Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 9 / b

The monotonic horizontal coordinate allows the path to be written as a graph. Its arc length element is $ds=\sqrt{1+y^{\prime 2}}dx$, so the variational density is $f(y,y^{\prime })=g(y)\sqrt{1+y^{\prime 2}}$. It has no explicit horizontal-coordinate dependence. The Beltrami identity becomesConsequently every smooth minimizing path satisfiesThis is a first integral of the weighted arc-length problem; fixed endpoint values remove the endpoint variation terms.

Interpret the density $1/y$ on the physical region $y>0$. It is singular at zero and would not describe a nonnegative density for $y<0$. The first integral gives $y\sqrt{1+y^{\prime 2}}=R$ for a positive constant $R$. To handle the horizontal tangent without dividing by $y^{\prime }$, use the full Euler-Lagrange equation:Two integrations give $y^2=-(x-x_0{)}^2+R^2$. The equal endpoint heights force $x_0=0$, and either endpoint then gives $R^2=2$. ThusThe path is the upper circular arc. A constant horizontal path solves the differentiated first integral trivially but fails $y{y}^{\prime \prime }+y^{\prime 2}+1=0$, so is not another extremum.

One can prove global minimality, rather than only stationarity. Define $\Phi (x,y)=\mathrm{arsinh}(x/y)$ on the upper half-plane. ThenThe Cauchy-Schwarz inequality gives, along any admissible path from $A$ to $B$,Our circle has tangent $(1,-x/y)$ for increasing $x$, so equality holds pointwise. This calibration for semicircular hyperbolic geodesics provesThe geometry is that of a hyperbolic geodesic in the Poincaré half-plane model. The proof even allows competing upper-half-plane paths beyond the class of monotonic graphs.

The path rises from $(-1,1)$ to its maximum height $\sqrt{2}$ at $x=0$, then falls symmetrically to $(1,1)$. Its radius is $\sqrt{2}$, and its endpoint polar angles are $3\pi /4$ and $\pi /4$, giving angular span $\pi /2$. Hence its ordinary arc length isEquivalently, $\sqrt{1+y^{\prime 2}}=\sqrt{2}/\sqrt{2-x^2}$ and integrating over $[-1,1]$ gives the same value. This Euclidean length differs from the weighted destruction integral $D_{\min }\simeq 1.76275$: the slightly longer route saves undergrowth by travelling at greater height.