The guassian curvature of a surface is . Basically

i am doing a question where i found that , so i got .

i was wondering if anyone knows what this means for guassian curvature or the surface as it is divided by zero.

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- Mar 10th 2009, 01:18 PMdopiguassian curvature over zero
The guassian curvature of a surface is . Basically

i am doing a question where i found that , so i got .

i was wondering if anyone knows what this means for guassian curvature or the surface as it is divided by zero. - Mar 11th 2009, 07:24 AMLaurent
I would say this means you made a mistake, or that your surface is not smooth... If the surface is given locally by , then applying Cauchy-Schwarz inequality to gives , and the equality implies that the vectors are colinear. However, this is supposed not to happen for a parameterization of a surface (the differential must be injective: the partial derivatives span the tangent plane).

- Mar 11th 2009, 05:15 PMdopi
**here is my original quesiton**

alpha(u) is a unit speed space curve, and use this curve to construct a tangent developable surface with where

where

so i want to work out the guassian curvature of this surface.

**my solution just for E, F , G**

and

so

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 - Mar 12th 2009, 02:33 AMLaurent
- Mar 12th 2009, 08:28 PMdopi
thanks for that it makes more sense now, could u check my answer for L,N,M and M^2

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 - Mar 13th 2009, 09:45 AMLaurent
Hi,

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?