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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.

3 Answers

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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