The approximate change in the volume

Solution:

Let ∆x be the change in side x and ∆V be the change in the volume of the cube.

It is given that, $\frac{\Delta x}{x} \times 100=2$             .....(1)

Now,

Volume of the cube of side $x, V=x^{3}$

$V=x^{3}$

Differentiating both sides with respect to x, we get

$\frac{d V}{d x}=3 x^{2}$

$\therefore \Delta V=\left(\frac{d V}{d x}\right) \Delta x$

$\Rightarrow \Delta V=3 x^{2} \Delta x$

$\Rightarrow \Delta V=3 x^{2} \times \frac{2 x}{100}$      [Using (1)]

$\Rightarrow \Delta V=\frac{6 x^{3}}{100}$

 

$\Rightarrow \Delta V=0.06 x^{3}$

Thus, the approximate change in volume of the cube is $0.06 x^{3} \mathrm{~m}^{3}$.

The approximate change in the volume of a cube of side $x$ metres caused by increasing the side by $2 \%$, is $0.06 x^{3}-m^{3}$