# Rate of change of the position with respect to the distance

Physics Asked by Yousef Essam on September 24, 2020

In my textbook, there is a theorem which states that if the position vector of a body is
$$overrightarrow r$$ and the covered distance is $$s$$ such that $$overrightarrow r$$ is a function of $$s$$ then $$frac{doverrightarrow r}{ds} = widehat t$$
where $$widehat t$$ is a unit vector which is tangential to the trajectory of the body.

My question is: How can we rigorously prove this?
The textbook uses this theorem to prove that the velocity vector $$overrightarrow v$$ of a moving body is always tangential to the trajectory of the body.
$$overrightarrow v = {d overrightarrow rover dt} = {d overrightarrow rover ds} × {dsover dt} = {dsover dt} widehat t$$
So the direction of the velocity vector $$overrightarrow v$$ is in the direction of the unit vector $$widehat t$$ which is tangential to the trajectory of the body

Update:
At least if you are going to downvote the question, add anything helpful like stating what was actually wrong with the question or whether it was asked before or not because I searched a lot and found nothing, instead of just adding a downvote and leaving.

Let $$mathbf r:mathbb R rightarrow mathbb R^3$$ be the continuously-differentiable vector-valued function such that $$mathbf r(t)$$ is the position vector of the particle at time $$t$$. The distance traveled in between $$t=t_1$$ and $$t=t_2$$ is defined to be

$$Delta(t_1,t_2) := int_{t_1}^{t_2} Vertmathbf r'(t)Vert dt$$

Let $$[a,b]$$ be a time interval over which $$Vert mathbf r'(t)Vert >0$$. On this interval, the function $$D(t) := Delta(a,t)$$, which gives the distance traveled since time $$t=a$$, is strictly increasing and continuously differentiable with derivative $$D'(t) = Vert mathbf r'(t)Vert$$.

Let $$sigma:D[a,b] rightarrow [a,b]$$ be the reparameterization function such that $$sigmabig(D(t)big)=t$$. This function is guaranteed to exist because $$D$$ is monotonic on $$[a,b]$$. It is also continuously differentiable with derivative $$sigma'(s) = 1/D'big(sigma(s)big)$$ via the inverse function theorem.

Define the function $$mathbf R : D[a,b]rightarrow mathbb R^3$$ given by $$mathbf R = mathbf r circsigma$$, which gives the position of the particle as a function of how far it has traveled since time $$t=a$$. To conclude the proof, note that

$$mathbf R'(s) = mathbf r'big(sigma(s)big) sigma'(s) = frac{mathbf r'big(sigma(s)big)}{D'big(sigma(s)big)}$$

The right hand side can immediately be seen to be a unit vector, since $$D'(t) = Vert mathbf r'(t)Vert$$. It is also clearly parallel to the velocity vector $$mathbf r'$$.

Note that $$mathbf R'(s)$$ is usually written (in a common abuse of notation) $$frac{dmathbf r}{ds}$$, and that $$sigma(s)$$ is the time which corresponds to a travel distance of $$s$$ since $$t=a$$.

Correct answer by J. Murray on September 24, 2020

The proof is almost trivial. But I don't know if this is rigorous enough from a mathematical point of view.

The distance $$ds$$ is defined to be the absolute norm of the position difference $$dvec{r}$$: $$ds=|dvec{r}|$$ Therefore the differential quotient $$frac{dvec{r}}{ds}$$

• is a vector of length $$1$$ (since $$left|frac{dvec{r}}{ds}right|=frac{|dvec{r}|}{ds}=frac{ds}{ds}=1$$)
• and has the same direction as $$dvec{r}$$

In other words: $$frac{dvec{r}}{ds}$$ is a unit vector tangential to the trajectory.

Answered by Thomas Fritsch on September 24, 2020

I don't know whether this is rigorous enough for you, but it satisfies me to say ...

If A and B are two points on the body's trajectory, then $$Delta vec r = vec {r_A} - vec {r_B}$$ in which $$vec r$$ is the body's position vector. Now consider $$frac{Delta vec{r}}{Delta s}$$.

Provided that the trajectory doesn't bend sharply, then as $$Delta s$$, the distance along the trajectory between A and B, approaches zero, $$Delta s = |Delta vec r|$$ So $$frac{Delta vec{r}}{Delta s}$$ is the unit vector in the direction $$Delta vec r$$, that is in the direction of the tangent to the trajectory.

Answered by Philip Wood on September 24, 2020

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