Let $\alpha:I\to R^3$ be a regular parametrized curve (not necessarily by arc length) and let $\beta: J\to R^3$ be a reparametrization of $\alpha(I)$ by the arc length $s=s(t)$, measured from $t_0\in I$. Let $t=t(s)$ be the inverse function of $s$ and set $\frac {d\alpha}{dt}=\alpha'$, $\frac {d^2\alpha}{dt^2}=\alpha''$, etc. Prove that:
a. $\frac {dt}{ds}=\frac {1}{\vert \alpha'\vert}$
My attempt at a solution:
First, I renamed the function $t$ to $\tau$ to distinguish it from the real number $t$, and the real number $s$ to $l$ to distinguish it from the function $s$, to avoid mixing things up. That is:
$$s:I\to \mathbb R, \space s(t) = \int_{t_0}^t{\vert \alpha'(\sigma)\vert\space d\sigma}$$ $$\tau: J\to \mathbb R, \space \tau(l)=s^{-1}(l)$$
The derivative of $\tau$ at $l$ is:
$$\tau'(l)=(s^{-1})^{'}(l)=\frac {1}{s'(s^{-1}(l))}=\frac {1}{s'(\tau(l))}$$
The derivative of $s$ at $t$ is:
$$s'(t)=\vert\alpha'(t)\vert$$
Therefore:
$$\tau'(l)=\frac {1}{\vert\alpha(\tau(l))\vert}$$
The way the exercise is stated makes it seem like $\frac {dt}{ds}(x)=\frac {1}{\vert \alpha'(x)\vert}$ for every $x$, even though those two functions don't even have the same domain, while my solution states it as $\frac {dt}{ds}(x)=\frac {1}{\vert \alpha'(y)\vert}$, for $y=\tau(x)$. Is this a standard way of referring to derivatives and inverses of functions? I'm confused if my solution missed something or if I'm simply not understanding notation.
Thanks.