# Ngô Quốc Anh

## July 30, 2010

### Curvature of a curve

Filed under: Riemannian geometry — Ngô Quốc Anh @ 16:16

Consider a car driving along a curvy road. The tighter the curve, the more difficult the driving is. In math we have a number, the curvature, that describes this “tightness”. If the curvature is zero then the curve looks like a line near this point. While if the curvature is a large number, then the curve has a sharp bend.

Let assume $\gamma : I \to \mathbb R^3$ be a curve. We also adopt the following notation

$r(t)=x(t)\vec i+y(t)\vec j+z(t)\vec k, \quad t \in I$.

More formally, if $T(t)$ is the unit tangent vector function then the curvature is defined at the rate at which the unit tangent vector changes with respect to arc length $s$.

Definition. Curvature is given by

$\displaystyle k(t) = \left\|\frac{d}{ds} (T(t)) \right\| = \|r''(s)\|$

Example. Let consider an example of a circle of radius $r$ given by

$\alpha(t) = (r \sin(t), r \cos(t))$.

We then see that

$\alpha '(t) = (r \cos t, -r \sin t)$

which implies $|\alpha'(t)| = r$. Thus, $s=rt$. Our circle is now parameterized by arc length as follows

$\displaystyle\alpha(s) = \left(r \sin\left(\frac{s}{r}\right), r \cos \left(\frac{s}{r}\right) \right)$.

The curvature vector at a given length $s$ is then

$\displaystyle \alpha ''(s) = \left( { - \frac{1}{r}\sin \left( {\frac{s}{r}} \right), - \frac{1}{r}\cos \left( {\frac{s}{r}} \right)} \right)$

and that

$\displaystyle \|\alpha ''(s)\| = \frac{1}{r}$.