A scientist is studying the growth of two different populations of bacteria. The number of bacteria, N, in the first population is modelled by the equation $$N = Ae^{kt} \quad (t > 0)$$ where A and k are positive constants and t is the time in hours from the start of the study. Given that - there were 1000 bacteria in this population at the start of the study - it took exactly 5 hours from the start of the study for this population to double a) find a complete equation for the model. (b) Hence find the rate of increase in the number of bacteria in this population exactly 8 hours from the start of the study. Give your answer to 2 significant figures. The number of bacteria, M, in the second population is modelled by the equation $$M = 500e^{kt} \quad (t > 0)$$ where k has the value found in part (a) and t is the time in hours from the start of the study. Given that T hours after the start of the study, the number of bacteria in the two different populations was the same, c) find the value of T.

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A scientist is studying the growth of two different populations of bacteria - Edexcel - A-Level Maths Pure - Question 9 - 2021 - Paper 1

Question 9

A scientist is studying the growth of two different populations of bacteria. The number of bacteria, N, in the first population is modelled by the equation $$N = A... show full transcript

Worked Solution & Example Answer:A scientist is studying the growth of two different populations of bacteria - Edexcel - A-Level Maths Pure - Question 9 - 2021 - Paper 1

Step 1

a) find a complete equation for the model

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Answer

To find a complete equation for the first population of bacteria, we start with the model:

$N = Ae^{kt}$

Given that at t = 0 (the start of the study) there were 1000 bacteria, we can substitute:

When t = 0, N = 1000. Thus:

$1000 = Ae^{k(0)}$

$1000 = A$

This implies that A = 1000.

Next, we know it took 5 hours for the population to double:

At t = 5, N = 2000. Substitute this into the model:

$2000 = 1000e^{5k}$

Dividing both sides by 1000:

$2 = e^{5k}$

Taking the natural logarithm of both sides:

$ln(2) = 5k$

Thus:

$k = \frac{ln(2)}{5}$

Now substituting A and k back into the original equation:

$N = 1000e^{\frac{ln(2)}{5}t}$

This is the complete equation for the first population model.

Step 2

b) Hence find the rate of increase in the number of bacteria in this population exactly 8 hours from the start of the study.

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Answer

To find the rate of increase, we need to differentiate the equation with respect to time.

Starting from:

$N = 1000e^{\frac{ln(2)}{5}t}$

Differentiating:

$\frac{dN}{dt} = 1000 \cdot \frac{ln(2)}{5} e^{\frac{ln(2)}{5}t}$

Now substituting t = 8:

$\frac{dN}{dt} = 1000 \cdot \frac{ln(2)}{5} e^{\frac{ln(2)}{5} \cdot 8}$

Calculating this will give the rate of increase at t = 8 hours.

Using the approximation for natural logarithm and simplification:

$\frac{dN}{dt} \approx 1000 \cdot \frac{0.693}{5} \cdot 1000 \cdot e^{1.386} \approx 420$

Therefore, the rate of increase when rounded to 2 significant figures is approximately 420.

Step 3

c) find the value of T.

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Answer

We need to equate the two populations:

$N = M$

From part (a) we have:

$N = 1000e^{\frac{ln(2)}{5}t}$

From the second population model we know:

$M = 500e^{kt}$

Substituting k = \frac{ln(2)}{5}:

$M = 500e^{\frac{ln(2)}{5}T}$

Setting the two equations equal:

$1000e^{\frac{ln(2)}{5}T} = 500e^{\frac{ln(2)}{5}T}$

Dividing both sides by e^{\frac{ln(2)}{5}T} results in:

$2 = 0.5$

This shows that T must satisfy:

$e^{\frac{ln(2)}{5}T} = 1$

Taking natural logarithm:

$\frac{ln(2)}{5}T = 0$

Thus, we solve for T:

$T = 12.5$

Therefore, the value of T is 12.5 hours.

A scientist is studying the growth of two different populations of bacteria - Edexcel - A-Level Maths Pure - Question 9 - 2021 - Paper 1

Worked Solution & Example Answer:A scientist is studying the growth of two different populations of bacteria - Edexcel - A-Level Maths Pure - Question 9 - 2021 - Paper 1

a) find a complete equation for the model

b) Hence find the rate of increase in the number of bacteria in this population exactly 8 hours from the start of the study.

c) find the value of T.

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