Arc length of a particles path.

Question

If there is a particle, which travels in a three dimensional plane, with the position of the particle; $$\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+z(t)\mathbf{k},$$

Then the speed of the particle, at a given time is given by;

$$v=\vert \mathbf{v} \vert= \sqrt{(\dot{x(t)})^2+(\dot{y(t)})^2+(\dot{z(t)})^2}?$$

I also read that the bottom equation is also the 'derivative' of the arc length, measured from some initial point. Is that the distance the particle has traveled, from $t=0$ to $t=t?$

That is, $$\sqrt{(\dot{x(t)})^2+(\dot{y(t)})^2+(\dot{z(t)})^2}= \frac{dS_a}{dt}$$

I don't understand why this would be the case?

Answer 1

There is a typo in your definition. It should be: $$|\vec{v}|=\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2 }$$

Consider then that $\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}$ are changes of the $x,y,z$ coordinates with respect to time and that $\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2 }$ is the length of the line connecting the starting point to the ending point. If we were to sum all infinitesimal lengths together, we'd have the total arc length. That is: $$\text{arc length} = S =\int_{t_0}^{t_1} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2 } dt$$

Then:

$$\frac{dS}{dt} = \frac{d}{dt} \int_{t_0}^{t} \sqrt{\left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dy}{d\tau}\right)^2 + \left(\frac{dz}{d\tau}\right)^2 } d\tau = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2 }$$