A2 June 2021 Paper 2 Q10
10 In a colony of seabirds, there are \(y\) birds at time \(t\) years.
In one year the reduction in the number of birds due to birds dying or leaving the colony is equal to 16% of the number of birds at the start of the year.
If no birds are born or join the colony, find the constant \(k\) such that
\[\frac{\mathrm{d}y}{\mathrm{d}t} = -ky\]Give your answer to three significant figures. [4 marks]
The rate of reduction in the number of birds due to birds dying or leaving the colony is the same as in part (a), but in addition:
- The rate of increase in the number of birds due to births is \(20t\) per year.
- The wildlife protection group brings 45 birds into the colony each year.
Write down a first-order differential equation for \(y\) and \(t\) [2 marks]
Solve your differential equation from part (b) to find \(y\) in terms of \(t\) [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Separates the variables | M1 | 1.1a |
| Deduces exponential form | M1 | 2.2a |
| Forms correct equation | M1 | 3.3 |
| Obtains correct answer | A1 | 1.1b |
| (4) |
Typical solution
\[\frac{\mathrm{d}y}{\mathrm{d}t} = -ky\]\[\int \frac{1}{y}\,\mathrm{d}y = \int -k\,\mathrm{d}t\]So \(y = y_0\mathrm{e}^{-kt}\)
\[0.84y_0 = y_0\mathrm{e}^{-k}\]\[k = 0.174\]| Scheme | Marks | AO |
|---|---|---|
| Forms DE with three terms and \(\frac{\mathrm{d}y}{\mathrm{d}t}\) | M1 | 3.3 |
| Obtains correct DE (consistent with their answer to part (a)) may use \(k\) in place of 0.174 | A1F | 3.3 |
| (2) |
Typical solution
\[\frac{\mathrm{d}y}{\mathrm{d}t} = -0.174y + 45 + 20t\]| Scheme | Marks | AO |
|---|---|---|
| Splits into CF and PI Or Finds an integrating factor | M1 | 3.1a |
| Obtains PI of the form \(p + qt\) Or Uses integration by parts to integrate the \(t\mathrm{e}^{kt}\) term | M1 | 1.1a |
| Obtains correct general solution from their value of \(k\) | A1F | 1.1b |
| Uses initial conditions to find the unknown constant for a general solution that includes an exponential term | M1 | 3.3 |
| Obtains correct solution | A1 | 1.1b |
| (5) |
Typical solution
\[\frac{\mathrm{d}y}{\mathrm{d}t} + 0.174y = 45 + 20t\]CF: \(y = A\mathrm{e}^{-0.174t}\)
PI: \(y = p + qt\)
\[\dot{y} = q\]\[q + 0.174p + 0.174qt = 45 + 20t\]\[q = \frac{20}{0.174} = 115\]\[p = \frac{45 - q}{0.174} = -402\]General solution:
\[y = A\mathrm{e}^{-0.174t} - 402 + 115t\]\[t = 0 \Rightarrow y = 340\]So \(A = 742\)
\[y = 742\mathrm{e}^{-0.174t} - 402 + 115t\]| Scheme | Marks | AO |
|---|---|---|
| One reasonable limitation of their model for limitation 1 | E1 | 3.5b |
| One reasonable limitation of their model for limitation 2 | E1 | 3.5b |
| (2) | ||
| (13 marks) |
Typical solution
Births would occur at a particular time of year, not at a steady rate
Over a long period of time the population would increase indefinitely according to the model