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}.

As we stated previously, this is not a practical definition, since parameterizing by arc length is typically impossible. Instead we use the chain rule to get

\displaystyle\left\| {\frac{d}{{ds}}(T(t))} \right\| = \left\| {T'(t)\frac{{dt}}{{ds}}} \right\| = \frac{{\left\| {T'(t)} \right\|}}{{\left\| {\frac{{ds}}{{dt}}} \right\|}} = \frac{{\left\| {T'(t)} \right\|}}{{\left\| {r'(t)} \right\|}}.

This formula is more practical to use, but still cumbersome. T'(t) is typically a mess. Instead we can borrow from the formula for the normal vector

\displaystyle T(t) = \frac{r'(t)}{\|r'(t)\|}

to get the curvature

\displaystyle k(t) = \frac{{\left\| {r'(t) \times r''(t)} \right\|}}{{{{\left\| {r'(t)} \right\|}^3}}}.

Curvature of a plane curve. If a curve in the xy-plane is defined by the function y = f(t) then there is an easier formula for the curvature. We can parameterize the curve by

r(t) = t \vec i + f(t) \vec j

We have

r'(t) = \vec i + f '(t)\vec j

and

r''(t) = f ''(t) \vec j .

Their cross product is just

r'(t) \times r''(t) = f ''(t) \vec k

which has magnitude

\|r'(t) \times r''(t)\| = |f ''(t)|

Thus we have

Definition. Curvature is given by

\displaystyle k(t) = \frac{{|f''(t)|}}{{{{\left( {1 + {{\left( {f'(t)} \right)}^2}} \right)}^{\frac{3}{2}}}}}.

Example. The curve defined by y=\sin x has

\displaystyle k(t) = \frac{{| - \sin t|}}{{{{\left( {1 + \cos {t^2}} \right)}^{\frac{3}{2}}}}}

as its curvature.

See also: http://www.ltcconline.net/greenl/courses/202/vectorFunctions/curvat.htm

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: