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Thread: Tangent planes anyone?

  1. #1
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    Exclamation Tangent planes anyone?

    Find an equation for the tangent plane to the surface x^2+y^2-z=0 at the point (1,1,2)

    Any ideas? thanks
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  2. #2
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    Find an equation for the tangent plane to the surface x^2+y^2-z=0 at the point (1,1,2).


    Your solution will be of the form $\displaystyle a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0 $, where $\displaystyle a,b,c $ are constants.

    Note that $\displaystyle x_{0}, y_{0}, \text{and } z_{0} $ are simply the coordinates of $\displaystyle P=(1, 1, 2) $. So, your solution will look something like this:
    $\displaystyle a(x-1) + b(y-1) + c(z-2) = 0 $.

    We want to find the constants a, b, and c. How? We first compute the partial derivatives of our surface equation $\displaystyle x^2+y^2-z=0 $.

    $\displaystyle \frac{\partial f}{\partial x} = 2x $
    $\displaystyle \frac{\partial f}{\partial y} = 2y $
    $\displaystyle \frac{\partial f}{\partial z} = -1 $

    The gradient vector is given by:
    $\displaystyle \nabla f(x,y,z) =(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}) $.

    So, $\displaystyle \nabla f(x,y,z) = (2x, 2y, -1) $.

    Then, plugging in the coordinates of $\displaystyle P=(1,1,2) $, we get:

    $\displaystyle \nabla f(1,1,2) = (2, 2, -1). $

    So0oo, the constants $\displaystyle a, b, \text{and } c \text{ are } 2, 2, \text{and } -1 $, respectively.

    Soooooooo, the equation for the tangent plane to the surface $\displaystyle x^2+y^2-z=0 \text{ at the point } (1,1,2) $ is
    $\displaystyle 2(x-1) + 2(y-1) + -1(z-2) = 0 $.

    Rearrange this if you so wish.

    Enjoy,
    -Andy
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