Question

) Consider point p = (0, 0, 0) of paraboloid z = x 2 + ky2 , k > 0. (a) Show that the unit vectors of x-axis and y-axis are eigenvectors of −dNp, with eigenvalues 2 and 2k, respectively. Here we choose the orientation of the surface, N, to be the one pointing inwards from the region bounded by the paraboloid. (b) For p = (0, 0, 0), find Gaussian curvature K and mean curvature H at this point.

Answer 1

Answer:

Answer: K = 4k and H=(2+2k) = k+1

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Step-by-step explanation:

Part(a) :Let's parametrize the given surface-

X (u, v) = ( u²,v²,u²+kv² )

Now : Xu = (1,0,2u) and Xv = (0,1,2kv)

using below formula:

N (u, v) = Xu x X v

               [ Xu x Xv]

N(u,v) = (-2u,-2kv , 1)

          √ {4u²+4k²v²+1}

Now value of normal vector at p=(0,0,0) is N(p)

N(p) = N(0,0 )= (0,0,1)

so finally at p=(0,0,0) : Xu = (1,0,0) , Xv = (0,1,0), Np = (0,0,1)

Clearly Tp(S) is spanned by (1,0,0),(0,1,0)

 dN.Xu = d/dt N(t,0)/ t=0 = (-2,0,0)

 dN.Xv = d/dr N(0,r)/r=0=(0,-2k,0)

since Xu ,Xv ∈Tp(S)

aXu + bXv ∈Tp(S)

-dNp (aXu + bXv) =(2a,2kb,0)

and represent above in matrix form:

   2    0              a

   0    2k           b

-dNp  is symmetric matrix with eigen values (LAMDA) ∧1=2 and ∧2 =2k

Part (b):

 

Product of Eigen Values is called Gaussian Curvature(K).

Mean of Eigen Values is called Mean Curvature(H).

Therefore K = 4k and H=(2+2k) = k+1

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