A circular ornamental garden pond, of radius 2 metres, has weed starting to grow and cover its surface - AQA - A-Level Maths Pure - Question 3 - 2019 - Paper 3

Given that at time $t = 0$, the area $A = 0.25$ m², we find:

$B = 0.25.$

Using the information that at $t = 20$, the area is $0.5$ m², we set up the equation:

$0.5 = 0.25 e^{20k}.$

Dividing both sides by $0.25$ gives:

$2 = e^{20k}$

Taking the natural logarithm on both sides yields:

$\ln 2 = 20k \Rightarrow k = \frac{\ln 2}{20}.$

Substituting $k$ back into the original equation gives:

$A = 0.25 e^{\frac{t \ln 2}{20}} = \frac{1}{4} \times 2^{\frac{t}{20}}.$

To find the time when the area $A$ equals half the surface of the pond, we first calculate the total surface area of the pond:

$\text{Area}_{pond} = \pi r^2 = \pi (2)^2 = 4\pi.$

Half of this area is $2\pi$. Setting $A = 2\pi$ in the model:

$2\pi = \frac{1}{4} \times 2^{\frac{t}{20}}$

Multiplying both sides by 4 gives:

$8\pi = 2^{\frac{t}{20}}.$

Taking logarithms:

$\frac{t}{20} = \log_2 (8\pi) = 3 + \log_2 (\pi).$

Therefore, we find:

$t = 20(3 + \log_2 (\pi)) \approx 93.03 \text{ days}.$

One limitation of the model is that it assumes the growth of the weed continues indefinitely and does not account for environmental factors such as competition for resources, which may limit the growth.

A refinement could include introducing a limiting factor such as a maximum growth rate as the surface area becomes increasingly covered, which would lead to a decrease in the rate of growth of the weed.