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Math Help - 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   a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0 , where  a,b,c  are constants.

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

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

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

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

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

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

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

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

    Rearrange this if you so wish.

    Enjoy,
    -Andy
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