# Assume we have an arc-length parameterized curve &#x03B2;<!-- β --> : I &#x2192;<!-- →

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 $\beta \left(s\right)\cdot {\beta }^{\prime }\left(s\right)=0$ 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 $\left(T,N\right)$, the curvature is: $\kappa ={T}^{\prime }\cdot N$. So to be constant, it needs to be:
${\kappa }^{\prime }=0={T}^{″}\cdot N+{T}^{\prime }\cdot {N}^{\prime }$
I was thinking about using the formulas $T={\beta }^{\prime }$, ${T}^{\prime }=\kappa N$ and ${N}^{\prime }=-\kappa T$. But together with the given condition: $\beta \left(s\right)\cdot {\beta }^{\prime }\left(s\right)=0$, I got stuck.
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Mekjulleymg
Note that a curve lies on a circle if $\beta \left(s\right)\cdot \beta \left(s\right)=c={r}^{2}$ for some (necessarily positive) constant c. Since I is connected, it suffices to check that the derivative of $\beta \left(s\right)\cdot \beta \left(s\right)$ is 0. However the derivative of $\beta \left(s\right)\cdot \beta \left(s\right)$ is $2\beta \left(s\right)\cdot {\beta }^{\prime }\left(s\right)$, and we already know this is 0 by the given information.

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

Yesenia Obrien
Since $||\beta \left(s\right)|{|}^{2}=\beta \left(s\right)\cdot \beta \left(s\right)$ you can use the product rule to show $\frac{\mathrm{\partial }}{\mathrm{\partial }s}||\beta \left(s\right)|{|}^{2}=0$.