Let $\vec{r}(t)$ be a spacecurve such that $|{\vec{r'}(t)}|=1$ everywhere. Consider a new spacecurve such that $\vec{u}(t)=\vec{r'}(t)$.
Prove that the curvature of the new curve equals $\sqrt{1+\frac{\tau^2}{\kappa^2}}$ , where $\tau$ and $\kappa$ are the torsion and curvature of the original curve.
the equation for torsion is $\tau=\frac{[ r' r'' r''']}{|r'X r''|^2}$ and that of curvature of original curve is $\kappa=\frac{[ r' X r''|}{r'^3}$