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Thread: Curvature

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    148

    Curvature

    Hello all,

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

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

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

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

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

    I am asked to:

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


    I have done the following:

    1)

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

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

    2)

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


    Could someone verify 1) and 2)?


    Thanks.
    Last edited by surjective; May 27th 2017 at 10:15 AM.
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  2. #2
    Member
    Joined
    Feb 2010
    Posts
    148

    Re: Curvature

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

    Suggestions would still be appreciated.
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