| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| 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 with standard integration and application of boundary conditions. Part (a) involves routine separation and integration of basic functions (1/y and sin t), while part (b) applies the result and uses standard second derivative test for maxima. The steps are mechanical and well-practiced in C4, making it slightly easier than average but not trivial due to the multi-part structure and need for careful algebraic manipulation. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \dfrac{dy}{y} = \int \sin t \, dt\) | M1 | Attempt to separate and integrate |
| \(\ln y = -\cos t + C\) | A1, A1 | A1 for \(\ln y\); A1 for \(-\cos t\); condone missing \(C\) |
| \(y = Ae^{-\cos t}\) | A1 (4 marks) | \(A\) present; or \(y = e^{-\cos t + C}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 50,\ t = \pi\): \(\quad 50 = Ae^{-\cos\pi} = Ae\) | M1, A1 | Substitute \(y=50\), \(t=\pi\) to find constant. Can have \(50 = e^{1+c}\) if substituted in above. \(e^c = \frac{50}{e}\) |
| \(y = 50e^{-1}e^{-\cos t}\) | A1 (3 marks) | AG (convincingly obtained) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 6\): \(y = 50e^{-1}e^{-\cos 6} = 7.0417... \approx 7\text{ cm}\) | M1A1 (2 marks) | Degrees 6.8 SC1. 7 or 7.0 for A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = \pi \Rightarrow (\sin t = 0 \Rightarrow) \dfrac{dy}{dt} = 0\) | B1 | Condone \(x\) for \(t\) |
| \(\dfrac{d^2y}{dt^2} = y\cos t + \dfrac{dy}{dt}\sin t\) | M1 | For attempt at product rule including \(\dfrac{dy}{dt}\) term; must have \(\dfrac{d^2y}{dt^2}=\) |
| A1 | ||
| \(\dfrac{d^2y}{dt^2} = y\cos\pi + \dfrac{dy}{dt}\sin\pi = -50 \Rightarrow \max\) | A1 (4 marks) | Accept \(= -y\), with explanation that \(y\) is never negative |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{dy}{dt} = \dfrac{50}{e}e^{-\cos t} \times \sin t = 0\) at \(t = \pi\) | (B1) | |
| \(\dfrac{d^2y}{dt^2} = \dfrac{50}{e}e^{-\cos t}\times\cos t + \dfrac{50}{e}e^{-\cos t}\times\sin^2 t\) | (M1)(A1) | Attempt at product rule. Correct. |
| Substitute \(t = \pi \rightarrow -50 \Rightarrow \max\) | (A1) |
# Question 8(a)(i):
$\int \dfrac{dy}{y} = \int \sin t \, dt$ | M1 | Attempt to separate and integrate
$\ln y = -\cos t + C$ | A1, A1 | A1 for $\ln y$; A1 for $-\cos t$; condone missing $C$
$y = Ae^{-\cos t}$ | A1 (4 marks) | $A$ present; or $y = e^{-\cos t + C}$
---
# Question 8(a)(ii):
$y = 50,\ t = \pi$: $\quad 50 = Ae^{-\cos\pi} = Ae$ | M1, A1 | Substitute $y=50$, $t=\pi$ to find constant. Can have $50 = e^{1+c}$ if substituted in above. $e^c = \frac{50}{e}$
$y = 50e^{-1}e^{-\cos t}$ | A1 (3 marks) | AG (convincingly obtained)
---
# Question 8(b)(i):
$t = 6$: $y = 50e^{-1}e^{-\cos 6} = 7.0417... \approx 7\text{ cm}$ | M1A1 (2 marks) | Degrees 6.8 SC1. 7 or 7.0 for A1
---
# Question 8(b)(ii):
$t = \pi \Rightarrow (\sin t = 0 \Rightarrow) \dfrac{dy}{dt} = 0$ | B1 | Condone $x$ for $t$
$\dfrac{d^2y}{dt^2} = y\cos t + \dfrac{dy}{dt}\sin t$ | M1 | For attempt at product rule including $\dfrac{dy}{dt}$ term; must have $\dfrac{d^2y}{dt^2}=$
| A1 |
$\dfrac{d^2y}{dt^2} = y\cos\pi + \dfrac{dy}{dt}\sin\pi = -50 \Rightarrow \max$ | A1 (4 marks) | Accept $= -y$, with explanation that $y$ is never negative
**Alternative:** $y = 50e^{-(1+\cos t)} = \dfrac{50}{e}e^{-\cos t}$
$\dfrac{dy}{dt} = \dfrac{50}{e}e^{-\cos t} \times \sin t = 0$ at $t = \pi$ | (B1) |
$\dfrac{d^2y}{dt^2} = \dfrac{50}{e}e^{-\cos t}\times\cos t + \dfrac{50}{e}e^{-\cos t}\times\sin^2 t$ | (M1)(A1) | Attempt at product rule. Correct.
Substitute $t = \pi \rightarrow -50 \Rightarrow \max$ | (A1) |8
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Solve the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} t } = y \sin t$ to obtain $y$ in terms of $t$.
\item Given that $y = 50$ when $t = \pi$, show that $y = 50 \mathrm { e } ^ { - ( 1 + \cos t ) }$.
\end{enumerate}\item A wave machine at a leisure pool produces waves. The height of the water, $y \mathrm {~cm}$, above a fixed point at time $t$ seconds is given by the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} t } = y \sin t$$
\begin{enumerate}[label=(\roman*)]
\item Given that this height is 50 cm after $\pi$ seconds, find, to the nearest centimetre, the height of the water after 6 seconds.
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }$ and hence verify that the water reaches a maximum height after $\pi$ seconds.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q8 [13]}}