun. Let a, b fine and coline of
BZ (PO— PZ); x, y = fine and cofine of PO +
PZ, c = cof. P, d = cof. ZO.
fchol.) ay+bx=S.PO, by — ax
=S.PZ, by + ax cof. PZ.
OPZ (Trig. III. 38.) ay+bx × ay—bx × c—by—ax
Xby+ax=d; that is, canyy—cbhxx+bbyy—aaxx=d,
but yy 1-xx, therefore caa-caaxx-cbbxx+bb
=
-bbxx-aaxxd, but aa+bb1, whence caa --
cxx+bb-xx=d, and xxcxx-caa+bb-d, and
F f 4
91.
1
440
B. II.
SPHERICAL TRIG.
Fig.
caa + bbd
x = √
Whence PO (or PB), and
91.
I
с
PZ are had.
PROB.
CXLIX.
92. In the fpherical triangle VAC, there is given the per-
pendicular AB, the angle A, and the baſe VC's to
find the fegments.
Put b
* =
10.
S.AB, t=tang. VAC,
tang. BC. In the triangle BAC, b:::*:
tang. VC,
1
1 :
= tang. BAC, and (Trig. I. 9.) 1 +cx :
t =
C
X
X :
I + cx
x tb X
bb + tx
tx
tang. VB, and I + : I:
Ъ
tang. VAB. And in the tri-
tb
C
X.
:
Whence
b + tx
1 + cx
angle VAB, 1:b::
C
x
tb
=
i + cx b + tx
Ђс
b. And multiplying,
bx+tcx-txx=tb²-bx+bbctx-cbxx, and re-
—
duced, tbc.xx + bbct — tc.x = bc tbb.
PROB. CL.
Travelling in an unknown part of the world, I found
by chance an old horizontal dial, whofe hour lines
were fo decayed by length of time, that I could only
difcover thofe of 4 and 5; whofe distance I found
juſt 21 degrees; to find the latitude of the place.
Let b tang. 60 the hour arch of 4.
d
tang. 75 the hour arch of 5.
t
tang. of their difference 21.
* = S. latitude.
Then
Se&t. VIII.
44!
PROBLEMS.
Then by the known proportion of dialling, rad. Fig.
tang. hour arch: tang. hour angle, that is,
S. lat.
1 :x :: b: bx = tang. hour
I
I:x::d: dx = tang. hour
But (Trig. I. 8.) 1—btx : 1 :: t + bx
of 4.
of 5.
dx, whence
t + bx = dx — bdtxx, and bdtxx + b = d.x +t=o,
2x = -.383864.
In numbers 2.48133**
and .49084 S.29 24 the lat.
x
pr x = .31518 = S.18 22 the lat.
PROB. CLI.
There are given the latitudes of three places lying in the
arch of a great circle; and the diff. longitude are
equal, between the middle one and each of the ex-
tream places; to find their diftances.
Let P be the pole, AE the equinoctial; G, V, 93,
M, the three places. Put b-tang. MD, c=tang.
VB, dtang. CG, ≈≈ S.AB, y S.CB or BD.
Then (Trig. I. 5.) x√1-yy +y√1−xx
and (Trig. I. 6.) x√1-yyy√1-xx
XV
≈:c
S.AD;
S.AC.
Then (Trig. III. 27. cor. 1.) x : c : : x√ I −yÿ
+√1-xx: b; and x: c :: x-yy — y√I—xx
bx
: d.
Whence
x1-jy + y√1−xx, and
ૐ
dx
= x√1–93 — 3√1−xx.
Then adding and
C
'bx+ dx
C
fubtracting theſe laft equations;
bx-dx
2xV1-yy, and
C
b + d
mer √1-yy =
20
21-xx by the for-
2
મ
= cof. CB or BD. Hence
y is known. Therefore in the triangles GPV, VPM,
twa
442
B. II.
SPHERICAL TRIG.
}
Fig. two fides and the included angle are given, to find
93. GV, VM, the diſtances required.
94.
Or thus,
I
ปี
(By Trig. III. 44. cor.) As rad: tang. PV ::
tang. GC+tang. MD : cof. GPV or MPV.
PROB. CLII.
In the spherical triangle ABC, we have given the an-
gles ADC, CDB, BDA, about the point D; and
the triangle ABC itſelf; to find the distances AO,
BO, CO.
Lets, c be the fine and cofine of A; p, q=fine
and cof. B, a=S.CDB, ƒ=S.CB, b=S.CDA, d=
S.AC, b=cof. AB, #cof. ADB, x, v=fine and
cof. DAB; y, z= fine and cof. DBA.
Z
Then (Trig. I. 6.) sv-cx=S.CAD, and pz---
gy=S.CBD, and in the triangles CAD, BAD, b:
dsv-dcx
d: : sucx :
= S.CD; and a:f::pz-
b
adcx=
gy:
fpz-fqy
a
bfpz — hfqy.
=S.CD; therefore adsu
In the triangle ADB (Cafe ro. Trig.) bảy0z=
But v = VIXX, % = √I
Ι
KK, Z
adsvi XX adcx = bfpvi
yy, therefore
ولا لا
gy bfqy.
and bxy-√xx × √1 — yy = n.
X I --
I
From theſe two equations, the roots may be
eaſieſt found by problem xcv; otherwiſe it will
afcend to a high equation: or if you pleaſe you
may proceed by rule 5, prob. xcii.
PROB
*
}
1
}
I
!
1
Fig.79,
D
T
F
A
80.
E
B
A
C
D
K
A
82.
83
D
*
F
B
2
E
Z
85
E
R
E
D
C I
z 88
H
B
{
E
81.
F
2
84.
C
H
B
A
D
B
86
D
27
P
87
2
P
B
P
89
2
90
B
P
H
A
91
C
D C
HA
P
A
92
93
M
V
B
A
C
C
A
B
E
D
B
94
B
1
P1.VIII. på..440
つ
​Sect. VIII.
443
PROBLEM S.
PROB.
CLIII.
Fig.
Given the difference of the azimuths of three known
Stars; to find their altitudes.
Let Z be the zenith; A, B, C, the ſtars. Since
95.
their places are given, the triangle ABC will be gi-
ven. Put s, fine and cof. ABC, a ≈´S.AB,
b=S.BC, n=cof. AC, q cof. AZC, d=S.AZB,
CBZ-ABZ
C
e=S.CZB; x,y=fine and cof.
2
Then
sy+cx=S.CBZ=v, and sy-cx=S.ABZ=z. And
in the triangles ABZ, CBZ, d: a :: z:
b
az
d
= S.CZ ; and in the tri-
ve
S.AZ, and b : e : : v :
angle AZC (Trig. III. 38.)
x √1vv = n; and
vv=n;
qazve
dò
tranfpofing,
+
IZZ
-ZZ X
qaz ve
I
טים.
-vv=n
; a
bd
and fquaring, I
ZZ -
vv + VVZZ - NN
vuzz
2nqae
bå
qqaaee
Vz +
v²x²; and
bbat
tranfpofing, I- nn — zz + vv —
2ngae
VZ +
bb dd
vz = ssyy·
bd
qqaæee-bbdd ¿²x², but z² + v² = 2ssy²+2ccxx, and
ร
CCXX. Therefore
-nn2ssyy+2ccxx
2nqaess
2nqaecc
qqaaee
yy +
xx +
bbdd
X
bd
bd
bbdd
$434
2
'<s² c²x²y² + c+x+. Let 1
2nqae = f, 2+ 2nqae
bd
bd
=g
2
then mmfssy + ccgxx+ps4y+~~2 pc²s*y²x² + pc + x^,
but yy-xx; therefore expunging y, and redu-
cing, px4 +gcc -ƒss — 2pss × xx=mm -fss—ps"。
PROB.
nn = 141 117,
q q z ze e
P:
bbat
D
444
SPHERICAL TRIG. B. IL
1
Fig.
96.
PROB.CLIV.
Given the fun's declination, the difference of two
altitudes, the difference of azimuths, and the dif-
ference of times; to find the altitudes, and the la-
titude,
Let P be the pole, Z the zenith; A, B, the
places of the fun. Put d S. fun's declination,
s=S. APB, c=cof. AZB; p, q=fine and cofine
BZ-AZ
2
; x, y fine and cof.
BZ+AZ
2
; then
(Trig. 1. 6. fchol. 2.) qx+py S.ZB, qx-py=
SZA, qy-px=cof. ZB, gy+px=cof. ZA. Then
in the triangle API, 1: d::s: ds S. half AD,
and (Trig. I. 2. fchol.) 1-2 ddss cof. AB; and
in the triangle AZB (Trig. III. 38.) qqxx-ppyy Xc
+qqyyppxx = 1-2ddss; but pp+qq1, and
xxyy1; therefore cqqxx-cpp + cppxx+qq-
qqxx-ppxx=1—2ddss, or cqqxx + cppxxqqxx-
ppxx = 1-2ddss+cpp-qq; that is, cxx-xx =
1-2 ddss+cpp-1+pp=cpp + pp2ddss, whence
Then the fides ZA, ZB
X = T
c+ 1.pp
2ddss
are known. In the triangle PAI, find the <A;
and in the triangle ZAB, find the angle A, and
their difference ZAP; then in the triangle ZAP,
there's given two fides and the included angle A;
to find ZP, the co-latitude.
PROB. CLV.
Given two altitudes of the fun or a star, and the times
of obfervation; to find the declination, and latitude.
Let Z be the zenith, P the pole; B, A, the
97 places of the fun or ftar. Let
= cof. BZ,
f =
Se&t. VIII.
445
PROBLEM S.
f=col. AZ, b=cof. ZPB, d=cof. ZPA, x=cof.
BPZP, y cof. BP + ZP.
I
Then (Trig. III. 42. cor. 1.) x-y: 2 :: x-c
: 1-5, and x-y: 2 :: x-f: 1-d; therefore
x-c: 1-b:: x-f: 1-d; therefore x-dx-
+cd=x-f―bx+bf; and bx-dx=c—f+bf—cd,
bf-cd+c-f. Allo 16xx-y
whence x =
b-d
=2x-2c, or by-yxbx
I- b
x+bx-20 20 1+b bf—cd+f—c
2c, and y=
=
x=
bI
1-6
I-b
bd
CLVI.
Fig.
97.
PROB.
Given two altitudes of the fun, the difference of times,
and difference of azimuths; to find the latitude and
declination.
Suppofe P the pole, Z the zenith; B, A, the
98
places of the fun. Put c=cof. BZA, s=cof. BPA,
y=S.BP or PA; b, d=fine and cof. ZB; ƒ, p =
fine and cof. ZA.
×
Then (Trig. III. 38.) in the triangle BZA,
bcf + dp = cof. BA; and in the triangle BPA,
syy + √I—yy XV y cof. BA; therefore,
syy+1—yy=bcf+dp, and s—1.yy = bcf + dp — 1,
=S.BP the declination.
and y =
1-bcf-dp
I-S
Then in the triangle BPA, find the angle B, and
in the triangle ZBA find the angle B, and then
the difference ZBP. Then in the triangle ZBP
there are given two fides ZB, BP, and the included
angle B; to find ZP, the co-latitude.
PROB.
446
B. II.
SPHERICAL TRIG.
2
Fig.
98.
PROB. CLVII.
Given the fun's declination, two altitudes, and the time
between them; to find the latitude.
Let a, b fine and cof. PA; d,´f=fine and cof.
PB; n=cof. ZB, m=cof. ZA; s, c=fine and co-
ZPA+ZPB
fine and cof.
Z
fine BPA; y, z
v, xfine and cof. ZP.
;
2
Then (Trig. I. 6. fchol.) cz-sy cof. ZPA
and cz+sy cof. ZPB, and (Trig. III. 38.) aczv
—asyv + bx=m, and dezv+ dsyv +fx, and
av
n-fx and by
adding and fubtracting, 2cx = +
•
m-hx
CZ sy =
and cz + sy =
du
n-fx
m-bx
an-afx+dm-dbx
adv
alfo 25y =
dv
n-fx
av
m-bx
do
an
afx
dm + dbx
adv
an+dm-afx-dbx
2cadv
and y =
Whence z =
an-dm+dbx-afx
2sadv
an + dm
an dm
Put
=P,
2cad
2sad
q,
db + af = 1
=r,
2cad
¿b-af
p-rx
=t. Then z =
and y =
q+tx
2sad
V
บ
But
yy, and v√xx; then
vi
XX
then ✔I-yy
p-rx, and
and I
D-rx\
Σ
yy =
and tranſpofing
Z
VV
D=rx1
Jy = I-
ซบ
พบ
= I-xx-P-rx |²
And yy =
q + tx
Sect. VIII.
447
PROBLEM S.
2
q + tz
VV
Whence 1-xx—pp+2prx—rrxx=qq +
2qtx+ttxx; and reduced
Fig.
9.8.
ttxx + 2tqx + qq = 0;
+ rr
+ I
2pr + PP.
I
Or thus,
In the triangle BPA, 2 des and the included
angle are given, to find <B, and BA. In the
triangle BZA, all the fides are given, to find the
angle B, and from thence ZBP. Then in the tri-
angle ZPB, two fides and the included <B, are
given; to find ZP the comp. of the latitude.
PROB.
CLVHI.
Given three altitudes of the fun in one day, and the times,
between them; to find the latitude, &c.
Lets, a fine and cof. APB; t, b = fine and 99.
cof. ABC, d=cof. AZ, ƒ=cof. BZ, g=col. CZ,
≈ — S.ZPA, y=S.AP, BP or CP; ≈≈ S.ZP.
z
Then (Trig. I. 5. cor. 1.) c√1—xx — sx = cof.
c№ XX
ZPB, and bi-xx-tx-cof. ZPC. And (Trig.
III. 38,) zyVi-t vì vị gia
zy√1 xx + √i ZZ X
czyvi
I -XX szyx + I--ZZ X xi-y=f, and
byz√ I~~xx + tyzx +√ ZZ -yyg. By
tranfpofition, I-22 XVI
Vi
-yy=d—zyvi
[--XX.
Then czy-xx-szyx+d—zvvxx=f, and
bzy✓i—xx—tzyx+d—zy√1−xx=g. From the
former of theſe two laft equations we get zy =
d-f
I
xx + sx
d-g
I-W 1−xx + tx
and from the latter y
and making theſe laft equa-
1
tions
*48
B. it.
SPHERICAL TRIG:
Fig.
99.
Xx
I-XX
tións equal, and reducing, we have
d-fx i-b-d-gx1-0
d-gxs-d-fxt
tang. ZPA. Then
will be known, and alſo ży.
But Izz X √T—yy=d—ży√1—xx.
zy=r, then
Put
yy—żż+zzyy—dd-2 dzy√1—xx
+zzyy — zzyýxx, and tranfpofing, yy+zz-1-dd
+ 2dr√1−xx + rrxxp by fubftitution; ther
jj+2yz+zz=p+2r, and yy-2ýz+zz = p-21;
and y + z =
2r, and y 2= P 2r,
whence y =
p + 2r + √ p— 2r
2
and z =
p+2r-√p—27
2
PROB. CLIX.
Having at one inftant the altitudes of two known ftats &
to find the latitude.
100. Let Z be the zenith, P the pole; F, A, thê
ftars. In the triangle APF, there is given the
fides AP, FP the co-declinations, and angle P, thẻ
diff. right-afcenfions, to find AF, and <F. Then
in the triangle ZFA, all the fides are given, to
find the angle F; then <ZFP will be known.
Let c=cof. ZFP; a, b fine and cof. ZF; d, f=
fine and cof. FP; x=cof. ZP. Then (Trig. IH. 38.)
adc + bf = x = S. latitude.
PROB. CLX.
If there be two known ftars in one azimuth, and having
the altitude of either given; to find the latitude of
the place.
101. Let Z be the zenith, P the pole; F, A the
two ſtars in the azimuth circle ZFA, and ZF is
1
given
Sect. VIII.
449
PROBLEM S
{
given. In the triangle APF, two fides and the in- Fig.
cluded angle P, are given; to find the angle F. 101.
Put c cof. <ZFP; a, b≈fine and cofine of ZF;
d, ffine and cofine of FP; x = cof. ZP. Then
in the triangle ZFP (Trig. III. 38.) adc + bf = x
the fine of the latitude.
If ZA is given, you must find the
and put a, b S. and cof. ZA; d, f
AP, &c.
