110 On the Solution of a Problem in the Calculus of Variations. 



Now, since y and y x are quantities to be determined, neither 8y 

 nor &/j is equal to zero. Hence 



Pl +1=0 and —^— --1 = 0; 



that is, j» = + oo and ;j, = — x . Hence if C and D be the ex- 

 tremities of the ordinates, the curve is at these points continuous 

 with the ordinates; also the equation (1) shows, by putting 

 jo = oo , that for each of the points y* = b. 



The equation (2) between y and s being put under the form 



y* = 2a' 2 + b-2a ^a^ + bcos (^ + *), . . . (4) 



it will be seen that in the case in which 6 = 0, and consequently 

 by (3) cos£=l and k = 0, we shall have 



2/ 2 =2« 2 (l-cos^), 



ory=2a 8 in^. 



The curve is therefore a semicircle, the radius of which is 2a. 

 Thus it appears that this is a particular instance contained in the 

 general solution. 



The equation (4) shows that for any other value of b the curve 

 differs from a semicircle. Its length from C to D is readily 

 found. For since y —y^, it follows from (3) that 



cos #=cos( — -\-k J. 



Hence — = 27r, or s,=27ra = half the circumference of a circle 



a ' 



whose radius is 2a. 



Also the area enclosed by the curve, the ordinates AC and BD, 

 and the base AB, can be exactly determined. For from the 

 equation 



we obtain 



(b-y z )ydy 

 V X ~ </4 a y-{b-yY 



which being integrated from A to B, gives for the above-men- 

 tioned area 



2a\/b + 27ra 2 . 



If the equation (1) were integrable, the values of the three arbi- 

 trary constants might be found from the given value of the en- 



