Problem

Figure 1 shows a cube which is increasing in size.

At time $t$ seconds,

• the length of each edge of the cube is $x$ cm
• the surface area of the cube is $S$ cm$^2$
• the volume of the cube is $V$ cm$^3$

Given that the surface area of the cube is increasing at a constant rate of $4$ cm$^2$ s$^{-1}$

(a) show that

$$\frac{dx}{dt}=\frac{k}{x}$$

where $k$ is a constant to be found,

(b) show that

$$\frac{dV}{dt}=V^p$$

where $p$ is a constant to be found.