Otherwife thus,
Let the altitude of A be given; s, c
cof. APF; d, f
angle FAP,
S. and cof.
fine and cof. AP; m,
fine and
n = fine
x = cof.
and cof. FP, a, b fine and cof. ŻA,
ZP. Then (Trig. cafe 7. fpherical triangles) co-
dn-mfc
tang. A =
and tang. A =
ms
ms
dn-mf.c
by fubftitution; and (Trig. I. 1. fchol.) cof. A =
I
Vitit
And (Trig. III. 38.)
the S. latitude.
This is a uſeful problem.
ad
+ bf = x
Vi+tt
$
##
#
#
!
G g
SECT.
450
B. II.
Fig.
SECT. IX.
Geometrical Loci, and Problems relating thereto,
line
102. TF the right line AP be drawn from a given
If
point A, and any number of right lines PM,
PM, &c. be drawn thereon, parallel to one ano-
ther, or making any given angle with AP. And
if the relation of the indetermined quantities AP,
PM be denoted in general by fome equation; and
if the lengths of PM be every where, fuch as that
equation gives; then the curve paffing through
all the points M, is called the Locus of the points
M, or the locus of that equation. And that equa-
tion declares the nature of the curve MM.
The degrees of the Loci are denominated from
the degrees of the equations, by which they are de-
noted. Thus a locus, of the first degree is that
where the indetermined quantities rife to one di-
menſion; of the fecond degree, when they ariſe to
two dimenfions; of the third degree, when they
rife to three dimenfions, &c.
Right lines are faid to be given in pofition, when
they make given angles with one another.
PROB. CLXI.
103. If the pole AC or BC revolves about the center C,
and the weight D, and ſtring BD hangs at the end
of it; to find the nature of the curve GD, defcribed
by the weight D.
Take AG, and CF,
BD. Then fince CF,
BD are equal and parallel, therefore (Geom. I. 5.
cor.
Sect. IX.
451
PROBLEMS of the Loci.
cor. 3.) CB, FD, are equal, or FD = CB=CA. Fig.
And fince FCGA, add CG, and then FGCA; 103.
whence FD=FG. Therefore GD is a circle whofe
center is F, the fame with AB, but in a lower po-
fition.
PRO B.
CLXII.
Suppofe ACD, acd, &c. to be right-angled triangles, 104.
one of whofe angles falls upon the fixed point A, the
other in the line AE; and if the fegments BD, bd,
be given; to find the nature of the curve paffing
through all the right-angles C, c, &c.
Let AB=x, BC=y, BD=a; then in the right-
angled triangle ACD; AB (x): BC (y) :: BC
(y): BD (a); whence ax = yy, for the nature of
the curve. Therefore the curve Cc is a parabola,
whofe latus rectum is BD or bd, and A the
vertex.
PROB. CLXIII.
A is a fixed point, AB a given line, ABD a given 105.
angle; then ſuppoſe the curve AMB to be generated
after fuch a manner that drawing any line AC, it
may be, as BC to BP :: as r: to s to find the
nature of the curve, or the locus of all the points M.
Draw MP, mp parallel to DB, and let AP =%,
PM y, AB = a, Then by fimilar triangles AP
(x) : PM (y) : : AB (a) : BC =
the problem, r: s :: BC (4)
say
Therefore =ra
ra rx,
X
*
GN
A
And by
: BP (a-x).
and y == Xax-xx, for
sa
the nature of the curve; and it paffes through A;
becauſe when x is
o, y iso.
G g 2
1
!
PROB.
452
PROBLEMS of B. II.
Fig.
PROB.
CLXIV.
2
106. Given the triangle ABC, and drawing PD parallel
to BC; ſuppoſe it be always PM PD-BC2;
to find the nature of the curve BM.
Put AB=a, BC=b, AP=x, PM=y, then by
bx
fimilar triangles, a b x: =PD, and by
a
bbxx
the queſtion,
-bb-yy, or aayy-bbxx-bbaa,
aa
b
and y =
xx-aa.
And the curve paffes
a
through B.
PROB. CLXV.
107.The triangle ABC is given being right-angled at B,
and drawing FM parallel to CB, it be every
where PF2+ PM
2
—
BC; to find the nature of
the curve BM.
Let ABa, BC=b, AP=x, PM-y. Then
bx
by fimilar triangles, a:b: :x :
=PF. And
a
bbxx
by the queſtion,
bbxx
b
and
aa
aa
✓aa-
y =
the points M.
or AD = CB.
a
aa-xx, the equation for all
And in A, where x is o, y = b,
+yy=bb, whence yy=bb-
PROB.
Sect. IX.
453
the LOC I
Fig.
PROB. CLXVI.
If AB, CF, AC be right lines given in pofition; 108.
and PD, pd be always parallel to AC; and if
PM be every where equal to CD; to find the
locus of the point M.
Since AC, PD are parallel; AP will be to CD
in a given ratio (Geom. II. 12. cor. 2.); put
AP=x, PM=y, and let a b: AP (x) :
bx
CD; therefore y=
bx
a
a right line, paffing through A.
:
Therefore AMm is
PROB. CLXVII.
If the three lines CA, CB, AB be given in pofition; 109.
and PM be always drawn parallel to AB; and it
be every where AP×PD=PM²; to find the nature
of the curve Mm.
Let CA=a, AB=b, AP=x, PM=y, then by
fimilar triangles, a:b::a+x: b = PD, then
a + x
a
a + x
by the question,
bx
=
yy,
yy,
and
and y =
a
b
a
Xax + xx; and the curve paffes through A,
ſince both x and y are o, at once.
But if CB be parallel to CA, then PD
and bxyy, or y=x.
AB,
And if C lie on the
other fide of A, then PD = ^=-x k, and y =
X ax
XX.
a
a
In which cafe, when x Ba
then yo, and the curve paffes there through
the axis C.
G g 3
When
454
PROBLEMS of B. II.
?
*
Fig. When is greater than a, y is the fquare root
109. of a negative quantity, which is impoffible; and
therefore the curve goes no further than A.
PROB. CLXVIII.
110.There is given the right-angled triangle ABD; and
drawing PM, pm" always parallel to BD; and
making PM every where equal to BF; to find
the nature of the curve DMm.
Put AB=a, BD=b, BP=x, PM-y.
by fimilar triangles, ab: a +x :
<="PF, and BF² = xx
aaxx+bbaa + zbbax + bb xx
aa
Then
a + x
b
Q
bb × a + x²
aa
;
therefore
266x
aa+bb
y = √ bb +
+
xx, for the nature
aa
of the curve.
1
PROB.
CLXIX.
111. BA is a given line, draw PM perpendicular to BAP;
and let AM be always a mean proportional between
AB and AP; to find the nature of the curve
AmM.
Let ABa, APx, PMy, then AM =
'xx+yy, and per queſt. a: √xx+yy:: √xx+yv
x, and ax = xx+yy, whence y = ax--xx.
PROB
}
},
I
}
1
1
T
#
Fig. 95
Ꮓ
96
B
Z
B
1
A
१
B
+
Z
97
B
A
Z
D
98.
A
99.
F
B
100
C
F
Z
101
104
C
M
M
P.
M
A P P P
A
B
A DE
102
t
拾
​103
105
MC
D
D
172
M
106
$
PP B
В Р Р
C
B
M
#
Τ
P
107 A
108
m
M
DI
F
109
C
M
M
A P
P
B
B
D
D
a
Pl. IX. pa. 454
N
OF
7
Sect, IX.
455
the LOC I.
Fig.
3
PROB. CLXX.
The line CF and point A being given; from any 112.
point D in that line, through A, draw DAM,
and make DA´× AM always equal to a given
Square; to find the locus of M.
x
Draw BAP perpendicular to CF, and PM per-
pendicular to AP, and put AB = a, AP = ≈
PM=y, then AM√xx+yy, bb the given
ſquare. By fimilar triangles x : √xx+yy :: a :
a√xx+yy =AD. Then per queſt. a
X
x xx+yy
X
bbx
bbx
bb, and xx+yy=
>
whence y=
-xx.
20
a
PROB. CLXXI.
AD is a circle, C its center; draw AB perpendicu-113.
lar to CAP. Then draw any line CB, and BM
parallel to AP; and make always BM = DB; to
find the nature of the curve mM.
=
Draw MP perpendicular to CP, and let CA=r,
AP = x, PM ≈y, then CB√rr+yy, and
BD = √rr+yy—r. But AP or BM BD, that
is, x =√rr+yy—r, and √rr+yy =r+x, and
fquaring, rr+yyrr + 2rx + xx, whence y
✔2rx
2rx + xx.
PROB.
Gg.4
456
B. II.
PROBLEMS of
Fig.
114.
115.
PROB.
CLXXII.
CAD is a given angle, CD a given line, M a given
point in it. Let this line fo move in the angle
CAD, that the ends D, C may always touch the
fides AD, AC; to find the curve defcribed by the
point M.
Draw MP, MB parallel to AC, AD, and put
DM a, CM, cof
=
<Ac, AP = x,
PMy. By fimilar triangles, CM (b) : BM (x)
:: DM (a): DP =
ax
But in the triangle
b
aaxx 2acxy
= ao
b
MPD (Trig. cafe 5.) yy +
or yy=
aa 2acxy anxx
+
>
bb
the curve.
PROB.
bb
for the equation of
CLXXIII.
The line CA is perpendicular to AP, ABM is a ſquare.
whofe fides CB, BM are given; to find the curve
defcribed by the point M, whilst the B, and end
C, move along PA, AC.
Draw MP perpendicular to AP, CB=a, BM=b,
APx, PM y; the triangles CAB, BMP are
cy
fimilar, and by::a:
AB. And BP =
{
b
ay
✔bb
bbyy, therefore
ō + √ bb →→yy=x, and
√bbyyx,
1
ay
bb_xy = x —
b
and fquaring bb-yy — xx --
2axy naviy
aa + bb
+
And reduced
h
66
bb
vy = bb
2ayx
f
XX.
PROB.
Sect. IX.
the LOC I.
457
PROB.
CLXXIV.
The lines AV, AB, AE are given in pofition, the
point V is given; if the line VE be always drawn
through V, and the part intercepted CE be divided
in a given ratio at M; to find the locus of the
points M, m.
Draw MP parallel to AB, and DM, BE
pa-
rallel to AP, and put AP≈≈, PM=y, AV—a.
r to s as CM to ME, S. <BAE=p, S. <EAP
or BEA = q.
+
The triangles VAC, VPM, DCM, BCE are
+x:x::a:
ay
a+x
Fig.
116.
fimilar, and
= AC, and a+x
y: (DM) x:
xy
a+x
- DC. Alfo r:s:: CM
xy
: ME::CD
: DB =
a + x
sxy
ra+rx
And CM
r+s
:CE::r:rts
r
angle BAE,
: BA =
r+s
sxy
qx = y +
'
and r+sx qx × a + x
pr
rxa +x
xBE, and in the tri-
да
BAE, p : 9 :: BE ("+s
p: q: : BE ('+ x)
):
= pry × a + x + psxy, which reduced is (putting
r+s=d) dqxx + dqax = pray +dpxy, and y =
daxx + dqax
dq
ax+xx
X
pra + dpx
P
ar + dx
PROB.
CLXXV.
BCD is a given angle, D a fixed point, BM parallel
to CP; and BM to MD are always in a given
ratio; to find the locus of M.
Draw MP parallel to BC; and put CA = a,
AD=b, cof. <P = c, AP_=x, FMy; then
PD=
117.
458
PROBLEMS of B. II.
Fig. PD-b-x, and BM=a+*; then (Trig. caſe 5.)
117.
2
MD = √√yy + b—x|
yy + b—x — 2cy × -x. Whence it
2
is, aba+x: yy + b = x[ -2cyxb-x;
which fquared and multiplied, bbaa+2bbax+bbxx
aayyaabb-2aabx+aaxx—2caaby+2caayx, and
reduced aayy+2caayx2caaby=bbxx +aaxx +
2bbax + zaabx.
PROB.
CLXXVI.
118. C, D are two fixed points in the line CD; and CM
Square is every where to MD Square, in the given
ratio of r to s. To find the locus of M.
X.
Draw MP perpendicular to CD, and let CA=a,
ADd, AP, PM =y, then CP = a + %,
PD = b - x. Therefore it is rs: aa: bb::
(CM²) yy + a + x|² : (MD²) yy + b—xl; there-
fore bbyy +bbaa + 2bbax + bbxx = aayy + aabb--
2aabx+aaxx. Whence aayy-bbyy bbxx-aaxx
+ 2bbax + 2aabx; and yy =
a+b
aa-bb
X 2abx
** =
XX
zab
a-b
X XX.
PROB. CLXXVII.
The lines AB, AD are given by pofition; the points
119.
P, B, and angles CPD, CBD are given. Now if
the angles CPD, CBD move about the centers P, B,
whilst the interjection D (of the fides PD, BD) runs
along the line AD; to find the curve which the in-
terfection C, of the other fides, defcribes.
Draw CS, DF perpendicular to AB; and put
AP-a, Pbb, tang. <PAD=1, tang, CPD=p,
tang.
Sect. IX.
LOC I.
the
459
tang. CBD = 4, PF = v, PS = x, SC=y, and Fig.
BF bv, BS = b — x.
Then by Trigonometry, 1:a+v:;t: ta+to
DF.
And v: 1 :: ta + tv
ta + tv
tang. DPF.
v
y: =tang, CPS.
1199
Alfo : I: Y
y
X
ta+tv
And b-v: 1 :: ta+tv:
I
b-v
tang. DBF.
y
And b-x: 1 :: y
y:
b-x
tang. CBS.
ta+tv
ta+tv
But (Trig. I. 8.) 1—
y
y: 1 ::
+
VX
บ
ta+tv
And I
يو
ta+tv
y
X
: I ::
+
b-x
bv b-x
and multiplying the extreams and means,
pux-ptay
ptvy tax + tvx + vy
= taxtvx+vy
And qx b-vxb-x-tagy-tqvy=ta+tv × b-x
+ by-vy; that is, gb x b-x-qvxb-x
taqy
tqvy=tab—tax+tv× b—x+by—vy; and tranf-
pofing, tv+qux b−x + tqvy — vy = qb xbx
×
tabby. By this and the former
9bbqbattaqy—tab+taxby
- tagy + tax
equation, v=
qb-
b-qxtqy+tb-tx-y
tax +ptay
px-tx-pty-y
And fubftituting for the known
compound quantities,
cx+dy+f
tax + tapy
; reduced
-gx+by+ l
nx-sy
sd
taphyy tagxx
C12 +
tapgxy + telx + taply = o.
fiz + sf
tab
+ sc
dn
PROB.
ト
​3
460
PROBLEMS of B. II.
Fig.
PRO B. CLXXVIII.
120. To find the figure for the ſection of a groin, the bafes
of the two folids being AFL a femicircle, and
ABC, a right-angled ifoceles triangle.
Groining in joinery is fitting two prifmatic fo-
lids to join at right angles, fo that the furfaces of
both may coincide, no part of one being higher
than the other, and the ends of both of them muſt
be cut away to a certain figure, or elfe they can
never join truly.
Let the perpendicular ſections AFL, ABC of
the folids be perpendicular to the plane LACD, on
which the figure is to be drawn. And ſuppoſe
AMD to be the figure; draw MI, MP parallel
to AC, AL; at I, P, draw the ordinates IF, PO,
perpendicular to AL, AC. Now the nature of
the groin requires that the lines FI, and PO, which
are to coincide, muſt be equal. Therefore compute
FI, OP in both figures, and put them equal to one
another.
X
Let AL or AC≈a, AP=x, PM=y. Then
IF = √AI × IL = a-yxy; and fince ABC
is a right angle and AB = BC, OP will = AP;
therefore OP=x, whence x✔ay-yy. Whence
AM is an arch of a circle equal to AF. And for
the ſame reaſon, the part at D of AMD is a like
arch, and the whole curve AMD confifts of two
quadrants of the circle AFL, meeting in the mid-
dle, and turning contrary ways. Therefore if the
ends of the two folids, be cut into the figures
ELDMAF, and BCDMAB; they will exactly fit
one another.
PROB.
Se&t. IX.
461
the LOC I
Fig.
PROB. CLXXIX.
To find the figure of a groin, when the bafes or ends 121.
of the bodies are AFL a femicircle, and ABC the
Segment of a circle.
