Gaussian Curvature of the Unit Sphere
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Compute the curvature of the unit sphere in R^3.
I started the problem with r = [sin u cos v
sin u sin v
cos v. ]
From there, how do you get the principal curvatures k_1 and k_2?
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Solution Summary
We use the first and second fundamental forms to show that the Gaussian curvature of the unit sphere is equal to 1; this solution includes a detailed step-by-step process for solving this problem as well as references to consult for further questions.
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The Gaussian curvature of a surface is given by
K = (LN - M^2)/(EG - F^2),
where
E = r_u . r_u
F = r_u . r_v
G = r_v . r_v
and
L = n . r_uu
M = n . r_uv
N = n . r_uv
where n = r_u X r_v / |r_u X r_v| is the unit normal vector to the surface. We have
( cos u cos v )
r_u = ( cos u sin v )
( -sin u )
( -sin u sin v )
r_v = ( sin u cos v )
( 0 )
whence
E = r_u . r_u = cos^2 u (cos^2 v + sin^2 v) + sin^2 u = cos^2 u + sin^2 u = 1,
F = r_u . ...
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