70 | Solution of a Problem in Fluxions. 
is wholly situated; (since x, y, z, are common tos and (/).) By as- 
suming (7) for the plane of x, y, I have z=0.°.é=0, and [= and 
(C) does not exist: also (A), (B), become the same as in the last 
Journal, (on the suppositions there made,) as they evidently ought to 
be. Put rsin..j=p the perpendicular from the centre of force to 
C’ ds 
the tangent tos at the place of the particle; gets qa. the veloci- 
ty; V’= the velocity of a particle describing a circle, about the same 
centre of force, at the distance r; D= the distance fallen through 
to acquire V from rest, by a particle, when continually acted on by 
d: 
a constant force, equal to F, (at the distance rap Ri half the 
chord of the equicurve circle, with s at the place of the particle, 
estimated on r. It is evident that ([) can be changed to . 
ee 
ae a= sin. =z) ae “3 ae ~ pdr Br ™ age : 9 
dr dr 
C/2 C/2 Vr ys 
=>R Ren?) > ap (L)3 and that (H) can be 
C? | jridv? +dr? ds? Vav 
changed toad ridy? )=-a(qa)=- - =F (M); 
eae 2dr 
“.VdV +Fdr=0 (N). : 
it Fdr is integrable, [ may change d into 6, and (N) becomes 
VoV +-For=0 (0); (@ being the characteristic of variations.) J 
shall suppose that Fdr is integrable; that is, that F is constant, or a. 
function of r only; and shall put its integral SFdr=or. Then the 
integral of (N) is V? =D —2or (1); (D= const.), (1) corrected is 
V2 =V"?4-2(9'r—or) (2); V”, o’r, being the values of V, or, at 
some given point of s. 
From (2) it appears that V depends on V”, 9’r, or; suppose then 
two spherical surfaces to be described from the centre of force (as 
centre) through the distances corresponding to ¢’r, gr; then it is evi- 
dent that if the particle passes through the first of these surfaces (in 
any point whatever) with the velocity V”, it will always arrive at the 
second with the same velocity, V, and that in whatever point it may 
meet it, and whether it moves in a straight line, a curve, or curved 
surface ; provided the curvature be continued, so that the direction 
