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Math Help - Max error and relative error

  1. #1
    Member mybrohshi5's Avatar
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    Max error and relative error

    The circumference of a sphere was measured to be cm with a possible error of cm. Use differentials to estimate the maximum error in the calculated surface area.

    Estimate the relative error in the calculated surface area.

    This is what i tried but i got it wrong?

    Circumference of a sphere = 2πr
    so r = 11.61831

    SA=4πr^2

    SA=1696.2731
    SAmax=1845.4147

    Maximum error = 149.1416 ? this is wrong

    Relative error = change in SA / SA
    f'sa = 8πr so change in SA = 8π(11.61831)(.5) = 145.9999

    relative error = 145.9999/1696.2731 = .08607

    can i get some help with what to do because i guess i am doing it wrong

    thank you
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  2. #2
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
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    408

    Use differentials

    What you are doing is correct, but you seem to have made some arithmetic errors. Consider:

    C=2\pi r=73\pm .5=C\pm\Delta C

    S=4\pi r^2=\frac{C^2}{\pi}

    So S=\frac{73^2}{\pi}=1696.27 but can be in the range \frac12\left(\frac{73.5^2}\pi-\frac{72.5^2}\pi\right)=\pm23.24

    Here is how to reach the same answer using differentials:

    S=\frac{C^2}{\pi}

    dS=\frac{2C}{\pi} dC=\frac{2*73}{\pi}(.5)=\pm23.24

    So the surface area is S=1696.27\pm23.24, with a relative error of 1.37%
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  3. #3
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
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    That was very helpful and clearly showed what i did wrong.

    i tried to get help from my professor but he just confused me more and your work was much more clear than his.

    thanks again
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