# Calculus

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The circumference of a sphere was measured to be 88000 cm with a possible error of 050000 cm. Use linear approximation to estimate the maximum error in the calculated surface area.

Estimate the relative error in the calculated surface area.

• Calculus -

Oops the 88000 is supposed to be 88 cm and 0.5 cm

• Calculus -

Circumference, C=88 cm
C = 2πr

Area, A=πr²
Differentiating:
dA/dr = 2πr
Using linear approximation:
δA = 2πr δr
=2πr δr
=C δr
=88 cm * 0.5 cm
=44 cm²

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