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Math Help - tangent plane problem (almost solved)

  1. #1
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    tangent plane problem (almost solved) [SOLVED]

    question:Find the tangent plane that is horizontal to the surface Q: 6x^2+3y^2+3z^2-6yz-2y-2z=0. Hint this is the lowest point on the surface

    work: I get the partial derivatives f_x=\frac{-12x^2}{6z-6y-2} and f_y=\frac{-6y-6z-2}{6z-6y-2}. I then equate them with zero and from f_x I get x=0 and from f_y I get z=y-1/3
    The formula of the tangent plane is z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\Longrightarrow3y^2+3(y-\frac{1}{3})^2-6y^2-2y-2(y-\frac{1}{3})=0 and solving for y I get y=\pm\frac{\sqrt{11}}{2}-\frac{3}{2} So now I'm stuck because I was thinking I could plug this into my original equation but I can't since I have two possibilities.
    Last edited by superdude; October 16th 2009 at 11:08 PM. Reason: solved
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  2. #2
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    Quote Originally Posted by superdude View Post
    question:Find the tangent plane that is horizontal to the surface Q: 6x^2+3y^2+3z^2-6yz-2y-2z=0. Hint this is the lowest point on the surface

    work: I get the partial derivatives f_x=\frac{-12x^2}{6z-6y-2} and f_y=\frac{-6y-6z-2}{6z-6y-2}. I then equate them with zero and from f_x I get x=0 and from f_y I get z=y-1/3
    The formula of the tangent plane is z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\Longrightarrow3y^2+3(y-\frac{1}{3})^2-6y^2-2y-2(y-\frac{1}{3})=0 and solving for y I get y=\pm\frac{sqrt{11}}{2}-\frac{3}{2} So now I'm stuck because I was thinking I could plug this into my original equation but I can't since I have two possibilities.
    The original equation can be written as 6x^2+3(y-z)^2-2y-2z=0. Plug the information in red into that and it becomes 3\bigl(\tfrac13\bigr)^2-2y-2\bigl(y-\tfrac13\bigr)=0, which you can easily solve.
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  3. #3
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    Thank you, I've got it.
    But how did you recognize that the original equation could be rewritten like that?
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