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Thread: Curvature

  1. #1
    Member
    Joined
    Feb 2010
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    144

    Curvature

    Hello all,

    I have a regular parametrized curve \gamma: \mathbb{R} \rightarrow \mathbb{R}^{3} such that  \Vert\gamma''(t)\Vert =1 for all t \in \mathbb{R}.
    Assume that \gamma(t) has constant curvature k \neq 0 and constant torsio \tau=\frac{1}{\sqrt{2}}. We also assume that:

    \gamma(0)=\left( \frac{1}{\sqrt{2}},0,0  \right)

    \gamma'(0)=\left( 0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right)

    \mathbf{b}(t)=\frac{1}{\sqrt{2}} \left(  \sin(t),-\cos(t),1  \right)

    where \mathbf{b}(t) is the binormal of \gamma(t)

    I am asked to:

    1) find the curvature K
    2) find the \gamma(t) explicitely


    I have done the following:

    1)

    Using the Frenet equations I have that \mathbf{n'}=-k\mathbf{t}+\tau\mathbf{b}.
    Differentiating on both sides yields \mathbf{n''}=-k\mathbf{t'}+\tau\mathbf{b'} (*).

    Using \mathbf{b'}=-\tau\mathbf{n} and \mathbf{t'}=k\mathbf{n} and substituting in (*) gives k=\frac{\sqrt{2}}{2}.

    2)

    I am still not done with this one but my suggestion is to use that \mathbf{n}=\frac{\gamma''(t)}{\Vert \gamma''(t)  \Vert} and integrate together with the fact that  \Vert\gamma''(t)t\Vert =1 .


    Could someone verify 1) and 2)?


    Thanks.
    Last edited by surjective; May 27th 2017 at 11:15 AM.
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  2. #2
    Member
    Joined
    Feb 2010
    Posts
    144

    Re: Curvature

    Sorry but 2) is wrong since \Vert \gamma''(t) \Vert \neq 1.

    Suggestions would still be appreciated.
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