A walking path in a nature reserve follows the curve y = x² on a map, with x and y in kilometres…

Step 1. The function. $D(x) = x^2 + (x^2 - 3)^2 = x^4 - 5x^2 + 9$.

Step 2. The derivative. $D' = 4x^3 - 10x = 2x(2x^2 - 5)$. The critical points are $x = 0$ and $x = \pm\sqrt{2.5}$.

Step 3. The point with $x > 0$. $x = \sqrt{2.5} = 1.581\ \text{km}$, and $y = x^2 = 2.5\ \text{km}$.

Step 4. The distance. $D = 2.5 + (2.5 - 3)^2 = 2.75$, so the distance is $\sqrt{2.75} = 1.658\ \text{km}$.

Step 5. The point $x = 0$. Here $D(0) = 9$, which is a distance of $3\ \text{km}$. The second derivative $D'' = 12x^2 - 10$ is $-10 < 0$ at $x = 0$, so it is a local maximum of the distance, not a minimum.

If the hut is at $(0, 0.4)$. Now $D = x^2 + (x^2 - 0.4)^2 = x^4 + 0.2x^2 + 0.16$, and $D' = 4x(x^2 + 0.1)$ is $0$ only at $x = 0$. The closest point is the vertex of the path, at a distance of $0.4\ \text{km}$.