Apostol: Why can we write unit velocity vector as $\hat{T}(t)=\frac{d\vec{r}}{ds}$, and what is $\frac{d\vec{r}}{ds}$ intuitively? I've been learning about a few topics related to vector calculus, and I just need to bring them all together in my understanding.
To cut to the chase, my question is 1) why can we write
$$\hat{T}(t)=\frac{d\vec{r}}{ds}\tag{1}$$
where $\hat{T}$ is the unit velocity vector, $\vec{r}$ is a position vector (ie, a parameterized curve), and $s$ is the arc length function, and 2) what is an intuitive explanation for why $\frac{d\vec{r}}{ds}$ in fact is the unit velocity vector.
By intuitive, I mean something like: $\frac{\vec{v}}{v}$ is the unit velocity vector simply because we take the velocity vector and divide it by its length.
I saw (1) in a section in Apostol's Calculus, Volume II, in Chapter 10 on Line Integrals.
I'm going to go through the ideas in that section step by step just to make sure I have it all down. I hope it's not too prolix.
Apostol defines line integrals with respect to arc length as follows (note that he names things differently than I do, though I copied everything verbatim except that I chose to use the vector notation with the arrows instead of using a bold font):

Let $\vec{\alpha}$ be a path with $\vec{\alpha}'$ continuous on an
interval $[a,b]$. The graph of $\vec{\alpha}$ is a rectifiable curve.
In Volume I we proved that the corresponding arc-length function $s$
is given by the integral
$$s(t)=\int_a^t \Vert\vec{\alpha}'(u)\Vert du$$
The derivative of the arc length is given by
$$s'(t)=\Vert\vec{\alpha}'(t)\Vert$$
Let $\phi$ be a scalar field defined and bounded on $C$, the graph of
$\vec{\alpha}$.
The line integral of $\phi$ with respect to arc length along $C$ is
denoted by the symbol $\int_C \phi ds$ and is defined by the equation
$$\int_C \phi ds = \int_a^b \phi[\vec{\alpha}(t)]s'(t)dt$$
whenever the integral on the right exists.

Then he says

Now consider the scalar field $\phi$ given by
$\phi[\vec{\alpha}(t)]=\vec{f}[\vec{\alpha}(t)]\cdot \vec{T}(t)$, the
dot product of a vector field $\vec{f}$ defined on $C$ and the unit
tangent vector $\vec{T}(t)=\frac{d\vec{\alpha}}{ds}$. In this case,
the line integral $\int_C \phi ds$ is the same as the line integral
$\int_C\vec{f}\cdot d\vec{\alpha}$ because
$$\vec{f}[\vec{\alpha}(t)]\cdot\alpha'(t)=\vec{f}[\vec{\alpha}(t)]\cdot\frac{d\vec{\alpha}}{ds}\frac{ds}{dt}$$
$$=\vec{f}[\vec{\alpha}(t)]\cdot\vec{T}(t)s'(t)$$
$$=\phi[\vec{\alpha}(t)]s'(t)$$

My main question boils down to why $\vec{T}(t)$ equals $\frac{d\vec{\alpha}}{ds}$?
Here is my understanding of the ideas above (using notation that I am used to):
We have a position vector $\vec{r}(t)$ which is a parameterized curve.
$$\frac{d\vec{r}(t)}{dt}=\vec{v}(t)$$
is the velocity vector.
$$\phi(\vec{r}(t))=\vec{f}(\vec{r}(t))\cdot \hat{T}(t)$$
is a scalar field (the magnitude of the component of $\vec{f}$ in the direction of $\vec{v}$), where $\hat{T}(t)=\frac{\vec{v}(t)}{v(t)}$ is the unit vector in the direction of the velocity vector.
Now, using Apostol's definition of line integral with respect to arc length we have
$$\int_C \phi ds = \int_{t_i}^{t_f} \vec{f}(\vec{r}(t))\cdot\hat{T}(t) v(t)dt$$
$$= \int_{t_i}^{t_f} \vec{f}(\vec{r}(t))\cdot\frac{\vec{v}(t)}{v(t)}) v(t)dt$$
$$= \int_{t_i}^{t_f} \vec{f}(\vec{r}(t))\cdot\vec{v}(t)dt$$
$$=\int_C \vec{f}\cdot d\vec{r}\tag{2}$$
(2) is the result from the second snippet from Apostol above.
Now, specifically regarding $\hat{T}$, we have
$$\hat{T}(t)=\frac{\vec{v}(t)}{v(t)}=\frac{\vec{v}(t)dt}{ds}$$
$$=\frac{d\vec{r}}{ds}$$
I obtained this by either

*

*multiplying numerator and denominator by $dt$ or


*writing $v(t)=\frac{ds}{dt}$
It may seem equivalent, but it affects my understanding of what exactly is happening.
Ultimately, I would also like to know what $\frac{d\vec{r}}{ds}$ means exactly.
It seems to mean the rate of change of the position vector as we change the arc length. But why does Apostol define the velocity unit vector this way? I find it much more intuitively clear to think about $\frac{\vec{v}}{v}$.
What is an intuitive way to reach the conclusion that the unit velocity vector is in fact $\frac{d\vec{r}}{ds}$?
 A: We can use the parameterization of $\vec{\alpha}$ to get the function $\vec{r}(s)$. First, we know that $\vec{r} = \vec{\alpha}(t)$ for some $t$. Additionally, since the arc length function $s_\alpha$ is an increasing function of $t$, it is invertible. (I'm giving it a subscript to distinguish the function from the parameter). Therefore we have $\vec{r}(s) = \vec{\alpha}(s_\alpha^{-1}(s))$. We can then differentiate this using the chain rule and inverse function rule
$$
\frac{d\vec{r}}{ds} = \vec{\alpha}'(s_\alpha^{-1}(s))\frac{ds_\alpha^{-1}}{ds} = \frac{\vec{\alpha}'(s_\alpha^{-1}(s))}{s'(s_\alpha^{-1}(s))} = \frac{\vec{\alpha}'(s_\alpha^{-1}(s))}{||\vec{\alpha}'(s_\alpha^{-1}(s))||} = \hat{T}(s_\alpha^{-1}(s))
$$
Now, since $s_\alpha^{-1}(s)$ is just the $t$ value corresponding to an arc length of $s$ along $\vec{\alpha}$, it makes some sense to abbreviate it to $t$, giving
$$
\frac{d\vec{r}}{ds} =\hat{T}(t)
$$
It's worth noting that "the parameterization that always has unit speed" is essentially the definition of arc length. After all, if your speed is always 1, then the distance you travel is always equal to the time elapsed.
