OCR MEI C2 (Core Mathematics 2)

1 Oskar is designing a building. Fig. 12 shows his design for the end wall and the curve of the roof. The units for \(x\) and \(y\) are metres. \begin{figure}[h] \end{figure}

1. Use the trapezium rule with 5 strips to estimate the area of the end wall of the building.
2. Oskar now uses the equation \(y = - 0.001 x ^ { 3 } - 0.025 x ^ { 2 } + 0.6 x + 9\), for \(0 \leqslant x \leqslant 15\), to model the curve of the roof.
   (A) Calculate the difference between the height of the roof when \(x = 12\) given by this model and the data shown in Fig. 12.
   (B) Use integration to find the area of the end wall given by this model.

2 Fig. 7 shows a curve and the coordinates of some points on it. \begin{figure}[h] \end{figure} Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve and the positive \(x\) - and \(y\)-axes.

3 A farmer digs ditches for flood relief. He experiments with different cross-sections. Assume that the surface of the ground is horizontal.

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_432_640_410_745} \captionsetup{labelformat=empty} \caption{Fig. 9.1}
   \end{figure} 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.
2. Another ditch is 0.9 m wide, with cross-section as shown in Fig. 9.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_579_813_1336_656} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} With \(x\) - and \(y\)-axes as shown in Fig. 9.2, the curve of the ditch may be modelled closely by \(y = 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x\).
   (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 \(x\).
   (B) Find \(\int \left( 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x \right) \mathrm { d } x\). Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.

4

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-4_765_757_203_764} \captionsetup{labelformat=empty} \caption{Fig. 11.1}
   \end{figure} A boat travels from P to Q and then to R . As shown in Fig. 11.1, Q is 10.6 km from P on a bearing of \(045 ^ { \circ }\). R is 9.2 km from P on a bearing of \(113 ^ { \circ }\), so that angle QPR is \(68 ^ { \circ }\). Calculate the distance and bearing of R from Q .
2. Fig. 11.2 shows the cross-section, EBC, of the rudder of a boat. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-4_527_1474_1452_404} \captionsetup{labelformat=empty} \caption{Fig. 11.2}
   \end{figure} BC is an arc of a circle with centre A and radius 80 cm . Angle \(\mathrm { CAB } = \frac { 2 \pi } { 3 }\) radians.
   EC is an arc of a circle with centre D and radius \(r \mathrm {~cm}\). Angle CDE is a right angle.
   1. Calculate the area of sector ABC .
   2. Show that \(r = 40 \sqrt { 3 }\) and calculate the area of triangle CDA.
   3. Hence calculate the area of cross-section of the rudder. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-5_695_1012_271_600} \captionsetup{labelformat=empty} \caption{Fig. 12}
      \end{figure} A water trough is a prism 2.5 m long. Fig. 12 shows the cross-section of the trough, with the depths in metres at 0.1 m intervals across the trough. The trough is full of water.
   4. Use the trapezium rule with 5 strips to calculate an estimate of the area of cross-section of the trough. Hence estimate the volume of water in the trough.
   5. A computer program models the curve of the base of the trough, with axes as shown and units in metres, using the equation \(y = 8 x ^ { 3 } - 3 x ^ { 2 } - 0.5 x - 0.15\), for \(0 \leqslant x \leqslant 0.5\). Calculate \(\int _ { 0 } ^ { 0.5 } \left( 8 x ^ { 3 } - 3 x ^ { 2 } - 0.5 x - 0.15 \right) \mathrm { d } x\) and state what this represents.
      Hence find the volume of water in the trough as given by this model.