# Calculating the gaussian curvature of a surface

1. Nov 3, 2011

### demonelite123

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 $K_y = \frac {1}{c^2}K_x$ where $K_x$ is the gaussian curvature calculated using x(u, v) and $K_y$ is the gaussian curvature calculated using y(u, v).

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

where S is the shape operator.

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

but this seems contradictory since i got $c^2 S(x_u) \cdot x_u$ 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. Nov 3, 2011

### lavinia

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.