A circular ornamental garden pond, of radius 2 metres, has weed starting to grow and cover its surface. As the weed grows, it covers an area of $A$ square metres. A simple model assumes that the weed grows so that the rate of increase of its area is proportional to $A$. 3 (a) Show that the area covered by the weed can be modelled by $$A = Be^{kt}$$ where $B$ and $k$ are constants and $t$ is time in days since the weed was first noticed. 3 (b) When it was first noticed, the weed covered an area of 0.25 m². Twenty days later the weed covered an area of 0.5 m². 3 (b) (i) State the value of $B$: 3 (b) (ii) Show that the model for the area covered by the weed can be written as $$A = rac{1}{2} imes 2^{rac{t}{20}}$$ 3 (b) (iii) How many days does it take for the weed to cover half of the surface of the pond? 3 (c) State one limitation of the model. 3 (d) Suggest one refinement that could be made to improve the model.

A-Level AQA Maths Pure Exponential & Logarithms: A circular ornamental garden pon

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

Question 3

A circular ornamental garden pond, of radius 2 metres, has weed starting to grow and cover its surface. As the weed grows, it covers an area of $A$ square metres. A... show full transcript

Worked Solution & Example Answer: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

Step 1

Show that the area covered by the weed can be modelled by A = Be^{kt}

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Answer

To show that the area can be modelled by the function, we start with the assumption that the rate of growth of the area $A$ is proportional to its current area:

$\frac{dA}{dt} = kA$

This is a separable differential equation. We can rearrange it as:

$\frac{dA}{A} = k \, dt$

Integrating both sides gives:

$\int \frac{dA}{A} = \int k \, dt$

Thus,

$\ln(A) = kt + C$

Exponentiating both sides results in:

$A = e^{kt + C} = e^C e^{kt} = Be^{kt}$

where $B = e^C$ is a constant.

Step 2

State the value of B:

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Answer

$B = 0.25$ m², since at $t = 0$ , $A = 0.25$ m².

Step 3

Show that the model for the area covered by the weed can be written as A = rac{1}{2} imes 2^{rac{t}{20}}

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Answer

From the information given:

When $t = 0$ , $A = 0.25$ m².
When $t = 20$ , $A = 0.5$ m².

We can express $A$ as:

$A = 0.25 \cdot 2^{\frac{t}{20}}$

Therefore, since $A$ jumps from $0.25$ to $0.5$ , we can linearize this as:

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

Step 4

How many days does it take for the weed to cover half of the surface of the pond?

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Answer

The surface area of the pond is: $\text{Area} = \pi r^2 = \pi (2)^2 = 4\pi \text{ m}^2$

Half of this area is: $2\pi \text{ m}^2$

We set the equation $A = \frac{1}{2} \times 2^{\frac{t}{20}}$ equal to $2\pi$ and solve for $t$ :

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

This simplifies to: $2^{\frac{t}{20}} = \pi$

Taking the logarithm of both sides: $\frac{t}{20} = \log_2(\pi)$

Thus, $t = 20 \log_2(\pi) \approx 20 \times 1.585 = 31.7\text{ days}.$

Step 5

State one limitation of the model.

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Answer

The model assumes unlimited resources for the weed growth, which might not hold true in a real-world scenario.

Step 6

Suggest one refinement that could be made to improve the model.

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Answer

One refinement could be to include a limiting factor such as nutrient availability or competition from other plants, which typically slows the growth rate as the area covered increases.

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

Worked Solution & Example Answer: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

Show that the area covered by the weed can be modelled by A = Be^{kt}

State the value of B:

Show that the model for the area covered by the weed can be written as A = rac{1}{2} imes 2^{ rac{t}{20}}

How many days does it take for the weed to cover half of the surface of the pond?

State one limitation of the model.

Suggest one refinement that could be made to improve the model.

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Show that the model for the area covered by the weed can be written as A = rac{1}{2} imes 2^{rac{t}{20}}