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Math Help - Finding a tangent plane of a surface z = f(x,y)

  1. #1
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    Finding a tangent plane of a surface z = f(x,y)

    Assume the tangent plane of the surface z = f(x,y) in the point (0,0,2) is given by z = Ax + By + 2. Find the equation for the tangent plane of the surface z = 1 / f(x,y) in the point (0,0,1/2).

    No clue how to go about it. A hint or two, not a complete solution, would be appreciated.
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  2. #2
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    Re: Finding a tangent plane of a surface z = f(x,y)

    For any surface defined by f(x,y,z)= constant, \nabla f is perpendicular to the surface at each point. You know how to find a plane given a point and normal vector?
    Last edited by Ackbeet; August 15th 2011 at 06:11 AM. Reason: Changed itex to tex.
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  3. #3
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    Re: Finding a tangent plane of a surface z = f(x,y)

    Quote Originally Posted by HallsofIvy View Post
    For any surface defined by f(x,y,z)= constant, [itex]\nabla f[/itex] is perpendicular to the surface at each point. You know how to find a plane given a point and normal vector?
    I don't know if off-hand since I have done any linear algebra in a while, but I've done it before, so Google will help me out in that regards. Now, if I only could figure out what you meant to write there, "[itex]\nabla f[/itex]...
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    Re: Finding a tangent plane of a surface z = f(x,y)

    Quote Originally Posted by Sabo View Post
    I don't know if off-hand since I have done any linear algebra in a while, but I've done it before, so Google will help me out in that regards. Now, if I only could figure out what you meant to write there, "[itex]\nabla f[/itex]...
    See HallsofIvy's post again. I have corrected the code to [tex]\nabla f[/tex].
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  5. #5
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    Re: Finding a tangent plane of a surface z = f(x,y)

    Thanks.
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    MHF Contributor chisigma's Avatar
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    Re: Finding a tangent plane of a surface z = f(x,y)

    Quote Originally Posted by Sabo View Post
    Assume the tangent plane of the surface z = f(x,y) in the point (0,0,2) is given by z = Ax + By + 2. Find the equation for the tangent plane of the surface z = 1 / f(x,y) in the point (0,0,1/2).

    No clue how to go about it. A hint or two, not a complete solution, would be appreciated.
    A simple approach that requires only the basic concepts of partial derivative: if the tangent plane of the surface z=f(x,y) in the point (x_{0}, y_{0}, z_{0}) is given by...

    z=z_{0} + x\ \frac{\partial{f}}{\partial{x}} (x_{0}, y_{0}) + y\ \frac{\partial{f}}{\partial{y}}  (x_{0}, y_{0}) (1)

    ... then the tangent plane of the surface z=\frac{1}{f(x,y)} in the point (x_{0}, y_{0}, \frac{1}{z_{0}}) is given by...

    z=\frac{1}{z_{0}} + \frac{1}{f^{2}(x_{0}, y_{0})}\ \{ x\ \frac{\partial{f}}{\partial{x}}  (x_{0}, y_{0})+ y\ \frac{\partial{f}}{\partial{y}}  (x_{0}, y_{0})\} (2)

    Kind regards

    \chi \sigma
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  7. #7
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    Re: Finding a tangent plane of a surface z = f(x,y)

    Thanks. The partial derivative approach was what I was looking for.
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