| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (polynomial/exponential x-side) |
| Difficulty | Standard +0.3 This is a standard separable differential equations question requiring routine integration techniques (separation of variables, exponential/logarithm manipulation) and substitution of initial conditions. The steps are mechanical and follow textbook procedures, though the algebra with exponentials requires care. Slightly above average difficulty due to the multi-step nature and need for algebraic fluency, but no novel insight required. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\int \frac{dx}{e^{\frac{1}{2}x}} = \int -kt\, dt\) | B1 | Separate; condone missing integral signs |
| \(-2e^{-\frac{1}{2}x} = -k\frac{t^2}{2}\ (+C)\) | B1, B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(-2e^{-\frac{1}{2}x} = -k\frac{t^2}{2} - 2e^{-3}\) | M1 | Use \((6,0)\) to find constant |
| \(\ln\left(e^{-\frac{1}{2}x}\right) = \ln\left(k\frac{t^2}{4} + e^{-3}\right)\) | M1 | Take logarithms correctly; condone one side negative. Must have a constant |
| \(-\frac{1}{2}x = \ln\left(k\frac{t^2}{4} + e^{-3}\right)\) | ||
| \(x = -2\ln\left(\frac{kt^2}{4} + e^{-3}\right)\) | A1 | Answer given; CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(t=10\): \(x = -2\ln\left(\frac{0.004 \times 10^2}{4} + e^{-3}\right)\) | M1 | |
| \(= 3.8 \Rightarrow 3800\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(x=0\): \(\frac{0.004 \times t^2}{4} + e^{-3} = 1\) | M1 | |
| \(t = 30.8\) | A1 | CAO; Treat 0.04 or 0.0004 as misread \((-1)\) |
# Question 7:
## Part (a)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\int \frac{dx}{e^{\frac{1}{2}x}} = \int -kt\, dt$ | B1 | Separate; condone missing integral signs |
| $-2e^{-\frac{1}{2}x} = -k\frac{t^2}{2}\ (+C)$ | B1, B1 | | **Total: 3** |
## Part (a)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $-2e^{-\frac{1}{2}x} = -k\frac{t^2}{2} - 2e^{-3}$ | M1 | Use $(6,0)$ to find constant |
| $\ln\left(e^{-\frac{1}{2}x}\right) = \ln\left(k\frac{t^2}{4} + e^{-3}\right)$ | M1 | Take logarithms correctly; condone one side negative. Must have a constant |
| $-\frac{1}{2}x = \ln\left(k\frac{t^2}{4} + e^{-3}\right)$ | | |
| $x = -2\ln\left(\frac{kt^2}{4} + e^{-3}\right)$ | A1 | Answer given; CSO | **Total: 3** |
## Part (b)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $t=10$: $x = -2\ln\left(\frac{0.004 \times 10^2}{4} + e^{-3}\right)$ | M1 | |
| $= 3.8 \Rightarrow 3800$ | A1 | CAO | **Total: 2** |
## Part (b)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $x=0$: $\frac{0.004 \times t^2}{4} + e^{-3} = 1$ | M1 | |
| $t = 30.8$ | A1 | CAO; Treat 0.04 or 0.0004 as misread $(-1)$ | **Total: 2** |
---7
\begin{enumerate}[label=(\alph*)]
\item A differential equation is given by $\frac { \mathrm { d } x } { \mathrm {~d} t } = - k t \mathrm { e } ^ { \frac { 1 } { 2 } x }$, where $k$ is a positive constant.
\begin{enumerate}[label=(\roman*)]
\item Solve the differential equation.
\item Hence, given that $x = 6$ when $t = 0$, show that $x = - 2 \ln \left( \frac { k t ^ { 2 } } { 4 } + \mathrm { e } ^ { - 3 } \right)$.\\
(3 marks)
\end{enumerate}\item The population of a colony of insects is decreasing according to the model $\frac { \mathrm { d } x } { \mathrm {~d} t } = - k t \mathrm { e } ^ { \frac { 1 } { 2 } x }$, where $x$ thousands is the number of insects in the colony after time $t$ minutes. Initially, there were 6000 insects in the colony.
Given that $k = 0.004$, find:
\begin{enumerate}[label=(\roman*)]
\item the population of the colony after 10 minutes, giving your answer to the nearest hundred;
\item the time after which there will be no insects left in the colony, giving your answer to the nearest 0.1 of a minute.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q7 [10]}}