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Math Help - Percentage Error.

  1. #1
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    Percentage Error.

    Hello , please show me how to solve this problem.


    The circumference of a sphere was measured to be 84cm with a possible error of 0.5
    a. Use differentials to estimate the maximum error in the calculated surface area. What is the relative error?
    b. Do the same thing calculate volume.
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  2. #2
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    I really don't understand, can we measure the circumference of the sphere?
    what is the formula?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    Hello , please show me how to solve this problem.


    The circumference of a sphere was measured to be 84cm with a possible error of 0.5
    a. Use differentials to estimate the maximum error in the calculated surface area. What is the relative error?
    if the radius of the sphere is r, then it's surface area is 4pi*r^2. the error in the measurement of r we can call dr (the change in r). then we can calculate the error in the Area A, by finding dA. but first we have to find how the error in the circumference C affects the error in the radius. here goes.

    C = 2pi*r
    => dC = 2pi dr
    => 0.5 = 2pi dr
    => dr = 1/4pi

    now with the error:
    C = 2pi*r
    => r = C/2pi = 84/2pi = 42/pi

    Now A = 4pi*r^2
    => dA = 8pi*r dr
    => dA = 8pi*(42/pi)(1/4pi)
    => dA = 336/pi ...........the maximum error in surface area.

    to find the relative error, we divide the error in area by the total area.

    dA/A = 8pi*r dr/4pi*r^2 = 2 dr/r = 2[(1/4pi)/(42/pi)] = 2(1/168) = 0.0119 which is about 1.2% relative error
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    Hello , please show me how to solve this problem.


    The circumference of a sphere was measured to be 84cm with a possible error of 0.5
    a. Use differentials to estimate the maximum error in the calculated surface area. What is the relative error?
    b. Do the same thing calculate volume.
    now part b is done in a similar way

    we have r = 42/pi and dr = 1/4pi

    now V = (4/3)pi*r^3
    => dV = 4pi*r^2 dr
    => dV = 4pi*(42/pi)^2 * (1/4pi)
    => dV = 1764/pi^2 = 178.73 approx ...maximum error in calculated volume

    So relative error is given by
    dV/V = (4pi*r^2)/(4/3)pi*r^3 = 3 dr/r = 3[(1/4pi)/(42/pi)] = 3(1/168) = 0.018 or about 1.8 % relative error
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