OCR MEI C4 (Core Mathematics 4)

1 Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve $$x = 2 t ^ { 2 } , y = 4 t , \quad - \sqrt { 2 } \leqslant t \leqslant \sqrt { 2 } .$$ \(\mathrm { P } \left( 2 t ^ { 2 } , 4 t \right)\) is a point on the curve with parameter \(t\). TS is the tangent to the curve at P , and PR is the line through P parallel to the \(x\)-axis. Q is the point \(( 2,0 )\). The angles that PS and QP make with the positive \(x\)-direction are \(\theta\) and \(\phi\) respectively. \begin{figure}[h] \end{figure}

1. By considering the gradient of the tangent TS, show that \(\tan \theta = \frac { 1 } { t }\).
2. Find the gradient of the line QP in terms of \(t\). Hence show that \(\phi = 2 \theta\), and that angle TPQ is equal to \(\theta\).
   [0pt] [The above result shows that if a lamp bulb is placed at Q , then the light from the bulb is reflected to produce a parallel beam of light.] The inside surface of the headlight has the shape produced by rotating the curve about the \(x\)-axis.
3. Show that the curve has cartesian equation \(y ^ { 2 } = 8 x\). Hence find the volume of revolution of the curve, giving your answer as a multiple of \(\pi\).

2 Fig. 3 shows part of the curve \(y = 1 + x ^ { 2 }\), together with the line \(y = 2\). \begin{figure}[h] \end{figure} The region enclosed by the curve, the \(y\)-axis and the line \(y = 2\) is rotated through \(360 ^ { \circ }\) about the \(y\)-axis. Find the volume of the solid generated, giving your answer in terms of \(\pi\).

3 Fig. 7 shows the curve BC defined by the parametric equations $$x = 5 \ln u , y = u + \frac { 1 } { u } , \quad 1 \leqslant u \leqslant 10$$ The point A lies on the \(x\)-axis and AC is parallel to the \(y\)-axis. The tangent to the curve at C makes an angle \(\theta\) with AC, as shown. \begin{figure}[h] \end{figure}

1. Find the lengths \(\mathrm { OA } , \mathrm { OB }\) and AC .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(u\). Hence find the angle \(\theta\).
3. Show that the cartesian equation of the curve is \(y = \mathrm { e } ^ { \frac { 1 } { 5 } x } + \mathrm { e } ^ { - \frac { 1 } { 5 } x }\). An object is formed by rotating the region OACB through \(360 ^ { \circ }\) about \(\mathrm { O } x\).
4. Find the volume of the object.

4 Fig. 2 shows the curve \(y = \sqrt { 1 + x ^ { 2 } }\). \begin{figure}[h] \end{figure}

1. The following table gives some values of \(x\) and \(y\).

   \(x\)	0	0.25	0.5	0.75	1
   \(y\)	1	1.0308		1.25	1.4142

   Find the missing value of \(y\), giving your answer correct to 4 decimal places. Hence show that, using the trapezium rule with four strips, the shaded area is approximately 1.151 square units.
2. Jenny uses a trapezium rule with 8 strips, and obtains a value of 1.158 square units. Explain why she must have made a mistake.
3. The shaded area is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the exact volume of the solid of revolution formed.

5 Fig. 4 shows the curve \(y = \sqrt { 1 + \mathrm { e } ^ { 2 x } }\), and the region between the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\). \begin{figure}[h] \end{figure}

1. Find the exact volume of revolution when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
2. 1. Complete the table of values, and use the trapezium rule with 4 strips to estimate the area of the shaded region.

      \(x\)	0	0.5	1	1.5	2
      \(y\)		1.9283	2.8964	4.5919

   2. The trapezium rule for \(\int _ { 0 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { 2 x } } \mathrm {~d} x\) with 8 and 16 strips gives 6.797 and 6.823, although not necessarily in that order. Without doing the calculations, say which result is which, explaining your reasoning.

6 Fig. 8 illustrates a hot air balloon on its side. The balloon is modelled by the volume of revolution about the \(x\)-axis of the curve with parametric equations $$x = 2 + 2 \sin \theta , \quad y = 2 \cos \theta + \sin 2 \theta , \quad ( 0 \leqslant \theta \leqslant 2 \pi ) .$$ The curve crosses the \(x\)-axis at the point \(\mathrm { A } ( 4,0 )\). B and C are maximum and minimum points on the curve. Units on the axes are metres. \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\).
2. Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 6 } \pi\), and find the exact coordinates of B . Hence find the maximum width BC of the balloon.
3. (A) Show that \(y = x \cos \theta\).
   (B) Find \(\sin \theta\) in terms of \(x\) and show that \(\cos ^ { 2 } \theta = x - \frac { 1 } { 4 } x ^ { 2 }\).
   (C) Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }\).
4. Find the volume of the balloon.