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The circumference of a sphere was measured to be 88000 cm with a possible error of 050000 cm. Use linear approximation to estimate the maximum error in the calculated surface area.

Estimate the relative error in the calculated surface area.

  • Calculus - ,

    Oops the 88000 is supposed to be 88 cm and 0.5 cm

  • Calculus - ,

    Circumference, C=88 cm
    C = 2πr

    Area, A=πr²
    dA/dr = 2πr
    Using linear approximation:
    δA = 2πr δr
    =2πr δr
    =C δr
    =88 cm * 0.5 cm
    =44 cm²

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