Central Forces. 335 
of r, p (or r sin.J,): if Ac’? r=p at the origin, c==0, and the 
particle evidently describes a circle whose centre is at the centre of 
force, and its radius = the initial value of r; but if J is not a right 
A 
angle at the origin and c’* cosec.2L.=A, c=0, .*. cosec. ave 
const. at all points of the curve, which shows the curve to be the loga- 
rithmic spiral, its centre being at the centre of force. c’# cosec.2b=A 
c/2 
c/a’? hence and because c=0, (5), (6) become 
gives tn.* = 
dys: —tan.Ldr 
s 
> dt= 
—tan.Lrd 
“ ir fs supposing 7 to decrease ; hence by 
taking the integrals on the 9 as ie v, t commence when 
R 
r=R, I have v=tan.p xhl—s oe y (Re? )3 or r= 
= 
VR? —2¢/ a ee or t= 
e’cot. xt, v = tan. X “AR? — 2c/ cot. b xt? 
R? 
Weot.p * (1 ements), h.l.e=1; the values of r and v in terms 
fe 
os Bs ac R?* tan 
oft will give r, » at any time which is less tha aoe os the time 
fom the extremity of R to the centre of the spiral, as is evident by 
making r=0 in the value of t, at which time v becomes infinite. If J 
isnot a right angle at the origin, but A=c’?, then by (3) r? tan.?b= 
‘ss 
=const. which shows the curve tobe the hyperbolic spiral. (Vince’s 
cad 
Fluxions, se 129, ex.2.) In this case (5), (6) become sitet BR 
a 
Fi dt=—, Supposing r to decrease ; by taking the integrals on 
a 
Ror 
c’ 
the supposition that ¢, » commence when r=R, I babe vax Rr? 
ir 
_R-r t ; 
= =R— _ %—————, and the time from the 
aa > orr=R tv c, Poss We an 
“xtremity of R to the centre of the spiral = ma hence the values 
R 
ofr, are easily found at any time which is less than = ig By ta- 
rd teal 
king the value of the integral of ove x —— between r and r in- 
