229 
therefore 7 is constant for all points, and 
ds T 
1 ; : 
P-5 av sin’ pb 
therefore 2.s= STs eS (ak Gat Oe a 
eo es oP Sint 1+(1-$). tan? 
2 p 2p p 
In the integration indicated by the last expression, two 
: : ; lei ' 
cases arise according as p is greater or less than kaa that is, 
according as the power of growth is more or less than sufficient 
to balance the direct resistance of the current. 
In the first case take a a subsidiary angle such that 
4 ee OU. 
sino ——— 
2p 
p d.tand 
—-  6¢= Ca SO = 
T 1+cos’a.tan’d cos 
then . tan” (cosa. tan ) ; 
no constant is required if s and @ vanish together. 
Also if «, y be the rectangular coordinates of P, the axis 
of x being parallel to BC, 
dz dx ds ee 
dp ds ° db Pe p (1 —sin*a. sin” )’ 
Ag Nee =/[—3 __d.singd 1 i 1+sing.sing 
Y=} {osint'a. sin’ 2sina °l—sina.sind’ 
on p sin 6 df - | d.cos 
Bee =e — sin’ a. sin’ cos? a + sin’a . cos’ d 
—l1 
= tale (tale COs) 5 
sin &.COS a : $) 5 
no constants are required if # vanishes when ¢=0, and y when 
Tv 
b= 5: 
