gaiaecologicaq2

2022-07-11

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 $\beta (s)\cdot {\beta}^{\prime}(s)=0$ 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: $\kappa ={T}^{\prime}\cdot N$. So to be constant, it needs to be:

${\kappa}^{\prime}=0={T}^{\u2033}\cdot N+{T}^{\prime}\cdot {N}^{\prime}$

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: $\beta (s)\cdot {\beta}^{\prime}(s)=0$, I got stuck.

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: $\kappa ={T}^{\prime}\cdot N$. So to be constant, it needs to be:

${\kappa}^{\prime}=0={T}^{\u2033}\cdot N+{T}^{\prime}\cdot {N}^{\prime}$

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: $\beta (s)\cdot {\beta}^{\prime}(s)=0$, I got stuck.

Mekjulleymg

Beginner2022-07-12Added 14 answers

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 $2\beta (s)\cdot {\beta}^{\prime}(s)$, 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.

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

Yesenia Obrien

Beginner2022-07-13Added 5 answers

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

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