Question 7 - A-Level Maths

OCR MEI C4 2007 June — Question 7 20 marks

Exam Board	OCR MEI
Module	C4 (Core Mathematics 4)
Year	2007
Session	June
Marks	20
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Differential equations
Type	Verification of solutions
Difficulty	Moderate -0.3 This is a multi-part question testing standard C4 techniques: verifying a solution satisfies a differential equation (routine differentiation), partial fractions (standard A-level method), separating variables and integrating (core C4 skill), and parametric differentiation. All parts follow textbook procedures with no novel insight required. The partial fractions and integration are straightforward, and the verification is mechanical. Slightly easier than average due to the guided structure and routine nature of each step.
Spec	1.02y Partial fractions: decompose rational functions 1.05a Sine, cosine, tangent: definitions for all arguments 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates 1.08k Separable differential equations: dy/dx = f(x)g(y)4.04c Scalar product: calculate and use for angles

7 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.

Suppose that the number of cases, $P$ thousand, after time $t$ months is modelled by the equation $P = \frac { 2 } { 2 - \sin t }$. Thus, when $t = 0 , P = 1$.
1. By considering the greatest and least values of $\sin t$, write down the greatest and least values of $P$ predicted by this model.
2. Verify that $P$ satisfies the differential equation $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t$.
An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, $P = 1$ when $t = 0$.
1. Express $\frac { 1 } { P ( 2 P - 1 ) }$ in partial fractions.
2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P - 1 } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give $P = \frac { 1 } { 2 - \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }$.
3. Find the greatest and least values of $P$ predicted by this model. \begin{figure}[h]
  \includegraphics[alt={},max width=\textwidth]{9296c786-a42a-4aa5-b326-39adbb544cbc-05_609_622_301_719} \captionsetup{labelformat=empty} \caption{Fig. 8}
  \end{figure} In a theme park ride, a capsule C moves in a vertical plane (see Fig. 8). With respect to the axes shown, the path of C is modelled by the parametric equations $$x = 10 \cos \theta + 5 \cos 2 \theta , \quad y = 10 \sin \theta + 5 \sin 2 \theta , \quad ( 0 \leqslant \theta < 2 \pi )$$ where $x$ and $y$ are in metres.
  1. Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { \cos \theta + \cos 2 \theta } { \sin \theta + \sin 2 \theta }$. Verify that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ when $\theta = \frac { 1 } { 3 } \pi$. Hence find the exact coordinates of the highest point A on the path of C .
  2. Express $x ^ { 2 } + y ^ { 2 }$ in terms of $\theta$. Hence show that $$x ^ { 2 } + y ^ { 2 } = 125 + 100 \cos \theta$$
  3. Using this result, or otherwise, find the greatest and least distances of C from O . You are given that, at the point B on the path vertically above O , $$2 \cos ^ { 2 } \theta + 2 \cos \theta - 1 = 0$$
  4. Using this result, and the result in part (ii), find the distance OB. Give your answer to 3 significant figures. \section*{ADVANCED GCE UNIT MATHEMATICS (MEI)} Applications of Advanced Mathematics (C4) \section*{Paper B: Comprehension} \section*{THURSDAY 14 JUNE 2007} Afternoon
    Time: Up to 1 hour
    Additional materials:
    Rough paper
    MEI Examination Formulae and Tables (MF2) Candidate
    Name □
    Centre
    Number sufficient detail of the working to indicate that a correct method is being used. 1 This basic cycloid has parametric equations $$x = a \theta - a \sin \theta , \quad y = a - a \cos \theta$$
    \includegraphics[max width=\textwidth, alt={}]{9296c786-a42a-4aa5-b326-39adbb544cbc-10_307_1138_445_411}
    Find the coordinates of the points M and N , stating the value of $\theta$ at each of them. Point M Point N 2 A sea wave has parametric equations (in suitable units) $$x = 7 \theta - 0.25 \sin \theta , \quad y = 0.25 \cos \theta$$ Find the wavelength and height of the wave.
    3 The graph below shows the profile of a wave.
    1. Assuming that it has parametric equations of the form given on line 68 , find the values of $a$ and $b$.
    2. Investigate whether the ratio of the trough length to the crest length is consistent with this shape. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-11_312_1141_623_415}
    3. $\_\_\_\_$
    4. $\_\_\_\_$ 4 This diagram illustrates two wave shapes $U$ and $V$. They have the same wavelength and the same height. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-12_423_1552_356_205} One of the curves is a sine wave, the other is a curtate cycloid.
    5. State which is which, justifying your answer.
    6. $\_\_\_\_$ The parametric equations for the curves are: $$x = a \theta , \quad y = b \cos \theta ,$$ and $$x = a \theta - b \sin \theta , \quad y = b \cos \theta .$$
    7. Show that the distance marked $d$ on the diagram is equal to $b$.
    8. Hence justify the statement in lines 109 to 111: "In such cases, the curtate cycloid and the sine curve with the same wavelength and height are very similar and so the sine curve is also a good model."
    9. $\_\_\_\_$
    10. $\_\_\_\_$ 5 The diagram shows a curtate cycloid with scales given. Show that this curve could not be a scale drawing of the shape of a stable sea wave. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-13_289_1310_397_331}

