Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 14D (Methods)

For variations $y+ϵ\eta$ with $\eta (x_1)=\eta (x_2)=0$, integration by parts gives $\delta I={\int }_{x_1}^{x_2}[f_y-(d/dx)f_{y^{\prime }}]\eta dx$. Since this vanishes for every such variation, the Euler-Lagrange equation isWhen $f$ is independent of $y$, it has the first integral $\boxed{\partial f/\partial y^{\prime }=C}$.

For a ray represented as a graph traversed with increasing $x$, the travel time is $T={\int }_{x_1}^{x_2}\sqrt{1+y^{\prime 2}}/c(x)dx$, with $c(x)>0$. By Fermat principle, use $f=\sqrt{1+y^{\prime 2}}/c(x)$. The first integral becomesThe sign of $y^{\prime }$ is constant, and $y_2>y_1$ makes it positive. Write $C=1/k$ with $k>0$ and solve to obtainA finite slope requires $k>c(x)$ in the interior; a limiting vertical tangent is possible at an endpoint.

For $c(x)=ax$ and $k=a{x}_2$, positive speed on the whole interval and $k\ge c(x)$ select $a>0$ and $0<x_1<x_2$. The slope is $x/\sqrt{x_2^2-x^2}$. Integrating backwards from $(x_2,y_2)$ givesso $\boxed{(y_2-y{)}^2=x_2^2-x^2}$, with the branch $y\le y_2$. The endpoint integral is finite despite the vertical tangent. The specified value $k=a{x}_2$ fits the first endpoint precisely when $y_2-y_1=\sqrt{x_2^2-x_1^2}$; for other endpoint data the preceding integral determines a different $k$.