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Thread: Worst Case Error

  1. #1
    Oct 2012

    Worst Case Error

    A metal pipe with outer radius of 10cm ,inner radius of 8cm and length of 20 have allowed errors of 1cm, 0.5cm and 2cm respectively. If need to estimate the worst-case error in volume.

    So I know that volume = (pi*r1^2*l) -(pi*r2^2*l)
    Then using partial differentiation I got
    Vr1 = 2pi*r1*l
    Vr2 = -2pi*r2*l
    Vl = pi *r1^2 - pi*r2^2

    V(10,8,20) = 720pi

    is this the correct formula to calculate error
    abs(e) = abs(Vri *E1) +abs(Vr2 *E2) +abs(Vl *E3)

    Im not sure which error terms I should sub in, should i convert allowed error into a decimal or simply sub in as they are?

    Any help would be great
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  2. #2
    MHF Contributor

    Apr 2005

    Re: Worst Case Error

    Personally, I wouldn't use differentiation to answer this question. If, in the 'worst case' scenario we over measure the outer measure and and length by the maximum possible error, and under measure (because we will be subtracting it) the inner radius, we would get outer radius of 10+ 1= 11 cm, length of 20+ 2= 22 cm and inner radius 8- .5= 7.5 cm we would get a volume of \pi (11^2- 7.5^2)(22)= 4475 cubic centimeters. That is 4475- 2262= 2213 cubic centimeter larger than the "actual value".

    If we under measure the outer radius and lengthby the maximum possible error, and over measure the inner radius, we would have outer radius 10-1= 9 cm, length 20- 2= 18 cm, and inner radius 8+ .5= 8.5 and get a volume of [tex] \pi(9^2- 8.5^2)(18)= 495 cubic centimeters That is 2262- 495= 1767 cubic centimeters less than the "actual value'.

    The larger or fthose "errors" is 2213 cubic centimeters.
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