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Assume we have an arc-length parameterized curve $\beta: I \to \mathbb{E}^2$ with $I$ a random interval. I want to show that if $\beta(s)\cdot\beta'(s) = 0$ for arc-length parameter $s$, then $\beta$ is (part of) a circle.

I was thinking about using the Frenet-formulas for proving this. If we want to show something is a circle, one needs to show that the curvature is constant. In the Frenet-frame $(T,N)$, the curvature is: $\kappa = T' \cdot N$. So to be constant, it needs to be: $$\kappa' = 0 = T''\cdot N+ T'\cdot N'$$ I was thinking about using the formulas $T = \beta'$, $T' = \kappa N$ and $N' = -\kappa T$. But together with the given condition: $\beta(s)\cdot\beta'(s) = 0$, I got stuck.

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  • $\begingroup$ How about just differentiating $\beta(s)\cdot\beta(s)$? $\endgroup$ – Angina Seng Jun 22 '19 at 15:31
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Note that a curve lies on a circle if $\beta(s)\cdot \beta(s) = c = r^2$ for some (necessarily positive) constant $c$. Since $I$ is connected, it suffices to check that the derivative of $\beta(s)\cdot \beta(s)$ is $0$. However the derivative of $\beta(s)\cdot \beta(s)$ is $2\beta(s)\cdot \beta'(s)$, and we already know this is $0$ by the given information.

In general, if the curve lies in $\Bbb{R}^n$, this shows that the curve lies on a sphere.

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  • $\begingroup$ why lies a curve on a circle if $\beta(s)\beta(s)$ =c $\endgroup$ – questmath Aug 20 '20 at 19:14
  • $\begingroup$ @mathmath Because that says that the length of $\beta$ is constant, so all points of the curve lie on a circle of radius $\sqrt{c}$. $\endgroup$ – jgon Aug 20 '20 at 20:47
  • $\begingroup$ yes but isn't it the curve that needs to be constant or am i wrong? $\endgroup$ – questmath Aug 20 '20 at 20:48
  • $\begingroup$ @mathmath What do you mean the curve needs to be constant? I'm not sure I understand your question. $\endgroup$ – jgon Aug 20 '20 at 20:50
  • $\begingroup$ a curve lies on a sphere is the curvature is constant i mean $\endgroup$ – questmath Aug 20 '20 at 20:51
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Hint: Since $||\beta(s)||^2=\beta(s)\cdot\beta(s)$ you can use the product rule to show $\frac{\partial}{\partial s}||\beta(s)||^2=0$.

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