Unit speed reparametrization of curve I am learning Elementary Differential Geometry by O'Neill and having a hard time with this exercise.
Suppose that $\beta_1$ and $\beta_2$ are unit-speed reparametrizations of the same curve $\alpha$.
Show that there is a number $s_0$ such that $\beta_2(s)=\beta_1(s+s_0)$ for all $s$.
My questions are
1 The two curves may not be defined on the same interval and each of them may traverse part of the curve $\alpha$. What is the meaning of for all $s$?
2 Why must the two beta curves have the same orientation?
3 How to prove the statement rigorously and completely?
 A: Let's recall some facts and definitions:  Let $\alpha$ be defined on some interval $I$ and define for $t_0\in I$
$$\lambda_{t_{0}}(t)=\int_{t_0}^t\|\alpha'(x)\|\,dx.$$ 
Any unit-speed-reparametrization (usp) $\phi_{t_{0}}$ is then given by 
$$\phi_{t_{0}}=\lambda_{t_{0}}^{-1}.$$
That's for the facts.
Now let $i\in\{1,2\}$ and choose some usp $\phi_{t_{i}}$  such that  $\beta_i=
\alpha\mathop{\circ}\phi_{t_i}$.  Let $t_0\in I$.  Then there exists $s_{i}$ such that $t_0=\phi_{t_i}(s_i)$.  So that gives
$$s_i=\phi_{t_i}^{-1}(t_0)=\lambda_i(t_0)=\int_{t_i}^{t_0}\|\alpha'(x)\|dx\,$$
and that means that $s_1$ and $s_2$ only differ by constant, namely
$$
s_0=\int_{t_1}^{t_2}\|\alpha'(x)|\,dx,
$$ 
i.$\,$ e., $|s_0|=$ length of $c$ between $t_0$ and $t_1$.  Extremely intuitive.  It then follows that $\beta_1(s)=\beta_2(s+s_0)$.
Ad (2): By definition of ups: it preserves orientation.
Michael
A: Yet another approach: let $\beta_2(s_2)=\beta_1(s_1)$.  Then there exists a differentiable map $\gamma$ such that $s_2=\gamma(s_1)$ (why?).  Thus
$$\beta_2\bigl(\gamma(s_1)\bigr)=\beta_1(s_1)\Rightarrow\beta_2'\bigl(\gamma(s_2)\bigr)\gamma'(s_2)=\beta_1'(s_1).$$
Taking lengths on both sides gives $|\gamma'(s_1)|=1$, thus $\gamma(s_1)=\pm s_1+s_0$. Since an usp preserves orientation, we must have $\gamma(s_1)=s_1+s_0$.
Michael
