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Calculating the gaussian curvature of a surface

  1. Nov 3, 2011 #1
    Let x(u,v) be a coordinate patch. Define a new patch by y(u, v) = c x (u, v) where c is a constant. Show that [itex] K_y = \frac {1}{c^2}K_x [/itex] where [itex] K_x [/itex] is the gaussian curvature calculated using x(u, v) and [itex] K_y [/itex] is the gaussian curvature calculated using y(u, v).

    my book expects us to use the formulas [itex] K = \frac {ln-m^2}{EG-F^2} [/itex] where [itex] E = x_u \cdot x_u, F = x_u \cdot x_v, G = x_v \cdot x_v, l = S(x_v) \dot x_v, m = S(x_u)\cdot x_v, n = S(x_v)\cdot x_v [/itex].

    where S is the shape operator.

    i can see that [itex] y_u = c x_u [/itex] and [itex] y_{uu} = c x_{uu} [/itex]. so i tried calculating [itex] l = S(y_u) \cdot y_u = S(cx_u) \cdot x_u = c^2 S(x_u) \cdot x_u [/itex]. but a result from my book states that [itex] S(x_u) \cdot x_u = U \cdot x_{uu} [/itex] where U is the unit vector created by taking the cross product [itex] x_u \times x_v [/itex] and dividing by its length. when calculated this way, i get that [itex] U \cdot y_{uu} = c(U \cdot x_{uu}) = c(S(x_u) \cdot x_u)) [/itex]

    but this seems contradictory since i got [itex] c^2 S(x_u) \cdot x_u [/itex] earlier and now i only get one factor of c even though the 2 are supposed to be equal. i am confused on what is going on here. help will be greatly appreciated.
     
  2. jcsd
  3. Nov 3, 2011 #2

    lavinia

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    I think that the derivative of the unit normal with respect to the parameters is the same for both the patch,X and the patch, cX. So the factor is c not c^2. Does this seems right? I think it is clear from the Chain Rule. Intuitively the same deviation in the unit normal along a parameter curve in the surface cX(u,v) occurs along a curve that is c longer than the the corresponding curve in the surface, x(u,v).

    Try this out for a sphere centered at the origin of 3 space.
     
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