i88.] MOTION ON A FIXED SURFACE. IO i 



187. The constants of integration h, c can now be determined if the 

 initial values of 0, 0, <j> are given. 



Eliminating dt between the equations (10) and (n), we find the 

 differential equation of the path 



,, cdB 



, (12) 



sin B V2 r 2 sin 2 0(gr co$0 + h)-<? 



whose integration gives the equation of the path in the spherical co- 

 ordinates 6, (f> (colatitude and longitude). 



On the other hand, if <j> be eliminated between (10) and (n), we 

 find the relation 



d _ r* sin dQ _ , v 



" V2 r 2 sin 2 0(rcos + h)- / 



which, upon integration, determines the time as a function of 0, or the 

 position of the point in its path at any given time. 



The equations (12) and (13) contain therefore the complete solution 

 of our problem, with the exception of the determination of the resist- 

 ance N. Their discussion cannot here be given, as they lead to elliptic 

 integrals. 



188. To find the resistance N t multiply the equations (6) by x, y t z 

 and add ; this gives 



m (xx + yy + zz) = mgz Nr. ; ; (14) 



Differentiating twice the equation of the sphere (5), we find 



x'x +yy + zz + x 2 +y + 2 = 0, 



or since x 2 +/ + z 2 = z/ 2 , 



xx -\-yy + zz = v 2 . 

 Substituting this value in (14), we find 



or, by (8), N=-($gz + v<?- 2gz ). (15) 



If the constraint be one-sided as in the case of a string pendu- 

 lum, the particle will leave the sphere whenever in its upper half z 

 becomes ^ f z v<?/$g. 



