PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 615 



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

 (/ the force of gravity ; then, by the ordinary formula : 



Time = f, ^ ^ ; 



J V Zg X cos V 

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

 origin of motion to the synchronizing curve. 

 Hence, by our formula, 



Time = / dx x-^ . , r 



Jo V 2^ COS a 



d-^ 1 



2Jz 



\/2g cos 6 



Z OC COS Q 



constant 



and the synchronizing 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 



7f^z dx r^ dx f X , 



— ==^==— cc / [z — xY^ 



\/2g{z — x)(io»6 Jo V cosO 



Hence p in the formula is -^' ^iid (a; + «) = , — g 



X 



dx , , , d^'^ 



[z — xY^ a 



\/cos 6 dz ^- a/ cos 6 



zi 



Vcos 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 gravity. Retaining the notation of the last problem, measuring from the 



lowest point, 



d s , 

 — ax 

 dx 



pz dx 



~Jo \/2g{z-x) 

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



ds 



Let, therefore, -j^ = («) 



and t = 



Jo \/2g 



(p (x) dx 



\/2g Vz — x 

 VOL. XIV. PART II. 5 U 



