Resolved: "The circumference of a sphere was measured to..."

Lennie Carroll

Answered question

2021-08-06

The circumference of a sphere was measured to be 84 cm with a possible error of 0.5 cm.
Use differentials to estimate the maximum error in the calculated surface area. What is the relative error?

Answer & Explanation

Bertha Stark

Skilled2021-08-07Added 96 answers

$C = 2 π r$
$\frac{d C}{d r} = 2 π$
$\frac{△ C}{△ r} = 2 π$
$\frac{△ C}{2 π} = △ r$
$△ r = \frac{△ C}{2 π}$
$△ r = \frac{0.5}{2 π}$
$△ r = \frac{1}{4 π}$
$S (r) - 4 π r^{2}$
$\frac{d S}{d r} = 8 π r$
$\frac{△ S}{△ r} = 8 π r$
$△ S = 8 π r △ r$
$△ S = 4 \cdot 2 π r △ r$
$△ S = 4 C △ r$
$△ S = 4 \cdot 84 \cdot \frac{1}{4 π} = \frac{84}{π} \approx 27 {cm}^{2}$
Relative Error: $= \frac{△ S}{S} = \frac{\frac{84}{π}}{4 π r^{2}} = \frac{\frac{84}{π}}{4 π {(\frac{C}{2 π})}^{2}} = \frac{\frac{84}{π}}{4 π {(\frac{84}{2 π})}^{2}} = \frac{1}{84} \approx 0.012$

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