| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| 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 standard separable variables question with routine integration techniques (√x and sin(t/2)), followed by applying initial conditions and straightforward algebraic manipulation. The context application in part (b) requires only substitution into the derived formula. Slightly easier than average due to the straightforward separation and integration steps. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int x^{-1/2}\,dx = \int \sin\!\left(\frac{t}{2}\right)dt\) | M1 | Separate variables correctly |
| \(2\sqrt{x} = -2\cos\!\left(\frac{t}{2}\right) + c\) | A1 | Correct integration both sides |
| \(\sqrt{x} = -\cos\!\left(\frac{t}{2}\right) + c\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=1, t=0\): \(1 = -\cos 0 + c \Rightarrow c = 2\) | M1 | Apply initial condition |
| \(\sqrt{x} = 2 - \cos\!\left(\frac{t}{2}\right)\) | A1 | |
| \(x = \left(2 - \cos\!\left(\frac{t}{2}\right)\right)^2\), so \(a=2, b=\frac{1}{2}\) | A1 | Both constants correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Maximum when \(\cos\!\left(\frac{t}{2}\right) = -1\), so \(\sqrt{x} = 3\) | M1 | |
| Greatest height \(= 9\) metres | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5 = \left(2-\cos\!\left(\frac{t}{2}\right)\right)^2 \Rightarrow \cos\!\left(\frac{t}{2}\right) = 2-\sqrt{5}\) | M1 | |
| \(t = 2\arccos(2-\sqrt{5}) \approx 4.5\) seconds | A1 |
## Question 7:
**Part (a)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int x^{-1/2}\,dx = \int \sin\!\left(\frac{t}{2}\right)dt$ | M1 | Separate variables correctly |
| $2\sqrt{x} = -2\cos\!\left(\frac{t}{2}\right) + c$ | A1 | Correct integration both sides |
| $\sqrt{x} = -\cos\!\left(\frac{t}{2}\right) + c$ | A1 | |
**Part (a)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=1, t=0$: $1 = -\cos 0 + c \Rightarrow c = 2$ | M1 | Apply initial condition |
| $\sqrt{x} = 2 - \cos\!\left(\frac{t}{2}\right)$ | A1 | |
| $x = \left(2 - \cos\!\left(\frac{t}{2}\right)\right)^2$, so $a=2, b=\frac{1}{2}$ | A1 | Both constants correct |
**Part (b)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Maximum when $\cos\!\left(\frac{t}{2}\right) = -1$, so $\sqrt{x} = 3$ | M1 | |
| Greatest height $= 9$ metres | A1 | |
**Part (b)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5 = \left(2-\cos\!\left(\frac{t}{2}\right)\right)^2 \Rightarrow \cos\!\left(\frac{t}{2}\right) = 2-\sqrt{5}$ | M1 | |
| $t = 2\arccos(2-\sqrt{5}) \approx 4.5$ seconds | A1 | |
---7
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Solve the differential equation $\frac { \mathrm { d } x } { \mathrm {~d} t } = \sqrt { x } \sin \left( \frac { t } { 2 } \right)$ to find $x$ in terms of $t$.
\item Given that $x = 1$ when $t = 0$, show that the solution can be written as
$$x = ( a - \cos b t ) ^ { 2 }$$
where $a$ and $b$ are constants to be found.
\end{enumerate}\item The height, $x$ metres, above the ground of a car in a fairground ride at time $t$ seconds is modelled by the differential equation $\frac { \mathrm { d } x } { \mathrm {~d} t } = \sqrt { x } \sin \left( \frac { t } { 2 } \right)$.
The car is 1 metre above the ground when $t = 0$.
\begin{enumerate}[label=(\roman*)]
\item Find the greatest height above the ground reached by the car during the ride.
\item Find the value of $t$ when the car is first 5 metres above the ground, giving your answer to one decimal place.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2011 Q7 [10]}}