Let AMD be the curve; draw MP, MI pa-
rallel to AL, AC; and IF, PO perpendicular to
AL, AC.
ža
I
Put AL a, AC≈b, AP=x, PM=y. Then
becauſe the figures APL and ABC muft always
be of equal height, therefore a the height of
ABC. Then to find the diameter of ABC, we
fhall have bb+aa divided by a, for the diame-
ter; put D diameter, then D-a the diſtance
of the chord AC from the center, put D—a=c,
and PO or 1F = v. Then by the nature of the
circle (in the figure ABC), 2cv + vv = bx-XX ;
and ay-yy vv, in the figure AFL. Therefore
2cv ay-yy=bx-xx, and 2cv=yy+bx-xx-ay,
alfo v = √ ay-yy, and 2cv=2cay-yy yy -
ay+bx-xx; which fquared and reduced gives an
equation of the fourth power for the locus of M.
PROB. CLXXX.
To find a general equation to the ellipfis, referred to any
line as an axis; which ellipfis will therefore be the
locus of that equation.
Let BFDG be the ellipfis, C the center: Let 1222
the point A be given, and any line AL, given in
pofition, for the axis. Take the angle KAL at
pleaſure, and through C, draw the diameter BD,
parallel to AK, and FCG the conjugate to it,
and AN, PM, LK parallel to FG. Put BC or
CD=t,
462
B. II.
PROBLEMS of
$
Fig. CDt, p=parameter belonging to BD, AL-a,
122.LK = 6, AK = c, CN=ƒ, AN=n, AP =*,
b,
PM = y.
By the fimilar triangles ALK, API; a:b::x
123.
bx
CX
:
PI; and a:c::x:
- AI. Then PR
a
a
bac
bx
= n +
a
a
CX
CX
RM=y-n- CR = —f, BR
= 1−f + C*, RD = 1+ f — c.
a
a
And by the pro-
perty of the ellipfis 2t: p:: BRD: RM² :: tt --
f+
2cfx
a
26xy 2bn x
a
bbxx
CCXX
: yy any + nn +
aa
aa
+ ; and multiplying extreams and
a
means, and reducing,
2aatyy—-—4abtxy+2tbbxx—4aatny+4nbtnx+2aatnn=0
+pcc
-zapcf-paatt
+paaff
An equation to the ellipfis FD referred to the axis
AL. ´Where note, yy and xx have the fame fign.
And if xy is in the equation, the fquare of half its
coefficient is lefs than the coefficient of xx multi-
plied by the coefficient of yy. And if xy be want-
ing, xx and yy have the fame fign.
PROB. CLXXXI.
To find a general equation to the hyperbola, referred
to any line as an axis; and which hyperbola will
confequently be the locus of that equation.
Make
Let DM be a hyperbola, C the center, AL
any line drawn from the given point A.
LAK any given angle; and through C draw the
diameter BD, parallel to AK, and FCG its con-
jugate,
Sect. IX.
463
the LOCI
jugate, and draw AN, PM, LK parallel to FG. Fig.
Put BC or CD=t, p=parameter of BD; AL-a, 123.
LK = b, AK = c, CN = ƒ, ANn, AP = x,
PM = y.
From the fimilar triangles ALK,
get (as in the laſt problem,) PI =
API, we fhall
bx
Al=
сх
a
a
bx
bx
whence PR n +
, RM-y--
and CR
a
a
CX
-f-t, and BR =
CX
a
a
=-f, and DR =
a
f+t. And by the
(BR × DR)
p:
2t: p
-
nature of the hyperbola,
2cfx
(MR) yy 2ny + nn -
CCXX
aa
a
+f-tt:
2bxy
+
2bnx bbxx
+
a
a
aa
And the means and extreams multiplied, and
then reduced,
2taayy—4btaxy + 2tbbxx➡4tnaay + 4tnbax+2taann
pcc
+2acpf -paaff
+paatt
Note, when xy is not in the equation, yy and
xx have different figns. And if xy be there, the
ſquare of half its coefficient is greater than the co-
efficient of xx multiplied by the coefficient of yy.
SECT!
464
1
"3
B. II.
Fig.
F
SECT. X.
Mechanical Problems.
PROB.
CLXXXII.
124. If the weight P break the beam DE, when fupported
looſe at A, B; to find what weight will break it,
when the ends D, E, are fixed, that they cannot rife.
SUP
=
>
UPPOSE DA AG, and BE BC. Sup-
pofe the beam cut through at C, and let P
be laid upon D, whilft P remains at C; then the
preffure at A will be P, therefore the beam will
alfo break at A, having the fame ſtreſs there as it
had at C. For the fame reaſon, if P be applied
to E, CE will break at B. Confequently, if 2P
be applied to C, the beam being whole; and the
ends D, E fixed; the beam will break at A, C,
and B; and therefore it bears twice the weight
of 2P, at C, before it breaks.
PROB. CLXXXIII.
125. The Strength of a beam AB, being given; to find its
Strength when a hole (ac) is cut out of the middle,
and alfo an equal one (rn) in the fide.
By the principles of mechanics, (Mechan. 4to.
Prop. LXX. and Cor.) the ftrength of the beams
whoſe thickneffes are db, da, dc, will be as
db², da", and de. Now as the ftrength of all
the particles between b and d, is denoted by db²,
and the ſtrength of all the particles between a
and
C
1
1
}
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Fig no.
712
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B
116.
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117.
115
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118
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P
F
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Ay
Sect. X.
465
MECHANICAL PROBLEMS.
and d, by ad'; therefore the ftrength of all the Fig.
particles between b and a, (the point D being 125.
fixed) will be db-da", add the ftrength between
and d, which is cd; and the strength of ba
and cd, that is, the ftrength of the hollow beam
is db-da+cd. But at the fection r the ftrength
is fn".
2
2.
Whence if nrac, the ftrength at b to the
ftrength at r is as db-da+cd to db-ca; that
is, as db2dc X ca-ca² to db dbx ca+ca².
Therefore if dh' be the ftrength of the whole
db?
beam, 2dc + ca x ca will be the
be the defect of
ftrength of the hollow beam, when it breaks at b;
and 2db-ca × ca, the defect of ftrength when it
breaks at n or f, which is greater than the former.
And for the fame reafon the defect of ftrength to
break at d, will be 2ba + ac X ca.
PROB.
CLXXXIV.
To Support a long prifmatic body horizontally by two 126.
props A, B; that it fhall as foon break in A or B
as in C.
AFGB
BE=y, CF=CG=x,
Let DA
DC CE=n, then n 2y+x.
=
The parts AF and BG lay no ftrefs upon C,
being balanced by the contrary weights DA, BE,
equal to them. Therefore the ftrefs at C, arifes
from the weight FG; and must be equal to the
ftrefs at A, arifing from the weights AD, AF.
The ftrefs at A by the weight DF is DFX DF
or 2yy, (Mechan. 70. and cor.) and the ſtreſs (by
FG fufpended) at C is AB x FG, or 2y+4x X 2.
But (ib. cor. 5.) 2 AC (2y+2x): AF+AC (23+x)
:: ftrefs at C, by FG fufpended at C (2y+2xx 2x):
to the ſtreſs at C, in the poſition FG
Hh
2y+*X 2x.
Therefore
1
1
1
466
B. II.
MECHANICAL
Fig. Therefore 23y = 2y+-x×2x. Or yy=2xy+xx =
126.nxnxn—2y, and
2y, and yy+2n1y
yy+2ny = nn. Whence y=
And xnx 3—2√2•
}
PROB. CLXXXV.
127. If two weights P, T keep one another in equilibrio,
on the two wheels whofe radii are AB, CB; the
Strait tooth AB of the one, acting on the crooked
tooth BD of the other; to find the proportion of
the weights P, T.
Draw EBF perpendicular to OD, EH perpen-
dicular to AB, and FG perpendicular to BC. The
point B of the end AB, is acted upon by three
forces: 1. in direction AB; 2. in direction BE,
3. in direction EH by the weight P; and theſe
forces are as BH, BE, EH.
Again, the point B of the tooth BD is acted on
by theſe three forces: 1. in direction BC ; 2. in
direction FB; 3. in direction FG by the weight T,
and theſe forces are as BG, BF and FG. But the
action and reaction at the point B, being equal;
we have BE BF, and in the right-angled triangle
BHE, rad. (1): EB:: S.ABE: HE EBXS.ABE.
EB::S.ABE
And in the triangle BGF, rad. (1): BF or EB::
S.FBG : GF = EB × S.FBG. Whence P: T::
HE: GF:: EBXS.ABE: EBXS.FBG; that is,
P:T:: cof. ABD: cof, CBD, when the weights
are in equilibrio.
Whence if ABC is a right line, P=F; and if
<CBD≈o, then P: T:: cof. ABO : radius.
PROB
Sect. X.
457
PROBLEM S.
Che
PRQ B.
CLXXXVI.
To find proper numbers for the wheels and pinions of
a clock, to go eight days; and to ſhew hours by the
great wheel, minutes by the ſecond wheel, feconds by
the balance wheel, and to beat feconds.
For the moving part.
Fig.
Suppoſe four wheels in the moving part A, B, 128,
C, D, and let the numbers for the wheels and pi-
nions be denoted as in the figure, and let fi2,
bheight the weight defcends, t time of going
down in hours.
A
It is plain
number of revolutions B has
P
B
for one of A, and
number of revolutions C
9
AB
has for one of B; whence
number of revo-
Pq
ABC
lutions C has for one of A.
And likewiſe
pqr
number of revolutions D has for one of A.
1
Since the arbor of D carries an index to fhew
feconds, therefore D30, becauſe for every tooth
there are two beats, and 2D = 60,
Becauſe the arbor of B carries an index to ſhew
minutes, and of A to fhew hours; confequently A
goes about in 12 hours, and B in 1, whence
A
P
12. And becaufe D goes 60 times round for B's
BC
once, therefore
= 60.
qr
BC
qr
Therefore the two equations.
A
12, and
P
60, will refolve the queftion; which being
Hh 2
un-
468
B. H.
MECHANICAL
Fig. unlimited, many of them may be taken at pleaſure,
128. provided they be all whole numbers.
Suppoſer 6, p=8, q=8; then A=96, and
BC=6x8x60, and if Boo, then C 48; or
B=72, and C=40. It will be better if B and q,
C and r be prime to one another.
To find the diameter of the wheel for the rope,
it will be t:b:: f:
fb
t
circumference, and
fb
3.1416t
= diameter.
For the striking part.
Let L be the fly, K the warning-wheel, I the
detent wheel, H the pin-wheel, G the great wheel,
F the count wheel, their teeth and pinions as in
the figure; n = number of ſtriking pins, and there
are 78 ftrokes in 12 hours: F goes round in 12
hours, I goes round for every ftroke of the clock.
number of revolutions of H in 13
Now
78
n
FG
hours, and
ab
number of revolutions of H
to
FG
ub
one of F, that is, in 12 hours; therefore
78
12
Again, I goes round 2 times for H's once, and
therefore =n. Therefore from theſe two equations
FG
ab
78
n
H
C
>
and Hcn, all the requifites may
be found; but being unlimited moſt of the num
bers may be taken at pleaſure, fo as they be all
convenient whole numbers.
Becauſe the pin in the warning-wheel muſt al-
ways come to the fame place when the clock has
ftruck
Sect. X.
469
PROBLEM S.
+
}
I
ftruck out, therefore
a whole number.
L
Fig.
d
128.
phenomenon to be fhewn by them.
SABC
p q r
then H=72, and FG=
78×48
=4×78, there-
1 2
FG
13 and G = 24.
But note
ab
and e may be any numbers, becauſe there is
The train or beats in an hour is
Suppofe 12, a=6, b=8, c=6;
ʼn =
no
ABC
60
X
p q r
12
=
fore F may be
may be put into one wheel or more as one pleaſes.
If the ſtring go about the axis of F, its diameter
is found as in the other. But if it go round the
axis of G, it must be made lefs in proportion as a
to F. If one weight carry both parts, their dia-
meters muſt be but half the former quantities.
PROB.
CLXXXVII.
Suppofing with BORELLI part. I. prop. 22. de motu
animalium), that a ſtrong man can but bear 26 lb.
at arm's end, and that the weight of his whole arm
is equivalent to 4 lb at arm's end; from the length
of his arm given; to find the dimenfions of that man's
arm, that can bear no more than its own weight.
Suppofe 4 lb. at arm's end equivalent to 8, the
weight of the arm. And fuppofe the two arms, fi-
milar folids, and the arm half the length of the
body. Put a length of a common man's arm,
b = 4 lb. w = 261b, x = length of the great man's
arm.
The weight of like bodies are as the cubes of
weight of the
the fides, ax³:: 26:
26x3
H h 3
AL
1
great
470
B. II.
MECHANICAL
1
Fig.
6x3
great man's arm, and
as
the weight at arm's
end, producing the fame ftrefs.
And the ftrefs being as the length and weight,
we have w+6 × a = trefs of the common man's
bx3
arm; and xftrefs of the great man's arm.
4
But (by mechanics) the ftrefs in this cafe, is as the
ftrength, that is, as the cubes of the
Therefore a³ : x³ : : w + bx c :
a³:
like fides,
bx4
935
whence bx+
= w + b x ax³, or x =
w+b
b
304
a =
= 71a.
4
}
=
Now if a 1 yard; then if there be a man whoſe
height is above 15 yards; he will not be able to
ftretch out his arm.
PROB.
CLXXXVIII.
129. Given the length and poſition of the beam AD, leaning
against the wall DE; to find the pofition of the
plane BE, on which it may stand without moving.
Let G be the center of gravity of the beam to-
gether with any weight it carries. Through G,
draw the horizontal line BH. And fuppofe DA
put into the pofition da, infinitely near the former.
Now fince the beam is to have no inclination of
moving from the pofition DA, or da; the center
of gravity G, g muit be in the horizontal line BH,
by the principles of mechanics. Draw Gn, dm,
Ar perpendicular to ad or AD. And let DG=b,
AG=c, b=S.DHG, p, q=fine and cof. ADH;
s, ƒ fine and coline DGH, x tang. DAE,
f
v = DF.
=
Since DG dgmn, and AG=ag=rn, there-
fore Dmngar. In the triangle Ddm, SmdD
(g)
Sect. X.
471
PROBLEM S.
Р
(9) : S.mDd (p) : : mD : md = 2 × mD or
x gn.
9
Fig.
q 129.
And in the triangle Ggn, S.gGn (f):
S
: gn: Gn = x gn. By the
F
S.Ggn (s) :: gn
fimilar triangles Fdm, FGn, Fd
~)
(b-v
Р
må (2 × gn)
(×gn), whence
(v):
(v) : FG
: n G
v) : : : má
q
So
- pb-pv
and
qsv =
f
9
pbf —pfv, and v =
pbf
But (Trigon. I. 5.)
pf+qs
pf+qsb. Therefore v
pbf. In the triangle
b
=
Aar, I ar or ng : :x : rA≈≈× ng. And in
the fimilar triangles FDm, FAr, Fd (v) : md
::FA(b+c− 7) :TA (Xng);
×gn) :
( 2 x
P
q
therefore vx = × b+c—v, and vqx=pb+pc
-pv, and vqx+pv=pb+pc, and fubftituting the
value of v, qx + px =pb+pc, and bfqx +
pbf
b
bb+ch-bfp
bfp=bb+ch, whence x =
bfq
P
Whence
9
b + c { ?
bjq
1. If DH be perpendicular to the horizon, 1,
s=q, f= p, and x =
2. If DH nearly coincides
b + c
P.
9
bpq
with DA, bs, p=0,
b+c
b+c S
q=1, then x=2 X
b
or
X tang
DGH.
H h 4
PROB.
:
472
B. II.
MECHANICAL
Fig.
PROB
CLXXXIX.
Having given, the specific gravity of two things, and
likewife the Specific gravity of a mixture of them;
to find the proportion of the things mixed.
Let A, B be the two things, and M the mix-
ture, a, b, m, the ſpecific gravity of A, B, M; A,
B, M their magnitudes. Then fince the abfolute
weight is as the magnitude and fpecific gravity;
therefore a A, B, mM will be the weight of A,
B, M. And aA+bB=mM=mxA+B, and tranf-
pofing aAmAmB-bB. Whence m-b:
a m:: A: B.
PROB. CXC.
