| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (polynomial/exponential x-side) |
| Difficulty | Moderate -0.3 This is a straightforward separable variables question requiring standard integration techniques (cos 2t and power rule), followed by routine substitution into the solution. The context application in part (b) involves simple evaluation and inverse calculation with no conceptual challenges beyond the basic method. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int x\ dx = \int 150\cos 2t\ dt\) | B1 | Correct separation; condone missing \(\int\) signs; must see \(dx\), \(dt\) |
| \(\frac{1}{2}x^2 = 75\sin 2t \quad (+C)\) | B1B1 | Correct integrals; accept \(\frac{1}{2}\times 150\) |
| \(\left(20, \frac{\pi}{4}\right)\): \(\frac{1}{2}\times 20^2 = 75\sin\left(2\times\frac{\pi}{4}\right) + C\) | M1 | \(C\) present. Use \(\left(20, \frac{\pi}{4}\right)\) to find \(C\) |
| \(C = 125\) | A1F | F on \(x^2 = k\sin 2t\) |
| \(x^2 = 150\sin 2t + 250\) | A1 (6 marks) | Correct integrals and evaluation of \(C\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 13\): \(x^2 = 150\sin 26 + 250\ (= 364.38)\); \(x = 19.1\) (cm) | M1 A1 (2 marks) | Evaluate \(x^2 = f(13)\); \(x^2 = k\sin 2t + c\) with numerical \(k\) and \(t\); AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 11\): \(\sin 2t = -\frac{129}{150}\ (= -0.86)\); or \(2t = -1.035...,\ 4.176...\) | M1 | |
| \(t = 2.1\) (seconds) | A1 (2 marks) | AWRT |
# Question 8(a):
$\int x\ dx = \int 150\cos 2t\ dt$ | B1 | Correct separation; condone missing $\int$ signs; must see $dx$, $dt$
$\frac{1}{2}x^2 = 75\sin 2t \quad (+C)$ | B1B1 | Correct integrals; accept $\frac{1}{2}\times 150$
$\left(20, \frac{\pi}{4}\right)$: $\frac{1}{2}\times 20^2 = 75\sin\left(2\times\frac{\pi}{4}\right) + C$ | M1 | $C$ present. Use $\left(20, \frac{\pi}{4}\right)$ to find $C$
$C = 125$ | A1F | F on $x^2 = k\sin 2t$
$x^2 = 150\sin 2t + 250$ | A1 (6 marks) | Correct integrals and evaluation of $C$
# Question 8(b)(i):
$t = 13$: $x^2 = 150\sin 26 + 250\ (= 364.38)$; $x = 19.1$ (cm) | M1 A1 (2 marks) | Evaluate $x^2 = f(13)$; $x^2 = k\sin 2t + c$ with numerical $k$ and $t$; AWRT
# Question 8(b)(ii):
$x = 11$: $\sin 2t = -\frac{129}{150}\ (= -0.86)$; or $2t = -1.035...,\ 4.176...$ | M1 |
$t = 2.1$ (seconds) | A1 (2 marks) | AWRT8
\begin{enumerate}[label=(\alph*)]
\item Solve the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 150 \cos 2 t } { x }$$
given that $x = 20$ when $t = \frac { \pi } { 4 }$, giving your solution in the form $x ^ { 2 } = \mathrm { f } ( t )$. (6 marks)
\item The oscillations of a 'baby bouncy cradle' are modelled by the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 150 \cos 2 t } { x }$$
where $x \mathrm {~cm}$ is the height of the cradle above its base $t$ seconds after the cradle begins to oscillate.
Given that the cradle is 20 cm above its base at time $t = \frac { \pi } { 4 }$ seconds, find:
\begin{enumerate}[label=(\roman*)]
\item the height of the cradle above its base 13 seconds after it starts oscillating, giving your answer to the nearest millimetre;
\item the time at which the cradle will first be 11 cm above its base, giving your answer to the nearest tenth of a second.\\
(2 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q8 [10]}}