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. (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. (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. (i) State the value of B. (ii) Show that the model for the area covered by the weed can be written as $$A = \frac{1}{2} \cdot 2^{\frac{t}{20}}$$ (iii) How many days does it take for the weed to cover half of the surface of the pond? (c) State one limitation of the model. (d) Suggest one refinement that could be made to improve the model.

A-Level AQA Maths Mechanics Quantities, Units & Modelling: 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 Mechanics - 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 s... 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 Mechanics - 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 model the growth of the weed, we start with the assumption that the rate of increase of the area A is proportional to A itself: $\frac{dA}{dt} = kA$ where k is a constant. We can separate the variables: $\frac{1}{A} dA = k \, dt$ Integrating both sides gives: $\ln A = kt + C$ Exponentiating both sides leads to: $A = e^{kt + C} = e^C e^{kt}$ Letting (B = e^C), we rewrite this as: $A = Be^{kt}$

Step 2

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.

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Answer

From the problem statement, when t = 0, A = 0.25 m². Thus, $B = 0.25$ Therefore, we have: $B = 0.25$

Step 3

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

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Answer

Using the two points of data:

When t = 0, A = 0.25.
When t = 20, A = 0.5. Now substituting these values: For t = 0: $0.25 = 0.25e^{0}$ This fits our model. Now for t = 20: $0.5 = 0.25e^{20k}$ Dividing both sides by 0.25: $2 = e^{20k}$ Taking natural logs gives: $\ln 2 = 20k \Rightarrow k = \frac{\ln 2}{20}$ Thus, we can rewrite the model as: $A = 0.25e^{kt} = 0.25e^{\frac{t \ln 2}{20}} = \frac{1}{2} \cdot 2^{\frac{t}{20}}$

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, which is circular with a radius of 2 m, is: $\text{Area} = \pi r^2 = \pi (2)^2 = 4\pi \approx 12.57 m²$ To find out when the weed covers half of this area: $A = \frac{1}{2} \cdot 12.57 = 6.285 m²$ Setting our model equal to this area: $\frac{1}{2} \cdot 2^{\frac{t}{20}} = 6.285$ Therefore, $2^{\frac{t}{20}} = 12.57$ Taking logs: $\frac{t}{20} \ln(2) = \ln(12.57)$ Solving for t: $t = 20 \cdot \frac{\ln(12.57)}{\ln(2)} \approx 93.03 \, \text{days}$

Step 5

State one limitation of the model.

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Answer

One limitation of the model is that it assumes constant growth rates, which may not be realistic over time due to environmental factors.

Step 6

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

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Answer

One refinement could be to introduce a limiting factor such as nutrient availability, which decreases as the area covered by the weed increases.

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

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

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.

Show that the model for the area covered by the weed can be written as A = \frac{1}{2} \cdot 2^{\frac{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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