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How does regularity of curves prevent "cusps"?

  1. Feb 16, 2015 #1
    A regular curve on a manifold ##M## is a curve ##\gamma:I \to M## such that ##\dot \gamma(t) \neq 0## for any ##t \in I##. In John Lee's "Introduction to Curvature" he says that this intuitively means that we prevent the curve from having "cusps" and "kinks".

    How can I see that this is the case? I.e. how could ##\dot \gamma(t) = 0## for some t lead to a "kink" or "cusp"?
     
  2. jcsd
  3. Feb 16, 2015 #2

    wabbit

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    It doesn't. ##\dot \gamma(t) = 0## means a stationary point and can happen just from parametrization. But cusps or kinks do require ##\dot \gamma(t) = 0##, so forbidding it ensures you get none.

    ##\dot \gamma(t) \neq 0## is needed for two other reasons : it ensures ##\dot \gamma(t) ## gives you the direction of the tangent vector to the curve at every point, and that the parametrization is locally one-to-one / doesn't backtrack.
     
    Last edited: Feb 16, 2015
  4. Feb 17, 2015 #3
    But why -- intuitively -- do cusps or kinks require ##\dot gamma = 0##?
     
  5. Feb 17, 2015 #4

    wabbit

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    One way to see it is to think of the curve as traced by a moving point at ## \gamma(t) ##; then ## \dot\gamma(t) ## is the velocity of that point. If it's not zero it has a direction. What is that direction at a kink ?
     
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