Long-term behaviour analysis

1. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

A population of fruit flies is being studied.
The number of fruit flies, \(F\), in the population, \(t\) days after the start of the study, is modelled by the equation $$F = \frac { 350 \mathrm { e } ^ { k t } } { 9 + \mathrm { e } ^ { k t } }$$ where \(k\) is a constant.
Use the equation of the model to answer parts (a), (b) and (c).

1. Find the number of fruit flies in the population at the start of the study. Given that there are 200 fruit flies in the population 15 days after the start of the study,
2. show that \(k = \frac { 1 } { 15 } \ln 12\) Given also that, when \(t = T\), the number of fruit flies in the population is increasing at a rate of 10 per day,
3. find the possible values of \(T\), giving your answers to one decimal place.

9. \begin{figure}[h] \end{figure} The population of a species of animal is being studied. The population \(P\), at time \(t\) years from the start of the study, is assumed to be $$P = \frac { 9000 \mathrm { e } ^ { k t } } { 3 \mathrm { e } ^ { k t } + 7 } , \quad t \geqslant 0$$ where \(k\) is a positive constant.
A sketch of the graph of \(P\) against \(t\) is shown in Figure 2 .
Use the given equation to

1. find the population at the start of the study,
2. find the value for the upper limit of the population. Given that \(P = 2500\) when \(t = 4\)
3. calculate the value of \(k\), giving your answer to 3 decimal places. Using this value for \(k\),
4. find, using \(\frac { \mathrm { d } P } { \mathrm {~d} t }\), the rate at which the population is increasing when \(t = 10\) Give your answer to the nearest integer.

The growth of a tree is modelled by the differential equation $$10\frac{dh}{dt} = 20 - h,$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).

1. Write down the value of \(h\) for which \(\frac{dh}{dt} = 0\), and interpret this in terms of the growth of the tree. [1]
2. Verify that \(h = 20(1 - e^{-0.1t})\) satisfies this differential equation and its initial condition. [5]

The alternative differential equation $$200\frac{dh}{dt} = 400 - h^2$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).

1. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac{20(1 - e^{-0.2t})}{1 + e^{-0.2t}}.$$ [9]
2. What does this solution indicate about the long-term height of the tree? [1]
3. After a year, the tree has grown to a height of 2 m. Which model fits this information better? [3]

The growth of a tree is modelled by the differential equation $$10\frac{dh}{dt} = 20 - h$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).

1. Write down the value of \(h\) for which \(\frac{dh}{dt} = 0\), and interpret this in terms of the growth of the tree. [1]
2. Verify that \(h = 20(1 - e^{-0.1t})\) satisfies this differential equation and its initial condition. [5]

The alternative differential equation $$200\frac{dh}{dt} = 400 - h^2$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).

1. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac{20(1 - e^{-0.2t})}{1 + e^{-0.2t}}$$ [9]
2. What does this solution indicate about the long-term height of the tree? [1]
3. After a year, the tree has grown to a height of 2m. Which model fits this information better? [3]