The largest percentage error you can allow in the measurement of the radius of a sphere if you want the error in the calculation of its surface area using the formula $S = 4 \pi r ^ { 2 }$ to be no greater than 6 percent is about:
A)2%
B)6%
C)3%
D)1%

Question

Quizplus · Accepted Answer

The surface area $S$ of a sphere is given by $S = 4\pi r^2$. If the radius $r$ changes by a small amount $\Delta r$, the change in surface area $\Delta S$ can be approximated by the derivative of $S$ with respect to $r$, which gives $\Delta S \approx 8\pi r \Delta r$. The percentage error in the surface area is then $\frac{\Delta S}{S} \times 100\% = \frac{8\pi r \Delta r}{4\pi r^2} \times 100\% = \frac{2\Delta r}{r} \times 100\%$. If we want the error in the surface area to be no greater than 6%, we set this expression equal to 6% and solve for the percentage error in $r$, which gives $\frac{2\Delta r}{r} \times 100\% = 6\%$, or $\frac{\Delta r}{r} \times 100\% = 3\%$. Therefore, the largest percentage error allowed in the measurement of the radius to keep the error in the surface area calculation below 6% is about 3%.

The Largest Percentage Error You Can Allow in the Measurement S=4πr2S = 4 \pi r ^ { 2 }S=4πr2

The Largest Percentage Error You Can Allow in the Measurement $S = 4 \pi r ^ { 2 }$