| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Gradient condition leads to trig equation |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question with routine techniques: finding stationary points by setting dy/dx = 0, using the chain rule for dy/dx = (dy/dθ)/(dx/dθ), and applying R cos(θ + α) form for harmonic equations. All parts follow predictable methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At A, \(y=0 \Rightarrow 4\cos\theta = 0\), \(\theta = \pi/2\) | B1 | |
| At B, \(\cos\theta = -1 \Rightarrow \theta = \pi\) | B1 | |
| \(x\)-coord of A \(= 2\times\pi/2 - \sin\pi/2 = \pi - 1\) | M1 | for either A or B/C |
| \(x\)-coord of B \(= 2\times\pi - \sin\pi = 2\pi\) | A1 | for both A and B/C |
| \(OA = \pi-1\), \(AC = 2\pi - \pi + 1 = \pi+1\) | ||
| ratio is \((\pi-1):(\pi+1)\) | E1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{d\theta} = -4\sin\theta\) | B1 | either \(dx/d\theta\) or \(dy/d\theta\) |
| \(\frac{dx}{d\theta} = 2 - \cos\theta\) | ||
| \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{4\sin\theta}{2-\cos\theta}\) | M1, A1 | |
| At A, gradient \(= -\frac{4\sin(\pi/2)}{2-\cos(\pi/2)} = -2\) | A1 [4] | www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 1 \Rightarrow -\frac{4\sin\theta}{2-\cos\theta} = 1\) | M1 | their \(dy/dx = 1\) |
| \(-4\sin\theta = 2 - \cos\theta\) | ||
| \(\cos\theta - 4\sin\theta = 2\) | E1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos\theta - 4\sin\theta = R\cos(\theta+\alpha) = R(\cos\theta\cos\alpha - \sin\theta\sin\alpha)\) | M1 | corr pairs |
| \(R\cos\alpha = 1\), \(R\sin\alpha = 4\) | ||
| \(R^2 = 1^2 + 4^2 = 17\), \(R = \sqrt{17}\) | B1, M1, A1 | accept \(76.0°\), 1.33 radians |
| \(\tan\alpha = 4\), \(\alpha = 1.326\) | ||
| \(\sqrt{17}\cos(\theta + 1.326) = 2\) | ||
| \(\cos(\theta+1.326) = 2/\sqrt{17}\) | M1 | inv cos \((2/\sqrt{17})\) ft their \(R\) for method |
| \(\theta + 1.326 = 1.064, 5.219, 7.348\) | ||
| \(\theta = (-0.262), 3.89, 6.02\) | A1, A1 [7] | \(-1\) extra solutions in the range |
# Question 8:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At A, $y=0 \Rightarrow 4\cos\theta = 0$, $\theta = \pi/2$ | B1 | |
| At B, $\cos\theta = -1 \Rightarrow \theta = \pi$ | B1 | |
| $x$-coord of A $= 2\times\pi/2 - \sin\pi/2 = \pi - 1$ | M1 | for either A or B/C |
| $x$-coord of B $= 2\times\pi - \sin\pi = 2\pi$ | A1 | for both A and B/C |
| $OA = \pi-1$, $AC = 2\pi - \pi + 1 = \pi+1$ | | |
| ratio is $(\pi-1):(\pi+1)$ | E1 [5] | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{d\theta} = -4\sin\theta$ | B1 | either $dx/d\theta$ or $dy/d\theta$ |
| $\frac{dx}{d\theta} = 2 - \cos\theta$ | | |
| $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{4\sin\theta}{2-\cos\theta}$ | M1, A1 | |
| At A, gradient $= -\frac{4\sin(\pi/2)}{2-\cos(\pi/2)} = -2$ | A1 [4] | www |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 1 \Rightarrow -\frac{4\sin\theta}{2-\cos\theta} = 1$ | M1 | their $dy/dx = 1$ |
| $-4\sin\theta = 2 - \cos\theta$ | | |
| $\cos\theta - 4\sin\theta = 2$ | E1 [2] | |
## Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos\theta - 4\sin\theta = R\cos(\theta+\alpha) = R(\cos\theta\cos\alpha - \sin\theta\sin\alpha)$ | M1 | corr pairs |
| $R\cos\alpha = 1$, $R\sin\alpha = 4$ | | |
| $R^2 = 1^2 + 4^2 = 17$, $R = \sqrt{17}$ | B1, M1, A1 | accept $76.0°$, 1.33 radians |
| $\tan\alpha = 4$, $\alpha = 1.326$ | | |
| $\sqrt{17}\cos(\theta + 1.326) = 2$ | | |
| $\cos(\theta+1.326) = 2/\sqrt{17}$ | M1 | inv cos $(2/\sqrt{17})$ ft their $R$ for method |
| $\theta + 1.326 = 1.064, 5.219, 7.348$ | | |
| $\theta = (-0.262), 3.89, 6.02$ | A1, A1 [7] | $-1$ extra solutions in the range |8 Part of the track of a roller-coaster is modelled by a curve with the parametric equations
$$x = 2 \theta - \sin \theta , \quad y = 4 \cos \theta \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi$$
This is shown in Fig. 8. B is a minimum point, and BC is vertical.
\begin{figure}[h]
\begin{center}
\includegraphics[width=\textwidth]{5c149cb5-7392-4219-b285-486f4694aa6f-4_602_1447_488_351}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Find the values of the parameter at A and B .