Having given the weights and velocities of two fphe-
rical bodies perfectly eiaftic, meeting one another in
a right line; to determine their velocities, after re-
flexion.
Let A, B, be the weights of the bodies, a, b,
their velocities towards different parts, x and y their
velocities the contrary way, after reflexion. Then
Aa, Bb are the quantities of motion in their re-
ſpective directions, before reflexion; and Ax, By
after. As the bodies are elatlic, they will recede
from one another, with the fame relative velocity
they met, whence a+by+x. And (by mecha-
nics) the difference of the motions, moving the
fame way, will remain the fame after as before the
ſtroke, therefore Aa-Bb-By-Ax, but y=a+b
x, therefore Aa-Bb-Ba+Bb→Bx—Ax; and
tranfpofing, Ax+Bx=aB÷bB—Aa+bB, and ≈≈
B-AXa+2Bb
A+B
and y=
A−Bxb+2 Â a
A+B
PROB
1
Sect. X.
473
PROBLEM S.
PROB. CXCI.
¿
Fig.
ACDB is a thread fixed at. A and B, at the points
C, D of this thread are fixed the two threads CE,
DF, with the weights E, F; having given the
weight F, and the pofition of the points C, D, to
find the weight E.
}
Let the weight Fw, weight E=x, S.<CDB 130.
=s, S.FDB=t, S DCA = p, S.ECA =q.
The point D is kept in equilibrio by forces.
in directions DB, DC, DF, which are to one ano-
ther, as the fines of the angles they pafs through
(Mechan. 8. cor. 2.): therefore S. CDs (s) :
force at F (w) :: S.FDB (t) force in DC =
tw
کو
= force in CD, becauſe action and reaction are
equal and contrary.
Again, the point C is drawn by three forces, in
directions CD, CA, CE; therefore, S.ECA (q):
force CD (
CD (120)
(tw) :: S.ACD (p) : force at E (x),
S
therefore qx =
ptw
and x = pt w.
S
PRO B.
#
qs
CXCII.
Three points of the cieling, A, B, C are given, to 131.
which are fixed the threads AF, BF, CF whofe
lengths are given; to thefe is fixed the thread FD,
with the given weight D; to find the tenfion of all
the threads.
Becauſe the triangle ABC is given, and the
lengths of the threads; the point O will be given,
where DF produced cuts the cieling. Produce AQ
to E, and draw EF, which will be = √FO²+OE².
All
1
474
MECHANICAL, &c. B. II.
}
Fig. All the fides of the triangles CFE, EFB, are gi
31. ven, and confequently the angles. Now inftead
of the threads FC, FB, fuppofe the thread FE to
fuftain the weight. And then the whole is fuſtain-
ed by the two threads AF, FE acting in the per-
pendicular plane AOEF. Draw OL parallel to
AF, in the plain AEF, and LG parallel to CF in
the plane CFB.
Put AE, AF÷b, EO=c, AQ=d, EF=ƒ,
OF = b, S.<CFB = p, S.CFE = 9, S.EFB = ș.
Then (Mechan. &.) the tenſion of the threads
DF, AF, EF, will be denoted by OF, OL, LF;
and taking away the thread FE, the tenfion of the
threads CF, BF, will be LG, GF. Then to find
each, By fimilar triangles, EA (a) AF (b) ::
bc
:
AO (d) ::
EO (c): OL =
And EA (a)
a
fd
EF (ƒ): LF =
And in the triangle FLG,
¿
3
t
a
S.LGF (p) : LF (fd
a
:: S.LFG (s): LG =
S.FLG (q) : FG =
sfd
and
ap
are refpectively as
qd f
pa
Therefore the tenfions of DF, AF, CF, BF,
bc sfd qfd
b, a
ap ap
SECT
1
·
1
S E C T. XI
Philofophical or Phyfical Problems.
475
Fig.
PROB. CXCIII.
Required the height of the tower, from the top of
which a stone falling to the bottom, the found will
reach the ear at the top, in the time of the fall.
T2
UT b-16 feet, the height a body falls in
a ſecond.
PUT
1142 feet, the ſpace found moves through
in a fecond.
a = time of the body's falling.
Then :c::a: cafpace found moves in the time
I
And 1: aa :: b: baa height the ball defcends.
Therefore per qu, baa—ca, and a=
CC
C
b
=71 feconds.
And baa = ca = 81088, the height.
b
PROB. CXCIV.
There is a round tower, whofe circumference is 100
yards, a ſpiral tube runs about, from bottom to top, .
at an elevation of 61° : 5'. A ball put in at the
top of this tube will run down to the bottom in 8 fe-
conds; to find the height.
Let ABD be 61° : 5', AC perpendicular to 132.
BD, and BC perpendicular to AB. Then whilſt
a body falls through AC, another would defcend
M
through
"
476
PHYSICAL
B. II.
Fig. through AB in the fame time (Mechan. Prop 34.
132.cor. I. 4to.) Fut b 16.1 feet, d=8", s=S.ABD.
Then by the laws of falling bodies, Ib:: dd:
bdd height fallen in 8"-AC. And rad. (1): AC
(bdd) :: S.C (s) : bdds = AB. And rad. (1) :
AB (bdds): S.ABD (s): bddss AD, the height
required 789.
PROB. CXCV.
and
Given the distance of the earth and the moon,
their quantities of matter; to find the place where
a body will be attracted to neither of them.
t
Let ddiftance of their centers, matter in
/=
the moon, matter in the earth, x = diftance
from the earth where the body is, then d-xits
diſtance from the moon.
Then fince the force of attraction is as the mat-
ter directly, and the ſquare of the diſtance inverſe-
ly; therefore we have earth's attraction, and
Z
t
XX
moon's attraction; but per queſt. theſe
A
d-x1
t
1
are equal, therefore
XX
dd—2dx+xx
reduced is t—l.xx
t-
2dtx + ddto.
which
PROB. CXCVI.
A clock that keeps true time on the furface of the
earth; being carried to the top of a certain moun-
tain, loft 2 minutes in a day. What was the
mountain's height?
Let r earth's radius=6982000 yards, b=1440
minutes, 2 minutes, a height of the mountain.
But
Sect. XI.
477
PROBLEMS.
But (Mechan, prop. 40. cor. 6. 4to.) the length Fig.
of a pendulum is as the force of gravity, and the
fquare of the time of vibration; and the length be-
ing given, the force of gravity is reciprocally as the
fquare of the time of vibration
But the force of gravity is alfo as the fquare of
the diſtance from the earth's center; therefore the
time of vibration of the fame pendulum, is as the
diſtance from the earth's center and the number
of vibrations in a given time, reciprocally as that
diſtance. Therefore r : ::rta:
I
I
and
?
b
b-c
rta
b
r
br
and r+a =
Whence
b-c
b-c°
br
cr
$ =
=9697.
b-c
b-c
PRO B. CXCVII.
A ball projected from the top of a tower, at an ele-
133.
vation of 31 deg. above the horizon, did in 9% ſen
conds fall 2000 feet from its bafe; to find the
height.
Let x=VB the tower's height, BA=d the di-
Atance, tang. DVC = 31°, t=9% the time,
b
f16.1; then
In the time t, the ball without gravity would
arrive at D, and in the fame time it would defcend
through DA. Whence 1:f: : tt: ttf = DA by
the laws of falling bodies.
And in the triangle DVC, 1 b::d: db DC,
and DC+CA=DA, or db + x = ttf, and x=ttf
db = 149.
PROB.
478
B. II.
PHYSICAL
Fig.
PROB.
CXCVIII.
If a ball be dropped from the top of a tower a mile high,
on the fide facing the east, in latitude 51; where
will it fall?
134. Let the body fall at D, whilft the tower by the
rotation of the earth is carried to IC. Now by
the laws of centripetal force, the area AIE, which
the body, moving in the circle AIF defcribes; is
equal to the area AGDE, which the body moving
in the curve AGD defcribes in the fame time, that
is, in the time of falling through AB. Hence the
área AGIarea EGD; and AGDF-EIF. But
by reafon of the fmall diſtance BD, the curve
AGD. (which otherwife would be an ellipfis) is
nearly a parabola; and the area of AFD=÷AFX
AB=FIXAE, the area of the ſector EIF. Firſt,
let A be a place in the equinoctial.
Put BE21000000 feet, AB=m=5280,
f=16.1, t=24 hours 86400", c=3.1410, a =
DC, p col. 51%. Then by the laws of falling
hours=86400″,
bodies, f: 1 :: √m :
:
ول
f
time of falling
m
: BC =
through AB. And t: 2rc::
M =d, and BD = a+d; alfo by fimilar
2rc
t
f
m+r
fectors, rr+m::a+d:
Xa+d = AF.
до
And ra: :r+m:
r+m
a FI. Therefore
r
r+m
r+m
x a + d x m
axr+m; therefore
3r
2r
m
r+m
a + d x
a, and 2am + 2dm = 3ra +
3
2
3ma, and reduced a =
2dm
3r+m
4.64; and pa=
2.88 for the lat. 51.
PROB.
7
Sect. XI.
479
PROBLEMS.
PROB.
CXCIX.
There are two iflands A, C; at C is a caftle. A fhip
from A to C keeps pace with the waves of the fea,
100 in number, from A to B. At В he fires a
gun, which ecchoes back from the castle to B, in 3
Seconds; and the time of failing from В to C was
3 minutes; to find the distance AC.
Let b=100, c=3″, d=3', ƒ=39.2 in. the length
of a fecond pendulum, a=1142 feet, the velocity
of found in a fecond, AC=y, and breadth of
a wave.
Then by the motion of pendulums, √ƒ: 1 ::
-time of vibration of the pendulum x.
√x: √
And
✓
f
"
:*:: I': √fx = a fpace. But (by
the principles of philofophy) while the pendulum x
vibrates once, the fhip or a wave runs through the
breadth ; or in fecond runs through the ſpace
√fx.
6a
2
And "fx:: d": dfx = CB.
Alfo by the motion of found r″ : a ::
I
• c :
2
།e
CB, for the eccho returns with the fame velo-
city the found went. Therefore d/fx
ca
and
,
2
therefore AB =
ccaa
ccaa
ddfx =
and x =
;
4
4ddf
bccad
bx =
4ddf
ca
and y =
y:
+
2
4adj
bccaa
1.g.
135.
"
PROB.
480
B. II,
PHYSICAL
}
,
Fig.
136.
PROB.
CC.
*
Suppoſing a planet and its fatellite to move in circular
orbits. it is required to find, whether the path of a
fatellite is concave or convex to the fun, when it is
in a line between the fun and its primary.
At the time of the conjunction, if the planet
and fatellite, both defcribe very fmall arches in the
fame time, whofe verfed fines are equal, the fatel-
lite will then move in a right line. Let ABC be
the orbit of the planet, EF that of the fatellite
whilſt the planet moves through AB, the orbit of
the fatellite EF is moved into the pofition ef, and
the fatellite has moved from e to o. Draw BD,
on perpendicular to AE, Be; and CG perpendicu-
lar to AG. Now put AD ne, then fince the
center (of the orbit EF,) A is advanced to B,
nearer to the line CG, by the diſtance AD; and the
point o is receded from the fame line CG, by the
diſtance en equal to AD; it is plain, E and o are
equidiftant from GC, and Eo is a right line, or
the fatellite E, o, at that time moves in a right
line.
Let aeo, b=AB, r, s≈ the radii of eo, AB.
Hence AD =
=
bb
aa
bb
a a
>
and en =
and
>
25
27
25
27
Put p periodic time of the fatellite, 9 that of
the primary; c=3.1416×2. Then crp: a :
:
time of de-
pa
time of defcribing a; and
b q
cr
CS
ра
дв
==
qb, and
ppaa
qqbb
r
S
27r
fcribing b, then
vide this by the former equation, and
pp = rqq.
S
PP
r
25.5
?
99
di-
or
Therefore as pp is greater, equal, or
leffer
Sect. XI.
481
PROBLEM S
leffer than 2, then the fatellite's orbit is con- Fig.
136.
cave, ftreight, or convex towards the fun, in its
conjunction.
PROB.
CCI.
To find the divifions of a monochord, to found all the
half notes, according to equal intervals of found;
and alfo to find the variations between thefe and the
Strict harmonic divifions.
It is well known an octave is divided into 6137.
whole tones, or 12 femitones Let BA be the
monochord or vibrating flring, C the middle
point; then BC will be an octave above BA. Les
Bd, Be, Bf, Bg, &c. be the ſeveral lengths of the
ftrings founding the half notes, gradually afcend-
ing, above AB, by equal degrees of found. Then
will Ad, de, ef, &c. be all unequal in length; and
whatever part Bd is of BA, the fame part will Be
be of Bd, and Bf of Be, and Bg of Bf, &c. to
make the ſeveral founds afcend equally. There-
fore BA, Bd, Be, Bf, &c. are a fet of geometri-
cal proportionals decreafing, continued to 13 terms,
the laft of which is BC, Alfo Ad, de, ef, &c.
are a fet of geometrical proportionals in the fame
ratio.
Put BA, BC, Bd=x.
: Bd (x): Bexx;
Bd (x):
Bg, &c. and BC¹.
•9439.
Then BA (1):
likewife Bƒ = x³,
Bf
x=
And x = 12/1
=
2
Or, put X=log: *. Then X =
log:
12
2
1.9749142, confequently 2X, 3X, 4X, &c. =
logarithms of ², x³, x+, &c. Therefore x =
I i
•9439,
៩
482
B. 11.
PHYSICAL
137.
9439, 2.8909 for a mean tone, &c. and the
reft are as in the following table.
The harmonic divifions of the monochord, to
found the pure concords will be, as follows; the
leffer third, greater third, fourth, fifth,
leffer fixth, greater fixth, eight; which fee
in the following table, in decimals.
Equal
Errors.
Names of Pure
the chords. concords. divifions.
whole ftring 1.0000
1.0000
0.
b fecond
•9439*
# fecond
.8909
leffer third
8333
.8409
bxJ
greater third
.8000
•7937
$15
fourth
..7500
.7492
#
J
# fourth
.7071
fifth
.6666
.6674
6700
leffer fixth
.6250
.6300
bis
IS
greater fixth
.6000
•5946
I
13
b feventh
.5612
#feventh
•5297
Eight
.5000
•5000
O.
138.
Then to find the errors or variation of the cor-
reſpondent cords. Let Bt cord by column 2d,
Br cord by column 3d, rp = a whole tone, n =
number of mean proportionals between Br and Bp,
then will be the error, for it fhews what part rt
I
n
is of the whole note rp.
Here then
Bin
Brπ-1
- Bp
Brx.8909. For .8909 being a whole note for
the
Sec. XI.
483
PROBLEM S:
༣
the ſtring 1, Br x-8909 will be a note for the Fig.
Aring Br. Therefore
BA" n
138.
.8909. And nxlog:
Br
Bt
= log : .8909 =
1.94983; and # =
Br
-I 94983
.05017
As in a fifth,
log: B-log: Br
.05017
log: Bt-log: Br
log:Bt-log:Br = .000500, and n
100; whence the error -
I
100
.000500
But as this variation bears but a fall propor
tion to the length of the ftring, there will be no
heed to make ufe of logarithms. For fince
1-8909.1091 is the length of a note when the
ftring is ; therefore .1c91 x Bt
rt
ftring Br. Whence or
rp
rt
a note for the
the error, or
tp
Br Bt
which is the fame thing
= the error.
•1091 Bt
As in the fifth, Br-Bt.6674-.66663.0007,
.0727
and .1091XBt.0727, and
=100, nearly,
0007
7
I
or
727
100
= the error.
Or fhorter thus. Since Br-Bt-twice the dif-
ference of two adjoining numbers in col. 3. or =
difference of two numbers 2 degrees diſtant, tak-
ing one greater and the other lefs than the proper
Br-Bt
note; therefore
Br-Bp
= the error.
As in the fifth,
6674-6666
8
7071-6300
771
97
the error. And in a greater third,
.8000-7937
.8409-7492
.0063
T
11
the error.
.0917
$4
2
I i2
The
484
B. II.
PHYSICAL
Fig.
The errors for each concord being thus compu-
ted, are fet down in the fourth column, which
138 fhews the error of the third column, as it differs
from the fecond; thoſe below denoted by (b), theſe
above, by (#).
