Is it possible to find an explicit relation between these two, meaning that if one is given we can find another? We know that the motion of a point on a space can be given as $\vec r = \vec r(t)$, where the vector is the position vector of the point or as $s = s(t)$ where $s$ is the coordinate of an arc on the trajectory of the point.
Question: Is it possible to find an explicit relation between these two, meaning that if one is given we can find another?
Professor gave us this question a week ago, but I could not succeed. Actually, I don't know how to search for that relation.
 A: Welcome to the world of differential geometry.

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*The first task to go from $\vec{\rm pos}(t)$ to $s(t)$ is rather straight forward as it requires that $\vec{\rm pos}$ be a differentiable function only.
Given $\vec{\rm pos}(t)= (x(t),y(t),z(t))$
$$ {\rm d}s = \sqrt{ \left(\tfrac{{\rm d}}{{\rm d}t} x\right)^2 + \left(\tfrac{{\rm d}}{{\rm d}t} y\right)^2 + \left(\tfrac{{\rm d}}{{\rm d}t} z\right)^2}\,{\rm d}t $$
And thus $s(t) = \int {\rm d}s$.
This is identical to defining $$\vec{\rm vel}(t) = \tfrac{\rm d}{{\rm d}t} \vec{\rm pos}(t)$$
and setting $s(t) = \int \| \vec{\rm vel}(t)\,{\rm d} t$


*The second task requires prior knowledge of the path, given some arbitrary parameter $\alpha$, again in the form of $\vec{\rm pos}(\alpha)$. And then converting the path into a function of time. Use the chain rule and the speed profile
$$ \vec{\rm vel}(t) = \left( \tfrac{{\rm d}}{{\rm d}\alpha} \vec{\rm pos}(\alpha) \right) \dot{\alpha}$$
$$ \tfrac{\rm d}{{\rm d}t} s(t) = \| \vec{\rm vel}(t) \| = \| \tfrac{{\rm d}}{{\rm d}\alpha} \vec{\rm pos}(\alpha) \| \,\dot{\alpha} $$
$$ \Rightarrow \dot{\alpha}(t) = \frac{ \frac{\rm d}{{\rm d}t} s(t) } { \| \tfrac{{\rm d}}{{\rm d}\alpha} \vec{\rm pos}(\alpha) \| } $$
Now integrate $\dot{\alpha}$ to get $\alpha(t)$ for use in $\vec{\rm pos}(\alpha)$
$$ \alpha(t) = \int \dot{\alpha} \, {\rm d} t $$
and
$$ \vec{\rm pos}(t) = \vec{\rm pos}( \alpha(t) ) $$
