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}.


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