284 Solution of a Problem in Fluxvons. 
d?y dz xd? x --yd*y+-2d?z ge 
Ben gr PO ge rx dt? ree 
ly —yd zde —ad. 
(a —_)= aE”, -(¢} 5 tos. vxd(—G vactarb ie ) sin. v 
Say yee a 
zd d 
xd( aE : 7 —)= —F” (d); s= the portion of the curve de- 
dt 
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 de= the elementary angle contained by r and r+dr; 
then rd@ and dr are the legs of a right angled triangle, whose 
hypothenuse is ds; .° . ds? —dr? =r2dp?, (e); but rdg is the 
hypethenuse of ariother tight angled triangle, whose legs are 
r cos. 9 dv and rdé .". r2d92 =r2cos, 29dvu2 +-r2do*, (f). The 
second differential of ( it) eens tof ¥ joes dt as con- 
— dx? sa ia ne BBE yd? yted?z — 
stant,) gives = di? 
(g¢); but it is evident that dx? dg? = =ds?, .’. by substi- 
dr 24 rd?r ds? 
tution in (g) of ds? and by (6), I have —j5-—=7ia— rF, 
rd*r ds? —dr? 
Fas =— ga ~— TF; or by (e) and (f) I heve 
r cos, *6dv* +rdo? — d*r : 
are is (A). Multiply the differential 
r? COS. 20d 
it 
of (z), (taken relatively to z,) by 22, and I have —~- 
ais ak the differential of this gives d =H Nac ee 
“, As 
qtr arias (B), by (9. Multiply the differen- 
zd. dd 
tials of (3) and iy), by z*, and there results ane dz 
_— cos. — sin. D COS. 9 SIN. 7 zdy — ~ydz_ : sin. vd6 
sae mi 
a dt 
ar r? Cos. v Cos. 6 sin. 6 | : 
rE at > multiply the differential of the first of 