Show mark scheme Show mark scheme source

Question 7(a)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Greatest value of $\sin t = 1 \Rightarrow P = \frac{2}{2-1} = 2$	B1
Least value of $\sin t = -1 \Rightarrow P = \frac{2}{2+1} = \frac{2}{3}$	B1

Question 7(a)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$P = 2(2-\sin t)^{-1}$	M1
$\frac{dP}{dt} = -2(2-\sin t)^{-2}\cdot(-\cos t) = \frac{2\cos t}{(2-\sin t)^2}$	A1 M1	Chain rule
$\frac{1}{2}P^2\cos t = \frac{1}{2}\cdot\frac{4}{(2-\sin t)^2}\cdot\cos t = \frac{2\cos t}{(2-\sin t)^2}$ ✓	M1 A1	Showing equivalence

Question 7(b)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\frac{1}{P(2P-1)} = \frac{A}{P} + \frac{B}{2P-1}$	M1
$1 = A(2P-1) + BP$
$P=0$: $A = -1$	A1
$P=\frac{1}{2}$: $B = 2$	A1
$\frac{1}{P(2P-1)} = \frac{-1}{P} + \frac{2}{2P-1}$	A1

Question 7(b)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\frac{dP}{\frac{1}{2}(2P^2-P)} = \cos t \, dt$	M1	Separating variables
$\frac{2}{P(2P-1)}dP = \cos t \, dt$
$\int\left(\frac{-2}{P} + \frac{4}{2P-1}\right)dP = \int\cos t \, dt$	M1	Using partial fractions
$-2\ln P + 2\ln(2P-1) = \sin t + c$	A1 A1
$t=0, P=1$: $c = 0$	M1	Applying initial conditions
$\ln\left(\frac{2P-1}{P}\right)^2 = \sin t$... leading to $\ln\left(\frac{2P-1}{P}\right) = \frac{1}{2}\sin t$	A1

Question 7(b)(iii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$P = \frac{1}{2 - e^{\frac{1}{2}\sin t}}$
Greatest: $\sin t = 1 \Rightarrow P = \frac{1}{2-e^{1/2}} = \frac{1}{2-\sqrt{e}}$	M1 A1
Least: $\sin t = -1 \Rightarrow P = \frac{1}{2-e^{-1/2}} = \frac{1}{2-1/\sqrt{e}}$	M1 A1

## Question 7(a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Greatest value of $\sin t = 1 \Rightarrow P = \frac{2}{2-1} = 2$ | B1 | |
| Least value of $\sin t = -1 \Rightarrow P = \frac{2}{2+1} = \frac{2}{3}$ | B1 | |

## Question 7(a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = 2(2-\sin t)^{-1}$ | M1 | |
| $\frac{dP}{dt} = -2(2-\sin t)^{-2}\cdot(-\cos t) = \frac{2\cos t}{(2-\sin t)^2}$ | A1 M1 | Chain rule |
| $\frac{1}{2}P^2\cos t = \frac{1}{2}\cdot\frac{4}{(2-\sin t)^2}\cdot\cos t = \frac{2\cos t}{(2-\sin t)^2}$ ✓ | M1 A1 | Showing equivalence |

## Question 7(b)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{P(2P-1)} = \frac{A}{P} + \frac{B}{2P-1}$ | M1 | |
| $1 = A(2P-1) + BP$ | | |
| $P=0$: $A = -1$ | A1 | |
| $P=\frac{1}{2}$: $B = 2$ | A1 | |
| $\frac{1}{P(2P-1)} = \frac{-1}{P} + \frac{2}{2P-1}$ | A1 | |

## Question 7(b)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dP}{\frac{1}{2}(2P^2-P)} = \cos t \, dt$ | M1 | Separating variables |
| $\frac{2}{P(2P-1)}dP = \cos t \, dt$ | | |
| $\int\left(\frac{-2}{P} + \frac{4}{2P-1}\right)dP = \int\cos t \, dt$ | M1 | Using partial fractions |
| $-2\ln P + 2\ln(2P-1) = \sin t + c$ | A1 A1 | |
| $t=0, P=1$: $c = 0$ | M1 | Applying initial conditions |
| $\ln\left(\frac{2P-1}{P}\right)^2 = \sin t$... leading to $\ln\left(\frac{2P-1}{P}\right) = \frac{1}{2}\sin t$ | A1 | |

## Question 7(b)(iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = \frac{1}{2 - e^{\frac{1}{2}\sin t}}$ | | |
| Greatest: $\sin t = 1 \Rightarrow P = \frac{1}{2-e^{1/2}} = \frac{1}{2-\sqrt{e}}$ | M1 A1 | |
| Least: $\sin t = -1 \Rightarrow P = \frac{1}{2-e^{-1/2}} = \frac{1}{2-1/\sqrt{e}}$ | M1 A1 | |

