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For variations $y+ϵ\eta$ with $\eta (a)=\eta (b)=0$, integration by parts gives $\delta I={\int }_a^b[f_y-d(f_{y^{\prime }})/dx]\eta dx$. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationFor the ramp, use conservation of energy from rest at the rim. The vertical drop is $R\cos \theta$, so $v=\sqrt{2gR\cos \theta }$. The spherical surface arc length is $ds=R\sqrt{d{\theta }^2+{\sin }^2\theta d{\varphi }^2}$. On a monotone segment described by $\varphi (\theta )$, the travel time is consequentlyFor increasing $\theta$ this has the printed limits ${\theta }_0$ to ${\theta }_1$; for a descent, reverse the decreasing angular limits to keep time positive. The azimuth is a cyclic coordinate of this integrand. Its conserved derivative gives the first integral for the brachistochrone on a hemispherical ramp:Solving for its square yieldsFor a regular stationary path with finite $C$, this tends to zero as the rim $\theta \to \pi /2$ is approached. The ratio of azimuthal to meridional displacements is $\sin \theta {\varphi }^{\prime }$, which also tends to zero. Hence the initial direction is along a meridian toward the bottom. This conclusion concerns the initial heading; the later route can turn before reaching the bottom when its rim endpoints require a nonzero azimuthal displacement.