260 CENTRAL FORCES. 



then y = -^ and the curve B L F is quadrable. If we seek 



the circle V X Y by rectangular co-ordinates X 0, V, we find 

 the equation to obtain V = D in terms of x, is of the form] 



~Q 2 

 2^ 2 \/ jydx -^ 



Q cfdx 



(c being the constant introduced by integrating f y d x). 

 Now there is no possibility of integrating these two quantities 

 otherwise than by sines, and we thus obtain, nor can we do 

 more, the following equation to D in terms of x ; 



a-D V~2 . Q . a 2 V n + 2 Q 2 



C a 2 arc sin. - = - x arc cos. - - - . 



And if we get D from this, in terms of cos. x, we have then to 

 obtain P C by similar triangles, and from I P C being right- 

 angled and 1C = x, to obtain P I, in order to have the curve 

 VIK. 



But if we proceed otherwise, and instead of working by 

 quadratures, take v the velocity of the body at I, or in the 



Q 



straight line at D, and make the area described in a 

 second, and the angle V C I, we obtain as a polar equation 



f* fi T* 



to V I K, d 6 = (x being in this case both C D 



/ j 9 9 9 V. O 



x v 4 x v c 

 and the radius vector). Then, to apply this general equation 



to the case of the centripetal force being as 5, let the force 



3C 



at the distance 1 be put equal to unity, and supposing the 

 velocity of projection to be that acquired in falling from an 

 infinite height, the equation to the trajectory becomes 



c (I cr c* x 



de = - - . and integrating, = -- - x log. . 

 #V4-c 2 -v/4-c 2 



