Proof for Cauchy's Definition of Centre of Curvature.

Question

Cauchy defined the center of curvature as the intersection point of normals drawn to two infinitely close points on a curve. Is there any way to prove this? I am unable to get a starting point for this.

Answer 1

I'll assume you are dealing with biregular plane curve $\gamma=(x,y)\colon(-\epsilon,\epsilon)\to\mathbb{R}^2$ with unit speed. The normal line at $\gamma(0)=(x_0,y_0)$ is (self-explanary notation) $$ (\dot x_0,\dot y_0)\cdot(x,y)=x_0\dot x_0+y_0\dot y_0 $$ and normal line at $\gamma(h)$ is $$ (\dot x_h,\dot y_h)\cdot(x,y)=x_h\dot x_h+y_h\dot y_h $$ so their intersection is $$ (x,y)=(x_0,y_0)+(-\dot y_0,\dot x_0)\lambda $$ where $$ \lambda=\frac{(x_h-x_0)\dot x_h+(y_h-y_0)\dot y_h}{(-\dot y_0,\dot x_0)\cdot(\dot x_h,\dot y_h)}. $$ But the numerator is $h(\dot x_0^2+\dot y_0^2)+o(h)=h+o(h)$ and the denominator is $h(\dot x_0\ddot y_0-\ddot x_0\dot y_0)+o(h)$, so $\lambda$ agrees with the (signed) radius of curvature in the limit and you recover the formula for the center of curvature.