Browse Questions

# The approximate change in the volume of a cube of side $x$ metres caused by increasing the side by $3\%$ is

$(A)\; 0.06 x^3 \: m^3 \quad (B)\; 0.6 x^3 \: m^3 \quad (C)\; 0.09 x^3 \: m^3 \quad (D)\; 0.9 x^3 \: m^3$

Toolbox:
• Let $y=f(x)$
• $\Delta y=f(x+\Delta x)-f(x)$
• $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
• Volume of the cube=$x^3m^3$
Step 1:
Side of a cube =$x$
Increase in the side=3%
Volume of the cube=$x^3m^3$
$\Delta x$=3% of $x=0.03x$
$\Delta V$ be the change in volume.
Step 2:
$\Delta V=\large\frac{dV}{dx}$$\times \Delta x----(1) V=x^3 \large\frac{dV}{dx}=$$3x^2$[Differentiating with respect to x]
Step 3:
Substitute the value of $\large\frac{dV}{dx}$ in equation (1) we get,
$\Delta V=3x^2\times 0.03x$
$\quad\;\; =0.09x^3m^3$
$\Delta V=0.09x^3m^3$
Part (C) is the correct answer.
edited Aug 7, 2013