= Solution
The path rises from $(-1,1)$ to its maximum height $\sqrt2$ at $x=0$, then falls symmetrically to $(1,1)$. Its radius is $\sqrt2$, and its endpoint polar angles are $3\pi/4$ and $\pi/4$, giving angular span $\pi/2$. Hence its ordinary <arc length> is
$$
\boxed{L=\sqrt2\,\frac\pi2=\frac\pi{\sqrt2}\simeq2.22144.}
$$
Equivalently, $\sqrt{1+y'^2}=\sqrt2/\sqrt{2-x^2}$ and integrating over $[-1,1]$ gives the same value. This Euclidean length differs from the weighted destruction integral $D_{\min}\simeq1.76275$: the slightly longer route saves undergrowth by travelling at greater height.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1-q9-minimum-destruction-path.png]
{title=The least-destruction path bends towards lower density}
{description=Original exact semicircular minimizer and its horizontal competitor. The shaded background indicates the density $1/y$ in the upper half-plane. The minimizer has Euclidean length $\pi/\sqrt2$ but weighted cost $2\ln(1+\sqrt2)$, smaller than the straight path's cost two.}
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