June 2024 Paper 2 Q5
5 A scientist is monitoring the decline in the population of a certain endangered species of animal in an area where their natural habitat has been damaged.
As a model, the scientist proposes that the rate of decline per year of the population is given by \(\dfrac{1}{80}P^2\), where \(P\) is the size of the population \(t\) years after the start of the modelling.
The scientist notes that at the start of the monitoring the population is 120.
The model predicts that the population will never reach zero.
| Scheme | Marks | AO |
|---|---|---|
| (The rate of change with respect to time is \(\frac{\mathrm{d}P}{\mathrm{d}t}\)), which is \(-\frac{1}{80}P^2\) because the population is in decline. | B1 | 3.3 |
| [1] |
Notes
B1: Must see use of decrease/decline linked to the negative sign.
Condone answers that do not refer to \(\frac{\mathrm{d}P}{\mathrm{d}t}\)
Examples:
- “it’s \(-\frac{1}{80}P^2\) because decreasing” B1
- “decreasing” [and nothing further] B0
- “decline implies negative sign” B1
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}P}{P^2} = -\dfrac{\mathrm{d}t}{80}\) oe | M1* | 3.4 |
| \(-\dfrac{1}{P} = -\dfrac{t}{80}\,(+c)\) oe | A1 | 1.1 |
| \(c = -\dfrac{1}{120}\) oe \(\left(\dfrac{1}{P} = \dfrac{3t + 2}{240}\right)\) | M1 dep* | 2.1 |
| \(P = \dfrac{240}{3t + 2}\) oe | A1 | 1.1 |
| [4] |
Notes
M1*: Attempt to separate variables, must see \(P\) and d\(P\) on the same side
May see \(\frac{\mathrm{d}t}{\mathrm{d}P} = \frac{-80}{P^2}\) as a first step. In this case M1 is implied by a subsequent attempt to integrate the RHS w.r.t. \(P\) (so do not give M1 for this statement alone).
A1: May see \(\frac{1}{P} = \frac{t}{80}\,(+c)\) etc. Allow without \(+c\)
M1 dep*: Substitute (0, 120) and attempt to find \(c\) (which may not be this value). Must reach a value of \(c\) for this mark.
A1: Must be in terms of \(P\) (\(P\)=…) but need not be simplified.
isw incorrect attempts to simplify following \(P\)=…
| Scheme | Marks | AO |
|---|---|---|
| \(10 = \dfrac{240}{3t + 2}\) or \(t = \dfrac{80}{P} - \dfrac{2}{3}\) | M1 | 3.4 |
| \(t = 7\tfrac{1}{3}\) (years) oe | A1 | 1.1 |
| [2] |
Notes
M1: Either substitute \(P = 10\) into their equation from (b) or rearrange to make \(t\) the subject.
A1: Accept 7.33 (3sf), ignore units.
| Scheme | Marks | AO |
|---|---|---|
| ‘for large \(t\), \(P\) is small’ or \(t \geqslant 160 \Rightarrow P \lt 1\) (\(P\) is modelled as continuous but in fact the number of animals is discrete), at this time the actual population would be 0. | B1 | 3.5a |
| [1] |
Notes
B1: Allow \(t = 160 \Rightarrow P = 0.5\)
(Use of the value \(t\)=160 not required)
Any correct statement in context relating to the model predicting non-integer values between 0 and 1. (Note that the statement ‘will never reach zero’ is given in the question so gains no credit).
Acceptable examples (for a comment, in combination with the statement that \(P\) is small for large \(t\)):
- “cannot have part of an animal”
- “the actual population has reached 0”
- “not possible to have a population of 0.49(8)”
- “the population will go extinct when less than 1”