I have the following question:
Let be $ \alpha : I \rightarrow \mathbb R^3 $ a regular curve parameterized by its arc length. Prove that it is verified that:
$\alpha' \cdot \alpha'''' = -3 \kappa \frac{d\kappa}{ds} $
Where $\kappa$ denotes the curvature of $\alpha $.
My attempts were unsuccessful. Thank you for your help.
Note that $\alpha'\cdot \alpha' =1$ since this is an arc-length parametrisation. Then $$0=\frac{d^3}{ds^3} \alpha' \cdot \alpha' = \frac d{ds}[2\alpha'\alpha^{(3)}+2\alpha''\alpha''] = 2\alpha''\alpha^{(3)} + 2\alpha'\alpha^{(4)}+4\alpha''\alpha{(3)}=2(3\alpha''\alpha^{(3)}+\alpha'\alpha^{(4)}).$$ Look more closely at $\alpha''\alpha^{(3)}$, this is $\frac12 \frac d{ds}\alpha''\cdot \alpha''=\frac 12\frac d{ds}\|\alpha''\|^2 = \frac12\frac d{ds}\kappa^2=\kappa \frac {d\kappa}{ds}$. Combine these two equations to get the desired equation.