AS June 2019 Q3
3. Julie decides to start a business breeding rabbits to sell as pets.
Initially she buys 20 rabbits. After \(t\) years the number of rabbits, \(R\), is modelled by the differential equation
\[\frac{\mathrm{d}R}{\mathrm{d}t} = 2R + 4\sin t \qquad t \gt 0\]Julie needs to have at least 40 rabbits before she can start to sell them.
Use two iterations of the approximation formula
\[\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)_n \approx \frac{y_{n+1} - y_n}{h}\]to find out if, according to the model, Julie will be able to start selling rabbits after 4 months.
(7)
| Scheme | Marks | AO |
|---|---|---|
| {The population after 4 months is required over two iterations} \(\Rightarrow h = \dfrac{1}{6}\) | B1 | 3.3 |
| \(\{t_0 = 0,\ R_0 = 20 \Rightarrow\}\ \left(\dfrac{\mathrm{d}R}{\mathrm{d}t}\right)_{0} = 2(20) + 4\sin 0\ \{= 40\}\) | M1 | 3.4 |
| \(\left\{\dfrac{R_1 - 20}{\text{“}(\tfrac{1}{6})\text{”}} = \text{“}40\text{”} \Rightarrow\right\}\ R_1 = 20 + \text{“}\tfrac{1}{6}\text{”}\text{“}(40)\text{”}\) | M1 | 1.1b |
| \(R_1 = \dfrac{80}{3}\) or awrt 26.7 or \(20 + (\text{their } h)(40)\) | A1ft | 1.1b |
| \(\left(\dfrac{\mathrm{d}R}{\mathrm{d}t}\right)_{1} = 2(\text{“}R_1\text{”}) + 4\sin(\text{“}h\text{”}) = 2\left(\dfrac{80}{3}\right) + 4\sin\left(\dfrac{1}{6}\right)\ \{= 53.9969\ldots\}\) | M1 | 1.1b |
| \(R_2 = R_1 + h\left(\dfrac{\mathrm{d}R}{\mathrm{d}t}\right)_{1} = \dfrac{80}{3} + \dfrac{1}{6}(53.9969\ldots) = 35.666\ldots = 35\) or \(36\) rabbits | A1 | 1.1b |
| \(R_2 = 35.666\ldots \approx 35\) or \(36 \lt 40\) Julie will not be able to start to sell her rabbits after 4 months. | B1ft | 3.2a |
| (7) | ||
| (7 marks) |
Notes
B1: Translates the situation given to state (or use) the correct value for the step length \(h\)
M1: Uses the model to find the initial value of \(\dfrac{\mathrm{d}R}{\mathrm{d}t}\) using the initial condition \(t_0 = 0,\ R_0 = 20\)
M1: Applies the approximation formula with \(R_0 = 20\), their stated \(h\), their \(\left(\dfrac{\mathrm{d}R}{\mathrm{d}t}\right)_{0}\) to find a numerical expression for \(R_1\)
A1: depends on both previous M marks
At 2 months, finds the approximation for \(R\) as \(\dfrac{80}{3}\) or awrt 26.7
Note: Only give the following follow through. i.e. Allow A1ft for \(20 + (\text{their } h)(40)\) for their stated \(h\)
M1: Attempts to find a numerical expression for \(\left(\dfrac{\mathrm{d}R}{\mathrm{d}t}\right)_{1}\) with their \(\dfrac{80}{3}\) and \(t_1 = \) their \(h\)
A1: Applies the approximation formula for a second time to give \(R_2\) as a truncated 35 or a value in the interval \([35.5, 36]\)
B1ft: Attempts two iterations of their \(R_{n+1} = R_n + h\left(\dfrac{\mathrm{d}R}{\mathrm{d}t}\right)_n\) to find a value for \(R_2\).
Compares their value of \(R_2\) with 40 (which can be implied) and draws a conclusion about whether Julie will be able to start to sell her rabbits after 4 months.
Note: Give final B0 for applying more than or fewer than two iterations before comparing
Note: Using \(h = \dfrac{1}{12}\) yields \(R_1 = 23.3333\ldots,\ R_2 = 27.2499\ldots,\ R_3 = 31.8469\ldots,\ R_4 = 37.2372\ldots\)
Note: Give special case final A1 for giving \(R_4\) as a truncated 37 or a value in the interval \([37, 37.4]\)
Note: Therefore, using \(h = \dfrac{1}{12}\) with four iterations can gain a maximum B0 M1 M1 A1 M1 A1 B0
Note: Answers in the range \([35.5, 36]\) can follow from an incorrect method. E.g. Give final M0 A0 for using \(h = \dfrac{1}{6}\), \(\left(\dfrac{\mathrm{d}R}{\mathrm{d}t}\right)_{1} = 2\left(\dfrac{80}{3}\right) + 4\sin(\underline{0.1}) = 53.73266\ldots \Rightarrow R_2 = \dfrac{80}{3} + \dfrac{1}{6}(53.73266\ldots) = 35.622\ldots\)