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Thread: Percent Error Calculation (sphere)

  1. #1
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    Percent Error Calculation (sphere)

    What i know:

    Volume of Sphere = 4/3 pi r ^3
    dv/dr = 4 pi r^2

    The question

    When the radius of a sphere is measured , the measurement has a 1 percent error. What is the resulting percentage error when the volume of sphere is computed using this inaccurate measurement of the radius?


    Help!
    I'm confused. I googled the answer, and there was work shown but i had no idea how the answer was 6%. Can someone please go through the steps? I don't understand how taking the derivative helps us find the percent error in volume.

    I also know that the percent error is (measured value)/(actual value) * 100
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  2. #2
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    Re: Percent Error Calculation (sphere)

    V = kr^3 where k = \dfrac{4}{3}\pi

    \begin{align*}\dfrac{\Delta V}{V} & = \dfrac{k\left(r+\Delta r\right)^3 - kr^3}{kr^3} \\ & = \dfrac{kr^3 + 3kr^2\Delta r + 3kr\left( \Delta r\right)^2 + k\left( \Delta r \right)^3 - kr^3 }{kr^3} \\ & = \dfrac{3\cancel{kr^3} \left( \dfrac{ \Delta r}{r} + \left( \dfrac{ \Delta r}{r} \right)^2 + \dfrac{1}{3}\left( \dfrac{ \Delta r }{r} \right)^3 \right)}{\cancel{kr^3}} \\ & = 3\left(0.01 + 0.01^2 + \dfrac{1}{3}0.01^3\right) \\ & = .030301\end{align*}


    So, 3.0301% error.
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  3. #3
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    Re: Percent Error Calculation (sphere)

    Hi,

    Thanks for the response, but can you explain what you did cause im really confused with what's happening and why you took change in v over v and plugged in the change in r?

    is this also correct?vvvvv
    https://imgur.com/a/RwctV
    Last edited by lc99; Nov 10th 2017 at 03:25 PM.
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  4. #4
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    Re: Percent Error Calculation (sphere)

    What he did was calculate the "correct" volume using "r" as the "correct" radius, then calculate the "incorrect" radius using $r+ \Delta r= r+ 0.01r= 1.01r$ where $0.01r$ is the "1 percent error". He subtracted them to find the actual error. Finally, the "percent" error in the volume is that actual error divided by the volume.

    The method you link to is not the actual percent error but an approximation to it using the derivative rather than the actual difference (the derivative, remember, is the limit of the "difference quotient" so can be used to approximate the difference). Did the problem actually say "what is the percent error" or did it say "approximate the percentage error"?
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  5. #5
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    Re: Percent Error Calculation (sphere)

    Oh i think i understand now. So, the approximated error is just the V'(a)(change in x). But there is also the actual error which is just v(x+change) - v(actual)

    and if we want percent error, it's just actual error - measured all divided by actual error

    I see in my calc notes that the actual change/error is given by v(a + change in x) - v(a) and the approximated error is just v'(a)*change in x. So the percent error is the different divided by actual change/error
    Last edited by lc99; Nov 11th 2017 at 03:14 PM.
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