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Show that if $(\frac{1}{k})^{'} \neq 0$ and $(\frac{1}{k})^2 + ((\frac{1}{k})^{'}\frac{1}{\tau})^2$ is a constant, then a unit speed curve $\alpha$ lies on a sphere.

Using the following formulas given by a Frenet Frame:

$\mathbf T = f^{\prime}$

$\mathbf B = \mathbf T \times \mathbf N$

$\mathbf T^{\prime} = \kappa \mathbf N$

$\mathbf N^{\prime} = -\kappa \mathbf T + \tau\mathbf B$

$\mathbf B^{\prime} = -\tau\mathbf N$

Let $\beta(s)=\alpha(s) + \frac{1}{\kappa} \mathbf N+ \left ( \frac{1}{\kappa} \right )^{'} \left (\frac{1}{t} \right )\mathbf B$

How do you show $\beta^{'}(s)=0$?

I know $\alpha^{'}(s)$ = $\mathbf{T}$. As for the succeeding terms, I know their sum should be equal to $\mathbf{-T}$ after differentiation, but I am not sure how to get there.

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  • $\begingroup$ The equation $\beta(s) = \alpha(s) + \kappa^{-1} + (\kappa^{-1})' \tau^{-1}$ is problematic because $\alpha, \beta$ are vectors whilst $\kappa, \tau$ are scalars. Can you clarify? $\endgroup$ – Robert Lewis Jan 23 '14 at 8:45
  • $\begingroup$ $\kappa$ and $\tau$ are both functions of s, which indicates arc length. $\kappa$ is curvature and $\tau$ is torsion. $\endgroup$ – KangHoon You Jan 23 '14 at 9:02
  • $\begingroup$ In fact you can't because you are missing important details of the question. Should $\alpha$ lie in a sphere? Do you know more about the torsion and the curvature? Please quote the whole question. $\endgroup$ – Michael Hoppe Jan 23 '14 at 9:19
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Hint: $\beta(s)=\alpha(s) + \gamma(s)$ where $\gamma$ has constant length. So $<\gamma',\gamma>=0$.

Now after differentiating $\beta$ you get $\beta(s)'=u(s)\mathbf{B}$. So $\gamma'=u(s)\mathbf{B}-\mathbf{T}$. Combining this with $<\gamma',\gamma>=0$ gives $u(s)=0$.

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