| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Logistic/bounded growth |
| Difficulty | Standard +0.8 This is a logistic differential equation requiring separation of variables and partial fractions decomposition. While the setup (part i) is straightforward, part (ii) demands careful algebraic manipulation of partial fractions and logarithms over multiple steps. The 8-mark allocation reflects substantial working, placing it above average difficulty but within reach of well-prepared A-level students. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks |
|---|---|
| (i) State or imply \(\frac{dN}{dt} = kN(1 - 0.01N)\) and obtain the given answer \(k = 0.02\) | B1 |
| Total: 1 mark | |
| (ii) Separate variables and attempt integration of at least one side | M1 |
| Integrate and obtain term 0.02t, or equivalent | A1 |
| Carry out a relevant method to obtain \(A\) or \(B\) such that \(\frac{1}{N(1 - 0.01N)} = \frac{A}{N} + \frac{B}{1 - 0.01N}\), or equivalent | M1* |
| Obtain \(A = 1\) and \(B = 0.01\), or equivalent | A1 |
| Integrate and obtain terms \(\ln N - \ln(1 - 0.01N)\), or equivalent | A1√ |
| Evaluate a constant or use limits \(t = 0, N = 20\) in a solution with terms \(a\ln N\) and \(b\ln(1 - 0.01N)\), \(ab \neq 0\) | M1(dep*) |
| Obtain correct answer in any form, e.g. \(\ln N - \ln(1 - 0.01N) = 0.02t + \ln 25\) | A1 |
| Rearrange and obtain \(t = 50\ln(4N/(100 - N))\), or equivalent | A1 |
| Total: 8 marks | |
| (iii) Substitute \(N = 40\) and obtain \(t = 49.0\) | B1 |
| Total: 1 mark |
**(i)** State or imply $\frac{dN}{dt} = kN(1 - 0.01N)$ and obtain the given answer $k = 0.02$ | B1 |
| Total: 1 mark |
**(ii)** Separate variables and attempt integration of at least one side | M1 |
Integrate and obtain term 0.02t, or equivalent | A1 |
Carry out a relevant method to obtain $A$ or $B$ such that $\frac{1}{N(1 - 0.01N)} = \frac{A}{N} + \frac{B}{1 - 0.01N}$, or equivalent | M1* |
Obtain $A = 1$ and $B = 0.01$, or equivalent | A1 |
Integrate and obtain terms $\ln N - \ln(1 - 0.01N)$, or equivalent | A1√ |
Evaluate a constant or use limits $t = 0, N = 20$ in a solution with terms $a\ln N$ and $b\ln(1 - 0.01N)$, $ab \neq 0$ | M1(dep*) |
Obtain correct answer in any form, e.g. $\ln N - \ln(1 - 0.01N) = 0.02t + \ln 25$ | A1 |
Rearrange and obtain $t = 50\ln(4N/(100 - N))$, or equivalent | A1 |
| Total: 8 marks |
**(iii)** Substitute $N = 40$ and obtain $t = 49.0$ | B1 |
| Total: 1 mark |
---The population of a country at time $t$ years is $N$ millions. At any time, $N$ is assumed to increase at a rate proportional to the product of $N$ and $(1 - 0.01N)$. When $t = 0$, $N = 20$ and $\frac{dN}{dt} = 0.32$.
\begin{enumerate}[label=(\roman*)]
\item Treating $N$ and $t$ as continuous variables, show that they satisfy the differential equation
$$\frac{dN}{dt} = 0.02N(1 - 0.01N).$$ [1]
\item Solve the differential equation, obtaining an expression for $t$ in terms of $N$. [8]
\item Find the time at which the population will be double its value at $t = 0$. [1]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2014 Q9 [10]}}