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Math Help - Finding maximum error

  1. #1
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    Finding maximum error

    We measure the radius r of a circle and obtain that that r=9 centimeters. We use this value to compute the area of the circle. We know that the error in our measurement is at most 1.6%. Use the linear approximation to the area A(r)=pi*r^2, at r=9 to estimate the maximum error in our computed value for the area.

    I get (162/125) * pi as my answer. :/
    Is the equation not 2 * pi * 9 * .016?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Velvet Love View Post
    We measure the radius r of a circle and obtain that that r=9 centimeters. We use this value to compute the area of the circle. We know that the error in our measurement is at most 1.6%. Use the linear approximation to the area A(r)=pi*r^2, at r=9 to estimate the maximum error in our computed value for the area.

    I get (162/125) * pi as my answer. :/
    Is the equation not 2 * pi * 9 * .016?
    The area of a circle of radius r is:

    A(r)=\pi r^2

    for a small error \delta in the radius we have:

    A(r+\delta)\approx A(r)+\delta \frac{dA}{dr}=A(r)+\delta (2\pi r)

    Thus the error is the area is:

    A(r+\delta)-A(r)=2\pi r \delta

    In this case r=9 and \delta=0.016 \times r=0.016 \times 9 cm (it's 1.6\% of r )

    CB
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  3. #3
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    A slightly different way of looking at this. A(r)= \pi r^2 so the derivative is A'(r)= 2\pi r. At r= 9, A(9)= 81\pi and [tex]A'(9)= 18\pi[itex]. The tangent line at r= 9, [tex]18\pi(r- 9)+ 81\pi[/itex] gives the linear approximation. The "change" is 18\pi (r- 9). 1.6% of 9 is 0.144 so evaluate 18\pi (r-9) for r= 9.144.

    There is, by the way, an engineers "rule of thumb" that is based on this: When adding (or subtracting) quantities add their errors. When multiplying (or dividing) add their percentage errors. Here we are multiplying r times r. since the percentage error for r is 1.4%, the percentage error for r^2, and so for \pi r^2 would be 1.4+ 1.4= 2.8%
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