Why is Parametric/Vector arc length the integral of speed?

Question

$$\mathrm{Speed}=\lvert\mathrm{velocity}\rvert$$

$$\mathrm{speed}=\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^{\!2}+\left(\frac{\mathrm dy}{\mathrm dt}\right)^{\!2}}$$

The parametric speed formula is based on the Pythagoras' theorem using a triangle formed by $\mathrm dy$ and $\mathrm dx$, with the length/magnitude of the hypotenuse being the speed. I think that makes sense to me.

But, what is the conceptual idea behind arc length (total distance) being the integral of the speed? What exactly are you adding up? Tiny slices of speed hypotenuses?

Ok, so the definite integral is based on t. Starting and ending t. For each t, you have a velocity vector distance (speed). The integral is adding up all the speeds to get a distance traveled?

Answer 1

The time integral of speed is distance or length because $d=s\times t$. You're adding up small hypotenuses of speed multiplied by $\mathrm{d}t$, which gives small slices of distance. Loosely speaking.

You can rigorously prove the following for all $C^1$ functions $\gamma:I\to\Bbb R^n$: $$\ell(\gamma)=\int_0^1\|\gamma’\|$$For a suitable definition of length $\ell$, which is done in the first few pages here.

There is a further equivalence of this with the Hausdorff 1-measure of the trace $\gamma(I)$.