# 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.

Welcome to the world of differential geometry.

• 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

$$\vec{\rm pos}(t) = \vec{\rm pos}( \alpha(t) )$$

• Thank you! I am a student majoring in math so this answer is really appreciated. – VIVID Mar 26 at 19:37