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}