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Math Help - Accuracy question

  1. #1
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    Accuracy question

    I have done similar questions but this one does not give a value for r. Should I give it any value?
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  2. #2
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    Quote Originally Posted by Stuck Man View Post
    I have done similar questions but this one does not give a value for r. Should I give it any value?
    No need...

    relative\ error=\frac{absolute\ error}{exact\ value\ of\ that\ being\ measured}

    =\displaystyle\huge\frac{{\pi}r^2-{\pi}\left(r-\frac{r}{400}\right)^2}{{\pi}r^2}=\frac{{\pi}r^2-{\pi}r^2\left(1-\frac{1}{400}\right)^2}{{\pi}r^2}

    or

    \displaystyle\huge\frac{{\pi}\left(r+\frac{r}{400}  \right)^2-{\pi}r^2}{{\pi}r^2}=\frac{{\pi}r^2\left(1+\frac{1}  {400}\right)^2-{\pi}r^2}{{\pi}r^2}

    which may be factorised, removing r and \pi} completely.
    Those parameters are needed for the "absolute" error,
    but not for the "relative" error.
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  3. #3
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    Ok. In general is the absolute error either +ve or -ve? Is an error bound always positive? Wikipedia seems to muddle these things up.
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  4. #4
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    Quote Originally Posted by Stuck Man View Post
    Ok. In general is the absolute error either +ve or -ve? Is an error bound always positive? Wikipedia seems to muddle these things up.
    It would be positive,
    however that's because the modulus is taken,
    which doesn't mean we get the same error here if the radius is lower by 0.25%
    or if it's higher by 0.25% from it's nominal value.
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  5. #5
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    To calculate the absolute error you've taken the minimum value of r so you must also take the minimum value of pi which is 3.1415. This will get the answer in the book. Part b is done similarly for pi=3.14159.
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  6. #6
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    There is much more work than this. I know how to do it though.
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