66 Central Forces. 
cos. (v = r v 
1, arp dais ats Une | eres v-+-qsin. vtan.d= 5=q(1+ tan.*5)? 
Pee: — cos. v) q - 2 
or qtan. = ge = \3 tan.5+tan.*5 1 OF by (9) 
q tan. d= saat agrees with Newton’s construction, (Prin. 
\ A 
Vol. I, Sec. VI, prop. 30.) his M=3—, Op and qgtan.6= his GH= 
3M. Let oo the semi ancl of acircle rad. being 1, then if — 
Pp? E tA _ 
_P © wth = 
v=9) tan. =I, and A becomes : 3 > hence 2p! ; 
3A 
which agrees wtih Newton’s eae in cor. 1. and gtan.d=>5;5 5) 
p 
= Fy gives (4 i. ) ee = Ze the velocity of the par- 
| ‘i 
d(q tan. 4 
ticle at the vertex ;) or aoe 
:V¥::3:8, which is his propor- 
tion in cor. 2. and his cor. 3 is evident by (10) which are assumed 
on the supposition that a circle is described through the points A, S, 
P, in his figure ; its centre H being at the intersection of GH and a 
perpendicular erected at the middle of SP or 7, its radius SH=R, 
GS=q, HSG=4, ASP=», ...HSP=v—é 
/ 
Let r, v, ¢, A, become 2”, v’, t’, A’; then by (8) r-=-~ Pp = 
2cos.*5 
r B (1+-um. 5) (12); and by (9) ae tan. © tan. 5) oo Sig 
ee by subtracting (9) from aay. resolving the remainder into 
: v v’ ‘ 
factors, and putting for 1+ tan.? 9’ 1-+tan.?-3> their respective 
Qr 2r’ ; p! iy 
equals a” pl as given by (8) and (12); there results “> tan.g — 
| p’ vy’ v 
tan. Ae (marge + tan. 5 tan. 5) = 3(A/ —A) (14). Puto'- 
v=2u and let c= the chord connecting the extremities of 7’, 7, and 
r’+r+e=2m,r+r—c=2n.’.r’+r=m-+n; (by trig.) ctr! + 
7? —2r’r cos. 2u, or since cos. 2u=2cos.2u—1, c?=(r'-+r)? = 
