| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2015 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Deriving the differential equation |
| Difficulty | Moderate -0.8 This is a straightforward modelling question requiring translation of a word problem into a differential equation (dN/dt = k(N-150)), then solving using separation of variables with given initial/boundary conditions. The mathematics is routine for P3 level—no integrating factor needed despite the topic tag, just basic exponential growth model. The multi-part structure and context add length but not conceptual difficulty. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| State \(\frac{dN}{dt} = k(N - 150)\) | B1 | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute \(\frac{dN}{dt} = 60\) and \(N = 900\) to find value of \(k\) | M1 | |
| Obtain \(k = 0.08\) | A1 | |
| Separate variables and obtain general solution involving \(\ln(N - 150)\) | M1* | |
| Obtain \(\ln(N - 150) = 0.08t + c\) (following their \(k\)) or \(\ln(N - 150) = kt + c\) | A1* | |
| Substitute \(t = 0\) and \(N = 650\) to find \(c\) | dep M1* | |
| Obtain \(\ln(N - 150) = 0.08t + \ln 500\) or equivalent | A1 | |
| Obtain \(N = 500e^{0.08t} + 150\) | A1 | [7] |
| Answer | Marks |
|---|---|
| Substitute \(t = 15\) to find \(N\) or solve for \(t\) with \(N = 2000\) | M1 |
| Obtain either \(N = 1810\) or \(t = 16.4\) and conclude target not met | A1 |
| [2] |
**(i)**
State $\frac{dN}{dt} = k(N - 150)$ | B1 | [1]
**(ii)**
Substitute $\frac{dN}{dt} = 60$ and $N = 900$ to find value of $k$ | M1 |
Obtain $k = 0.08$ | A1 |
Separate variables and obtain general solution involving $\ln(N - 150)$ | M1* |
Obtain $\ln(N - 150) = 0.08t + c$ (following their $k$) or $\ln(N - 150) = kt + c$ | A1* |
Substitute $t = 0$ and $N = 650$ to find $c$ | dep M1* |
Obtain $\ln(N - 150) = 0.08t + \ln 500$ or equivalent | A1 |
Obtain $N = 500e^{0.08t} + 150$ | A1 | [7]
**(iii)**
**Either**
Substitute $t = 15$ to find $N$ or solve for $t$ with $N = 2000$ | M1 |
Obtain either $N = 1810$ or $t = 16.4$ and conclude target not met | A1 |
| [2] |10 Naturalists are managing a wildlife reserve to increase the number of plants of a rare species. The number of plants at time $t$ years is denoted by $N$, where $N$ is treated as a continuous variable.\\
(i) It is given that the rate of increase of $N$ with respect to $t$ is proportional to ( $N - 150$ ). Write down a differential equation relating $N , t$ and a constant of proportionality.\\
(ii) Initially, when $t = 0$, the number of plants was 650 . It was noted that, at a time when there were 900 plants, the number of plants was increasing at a rate of 60 per year. Express $N$ in terms of $t$.\\
(iii) The naturalists had a target of increasing the number of plants from 650 to 2000 within 15 years. Will this target be met?
\hfill \mbox{\textit{CAIE P3 2015 Q10 [10]}}