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The circumference of a sphere was measured to be cm with a possible error of cm. Use linear approximation to estimate the maxim...Asked by Salman
The circumference of a sphere was measured to be 74 cm with a possible error of 0.5 cm. Use linear approximation to estimate the maximum error in the calculated surface area.
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Answered by
drwls
The (small) relative error in a linear dimension, which in this case is 0.676%, gets doubled when you are calculating the relative error in an area dimension. That is the "linear approximation".
Here's why:
Area = C * r^2 * (1 + e)^2
= C r^2 (1 + 2e + e^2)
where e is the relative error in radius (or circumference). For e <<1, you can neglect e^2. 2e becomes the area realtive error. C is a constant. The same rule can be applied to any shape or figure. The relative error in a volume dimension is 3e.
Here's why:
Area = C * r^2 * (1 + e)^2
= C r^2 (1 + 2e + e^2)
where e is the relative error in radius (or circumference). For e <<1, you can neglect e^2. 2e becomes the area realtive error. C is a constant. The same rule can be applied to any shape or figure. The relative error in a volume dimension is 3e.
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