*will*be working this week-end: proof.

## 21 October 2016

## 10 October 2016

### 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:

\[C = \frac{\mathrm{d}\alpha}{\mathrm{d}s} = \frac{f''(x)}{\left [ 1 + f'(x)^2\right ]^{3/2}}\]
Labels:
derivation,
equation,
geometry

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