| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - non-standard rate function |
| Difficulty | Standard +0.3 This is a straightforward differential equations question requiring separation of variables and integration of a standard exponential form. Part (i) involves routine integration and substitution to verify a given answer, while part (ii) requires finding the limit as tââ, which is a direct application of exponential decay properties. The question is slightly easier than average as it guides students through verification rather than requiring independent problem-solving. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\int \frac{1}{P} dP = \int 0.05e^{-0.05t} dt\) | M1 | |
| \(\ln | P | = -e^{-0.05t} + c\) |
| \(t = 0, P = 9000 \Rightarrow \ln 9000 = -1 + c, \quad c = 1 + \ln 9000\) | M1 A1 | |
| \(\ln | P | = 1 + \ln 9000 - e^{-0.05t}\) |
| \(t = 10 \Rightarrow \ln | P | = 1 + \ln 9000 - e^{-0.5} = 9.498\) |
| \(P = e^{9.498} = 13339 = 13300 (3sf)\) | A1 | |
| (ii) \(t \to \infty, \ln | P | \to 1 + \ln 9000\) |
| \(\therefore P \to e^{1 + \ln 9000} = 9000e = 24465 = 24500 (3sf)\) | M1 A1 |
**(i)** $\int \frac{1}{P} dP = \int 0.05e^{-0.05t} dt$ | M1
$\ln|P| = -e^{-0.05t} + c$ | M1 A1
$t = 0, P = 9000 \Rightarrow \ln 9000 = -1 + c, \quad c = 1 + \ln 9000$ | M1 A1
$\ln|P| = 1 + \ln 9000 - e^{-0.05t}$ | A1
$t = 10 \Rightarrow \ln|P| = 1 + \ln 9000 - e^{-0.5} = 9.498$ | M1
$P = e^{9.498} = 13339 = 13300 (3sf)$ | A1
**(ii)** $t \to \infty, \ln|P| \to 1 + \ln 9000$ | M1 A1
$\therefore P \to e^{1 + \ln 9000} = 9000e = 24465 = 24500 (3sf)$ | M1 A18. A small town had a population of 9000 in the year 2001.
In a model, it is assumed that the population of the town, $P$, at time $t$ years after 2001 satisfies the differential equation
$$\frac { \mathrm { d } P } { \mathrm {~d} t } = 0.05 P \mathrm { e } ^ { - 0.05 t }$$
(i) Show that, according to the model, the population of the town in 2011 will be 13300 to 3 significant figures.\\
(ii) Find the value which the population of the town will approach in the long term, according to the model.\\
\hfill \mbox{\textit{OCR C4 Q8 [10]}}