OCR MEI C2 — Question 9

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeApplied context: real-world solid
DifficultyStandard +0.3 This is a straightforward application question requiring: (A) completing the square to find maximum height, (B) explaining and computing a simple integral of a quadratic, and (ii) applying trapezium rule with given strips then multiplying by length. All techniques are routine C2 level with no novel problem-solving required, making it slightly easier than average.
Spec1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration
9
  1. A tunnel is 100 m long. Its cross-section, shown in Fig. 9.1, is modelled by the curve $$y = \frac { 1 } { 4 } \left( 10 x - x ^ { 2 } \right) ,$$ where \(x\) and \(y\) are horizontal and vertical distances in metres. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{1e43ddbe-ae95-467b-a527-351ab8a4c4fe-004_506_812_676_653} \captionsetup{labelformat=empty} \caption{Figure 9.1}
    \end{figure} Using this model,
    (A) find the greatest height of the tunnel,
    (B) explain why \(100 \int _ { 0 } ^ { 10 } y \mathrm {~d} x\) gives the volume, in cubic metres, of earth removed to make the tunnel. Calculate this volume.
  2. The roof of the tunnel is re-shaped to allow for larger vehicles. Fig. 9.2 shows the new crosssection. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{1e43ddbe-ae95-467b-a527-351ab8a4c4fe-004_513_1256_1894_575} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
    \end{figure} Use the trapezium rule with 5 strips to estimate the new cross-sectional area.
    Hence estimate the volume of earth removed when the tunnel is re-shaped.
A farmer digs ditches for flood relief. He experiments with different cross-sections. Assume that the surface of the ground is horizontal.
(i) Fig. 9.1 shows the cross-section of one ditch, with measurements in metres. The width of the ditch is 1.2 m and Fig. 9.1 shows the depth every 0.2 m across the ditch.
Use the trapezium rule with six intervals to estimate the area of cross-section. Hence estimate the volume of water that can be contained in a 50-metre length of this ditch. [5]
(ii) Another ditch is 0.9 m wide, with cross-section as shown in Fig. 9.2.
With x- and y-axes as shown in Fig. 9.2, the curve of the ditch may be modelled closely by \(y = 3.8x^4 - 6.8x^3 + 7.7x^2 - 4.2x\).
(A) The actual ditch is 0.6 m deep when \(x = 0.2\). Calculate the difference between the depth given by the model and the true depth for this value of \(x\). [2]
(B) Find \(\int (3.8x^4 - 6.8x^3 + 7.7x^2 - 4.2x) \, dx\). Hence estimate the volume of water that can be contained in a 50-metre length of this ditch. [5]
A farmer digs ditches for flood relief. He experiments with different cross-sections. Assume that the surface of the ground is horizontal.

(i) Fig. 9.1 shows the cross-section of one ditch, with measurements in metres. The width of the ditch is 1.2 m and Fig. 9.1 shows the depth every 0.2 m across the ditch.

Use the trapezium rule with six intervals to estimate the area of cross-section. Hence estimate the volume of water that can be contained in a 50-metre length of this ditch. [5]

(ii) Another ditch is 0.9 m wide, with cross-section as shown in Fig. 9.2.

With x- and y-axes as shown in Fig. 9.2, the curve of the ditch may be modelled closely by $y = 3.8x^4 - 6.8x^3 + 7.7x^2 - 4.2x$.

(A) The actual ditch is 0.6 m deep when $x = 0.2$. Calculate the difference between the depth given by the model and the true depth for this value of $x$. [2]

(B) Find $\int (3.8x^4 - 6.8x^3 + 7.7x^2 - 4.2x) \, dx$. Hence estimate the volume of water that can be contained in a 50-metre length of this ditch. [5]
9
\begin{enumerate}[label=(\roman*)]
\item A tunnel is 100 m long. Its cross-section, shown in Fig. 9.1, is modelled by the curve

$$y = \frac { 1 } { 4 } \left( 10 x - x ^ { 2 } \right) ,$$

where $x$ and $y$ are horizontal and vertical distances in metres.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{1e43ddbe-ae95-467b-a527-351ab8a4c4fe-004_506_812_676_653}
\captionsetup{labelformat=empty}
\caption{Figure 9.1}
\end{center}
\end{figure}

Using this model,\\
(A) find the greatest height of the tunnel,\\
(B) explain why $100 \int _ { 0 } ^ { 10 } y \mathrm {~d} x$ gives the volume, in cubic metres, of earth removed to make the tunnel. Calculate this volume.
\item The roof of the tunnel is re-shaped to allow for larger vehicles. Fig. 9.2 shows the new crosssection.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{1e43ddbe-ae95-467b-a527-351ab8a4c4fe-004_513_1256_1894_575}
\captionsetup{labelformat=empty}
\caption{Fig. 9.2}
\end{center}
\end{figure}

Use the trapezium rule with 5 strips to estimate the new cross-sectional area.\\
Hence estimate the volume of earth removed when the tunnel is re-shaped.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C2  Q9}}
This paper (3 questions)
View full paper
Q7 Q9 Q11