ON SUSPENSION BRIDGES. 369 



iHere e = o, and the equation ( i\ ) becomes 



Differentiating, making dy constant, and sub- 

 stituting ^(dv^ 4- dif) for dz gives 



Putting dx = wdy, d'x = dwdy. Substituting 

 these in the last equation gives 



adwdy = hdy^ ^ (^vo^ + 1 ) + c%», 

 whence adw = bdy V (w* +1 ) 4- (^dy^ 



dddl ;tj1< a dw 



And dx = wdy = 



6 v(«,«+i;+-^ 



6 ^/(^> 4. !)+-£. 



To integrate the latter equation put w^ + i 

 = v', then wdw = i?c/i;. Substituting we obtain 



a vdv 



b ' c 



Integrating gives i = A[^„_f., log. (,;+£. j-j ^ q^^^^ 

 Substituting for v its value v/{u;*-f n 



ar=|-[v(«.'+l)-y.log.(v/(»'+l) + y)]+Const. 



But when x = o.«;=^= „ .-. Constant = --I- 

 3 A 



