Consider a plane with the surface patch σ(u, v) = (1+2u+3v, u-v,-2+u-4v). Verify the Gauss equations for σ.
K=LN−M2EG−F2E=σu⋅σuF=σu⋅σvG=σv⋅σvL=σuu⋅nM=σuv⋅nN=σvv⋅nK = \dfrac{LN-M^2}{EG-F^2} \\ E = \sigma_{u} \cdot \sigma_{u} \\ F = \sigma_{u} \cdot \sigma_{v} \\ G = \sigma_{v} \cdot \sigma_{v} \\ L = \sigma_{uu} \cdot \mathcal{n} \\ M = \sigma_{uv} \cdot \mathcal{n} \\ N = \sigma_{vv} \cdot \mathcal{n}K=EG−F2LN−M2E=σu⋅σuF=σu⋅σvG=σv⋅σvL=σuu⋅nM=σuv⋅nN=σvv⋅n
σu=(2,1,1)σv=(3,−1,−4)σuu=(0,0,0)σuv=(0,0,0)σvv=(0,0,0)\sigma_u = (2,1,1) \\ \sigma_v = (3,-1,-4) \\ \sigma_{uu} = (0,0,0) \\ \sigma_{uv} = (0,0,0) \\ \sigma_{vv} = (0,0,0)σu=(2,1,1)σv=(3,−1,−4)σuu=(0,0,0)σuv=(0,0,0)σvv=(0,0,0)
E=σu⋅σu=2(2)+1(1)+1(1)=6E = \sigma_{u} \cdot \sigma_{u} = 2(2)+1(1)+1(1) = 6E=σu⋅σu=2(2)+1(1)+1(1)=6
F=σu⋅σv=2(3)+1(−1)+1(−4)=1F = \sigma_{u} \cdot \sigma_{v} = 2(3) +1(-1)+1(-4) = 1F=σu⋅σv=2(3)+1(−1)+1(−4)=1
G=σv⋅σv=3(3)+(−1)(−1)+(−4)(−4)=26G = \sigma_{v} \cdot \sigma_{v} = 3(3)+(-1)(-1)+(-4)(-4) = 26G=σv⋅σv=3(3)+(−1)(−1)+(−4)(−4)=26
L=0M=0N=0K=0(0)−06(26)−1=0L = 0 \\ M =0 \\ N = 0 \\ K = \dfrac{0(0)-0}{6(26)-1} = 0L=0M=0N=0K=6(26)−10(0)−0=0
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