Find the approximate change in the volume of a cube of side $x$ meters caused by increasing the side by 3%.

The correct answer is $0.09 x^2 m^3$

Volume of a cube = $V$ = $x^3$

⇒ $\dfrac{dV}{dx}$ = $3x^2$ and $\triangle{x} = 3$% of $x$

∴ $\triangle{V} \approx (3x^2) \triangle{x}$

= $(3x^2) \left( \dfrac{3x}{100} \right)$

= $0.09x^3m^3$


∴ $\triangle{V} \approx 0.09 x^3 m^3$