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Math Help - Equation of the tangent plane to the given surface

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    Equation of the tangent plane to the given surface

    Find an equation of the tangent plane to the given surface at the specified point.
    z = ex2 - y2, (1, -1, 1)
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    Quote Originally Posted by acg716 View Post
    Find an equation of the tangent plane to the given surface at the specified point.
    z = ex2 - y2, (1, -1, 1)
    What does ex2 mean? You have to be a whole lot clearer in stating your equations if you expect to get effective help.
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    i'm really sorry! when i posted it it looked correct, but e is raised to x^2-y^2
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    Quote Originally Posted by acg716 View Post
    i'm really sorry! when i posted it it looked correct, but e is raised to x^2-y^2
    First get the partial derivatives \frac{\partial z}{\partial x} and \frac{\partial z}{\partial y}.


    Now construct the vector \frac{\partial z}{\partial x} \, \vec{i} + \frac{\partial z}{\partial y} \, \vec{j} - \vec{k}.


    Substitute x = 1 and y = -1 into this vector. This gives a normal of the form a \vec{i} + b \vec{j} - \vec{k} to the surface at (1, -1, 1).

    The equation of the tangent plane is therefore ax + by  - z = d.

    To get the value of d, substitute x = 1, y = -1 and z = 1.
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