I read in my textbook that if you parameterize a vector function's derivative in terms of that function's arc-length parameter by writing $r'(t)$ as $r'(s)$ then it will always be the case that $||r'(s)||= 1$.
Why is this so? The little discussion of this I found online was too difficult for me to penetrate. I've also tried experimenting with different expressions of arc-length to no avail.
First, let us see how do we "reparametrize" your vector valued function.
If $r: I \rightarrow \mathbb{R}^n$ is a given function, where $I$ is an interval in $\mathbb{R}$, then the arc-length can be seen as a function $s: I \rightarrow J$, where $J$ is another interval in $\mathbb{R}$ and,
$$s \left( t \right) = \int\limits_{t_0}^t \left| \left| r' \left( u \right) \right| \right| \ du$$
$t_0 \in I$ is fixed and all other notations have their usual meaning.
Notice that $s$ defined in such a way is one-one and continuous so that $s: I \rightarrow s \left( I \right)$ is a continuous bijection. Since $I$ is an interval, $s \left( I \right)$ is also an interval, so that it is not dangerous to assume $J = s \left( I \right)$. Also, we have an inverse $s^{-1} : J \rightarrow I$, which we can denote by $t$.
Note that this is pure abuse of notations! Since $s$ is a function of $t$ and takes every $t$ in $I$ to some point in $J$, we can say that $t$ is a function of '$s$', which takes every $s$ in $J$ to some point in $I$. Keeping this in mind, you can also prove that $t$ is a continuous function of $s$ so that $s$ becomes a homeomorphism and we have our reparametrization as
$$r \left( s \left( t \right) \right) = r \left( t \right)$$
By the definition of $s$, we have
$$\dfrac{ds}{dt} = \left| \left| r' \left( t \right) \right| \right|$$
so that by applying the chain rule on our reparametrized function, we get
$$r' \left( s \left( t \right) \right) \dfrac{ds}{dt} = r' \left( t \right)$$
From all the above equations, we can conclude that
$$\left| \left| r' \left( s \left( t \right) \right) \right| \right| = 1$$
Now, notice that since $s$ is a homeomorphism, $s \left( t \right)$ is itself a parameter, and the proof is done!
We can infact observe one important "fact" from our proof. In day-to-day lives, if we have a curve (considered as a path of a moving particle), we can always change the measurements so that the particle covers unit distance in unit time. That is to say, if we choose appropriate measurement systems, every particle can have "unit speed".