guassian curvature over zero

• March 10th 2009, 12:18 PM
dopi
guassian curvature over zero
The guassian curvature of a surface is $(LM - N^2) / (EG - F^2)$. Basically

i am doing a question where i found that $(EG - F^2)=0$ , so i got $(LN - M^2)/ 0$.

i was wondering if anyone knows what this means for guassian curvature or the surface as it is divided by zero.
• March 11th 2009, 06:24 AM
Laurent
Quote:

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

I would say this means you made a mistake, or that your surface is not smooth... If the surface is given locally by $(u,v)\mapsto X(u,v)$, then applying Cauchy-Schwarz inequality to $\partial_u X\cdot \partial_v X$ gives $EG\geq F^2$, and the equality implies that the vectors $\partial_u X,\ \partial_v X$ 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).
• March 11th 2009, 04:15 PM
dopi
Quote:

Originally Posted by Laurent
I would say this means you made a mistake, or that your surface is not smooth... If the surface is given locally by $(u,v)\mapsto X(u,v)$, then applying Cauchy-Schwarz inequality to $\partial_u X\cdot \partial_v X$ gives $EG\geq F^2$, and the equality implies that the vectors $\partial_u X,\ \partial_v X$ 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).

here is my original quesiton

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

$x(u,v) = alpha(u) + v*(alpha'(u)), (u,v) in U$

where $U = {(u,v) in R^2 : -infinity < u < infinity, v>0}$

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

my solution just for E, F , G

$E = Xu . Xu = 1 + (v^2)*(alpha''(u)^2) +2*v*t*alpha''(u)$

$F = Xu . Xv = ! + v*t*alpha''(u)$

$G = t.t =1$
and
$F^2 = 1 + (v^2)*(alpha''(u)^2) +2*v*t*alpha''(u)$

so $EG - F^2 = 0$

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
• March 12th 2009, 01:33 AM
Laurent
Quote:

Originally Posted by dopi
here is my original quesiton

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

$x(u,v) = alpha(u) + v*(alpha'(u)), (u,v) in U$

where $U = {(u,v) in R^2 : -infinity < u < infinity, v>0}$

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

my solution just for E, F , G

$E = Xu . Xu = 1 + (v^2)*(alpha''(u)^2) +2*v*t*alpha''(u)$

$F = Xu . Xv = ! + v*t*alpha''(u)$

$G = t.t =1$
and
$F^2 = 1 + (v^2)*(alpha''(u)^2) +2*v*t*alpha''(u)$

so $EG - F^2 = 0$

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

There are mistakes in your final computation. Here are the details:
You have $X_u=T+v\kappa N$ and $X_v=T$. Since $(T,N,B)$ is orthonormal, $\|X_u\|^2=1+(v\kappa)^2$, $\|X_v\|^2=1$ and $X_u\cdot X_v=1$.
Then $EG-F^2=(1+\kappa^2v^2)-1=\kappa^2v^2$.

Note: you can notice that $EG-F^2=\|X_u\times X_v\|^2$. This is always true, so that your previous question already gave you the result...
• March 12th 2009, 07:28 PM
dopi
Quote:

Originally Posted by Laurent
There are mistakes in your final computation. Here are the details:
You have $X_u=T+v\kappa N$ and $X_v=T$. Since $(T,N,B)$ is orthonormal, $\|X_u\|^2=1+(v\kappa)^2$, $\|X_v\|^2=1$ and $X_u\cdot X_v=1$.
Then $EG-F^2=(1+\kappa^2v^2)-1=\kappa^2v^2$.

Note: you can notice that $EG-F^2=\|X_u\times X_v\|^2$. This is always true, so that your previous question already gave you the result...

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
• March 13th 2009, 08:45 AM
Laurent
Quote:

Originally Posted by dopi
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

Hi,
I don't understand how you got $L=0$. You didn't give your formula for $n$, and that's probably where there's a problem.
Using Serret-Frenet coordinates (like in my previous post), we see that the tangent vectors $\partial_u X$ and $\partial_v X$ are linear combinations of $T$ and $N$. Therefore, $n=\pm B$ (the sign depends on the sign of the curvature, I let you find out).

After computations, I find $L=v\kappa\tau$ (where $\tau$ is the torsion), and $N=M=0$. Can you find the same?