Why do we need that $\alpha$ is for regular to existence of arc length as integral?

Question

Take $\alpha:I\rightarrow R^n$ a regular curve. For $t_0\in I$ Define $$s(t)=\int_{t_0}^t|\alpha'(x)|dx$$ as the arc lenth.

In Differential Geometry of Curves and sufaces of Do Carmo is written that "as $\alpha'(t)\not=0$ then $\displaystyle\frac{ds(t)}{dt}=|\alpha'(t)|$". Why the condition $\alpha'(t)\not =0$ is necessary?

Can why find a not regular curve $\alpha$ (but smooth) such that

$$\displaystyle\frac{ds(t)}{dt}=|\alpha'(t)|$$

is not true for some $t\in I$?

I think this is not possible because $\alpha'$ is always continue, so is $|\alpha'|$. Then by the fundamental theorem of calculus the equation $ds/dt=|\alpha'|$ holds always. Even if $\alpha$ is not regular (but smooth).

I'm right?

Thanks!

Answer 1

The condition $\alpha'(t) \not=0$ isn't necessary for the formula $s'(t)=|\alpha'(t)|$ to hold.

However, if you allow $\alpha'(t_0)={\bf 0}$, then $s'(t_0)=0$ and we no longer have that $s(t)$ is necessarily strictly increasing. So solving for $s$ in terms of $t$ comes into doubt. This means we may no longer have the (unique) parameterization with respect to arc length.

$\alpha'(t_0)={\bf 0}$ (i.e., $s'(t_0)=0$) also means that we don't have a well defined tangent line at $r(t_0)$.

Some of the theory that follows could be repaired/extended to curves with $\alpha'(t)=0$, but everything gets a bit more ugly!