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Math Help - Maximum Percentage error

  1. #1
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    Question Maximum Percentage error

    The resonant frequency of an oscillation in electrical circuits is given by the formula f = 1/2pLC. If the error in measuring L is 4%, and that measuring C is 2%, calculate the maximum percentage error in calculating f?

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  2. #2
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    Hello, Stazzer5!

    I think I've solved it.
    Someone please check my reasoning and my work.


    The resonant frequency of an oscillation in electrical circuits
    . . is given by the formula: . f \:=\:\tfrac{\pi}{2}\sqrt{LC}

    If the error in measuring L is 4%, and that measuring C is 2%,
    . . calculate the maximum percentage error in calculating f?
    We are given: . \begin{Bmatrix}\dfrac{dL}{L}\:=\: 0.04 \\ \\[-3mm]\dfrac{dC}{C} \:=\: 0.02 \end{Bmatrix} . [1]


    We have: . f \:=\:\tfrac{\pi}{2}(LC)^{\frac{1}{2}}

    Take differentials: . df \;=\;\tfrac{\pi}{2}\cdot\tfrac{1}{2}(LC)^{-\frac{1}{2}}(L\,dC + C\,dL) \;=\;\frac{\pi(L\,dC + C\,dL)}{4\sqrt{LC}}

    Divide by f\!:\;\;\frac{df}{f} \;=\;\frac{\frac{\pi(L\,dC + C\,dL)}{4\sqrt{LC}}}{\frac{\pi}{2}\sqrt{LC}} \;=\;\frac{L\,dC + C\,dL}{2LC} \;=\;\frac{L\,dC}{2LC} + \frac{C\,dL}{2LC}<br />

    . . Hence: . \frac{df}{f} \;=\;\frac{1}{2}\left(\frac{dC}{C} + \frac{dL}{L}\right)


    Substitute [1]: . \frac{df}{f} \;=\;\tfrac{1}{2}(0.04 + 0.02) \;=\;0.03


    Therefore, the maximum percentage error in calculating f is \boxed{3\%}

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  3. #3
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    Hello everyone

    I don't know enough about electrical circuits to say for sure, but, according to this Wikipedia article, it looks as though the formula should be:
    f= \frac{1}{2\pi\sqrt{LC}}
    It's easiest to deal with percentage errors with this type of equation if you take logs before differentiating:

    So, writing f as:
    f=(2\pi)^{-1}L^{-\tfrac12}C^{-\tfrac12}
    we get:
    \ln(f) = -\ln(2\pi)-\tfrac12\ln(L) - \tfrac12\ln(C)
    Then on differentiating we get:
    \frac{df}{f}=-\frac12\left(\frac{dL}{L}+\frac{dC}{C}\right)
    You can then substitute in the values for \frac{dL}{L} and \frac{dC}{C} as Soroban has done. (You get the same answer, 3%, but with a negative sign, indicating that if L and C are larger than the true value, f will be smaller.)

    Note that any formula where you have powers of a number of variables can be dealt with in this way:
    y = k\,a^p\,b^q\,c^r...

    \Rightarrow \ln(y) = \ln(k) + p\ln(a)+q\ln(b)+r\ln(c) + ...

    \Rightarrow \frac{dy}{y}= p\frac{da}{a}+q\frac{db}{b}+r\frac{dc}{c}+ ...
    Grandad
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  4. #4
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    hi there,

    Thank you very much for your help.

    stazzer5
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