# The volume of a cube is increasing at the rate of 8 cm3/s.

Question:

The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Solution:

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 $S=6 x^{2}$ where $x$ is a function of time $t$.

It is given that $\frac{d V}{d t}=8 \mathrm{~cm}^{3} / \mathrm{s}$.

Then, by using the chain rule, we have:

$\therefore 8=\frac{d V}{d t}=\frac{d}{d t}\left(x^{3}\right)=\frac{d}{d x}\left(x^{3}\right) \cdot \frac{d x}{d t}=3 x^{2} \cdot \frac{d x}{d t}$

$\Rightarrow \frac{d x}{d t}=\frac{8}{3 x^{2}}$  ...(1)

Now, $\frac{d \mathrm{~S}}{d t}=\frac{d}{d t}\left(6 x^{2}\right)=\frac{d}{d x}\left(6 x^{2}\right) \cdot \frac{d x}{d t}$  [By chain rule]

$=12 x \cdot \frac{d x}{d t}=12 x \cdot\left(\frac{8}{3 x^{2}}\right)=\frac{32}{x}$

Thus, when $x=12 \mathrm{~cm}, \frac{d S}{d t}=\frac{32}{12} \mathrm{~cm}^{2} / \mathrm{s}=\frac{8}{3} \mathrm{~cm}^{2} / \mathrm{s}$.

Hence, if the length of the edge of the cube is $12 \mathrm{~cm}$, then the surface area is increasing at the rate of $\frac{8}{3} \mathrm{~cm}^{2} / \mathrm{s}$.