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Math Help - linearization?

  1. #1
    Junior Member
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    linearization?

    Please help. Don't know how to do this...

    The circumference of a sphere was measured to be 88 cm with a possible error of 0.5 cm.

    (a) Use differentials to estimate the maximum and relative error in the calculated surface area. (Round maximum error to the nearest whole number and relative error to three decimal places.)
    Maximum error=? cm^2
    Relative error= ?
    (b) Use differentials to estimate the maximum and relative error in the calculated volume. (Round maximum error to the nearest whole number and relative error to three decimal places.)
    Maximum error= ? cm^3
    Relative error=?
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  2. #2
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    Hello, kwivo!

    The circumference of a sphere was measured to be 88 cm with a possible error of 0.5 cm.
    Since C = 2\pi r, we have: . 2\pi r = 88\quad\Rightarrow\quad r \,=\,\frac{44}{\pi} cm

    We are told that: . \Delta r\:=\:dr\:=\:\pm0.5 cm



    (a) Use differentials to estimate the maximum and relative error
    in the calculated surface area.
    (Round maximum error to the nearest whole number
    and relative error to three decimal places.)

    Surface area: . A\;=\;4\pi r^2

    The base sphere has area: . A \:=\:4\pi\left(\frac{44}{\pi}\right)^2 \:=\:\frac{7744}{\pi} cm².

    The approximate error in the area is: . dA \;=\;8\pi r\,dr

    Hence: . dA \;=\;8\pi\left(\frac{44}{\pi}\right)(\pm0.5)


    The maximum error is: . dA \:=\:\pm176 cm²

    The relative error is: . \frac{dA}{A} \;=\;\frac{\pm176}{\frac{7744}{\pi}} \;\approx\;\pm 0.071




    (b) Use differentials to estimate the maximum and relative error
    in the calculated volume.
    (Round maximum error to the nearest whole number
    and relative error to three decimal places.)

    Volume of a sphere: . V \;=\;\frac{4}{3}\pi r^3

    The base sphere has volume: . V \;=\;\frac{4}{3}\pi\left(\frac{44}{\pi}\right)^3 \;=\;\frac{340,736}{3\pi^2} cm³

    The approximate error in the volume is: . dV \;=\;4\pi r^2\,dr

    Hence: . dV\;=\;4\pi\left(\frac{44}{\pi}\right)^2(\pm 0.5)


    The maximum error is: . \pm\frac{3872}{\pi} \;\approx\;\pm1232 cm³

    The relvative error is: . \frac{dV}{v}\;=\;\frac{\pm\frac{3872}{\pi}}{\frac{  340,736}{3\pi^2}} \;\approx\;\pm0.107

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