Consider the infinitesimal (3-d) version of the Pythagorean Theorem \begin{equation} ds^2 = dx^2 + dy^2 + dz^2 \end{equation} which implies that \begin{equation} \left({ds\over du}\right)^2 = \left({dx\over du}\right)^2 + \left({dy\over du}\right)^2 + \left({dz\over du}\right)^2 \end{equation} (just “divide” by $du^2$…) Thus, we see that the speed $v$ satisfies \begin{equation} v = {ds\over du} \end{equation} Since speed equals distance divided by time, the arc length $s$ between the points $\rr(a)$ and $\rr(b)$ is given by “adding up” the infinitesimal distances $ds$, that is, \begin{equation} s = \Int_a^b v\, du = \Int_a^b {\textstyle\sqrt{ \left( {dx\over du} \right)^2 + \left( {dy\over du} \right)^2 + \left( {dz\over du} \right)^2 }} \> du \end{equation} It is important to realize that this constuction is independent of the parameterization used, and depends only on the curve itself.

In principle one can always use this integral to reparameterize the curve in terms of arc length, i.e. to replace $u$ by $s$. However, it is often quite difficult in practice to actually evaluate the integral to obtain a formula for $s$.

The unit tangent vector to the curve is given by \begin{equation} \TT = {\vv\over v} \end{equation} which can now be written as \begin{equation} \TT = {d\rr/du\over ds/du} = {d\rr\over ds} \end{equation} For this reason, curves parameterized by arc length are sometimes called unit-speed curves.