A'oA

FLUXIONS.

of fluents; thus the fluent of x is x, or x + c, (art. 18.) whateverbe the value of the constant part c, called the correction.

To determine the correction, consider whether the fluent be-comes equal to nothing, or of a known value at the time it ought;if not such a quantity must be annexed as will bring it to its pro-per value. Or proceed by the following rule :

In the general equation of the fluent-, write for each quantitythe value it is known to have at any one certain time or position,and subtract the sides of the resulting equation from the corres-ponding side of the general one, ami the remainder will give thecorrection sought.

J.et » = 4ax 3 i, the general fluent is z = ax 4 , and let z = dwhen r = 6; then,

For the general equation z = ai 4 ; substitute d for and b for x,then d = «/A, and by subtracting z — d = ax* — ah*, or z = ax*— al>'-\-d the correct fluent, the correction being equal tod — ah' — c, it is evident that if d = 0, the correction isc = — id)', and if h = 0, it is c = d, if dzzO, and 6 = 0, the cor-rection vanishes, or is equal to nothing.

Prob. I. To fin’d the Area of any Curve.

(42.) Rule. Multiply the ordinate by the fluxion of the ab-sciss, and the product gives the fluxion of the figure, whose fluent,corrected if necessary, is the area of the figure.

Examp. 1. Fig. 17. Let the curve AllMIf he the commonparabola. Let u represent the area, and u its fluxion.

?/ = u z x 2 ; therefore u — Rm IIII (= yx) — a* x 2 x ■: whence u

= | X a 2 x 5 = •§«* x 2 X x = j yr = 2 x A15 x Bit: hence aparabola is ^ of a rectangle of the same base and altitude.

Examp. 2. Let the proposed curve (SDH (tig. 18,) be of sucha nature, that (supposing Alt unity) the sum ot the areas CS'l BCand CDC 15C, whose abscisses are AT and AC, shall he equalto the area CRN BC, w hose corresponding absciss AN is equalto AT x AtJ, the product of the measures of the two formerabscisses.

Fir-t, in order to determine the equation of the curve (whichmust he knot'n before the area can In- found,) let the ordinatesGl) and NR move parallel to themselves towards 11F ; and thenhaving put CD = y, NR = :, AT = u, AG = s, and AN = u,the fluxion of the area C DGB will be represented by i/.s, ami thatof the area CRNB bv ;« ; which two expressions must, bv thenature of the problem, be equal to each other ; because tjie latterarea (.STB, which is here considered as a constant quantity : andit is evident that two expressions that differ only L>y a constantquantity, must always have equal fluxions.

Since, therefore, ys is = zu, and u = as, by hypothesis, it fol-lows, that u = as, and that the first equation (by substituting forv) will become ys = azs, or y = uz, or lastly i/s = zu,s, that is,Cl) X AG=NRx AN: therefore, Gl): "N R : : AN : AG ;whence it appears, that every ordinate of the curve is reciprocallyas its corresponding absciss.

Now, to find the area of the curve so determined, put AH= 1,15C = 6, ami BG = x': then since AG (I 4-x): A.B (1) :: BC

b

(6):GD (») we have y = -, consequently u i=yx) =

1+x

hi-

-= 6 x : x — x.v -f- xfv — xfs -f-, Ac. Whence BGDC,

I -f"■*

X* I 3 X* X s

tlie area itself, will he = 6 x : x-1-1-, &c.

2 3 4 . 5

which was to be found.

Hi) Hence it appears, that as these areas have the same pro-perties as logarithms this series gives an easy method of comput-ing logarithms ; and the fluent may be found by means of attibleof logarithms; and every fluxion whose fluent agrees with anyknown logarithmic expression, may be found the same way.

Prob. II. To determine the Length of Curves,

(4.4.) Fig. 19. Because C cd is a right-angle, C 4* = C e* +d c- ; wherefore the fluxions of the absciss and. onlinat* being

squared, their sum gives the square of the fluxion of the curve ;whose root being extracted, and the fluent taken, gives the lengthof the curve.

Examp. To find the length of a circle from its tangent. Makethe radius AO, fig. 19, = a, the tangent of AC = t, and its so.cant = s, the curve = z, and its fluxion * ; because the trianglesOTC, OCS, are similar, OT : OC : : OC i OS ; whence OS =a % a■ a* a q lt

—, and SA = a -= a -; whose fluxion is - {

4 s */a 1 -f- F

and because the triangles OTC, d C e aresimilar, TC (=t) : TO (=

_ / a q ti \ a q i

v/it 4 -f- (’) i; C c = | - ] • C d = —-=r lluxiou of

V+^)v

the curve. Now by converting this into an infinite series we have

in in in

the fluxion of the curve = (- 4-, Ac. and come-

a 5 a* a 6

o t* r <’

quently s — t -1-1-, Sic. = AR.

3« e 5a 4 7a" 0a s

(45.) Other series may lie deduced from the versed sine, se-cant, Sec. and these are of use for finding fluents which cannot beexpressed in Unite terms.

I’rob. III. To find the Content of a Solid.

(4t>.) Let the surface of the generating plane be multiplied bythe space it passes through in any time, the product will give asolid which is the fluxion of the solid required : the surface mustbe computed in terms of x, which represents the line or axis onwhich it moves, ami by its motion on which the fluxion is to bemeasured, and the fluent found will give the content of the solid.

Examp. 1. Let it he proposed to find the content of a coneABC, fig. 20.

Put the given altitude (AD) of the cone = a, and the semi-diameter (HD) of it- base = 6, the solid = s, its fluxion = s andthe area of a circle, whose radius is unity, =p: then the dis-tance (AF) of the circle KH, from the vertex A, being denoted

bx

by ,r, we have by similar triangles, as a : 6 : : x : EF (y ) = —.

a

pb-x 3 x

Whence in this case, s (=pj*x) =-; and consequently *

ad

_ pb 3 l 3

—-—; (=p x III) 2 X 4 AD) for the content of the whold3a’

coni’ ABC : which appears from hence to be just -J- of a cylinder ofthe same base and altitude.

! Examp. 2. Let it be required to find the solid content of a sphe-roid, A FBI I, fig. 21. Let the axis AB, about which the solid isgenerated, be = a, p = area of a circle to the radius I, and theother axis F11 of the generating ellipsis = 6; then, from the pro-perty of the ellipsis, we havea 8 : 6*:: A D x BD (x x a — x) : DE

! 6 1 _ _ .

■ (y*), JIem:<*y 3 ■ = — X a x— jj ; and theftuxion of the solid, s

! a 8

, P A 8 _. . ph*

l(= p y 3 x) = — x 4x. x — x‘ x and the solidity s =3 — X

■•a xx — 4 x* = the segment ALE ; which, when AD (g) = AB/ p b* _

(a) becomes I-X \a 3 '~- 4a 5 )\p ab 3 * = the content of the

\ 01

whole spheroid. Where, if 6 (Eli) be takien = a (AB), we »ht»Balso get •$ p a for the true content of the sphere, whose diameter i*«. Hence a sphere or. spheroid is \ of its circumscribing cy«U J "

pb ' .he

der: for the area of the circle FH being expressed, by —> ,n

4

content