Rate of change of the position with respect to the distance

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 $mathbf r'$.

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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 $t=a$.