How to parametrize a curve by its arc length I am reading on Wikipedia that 
''...Any regular curve may be parametrized by the arc length (the natural parametrization) and...''
I know that if $a(t) = (x(t),y(t),z(t))$ is a curve (say, smooth) then it is regular iff for all $t$: $a' (t) \neq 0$. I also know the definition of arc length:
The arc length of a curve $a$ between $t_0$ and $t$ is defined as
$$ l = \int_{t_0}^t |a'(t)|dt$$
But what is the parametrization of $a$ using its arc lenght? 
 A: "Parameterization by arclength" means that the parameter $t$ used in the parametric equations represents arclength along the curve, measured from some base point. One simple example is 
$$
x(t) = \cos(t) \quad ; \quad y(t) = \sin(t) \quad (0 \le t \le 2\pi)
$$
This a parameterization of the unit circle, and the arclength from the start of the curve to the point $(x(t), y(t))$ is $t$.
In most cases, it's not possible to find simple formulas that give arclength parameterizations, so the whole approach is somewhat academic, in my view.
A: We may have a good notion of arc length even for some nonregular (or not even differentiable) curves; all we need (by definition) is that the curve $a\colon[0,b]\to \mathbb R^n$ is rectifiable, that is: 
$$ l(a):=\sup\{\,d(a(0),a(t_1))+d(a(t_1),a(t_2))+\ldots+d(a(t_{n}),a(b))\mid 0<t_1<\ldots<t_n<b\,\}$$
exists (is finite). For the case you mention, this supremum definition coincides with the integral definition, of course.
This $l$ gives us a map $\ell\colon[0,b]\to[0,\infty)$ given by $\ell(t)=l(a|_{[0,t]})$ (because automatically all these restrictions are also rectifiable). Then $\ell(0)=0$, $\ell(b)=l(a)$, $\ell$ is continuous and strictly increasing. Thus $\ell^{-1}\colon[0,l(a)]\to[0,b]$ exists and allows us to reparametrize our curve as $\hat a=a\circ\ell^{-1}\colon[0,l(a)]\to\mathbb R^n$. This is reparametrization by arc length. With this the arc length from $\hat a(t_1)$ to $\hat a(t_2)$ is always $t_2-t_1$ for $0\le t_1\le t_2\le l(a)$.
A: Simply if $$\alpha: I\to \mathbb{R}^3$$ is a regular curve and the arc length is
$$s(t)=\int_{t_0}^t |\alpha'|dt$$
Now we solve for $t$ as $t=t(s)$ to get the function $$t:J\to I$$So we define a new curve $\beta(s)=\alpha\circ t=\alpha(t(s))$ where
$$\beta: J\to I \to \mathbb{R}^3 \\ |\beta'(s)|=|\frac{d\beta}{ds}|=|\frac{d\beta}{dt}. \frac{dt}{ds}|=|\frac{d\alpha(t(s))}{dt}. \frac{1}{\alpha'(t)}|=1$$
