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Let $\epsilon>0$ and $\alpha:(-\epsilon,\epsilon) \rightarrow \mathbb{R}^2$ be a regular plane curve parametrized by arc-length. Suppose that $k(s) = k(-s)$ for all $s \in (-\epsilon,\epsilon)$. Prove that $M(\alpha(-s)) = \alpha(s)$ for all $s$ in $(-\epsilon,\epsilon)$, where $M: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is reflection in the plane about the normal line of $\alpha$ at $s = 0$.

I know I have to use the uniqueness part from the Fundamental Theorem of Curves in the plane. But I don't know how to start it. Does someone have a little tip????

Ivo Terek
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Ju Grigori
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1 Answers1

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Hint: To keep track of things, let $\beta(s) = M\alpha(-s)$. Then note that $\beta$ is also parametrized by arc-length, and $\kappa_\beta(s) = \kappa_\alpha(s)$ (because $s \mapsto -s$ changes the sign of the curvature once, and $M$ being a reflection changes this sign back to the original $+$). We also have $\beta(0) = \alpha(0)$ (clear), $T_\beta(s) = T_\alpha(s)$ (why?), and thus $N_\beta(0) = N_\alpha(0)$. Hence...?

Ivo Terek
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    Sorry, I don't have time to go over all of it (the actual writing is up to you), but I have one comment to simplify all of that: you said that $\alpha$ is a plane curve, so there's no need to look at torsion and binormal. The hint I gave is almost a full solution (give some justification for why $T_\beta = T_\alpha$ and quote the uniqueness part of the fundamental theorem after "Hence..."). – Ivo Terek Nov 09 '20 at 18:35