WAEC Further Mathematics 2016 Paper

With radius = 7cm, $\text{Area} = \pi r^{2} = \frac{22}{7} \times 7^{2}$

= $\mathrm{154\,cm^{2}}$

The next second, radius = 7.5cm, $\text{Area} = \pi r^{2} = \frac{22}{7} \times 7.5^{2}$

= $\mathrm{176\,cm^{2}}$

Change in area = $(176 - 154)\,\mathrm{cm}^{2} = 22\,\mathrm{cm}^{2}$

$\therefore$ The rate of increase = $22\,\mathrm{cm}^{2}\,\mathrm{s}^{-1}$

OR

$\text{Area }(A) = \pi r^{2} \Rightarrow \frac{dA}{dr} = 2\pi r$

Given $\frac{dr}{dt} = 0.5$

$\frac{dA}{dr} \times \frac{dr}{dt} = \frac{dA}{dt}$

$\frac{dA}{dt} = 2\pi r \times 0.5 = 2 \times \frac{22}{7} \times 7 \times 0.5$

= $22\,\mathrm{cm}^{2}\,\mathrm{s}^{-1}$