Radius of Curvature is an expression of how curved a curve is. In Intrinsic coordinates it is the rate of change of s with respect to ψ, that is ds/dψ. Geometrically, it represents the radius of an "osculating" (lit. kissing, apparently) circle- that is, the circle which is the best fit for the line at a given point. Hence the radius of curvature for a circle is simply its radius.

Otherwise, radius of curvature (or its reciprocal, "curvature", k = dψ/ds) is most easily found when the function is expressed in intrinsic coordinates. However, as conversion from other coordinate systems to intrinsic is often complicated, radius of curvature can be expressed for cartesian form directly.

For a cartesian curve y=f(x), at a point P(x,y) the gradient is dy/dx and is also the gradient of the tangent at P. From the definition of the intrinsic coordinates (s,ψ) of P, this gradient is equal to tan ψ. It should also be noted that dx/ds = cos ψ (from the expression for arc length).

As dy/dx = tan ψ, d2y/dx2 = d/dx (tan ψ) = sec2ψ . dψ/dx (implicit differentiation).

Rearrangement gives dx/dψ = sec2ψ / (d2y/dx2)

As dx/ds = cos ψ, ds/dx = 1/cos ψ

p, the radius of curvature, is found by ds/dψ, which by the chain rule equals ds/dx . dx/dψ

Therefore p= 1/cos ψ . sec2ψ / (d2y/dx2)
= sec3ψ / (d2y/dx2)

As 1 + tan2ψ = sec2ψ, sec ψ = {1+ tan2ψ}1/2

So sec3ψ = {1+ tan2ψ}3/2
= {1+ (dy/dx)2}3/2

So radius of curvature, p, for y=f(x) =

 
{1+ (dy/dx)2}3/2
----------------
(d2y/dx2)

Similarly, for a curve defined by parametric coordinates, x=f(t) and y=g(t) then p=

{ (dx/dt)2 + (dy/dt)2 }3/2
-------------------------------
(dx/dt)(d2y/dt2) - (dy/dt)(d2x/dt2)


Note, however, that once you move from the cartesian plane to higher dimensions, it becomes necessary to use vector calculus. In that case, it is easier to find the curvature and then take the reciprocal of that to get radius of curvature.

To achieve this, we find the unit tangent T to the curve C defined by vector function r(t) at a given point using T = r'(t)/||r'(t)|| (that is, find the tangent line as usual then normalise it). Then, we get curvature Κ by ||dT/ds||. By the chain rule this is ||dT/dt . dt/ds|| which by definition of arclength s is:

||dT/dt||          ||dT/dt||
----------      = ------------
||ds/dt||          ||r'(t)||
So our radius of curvature = 1/Κ is given by
||r'(t)||
---------
|| dT/dt||

Thanks to krimson for some insights into this last part. A year later, I've done enough (second semester of degree) maths to actually understand it!

Log in or registerto write something here or to contact authors.