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Finding the equation of the tangent plane

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the equations of the tangent plane and the normal line to the given surface at the specified point.

    xy+yz+zx=3 @ (1,1,1)


    2. Relevant equations

    z-z0=fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0)


    3. The attempt at a solution
    I feel like this should be really easy, but I'm not sure how to get started. the equation from the book for the tangent plane seems to imply I should isolate z from the equation and go from there, but that seems like it would make things messy and more complicated than necessary.
     
  2. jcsd
  3. Dec 12, 2008 #2
    ok, based on the solution to another problem, I've worked out:

    F(x,y,z) = xy+yz+zx = 0
    Fx = y+z = 2
    Fy = x+z = 2
    Fz = y+x = 2

    and the equation of the plane as

    2(x-1) + 2(y-1) + 2(z-1) = 0
    or
    x+y+z = 3

    How does that look for the equation of the tangent plane?
     
  4. Dec 13, 2008 #3

    Mark44

    Staff: Mentor

    A correction to the above. The partials are all functions, and are not constant.
    Fx(x, y, z) = y+z, Fx(1, 1, 1) = 2
    Fy(x, y, z) = x+z, Fy(1, 1, 1) = 2
    Fz(x, y, z) = y+x, Fz(1, 1, 1) = 2
    Looks fine to me.
    I approached this a different way, assuming that z was (implicitly) a function of x and y.
    From the original equation, I got
    [tex]z_x = \frac{-y - z}{x + y}[/tex], and
    [tex]z_y = \frac{-x - z}{x + y}[/tex]

    At the point of tangency,
    [tex]z_x(1, 1) = \frac{-y - z}{x + y}|_{(1, 1)} = -1[/tex], and
    [tex]z_y(1, 1) = \frac{-x - z}{x + y}|_{(1, 1)} = -1[/tex]

    From these I got the same equation for the plane, namely x + y + z = 3.

    As a check, the point (1, 1, 1) satisfies the plane's equation.
    The vector i + j + k is normal to the plane.
    The two first partials, expressed as vectors are i - k [tex](z_x at (1, 1, 1))[/tex])
    and j - k [tex](z_y at (1, 1, 1))[/tex]

    If you dot the latter two vectors with the plane's normal vector, you get zero, meaning that the slopes of the partial derivatives are each perpendicular to the plane's normal. That pretty well nails down the plane.
     
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