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Q.
If the radius of a sphere is measured as $8\, cm$ with an error of $0.03\, cm$, then the approximate error in calculating its volume is
Application of Derivatives
Solution:
Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius.
Then $r =8 \,cm$, and $\Delta r =0.03 \,cm$,
Now the volume $V$ of the sphere is given by
$V =\frac{4}{3} \pi r ^3$
$\frac{ dV }{ dr }=4 \pi r ^2 $
$ \Delta V =\left(\frac{ dV }{ dr }\right) \Delta r =\left(4 \pi r ^2\right) \Delta r$
$ =4 \pi(8)^2 \times 0.03=7.68 \, \pi cm ^3$