Q:

reparametrization of a regular curve

Let $\gamma: I \to \mathbb{R}^n$ be a regular curve and $t_0 \in I$.
I want to show that $\sigma(t) := \int_{t_0}^t ||\dot{\gamma}(\tau)||d\tau$ defines an orientation-preserving transformation of the parameter, and that the reparametrization is given by $\tilde{\gamma} = \gamma \circ \sigma^{-1}$, $||\dot{\tilde{\gamma}}|| = 1$.
Thanks in advance.  Working with curves is relatively new to me, so I've been struggling with this so far. 

A:

The moral of this result is that if you have a curve with nonvanishing derivative, then you can reparametrize it to be a unit speed curve, so it is sufficient to think of unit speed curves only when building up the theory.  For clarity I shall be introducing some new notations along the way. 
Let $\gamma:I:=]a,b[\subseteq\mathbb{R}\to\mathbb{R}^n$ be a regular curve (i.e., $\forall t\in I: \frac{d\gamma}{dt}(t)\neq 0$).  Let $t_0\in I$, and suppose we use the parameter $t$ exclusively for elements of $I$. Set $\sigma:I\to\mathbb{R},\sigma(t):=\int_{t_0}^t\left\|\frac{d\gamma}{dt}(\tau)\right\|d\tau$ (this is the arclength of $\gamma$).  $\sigma:I\to J:=\sigma(I)$ is invertible, as it is increasing: $\frac{d\sigma}{dt}=\left\|\frac{d\gamma}{dt}\right\|>0$ by the Fundamental Theorem of Calculus and the fact that $\gamma$ is regular.  It is also easy to see that the smoothness of $\sigma$ is inherited from the smoothness of $\gamma$, viz.,  if $\gamma\in C^n (n\geq 1)$ then $\sigma\in C^n$ as well, as $\frac{d\sigma}{dt}=\left(\left\langle\frac{d\gamma}{dt},\frac{d\gamma}{dt}\right\rangle\right)^{1/2}$. Let us use the parameter $s$ for the elements of $J$, which at this point we know to be an interval also.  The differentiability of $\sigma^{-1}$ follows again from the regularity of $\gamma$ and the chain rule.  Indeed, keeping in mind that $\sigma(t)=s$ and $\sigma^{-1}(s)=t$, we have:
$$ id_I=\sigma^{-1}\circ\sigma \implies 1=\frac{ds}{ds}=\frac{ds}{dt}\frac{dt}{ds}=\frac{d\sigma}{dt}\frac{d\sigma^{-1}}{ds} \implies \frac{d\sigma^{-1}}{ds}=\frac{1}{\frac{d\sigma}{dt}}=\frac{1}{\left\|\frac{d\gamma}{dt}\right\|}\in[0,\infty[.$$
By the same virtue, $\sigma$ is orientation-preserving: Set $\tilde\gamma:=\gamma\circ\sigma^{-1}$. Observe that this is nothing but considering $\gamma(t(s))=\tilde\gamma(s)$ instead of $\gamma(t)$. Then $\frac{d\tilde\gamma}{ds}=\frac{d\gamma}{dt}\frac{d\sigma^{-1}}{ds}$, where the last product of the RHS is positive (if you want to be more formal use you can also argue that the determinant of the Jacobian is positive by the same reasu).  Observe that my first remark says that, as we can use $\sigma$ for regular $\gamma$, wlog we can even consider $\gamma(s)$ instead of $\gamma(t(s))$. Finally we have that
$$\left\|\frac{d\tilde\gamma}{ds}\right\|=\left\|\frac{d\gamma}{dt}\right\|\left|\frac{d\sigma^{-1}}{ds}\right|=1.$$

