284 Solution of a Problem in Fluxions, 



d^y ,., d^z xd^x-\-tid"u-\-zd'^z 



^ixdy—ijdx\ izdx-xdz\ 

 ( Jt j=jP''(c)' COS. uXf/f j^ j+sin. D 



dt Jt 



fzdy—ydz\ 

 Xa( -^ j = — F'' {d)\ s= the portion of the curve de- 



Tit 

 scribed, (in the time ?,) its concavity being supposed to be 

 turned towards the origin of the co-ordinates ; it is evident 

 that r= the distance of the particle from the origin of (x, ?/, 

 z) ; let d^=^ the elementary angle contained by r and r-\-dr; 

 then 7-df and dr are the legs of a right angled triangle, whose 

 hypothenuse is ds ; .'. ds^ -~dr^=r^d^'^, (e) ; but rd^ is the 

 hypothenuse of another right angled triangle, whose legs are 

 r COS. dv andrde .'. r^d^'^ =r^cos. ^Odv^ -{-r'^de'^, (/). The 

 second differential of ( 1 ) relatively to t, (considering dt as con- 

 ^ . d7'"-\-rd'^r dx~ -\-dii^ -{-dz^ xd"X-^yd^y-\'zd"z 

 stant,)g.ves -^p— = J^ + -^g^/ • 



(g); but it is evident that<?a:2-j-f;y2_|_^2:2=(Zs2^ .'.bysubsti- 



dr^ A-rd^r ds^ 

 tution in {g) of ^s= and by (6), I have jr^ =-3— — rF, 



?'d"r ds^ —dr^ 

 or -^^= — -^2 rY, or by (e) and (/) I have 



r COS. ^edv^-i-rdo^ —d^r 



^p =F, (A). Multiply the differential 



V "^ COS. ^ (^dv 

 of (z), (taken relatively to t,) by :r-, and I have j: = 



xdy—ydx ,, jn^ .. 1 r ,. • • j /r^ cos. ^Odv 



dt 



-, the differential of this gives d 



(v COS. "oavs. 

 df—) ' 



dt 



"^ en- 



zdx — xdz 

 tialsof(3) and (y), by c^^ = , and there results j* = 



(r^ co^. vd^-\-r^ ^xn.vcos.esm.Qdv) zdy —ydz r^ sin. vdB 



~ ~dt dt ~~ dt 



r^ cos. ?j COS. sin, e 

 + -j: ™ ' multiply the differential of the first of 



