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Curvature of curve with arbitrary parametrization

  1. Nov 6, 2008 #1
    Hi, I'm preparing a little exposition of curvature and torsion for my Calculus class and so I need to include some simple proofs for the things I'll use to define curvature. So I'm looking for a proof of the formula for the curvature of an arbitrarily parametrized curve that doens't use the Frenet frame; it'll get too complicated for a 15 minute presentation. The closest thing to what I'm looking for is in Pogorelov's Differential geometry which goes like this:

    Say [tex]\textbf{r} = \textbf{r}(t)[/tex] is a vector valued function. He then goes on to take the derivative with respect to t, remembering that the arclength can be expressed as a function of time:
    [tex]\textbf{r}' = \textbf{r}_{s}' s'[/tex]
    What I understand by this notation is:
    [tex](x'(t),y'(t),z'(t)) = (x'(s)s'(t),y'(s)s'(t),z'(s)s'(t))[/tex],
    where the derivatives are take with respect to their arguments. He then goes on to say that from that formula, we can arrive at:
    This is the unclear point for me. The rest is pretty straight forward, he uses the first formula to find an expression for [tex]\textbf{r}_{s}'[/tex] in terms of [tex]\textbf{r}'[/tex], which he differentiates again with respect to [tex]t[/tex] and the proof if pretty much done. However, I don't understand how he jumps from the first to the second equation.

    Something else I've been looking for is an easy way to prove that the derivative of a parametrization by arclenght has unit norm.
    And one last thing, any books to be recommended? I'm looking for very basic DG books, nothing on manifolds or even surfaces, but it must treat space curves.

    Thanks to all!
  2. jcsd
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