Hence show that the ratio of the lengths OA and AC is $( \pi - 1 ) : ( \pi + 1 )$.\\
(ii) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$. Find the gradient of the track at A .\\
(iii) Show that, when the gradient of the track is $1 , \theta$ satisfies the equation
$$\cos \theta - 4 \sin \theta = 2 .$$
(iv) Express $\cos \theta - 4 \sin \theta$ in the form $R \cos ( \theta + \alpha )$.
Hence solve the equation $\cos \theta - 4 \sin \theta = 2$ for $0 \leqslant \theta \leqslant 2 \pi$.
www.ocr.org.uk after the live examination series.\\
If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.\\
OCR is part of the
\section*{ADVANCED GCE \\
MATHEMATICS (MEI)}
4754B\\
Applications of Advanced Mathematics (C4) Paper B: Comprehension
\section*{Candidates answer on the Question Paper}
OCR Supplied Materials:
\begin{itemize}
\item Insert (inserted)
\item MEI Examination Formulae and Tables (MF2)
\end{itemize}
\section*{Other Materials Required:}
\begin{itemize}
\item Rough paper
\item Scientific or graphical calculator
\end{itemize}
Wednesday 9 June 2010 Afternoon\\
\includegraphics[width=\textwidth]{5c149cb5-7392-4219-b285-486f4694aa6f-5_264_456_881_1361}
1 The train journey from Swansea to London is 307 km and that by road is 300 km . Carry out the calculations performed on the First Great Western website to estimate how much lower the carbon dioxide emissions are when travelling by rail rather than road.\\
2 The equation of the curve in Fig. 3 is
$$y = \frac { 1 } { 10 ^ { 4 } } \left( x ^ { 3 } - 100 x ^ { 2 } - 10000 x + 2100100 \right)$$
Calculate the speed at which the car has its lowest carbon dioxide emissions and the value of its emissions at that speed.\\[0pt]
[An answer obtained from the graph will be given no marks.]\\
3 (i) In line 109 the carbon dioxide emissions for a particular train journey from Exeter to London are estimated to be 3.7 tonnes. Obtain this figure.\\
(ii) The text then goes on to state that the emissions per extra passenger on this journey are less than $\frac { 1 } { 2 } \mathrm {~kg}$. Justify this figure.\\
(i) $\_\_\_\_$\\
(ii) $\_\_\_\_$\\
4 The daily number of trains, $n$, on a line in another country may be modelled by the function defined below, where $P$ is the annual number of passengers.
$$\begin{aligned}
& n = 10 \text { for } 0 \leqslant P < 10 ^ { 6 } \\
& n = 11 \text { for } 10 ^ { 6 } \leqslant P < 1.5 \times 10 ^ { 6 } \\
& n = 12 \text { for } 1.5 \times 10 ^ { 6 } \leqslant P < 2 \times 10 ^ { 6 } \\
& n = 13 \text { for } 2 \times 10 ^ { 6 } \leqslant P < 2.5 \times 10 ^ { 6 } \\
& n = 14 \text { for } 2.5 \times 10 ^ { 6 } \leqslant P < 3 \times 10 ^ { 6 } \\
& \ldots \text { and so on } \ldots
\end{aligned}$$
(i) Sketch the graph of $n$ against $P$.\\
(ii) Describe, in words, the relationship between the daily number of trains and the annual number of passengers.\\
(i)\\
\includegraphics[width=\textwidth]{5c149cb5-7392-4219-b285-486f4694aa6f-7_716_1249_1011_440}\\
(ii) $\_\_\_\_$\\
5 The FGW website gives the conversion factor for miles to kilometres to 7 significant figures.\\
"We got the distance between the two stations by road from theaa.com. We then converted this distance to kilometres by multiplying it by $1.609344 . "$
Suppose this conversion factor is applied to a distance of exactly 100 miles.\\
State which one of the following best expresses the level of accuracy for the distance in metric units, justifying your answer.
A : to the nearest millimetre\\
B : to the nearest 10 centimetres\\
C : to the nearest metre\\
\hfill \mbox{\textit{OCR MEI C4 2010 Q8 [18]}}