alpha(u) is a unit speed space curve, and use this curve to construct a tangent developable surface with where
so i want to work out the guassian curvature of this surface.
my solution just for E, F , G
where i used t = alpha'(u) from frenet serret equation, as txt =0 and t.t =1
im not what i done is right but thats how i got EG - f^2 for the guassian curvature as zero
L= n.Xuu = n.(alpha''(u) + v*alpha'''(u))
M= n.Xuv = n.alpha''(u)
N= n.Xvv = n.0 = 0
so therefore LN - M^2= -M^2 , as L=0
therefore -M^2 = -(n.alpha''(u))^2
so im assuming n.n =1 and therefore -M^2 = -alpha''(u)^2
thanks for the response
I don't understand how you got . You didn't give your formula for , and that's probably where there's a problem.
Using Serret-Frenet coordinates (like in my previous post), we see that the tangent vectors and are linear combinations of and . Therefore, (the sign depends on the sign of the curvature, I let you find out).
After computations, I find (where is the torsion), and . Can you find the same?