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Equation of the Tangent Plane

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Find equations of the following.

    x2-2y2+z2+yz=7, (5,3,-3)
    (a) the tangent plane
    (b) the normal line to the given surface at the point

    2. Relevant equations

    I know it involves fx, fy, fz

    3. The attempt at a solution
    I got 10x-15y-3z=7. Is this correct? Because its not true at the point (5,3,-3).

    I got it by f(5,3,-3)+fx(5,3,-3)(x-5)+fy(5,3,-3)(y-3)+fz(5,3,-3)(z+3)

    As for the normal line I know the answers are
    x=10t+5
    y=-15+3
    z=-3-3
     
    Last edited: Oct 6, 2009
  2. jcsd
  3. Oct 6, 2009 #2

    Mark44

    Staff: Mentor

    No. The point (5, 3 -3) has to satisfy both the equation of the plane and the equation of the surface.
    For one thing, this is not an equation, so there's no way to get an equation out of it. For another thing, if my memory is correct, the equation of the tangent plane is fx(5, 3, -3)(x - 5) + fy(5, 3, -3)(y - 3) + fz(5, 3, -3)(z - (-3)) = 0.

    You didn't show the partial derivatives that you calculated, so it might also be that you have an error in one or more of them.

     
  4. Oct 6, 2009 #3
    Ive got it now. I used the linear approximation and set it = to 0. I just had to take out the f(5,3,-3). I figured it was something stupid like that. Thanks for the help though.
     
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