Question

Consider a plane with the surface patch σ(u, v) = (1+2u+3v, u−v, −2+u−4v). Verify

the Gauss equations for σ.​

Answer 1

SOLUTION

GIVEN

Consider a plane with the surface patch

σ(u, v) = (1+2u+3v, u−v, −2+u−4v)

TO VERIFY

Gauss equations for σ.

EVALUATION

Here it is given that

A plane with the surface patch

σ(u, v) = (1+2u+3v, u−v, −2+u−4v)

Now

$\sf{ \sigma_{u} = (2,1,1)}$

$\sf{ \sigma_{v} = (3, - 1, - 4)}$

$\sf{ \sigma_{uu} =(0,0,0) }$

$\sf{ \sigma_{vv}=(0,0,0)}$

$\sf{ \sigma_{uv}=(0,0,0)}$

Now we find the value of E, F, G, L, M, N

$\sf{ E = \sigma_{u}.\sigma_{u} = 4 + 1 + 1 = 6}$

$\sf{F = \sigma_{u}.\sigma_{v} = 6 - 1 - 4 = 1 }$

$\sf{ G= \sigma_{v}.\sigma_{v} = 9 + 1 + 16 = 26}$

$\sf{ L = \sigma_{uu}.n = 0}$

$\sf{ M = \sigma_{uv}.n = 0}$

$\sf{ N= \sigma_{vv}.n = 0}$

Now find the value of K

$\displaystyle \sf{K = \frac{LM - {N}^{2}}{EG - {F}^{2} } }$

$\displaystyle \sf{ \implies \: K = \frac{0.0- {0}^{2}}{6.26 - {1}^{2} } }$

$\displaystyle \sf{ \implies \: K = \frac{0}{155} }$

$\displaystyle \sf{ \implies \: K = 0 }$

Hence Gauss equations for σ is verified

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