C3 June 2005 Q7
7. A particular species of orchid is being studied. The population \(p\) at time \(t\) years after the study started is assumed to be\[p = \frac{2800a\mathrm{e}^{0.2t}}{1 + a\mathrm{e}^{0.2t}}, \quad \text{where } a \text{ is a constant.}\]
Given that there were 300 orchids when the study started,
(a) show that \(a = 0.12\), (3)
(b) use the equation with \(a = 0.12\) to predict the number of years before the population of orchids reaches 1850. (4)
(c) Show that \(p = \dfrac{336}{0.12 + \mathrm{e}^{-0.2t}}\). (1)
(d) Hence show that the population cannot exceed 2800. (2)
| Scheme | Marks |
|---|---|
| Setting \(p = 300\) at \(t = 0 \Rightarrow 300 = \dfrac{2800a}{1 + a}\) | M1 |
| \((300 = 2500a)\); \(\quad a = 0.12\) (c.s.o) * | dM1 A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(1850 = \dfrac{2800(0.12)\mathrm{e}^{0.2t}}{1 + 0.12\mathrm{e}^{0.2t}}\); \(\quad\mathrm{e}^{0.2t} = 16.2\ldots\) | M1 A1 |
| Correctly taking logs to \(0.2t = \ln k\) | M1 |
| \(t = 14\) (13.9..) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Correct derivation: (Showing division of num. and den. by \(\mathrm{e}^{0.2t}\); using \(a\)) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| Using \(t \to \infty\), \(\mathrm{e}^{-0.2t} \to 0\), | M1 |
| \(p \to \dfrac{336}{0.12} = 2800\) | A1 |
| (2) | |
| (10 marks) |