OCR MEI C4 (Core Mathematics 4)

2 Fig. 6 shows the arch ABCD of a bridge. \begin{figure}[h] \end{figure} The section from \(B\) to \(C\) is part of the curve \(O B C E\) with parametric equations $$x = a ( \theta - \sin \theta ) , y = a ( 1 - \cos \theta ) \text { for } 0 \leqslant \theta \leqslant 2 \pi \text {, }$$ where \(a\) is a constant.

1. Find, in terms of \(a\),
   (A) the length of the straight line OE,
   (B) the maximum height of the arch.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). The straight line sections AB and CD are inclined at \(30 ^ { \circ }\) to the horizontal, and are tangents to the curve at B and C respectively. BC is parallel to the \(x\)-axis. BF is parallel to the \(y\)-axis.
3. Show that at the point B the parameter \(\theta\) satisfies the equation $$\sin \theta = \frac { 1 } { \sqrt { 3 } } ( 1 \quad \cos \theta ) .$$ Verify that \(\theta = \frac { 2 } { 3 } \pi\) is a solution of this equation.
   Hence show that \(\mathrm { BF } = \frac { 3 } { 2 } a\), and find OF in terms of \(a\), giving your answer exactly.
4. Find BC and AF in terms of \(a\). Given that the straight line distance AD is 20 metres, calculate the value of \(a\).

3 A curve has carlesian equation \(\mathrm { y } ^ { 2 } - \mathrm { x } _ { 2 } = 4\).

1. Verify that $$\boldsymbol { x } = \boldsymbol { t } - - ^ { 1 } \quad \boldsymbol { t ^ { \prime } } \quad \boldsymbol { y } = \boldsymbol { t } + \frac { 1 } { \boldsymbol { t } ^ { \prime } }$$ are parametric equations of the curve.
   (u) Show lhat \(\left. \underset { d x } { \mathbf { d y } } = \frac { ( t - I ) ( r } { 12 + 1 } + 1 \right)\). Hence find the coordinates of the staionary points of the curve.

4 The parametric equations of a curve are $$x = \sin \theta , \quad y = \sin 2 \theta , \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi$$

1. Find the exact value of the gradient of the curve at the point where \(\theta = \frac { 1 } { 6 } \pi\).
2. Show that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4 x ^ { 4 }\).

5 A curve is defined parametrically by the equations $$x = \frac { 1 } { 1 + t } , \quad y = \frac { 1 - t } { 1 + 2 t }$$ Find \(t\) in terms of \(x\). Hence find the cartesian equation of the curve, giving your answer as simply as possible.

6 A curve has parametric equations $$x = \mathrm { e } ^ { 2 t } , \quad y = \frac { 2 t } { 1 + t }$$

1. Find the gradient of the curve at the point where \(t = 0\).
2. Find \(y\) in terms of \(x\).