The volume of a cube is increasing at a rate of 9 cubic centimetres per second. How fast is the surface area increasing when the length of an edge is 10 centimetres?

$3.6 \text{ cm}^2/\text{s}$

The correct answer is Option (2) → $3.6 \text{ cm}^2/\text{s}$ ##

Let $x$ be the length of a side, $V$ be the volume and $S$ be the surface area of the cube. Then, $V = x^3$ and $S = 6x^2$, where $x$ is a function of time $t$.

Now $\frac{dV}{dt} = 9 \text{cm}^3/\text{s (Given)}$

Therefore $9 = \frac{dV}{dt} = \frac{d}{dt}(x^3) = \frac{d}{dx}(x^3) \cdot \frac{dx}{dt} \quad \text{(By Chain Rule)}$

$= 3x^2 \cdot \frac{dx}{dt}$

or $\frac{dx}{dt} = \frac{3}{x^2} \quad \dots (1)$

Now $\frac{dS}{dt} = \frac{d}{dt}(6x^2) = \frac{d}{dx}(6x^2) \cdot \frac{dx}{dt} \quad \text{(By Chain Rule)}$

$= 12x \cdot \left( \frac{3}{x^2} \right) = \frac{36}{x} \quad \text{(Using (1))}$

Hence, when $x = 10 \text{ cm}$, $\frac{dS}{dt} = 3.6 \text{ cm}^2/\text{s}$