---

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7 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.
\begin{enumerate}[label=(\alph*)]
\item Suppose that the number of cases, $P$ thousand, after time $t$ months is modelled by the equation $P = \frac { 2 } { 2 - \sin t }$. Thus, when $t = 0 , P = 1$.
\begin{enumerate}[label=(\roman*)]
\item By considering the greatest and least values of $\sin t$, write down the greatest and least values of $P$ predicted by this model.
\item Verify that $P$ satisfies the differential equation $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t$.
\end{enumerate}\item An alternative model is proposed, with differential equation

$$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$

As before, $P = 1$ when $t = 0$.
\begin{enumerate}[label=(\roman*)]
\item Express $\frac { 1 } { P ( 2 P - 1 ) }$ in partial fractions.
\item Solve the differential equation (*) to show that

$$\ln \left( \frac { 2 P - 1 } { P } \right) = \frac { 1 } { 2 } \sin t$$

This equation can be rearranged to give $P = \frac { 1 } { 2 - \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }$.
\item Find the greatest and least values of $P$ predicted by this model.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{9296c786-a42a-4aa5-b326-39adbb544cbc-05_609_622_301_719}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}

In a theme park ride, a capsule C moves in a vertical plane (see Fig. 8). With respect to the axes shown, the path of C is modelled by the parametric equations

$$x = 10 \cos \theta + 5 \cos 2 \theta , \quad y = 10 \sin \theta + 5 \sin 2 \theta , \quad ( 0 \leqslant \theta < 2 \pi )$$

where $x$ and $y$ are in metres.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { \cos \theta + \cos 2 \theta } { \sin \theta + \sin 2 \theta }$.

Verify that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ when $\theta = \frac { 1 } { 3 } \pi$. Hence find the exact coordinates of the highest point A on the path of C .
\item Express $x ^ { 2 } + y ^ { 2 }$ in terms of $\theta$. Hence show that

$$x ^ { 2 } + y ^ { 2 } = 125 + 100 \cos \theta$$
\item Using this result, or otherwise, find the greatest and least distances of C from O .

You are given that, at the point B on the path vertically above O ,

$$2 \cos ^ { 2 } \theta + 2 \cos \theta - 1 = 0$$
\item Using this result, and the result in part (ii), find the distance OB. Give your answer to 3 significant figures.

\section*{ADVANCED GCE UNIT MATHEMATICS (MEI)}
Applications of Advanced Mathematics (C4)

\section*{Paper B: Comprehension}
\section*{THURSDAY 14 JUNE 2007}
Afternoon\\
Time: Up to 1 hour\\
Additional materials:\\
Rough paper\\
MEI Examination Formulae and Tables (MF2)

Candidate\\
Name □\\
Centre\\
Number

sufficient detail of the working to indicate that a correct method is being used.

1 This basic cycloid has parametric equations

$$x = a \theta - a \sin \theta , \quad y = a - a \cos \theta$$

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{9296c786-a42a-4aa5-b326-39adbb544cbc-10_307_1138_445_411}
\end{center}

Find the coordinates of the points M and N , stating the value of $\theta$ at each of them.

Point M

Point N

2 A sea wave has parametric equations (in suitable units)

$$x = 7 \theta - 0.25 \sin \theta , \quad y = 0.25 \cos \theta$$

Find the wavelength and height of the wave.\\

3 The graph below shows the profile of a wave.\\
(i) Assuming that it has parametric equations of the form given on line 68 , find the values of $a$ and $b$.\\
(ii) Investigate whether the ratio of the trough length to the crest length is consistent with this shape.\\
\includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-11_312_1141_623_415}\\
(i) $\_\_\_\_$\\

(ii) $\_\_\_\_$\\

4 This diagram illustrates two wave shapes $U$ and $V$. They have the same wavelength and the same height.\\
\includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-12_423_1552_356_205}

One of the curves is a sine wave, the other is a curtate cycloid.\\
(i) State which is which, justifying your answer.\\
(i) $\_\_\_\_$\\

The parametric equations for the curves are:

$$x = a \theta , \quad y = b \cos \theta ,$$

and

$$x = a \theta - b \sin \theta , \quad y = b \cos \theta .$$

(ii) Show that the distance marked $d$ on the diagram is equal to $b$.\\
(iii) Hence justify the statement in lines 109 to 111: "In such cases, the curtate cycloid and the sine curve with the same wavelength and height are very similar and so the sine curve is also a good model."\\
(ii) $\_\_\_\_$\\

(iii) $\_\_\_\_$\\

5 The diagram shows a curtate cycloid with scales given. Show that this curve could not be a scale drawing of the shape of a stable sea wave.\\
\includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-13_289_1310_397_331}
\end{enumerate}
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI C4 2007 Q7 [20]}}

This paper (7 questions)

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Q1 7 Q2 4 Q3 6 Q4 4 Q5 7 Q6 8 Q7 20

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(\frac{1}{P(2P-1)} = \frac{A}{P} + \frac{B}{2P-1}\)	M1
\(1 = A(2P-1) + BP\)
\(P=0\): \(A = -1\)	A1
\(P=\frac{1}{2}\): \(B = 2\)	A1
\(\frac{1}{P(2P-1)} = \frac{-1}{P} + \frac{2}{2P-1}\)	A1