Curvature defined as $k(t) = \frac{\|T'(t)\|}{v(t)}$, explanation of the proof I have problems to understand how it's derived the formula for the curvature $k(t) = \frac{\|T'(t)\|}{v(t)}$.
The derivation starts with the equality $$\frac{dT}{ds}\frac{ds}{dt} = \frac{dT}{dt}$$
where $T$ is the unit velocity vector for the curve  and $s$ is the arc length function for the curve.
So $$\frac{dT}{ds} = \frac{T'(t)}{v(t)}$$
and from here it's all clear to me how to finish.

Initially I supposed it was just an application of chain rule, but if I rewrite the proof in Netwon's notation I obtain
$$[T(s(t))]' = T'(s(t))s'(t)= T'(s(t))v(t)$$
So to get the final equality I should have $[T(s(t))]' = T'(t)$, but is it always the case ?
I don't know if it is useful the fact that $T(s(t))$ and $T(t)$ have the same trajectory (in general they differ for a certain value of $t$ though). Is this enough? Why?
 A: Leibniz notation obscures what's going on. If everything is functions of $t$ and we reparametrize the curve by arclength, getting $\alpha(t) = \tilde\alpha(s(t))$, then $\alpha'(t) = \tilde T(s(t))\upsilon(t)$ and $T(t) = \dfrac{\alpha'(t)}{\|\alpha'(t)\|} = \dfrac{\tilde T(s(t))\upsilon(t)}{\upsilon(t)} = \tilde T(s(t))$. Now the chain rule gives
$$T'(t) = (\tilde T\circ s)'(t) = \tilde T'(s(t))\upsilon(t) = \tilde k(s(t))\upsilon(t),$$
so
$$k(t)=\tilde k(s(t)) = \dfrac{T'(t)}{\upsilon(t)}.$$
A: I think some of the confusion here is due to sloppy notation. $T(s)$ and $T(t)$ are two different functions (one takes in arc length parameter $s$, and the other takes in your arbitrary parameter value $t$) and the meaning of $T'$ will differ depending on which function you're differentiating.
If we make things more explicit, by naming the two functions $T(s)$ and $\tilde T(t)$ for instance, with
$$T(s[t]) = \tilde T(t)$$
then differentiating both sides by $t$ gives you the necessary identity,
$$\frac{dT}{ds} \frac{ds}{dt} = \frac{d\tilde{T}}{dt}.$$
