Why is $\frac{\vec{r}(t)}{ds}$ the unit velocity vector for a parametrized curve $\vec{r}(t)$ and arc-length function $s(t)$?

Question

To ask my question directly, what does

$$\frac{\vec{r}(t)}{ds}\tag{1}$$

mean intuitively, where $\vec{r}(t)$ is a position vector and $s(t)$ is an arc length function? In addition, why is this a unit velocity vector?

My calculations

It seems that I might be able to write

$$\vec{v}(t)=\frac{d\vec{r}(t)}{dt}\tag{2}$$

$$\frac{\vec{v}(t)}{v(t)}=\frac{d\vec{r}(t)/dt}{ds(t)/dt}\tag{3}$$

where $v(t)=\lVert \vec{v}(t)\rVert$.

Now, if I am allowed to cancel the terms in the notation in (3) I obtain

$$\hat{T}=\frac{\vec{v}(t)}{v(t)}=\frac{d\vec{r}(t)}{ds}\tag{4}$$

Though this operation from (3) to (4) still seems fishy and incomprehensible at this point to me and I don't know if it is allowed.

Context on why I am asking this

I am studying line integrals (Apostol's Calculus Vol II, Chapter 10, Section 10.7 "Line Integrals with Respect to Arc Length").

Here is the reasoning contained in that section (he uses $\vec{\alpha}$ for $\vec{r}$ but I prefer $\vec{r}$)

Let $\vec{r}$ be a path with $\vec{r}'$ continuous on an interval $[a,b]$. The graph of $\vec{r}$ is a rectifiable curve. In Volume I we proved that the corresponding arc-length function $s$ is given by the integral

$$s(t)=\int_a^t \lVert\vec{r}'(u)\rVert du$$

(By the way, technically, shouldn't it be $s(t)-s(a)$ here?)

The derivative of arc length is given by

$$s'(t)=\lVert\vec{r}'(t)\rVert$$

Let $\varphi$ be a scalar field defined and bounded on $C$, the graph of $\vec{r}$. The line integral of $\varphi$ with respect to arc length along $C$ is denoted by the symbol $\int_C\varphi ds$ and is defined by the equation

$$\int_C \varphi ds=\int_a^b\varphi[\vec{r}(t)]s'(t) dt$$

whenever the integral on the right exists.

Up to this point all good. Now is the part that my question concerns.

Now consider a scalar field $\varphi$ given by $\varphi[\vec{r}(t)]=\vec{f}[\vec{r}(t)]\cdot \vec{T}(t)$, the dot product of a vector field $\vec{f}$ defined on $C$ and the unit tangent vector $\vec{T}(t)=d\vec{r}/ds$. In this case the line integral $\int_C\varphi ds$ is the same as the line integral $\int_C\vec{f}\cdot d\vec{r}$ because

$$\vec{f}[\vec{r}(t)]\cdot\vec{r}'(t)=\vec{f}[\vec{r}(t)]\cdot\frac{d\vec{r}}{ds}\frac{ds}{dt}$$

$$=\vec{f}[\vec{r}(t)]\cdot\vec{T}(t)s'(t)$$

$$=\varphi[\vec{r}(t)]s'(t)$$

Answer 1

To ask my question directly, what does $\dfrac{\vec{r}(t)}{ds}$ mean intuitively

I guess this would mean something like "position divided by a small change in the length of the arc, but I doubt this expression with a single differential is what you're actually interested in.

I assume you actually want to know about $\dfrac{\mathrm d\vec{r}(t)}{\mathrm ds}=\hat{T}$. Intuitively, it takes a small vector difference of nearby points (which hugs the curve quite closely as the points are nearby), and then divides that by the length of that little part of the curve. Since the two points are so close together, the curve looks straight at that scale, and so we've basically divided by the length of the vector difference, to get a vector of length $1$ in the direction of that secant, which is basically a tangent.

This is illustrated in this image from Part 2 of Section 1.7 of the pre-alphaof "Multivariable Calculus Online" by Jeff Knisley:

In addition, why is this a unit velocity vector?

$\dfrac{\vec{r}(t)}{ds}$ is not a unit vector. Since it's like a finite vector divide by something infinitesimal, it's at least intuitively infinitely big.

But for $\dfrac{\mathrm d\vec{r}(t)}{\mathrm ds}=\hat{T}$, it's a unit vector and goes in the same direction as the velocity vector $\vec{r}'(t)$, so it's a "unit velocity vector.

this operation from (3) to (4) still seems fishy and incomprehensible at this point to me and I don't know if it is allowed.

It's not really any different from other applications of the chain rule you've already seen in single variable calculus.

(By the way, technically, shouldn't it be $s(t)-s(a)$ here?)

You have a different arc length function for each starting location, but they all differ by a constant (and hence have the same derivatives, etc.) so, in practice, it's not worth giving them separate names like "$s_3(t)=s_1(t)-s_1(3)$".

Up to this point all good. Now is the part that my question concerns.

I'm not sure if my earlier paragraphs alleviated your concerns about this, but for an approach that doesn't use $\mathrm ds$, see my answer to Why "A line integral of a scalar field is thus a line integral of a vector field where the vectors are always tangential to the line"?