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:

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

Electrical circuits and Euler's polyhedron formula

When solving an electrical circuit consisting of impedances and voltage sources, one needs to apply Kirchhoff's laws:

• The current law, yielding N-1 equations (with N the number of nodes)
•  The voltage law, for an additional L equations (where L is the number of elementary loops)

The unknowns are the currents flowing in each branch (B of them). We also assume that the branches do not cross. Note that writing the current law for the N-th node or the voltage law for a composite loop (consisting of several adjacent elementary loops) does not provide any further information, the resulting equations being linear combinations of the previous ones.

For the problem to be well-posed the number of unknowns and equations is equal, which we can write as:
\[ N + L - B = 1\] This relation is easily proven in plane geometry, but here I would like to show its intimate connection with Euler's formula, which states that, for a convex polyhedron, \[ V + F - E = 2\] where V, F and E are the numbers of vertices, faces and edges, respectively.

Let us start by establishing a correspondence between circuits and polyhedra, as shown in the figure below. Place a sphere on top of the (planar) circuit diagram and connect each node to the North pole by a line segment (this is known as a stereographic projection.) We define the vertices as the intersections of these segments with the sphere; the result is a convex polyhedron.

It is easily seen that, with the notations above, we have the straightforward equivalences V = N and E = B. The number of faces, however, F = L + 1, since the "topmost" face corresponds to the open area surrounding the circuit. Substitution in either of the equations above yields the other one.