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Math Help - Finding Tangent Plane

  1. #1
    Member CalcGeek31's Avatar
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    Finding Tangent Plane

    If a function z = g(x, y) has g(1, 2) = 10, gx(1, 2) = 7 and gy(1, 2) = -5, find the equation of the plane tangent to the surface z = g(x, y) at the point where x = 1 and y = 2.


    I understand how to do the question when an equation is given but without it, I am lost. Could someone help me get started through the problem.
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  2. #2
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    Quote Originally Posted by CalcGeek31 View Post
    If a function z = g(x, y) has g(1, 2) = 10, gx(1, 2) = 7 and gy(1, 2) = -5, find the equation of the plane tangent to the surface z = g(x, y) at the point where x = 1 and y = 2.


    I understand how to do the question when an equation is given but without it, I am lost. Could someone help me get started through the problem.

    The normal vector to the surface is given by  <g_x(1,2) , g_y(1,2), -1>

    All you need for a plane is a normal vector and a point. You have both:
     \overrightarrow{n}=<7, -5, -1> and your P is (1,2)

    Just plug these into the plane equation and you're done.
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  3. #3
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    Re: Finding Tangent Plane

    I know this is an old post, but it is what I am currently studying and I'm curious where the -1 value came from here.

    Would anyone mind shedding some light on it for me?
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  4. #4
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    Re: Finding Tangent Plane

    Since the equation for the surface is f(x,y,z) = g(x,y)-z = 0, you can do a linear estimate:

    f(x,y,z)\approx{f}(x_0,y_0,z_0)+f_x(x_0,y_0,z_0)(x-x_0)+
    f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0),

    which is the equation of the tangent plane. Of course f_z is just -1 in this case.

    - Hollywood
    Last edited by hollywood; March 25th 2013 at 06:22 PM. Reason: forgot a term
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