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Linearization Problem [Solved]

  1. Nov 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the linearization of the function ƒ(x,y) = sqrt(29 - 4x2 - 4y2) at the point (2,1)

    2. Relevant equations

    Point (a,b)
    L(x,y) = Linearization
    L(x,y) = ƒ(a,b) + ƒx(a,b)(x-a) + ƒy(a,b)(y-b)

    3. The attempt at a solution

    ƒ(2,1) = 3

    ƒx(x,y) = [1/2⋅-8x]/[sqrt(29-4x2-4y2]
    ƒx(2,1) = -8/3

    ƒy(x,y) = [1/2⋅-8y]/[sqrt(29-4x2-4y2]
    ƒy(2,1) = -4/3

    L(x,y) = 3 + [-8/3(x-2)] + [-4/3(y-1)]
    = 3 + (-8/3x + 16/3) + (-4/3y+4/3)
    = 3 -8/3x +16/3 - 4/3y + 4/3
    Simplifying gives me
    = -8x - 4y + 29

    Which is wrong according to my Webwork

    please any help would be amazing.
     
  2. jcsd
  3. Nov 10, 2014 #2

    BvU

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    3 -8/3x +16/3 - 4/3y + 4/3 gives (-8x - 4y + 29)/3 You can't just strike the /3 ....
     
  4. Nov 10, 2014 #3
    i multiplied everything by 3, can i not do that?
     
  5. Nov 10, 2014 #4

    BvU

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    Nope: f = x + 2 is something quite different from f = 3x + 6 !!

    Lean back a little, relax, take a breath or a break and it'll be obvious...

    :) If you don't believe me you can always evaluate f(2.1,1) and f(2,1.1) on a calculator or a spreadsheet...
     
  6. Nov 10, 2014 #5
    Thank you BvU appreciate the help, more questions may be on the way, its been a rough day.
     
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