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    bluedicius's Avatar
    bluedicius Posts: 20, Reputation: 1
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    #1

    Nov 24, 2004, 06:44 PM
    Percentage error
    The volume of a sphere is 3500 cm3 + 1.8%.
    What are the possible maximum and minimum lengths of the radius?
    Calculate the radius and its percentage error.
    drwls's Avatar
    drwls Posts: 14, Reputation: 2
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    #2

    Jan 7, 2005, 09:14 PM
    re: percentage error
    Your question was :
    <<The volume of a sphere is 3500 cm3 + 1.8%.
    What are the possible maximum and minimum lengths of the radius?
    Calculate the radius and its percentage error.>>

    Since V = (4/3) pi r^3,
    r = [3V/(4 pi)]^(1/3), which is the cube root of 3V/(4 pi)

    When V = 3500 cm^3, r = 9.419 cm
    When V = 3500 cm^3 (1.018), r(max) = 9.419*(1.018)^(1/3) = 9.475 cm
    When V = 3500 cm^3 (0.982), r(min) = 9.419*(0.982)^(1/3) = 9.362 cm

    The percentage error is 100 x (9.475-9.419)/9.419 = 0.6%
    Note that this is about 1/3 of the volume % error, which could have been deduced in far fewer steps, since the cube root of 1 + x = 1 + x/3, for x<<1

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