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Math Help - guassian curvature over zero

  1. #1
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    Unhappy 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.
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  2. #2
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    Quote Originally Posted by dopi View Post
    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).
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  3. #3
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    Quote Originally Posted by Laurent View Post
    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
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  4. #4
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    Quote Originally Posted by dopi View Post
    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...
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  5. #5
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    Question

    Quote Originally Posted by Laurent View Post
    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
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  6. #6
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    Quote Originally Posted by dopi View Post
    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?
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