FLUXIONS. 
493 
2 ah 
, l,en«,i==^+j«=2+.X •'*“*-*’* 
4 « ’ 4 a 4 « 
.’7 
yy 
I 
y f M 2 -\-y 2 fM 2 +7 2 jy y / M 2 +^ 2 V M 2 -)-^ 2 
g _ 9 y+ 4 ^ . vvhofe fluent, by M ‘ 
- A) > • ■ -- ,/jJ. ' 
+ 
U' 2 i 
4« 
by Art. 39. is z =. 9 --~- L~+C ; now when ;'=o, 2=0, 
274 
in which cafe, this equation becomes o=—-j-C, C— 
9 v—J—42}1 ? 8 a 
hence ;( b y pro P' i5-and Prop.17. Ex.8.) 
- M+./M 2 4-v 2 
2=v' M 2 +7 2 — MX h. 1 . ■ ■ y- --— -f-C; but when 
8a , 
— ; hence, 2= 
27 27/4 
2=0, jy=£, and we have o=y / M 2 -|-/> 2 —Mx h. 1 . 
M -U U M 2 -j-^ 2 
27 
Mb 
+ C; hence, C= — ^M 2 -^ 2 + M 
M4-t/M 2 +i 2 -- 
Ex. 2. Let ByA be a cycloid; to find its length. X h. 1 . -——-; therefore 2:2= fM z -\-y 2 — 
Put BD=a, B r=x, Bp'=2, yv=zz, vz—rq—x ; then bv __ 
the prop, of th<.circle, Br: B«:: B*: BD, B«*=BD „ h , M+V f +tl _ M X 1 ,. 1 . 
X -br=ax, and an—0,2x2 •, and by the prop, of the cy- M* 
M+y/M 2 4- y 2 
My 
— v'M 2 -f-y 2 — -v/M 2 +i 2 + M X h. 1 . 
■. C=o; confequently 2=2« 2 x 2 = zBrc. 
Ex. 3. Let AC be the common parabola ; to find its 
2yy . yy 
length. Here ax—y 2 , x— = (if a =.b) zL ■ hence, 
...fafsgfxj = (by 
b 2 
multiplying numerator and denominator by jyxj^-H 2 )' 2 ) 
« y’ijy.J^yy j ^ ly^yf-a^yy _ i ( ly^yflTyy 
1 x ~ *b x 2 b 1 y^fbf^k 
(by dividing the num. and den. of 
, 1 i>-yy 
+ T x 
2 b y 4 yb 2 yz]i 
the lad term by y) X y 4 -\-b 2 y' z \ 2 X zy z yArb 2 yy J rW 1 'A 
2b x y \ \y\ z=z—, and z— ^ 
— • now the fluent of the fir ft term is — X 
26 
Mp'+ 7 t/M 2 +^ 2 
\J M 2 -|-y 2 
Ex. 5. To find the length of a circular arc. By 
Art. 46. *=~= (by divifion) 
1 3 /5 ^7 
&c. hence, z~t - -1 -- -j- See. + C ; but 
3 a 1 5a 4 7 a 6 1 n 
when £ —o, 2—0, therefore C—o; hence, z—t -- Y 
3a 1 
t 5 t 7 
-7 4 - &c. Now 'if a=z 1, and 2 be an arc of 30% 
5 « 4 7a_ 
then /= 4/^=0,5773502, which being fubftituted for <, if 
we take 12 termsof this feries, we get 2=0,5235987, the 
length of an arc of 30°; which multiplied by 12 gives 
6,2831804 for the length of the circumference of a circle 
whole radius is unity. If we take the arc 2=45°, then 
will t—a ; hence, 2= a x 1 —f + I —7 + See. 
To find the LENGTHS of SPIRALS. 
Prop. XXIII .—To find the length of a fpiral SC. 
54. Let the ordinate SC=y, the curve SC=z, CY=f; 
then, by Art. 31. s 
Cj—z, E s=j> ; and 
by fan. triangles, t : 
y 4 -\-o 2 y 2 \i, by Art. 4®, and the fluent of the iaft term 
is \b x h. 1 . 7 + 7 2 -H 1 *, by Art. 45. Ex. 4. hence, 2= 
yy 
the fluent of —, 
2 b 
X y 4 -\-b 2 y 2 \ l -j- \b x h. 1 . yfi-y 2 -\-b 2 \h 4 -C; now when 
y = o, 2 = 0, in which cafe, the equation is oz=feb 
X h. 1 . b -j- C; hence, C =—ii h. 1 . b\ therefore 
(Algebra, vol. i. 308.) 2=—- X y 4 -{~b 2 y 2 r + 2 b x h. I. 
2 0 
y±f±!fiV. 
b 
Ex. 4. To find the length CD of any part of the 
logarithmic curve. (See Fig. p. 492.) Put AC —a t 
correfted if necef- 
fary. 
Ex. 1. Let SC be the logarithmic fpiral; to find Its 
length. Here t • y m : n, a conftant ratio; lienee, 
t = —, £= and 2=— + C ; but when jy=o, 
n vi vi 
ny y* 
2=0, .’. C=o; confequently 2= — = — ; therefore 
m t 
CY : CS :: CS : the length of the curve. 
Ex. 2. Let it be the fpiral of Archimedes; to find 
by 
its length. By Art. 32. Ex. r. t= 
vy 2 +b z 
; hence 
My » 
AB=x, BD=y, CD=z; then—- —x (Art. 49.), there 
yfy 2 - i-2> 2 
k=Lj__— which is the fame as the fluxion of the 
(M 2 / 2 
JV M 2 +i 2 
y 
length of the parabolic arc, Art. 53. Ex. 3. z— 
( by 
fore z= yf x 2 -bj 2 ~ v /—p- +j 2 
tnultiplying the numerator and denominator by v'M 2 -fy 2 ) 
Vol. VII. No. 444. 
2 b 
X h. 1 
y -Yf y 2 d-b 2 
$ K 
Ex. 