In tuning a harpsichord, fince the fifth must be
12 times repeated to make 7 octaves, therefore
the variation, by tuning by true fifths, will be
or about of a note, which is an error that
12
100
४.
a good ear can difcover; and being too fharp, the
fifths therefore ought to be tuned as flat as the ear
will bear.
Hence the equal diviſion of the notes in an octave
is the beſt ſyſtem, for the greateſt error is in the
leffer third and greater fixth, which only amounts
I
to of a note.
13
PROB. CCII.
To find the number of beats made in any imperfect
concord, in mufic.
I call that an imperfet concord that varies a little
from the perfect one, which is made by a harmo-
nial divifion of the monochord. Thus when the
lengths of the ftrings are 4 and 5, you have the
perfect cord (a greater third), but vary one length
as 4, making it 3.99, and you will have an im-
perfect cord attended with beats.
A beat is a jarring found made by the irregular
vibrations of two ftrings, founding together, when
the due period, or coincidence of their vibrations
is interrupted. Its noife is fuch as this waw, aw,
aw, aw, or yâ, yâ, yâ, yâ, yâ, Our buſineſs is to
find in how many vibrations this perturbation hap-
pens, or how many yaws in a ſecond of time.
Let
N
4
}
!
*
ឋ
}
{
}
Fig. 124
D
A
C
B
E
PQ
C
A
P
127
E
D
B
F
P
125
1
B
D
$
A
126
FCG
E
D
ア
​128
d K
TO
C
B
P
A
my
d
B
F
A
129
130
H
G
D
C
a
།
A
E
E
A
132
D
B
135
C
B
At
A
133
G
A
A d e f g
136
EB
A
A
}
H
F
131
F
G
G
B
3
D
T
F
G
C
134
E
с
137
B
rt
P
P 138
B.
1
Pl. XI. pa. 48.3
برا الم
486
B. II.
PHYSICAL
Fig.
139
P
9
length of its ftring on the monochord.
number expreffing the concord,
2
3
4
for the fourth, or for the fifth, &c.
n
3
I
Then AB =
, and AC = 2; alfo ab =
and ad = 2.
P
q
=dC.
22
r
n
Then AC-ad=
Then if dC ( (?
e/s
the time AC (2
1) be loft or gained in
:: AB will be loft in the
time
P
12
n
r
× AB =
9
pr
pr-qn
pr
X AB =
pr-pt
qu
× AB = X AB, and Bc will be loft fooner,
X
r-t
in proportion of AB to Bc, that is, in the time
y × Bc, which is the time of the period. But
-t
by the laws of vibration, r:::
I
I
:
::c: b₂
b
C
go
and r-t: r : : c-b: c; whence × BC=
C
E-b
r-t
× Bc = the periodic time of the beats. And
X
if AZ be divided by the periodic time, you will
have
c-b AZ
C
X = number of beats in a fecond.
Bc
AB
ᎪᏃ
But Bc =
AZ = n× AB, therefore
Bc
9
nh x AB
c-b
ng.
Whence
ng
AB
12 x 12 = = x²² =
хр
C
umber of beats in a fecond.
Hence
Sect. XI.
48.7
PROBLEM S
4
ร
Hence, from the length of the ftring or divifion Fig.
of the monochord, as given in the table of the last 139.
problem, and having alfo the number of vibra-
tions; the beats will be found, as in this table.
Where the ground or loweſt note is F the cliff-
note of the baſe.
Cords.
Vibrations.
b
f p. 1. Beats.
Eight
600
5000
g.
fixth
500
5946
1. fixth
480
6300
5000 I
6000 3.5
62505.8
•
2
13
18
fifth
450
6674
6666
2.3
1 b
fourth
400
7492
g. third
375
7937
7500 3 4
8000 4 · 5
•
I
I
1
1. third
360
8409
333 5.6
15
b
Bafe F
300
10000
10000
I
This table fhews the beats for all the concords,
reckoning upwards from F; when the inftrument
is tuned according to an equal afcent of notes;
where the flats and fharps (b, ) fhew whether
the upper note is lower or higher, than the true
concord in the laft column. In the octave above,
the beats will be twice as many; and in the
octave below, but half as many; being always
proportional to the number of vibrations of the
baſe note. The fifth is moſt ferviceable in tuning,
and the number of beats in one fecond, for the
fifth is
N
300
If it be fuppofed that the beat is not made at the
points X, Y, but at fome intermediate place, where
they fall thicker and more confufed; and that at
the points X, Y, there is the leaft imperfection.
Yet the periodic time will ftill be the fame, what-
ever part of the cycle XY it falls in. When the
1 i 4
cycle
488
B. 11:
PHYSICAL, &c.
Fig. cycle XY is very fhort, the fingle beats are im-
139. perceptible, and we hear nothing but a difagreea-
ble noife. All the concords beat, but being ex-
ceeding quick, they are not perceived fingly; and
being regular throughout, they exhibit an agreea-
ble harmony.
When the pitch of the two notes are not altered,
the beats fucceed one another in equal times, but
altering either of them nearer to a perfect har-
mony, the beats fucceed in longer times, and the
nearer the longer, till at last they vanifh, when
the concord is perfect. All the beats are heard
in organs; but only half of them are heard in
ftringed instruments.
:
C
1
SECT
"
SECT. XII.
Problems relating to Series.
498
Fig.
PROB. CCIII.
Given the diameter of a circle; to find the fide of any
regular poligon, infcribed in it.
L'
=
I
ET d diameter, n number of fides, 140.
x fide of the figure, EB. By Trigono
3.14159d
metry,
2n
=arch DE=a, by fubftitution.
And (Trig. I. 12.) half the fide, or EA=a
4aa
4aa
A
4aa
B
C&c. And 2EA or
6dd
20dd
42dd
EB or x = 20
4aa
4aa
4aa
A
B
C
Odd
20dd
42dd
4aa D &c.
72dd
Or thus,
By a table of natural fines, find the fine of
180
=s, then x = ds.
n
PROB.
2
3
}
CCIV.
31
Suppoſe x3 - cx² + x + bx* = dccx; to find x.
2
Divide by the leaft power of x, that is, by x,
5
44
2
and x- ·c + x + bxˆ - dccx Taker the near-
eft root, and put r+ex.
Then
}
490
B. II,
PROBLEMS of
Fig.
Then
S
मेरे
=
2을 ​+ 독
​C
+ ** =+5 of
+ bx += = br² +
decx-deer-
3
T
}³½e + 1 = ee
돌​을
​2
1e
S
645
be
3
5ee
720} = 0;
3bee
T
4r7
dece
3277
decee
z
2 x 12 +
87 2
r
That is, p + qe + see = 0, by ſubſtitution.
Whence
141.
H
2 + e
S
more exactnefs.
{
which may be repeated for
Or thus,
Seek the leaſt common dividend of the denomi-
nators of the indices of x, and reduce the equation,
which will become x
3
T2
2
T2
3
TZ
c + x + bx-dccx
T3
0. Pute¹²; then the equation becomes
e³°~~c+e² + be³-deceo, or eso dece + be³ +ee
=c, and the root extracted gives e, and confe-
quently x is had.
PROB. CCV.
Given the fides AC, CB, of the triangle ACB; and
the ratio of AB to the arch CE is given; to
find AB.
Let AC=r14, CBs 22, AB=x, and
AB: CE :: 10: 4, whence arch CEx. Let
y cof. CAB.
Then
Sect. XII.
SERIE S.
495
Then (Trig. cafe 5.) rr+xx—2rsy=ss, whence Fig,
ŷ=
rr+xx−ss
2rs
rr-ss+xx
25
cof. A to the radius I, and ry =
cof. A to the radius r. But (Trig.
I. 12. cor. 1.) cof. A=r
I
11. |+1
4x
200r
77-ss
25
4
4x
+
24.10000r³
XX
2473
aa
a4
+
&c. r
21
4x
&c.
720.100000ors
+ and tranfpofing,
"
25
42
+
Xx2
25
2007
44
240000r³
xx4 +
46
740000000rs
rr-ss
&c. = r —
да
or Ax² + Bx++ Cx³ &c. = b,
25
141,
by ſubſtitution, then (Prob. lxii. I.) xx =
b
B
2BB-AC
b² +
A
A³
As
b³ &c. =836.95, and
*=28.93.
PROB
CCVI.
Given the arch of the circle BHE, and the fine BD
to find the radius BC.
Let BHE=d=8, BD=s=3. Take an angle
p nearly equal to ACB, a fine, b its cofine,
rad.=.=.0174533, ¢ = 3.1415926, then up =
np
arch belonging to the angle p. Let p + x = true
angle ACB; then np+nx or np+x=correfpon-
dent arch, (putting xnx). And (Trig. I. 13.),
az2 bz3 az4
the fine of np+za+bz—
2
+ +
6 24
&c. = c — — xp — — z per queft, that is,
np -
1428
492
B. II.
PROBLEMS of
Fig. s
a
b
a
ZZ-
r
2
6
24
142. 2 + b . 2 - 2/2 zz - 12/22² + = x+ &c. = = ‹ —
S
d
np-a, or Az + Bzz + Cz³+Dz4 &c. = R.
Affume p
55°, then a and b will be known,
and R,0010292, and (Prob. lxii. I.) z or
R
11x =
A
B
A³
2BB-AC
R² +
As
R³ &c. =
001084, and x
x=.0621 degrees, which is
3′43″; therefore p+x=54° 56′′ 16″½=<ACB.
Hence BC
3.66513.
PROB.
CCVII.
143. The ordinate AC, and curve BC are given; and the
equation of the curve is 2
= hyp.log
z+√aa+zz
a
a
and a+x=√aa+zz; where AC=y, BC=2;
to find AB.
Let y=6=h, z=9=g, a=r+e, taking r the
neareft value of
Then hyp.
a.
hyp. log
log:
g+√rr+2re+ee+gg &c.
b
b
be
=
+
rte
r+e r
bee
r3
&c. And (putting frr+gg), by evolu
tion, log:
re
re
ee
8 + s + ² + " b
ƒ+8+ +
f
f f3
r+e
be bee
+
and
rr
доз
number of the hyp, log:
gg
ee
f3
&te
be
be²
+
2
nhe
= 12
+
go
rr
goz
nbee
ri
h. log: =) (Prob. lxxxv. I.)
nbb
2n+
ee (putting n
And
Sect. XII.
493
SERIE S.
& + ƒ+ +
88
And multiplying by r+e,
re
nhe
nbree
ee = rn
of
Fig.
143.
2ƒ³
2r3
nhee
+ ne +
rr
nhee
rr
And reduced
nb
+
F
}
r
nbb
nxe+
2f3
273
× ee=rn f―g.
Affume r3.7; then e-.00112, and a =
3.69885, and x√aa+zz-a=6.0316; ſubſti-
tute this value of a for r in the laft equation,
and the operation repeated, gives a ftill more
exact.
PRO B. CCVIII.
Given the length of a pendulum, and the arch it de
fcribes; to find the time loft by defcribing a greater
arch.
Let r length of the pendulum, c=cord of half
the arch it defcribes, C
of falling through 2r.
any other cord, time
3.1416.
P =
2
mechanics it is found that the time of
Then by
vibration,
is PXI +
CC
964
+
16rr
&c. for the cord c.
4574
'CC
And P:XI+
9C4
+
16rr
&c. for the cord C.
4574
and P X:
CC-cc
9
+
16rr
X C+ C+ &c. =
45r4
time loft in one vibration for the cord . But
when
६
494
B. II.
PROBLEMS of
Fig.
when is o, P is the time of one vibration, which
does not fenfibly differ from a vibration for the
cord c. Therefore fince 86400 the number of
86400
feconds in 24 hours, therefore
P
number
of vibrations for the cord c, in 24 hours. There-
fore
86400
P
x P x
CC-cc
16rr
&c. or 86400 X :
9
+
Iorr
4574
X C464 &c feconds loft in
: =
24 hours; that is,
CC — cc
·CC +
9
64rr
24 hours; and
5400
X:
× C4—C4 &c. feconds loft in
5400
gr
× CC—cc, is nearly = the
feconds loft in 24 hours.
It ſwings feconds, then 39.2, and the time
loft in 24 hours is nearly
3.52 X CC
Cor. If c be o, and Ccord of yo°.
dulum vibrating in the double arch of 90°,
4 b. 20 min, in 24 hours time.
x
cc.
A pen-
will loſe
And if co, then to find the length of a pen-
dulum vibrating in the arch of C in the fame
time. Let r pendulum vibrating in the very
fmall arch, pendulum vibrating in the arch
of C. Then the lengths being as the fquares of
the times of vibration, we fhall have in the firſt
caſe ✔r for t, and in the fecond for t;
3.1416
whence in the firft cafe P=
3.1416
2
2
✔r, in the
fecond P =
x. And the times being es
qual we fhall have P x or
3.1416
√r = P x
2
I +
Sect. XII.
SERIE S.
$495
CC
I +
3.1416
CC
or
16rr
√xx 1 +
2
› 6rr
or ✔ Fig.
CC
CC
=√xX1+
and 'rx+
which re-
16xx
8x
duced is rx-xx-÷CC; whence x will be found.
And on the contrary x being given, r will be
found.
PROB. CCIX.
Given the latitude failed from, the departure, and
difference of longitude; to find the difference of
latitude.
کم
Let d departure, 7 diff. longitude, arch
of latitude come to, zits mer. parts, a the
given lat. m its mer. parts.
Then by Mercator's Sailing; as diff. lat. (a-x)
: mer, diff latitude (m-z): :d: 1, whence allx=
dm — dz, and dz-lx=dm-al. But Dr. Halley's
Series for the meridional parts of x,
is
* + 1/3
ㅎ ​+
I
61
x3
x5 f
*7 &c. Therefore
24
5040
dx - lx +
ཁ།
d
6id
x3 + *5 +
X7 &c.
11
6
24
5040
dm-al. And by reverfion of feries (Prob. Ixii.)
* will be found; then 3438x latitude in mi-
nutes.
Or this,
Seek another latitude, by the table of meridio
nal parts, fuch, that the proper difference of la
utude divided by the mer. diff. latitude, will be
d
equal to the quotient which is easily done by
a few trials; and that is the other latitude.
PROB
*496
B. II.
PROBLEMS, &c.
F
A
Fig.
144.
PRO B. CCX.
The curve BMD is defcribed with a pair of compafles
upon the furface of a cylinder, which is afterwards
Stretched into a plane; to find the ordinate PM.
Let d diameter of the cylinder, a = AB the
extent of the compaffes, AP, PM y, v=
cord, whofe arch is y Then (Geom. II. 21.)
aa-xx-vù. But (Trig. I. 12. cor. 2.) v=y—
دو
✓aa-xx
yy
A
2.3dd
4.5dd
B &c. Whence ✔aa
yy
A
2.3dd
yy B
4.5dd
yy
C&c. and
6.7dd
by reverfion of feries y is had.
If the arch was in a given ratio to the chord,
the figure would be an ellipfis, but as this is not
fo, the curve will be a mechanical one.
**
* * * *
*
1
SECT
قدام
497
SECT. XIII.
Problems concerning exponential quantities.
Fig.
PROB. CCXI,
Some maids driving a flock of sheep, were afked, kow
many they had? To which they answered, that if
the flock was equally divided among them, the share
of each would be twice as many as there were maids.
And if the terms of this double progreffion 1, 2, 4,
8, &c. be counted, as often as there are maids; the
laſt term will be the number of ſheep.
L
ET afheep, emaids. Then
E
a
e
H
20,
&c.
and the eth term of the progreffion 1, 2, 4, 8,
=2-1 (Propor. 25.), therefore 2-1 a, per
queft. Whence a zee, and 2º 24, and ex-
punging a, 2º 4ee, or 2-2 = ee.
2e
e
=
Therefore
2 × log: 2 = 2 log: e, or .30103e—,50206 =
2 log: e, and.150515e-log: e.30103 Then to
find e (by Rule 5. Prob. xcii.), aſſume e6, then
•1505€-log:e=.125, which ſhould be .301, and
the error is -.176.
Again, affume e=7, then .1505e-log:e=.208,
and the error is —.092. Then
therefore e 8.1 nearly.
IX.092
.176.-.092
=LI;
8.1; then .1505e-log: e=.2106,
Suppoſe e
=
and the error
+ .0096.
+.0096.
And the correction
1.04, and
e-
7.996; and e=8
exact, and a =
128.
K k
PRO B.
8.11.04
"
1
498
B. II:
EXPONENTIAL
Fig.
