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

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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