In the text "O'neill-Elementary differential geometry (2nd edition)", there are following problems.
As we know, any unit speed-reparametrized curve can be viewed as an arc-length reparametrizated curve. Since the arc-length reparametrized curve must achieve the positive orientation, the problem 5 can be easily proved. But, I have a question for the problem 7. $\alpha$, $\bar{\alpha}$ are unit-speed curves. Then, why we have to admit the "minus sign"(opposite orientation) in the problem 7? I want to know the exact difference of two problems.
Best Answer
$$\alpha(s)=\bar{\alpha}(h(s))$$
Differentiate w.r.t. $s$
$$\alpha'(s)=\bar{\alpha}'(h(s))h'(s)$$
Take absolute value
$$|\alpha'(s)|=|\bar{\alpha}'(h)||h'(s)|$$
$|h'(s)|=1$
Here $h(s)$ should be continuous, otherwise $\alpha$ won't satisfy as a regular curve
$h'(s)=1$ or $-1$
$h(s)=\pm s +s_0$ for some constant $s_0$
Arclength parametrization only means $|\alpha'(s)|=1$ at each point on the curve, it doesn't tell us anyone about the orientation. Reversing the direction would still give us the arclength parametrization.