The arc-length is a differentiable function of $t$, and $\frac{ds}{dt}=\vert \alpha'(t)\vert$

Question

I'm reviewing curves using the introductory chapters of a differential geometry book (Manfredo's book). He says something like the following:

The arc length of a regular parametrized curve $\alpha:I\rightarrow\mathbb{R}^3$ from the point $t_0$ is defined to be $$ s(t)=\int_{t_0}^t \vert \alpha'(u)\vert du $$ with $\alpha'(t)=\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}$.
Since $\alpha'(t) \ne 0$ for all $t$,the arc-length is a differentiable function of $t$, and $\frac{ds}{dt}=\vert \alpha'(t)\vert$.

My problem:
Why does it follow from $\alpha'(t) \ne 0$ that $s$ is differentiable?

My attempt:
The (second) fundamental theorem of calculus states that

Let $f,g:\mathbb{R}\rightarrow \mathbb{R}$. If $f$ is integrable on $[a,b]$ and $f=g'$, then $$ \int_a^b f(x)dx=g(b)-g(a) $$

In our case, the function $\alpha'$ is continuous, and hence integrable. But, in order to apply the f.t.c, we need to see that its modulus $f=\vert \alpha' \vert$ is integrable, and also that $\vert \alpha' \vert = g'$ for some function $g$. Since $\alpha'$ is continuous and the modulus function is also continuous, $\vert \alpha'\vert$, being the composition of those, is continuous, and thus integrable. Also, $\vert \alpha'\vert=\frac{d}{dt} \vert \alpha\vert$ (Question 1: is this true? Why? Is this where I use that $\alpha \ne 0$?) So, using the f.t.c, we get $$ s(t)=\vert \alpha(t)\vert - \vert \alpha(t_0)\vert $$ so $$ \frac{ds}{dt}(t)=\vert \alpha'(t)\vert - \frac{d}{dt}\vert \alpha(t_0)\vert=\vert \alpha'(t)\vert $$

Answer 1

As you note, we'll be in good shape if we can conclude that the assignment $t\mapsto|\alpha'(t)|$ is continuous, because then the (first) fundamental theorem of calculus will allow us to conclude that \begin{equation} s(t):=\int_{t_0}^t |\alpha'(u)|du \end{equation} is differentiable, and that $s'(t)=|\alpha'(t)|$. Since both $\alpha\colon I\to\mathbb{R}^3$ and the modulus map $\mathbb{R}^3\to\mathbb{R}$ are continuous, so is their composition, so $s(t)$ is differentiable.

So the condition that $\alpha'(t)$ should never vanish isn't necessary for the arc length to be differentiable. Bill Cook points out in this answer that we include this condition because it guarantees that we always have $s'(t)>0$. This means that $s(t)$ is strictly increasing, and can thus be inverted. So requiring that $\alpha'(t)\neq 0$ allows us to obtain an arclength parametrization of our curve.

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Edit: We should also point out that $\frac{d}{dt}|\alpha(t)|\neq|\alpha'(t)|$. You can work this out by unwinding what the two sides are in the components of $\alpha(t)$, but it might be more enlightening to think geometrically. If $\alpha(t)$ takes all of its values in the unit sphere, then $|\alpha(t)|=1$, so $\frac{d}{dt}|\alpha(t)|=0$. But certainly there are curves in the sphere with nonzero tangent vector, so our equality can't be true.