Work

I will be working this week-end: proof.

Curvature of a planar curve

I have done this calculation several times over the years, so I might as well write it down in detail, in case it may be of use to someone else.

We are interested in the curvature $C = 1/R$ of a planar curve $y=f(x)$ at a given point A, where $R$ is the curvature radius at that particular point, defined with respect to the curvature center $O$ (intersection of the normals raised to the curve in A and its infinitesimal neighbor B.)

The angle subtending AB is: $\displaystyle \mathrm{d}\alpha = \mathrm{d}s/R \Rightarrow C = \frac{\mathrm{d}\alpha}{\mathrm{d}s}$

The length of the curve element AB is: $\displaystyle \mathrm{d}s = \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2} \Rightarrow \frac{\mathrm{d}s}{\mathrm{d}x } = \sqrt{1+ f'(x)^2}$

The derivative of $f$ is directly related to the angle $\alpha$: $\displaystyle f'(x) = \frac{\mathrm{d}y}{\mathrm{d}x} = \tan \alpha \Rightarrow \alpha = \arctan \frac{\mathrm{d}y}{\mathrm{d}x} = \arctan [f'(x)] \Rightarrow \frac{\mathrm{d}\alpha}{\mathrm{d}x} = \frac{1}{1+f'(x)^2} f''(x)$

Putting together the three relations above yields:

$$C = \frac{\mathrm{d}\alpha}{\mathrm{d}s} = \frac{f''(x)}{\left [ 1 + f'(x)^2\right ]^{3/2}}$$