Can Parametrized Plane Curves Have Constant Curvature?

[Mandelbroth]

Homework Statement

Suppose $\sigma:I\subseteq\mathbb{R}\to\mathbb{R}^2$ is a smooth plane curve parametrized by a parameter $t\in I$. Prove that if $\|\sigma(t_1)-\sigma(t_0)\|$ depends entirely on $|t_1-t_0|$, then the image of $I$ under $\sigma$ is a subset of either $S^1$ or a line.

The Attempt at a Solution

Embarrassingly enough, I'm having trouble setting up a proof here. I understand intuitively why this is true, but I can't see where to start. Can someone just nudge me in the right direction?

Thank you.

 

[pasmith]

Homework Statement

Suppose $\sigma:I\subseteq\mathbb{R}\to\mathbb{R}^2$ is a smooth plane curve parametrized by a parameter $t\in I$. Prove that if $\|\sigma(t_1)-\sigma(t_0)\|$ depends entirely on $|t_1-t_0|$, then the image of $I$ under $\sigma$ is a subset of either $S^1$ or a line.

The Attempt at a Solution

Embarrassingly enough, I'm having trouble setting up a proof here. I understand intuitively why this is true, but I can't see where to start. Can someone just nudge me in the right direction?

Thank you.

Suppose \|\sigma(t_1) - \sigma(t_0)\| = f(|t_1 - t_0|) for some real-valued f such that f(t) \geq 0 for all t with f(0) = 0. Then
<br /> \sigma(t_1) = \sigma(t_0) + \hat n(t_0,t_1) f(|t_1 - t_0|)\qquad(1)<br />
where \hat n is a unit vector. This can be rearranged to give
<br /> \sigma(t_0) = \sigma(t_1) - \hat n(t_0,t_1) f(|t_1 - t_0|)<br />
and swapping t_0 and t_1 and comparing with (1) yields
\hat n(t_0, t_1) = -\hat n(t_1,t_0)

I don't know whether this will lead anywhere. However I do see that if \hat n is constant then \sigma(I) will be a line segment.

EDIT: It also occurs to me that circles and lines are the only plane curves of constant curvature.

 

[Mandelbroth]

Suppose \|\sigma(t_1) - \sigma(t_0)\| = f(|t_1 - t_0|) for some real-valued f such that f(t) \geq 0 for all t with f(0) = 0. Then
<br /> \sigma(t_1) = \sigma(t_0) + \hat n(t_0,t_1) f(|t_1 - t_0|)\qquad(1)<br />
where \hat n is a unit vector. This can be rearranged to give
<br /> \sigma(t_0) = \sigma(t_1) - \hat n(t_0,t_1) f(|t_1 - t_0|)<br />
and swapping t_0 and t_1 and comparing with (1) yields
\hat n(t_0, t_1) = -\hat n(t_1,t_0)

I don't know whether this will lead anywhere. However I do see that if \hat n is constant then \sigma(I) will be a line segment.

EDIT: It also occurs to me that circles and lines are the only plane curves of constant curvature.

I realized that they are both the only plane curves with constant curvature, then I proceeded to cook up a proof.

Thank you.