Figure 1 shows a metal cube which is expanding uniformly as it is heated - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 7

We know from the problem that the volume V is increasing at a rate of ( \frac{dV}{dt} = 0.048 ) cm³s⁻¹.

Using the chain rule, we can express this as: $\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt}$

At ( x = 8 ), we substitute into the volume derivative: $\frac{dV}{dx} = 3(8)^2 = 192$

Now we set up the equation: $0.048 = 192 \cdot \frac{dx}{dt}$

Solving for ( \frac{dx}{dt} ), we have: $\frac{dx}{dt} = \frac{0.048}{192} = 0.00025 \, (cm/s)$.

The surface area S of a cube is given by: $S = 6x^2$

To find the rate of change of surface area with respect to time, we differentiate S: $\frac{dS}{dt} = \frac{dS}{dx} \cdot \frac{dx}{dt}$

Calculating ( \frac{dS}{dx} ): $\frac{dS}{dx} = 12x$

At ( x = 8 ): $\frac{dS}{dx} = 12(8) = 96$

Substituting ( \frac{dx}{dt} = 0.00025 ): $\frac{dS}{dt} = 96 \cdot 0.00025 = 0.024 \, (cm²/s)$.