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 / iii

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.