Question

If the surface area of a cube is increasing at a rate of $3.6{\text{cm}}^2/\text{sec}$, retaining its shape; then the rate of change of its volume (in ${\mathrm{cm}}^3/\sec$), when the length of a side of the cube is $10\text{cm}$, is

• A. $9$
• B. $10$
• C. $18$
• D. $20$

Answer 1

The correct option is A

$9$

Explanation for the correct option:

Step 1. Find the value of $\frac{da}{dt}$ at $a=10\text{cm}$.

Let $a$ be the side length of the cube, then the surface area is given as: $S=6a^2$. Now, surface area is increasing at a rate of $3.6{\text{cm}}^2/\text{sec}$, so $\frac{dS}{dt}=3.6$, now differentiate $S=6a^2$ and find the rate of change of its side length at $a=10\text{cm}$.

$$\begin{gathered} \frac{dS}{dt}=\frac{d}{dt}\left(6a^2\right) \\ \Rightarrow \frac{dS}{dt}=6\times 2a\frac{da}{dt} \\ \Rightarrow 3.6=12\times 10\frac{da}{dt} \\ \Rightarrow 0.03=\frac{da}{dt} \end{gathered}$$

Step 2. Find the rate of change of volume.

The volume of the cube of side length $a$ is $V=a^3$. Now differentiate it to find the rate of change of volume at $a=10\text{cm}$.

$\begin{array}{rcl}\frac{dV}{dt}& =& \frac{d}{dt}\left(a^3\right)\\ & =& 3a^2\frac{da}{dt}\\ & =& 3\times {10}^2\times 0.03\\ & =& 9\end{array}$

Hence, the correct option is (A).