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If the radius of a sphere is measured as $9 \mathrm{~cm}$ with an error of $0.03 \mathrm{~cm}$, then find the approximating error in calculating its volume.
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Verified Answer
The correct answer is:
$9.72 \pi \mathrm{cm}^{3}$
Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius. Then, $r=9 \mathrm{~cm}$ and $\Delta r=0.03 \mathrm{~cm}$
Let $V$ be the volume of the sphere. Then,
$\begin{array}{l}
\mathrm{V}=\frac{4}{3} \pi r^{3} \\
\Rightarrow \frac{\mathrm{dV}}{\mathrm{dr}}=4 \pi r^{2} \\
\Rightarrow\left(\frac{\mathrm{dV}}{\mathrm{dr}}\right)_{r=9}=4 \pi \times 9^{2}=324 \pi
\end{array}$
Let $\Delta \mathrm{V}$ be the error in $\mathrm{V}$ due to error $\Delta \mathrm{r}$ in $r$.
Then,
$\begin{array}{l}
\Delta \mathrm{V}=\frac{\mathrm{dV}}{\mathrm{dr}} \Delta r \\
\Rightarrow \Delta \mathrm{V}=324 \pi \times 0.03=9.72 \pi \mathrm{cm}^{3}
\end{array}$
Let $V$ be the volume of the sphere. Then,
$\begin{array}{l}
\mathrm{V}=\frac{4}{3} \pi r^{3} \\
\Rightarrow \frac{\mathrm{dV}}{\mathrm{dr}}=4 \pi r^{2} \\
\Rightarrow\left(\frac{\mathrm{dV}}{\mathrm{dr}}\right)_{r=9}=4 \pi \times 9^{2}=324 \pi
\end{array}$
Let $\Delta \mathrm{V}$ be the error in $\mathrm{V}$ due to error $\Delta \mathrm{r}$ in $r$.
Then,
$\begin{array}{l}
\Delta \mathrm{V}=\frac{\mathrm{dV}}{\mathrm{dr}} \Delta r \\
\Rightarrow \Delta \mathrm{V}=324 \pi \times 0.03=9.72 \pi \mathrm{cm}^{3}
\end{array}$
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