Solution of a Problem in Fluzions. 287 
dv? — d? 
same as oS and SP? xQT?=r‘dv?,). The use of 
tap om (E) is to determine the curve when the force 
s given, and it is obvious that it requires ie to be a func- 
cuts the curve at the same angle, .:. by H— erga) 2 
:.F varies as => ; for different points of the curve, (which 
agrees with SFO: 9th, sec. second, Prin.), if }= a right an- 
Bie the cosec. }=1, and the spiral becomes a circle, and 
—- =F=const. for the same circle, and for different cir- 
r3 
neg du*r* 
cles a for c’? its value —7,~’ it becomes 
= =F, b eeaV nithiegel 
an J=> = y putting “ae de =V =the ve ocity,(which 
aS with Prin. sec. second, Ae 4th, cor. Ist.) esr by 
ashing the finite agin of (D), relatively to c’? and 
12 
regarding r and oa as constant, I have —,—=DF,(D be- 
ing the characteristic of finite differences.) ‘f De’? is consid- 
ered as constant, DF varies as75 ; (which in prop. 44th, sec. 
9th, Prin.) I shall here leave Fes subject, as I suppose I have 
said enough and pea too much already. 
CORRECTIONS. 
P age 284, 6th line from bottom, dele (z) and insert ns 
eS oe LOM ee, insert dv sae sin. G3 in tor thus, sin. 6dr. 
Some smaller corrections, not deemed important, have been omitied.— Ed. 
