Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 9

Take endpoint-preserving variations $y+\epsilon \eta$, with $\eta (a)=\eta (b)=0$. The first variation of the functional is $\delta I={\int }_a^b(f_y\eta +f_{y^{\prime }}{\eta }^{\prime })dx$; integration by parts supplies the differential equation and its conserved quantity below.

The boundary term $[f_{y^{\prime }}\eta {]}_a^b$ vanishes, so stationarity and the fundamental lemma of the calculus of variations give the Euler-Lagrange equation

If $f$ is independent of $y$, then $f_y=0$ and the Euler-Lagrange equation reduces to $d{f}_{y^{\prime }}/dx=0$. Therefore the first integral isThis is the conserved generalized momentum for a cyclic coordinate.

By Fermat principle, a light ray makes optical travel time stationary. Multiplication by the constant vacuum speed of light does not change the stationary paths, so the functional can be taken to be the refractive index integrated over arc length. The constant denoted $c$ below is an integration constant, not that vacuum speed.

The integrand $f=n(x)\sqrt{1+(y^{\prime }{)}^2}$ has no explicit dependence on $y$, so part (i) givesFor positive refractive index and a regular graph with $n(x)>|c|$, squaring and then restoring the sign of $y^{\prime }$ yieldsIt is the continuous-medium version of Snell's law, expressing conserved horizontal optical momentum. The Euler-Lagrange equation establishes stationarity; on a monotone-height segment, convexity of the integrand in $y^{\prime }$ also gives a minimum among graphs with fixed endpoints.

For the displayed form take $c>0$ and $1+\beta x\ge c$. Substitute $s=(1+\beta x)/c$ in the first integral:Inverting the hyperbolic cosine gives the catenary light ray in a linear refractive indexThe minimum occurs at $y=y_0$ and $x=x_0=(c-1)/\beta$, so the integration constant has the specified meaning. The integral above describes the $y\ge y_0$ graph branch; the other sign of the square root continues it smoothly to the other side of the turning point in $x(y)$.

The signed-constant qualification matters: for $c<0$ in part (a), integration reverses horizontal direction, and the positive prefactor in the full catenary is $|c|/\beta$, with minimum height $(|c|-1)/\beta$. The printed expression presupposes $c>0$. The separate case $c=0$ gives $y$ constant, a vertical ray. At the turning point $dy/dx$ diverges, so $y(x)$ is not a single regular graph across it, although $x(y)$ is smooth.

For $c>1$, the catenary has a positive minimum height $x_0=(c-1)/\beta$ above the road. Its derivatives areThus the light ray is symmetric about $y=y_0$, descends towards a horizontal tangent at its minimum, and then rises. It never intersects the road. This smooth upward bending in the refractive index gradient produces an inferior mirage, without reflection at the road.