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Differential geometry

  1. Jul 2, 2014 #1
    Consider the surface S defined as the graph of a function z = 2x ^ 2 - y ^ 2
    i) find a basis of the tangent plane Tp surface S at the point p = (-1,2, -2)
    ii) find a non-zero vector w in Tp with the property that the vertical curvature at point p in the direction of vector w is zero


    for the i) question i found that the tangent plane has this form: Tp={(-1,2,-2)+t(1,0-4)+s(0,1,-4)}
    but how i will find a basis of this tangent plane? maybe is <(1,0,4),(0,1,-4)> ?

    and for the ii) how i will find this non zero vector?
     
  2. jcsd
  3. Jul 2, 2014 #2

    HallsofIvy

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    What is the "vertical curvature" at a point?
     
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