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Finding normal vector to a surface

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Given a parameterized surface:

    C(u,v) = (3 cos u sin v, 2 sin u sin v, cos v) 0<u<2pi, 0<v<pi

    I have to find a normal vector to that surface.

    2. Relevant equations



    3. The attempt at a solution

    So tangent vectors can be Tu = (dx/du, dy/du, dz/du) and Tv = (dx/dv, dy/dv, dz/dv)

    And I can take the cross of those to find a normal vector. But what points of u and v do i use? The cross product gave me:

    (-2sin^2 v cos u, 3sin^2 v sin u, 6sin^2 u sin^2 v - 6 cos^2 u sin^2 v)
     
  2. jcsd
  3. Mar 9, 2012 #2

    Simon Bridge

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    What difference does it make? (And why?)
     
  4. Mar 9, 2012 #3
    It shouldn't make a difference. Do i just plug in the endpoints of u and v? ie would (0,0) work? I get (0,0,0) if I use that point. Not a vector...
     
  5. Mar 9, 2012 #4

    Simon Bridge

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    Why shouldn't it make a difference?
    What happens if the surface is curved?
     
  6. Mar 9, 2012 #5
    Right. When the surface is curved the cross product of the tangents shouldn't be 0.
     
  7. Mar 9, 2012 #6

    Simon Bridge

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    You are saying that the cross product of the tangents to a plane (flat) surface are zero?
    Then how would you find the normal vector to a plane surface?
     
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