The circumference of a sphere was measured to be 82.000 cm with a possible error of 0.50000 cm. Use linear approximation to estimate the maximum error in the calculated surface area._________ Estimate the relative error in the calculated surface area.________

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`S=4pir^2`    If the `C=82cm, C=2pir` , So `r = 82/(2pi) = 41/pi`

`S = 4pi(41/pi)^2 = 4pi*41^2/pi^2 = 4*1681/pi = 6724/pi`

` S=4pi(C/(2pi))^2=4piC^2/(4pi^2)=C^2/pi`

` DeltaS=(2C)/pi DeltaC`

   is a linear approximation for the error in S

`DeltaS = (2(82))/pi (0.5) = 82/pi` cm  this an estimate of the...

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`S=4pir^2`    If the `C=82cm, C=2pir` , So `r = 82/(2pi) = 41/pi`

`S = 4pi(41/pi)^2 = 4pi*41^2/pi^2 = 4*1681/pi = 6724/pi`

` S=4pi(C/(2pi))^2=4piC^2/(4pi^2)=C^2/pi`

 

` DeltaS=(2C)/pi DeltaC`

   is a linear approximation for the error in S

`DeltaS = (2(82))/pi (0.5) = 82/pi` cm  this an estimate of the maximum error in S

The relative error is `(82/pi)/(82^2/pi)^2 = pi/82`

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