Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 9 / c / ii / Solution

The downward drop from the starting rim to angle $\theta$ is $R\cos \theta$. With no dissipative forces, conservation of energy gives $m{v}^2/2=mgR\cos \theta$, so $v=\sqrt{2gR\cos \theta }$. Dividing the line element by this speed yieldsThe endpoint singularities are integrable for the minimizing path. The integrand is independent of $z$; its Euler-Lagrange equation gives a constant $K$, after absorbing the overall prefactor:Solving for $z^{\prime }$ and taking the positive branch required by $z_2>z_1$ givesThe constant is fixed implicitly byThis integral is strictly increasing from zero to infinity as $K$ increases from zero to one, so every finite positive separation selects a unique $K$. The time integrand has second derivative $R^2/[\sqrt{\cos \theta }(R^2+z^{\prime 2}{)}^{3/2}]>0$ with respect to $z^{\prime }$. Its supporting-line inequality, integrated between fixed endpoints with constant first derivative with respect to $z^{\prime }$, proves this stationary path is the unique global minimum within the stated single-valued class.

As $\theta \downarrow -\pi /2$, $z^{\prime }\to 0$. Therefore the skater should initially head directly down the curved ramp, perpendicular to the cylinder axis, and gain axial motion later. The path differs from the constant-pitch shortest route because extra distance is cheaper in time near the faster, lower part of the ramp. This is the brachistochrone on a horizontal half-cylinder.