Curvature of a regular parametrization 
Prove that if $\mu: [a,b] \to \mathbb{R}^n$ is a regular
  parametrization of a curve then the curvature at $\mu(t)$ is given
  by: $$\kappa(t) =
 \left|\left|\left(\frac{\mu'(t)}{||\mu'(t)||}\right)'\frac{1}{||\mu'(t)||}\right|\right|.$$

A regular parametrization is when $\mu'(t)\ne 0$ for all $t \in \mathbb{R}$.
I know that I must use arc length to show this but I don't know how to apply it. 
 A: Use the chain rule. In particular, $ \frac{d}{ds} = \frac{dt}{ds} \frac{d}{dt} = \frac{1}{||\mu'(t) ||} \frac{d}{dt} $.
A: The way I learned it, the curvature $\kappa$ of a curve such as $\mu(t)$ is the magnitude of rate of change of the unit tangent vector $\mathbf{u}$  with respect to the distance or arc length $s$ along the curve.  Now in the present case, the vector $\mu'(t)$ is tangent to $\mu(t)$ at $t$, so the unit tangent vector is
$\mathbf{u} = \frac{\mu'(t)}{\Vert \mu'(t) \Vert}$,
whence
$\kappa = \Vert \frac{d \mathbf{u}}{ds} \Vert$;
but
$\frac{d \mathbf{u}}{ds} = \frac{d \mathbf{u}}{dt} \frac{dt}{ds}$,
and since $\mu'(t)$ is the tangent vector to the curve $\mu(t)$, we have
$\frac{ds}{dt} = \Vert \mu'(t) \Vert$,
from which follows that
$\frac{dt}{ds} = \frac{1}{\Vert \mu'(t) \Vert};$
note we have used the given property that the parametrization is regular, i.e. that
$\mu'(t) \ne 0$, hence $\Vert \mu'(t) \Vert \ne 0$, in this last step to (albeit implicitly) infer that the function $s(t)$ which takes the $t$ parameter to arc length $s$,
is invertible to $t(s)$, and that the derivative $\frac{dt}{ds} = (\frac{ds}{dt})^{-1}$.  We thus have, bringing it all together,
$\kappa = \Vert \frac{d\mathbf{u}}{ds} \Vert = \Vert \frac{d\mathbf{u}}{dt} \frac{1}{\Vert \mu'(t) \Vert} \Vert = \Vert (\frac{\mu'(t)}{\Vert \mu'(t) \Vert})' \frac{1}{\Vert \mu'(t) \Vert} \Vert$,
which is the desired result.  Note that here and throughout the prime symbol (') denotes differentiation with respect to $t$.
Hope this helps!
