Question

The volume of a cube is increasing at the rate of $18 \, \text{cm}^{3}$ per second. When the edge of the cube is $12 \, cm,$ then the rate in $cm^{2}/s$ at which the surface area of the cube increases, is

Answer 1

Let, $x$ be the length of an edge, $V$ be the volume and $S$ be the surface area of the cube.
$\frac{d V}{d t}=18,V=x^{3}\Rightarrow \frac{d V}{d t}=18=3x^{2}\frac{d x}{d t}\Rightarrow \frac{d x}{d t}=\frac{6}{x^{2}}\ldots \left(i\right)$
$S=6x^{2}$
$\Rightarrow \frac{d S}{d t}=12x\frac{d x}{d t}\ldots \left[\text{From} \left(i\right)\right]$
$\Rightarrow \frac{d S}{d t}=12x\frac{6}{x^{2}}$
$\Rightarrow \frac{d S}{d t}=12\times \frac{6}{x}=6 \, \left[\because \, \, \text{Given } x = 12\right]$
$\therefore \frac{d S}{d t} = 6 \, \text{cm}^{\text{2}} \text{/sec}$