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For fixed endpoints, the Euler-Lagrange equation is $F_y-\frac{d}{dx}{F}_{y^{\prime }}=0$. Since $F$ has no explicit dependence on $x$, differentiating givesThus $F-y^{\prime }{F}_{y^{\prime }}=C$, the Beltrami identity.

With $r^{\prime }=dr/d\theta$, the arc length element and the prescribed speed give the travel-time functionalFor the circular road $r=a$, this is $T_{\mathrm{circle}}=\pi /(2ka)$.

For the integrand $F=\sqrt{r^{\prime 2}+r^2}/r^2$, the Beltrami identity simplifies toThe candidate $r=a(\cos \theta +\sin \theta )$ has $r^{\prime 2}+r^2=2a^2$ and the required endpoints. To prove it really minimizes the travel time, rather than merely satisfies a first integral, put $u=1/r$. ThenThis is the Euclidean arc length integrand for the inverted curve $(X,Y)=(u\cos \theta ,u\sin \theta )$, whose endpoints are $(1/a,0)$ and $(0,1/a)$. Every connecting curve has length at least their Euclidean distance $\sqrt{2}/a$, by the triangle inequality. Equality is attained by the straight line segment $X+Y=1/a$. In polar form it gives $u(\cos \theta +\sin \theta )=1/a$, hence the proposed $r$. It stays strictly positive throughout the interval. ConsequentlyThis travel-time minimization with speed proportional to squared radius is globally solved by the inversion: its minimum is strictly less than $\pi /(2ka)$.