Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In my handout it is said that a circle $\Gamma(s)$ that is tangent to second order to a curve $\xi:[a,b]\to \mathbb R^2$ with unit speed and with curvature $\kappa$, then the radius of $\Gamma(s)$ is $1\over |\kappa (s)|$.

I don't quite understand what "tangent to second order" means. Is there a simple/intuitive way to see this?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

"Tangent to second order" means that, at the point of intersection $p$, the circle and the curve not only agree in velocity, but also in acceleration. If you take a Taylor expansion of both $\Gamma$ and $\xi$ at $p$, the first two terms will be the same.

Here's a little bit of intuition about curvature, in case you find it useful. I often think of a curve as a particle following a track in space. Intuitively, you want to think of "curvature" as how hard the particle has to turn to stay on its track, just like you want to think of "velocity" as how fast the particle is going. You measure velocity by noting: "If the track disappeared in an instant, the particle would fly off in a straight line. What is the speed and direction of that line?"

A straight line is a constant-velocity curve. What's a constant-turning curve? A circle. So to measure curvature, you ask, "If the track disappeared in an instant and the particle kept turning at the same rate it's turning now, how big a circle would it follow?" If it's a small circle, the particle has to turn very hard, so, high curvature. If it's a large circle, the particle doesn't have to turn very hard, so, low curvature. If it's a straight line, just think of it as a circle of infinite radius, with curvature "$1/\infty$"$=0$.

share|improve this answer
    
Thank you, Neal, your answer is perfect! –  Bees Mar 13 '12 at 13:58
    
You're welcome! –  Neal Mar 13 '12 at 14:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.