Solution of a Problem in Fluxions. 7t 



of the motion changes by insensible degrees, and that there is no fric- 

 tion ; also the law of force is supposed to be the same in all these 

 cases. (Prin. b. I. sec. viii. prop. 40.) I will now show the use of 

 the different forms of F, as given in (L), by a few examples. By 



the form F= — jj it appears that F is as . 7 at different points 



of the described curve. (Vince's Fluxions, art. 206.) Also by 



C'2 v^ . 1 V2 . 



F= — rR~~R ' F is as — n or as ^ • (Prin. b, I. sec. ii. prop. 6. 



cor's 2, 3.) The form F==— ^=--— gives V:V'::V — 



/dp V^ V2 



' V — (3); (Vince's Fluxions, art. 208) and F= p- =^- gives 



R C'^ 



D = ^ (4): (Vince's do. art. 209.) The form F=—^.—^ 



furnishes a very simple solution of prob. 2. prop. 7. sec. ii. b. I. Prin. 



VP 



(See Newton's figure.) For R=-^-5 r=SP, sin. 4' = sin. SPY = 



VP 



—sin. PAV (Euc. 3. 32.) =Tv' substitute these values and reject 



1 



the invariable quantities 20^^, AV- , and there results F as gpa vPVs ^ 



the same that Newton has found. The same form does also enable 

 me to demonstrate very easily the second and third corollaries to the , 

 same proposition. (See Newton's fig.) I shall suppose F to denote 

 the force to the centre S ; F' the force to the centre R. Then I 



have the equations F^ ^.j^^-^^ ,^ ^ ^'""T^R^iT^"'' (C'',r',R',4' 



being the same for R, as C, r, R, 4^ respectively for S.) Hence I have 

 „ _ C'2 C'^ ^ C'^^r^^R sin. 24. 



* ** •■r^Rsin. H * ^'''R'sin.H'' K' ^sm,'-^'' 



PV PT 



but R=-2"' ^'"=~2~' '"^SP, r'=RP, 4.:r=SPG=T (Euc. 3. 32.) ; 



and 4^'= the supplement of V (Euc. 3. 32) ; •'• sin. 4' = sin. T and 



sin. ^1- sin. T PV 

 sm. 4-'=sm.V .•.g-j^^=^j^^=py (by trig.); substitute tliese 



values and there results F : F': :C'- XRP^ xSP : C'^ xSG^ (5) 



SP^xPV^ 



(since — pp^ = SG^ by the sim. tri's TPV, SPGj) hence if 