PRO B. CCXII.
Two travellers at 150 miles
one another. In the feveral
rate, 5, 10, 20, 40, &c.
18, &c. miles; to find in
meet.
x
distance fet out to meet
days, A goes at this
B goes 6, 10, 14,
what time they will
Let days, (by Geom. Prog.) A's laft day,
will be 5 X 21; and his journey 5 X 2-56
And (Arith. Prog.) B's laft day is 6+ 4x4 or
12+4x-4
4x+2, and
2
Xx, or 2x²+4x= B's
journey. Whence 2* X 55+4x+2x²=150,
4
and 2* +
x² +
ΙΟ
8
10
४
log: 31
4
10
8
ΙΟ
8
10
x= 31. And 2* = 31
xx. And log: 2* or * × log: 2 =
found greater than
b=4, c=log: 2.
8
4
ΙΟ
XX. By trials x will be
4; let n = 4, and n + v=x,
8
Then cn+cvlog: 31- N
10
8
4
4
nn
nv
vv. But the number
10
ΙΟ
ΙΟ
10
Mcv²
belonging to cn + cv = cn × : 1 + mcv +
&c.
2
c²nmv²
(Prob. lxxxv.) whence cn + ccmnv +
&c.
2
8
4
8
4
=31- 12
nn
IQ
10
10
+
n x v
10
10
+19
บบ.
And
Sect. XIII.
499
PROBLEM S
com n
And reduced, ccmnv +
Fig.
vv = 3
3 I
2
+
+
{
8
ΙΟ
n
001000101
+
10
10
4
nn
10
cn
Which put into numbers, and reverting the feries
(Prob. lxii), v is had .32; then put new n for
x+v or 4.32, and repeat the operation; and at
laſt v =.3256, and x = 4.3256.
PROB.
CCXIII.
To find x in this equation, xx=123456789:
Here x will be found between 8 and 9. Put
n=8,n+v=x,b=log: n, c=log: 123456789; then
xlog:x=log: 123456789=c, or n+vxlog:n+v=c.
Μυ
But (Prob. lxxxiv.) log: n + v = b + พ
Mv² Mv³
323
+
2n²
nMv
= nb +
n
&c. and +x log: n+v
nMv2 nMv³
if
&c.
2222
323
+ bu +
and tranfpofing and reducing,
Mv²
Mv3
= c₁₂
&c.
12
2n2
Ъ
v2
V3
24
+IXU+
25
+
M
21 2.322² 3.423 4.512+
&c.
c-nb
=
And extracting the root (Prob. xciii.)
M
v=,64002, and x or new n 8.64002 for ano-
ther operation, which will give x=8.6400268.
Kk 2
PROB.
1
500
B. II.
EXPONENTIAL
Fig.
PROB. CCXIV.
To find the value of x in the equation
1000-x X log: 1000-x = x.
Put b=1000, *=a+v, b—a=g, p=log: g;
then gvxlog: g-v=a+v. And fubftituting
the logarithmic feries inftead of log: gv (Prob.
Ixxxiv.), g-vx:p—
บ
22
03
24
&c.
g 2gg 3g3 484
=a+v, which multiplied and reduced is, gp-a
VV
V3
602
-p+2.0+ + +
&c. =0.
28
V4
+
درح
+
76
12g3 4.584 5.6gs
Affume a=836, and extracting the root
(Prob. xciii.) v.05315; and 836,05315.
PROB. CCXV.
To find x in the equation
X
J.000
log: 1000-x = 1000-X
X
Put ʼn
n
1000, x = a+v, g=n-a, p=log: g.
Since x is nearly 860, affume a 860; then
=
A
a+v
n
the equation is log: g-v
v=
But
g บ
a+o
ข
VV
log: g-v=R-
a+v/². ·ng+nv
&c. Whence
g
2gz
v
VV
= P
&c.
a + v × 8-V
g
28g
303
Which equation reduced gives
a
a
agp
3av +
z² of
v3 &c,
0.
28
6gg
aa
+ ng + sp
ор
n
།
I
P
+
2
28
In
Sect. XIII
gor
PROBLEM S.
In numbers,
4626.3 + 7138v + 3.870202 —.01088v3o;
or 1 + 1.543v + .000836v² .0000023=0;
whence, by extracting the root, v=-,64822, and
x=859,35178.
Fig.
PROB. CCXVI.
Having given the equations x+v=y*, and
y¹t" =x"; to find x and y.
m
From the first equation y = x
m
x+y
fecond, y=x ; therefore x
equating the indices
x+y
x+y
22
x+y
n
,
=X
m
and from the
And
m
and x+y=
mn
m
✔mn.
Whence y = ×
#n
n
= x
Therefore
√
172
****
MR
n
=y" = x
>
n
x+y
by the first equation, x*+*
and again equating their indices, x+x
✓ m
= √mn:
Then x being had y is known from the equa-
m
n
tion y = x
ก
To find x put xv, then x
mn
"
✓.
#L
172
or x 1
R
and v+v =✔mn. And the root
may be extracted by logarithms.
Kk 3
PROB.
<
502
B. II.
EXPONENTIAL
Fig.
PROB. CCXVII.
I
*
To find the value of x in this equation, X'+X=
x in this equation, X²+X= -
X being the hyperbolic log: of x.
Here x is between 1 and 2,
x=1+v, then (Prob. lxxxiv. cor.
22
therefore put
1.) X = v
2
73
04
༣ 23
+
&c.
Whence v
+ ²² &c. +
'&c.l+
2
3
4
2
3
บ
+
།8
1
&c. =
and multiplying and
2
3
I
reducing v+ 3 v² — — v³ &c. = 1, and by re-
2
6
verfion (Prob. lxii.) v = .56, and x = 1.56.
But becauſe this does not converge faft enough;
put n 1.56, and +x, 1.4446858 hyp.
log: n = mx log: n; then (Prob. lxxxiv. cor. 2.)
x=1+
972
V
213
+
9
whence we ſhall have
n
2nn
32213
บ
บร
It
n
2NN
&c. +1+
ย
บบ
: Xn+v=L
N
2 nn
5
And when multiplied and reduced,
1+3
l + 1 x ml + 1 + 11² + 1×0+ vv &c. = 1.
In numbers,
N
1.0021921 +2.53180222 + 1.2465930² = 1; or
2.031v+vv=.0017586. Whence (Prob. 88.)
~.000$651, and n + v or = 1.5591339.
=
Otherwife thus,
x
Let 7.4446858 the h. log: 1.56, or n, as be-
fore, /+s=X; then the number (x) belonging
to
Se&t. XIII.
503
PROBLEM S.
to 1 + s or X = n x: 1+ st ss + /s² &c. Fig,
I
2
(Prob. lxxxv.); whence +++s: xn+ns +
I nss &c. I; and by reduction,
2
=
¿+1×?n+l+1|²+l×ns+÷ll+2}/+2×ns² &c.=1.
In numbers,
1.0021921+3.949611s +5.008515ss &c. = 1.
or .788585 + ss =
.00043768;
And extracting the root (Prob. lxxxviii.) s =
X
.0005549, and 1+s or X = .4441309, and m
.1928836 the com. log: x; or elſe
m
.0002410, and fince com. log: 1.56.1931246;
therefore 1931246.0002410.1928836 thẹ
common log: x. Whence x = 1.559134.
•
PROB. CCXVIII.
To find x in the equation x
2 2
= 123456789 = b.
Put b=123456789, and by a few trials you will
find x near 2.8, put n=2.8, n+v=x, l=log:n.
ml=hyp. log: n.
Then (Prob. lxxxv. cor. 6.) * = n² ×
x"
I + mlv + v. Put rn", e = n" × mlv + v₂
then x = r + e; let this be an in-
X
dex, then x
2
X
rte
X
irte
= n + v = (by the
rv
fame cor.) n XI + mle + =b per queft. Then
12
12
n
reſtoring the values of r and c, n x: 1 + min
Kk 4
X
R
504
B. II.
EXPONENTIAL
Fig.
n" v
× mlu + v +
b. Put g = mln" xml + 1,
n
n
then nXI+gv + n²¬¹v=b, and by reducing,
22
n
n
; here
12
here n = 97620000,
bn
~=
72
n" × g + n
N~~ I
ml=1.02962, 837.3368, n=6.3810, there-
fore
v =
b
25830000
4266000000 = .006054.
Or let =f. Then v =
.26466
44.7178
12
12
12
f-I
8+2"-1
n
.006054; then n+v or x=2,806954
nearly; or put n=2.806054 for another operation.
This problem is easily refolved by rule 5, pro-
blem xcii. by making feveral fuppofitions for the
value of x, and finding the correction every time;
and fo you will continually approximate to the true
value.
PROB. CCXIX.
If X be the log. x, it is required to find x, in the
equation xx+X* = 100.
Let n+v=x,1=log:n, 1+s=log:n+v, L-log:l
Then (Prob. lxxxv.) n+v=nnms &c. whence
n+wins)²+s+1+5ms 100. But (Prob. lxxxv.
cor. 6.) n + nms\
And s
in + nms
S
=
n² x: 1 + mls + ims.
ns
= l®ª × 1 + m²Lns +7'
Therefore
Sect. XIII.
505
PROBLEMS.
Therefore nlx1+2mls+l”×1+m²Lns+
¹×2mls
Or n + 1 x m² Lns +
And s=
d
ns
= 109.
Fig:
R
ns
= 100
100 — n'— l = d.
Ī
2mln' + nLl”m² + nl^¬"
X
To approach nearly to the value of X, we fhall
have X log:x or XX= log: x, and x log: X =
log: X*. Therefore num. of XX + num. of
x log: X = 100. By a few trials X is found be
tween 1.25 and 1.26, but nearer 1.26; therefore
ſuppoſe 1= 1.257, then 18.072, L=.09933,
n² = 38.02, 1" = 62.41, d=—.43, 2mln'=220.1,
nLl"m² = 594.0, nl”! = 897.4; Whence
s=
and x
-.43
1711.5
=000251, and X=1.256749,
18.0613.
Here we have fought the logarithm X, for vari
ety; but the number x might have been found,
after the manner of the laft problem.
PROB. CCXX.
X
I
**
X
Given x + x = x + x
= 200; to find x
Take n very near the root,
to be found by fre-
quent trials, and put n+vx, 1=log:n, r = n",
n4-v
n
n
↑
I
f=ml + 1, p = n, q = mirf + — t = n", a =
n
M²
I
nn
Then
1
1
F
506
EXPONENTIAL, &c. B. II.
Fig.
Then x
Then x = n+v+v =n”x1+mlv+v (Prob‹
1xxxv. cor. 2.) =r + frv.
ར
And x
ทบ
I + mlrfv+
22
I
Alfo =
I
r+rfu = n + v\+¹ƒv = n² ×
n+
=p+pqv, (ib. cor. 6.).
11
I
r+fro
N-V
nn
I-fv
And
X
I
n + v
Alfo x* = n2+v
(ib. cor. 6.) = t
N-Y
1212
72
mlv
n
X I
+
nn
t- atv.
V
Therefore writing for the feveral powers of x, their
reſpective values, we have
p + pqv + r + rfy +
fv-I
+t-tav = 200.
r
200
prt +
ཁ
reduced v
pq + rf +
n
ƒ
ta
t
It easily appears that x is greater than 2, and
trying 24, it will be found a little too fmall;
therefore aflume 2.27, whence there will come
out v.0009463, and therefore 2.26905372
which may be put for 2, for another operation.
SECT.
i
507
SECT. XIV.
Problems of Maxima and Minima.
Fig.
PROB. CCXXI.
The line AE, and the two points B, C, being given
in pofition; to find the point P, fo that BP + PC
may be the least poffible.
T
po
AKE the point p extreamly near P, and
draw Bp, Cp, and alfo pD+ to BP, and
to CP. Then pD is the increment of BP;
and PO the decrement of CP, therefore DP=OP,
by the nature of the queftion. And fince the hy-
pothenufe Pp is common, pD=p0. And <pPD
pPO, that is BPA CPE; whence the triangles
BAP, CEP, are fimilar. Put AE, AB = p,
CE=q, AP=x, EP-b-x; then AB (p): AP
(x) :: CE (9): EP (b-x) therefore qx-pb-px,
Db
bq
and b-x-
,
and px+qx=pb, and x=
P+ q
PROB. CCXXII.
P+q
145.
The lines ABC, and CE being given in pofition, and
146.
the points A, B, being given; to find the point
D in the line CE, where the angle ADB is the
greateſt poſſible.
About AB defcribe a circle to touch the line
CE; then the point of contact D is the point
required.
For
508
B. II.
MAXIMA and
Fig. For to any other point E, in the line CE, draw
146. AE, BE, and draw BF. Then the angle AEB is
lefs than AFB, or its equal ADB (Geom. IV. 13.)
Let BC=b, AC=d, CD=x. Then (Geom.
IV. 21. cor. 2.) xx = bd, and x = √bd.
PROB. CCXXIII.
147• To draw the shortest line poffible, through a given
point P, placed within the right angle ABC.
*
Let CPA be the fhorteft line. Draw PD pa-
rallel to AB, and PF parallel to CB, and let
PD=b, PFc, CD=x, Cce an extreamly fmall
quantity, PC=2.
By the fimilar triangles CDP, PFA, ≈ :z::
CZ
*
AP, and by the fimilar triangles CDP,
xe
¿HC, ≈:~::e = Hc. Alfo :b::e:
༧
be
Z
=HC. And by the fimilar triangles PCH, PGa,
be cz bce
2:
::
Z
X XX
= @G.
triangles CDP, aGA, ≈ : b : :
And by the fimilar
bce bbce
= AG.
ZX ZXX
xe
bbce
But Hc=AG, that is
or x =
bbc
,
and
Z
ZXX
XX
3
x³ bbc, whence x = bbc.
PROB.
CCXXIV.
148. Given the line EF, and two points A, B; to find
the point D, fo that a × AD+b × BD, may be the
leaft poffible; a, b being given numbers.
Take d infinitely near D, and draw
on which let fall the perpendiculars Dr, Df.
Ad, Bd;
Then
will
Sect. XIV.
50g
MINI M A.
will a×AD+b×BD=axAd+bxBd; and by fub- Fig,
traction ax AD—Ad—b × Bd-BD, or ax dr — 148.
bxdf. But in the triangles Ddr, Ddf, the hypo-
thenuſe Dd is common; therefore dr: df: :
S.dDr S.dDf:: cof. ADF: cof. BDE.
:
ab: cof, BDE: cof. ADF.
Whence
Let AF, BE be perpendicular to EF; and put
AF=c, BE=d, EF=n, DF≈≈, DE=v. Then
DA (√cc+xx): 1 :: DF (x) :
cof. ADF; then ba::
X
X
Cc+xx
√ cc+xx
=S.DAF
√aa + vv;
ax
ax
b
cof. BDES.DBE. And b
√cc+xx
√cc+xx
:v :: 1 : (BD)
(BD)
ax√ ad+vv
aa + vv;
therefore v
bv cc+xx
ax√dd
and boy cc + xx = ax√ dd + vv
And fquaring, bbccvv+bbxxvvaaddxx+aaxxvu,
but v=n-x, put p=bb—aa, then bbcc+pxxxvv
aaddxx, or bbcc +pxxxnn-2nx + xx = aaddxx;
reduced, px+-2pnx³+pnnxx-2nbbccx+b²c³n²=o,
-aadd
PROB.
CCXXV.
Three points A, B, C being given; to find a fourth
149
point D, fo that axAD+ b × BD + c × CD, may
be the least poffible; where a, b, c are given
numbers.
Let D be the point fought; with radius CD,
deſcribe the circle GDH. Take the point d in-
finitely near D, and draw Ad, Bd; on AD, BD,
let fall the perpendiculars dr, df. Then fuppo-
fing
510
MAXIMA and
B. II.
149.
Fig, fing CD to be given, ax AD + bx BD will be a
minimum. But ax Dr is the increment of ax AD,
and bx Df is the decrement of bx BD, therefore
axDr=bxDf. But in the right-angled triangles Ddr.
Ddf, Dr: Df:: S.Ddr : S.Ddf: : S.rDC, or ADC
: S.mdf or EDC. Therefore b: a:: S.ADC :
S.BDC.
150.
