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Math Help - Find the tangent plane to the surface

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    Exclamation Find the tangent plane to the surface

    Find the tangent plane to the surface, x^2 + 2xy - y^2 + z^2 = 7
    at the point (1, -1, 3). Using the tangent plane as an approximation to the surface, find the approximate height of the surface above the (x,y) plane at the point (1.1, -1.2). How does this result compare to the true height?
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    Quote Originally Posted by njr008 View Post
    Find the tangent plane to the surface, x^2 + 2xy - y^2 + z^2 = 7
    at the point (1, -1, 3). Using the tangent plane as an approximation to the surface, find the approximate height of the surface above the (x,y) plane at the point (1.1, -1.2). How does this result compare to the true height?
    The Cartesian equation of the plane is ax + by + cz = d.

    ai + bj + ck is a normal to the plane. If you know a normal, you know the values of a, b and c. To get the value of d, substitute the known point (1, -1, 3) into the Cartesian equation:

    a(1) + b(-1) + c(3) = d.

    Oh yes ...... the values of a, b and c are got from the normal. If the surface has the equation x = f(x, y), then a normal is:

    \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j - k.

    Substitute x = 1 and y = -1 to get a normal at (1, -1, 3).
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