= Solution
For variations $y+\epsilon\eta$ with $\eta(a)=\eta(b)=0$, <integration by parts> gives $\delta I=\int_a^b[f_y-d(f_{y'})/dx]\eta\,dx$. The <fundamental lemma of the calculus of variations> therefore gives the <Euler-Lagrange equation>
$$
\boxed{\frac d{dx}f_{y'}-f_y=0.}
$$
For 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 $ds=R\sqrt{d\theta^2+\sin^2\theta\,d\phi^2}$. On a monotone segment described by $\phi(\theta)$, the <travel time> is consequently
$$
\boxed{T=\sqrt{\frac R{2g}}\int\sqrt{\frac{1+\sin^2\theta\,\phi'^2}{\cos\theta}}\,|d\theta|.}
$$
For 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>:
$$
\boxed{\frac{\sin^2\theta\,\phi'}{\sqrt{\cos\theta}\sqrt{1+\sin^2\theta\,\phi'^2}}=C.}
$$
Solving for its square yields
$$
\phi'^2=\frac{C^2\cos\theta}{\sin^2\theta(\sin^2\theta-C^2\cos\theta)}.
$$
For 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 $\sin\theta\,\phi'$, which also tends to zero. Hence \b[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.
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