:
After the fame manner, fuppofing BD given,
we ſhall have a S. ADB: S.BDC. There-
fore when axAD+bxBD + cx CD = minimum;
a, b, c, are reſpectively as the fines of BDC, ADC,
ADB; or of BDm, ADm, BDr, which makes 180°.
Therefore if a triangle be made of the 3 lines a, b, c ;
the angles of this triangle will be equal to the an-
gles at D, viz, that oppofite to a mDB, to
bmDA, to c BDr. Therefore all the angles
about the point D being given; the distances AD,
BD, CD will be found by Prob. cxxxii.
PROB. CCXXVI.
Given the triangle ABD, and the circle CFK whose
center is A; to find the point F in the circumfe-
rence CFK, that the angle BFD may be the
eft poffible.
great.
Through the points B, D, defcribe the circle
BFD to touch the circle CFK in F, the point
required. For to any other point C, in the cir-
cle CK, draw DC cutting BFD in S, and draw
BS, BC. Then the BSD or BFD=<BCD +
CBS; therefore BFD is greater than BCD.
On BD let fall the perpendiculars, AH, FI,
GE; G being the center of the circle BFD, Then
to find its radius GF; let BE-ED=b, HE=c,
2
AH=p, AF≈r, GE=x.,Then AG= √√√ p+xl²+cc
Sect. XIV.
511
ΜΙΝΙΜΑ
and GF
Fig.
= √pp + cc + 2px + xx,
PP + cc + 2px + xx-r, and BG= √bb+xx, 150.
whence √pp+cc+2px+xx-r=√bb + xx, and
√pp+cc+2px+xx=r+√bb+xx, which fquared
is cc+pp+2px+xx=rr+bb+xx+2r × √ bb +xx :
√bb+xx:
Put sccpp-rr- bb; then s+2px 2rX
=
√bb+xx,¯and ss+4spx+4ppxx=4rrbb + 4rrxx ;
reduced, 4ppxx + 45px = 4rrbb.
4rr
SS.
Then x being found, it will be EG (x): rad.
(1) :: BE (b) : tang. BGE, or its fupplement
BFD.
PRO B. CCXXVII.
To find the greatest area contained under any number of 151.
right lines given, and another line unknown.
Let ABCDE be the figure; then fince ABE +
BCDE is a maximum; it is evident, whatever the
figure BCDE is, ABE muſt be a right-angled tri-
angle, right-angled at B.
Again, fince ABC + CDE + ACE is a maxi-
mum; it is evident whatever ABC and CDE are,
ACE muſt be a right angled triangle, right-an-
gled at C.
Alfo fince ABCD + ADE is a maximum; it is
plain, whatever the figure ABCD is, ADE muſt
be a right-angled triangle, right-angled at D. And
fo on if there were never fo many lines. And
therefore all the angles ABE, ACE, ADE, ſub-
tended by AE, mult be right-angles; and confe
quently the whole figure is infcribed in a femi-
circle, whofe diameter is AE, fo that the whole may
be a maximum.
Therefore if it be required to find the area,
we muſt find the diameter AE, and then find
the
512
B. II.
MAXIM A and
1
Fig. the area of the poligon ABCDE infcribed in a
femicircle.
PROB. CCXXVIII.
152. To find a line, which with three given lines, will con-
tain the greatest area poffible.
It is plain the line fought is the diameter of the
femicircle in which the three given lines are in-
fcribed.
Let ABCD, be the quadrangle, draw the dia-
gonals AC, BD, on which let fall the perpendi
culars CP, BF.
Let AB=b, BC=c, CD=d, diameter AD=y.
Then BD-Vyy-bb, and ACVyydd. But
(Geom. IV. 28.) CP =
fore b+ √
od
y
bc
d +
y
cd
bc
and BF=
There-
>
y
y
yy bb 2 area ABCD, and
y-dd 2 areab+
:
by+cd × √yy—bò = dy+bc
cd
✔yy-bb, and
y
-ad, and fquar-
bbccyy dayy
ing and multiplying, bby4 + 2bcdy³ + ccddyy-
b + yy - 2 b ³ cdy - bbccddddy++2bcdy
-2bcd³y
bbccdd. And reducing
bb y³ b4 y + 2bcd³ = 0.
dd bbcc
+ ccdd
+ d4
263cd
And dividing by bb - dd
y³ —bb y = 2bcd, and y
2bcd, and y being known, the
-cc
-dd
area is known from the foregoing ſteps.
PROB.
}
1
{
!
I
६
$
1
a
L
d e
b c
e f g h i
A
B
C D E F
X
Fig. 139
B
E
141
E
C-
140
A D
B
B
B
C
}
A
143
M
P
144
D
Y
B
142
H
Z
À
C
Ꭰ
E
A
145
D
P A
F
E
D
146
1
F
G
С
Ꭰ .
E
147
P
H
C
B
A
A a F B
D
B
149
E
D
H
150
A
E
B
M
F
A
D
G
&
B
148
151
D
Pl.XII. pa. 572
UN
L
Sect. XIV. MINIMA;
513
[Fig.
PROB.
CCXXIX.
SP is perpendicular to PM, and there is given SP, 153.
SN; and drawing NL, fo that the angle LDM
may be equal to SCP; to find CD, a maximum.
Draw NA perpendicular to CD, then CA=AD,
and CA is a maximum. Put SNb, SP =d,
SC=y, then CN=b—y, CP=√yy-dd. Then
by fimilar triangles, y: Vyy-dd :: by: CA=
b-y
yỳ-dd
yy-dd
max. and b—y² ×
y
yy
dd
max. = byl
xby'. Increaſe y by a
yy
very ſmall quantity e, then b-y-e
= b-yľ² 2e X
b-y.
dd
Alfo y + el
dit
A
yyye, and by divifion
2dde
dd
Whence byl
X
xy+2ye
yy
درحمہ
yy
dob
z dde
"{² = b
=byl
ze X b
y
X
yy
y3
-2ex by, and tranfpofing, 2e x by
2dde
2dde
× b −y +
хё
13
y, and dividing
yy
a'd
dd
ڈو
by 2e × b-y, I
y, I = x by + and multi-
plying by y³, we have y³ = ddb — ddy + dły, or
y' bdd, and y = Vbdit.
—
3
yy
L1
PROB
514
B. II.
MAXIMA and
Fig.
154.
PROB. CCXXX.
Given the fituation of the two places A, E, and the
river BD; and fuppofe a traveller going from
A to C, can travel 6 miles an hour on this fide
the river from A to C ; and 9 miles an hour an
the other fide from C to E; it is required to
know where he must cross the river BD, ſo that
be may go from A to E in the least time poffible.
Let AB, ED be perpendicular to BD; let
AB a, DE=b, BD=d, m=6, n=9, BC=x. Then
CD=d-x, AC= aa+xx, CE = √ bb+d=xl² •
mil. ho.
And per queft. m: 1 :: Vaa+xx :
mil. ho.
aa+xx
m-
= time in AC, and n: 1 :: dd + d -~1²:
2
bb + d―xl
=
time in CE.
Therefore
n
~
✔aa + xx
bb + dx}
+
minimum. Or
m
n
n
n√aa+xx +m√ bb+d—x
for x; then xxxx + 2xe, and
Xs
2e X d
m²
d. - x 1²
naa+xx+2xe + m√
√ aa + xx + m
Therefore
bb + d—x1²
✓ bb + d
bb + d
aa+xx+2xe = √ aa + xx +
xe
d-x-el²
aa + xx
min.
Write x+e
=
we have
ze X d-x
x1.
But
and
9
√66
bb + dx12
2e × d x =
ex.
Sect. XIV.
515
MINI M A.
ex dux
bb+ d—xl²
Therefore naa+xx+
+m√ bb + d―xl²
mexd-x
bb + d-x1²
nxe
Fig.
√aa+xx 154.
= n✓aa+xx
+ m√ bb + d—x. Therefore
m²
mexd-x
bb+d-xl
nxe
√aa+xx
aa+xx
=o. And multiplying,
0.
nx√bb+dd—2dx+xx = md - mx√ aa+xx; and
fquaring nnbbxx+nnddxx-2nndx³ +nnx4—mmddaa
-2mmdaax + mmaaxx+mmddxx-2mmdx3+mmx+.
And being reduced is,
nnx+-2nndx³+nnddxx+2mmdaax-m² d'a²o.
-mm+2mmd +nnbb
-mmdd
·mmaa
PROB. CCXXXI.
Within the given angle ACB, to cut off a given area 1550.
with the forteft line AB.
Let the area; s, c=fine and cof. C; CA=x,
CB=y, then per queft. sxy = b, and by Trigono-
metry AB√xxyy2cxy
min.; therefore
b
xx + yy — 2cxy min. but xy= and y
therefore xx +
bb
2cb
S
SSXX
S
= min. or xx +
2
b
SX
bb
SSXX
min. Put xe for x, then w+el²x² + 2ex,
I
I
2e
-2
and
or x + el
# + el
XX
2:3
1
Li2
Whence
xx +
516
B. II.
MAXIMA and
Fig. xx + 2ex +
£55.
bb
I
bb
2e
X
SS XX
SS X3
2bbe
bb
2 ex
$583
=0, and x =
o,
X =xx +
bb
SSXX
whence x4 =
and
2
bb
SSX3
SS
64
64
But y4=
bb
X
;
therefore
S4X4
54
bb
SS
34 = x4, and y=x
PROB.
S
CCXXXII.
156. To find the greatest parallelogram infcribed in a
triangle.
Let the parallelogram BDEF be infcribed in the
triangle ABC. Put AB=a, BC=b, DB=x, DE=y.
Then by the fimilar triangles ABC, ADE, a — x
ya: b, and ay ba bx, and y =
bx
b
a
—
But xy=max, or bx
ba-bx
a
bxx
= max.
С
Pute for the fmall increment of x, then the in-
crement of bx is be, and the decrement of xx is
* + el -xx=2xe, and the decrement of
bxx
a
157.
2 bxe
2bxe
2x
whence be=
and
I 1 =
and
>
a
a
a
x=a. Therefore y = b.
PROB. CCXXXIII.
Given the point P within the right angle ACB; to
draw the line APB, fo that AP x PB may be a
minimum.
Draw DP, PF parallel to CB, CA; and put
CF=6, CD=6, AD=x. Then by fimilar tri-
angles
Sect. XIV.
5'7
MINI M A.
bc Fig.
X
angles xbcy, and xy be, and y = • 157.
Then AP✓bb+xx, and PB =Vcc+yy; and
AP x PB = √bb + xxx √l cut
басс
bbcc
XX
min. and
fquaring, bbcc + + ccxx + bbcc min. and
XX
b+cc
ccxx +
= min.
Whence cc Xx + el² +
XX
b4cc
x + e²
b+cc
男
​ccxx +
or cc xxx + 2xe + b4cc X
XX
I
2e
b4cc
264cce
= ccxx +
Whence 2ccxe
X3
XX
X³
b4
37
XX
0, x =
x37
or xb4, and xb, whence y=c.
And AC=x+c=b+c, and CB=b+y= b + c.
Therefore AC = CB.
And if it be required to have AP+PB, a mini-
we ſhall have ^/«b + xx + √ cc + yy =
mum,
min. or bb+xx +
√bb + xx + √ cc +
XX
bbcc
= min. But
2
bb +x+el²
√bb + xx =
xe
= the
√bb + xx
increment of ✔bb+xx. And in like manner
-bbcce
bbcc
*√ect
is the increment of cc+
or
XX
bbcc
cc+
XX
bbce
xx√xx+bb
its decrement.
Therefore
xe
√bb + xx
3
bbce
xx
bb+xx
L 13
and x
bbc,
bbc, or
bbc, as in Prob. ccxxiii. by another method
PROB.
f
518
B. II.
MAXIMA and
Fig.
1
CCXXXIV.
PROB.
Given the fum of the legs of a right-angled triangle;
to find the legs, fo as to contain the greatest area
poffible.
x
Let a fum of the legs, one of them; then
xXa-x= 2 area = max. therefore x+exa-x
=xX a x, that is, ax-xx xe+ae-
xe
ax
e
109
xx, and ge
2xe=0, or 2xa, whence x = a,
and a x = 1/2.
12. Therefore the legs are equal.
And therefore when the area is given; the fum of
the legs will be the leaft, when they are equal.
PROB. CCXXXV.
Given the area of a right angled triangle; to find the
fides, when the perimeter is the leaft pofible.
Let a area, x=fum of the legs, v, y=the two
legs; then vv+yy+2vy — xx, but vy = 2a, and
vv+yy=xx-2vyxx-4a, and the hypothenufe
=√ vv +yy = √xx·
xx-4a; therefore x+xx-40
= perimeter = min.
min. write x + e for x; then
2
x + e 4a = √xx + 2x
– 40 =
XX*
4a
xe
;
whence x + e + √xx
40 +
a
ке
xe
≈≈ 4 ww-44, and e +
·4%,
42
XX
4a
44
+
CO. And e/xx
40ee
xxee, and 400 € o, and ee
xe, and eexx-4aee
or e=0.
And
40
therefore fince the increment of x is nothing;
therefore x is a minimum, and when x or the
fum
میم
Sect. XIV.
519
MINIM A.
५
fum is a minimum; then the legs are equal, by Fig.
the laft problem; therefore v = y=x, and
xx 2a, or x = √õa.
ŏa.
PROB. CCXXXVI.
Given the folidity of a Square pyramid DF; to find 158.
the flant fide AВ the leafi poffible.
≈
Let b = folidity, x=CB the height, y=2AC
the breadth, then AB =√xx+4yy. But xyy=b,
26
उ
and y = ; therefore AB = √xx +
X
minimum, and xx +
x, then xxxx + 2xe, and
36
4x
36
min.
Put x + e for
4%
36
3h
35
4x
4x+3e
4x
3le
;
whence xx+2xe +
2be
36
xx+
36
4xx
4x
4xx
4x
and 2xe
3 be
=0, and 8x3
8x³ = 35, whence
4xx
/36
I
8
1/36.
2
PROB. CCXXXVII.
Given the folidity of the Square pyramid DF, to find 158.
that which has the leaft jurface, excluding the bafe.
=
Let b folidity, x = CB the height, y = 2 AC,
or 2AD the breadth. Then AB =√xx + ¡y,
and ÷y × √xx+yy DBL, and 2y√xx+yy =
iurface.
But xyyb, and yy
36
Whence the
X
36
96b
4x
L 14
36
surface = 2√3 × √xx+
X
2√3bx+
4**
= maxi-
520
B. II.
MAXIMA and
Fig.
158.
maximum. And 36x+
936
4XX
maximum, or
1 ½ bx +
obb
max, write x+e for x, then
gbb
XX
XX
11
12be +
gb b
xx+2xe
gbb
y3b
186be
Therefore 12bx +
XX
X3
18bbe
gbb
= 12bx +
; and 12bc-
XX
203
XX
18bbe
18bb
= 0, or 12b =
,
and 12׳=186, and
203
X3
36
*3=
whence
3
*
V
3
2
26
PROB. CCXXXVIII.
159. To find the greatest cylinder, infcribed in a given cone.
BC or
or BF = b,
c=
Let axis AB = a,
3.1416, DB, DE or DG =y. Then by the
fimilar triangles ABC, ADE, a — x: y ::
a: b, and ay — ba
ba—
bx, or bx = ab
But cyyx maximum,
and
X x =
ab-ay
b
cyy X
ab-ay
b
abcyy-acy³
b
ac
азд
or
= max. that is ex
byyy max and byy-y
=
b
max. put y+e for
y, then byy becomes byy2bye, and y3 becomes
y³ + gy¹e. Whence byy+2 bye-y³-3y²e—byyy³.
byy+2bye—y³
And 2bye-2yyeo, or 2by 3yy, and y36.
ab-zab
Whence x =
I
..
3
PROB.
Sect. XIV.
521
ΜΙΝΙ Μ Α.
Fig.
PROB.
CCXXXIX.
Given the weights of two elastic bodies A, C; to
find the weight of the intermediate body B; ſo
that A Striking B at reft, and B with the mo-
tion acquired, Ariking C at reft, may make C's
motion the greateſt poſſible.
Let x weight of B, a
velocity of A.
y velocity of B, acquired by the ſtrokes
v velocity of C, by the ftroke.
