calculus
posted by jin .
The circumference of a sphere was measured to be 78.000 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.

The relative error in area is, to first approximation, twice the relative error in a linear dimension.
That is because
(delta A)/A = (1 + epsilon)^2  1
= 1 + 2 epsilon + epsilon^2  1
= 2 epsilon
when epsilon << 1
In your case, the answer is
2*0.5/78 = 1/78 = 1.3%
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