Curvature of curve with arbitrary parametrization

[ccosta]

Hi, I'm preparing a little exposition of curvature and torsion for my Calculus class and so I need to include some simple proofs for the things I'll use to define curvature. So I'm looking for a proof of the formula for the curvature of an arbitrarily parametrized curve that doens't use the Frenet frame; it'll get too complicated for a 15 minute presentation. The closest thing to what I'm looking for is in Pogorelov's Differential geometry which goes like this:

Say [tex]\textbf{r} = \textbf{r}(t)[/tex] is a vector valued function. He then goes on to take the derivative with respect to t, remembering that the arclength can be expressed as a function of time:
[tex]\textbf{r}' = \textbf{r}_{s}' s'[/tex]
What I understand by this notation is:
[tex](x'(t),y'(t),z'(t)) = (x'(s)s'(t),y'(s)s'(t),z'(s)s'(t))[/tex],
where the derivatives are take with respect to their arguments. He then goes on to say that from that formula, we can arrive at:
[tex]{\textbf{r}'}^2={s'}^2[/tex]
This is the unclear point for me. The rest is pretty straight forward, he uses the first formula to find an expression for [tex]\textbf{r}_{s}'[/tex] in terms of [tex]\textbf{r}'[/tex], which he differentiates again with respect to [tex]t[/tex] and the proof if pretty much done. However, I don't understand how he jumps from the first to the second equation.

Something else I've been looking for is an easy way to prove that the derivative of a parametrization by arclenght has unit norm.
And one last thing, any books to be recommended? I'm looking for very basic DG books, nothing on manifolds or even surfaces, but it must treat space curves.

Thanks to all!

 

[pat_connell]

For the first question, there are a few ways you can go about proving the equation \textbf{r}' = \textbf{r}_{s}' s'. One way is to use the chain rule. Since \textbf{r} = \textbf{r}(t), its derivative can be written as \textbf{r}' = \textbf{r}_{t}' t' (where \textbf{r}_{t}' is the derivative with respect to t). Since arclength s is a function of t, we can express s' as a function of t': s' = f(t'). Therefore, \textbf{r}' = \textbf{r}_{t}' t' = \textbf{r}_{s}' f(t') = \textbf{r}_{s}' s'. To prove that the derivative of a parametrization by arclength has unit norm, you can use the fact that the unit tangent vector of a parametrized curve is given by \textbf{T} = \textbf{r}'/\|\textbf{r}'\|. Since we already know that \|\textbf{r}'\| = s', we can substitute this expression into \textbf{T} and get \textbf{T} = \textbf{r}'/s', which implies that \|\textbf{r}'\| = s'. As for books, I would recommend Do Carmo's Differential Geometry of Curves and Surfaces. This book covers the basics of differential geometry, including space curves, but doesn't go too deep into the more advanced topics.