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Math Help - Calculus 4: differentials

  1. #1
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    Calculus 4: differentials

    Hello everyone,

    Could you please tell me if this looks right?

    Find the differential of f(x,y)=square root(x^2+y^3) (1,2)

    and estimate f(1.04, 1.98)

    fx=1/2(x^2+y^3)^-1/2(2x)
    fx(1)=.71

    fy=1/2(x^2+y^3)^-1/2(3y^s)
    fy(2)=1.7

    df=fx(a,b)dx+fy(a,b)dy

    df=.71dx+1.7dy

    f(1.04, 1.98)=square root(1.04^2+1.98^3)=2.

    Thank you very much
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  2. #2
    Super Member Aryth's Avatar
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    So, we have:

    f(x,y) = \sqrt{x^2 + y^3}

    And we want to estimate:

    f(1.04, 1.98)

    So, we have:

    f_x(x,y) = \frac{1}{2}(x^2 + y^3)^{-\frac{1}{2}}2x

    =x(x^2 + y^3)^{-\frac{1}{2}}

    And:

    f_y(x,y) = \frac{1}{2}(x^2 + y^3)^{-\frac{1}{2}}3y^2

    =\frac{3y^2}{2(x^2 + y^3)^{-\frac{1}{2}}}

    So the differential becomes:

    df = x(x^2 + y^3)^{-\frac{1}{2}}dx + \frac{3y^2}{2(x^2 + y^3)^{-\frac{1}{2}}}dy

    Now, how to estimate it... I can't quite remember. I'll get back to you on that if someone else doesn't get to it before me.
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  3. #3
    MHF Contributor
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    You just estimate the change in the function using the formula you posted, using x = 1, y = 2, dx = .04 and dy = -.02.
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