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Let $f: \mathscr{R} \rightarrow \mathscr{R}^n$ be a differentiable mapping with $f'(t) \neq \textbf{0}$ for all $\textbf{t} \in \mathscr{R}$. Let p be a fixed point not on the image curve of $f$ as in Fig. 2.4. If $\textbf{q} = f(t_0)$ is the point of the curve closest to p, that is, if $|\textbf{p} - \textbf{q}| \leqq |\textbf{p} - f(t)|$ for all $t \in \mathscr{R}$, show that the vector $\textbf{p} - \textbf{q}$ is orthogonal to the curve at q.

Hint: Differentiate the function $\varphi(t) = |\textbf{p} - f(t)|^2.$

Source: Advanced Calculus of Several Variables

This problem is somewhat similar perhaps to this one, but different enough for me to still be confused.

In order to show that $\textbf{p} - \textbf{q}$ is orthogonal to $f'(t_0)$, I would think the easiest way to do that would be to show that their dot product is zero. The hint suggests that we differentiate $\varphi(t) = |\textbf{p} - f(t)|^2$, which I think is $-2f'(t)|\textbf{p} - f(t)|$. It seems like the next step would be to evaluate this expression at $t_0$ giving $-2f'(t_0)|\textbf{p} - \textbf{q}|$, but I'm not sure where to go from there. Sadly, the book has no answer key.

1 Answers 1

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You are close. Remember that $f'(t)$ is a vector for each $t$. We have $$ \varphi'(t) = -2 f'(t)\cdot (\mathbf{p} - f(t)) $$ So $\varphi'(t_0) = -2 f'(t_0) \cdot (\mathbf{p} - \mathbf{q})$.

  1. What does the fact that $\mathbf{q}$ is the point on the curve closest to $\mathbf{p}$ tell you about $\varphi'(t_0)$ on the left?

  2. What vectors are being dotted on the right?