##### its in the second chapter in calculus 1

label Calculus
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The circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm.

(a) Use differentials to estimate the maximum error in the calculated surface area. (Round your answer to the nearest integer.)
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What is the relative error? (Round your answer to three decimal places.)
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(b) Use differentials to estimate the maximum error in the calculated volume. (Round your answer to the nearest integer.)
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What is the relative error? (Round your answer to three decimal places.)
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Jun 14th, 2015

S = 4*pi*r^2

delta S = 8*pi*r delta r = 8*pi*76*0.5 = 955 cm^2

relative error = dela S/S = 8 pi*r*deltar / (4*pi*r^2) =  2delta r/r = 1/76 = 0.013

b) V= 4/3 pi*r^3

dV = 4pir^2 dr  = 4*pi*76^2 *0.5 = 36292

relative error = 4pi r^2 dr / (4/3*pir^3) = 3dr/r = 1.5/76 =  0.020

Jun 14th, 2015

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Jun 14th, 2015
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Jun 14th, 2015
Aug 21st, 2017
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