Calculus

The circumference of a sphere was measured to be 74.000 cm with a possible error of 0.50000 cm. Use linear approximation to estimate the maximum error in the calculated surface area

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asked by Julie
  1. Duplicate post. See "CALC"

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    posted by drwls
  2. C = 2 pi r
    dr/dC = 1/(2pi)

    A = 4 pi r^2
    dA/dr = 8 pi r

    dA/dC = dA/dr * dr/dC
    dA/dC = 8 pi r *1/(2pi)
    = 4 r
    but r = C/(2 pi)
    so
    dA/dC = 4 C/(2 pi)
    = 2 C/pi
    dA = 2 * 74/pi * .5 = 74/pi

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    posted by Damon

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