Then Aa is the motion of both A and B, af-
ter the ftroke, as well as before; and a is the
difference of their velocities; therefore y-a
a is
the velocity of A after the ftroke. And fince
the fum of their motions remains the fame (Me-
chan. 10.), therefore xy+y-axAAa, or xy-Aa
Ay Aa, that is, xy + Ay = 2aA, and y =
2CA
A+*'
Again, xy is the motion of both B and C, and
y the difference of their velocities, as well after
as before the ftroke of B. Therefore v-y is the
velocity of B after its ftriking C. Whence
Cvv―yxx=xy, or Cv+xv=2yx. Whence =
2YX
C+x
H
ftion; or
4a Ax
A+xxC+x
X
A+xxC+x
A+xxC+x
X
= maximum per que
= maximum; or
= minimum, that is,
AC+A+C.x+xx
X
AC
minimum, or ACC
+
A+ C
522
B. II.
MAXIMA and
Fig. A +C+x= minimum; therefore + mi-
nimum; put x+e for x; then
ACe
AC
and therefore
XX
X
AC
*
AC
X
AC
A C
x + e
ACe
+x+e=
XX
+x, and throwing out the fuperfluous quan-
11
tities, e- : =
o, and xx AC, whence
ACe
XX
x = VAC.
To find x*
-
PRO B. CCXL.
the greatest poffible, fuppofing n
greater than m.
Write xe for
12
M-I
x, then x+e!" =x™+m e,
and xe = x² + nx
el
12
*+ el² = xm
NI
- nx e=x
772
12.-I
e; therefore x + el” →
112
R
" + mx-le — x*
92
x²;
; that is,
x².
And
by fubtraction,
MI
OF
M-I
mx
TIL
2-I
nx
11X
o, or
e = 0,
mx nx". And n being
have nx-mm, and x"
= nx
27-I
greater than m, we
m
720-772
; whence
n
7- M
m
30 =
PROB.
Sect. XIV.
523
MINIMA:
Fig.
PROB.
CCXLI.
To find the greateſt parallelogram infcribed in the gi- 160;
ven curve AMC.
Let MPBF be the greateſt parallelogram. To
the point M where it touches the curve, draw
the tangent TMD. Then if the fubtangent PT
be equal to the height of the parallelogram PB,
then MPBF is the greateft parallelogram, For
it is plain from Problem ccxxxii, that this paralle-
logram is the greateft that can be infcribed in the
triangle TDB; and as this is greater than any
other that can be infcribed in the triangle, fo,
much more, is it greater than any other that can
be infcribed in the curve, fince the angle M which
is in the curve, will in all other cafes fall fhort
of the tangent.
Therefore knowing the method of drawing a
tangent to the curve; you must feek the point P,
where the ordinate PM being erected, and the
tangent TM drawn, TP may be equal to PB.
Thus if AM be a parabola; put AB-a, AP=x,
then by the nature of the curve, AF=x, whence
TP=2x, PB=-x, therefore 2xa-x, 3x = a,
or x = a.
And the fame will hold good, if not in all, yet
in most curves which are convex to the axis. For.
fince the parallelogram is the greateft for the tri-
angle, it will allo be greateft for the curve,
fince the curve at that place coincides with the
tangent.
Otherwife thus,
Suppofe the nature of the curve be rx
where AP, PM=y, alío AB=a,
m
= 3*
BC = b,
Then
524
B. II
MAXIMA and
Fig. Then PBax, and a-xxy max.
160.
m
12
y = g
X
>
therefore a
m + n
n
maxim. put
m
ax
*
M
12
a. x + el
m+n
N
=x +
= ax
n
+
m
72
c
1
But
xxr X = max. or
*; then
put +e:
≈ = * ;
m
n
M-1
12
ax e, and x + el
m+m
m+n
12
X e.
Therefore
n
M-n
72
ax + ax ex
m
1
m+n
m
m
m+n
R
m+n
n
72
12
x
e=ax
n
712-72
whence
m
72
m+n
ax
N
12
772-12
M2
127
max = m + n xx
272-72
m
12
X € = 0, or
;
and dividing by
m-m +12
=m+nxx, whence
12
ma =m+ n x x
X
HIZ
m+n
2
Which is general for all parabolical
figures. Thus if m=1, n=2, as in the common
parabola, then xa, and if m2, n= 1, then
is xa, as in the fame parabola, with its con-
vexity towards the axis. If m=1, n = 1; then
a, for the triangle, as was proved before.
PROB. CCXLII.
161,Given the distance of the point A from the perpen-
dicular plane BC; to find the poſition of the plane
AC, through which a body shail defcend in the
fborteft time poffible to the plane BC.
Let AB be perpendicular to BC, AD parallel
to it, and CD perpendicular to AC. Put AB=b,
BC=x; then (Mechan. prop 34. cor. I. 4to.) in the
time
Se&t. XIV.
525
MINIM A:
time a body defcends thro' the inclined plane AC, Fig.
another body will fall perpendicularly through the 161.
ſpace AD. Therefore as the time in AC muſt be a
minimum, the time in AD muſt be a minimum,
and AD itſelf must be a minimum. By the fi-
milar triangles BAC, CAD, it is BC (x) :
CA (√bb + xx) :: CA
(√ob + xx): AD =
bb+xx
bb
minimum,
And
+x=min.write xe
X
hb
bb
bbe
bb
for x, then
11
;
therefore
x+e
XX
bbe
318
bb
bbe
+x, and
+e=o, or
XX
bb
and xx bb, or
xb, therefore
x =
1 =
XX
XX
+x+e=
BCBA.
Otherwife thus,
Deſcribe the circle AGC with the center B, and 162.
radius BA; draw AC and any other line AE,
and CGF parallel to it. Then (Mechan prop. 37.
cor. 1. 4to.) the times of a body's defcending through
GC, AC, are equal. And the times of defcend-
ing through the equal lines, of equal inclinations,
AE, FC are equal. But the time of defcending
through GC is less than the time of defcending
through FC. Therefore the time of defcending
through AC is lefs than the time of defcending
through any other line AE.
PROB. CCXLIII.
AB is a horizontal line, BD an inclined plane. It is 163.
required to find the position of the plane AD, through
which a body defcending from A fall arrive at the
plane BD, in the leaft time poffible.
Suppoſe AD to be the plane, draw AL per-
pendicular to AB, and DH perpendicular to AL,
and
}
#
526
B. II
MAXIMA and
C
Fig, and DF perpendicular to AD. And put b=AB,
163.s, fine and cof B, BD = x, AD=y. Then
by plain Trigonometry bb+xx-2cbx=y; and
}
2
AD (y) : S.B (s) : : BD (*)
SX
y
S.BAD
or ADH. And rad (1): AD (†) : : S. ADH
: AH = SX. And by fimilar triangles, AH
(1)
(sx): AD (y) :: AD (y) : AF =
yy
SM
11
bb+xx-2cbx But the time of falling through
SX
AF is equal to the time of defcending through
AD (Mech. prop. 34. cor. 1. 4to.). And this time
is a minimum, therefore AF is a minimum, that is,
bb+xx-2cbx
bb+xx-2cbx
bb
SX
min. and
bb
X
= min.
or +x-2cb= min. whence + x = min.
X
X
as in the laft problem. And therefore xb, or
AD AB.
Or thus,
On AF defcribe a femicircle ADF to touch the
line AL in D; draw AD, which will be the line
of ſhorteſt time. For the time of defcending
through all the cords in the femicircle will be
equal (Mechan. prop. 37, cor. 1. 4to.) to the time
in AD. But the time in any cord is fhorter than the
time in the fame chord when produced to the line
BL, which lies without the circle. And therefore the
time in AD is alfo fhorter than in any other line
drawn to BL.
PROB
Sect. XIV.
527
MINIMA
Fig.
PROB. CCXLIV.
To divide a given line AB, into three parts, x, y, z; 164.
So that xyyz³ may be the greatest product poſſible.
2.
First, fuppofe x+y=b a given quantity, to find
xyy a maximum, Then x-b-y, and by xyy
or byyy³ max. put ye for y, then bxy+el
y+e³ max. that is, byy+2bye-y³—3yye—byy
—y³, and 2bye-3yye=0, and 3y=2b, or y=3b,
and xb. Therefore y = 2x.
Again, let x+y+x=d, to find xyz³ = max.
Then by what is gone before, whatever z be,
y will be =2x. Whence 4x³z³max. But x+y
=d-z, or 3x=d-%, and 4x32³ = = xd — 21³,
4.
3
-
Xz³max, and d-vl³ X 23max. or d—zxz=
max. or dz zz = max. put ze for 2, then
dz + dez+el' = dzzz, or dz + de
2ze=dz-zz, or de-zzeo; whence 2z
x+y+ ≈, and z=x+y=3x.
Therefore
ZZ
d=
+ 2x
+3x or 6x=d, and xd, and y = (2x=) zd,
and z = (3x =) 3d.
POST-
POSTSCRIPT.
HAVING met with an account of my book of
Algebra in the Monthly Review, I intend to make
a few Remarks. of what has been faid, that the reader'
may not be miſled by the trifling Objections of ignorant
Pretenders, that fet up to be judges in things they have
no more than a fuperficial knowledge of.
In commenting upon the 73d Prob. pag. 209. the
Critic begins thus; Among the Problems in the Seventh
Section, is one of fo extraordinary a nature, that we ſhall beg
leave to lay it before the reader entire. Now I cannot think
why this Prob. is faid to be of an extraordinary Nature,
except it be that the Critic has not met with fuch a thing
in all his reading before. But I can affure the reader,
that there is not one of theſe things mentioned in that
Problem, but what he will frequently meet with in prac-
tice, if his defign is to uſe himſelf to Calculations, and
to the folving of Problems. And, I believe, I fhould
have been guilty of a great overfight, and therefore.
ſhould have done a manifeſt injury to my reader, if I had
not given him fome light in this affair, and ha' laid down
fome rules how to manage fuch expreffions when they
occur, rather than leave him in the dark without at all
explaining their nature.
After I had given fuch rules and obfervations as I
thought proper; I tell the reader in the Scholium annext,
that o in a mathematical fenfe never fignifies abfolute
nothing, but always nothing in refpect to the object under
confideration. To which this Objector replies, The
above elucidation, however true in that particular, is not, we
apprehend, fufficient to remove the difficulties that attend this
Problem. For fuppofe inſtead of our being employed in con-
fidering the area of a fuperficies, our attention had been en-
gaged in confidering the length of a line. It will then furely
follow, that when its length vanishes, it becomes a mathemati-
cal point or nothing. I know not what fort of elucidation.
may
POSTSCRIPT.
529
may be fufficient to remove difficulties out of fome peoples
heads. But it will furely follow, that this Objector does
not know what he is talking about. For the very in-
ftance he brings is fufficient to confute him. For when
the length of the line is vanifhed, it then becomes a ma-
thematical point, or nothing; that is, it becomes nothing
when compared to a line. But will he fay that a geo-
metrical point is abfolutely nothing? If fo, it would be
impoffible for Geometers to give a definition of a point
as a geometrical term, becauſe abfolute nothing has no
manner of exiſtence, and cannot be a term in any
ſcience. I faid a line is nothing when compared to a
furface, though it is fomething in itſelf. And for the
ſame reaſon, a point is nothing when compared to a
line, and yet is a thing real in itſelf, and not abſolutely
nothing.
He proceeds, But we cannot compare mathematical points
together, BECAUSE they are totally deftitute of parts; and
without parts there can be no comparison. One would think
that things are the moſt eaſily compared, when they are
under equal circumftances, and of the fame kind. For
points muſt either be equal or elfe unequal to one another;
and I take equality to denote a compariſon. But if one be
deftitute of parts, and the other has parts, there's an end
of any compariſon among theſe, as being different kinds
of things. But being deftitute of parts is not the reafon
why things cannot be compared. For example, fup-
poſe any one fhould try to compare a line with a fur-
face, and give this as a reaſon that they cannot be com-
pared, viz. that both of them are deftitute of parts;
I think he muſt expect to be laught at for affigning ſo odd
a reaſon, fince the true reafon is, that they are of dif-
ferent kinds; and for that reafon no compariſon can be
made. But whether we can or we cannot compare ma-
thematical points together, is a thing I have nothing to
do with; but Dr. Halley has taken fome pains to fhew
the proportion of mathematical points to one another.
And I leave him and this Critic to determine the matter
between them.
He goes on, befides (fays he) we have often Equations
where o fignifies abfolute nothing. Thus, if x=y, then
x-7=0. Yes, in abftracted Equations made at plea-
M m
}
fure
530
POSTSCRIPT.
fure among letters that have no meaning, or which re-
late to nothing; and then it is no wonder if it is fo.
That is, when his Problem is about nothing at all, the
refult may be abfolutely nothing. But I was writing
about practical things; and any body may understand,
that when a man is folving a Problem in earneft, it is
always in regard to fome fort of quantity, and then his
o will always be of the fame kind as the quantity con-
cerned, and has relation to nothing elſe. Even in
his Equation x-yo, x and y if they be any thing,
muſt be quantities of the fame kind, or no fubtraction
can be made; and then his o will unavoidably be of the
fame kind too.
What follows is very extraordinary; he fays, multiplica-
tion is nothing more than a number of additions, and diviſion a
number of fubtractions. Confequently, if we can neither aug-
ment nor leſſen a quantity by the addition or ſubtraction of 0;
we can neither augment nor leffen it by the multiplication or
divifion of o. For otherwife the very bafis of Arithmetie
would be defroy'd, &c. That is, for example, becauſe
0+9=9, therefore o x 9 muſt alſo be 9, for other-
wife the very bafts of Arithmetic will be deftroy'd. A moſt
excellent reaſon furely to deftroy the principles of Arith-
metic. Such fort of Critics as thefe, have I the luck to
be concern'd with.
Then he proceeds; In fact, the Cypher is ONLY the li
mit or boundary between negative and affirmative quantities
the point from which both begin; and through which they must
pals in order to change their denomination. This Objector
then feems to know nothing of the great ufe of the o in
all arithmetical Operations. And as little has he been.
uted to the folving Problems; or elfe he would ha' known
what frequently happens, that o may be one of the roots
of an equation; and then o has as real, intelligible, de-
terminate a value as 1, 2, 3, &c. and this is fomething
more than being the mere boundary between negative
and affirmative quantities; about which the Algebraist
troubles not his head, but attends only to the real uſe of
that root when it fo happens.
At laft he concludes, that the difficulties attending the
Ideas of infinity and nothing, ought rather to be imputed to
the folly of comparing things of different kinds. But this will
not
POSTSCRIPT.
531
not do, fin ce no body that ever difputed about theſe
things, was ever fo ignorant, as to think, that there can
be any compariſon made, by way of equality, among
heterogeneous quantities. For inftance, whoever was fo
fooliſh as to fay, that fuch a quantity of ſpace was equal
to fo much folidity, or equal to fuch a length of a line.
This therefore is not the reaſon why the ideas of o and
infinity are fo difficult to us. For my part. I know no
difficulty at all of comprehending them in a mathemati-
cal fenfe, and that's all I have to do with them as a
Mathematician.
I ought to take notice for the fake of the reader, that
he has made nonfenfe of the 5th Corollary. The true
reading is, Hence alfo 0° 1, or the infinitely fmall power,
of an infinitely fmall quantity, is infinitely near 1.
0°=1
The very next Prob. to this is of the like kind; which
is, to find the proportion of o's in particular cafes.
This he has not thought fit to meddle with, tho' of the
fame extraordinary nature, altho' it is rather more excep-
tionable than the former. The reafon he does not men-
tion it, I fuppofe is, becauſe feveral of the most eminent
Geometers have handled it, and he might be call'd to ac-
count for contradicting them; and therefore he chooſes to
let it pafs without any notice, not even fo much as to
tell his readers, that he never faw it folved by common
Algebra before.
Thus I think I have anſwered all his Objections, and
fhewn, that what I have afferted in that Prob. is folid
and true, and that his Objections are mere Cavils, and
amount to abfolutely nothing.
W. E.
*
FINI S.
ERRATA:
In Plate VI. Fig. 54, for A read B, for B read C, for C
read A.
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3
#
}
I
}
B
Fig. 152.
}
D
F
153-
P.
A
C
P
D
M
B
N
155
156
E
B
D
B
}
i
B
154
良
​C
D
157
P
B
158
B
C
F
159
M
160
F
A A
T
Р
B
161
E
B
162 1
B
163
A
164
T
A
D
λ
B
1
Pl.XIII, the Ent
OF
#
A 547090
DUPL