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

For fixed endpoint values, a stationary differentiable path obeys the Euler-Lagrange equationIt follows by taking the first variation of the functional and integrating the term in $\delta y^{\prime }$ by parts; $\delta y$ vanishes at the endpoints.

If $f$ has no explicit $x$ dependence, the chain rule givesConsequentlyon an Euler-Lagrange equation solution. Thus the Beltrami identity supplies the first integralThe opposite sign is an equally valid convention. The conserved value can be zero for particular integrands or paths; absence of explicit $x$ alone does not guarantee a nonzero numerical value. The useful statement is that this expression is constant along an extremal.

On the cylindrical surface $r=R$, the line element is $d{s}^2=R^{2}d{\theta }^2+d{z}^2$. For a single-valued path parametrized by increasing $\theta$,The integrand has no explicit $z$, so its Euler-Lagrange equation gives the constant conjugate quantity $z^{\prime }/\sqrt{R^2+z^{\prime 2}}$. Its derivative with respect to $z^{\prime }$ is strictly positive, hence $z^{\prime }$ itself is constant. The two endpoints determineThis is a constant-pitch helix on the cylinder. To verify global minimality, unroll the cylinder with planar coordinates $(R\theta ,z)$. The path becomes a straight line between points separated by $\pi R$ and $z_2-z_1$, so the triangle inequality for arc length givesThis is the standard description of geodesics on a circular cylinder.

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.