Radius of Curvature of a Function

If you have a continuous function y(x), what is the local radius of curvature at a particular point? It is given by

R = [1+y'(x)²]^3/2 /y"

where y'(x) = dy/dx at a given point x, and y"(x) = dy'/dx at that same point

Derivation:

If R is the radius of curvature, and you rotate an angle du about the center of rotation, the distance moved ds is given by:
R du = ds
so R = ds/du

We will come up with expressions for ds and du:

Being the hypoteneuse of a right triangle,
ds = sqrt( dx² + dy² )

Meanwhile, Calculus tells us that y'(x) is the slope of the curve at x, and this is just the rise over the run, which is the same as the definition of tangent, so
tan u = y'(x)
rearranging: u = atan( y' )

Taking the derivative: du = d( atan y' ) = [1/(1+y'²)] d(y')
now d(y') = y"dx, so:
du=(y"dx)/(1+y'²)

Plugging the above expressions for ds and du into the expression for R:
R = sqrt( dx² + dy² )/ [(y"dx)/(1+y'²)] = sqrt(1+dy/dx)(1+y'²)/y"

so R = (1+y'²)^3/2/y" QED

If the slope y' is small

R = 1/y"

Polar Coordinates

r = f(u) u=angle

R = (r²+r'²)^3/2 / (r² - rr' + 2r'²)