Space Curve Question: Showing Derivative of t wrt s

[noospace]

Homework Statement

Let $\alpha : I \to \mathbb{R}^3$ be a regular parametrized curve and let $\beta : J \to \mathbb{R}^3$ be a reparametrization in terms of arc length $s = s(t)$.

Show that $\frac{d^2t}{ds^2} = -\frac{\dot{\alpha} \cdot \ddot{\alpha}}{\abs{\alpha}^4}$

The Attempt at a Solution

I'm going to use dots and dashes to represent differentiation wrt time and arc-length, respectively.

$\beta'(s) = [\alpha (t(s))]' = \dot{\alpha} t'$
$\beta''(s) = \dot{\alpha}t'' + \ddot{\alpha} t'^2$ (*)

$t'(s) = 1/\dot{s}(t)$ by the inverse function theorem for 1 variable so
$\dot{\alpha} = \dot{s} \beta'(s)$. This is where things start to get messy.

I think I need to use the fact that $\dot{\alpha} = \alpha'$. Then the result follows by dotting both sides of (*) by $\alpha'$.

Can you convince me that $\dot{\alpha} = \alpha'$?

Thanks.

 

[noospace]

I figured this out nearly as soon as I posted it. Here it is for future use:

$\beta'(s) = \dot{\alpha} t'$ (1)
$\beta''(s) = \dot{\alpha} t'' + \ddot{\alpha} t'^2$
$\beta''(s) = \dot{\alpha} t'' + \ddot{\alpha} t'$.

Applying inverse function to (1) gives $\beta'(s)\dot{s} = \dot{\alpha}$ showing that $\dot{\alpha}$ is parallel to $\beta'$.

But $\beta'$ is a unit vector field so $\beta'\cdot\beta''=0 \implies \dot{\alpha} \cdot \beta''=0$. Hence

$0 = \dot{\alpha}^2 t'' + \dot{\alpha}\cdot\ddot{\alpha} t'^2$
$t'' = - \frac{\dot{\alpha}\cdot \ddot{\alpha}t'^2}{\dot{\alpha}^2} = -\frac{\dot{\alpha}\cdot\ddot{\alpha}}{\dot{\alpha}^4}$