History: » arclength

### Arc Length

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.