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

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

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?

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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 $\gamma(I)$.

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  • $\begingroup$ 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? $\endgroup$
    – JackOfAll
    Mar 10, 2023 at 15:25
  • $\begingroup$ Also, can you find a link or video that explains your concept? $\endgroup$
    – JackOfAll
    Mar 10, 2023 at 15:27
  • $\begingroup$ The integral is adding up all the (tiny) distances to get the distance traveled. $\endgroup$
    – Ned
    Mar 10, 2023 at 15:28
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    $\begingroup$ @JackOfAll the integral is "adding up" all the speeds multiplied by time changes to get an accumulation of distance travelled. The concept is simply $d=st$. $\endgroup$
    – FShrike
    Mar 10, 2023 at 15:44
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    $\begingroup$ To recap, the hypotenuse vector is the speed, since it's based on the 2 vector velocities in the x & y direction. The integral adds up the speed vector * time (dt) to get a distance. Right, because d=s*t. I think it just clicked, thanks again! $\endgroup$
    – JackOfAll
    Mar 12, 2023 at 0:26

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