578 Mr. H. Bateman on the Velocity of an 



general method * of treating brachistochronic problems we 

 have 



Ybr) + \rWe) = ^' (1) 



where is the angle which the radius r makes with the radius 

 through the origin of the disturbance. 



Putting r = xR, T = SR, the equation becomes 



We get a solution of this by putting ^- = a, a quantity 



(SHtf)H' ••.-.■>. oo 



ing 

 independent of x and 6, This gives 



'f^±\/^ f3) 



The negative sign must be chosen for the first half of the 

 path because x begins to decrease as S increases. This holds 

 until a point on the path is reached at which the radius is 

 cut at right angles. This radius is then a line of symmetry 

 of the path and the point in which it meets the path is a 

 vertex of the path. It is clear that after passing the vertex 

 x increases with S, and so the positive sign must be attributed 

 to the square root. 



Substituting the values of ^- and ^— in the equation 



dS = =r-^dd+ ^—dx. 

 ou ox 



we get 



1 /^2 



dS = add -f - \ / -* - a 2 . dx, 

 ~ x V it 



When = 0, x — 1; hence for the first half of the path we 

 have 



Ji x V tr 



(4) 



The equation of the path is obtained by putting -^- equal 



* Transactions of the Royal Irish Academy, 1833. I am following 

 here the analysis given in Prof. Knott's paper. 



