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 the initial area covered by the weed was 0.25 m² when t = 0,

we can substitute:

$A = 0.25 ext{ when } t = 0$

Thus, the equation becomes:

From the initial condition, we know:

$A = 0.25\text{ m²} ext{ when } t = 0,$ therefore,

$A = 0.25e^{kt}$.

After 20 days, the area is 0.5 m²:

$0.5 = 0.25e^{20k}$

Dividing both sides by 0.25 gives:

$2 = e^{20k}$

Taking the natural logarithm:

$\ln(2) = 20k \Rightarrow k = \frac{\ln(2)}{20}$

Substituting this value of k back into the area equation, we have:

$A = 0.25e^{\frac{\ln(2)}{20}t}$

This can be rewritten as:

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

The surface area of the pond is:

$A_{pond} = \pi r^2 = \pi (2)^2 = 4\pi \text{ m²}$

Half of the surface area is:

$\frac{1}{2} \times 4\pi = 2\pi ext{ m²}$

Setting the area equation equal to half the pond's area and solving for t:

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

Multiplying both sides by 4:

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

Taking logarithm:

$\frac{t}{20} \ln(2) = \ln(8\pi)$

Thus, we have:

$t = 20 \cdot \frac{\ln(8\pi)}{\ln(2)}$

Calculating the value gives approximately:

t ≈ 93.03 days.

One limitation of the model is that it assumes the rate of growth of the weed continues indefinitely, without considering factors such as resource limitations or the impact of competing vegetation.

A potential refinement could include incorporating a limiting factor, such as the decrease in the rate of growth as the area covered by the weed increases, to better reflect real-world scenarios.