OCR MEI C4 2010 June — Question 5 8 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Year2010
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeFind dy/dx expression in terms of parameter
DifficultyStandard +0.3 This is a standard C4 parametric equations question with routine techniques: finding dy/dx using the chain rule, locating stationary points where dx/dθ=0, and solving a trigonometric equation using R-formula. All steps are textbook exercises requiring no novel insight, making it slightly easier than average.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.10b Vectors in 3D: i,j,k notation1.10f Distance between points: using position vectors
  1. Verify that \(\overrightarrow { \mathrm { AB } } = \left( \begin{array} { l } 300 \\ 100 \\ 100 \end{array} \right)\) and find the length of the pipeline.
  2. Write down a vector equation of the line AB , and calculate the angle it makes with the vertical. A thin flat layer of hard rock runs through the mountain. The equation of the plane containing this layer is \(x + 2 y + 3 z = 320\).
  3. Find the coordinates of the point where the pipeline meets the layer of rock.
  4. By calculating the angle between the line AB and the normal to the plane of the layer, find the angle at which the pipeline cuts through the layer. 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]
    \includegraphics[alt={},max width=\textwidth]{5c149cb5-7392-4219-b285-486f4694aa6f-4_602_1447_488_351} \caption{Fig. 8}
    \end{figure}
  5. Find the values of the parameter at A and B . Hence show that the ratio of the lengths OA and AC is \(( \pi - 1 ) : ( \pi + 1 )\).
  6. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Find the gradient of the track at A .
  7. Show that, when the gradient of the track is \(1 , \theta\) satisfies the equation $$\cos \theta - 4 \sin \theta = 2 .$$
  8. 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:
    • Insert (inserted)
    • MEI Examination Formulae and Tables (MF2)
    \section*{Other Materials Required:}
    • Rough paper
    • Scientific or graphical calculator
    Wednesday 9 June 2010 Afternoon \includegraphics[max width=\textwidth, alt={}, center]{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
  9. 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.
  10. 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.
  11. \(\_\_\_\_\)
  12. \(\_\_\_\_\) 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. $$\begin{aligned} & n = 10 \text { for } 0 \leqslant P < 10 ^ { 6 } \\ & n = 11 \text { for } 10 ^ { 6 } \leqslant P < 1.5 \times 10 ^ { 6 } \\ & n = 12 \text { for } 1.5 \times 10 ^ { 6 } \leqslant P < 2 \times 10 ^ { 6 } \\ & n = 13 \text { for } 2 \times 10 ^ { 6 } \leqslant P < 2.5 \times 10 ^ { 6 } \\ & n = 14 \text { for } 2.5 \times 10 ^ { 6 } \leqslant P < 3 \times 10 ^ { 6 } \\ & \ldots \text { and so on } \ldots \end{aligned}$$
  13. Sketch the graph of \(n\) against \(P\).
  14. Describe, in words, the relationship between the daily number of trains and the annual number of passengers.
  15. \includegraphics[max width=\textwidth, alt={}, center]{5c149cb5-7392-4219-b285-486f4694aa6f-7_716_1249_1011_440}
  16. \(\_\_\_\_\)
Question 5:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{3}{(y-2)(y+1)} = \frac{A}{y-2} + \frac{B}{y+1}\)
\(3 = A(y+1) + B(y-2)\)M1 substituting, equating coefficients or cover up
\(y=2 \Rightarrow 3 = 3A \Rightarrow A = 1\)A1
\(y=-1 \Rightarrow 3 = -3B \Rightarrow B = -1\)A1 [3]
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{dy}{dx} = x^2(y-2)(y+1)\)
\(\int \frac{3\,dy}{(y-2)(y+1)} = \int 3x^2\,dx\)M1 separating variables
\(\int\left(\frac{1}{y-2} - \frac{1}{y+1}\right)dy = \int 3x^2\,dx\)
\(\ln(y-2) - \ln(y+1) = x^3 + c\)B1ft \(\ln(y-2) - \ln(y+1)\); ft their \(A, B\)
\(\ln\left(\frac{y-2}{y+1}\right) = x^3 + c\)B1 \(x^3 + c\)
\(\frac{y-2}{y+1} = e^{x^3+c} = e^{x^3}\cdot e^c = Ae^{x^3}\)M1, E1 [5] anti-logging including \(c\); www 
# Question 5:

## Part (i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{3}{(y-2)(y+1)} = \frac{A}{y-2} + \frac{B}{y+1}$ | | |
| $3 = A(y+1) + B(y-2)$ | M1 | substituting, equating coefficients or cover up |
| $y=2 \Rightarrow 3 = 3A \Rightarrow A = 1$ | A1 | |
| $y=-1 \Rightarrow 3 = -3B \Rightarrow B = -1$ | A1 [3] | |

## Part (ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = x^2(y-2)(y+1)$ | | |
| $\int \frac{3\,dy}{(y-2)(y+1)} = \int 3x^2\,dx$ | M1 | separating variables |
| $\int\left(\frac{1}{y-2} - \frac{1}{y+1}\right)dy = \int 3x^2\,dx$ | | |
| $\ln(y-2) - \ln(y+1) = x^3 + c$ | B1ft | $\ln(y-2) - \ln(y+1)$; ft their $A, B$ |
| $\ln\left(\frac{y-2}{y+1}\right) = x^3 + c$ | B1 | $x^3 + c$ |
| $\frac{y-2}{y+1} = e^{x^3+c} = e^{x^3}\cdot e^c = Ae^{x^3}$ | M1, E1 [5] | anti-logging including $c$; www |

---
(i) Verify that $\overrightarrow { \mathrm { AB } } = \left( \begin{array} { l } 300 \\ 100 \\ 100 \end{array} \right)$ and find the length of the pipeline.\\
(ii) Write down a vector equation of the line AB , and calculate the angle it makes with the vertical.

A thin flat layer of hard rock runs through the mountain. The equation of the plane containing this layer is $x + 2 y + 3 z = 320$.\\
(iii) Find the coordinates of the point where the pipeline meets the layer of rock.\\
(iv) By calculating the angle between the line AB and the normal to the plane of the layer, find the angle at which the pipeline cuts through the layer.

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[alt={},max 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[max width=\textwidth, alt={}, center]{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[max width=\textwidth, alt={}, center]{5c149cb5-7392-4219-b285-486f4694aa6f-7_716_1249_1011_440}\\
(ii) $\_\_\_\_$\\

\hfill \mbox{\textit{OCR MEI C4 2010 Q5 [8]}}
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