intuition behind natural parametrization

Question

So if we have a parameterized curve $\gamma: r=r(q), \text{where } q\in [a,b]$ and there's a function for the length of the curve

$$ s(q)=\int_{q_0}^{q}\sqrt{(r'(t))^2}dt,\ \text{where } q\in [a,b] $$

I'm interested in why substituting $q$ in $r(q)$ with the inverse of $s: q(s)$ will give a curve where the length of the tangent vector will equal $1$, i.e. $|r'(t)|=1$.

What is the intuition behind this?

Thanks in advance!

Answer 1

Note that $ds/dq = |dr/dq|,$ which is the speed of the given parametrization of the curve. Thus, in order to slow the curve down to have speed $1$, we need to multiply the speed by a factor of $1/(ds/dq) = dq/ds.$ Choosing $s$ as our new parameter does exactly this, as we see from the chain rule $$\frac{dr}{ds} = \frac{dq}{ds}\frac{dr}{dq}.$$

Answer 2

When you do that substitution, you are moving along the curve in such a way that after a time $t$ the length of the distance that you have travelled is also equal to $t$. That is, the scalar speed is equal to $1$.