Relation between Curvature and Radius of Curvature

Question

Why is the radius of curvature the reciprocal of the curvature? How to see this intuitively as well show it rigorously?

Answer 1

For intuition, circles have constant curvature. The bigger the circle, the more any continuous length along the circle resembles a straight line which has having zero curvature, so curvature varies inversely with radius. This can be proven rigorously.

The radius of curvature is the radius of the osculating circle, the radius of a circle having the same curvature as a given curve and a point. So the inverse relationship of a circle's curvature to its radius transfers to arbitrary curves.

By definition curvature is $\kappa=d\theta/ds$. Where $ds$ is the distance along a curve, and the $\theta$ is a measure of how much clockwise or counter-clockwise the direction of the curve changes. Imagine a right triangle where the vertical leg is $dy$ and the horizontal leg is $dx$. Then $dy/dx$ is the tangent of the angle adjacent to the hypotenuse.

So $\kappa=\frac{d(atan(dy/dx))}{\sqrt{dx^2+dy^2}}=\frac{\frac{daTan(dy/dx)}{dx}}{\frac{d\sqrt{1+y'^2}}{dx}}$

Some more intuition: Suppose you're on a bicycle driving straight ahead. Turn the handlebars so the front wheel makes a large angle deviating only slightly from straight, you now move in a circle. The smaller the angle deviates from straight ahead, the bigger the circle. Holding a constant angle as you move along yields a constant radius of curvature.