I wonder if some one can help me with the following problem:

Let $\gamma: \mathbb{R} \to \mathbb{R}^3$ be a regular curve parametrized by arclength. Show that if the tangent curve $\dot{\gamma}: \mathbb{R} \to \mathbb{R}^3$ is a non-constant geodesic on the unit sphere $S^2$, then the curvature $\kappa$ and torsion $\tau$ are constant.

If $\dot{\gamma}$ is a non-constant geodesic on $S^2$, then its second derivative $\gamma^{(3)}$, satisfies $\lVert \gamma^{(3)} \rVert^{\text{tan}}=0$ (it's tangential part is 0). Then it must be parallel to the normal (which on the unit sphere is just the coordinate) so $\gamma^{(3)}=r (x,y,z)$ for some $r$. Is this correct? If yes, how can I proceed from here?

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    $\begingroup$ You reasoning so far seems correct. Now try calculating $\frac{d \kappa}{d t}$. $\endgroup$ – humanStampedist Oct 12 '17 at 9:08
  • $\begingroup$ Thanks! I don't see how to do that...? $\endgroup$ – user202542 Oct 12 '17 at 9:46
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    $\begingroup$ What is $\kappa$, if your curve is parametrised by arclength? It just is $\kappa=\|\ddot{\gamma}\|$. $\endgroup$ – humanStampedist Oct 12 '17 at 9:51
  • $\begingroup$ Right, $\frac{d \kappa}{dt}=\lVert \gamma^{(3)} \rVert= r$ ? $\endgroup$ – user202542 Oct 12 '17 at 9:53
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    $\begingroup$ Maybe the calculation is easier if you derive $\kappa^2=\langle \ddot{\gamma},\ddot{\gamma}\rangle$ instead. Here $\langle\cdot,\cdot\rangle$ is the euclidean scalar product. $\endgroup$ – humanStampedist Oct 12 '17 at 10:10

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