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Q.
If the radius of a sphere is measured as $9\, cm$ with an error of $0.03 \,cm$, then find the approximating error in calculating its volume.
Application of Derivatives
Solution:
We have, $r = 9\, cm$ and $\Delta r = 0.03\, cm$
Let $V$ be the volume of the sphere. Then,
$V = \frac{4}{3}\pi r^{3}$
$\Rightarrow \frac{dV}{dr} = 4\pi r^{2}$
$\Rightarrow \left(\frac{dV}{dr}\right)_{r = 9} = 4\pi \times 9^{2} = 324\,\pi$
Let $\Delta V$ be the error in $V$ due to error $\Delta r$ in $ r$. Then,
$\Delta V = \frac{dV}{dr}\Delta r$
$\Rightarrow \Delta V = 324\pi \times 0.03 = 9.72\, \pi\, cm^{3}$
Thus, the approximate error in calculating the volume is $9.72\, \pi \, cm^{3}$.