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Tangent planes and normal vectors

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Find all points on the surface at which the tangent plane is horizontal

    z=x3y2

    Things I know:
    Tangent plane is horizontal then therefore the normal must be vertical in order to be perpendicular.

    Dot product of the tangent plane with normal is = 0

    Normal plane is given by the partial with respect to x,y,z evaluated at some points P(x0,y0,z0)

    Putting that together I get
    f(x,y,z)=x3y2-z
    fx=3x2y2
    fy=2x3y
    fz=-1

    Here is where I am completely stuck I know that I still don't have a normal vector because I do not have any points to plug in to the partials.



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 31, 2012 #2

    LCKurtz

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    Your original equation is of the form ##z = f(x,y)##. The tangent plane will be horizontal at any point ##(x,y,z)## where ##f_x(x,y)=0## and ##f_y(x,y)=0##.
     
  4. Oct 31, 2012 #3

    HallsofIvy

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    Or, same thing, The vector you got, [itex]<3x^2y^2, 2x^3y, -1>[/itex] must be of the form <0, 0, a> to be vertical. Obviously we can take a= -1 but we need [itex]3x^2y^2= 0[/itex] and [itex]2x^3y= 0[/tex]. That will be true if x= 0 or if y= 0. In other words above the x and y axes.
     
  5. Oct 31, 2012 #4
    hmm, so I can equal them together, and get

    x=(3/2)y and y=2/3x

    Don't really understand what the values represent?
     
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