June 2025 Paper 2 Q16
16 In this question you must show detailed reasoning
The population of puffins on a remote island is estimated to be \(P = 58.5\), where \(P\) is measured in thousands of birds.
Just after this estimate is made there is a serious oil spill in the area. It is noted that there is a rapid decline in the population of puffins on the island.
The situation is modelled by the differential equation
\(\dfrac{\mathrm{d}P}{\mathrm{d}t} = (72t - 108)\mathrm{e}^{-0.8t}\), where \(t\) is the time in years after the pollution incident.
| Scheme | Marks | AO |
|---|---|---|
| DR \(\int \mathrm{d}P = \int (72t - 108)\mathrm{e}^{-0.8t}\,\mathrm{d}t\) oe | B1 | 3.1a |
| \(\dfrac{-108}{-0.8}\mathrm{e}^{-0.8t}\) oe seen | B1 | 2.1 |
| \(72\left[t\dfrac{\mathrm{e}^{-0.8t}}{-0.8} - \displaystyle\int \dfrac{\mathrm{e}^{-0.8t}}{-0.8}\,\mathrm{d}t\right]\) oe | M1* | 1.1 |
| A1 | 1.1 | |
| \([P =]\ -90t\mathrm{e}^{-0.8t} - 112.5\mathrm{e}^{-0.8t} + 135\mathrm{e}^{-0.8t}\ [+c]\) oe | A1 | 1.1 |
| \(58.5 = -90\times 0\times \mathrm{e}^0 + 22.5\mathrm{e}^0 + c\) oe | M1dep* | 3.3 |
| \(P = 36 + 22.5\mathrm{e}^{-0.8t} - 90t\mathrm{e}^{-0.8t}\) | A1 | 1.1 |
| [7] |
Notes
DR: This question included the instruction: In this question you must show detailed reasoning.
B1: separation of variables; condone omission of integral signs or of \(\mathrm{d}t\); allow \(P\) instead of \(\int \mathrm{d}P\) on left hand side
M1*: integration by parts; allow sign errors only
A1: all correct;
A1: allow omission of \(+\,c\); allow unsimplified
M1dep*: may be awarded after rearrangement; must see constant of integration;
NB if expression incorrect, need to see full substitution for M1
A1: allow just \(c = 36\) as long as “\(P\) =” seen following completion of integration; allow eg \(P = 36 + 45\left(\frac{1}{2}e^{-0.8t} - 2te^{-0.8t}\right)\); like terms must be collected and fractions must be in lowest terms
Alternative method
| Scheme | Marks |
|---|---|
| \(\int \mathrm{d}P = \int (72t - 108)\mathrm{e}^{-0.8t}\,\mathrm{d}t\) oe | B1 |
| \(\dfrac{\mathrm{e}^{-0.8t}}{-0.8}\) oe seen | B1 |
| \((72t - 108)\dfrac{\mathrm{e}^{-0.8t}}{-0.8} - \displaystyle\int 72\dfrac{\mathrm{e}^{-0.8t}}{-0.8}\,dt\) | M1* A1 |
| \([P =]\ -90te^{-0.8t} - 112.5e^{-0.8t} + 135e^{-0.8t} + c\) | A1 |
| \(58.5 = -90\times 0\times \mathrm{e}^0 + 22.5\mathrm{e}^0 + c\) oe | M1dep* |
| \(P = 36 + 22.5\mathrm{e}^{-0.8t} - 90t\mathrm{e}^{-0.8t}\) | A1 |
B1: separation of variables; condone omission of integral signs or of \(\mathrm{d}t\); allow \(P\) instead of \(\int \mathrm{d}P\) on left hand side
B1: may be embedded in integration by parts
M1* A1: integration by parts; allow sign errors only
all correct
A1: all correct; allow omission of \(+\,c\); allow unsimplified
M1dep*: may be awarded after rearrangement; must see constant of integration
A1: allow just \(c = 36\) as long as “\(P\) =” seen following completion of integration; allow eg \(P = 36 + 45\left(\frac{1}{2}e^{-0.8t} - 2te^{-0.8t}\right)\); like terms must be collected and fractions must be in lowest terms
| Scheme | Marks | AO |
|---|---|---|
| DR as \(t \to \infty,\ (k)\mathrm{e}^{-0.8t} \to 0\) | M1 | 3.4 |
| so \(P \to 36\). [Hence the model predicts that the population of puffins] does not recover to 58.5 thousand | A1 | 1.1 |
| [2] |
Notes
M1: or \(\displaystyle\lim_{t\to\infty} (k)\mathrm{e}^{-0.8t} = 0\); \(k\) is a constant;
allow eg as \(t\) tends to infinity / as \(t\) becomes very large / in the long term, \((k)\mathrm{e}^{-0.8t}\) tends to / converges to / approaches zero oe
do not allow
eg substitution of \(t = \infty\)
A1: condone \(P = 36\)