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Parametrized Plane Curves

  • #1
612
23

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.
 

Answers and Replies

  • #2
pasmith
Homework Helper
1,738
410

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

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

EDIT: It also occurs to me that circles and lines are the only plane curves of constant curvature.
 
Last edited:
  • #3
612
23
Suppose [itex]\|\sigma(t_1) - \sigma(t_0)\| = f(|t_1 - t_0|)[/itex] for some real-valued [itex]f[/itex] such that [itex]f(t) \geq 0[/itex] for all [itex]t[/itex] with [itex]f(0) = 0[/itex]. Then
[tex]
\sigma(t_1) = \sigma(t_0) + \hat n(t_0,t_1) f(|t_1 - t_0|)\qquad(1)
[/tex]
where [itex]\hat n[/itex] is a unit vector. This can be rearranged to give
[tex]
\sigma(t_0) = \sigma(t_1) - \hat n(t_0,t_1) f(|t_1 - t_0|)
[/tex]
and swapping [itex]t_0[/itex] and [itex]t_1[/itex] and comparing with (1) yields
[tex]\hat n(t_0, t_1) = -\hat n(t_1,t_0)[/tex]

I don't know whether this will lead anywhere. However I do see that if [itex]\hat n[/itex] is constant then [itex]\sigma(I)[/itex] 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.
 

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