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Parametrized Plane Curves

  1. Sep 13, 2013 #1
    1. The problem statement, all variables and given/known data
    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.

    3. 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.
     
  2. jcsd
  3. Sep 14, 2013 #2

    pasmith

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    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: Sep 14, 2013
  4. Sep 14, 2013 #3
    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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