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Math Help - Taylor Expansion x^2+y^2+z^2=9

  1. #1
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    Taylor Expansion x^2+y^2+z^2=9

    Hi!

    I have a calculus exam tomorrow, and one of the questions in the 'Series Expansion for Functions of Two Variables' is confusing me a little...

    "A Sphere is defined by the equation x^2 +y^2+z^2=9 Determine the equation of the tangent plane on its surface when x=1, and y=2"

    the answer is given: x+2y+2z=9

    I assumed I had to jiggle the equation, so i have z=-x-y+3, and start my partial differentiations from there, but my numbers end up silly, and the answer is way wrong when I do that.

    Any help?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Welcome to the forum!

    Consider the function

    f(x,y,z)=x^2+y^2+z^2-9

    which has partial derivatives

    f_x(x,y,z)=2x,\ f_y(x,y,z)=2y,\ f_z(x,y,z)=2z

    So if we consider its first Taylor polynomial about the point (1,2,2) :

    P(x,y,z)=f(1,2,2)+f_x(1,2,2)(x-1)+f_y(1,2,2)(y-2)+f_z(1,2,2)(z-2)

    = 2(x-1)+4(y-2)+4(z-2)

    Now notice that your sphere is the solution space of f(x,y,z)=0; and the tangent plane at the point (1,2,2) will be the solution space of P(x,y,z)=0, i.e.

    2(x-1)+4(y-2)+4(z-2)=0
    or
    x+2y+2z=9.
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  3. #3
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    Thanks for the help!

    Just a couple of questions, what made you choose the point z=2? As the question only included (x,y), as in (1,2), not (1,2,2).

    Also, at the end, where did you get the 9 from? you went from
    2(x-1)+4(y-2)+4(z-2)=0
    to


    sorry if I'm being blind, its just I have to say this in front of a panel tomorrow! best to know what I'm talking about!
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    You are right - there are two points z corresponding to those values of x,y (namely z=2 and z=-2). In that sense, the problem is unclear.

    As for how I went from 2(x-1)+4(y-2)+4(z-2)=0 to x+2y+2z=9... I just rearranged the equation. Divide everything by two, and put the constants on the right hand side.
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