PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 



615. 



Let the line itself be called x; the angle which it makes with the vertical ; 

 g the force of gravity ; then, by the ordinary formida : 



Time = A^^; 



the limits of the integral being ^=0 and x:^z, where z is the distance from the 

 origin of motion to the synchronizing curve. 

 Hence, by oiu* formula, 



1 



Time 



-J:' 



V'2g cos Q 



'^^ ^z '\/2y COS 6 



2Jz 



- = constant 



\/2^^^cos1p 



Z CC COS 6 



and the sjmchronizing curve is the circle. 



We can solve this problem by another process, which beautifully illustrates 

 our formula. 



Let the origin of measure be the lowest point of the line ; then the expression 

 for the time is 



r ^'- _ ^ r dz ^^_^^_^ 



Jo \/2ff(z—z)cos6 Jo \/cos0 



Hence p in the formula is — j-, and (i){x + a)= , ^ 



^ V cos V 



"' dx , , , d 



r dx d-"- 1 



/ / ^ (Z — x)-''- K J- . . 



Jo V COS dz'*"- Vcosfl 



~* 



V cos 6 



which being constant by hypothesis s a cos ^ as before. 



Prob. 2. To find the tautochronous curve when a body descends by the ac- 

 tion of gi-avity. Retaining the notation of the last problem, measuring from the 

 lowest point, 



ds 



, dx 

 dx 

 t = 



~Jo V2g{z-x) 



Now, by the conditions of the problem, this is to be the independent of ; 



Let, therefore, -j— =</>(«) 



dx ^ 



and t=r ^y^fJL. 



Jo V2g Vz — x 

 VOL. XIV. PART II. 5 u 



