| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (applied/contextual) |
| Difficulty | Standard +0.3 This is a straightforward C4 differential equations question with standard separable equations and routine algebraic manipulation. Part (a) requires basic separation of variables and integration of a power function, part (b) is direct substitution, part (c)(i) is translating words into a differential equation, and part (c)(ii) involves simple logarithmic equation solving. All techniques are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\frac{dx}{\sqrt{x+1}} = \int\frac{1}{5}dt\) or \(\frac{dt}{dx} = -5(x+1)^{-\frac{1}{2}}\) | B1 | Correct separation; or \(\frac{dt}{dx} = -5(x+1)^{-\frac{1}{2}}\); Condone missing integral signs |
| \(2\sqrt{x+1} = -\frac{1}{5}t\) \((+C)\) | B1 B1 | Correct integrals; condone \(\frac{\sqrt{x+1}}{t}\) |
| \(x = 80\), \(t = 0\), \(C = 2\sqrt{81} = 18\) | M1 | Use \((0, 80)\) to find a constant \(C\) |
| \(x = \left(9 - \frac{1}{10}t\right)^2 - 1\) | A1F | 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 60\), \(x = f(60) = 8\) | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dA}{dt} = kA(9 - A)\) | M1 | \(\frac{dA}{dt} = \) product involving \(A\); \(k\) required; Condone terms in \(t\) |
| A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(4.5 = \frac{9}{1 + 4e^{-0.09t}}\) | M1 | Condone one slip in denominator |
| \(e^{-0.09t} = \frac{1}{4}\) | A1 | |
| \(-0.09t = \ln(\frac{1}{4})\) | m1 | Take ln correctly |
| \(t = \frac{\ln(\frac{1}{4})}{-0.09} = 15.4\) (hours) | A1 | 4 marks |
**8(a)**
$\int\frac{dx}{\sqrt{x+1}} = \int\frac{1}{5}dt$ or $\frac{dt}{dx} = -5(x+1)^{-\frac{1}{2}}$ | B1 | Correct separation; or $\frac{dt}{dx} = -5(x+1)^{-\frac{1}{2}}$; Condone missing integral signs
$2\sqrt{x+1} = -\frac{1}{5}t$ $(+C)$ | B1 B1 | Correct integrals; condone $\frac{\sqrt{x+1}}{t}$
$x = 80$, $t = 0$, $C = 2\sqrt{81} = 18$ | M1 | Use $(0, 80)$ to find a constant $C$
$x = \left(9 - \frac{1}{10}t\right)^2 - 1$ | A1F | 6 marks | OE; CSO; $x = $ correct expression in $t$
**8(b)**
$t = 60$, $x = f(60) = 8$ | M1 A1 | 2 marks | Evaluate $f(60)$, ie $x = \ldots$ (C not required); CSO
**8(c)(i)**
$\frac{dA}{dt} = kA(9 - A)$ | M1 | $\frac{dA}{dt} = $ product involving $A$; $k$ required; Condone terms in $t$
A1 | 2 marks
**8(c)(ii)**
$4.5 = \frac{9}{1 + 4e^{-0.09t}}$ | M1 | Condone one slip in denominator
$e^{-0.09t} = \frac{1}{4}$ | A1
$-0.09t = \ln(\frac{1}{4})$ | m1 | Take ln correctly
$t = \frac{\ln(\frac{1}{4})}{-0.09} = 15.4$ (hours) | A1 | 4 marks | CAO; condone more than 3sf if correct 15.40327068. Allow 15h 24m
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## Summary
**Total Marks: 75**\begin{enumerate}[label=(\alph*)]
\item Solve the differential equation
$$\frac{dx}{dt} = -\frac{1}{5}(x + 1)^{\frac{1}{2}}$$
given that $x = 80$ when $t = 0$. Give your answer in the form $x = f(t)$. [6 marks]
\item A fungus is spreading on the surface of a wall. The proportion of the wall that is unaffected after time $t$ hours is $x\%$. The rate of change of $x$ is modelled by the differential equation
$$\frac{dx}{dt} = -\frac{1}{5}(x + 1)^{\frac{1}{2}}$$
At $t = 0$, the proportion of the wall that is unaffected is 80%. Find the proportion of the wall that will still be unaffected after 60 hours. [2 marks]
\item A biologist proposes an alternative model for the rate at which the fungus is spreading on the wall. The total surface area of the wall is $9\text{ m}^2$. The surface area that is affected at time $t$ hours is $A\text{ m}^2$. The biologist proposes that the rate of change of $A$ is proportional to the product of the surface area that is affected and the surface area that is unaffected.
\begin{enumerate}[label=(\roman*)]
\item Write down a differential equation for this model.
(You are not required to solve your differential equation.) [2 marks]
\item A solution of the differential equation for this model is given by
$$A = \frac{9}{1 + 4e^{-0.09t}}$$
Find the time taken for 50% of the area of the wall to be affected. Give your answer in hours to three significant figures. [4 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2010 Q8 [14]}}