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Math Help - Differentiation to approximate the maximum percentage error

  1. #1
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    Differentiation to approximate the maximum percentage error

    I’m having trouble trying to differentiate this function (See attachment). I think I need to differentiate it with respect to m1m2 and then D. I’m not sure if I should consider m1m2 one variable of F or two. I have been trying it as one as I am only give one % for m1 and m2.

    I was also wondering if someone could help me thought the differentiation process with respect to m1m2 and D if that is the case.

    Thanks, Daddy_Long_Legs
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  2. #2
    MHF Contributor Calculus26's Avatar
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    I'm assuming you mean the max possible % errors are 5% and 4%

    dF = |km2dm1/D^2| + |km1dm2/D^2| + |-2km1m2dD/D^3|

    Note since we are intgerested in max possible % error we use abs values

    dF/F = dm1/m1 + dm2/m2 + 2dD/D

    dF/F = .05 +.05 +.08 = .18 or a max possible % error of 18%
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  3. #3
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    Calculus26,

    I was meaning do I differentiate it with respect to m1m2 and then D or m1,m2 and then D. I was wondering this because they only give you one % error for m1 and m2 and so I thought this may be a single function.

    Differentiating with respect to m1m2 and the D and then using the percentage errors given m1m2 %5 and the D 4% I found the percentage error in F to be 13%

    I seems what you have done is differentiate with respect to m1, m2 and then D, and used 5% error for m1 and m2 and then 4% error for D.

    This was my question. I'm not sure if I have to differentiate with respect to m1m2 and D or m1, m2 and D.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    you treat like the differential of a function of 3 variables

    where m1,m2, and D play the roles of x, y, and z-not sure


    what you mean by differentiating wrt m1m2 as if it were a single variable.
    If you want see the discussion on my website on differentials of f(x,y) and f(x,y,z)
    Tangent Planes/ Differential for f(x,y)
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  5. #5
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    Quote Originally Posted by Daddy_Long_Legs View Post
    Calculus26,

    I was meaning do I differentiate it with respect to m1m2 and then D or m1,m2 and then D. I was wondering this because they only give you one % error for m1 and m2 and so I thought this may be a single function.

    Differentiating with respect to m1m2 and the D and then using the percentage errors given m1m2 %5 and the D 4% I found the percentage error in F to be 13%

    I seems what you have done is differentiate with respect to m1, m2 and then D, and used 5% error for m1 and m2 and then 4% error for D.

    This was my question. I'm not sure if I have to differentiate with respect to m1m2 and D or m1, m2 and D.
    You are given errors for both m1 and m2- they just happen to be the same: 5%.

    [tex]df(m1,m2,D)= \frac{\partial f}{\partial m1}dm1+ \frac{\partial f}{\partial m2} dm2+ \frac{\partial f}{\partial D}dD.

    Now, you aren't actually given dm1, dm2 and dD but if you divide both sides by m1m2D you get \frac{df(m1,m2,D)}{m1m2D}= \frac{\frac{\partial f}{\partial m1}}{m2D}\frac{dm1}{m1}+ \frac{\frac{\partial f}{\partial m2}}{m1D}\frac{dm2}{m2}+ \frac{\frac{\partial f}{\partial D}}{m1m2}\frac{dD}{D}.

    And now, \frac{dm1}{m1}= \frac{dm2}{m2}= 0.05, \frac{dD}{D}= 0.04.
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  6. #6
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    Thanks Calculus26 and HallsofIvy,

    I wasn't to sure if both m1 and m2 were 5%, that's why was I was considering it as one varibale. I think I'll have no trouble finishing the question now.

    Thanks for all the help.
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