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Space curve question

  1. Aug 18, 2007 #1
    1. The problem statement, all variables and given/known data

    Let [itex]\alpha : I \to \mathbb{R}^3[/itex] be a regular parametrized curve and let [itex]\beta : J \to \mathbb{R}^3[/itex] be a reparametrization in terms of arc length [itex]s = s(t)[/itex].

    Show that [itex]\frac{d^2t}{ds^2} = -\frac{\dot{\alpha} \cdot \ddot{\alpha}}{\abs{\alpha}^4}[/itex]


    3. The attempt at a solution

    I'm going to use dots and dashes to represent differentiation wrt time and arc-length, respectively.

    [itex]\beta'(s) = [\alpha (t(s))]' = \dot{\alpha} t'[/itex]
    [itex]\beta''(s) = \dot{\alpha}t'' + \ddot{\alpha} t'^2[/itex] (*)

    [itex]t'(s) = 1/\dot{s}(t)[/itex] by the inverse function theorem for 1 variable so
    [itex]\dot{\alpha} = \dot{s} \beta'(s)[/itex]. This is where things start to get messy.

    I think I need to use the fact that [itex]\dot{\alpha} = \alpha'[/itex]. Then the result follows by dotting both sides of (*) by [itex]\alpha'[/itex].

    Can you convince me that [itex]\dot{\alpha} = \alpha'[/itex]?

    Thanks.
     
    Last edited: Aug 18, 2007
  2. jcsd
  3. Aug 18, 2007 #2
    I figured this out nearly as soon as I posted it. Here it is for future use:

    [itex]\beta'(s) = \dot{\alpha} t'[/itex] (1)
    [itex]\beta''(s) = \dot{\alpha} t'' + \ddot{\alpha} t'^2[/itex]
    [itex]\beta''(s) = \dot{\alpha} t'' + \ddot{\alpha} t'[/itex].

    Applying inverse function to (1) gives [itex]\beta'(s)\dot{s} = \dot{\alpha}[/itex] showing that [itex]\dot{\alpha}[/itex] is parallel to [itex]\beta'[/itex].

    But [itex]\beta'[/itex] is a unit vector field so [itex]\beta'\cdot\beta''=0 \implies \dot{\alpha} \cdot \beta''=0[/itex]. Hence

    [itex]0 = \dot{\alpha}^2 t'' + \dot{\alpha}\cdot\ddot{\alpha} t'^2[/itex]
    [itex]t'' = - \frac{\dot{\alpha}\cdot \ddot{\alpha}t'^2}{\dot{\alpha}^2} = -\frac{\dot{\alpha}\cdot\ddot{\alpha}}{\dot{\alpha}^4}[/itex]
     
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