Proof that Curvature is independent of Parametrization One of my online calculus lectures asked, to be completed as an exercise, why two arc length parametrization yield the same curvature. Specifically it asks, if $C$ is a curve and $\mathbf{r}_1(s)$ and $\mathbf{r}_2(t)$ are two arc length parameterizations of C, and $\mathbf{r}_1(s_0)$ and $\mathbf{r}_2(t_0)$ correspond to some point P on C, then
$$\| \mathbf{r}_1''(s_0) \| = \| \mathbf{r}_2''(t_0) \|.$$
I tried deriving relationships from the definition that an arc length function satisfies and was able to obtain that if s and t are functions of some u that
$$s'(u)=t'(u)$$
such that
$$s(u)=t(u)+c.$$
I've gone in multiple directions to try to prove something that seems so simple, and I feel as though I am missing a certain relationship between the two parameterizations that the problem implies. Can anyone point me in the right direction to try and prove this? It seems like I'm way overthinking this problem.
 A: Considering the last but one sentence of Unit speed reparametrization of curve you know that there exists a constant $c$ such that $ r(s+c)=r(t)$.
A: EDIT: Answer outdated, please see Michael Hoppe's suggestion.
Here's a proof that's missing one step of rigor, which I cooked up with the help of a suggestion from JacobCheverie. I'll use the product rule for vector-valued functions, which is:
$$\frac{d}{dx} \mathbf{f}(x) \cdot \mathbf{g}(x) = \mathbf{f}'(x) \cdot \mathbf{g}(x) + \mathbf{f}(x) \cdot \mathbf{g}'(x)$$
Assume that both $\mathbf{r_1}(s)$ and $\mathbf{r_2}(t)$ are differentiable. Then there must be some $f(s) = t$ (a way to convert between parametrizations) and because both are arc-length parametrizations
$$f(s_0) = \pm 1$$
(at the shared point, either they are the same, or going in opposite directions). I'm not sure what the nicest way to show this is.
Assuming that, starting from $||\mathbf{r_1}'(s)|| = ||\mathbf{r_2}'(t)||$, square both sides to get
$$
\begin{aligned}
\mathbf{r_1}'(s) \cdot \mathbf{r_1}'(s) &= \mathbf{r_2}'(t) \cdot \mathbf{r_2}'(t)\\
\mathbf{r_1}'(s) \cdot \mathbf{r_1}'(s) &= \mathbf{r_2}'(f(s)) \cdot \mathbf{r_2}'(f(s))\\
\frac{d}{ds} \left(\mathbf{r_1}'(s) \cdot \mathbf{r_1}'(s)\right) &= \frac{d}{ds} \left(\mathbf{r_2}'(f(s)) \cdot \mathbf{r_2}'(f(s))\right)\\
2(\mathbf{r_1}''(s) \cdot \mathbf{r_1}'(s)) &= 2f'(s)(\mathbf{r_2}''(f(s)) \cdot \mathbf{r_2}'(f(s)))\\
\mathbf{r_1}''(s) \cdot \mathbf{r_1}'(s) &= f'(s)(\mathbf{r_2}''(f(s)) \cdot \mathbf{r_2}'(f(s)))\\
||\mathbf{r_1}''(s)|| ||\mathbf{r_1}'(s)|| &= f'(s)^2||\mathbf{r_2}''(f(s))|| ||\mathbf{r_2}'(f(s)))||\\
||\mathbf{r_1}''(s)|| &= f'(s)^2||\mathbf{r_2}''(f(s))||\\
||\mathbf{r_1}''(s_0)|| &= f'(s_0)^2||\mathbf{r_2}''(t_0)||\\
||\mathbf{r_1}''(s_0)|| &= ||\mathbf{r_2}''(t_0)||\\
\end{aligned}
$$
Can anyone help me with $f'(s)$? I actually don't know why $f'(s) \ne \pm 1$ in general, for all $s$ -- if we're assuming differentiable arc-length $\mathbf{r_1}$ and $\mathbf{r_2}$, it seems like the they must either be the same or be opposite to each other.
If they weren't differentiable, I see how one could suddenly change direction any number of times (covering the same ground over and over again), but keep the same speed. But for arc-length parameterizations, does any such reversal in direction have to be non-differentiable?
