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Q.
The approximate change in the volume of a cube of side $x m$ caused by increasing the side by $3 \%$ is
Application of Derivatives
Solution:
We know that, the volume $V$ of a cube of side $x$ is given by $V=x^3$
$\Rightarrow \frac{d V}{d x}=3 x^2$
Let $\Delta x$ be change in side $=3 \%$ of $x=0.03 x$
Now, change in volume, $\Delta V=\left(\frac{d V}{d x}\right) \Delta x$
$=\left(3 x^2\right) \Delta x=\left(3 x^2\right)(0.03 x)$ (as $\Delta x=3 \%$ of $x$ is $\left.0.03 x\right)$
$=0.09 x^3 m ^3$
Hence, the approximate change in the volume of the cube is $0.09 x^3 m ^3$.
$\therefore$ The correct option is (c).