Brachistochrone on a hemispherical ramp (${\varphi }^{\prime 2}=C^2\cos \theta /[{\sin }^2\theta ({\sin }^2\theta -C^2\cos \theta )]$)

Measure polar angle $\theta$ from the downward vertical on the lower hemisphere of radius $R$. A particle released from rest at the rim and sliding without energy loss has speed $\sqrt{2gR\cos \theta }$. The spherical arc length along $\varphi =\varphi (\theta )$ is $R\sqrt{1+{\sin }^2\theta {\varphi }^{\prime 2}}|d\theta |$. Thus its travel time has integrand proportional to $\sqrt{(1+{\sin }^2\theta {\varphi }^{\prime 2})/\cos \theta }$. Since the azimuth is a cyclic coordinate, the Euler-Lagrange equation givesSolving for ${\varphi }^{\prime 2}$ gives the displayed first integral on a segment where $\theta$ is a valid path parameter. For finite $C$, approaching the rim $\theta \uparrow \pi /2$ gives ${\varphi }^{\prime }\to 0$. Thus a regular stationary path begins along the meridian toward the bottom. This is an initial direction condition; a path joining other rim points can turn before reaching the bottom. Monotone segments use positive $|d\theta |$ in the time integral, and a turning point requires another path parameter or a split into segments.