Bending Energy Continued - Finding K

In the previous post I showed how to find F'(s) of a a curve F(t) = <X(t),Y(t),Z(t)>

Now according to the definition of bending energy , it is the integrated sum of squares of curvature over length of the curve. That is we need to find the curvature K(s). It is always like this , Just getting better and better! .

Curvature is the rate of change magnitude of Unit tangent vector with respect to curve length
That is , it is

K(s)  = || dF'(t) / ds ||

Lets find dF'(t) / ds

This is (dF'(t)/ dt) * (dt/ ds)   using chain rule of diff.

=   || dF'(t) / dt ||
    -----------
      || ds/ dt      ||       we know ds/dt = || F'(t) ||  from previous post. and dF'(t)/dt is F''(t) .

K(t)  = || F''(t) || /  || F'(t) ||

we calculated the value of K(t).  For curves like circle , K(t) is same everywhere ,so just we can remove parameter t for circle. Also for circle there is exists an easier fomula : K = 1 / radius.