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Math Help - Linear approximation

  1. #1
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    Linear approximation

    Last one, guys...

    "Given the function f(x,y)=x^2cos(xy)-5xy, estimate the value of f(3,97,0.025)."

    My work:

    fx(x,y) = 2x(cos(xy))-y(x^2sin(xy)+5)
    fy(x,y) = -x(x^2sin(xy)+5)

    So, f(x,y) + fx(x,y)(x-a) + fy(x,y)(y-b)

    f(3.97,.025) + fx(3.97,.025)(3.97-3.95) + fy(x,y)(.025-.020)

    And, after plugging into this equation, my final answer was 15.32.

    Did I mess up anywhere? I'm pretty confident about the answer, since (3.97,.025) plugged into the original equation is 15.26, but I just want to make sure I'm not missing anything.
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  2. #2
    MHF Contributor matheagle's Avatar
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    You confused me with having 3 coordinates in f(x,y).
    I would use a=4 and b=0.

    you want f(x,y)\approx f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)
    Last edited by matheagle; March 1st 2011 at 11:14 PM. Reason: I pasted OP errors by accident
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by matheagle View Post
    You confused me with having 3 coordinates in f(x,y).
    I would use a=4 and b=0.

    you want f(x,y)\approx f(a,b) + f_x(x,y)(x-a) + f_y(x,y)(y-b)
    Or may be:

    f(x,y)\approx f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)

    (the whole point is to evaluate the function and its partials at easy values)

    CB
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