Is the shortest path to a curve orthogonal to that curve?

Question

I'm wondering if the following is true. In $\mathbb R^2$, let $\gamma(t)$ be a $\mathcal C^1$ curve parametrized over $\mathbb R$ and let \begin{align} d(p) &= \min_t\{\mathrm{dist}_\mathrm{eucl}(p,\gamma(t))\}\\ \end{align}

Let $t_0$ be one of the $t$ that solve the minimization of $d(p)$. Is it true that the segment $[\gamma(t_0),p]$ is orthogonal to $\frac{\mathrm d \gamma}{\mathrm d t}(t_0)$ ?

I'm not necessarily looking for a rigorous proof, but a hint of how to prove this (or a counter example). Thank you for any help.

Answer 1

Without further conditions on $p$, I believe a counter-example is easy to find. Take $\gamma(t) = (1+t, 1+t)$ parametrized over non-negative times $t$, and $p = (0,0)$. Then the minimisation problem is solved by $t_0 = 0$, but the segment $[\gamma(t_0), p] $, connecting the origin to $(1,1)$ is not orthogonal to $\frac{\mathrm{d}\gamma}{\mathrm{d}t}(t_0)$.

Answer 2

Hint. Assume the minimum is realized at an interior point $t_0$ of the parameter interval. Let $h(t) = ||p - \gamma(t)||^2$ and compute $h'(t_0)$, which must be zero.