Stationary points of curves

3 A curve has equation \(y = 2 \sin 2 x - 5 \cos 2 x + 6\) and is defined for \(0 \leqslant x \leqslant \pi\). Find the \(x\)-coordinates of the stationary points of the curve, giving your answers correct to 3 significant figures.

7 \includegraphics[max width=\textwidth, alt={}, center]{6bf7ba66-8362-4ac0-8e5c-3f88a3ccdf86-12_424_488_260_826} The diagram shows the curve with equation \(y = \sin 2 x + 3 \cos 2 x\) for \(0 \leqslant x \leqslant \pi\). At the points \(P\) and \(Q\) on the curve, the gradient of the curve is 3 .

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. By first expressing \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in the form \(R \cos ( 2 x + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), find the \(x\)-coordinates of \(P\) and \(Q\), giving your answers correct to 4 significant figures.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

7 \includegraphics[max width=\textwidth, alt={}, center]{1b410c91-2fe9-46cf-8478-631b4165f98d-12_424_488_260_826} The diagram shows the curve with equation \(y = \sin 2 x + 3 \cos 2 x\) for \(0 \leqslant x \leqslant \pi\). At the points \(P\) and \(Q\) on the curve, the gradient of the curve is 3 .

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. By first expressing \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in the form \(R \cos ( 2 x + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), find the \(x\)-coordinates of \(P\) and \(Q\), giving your answers correct to 4 significant figures.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

5. \begin{figure}[h] \end{figure} The curve shown in Figure 1 has equation $$x = 3 \sin y + 3 \cos y , \quad - \frac { \pi } { 4 } < y < \frac { \pi } { 4 }$$

1. Express the equation of the curve in the form \(x = R \sin ( y + \alpha )\), where \(R\) and \(\alpha\) are constants, \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\)
2. Find the coordinates of the point on the curve where the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) is \(\frac { 1 } { 2 }\). Give your answers to 3 decimal places.

8. \begin{figure}[h] \end{figure} Figure 1 shows the graph of the function \(\mathrm { h } ( x )\) with equation $$h ( x ) = 45 + 15 \sin x + 21 \sin \left( \frac { x } { 2 } \right) + 25 \cos \left( \frac { x } { 2 } \right) \quad x \in [ 0,40 ]$$

1. Show that $$\frac { \mathrm { d } h } { \mathrm {~d} x } = \frac { \left( t ^ { 2 } - 6 t - 17 \right) \left( 9 t ^ { 2 } + 4 t - 3 \right) } { 2 \left( 1 + t ^ { 2 } \right) ^ { 2 } }$$ where \(t = \tan \left( \frac { x } { 4 } \right)\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{a52911da-4b69-4d86-975e-d10e3a481e1d-16_581_1403_1263_331} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Source: \({ } ^ { 1 }\) Data taken on 29th December 2016 from \href{http://www.ukho.gov.uk/easytide/EasyTide}{http://www.ukho.gov.uk/easytide/EasyTide} Figure 2 shows a graph of predicted tide heights, in metres, for Portland harbour from 08:00 on the 3rd January 2017 to the end of the 4th January \(2017 { } ^ { 1 }\). The graph of \(k \mathrm {~h} ( x )\), where \(k\) is a constant and \(x\) is the number of hours after 08:00 on 3rd of January, can be used to model the predicted tide heights, in metres, for this period of time.
2. 1. Suggest a value of \(k\) that could be used for the graph of \(k \mathrm {~h} ( x )\) to form a suitable model.
   2. Why may such a model be suitable to predict the times when the tide heights are at their peaks, but not to predict the heights of these peaks?
3. Use Figure 2 and the result of part (a) to estimate, to the nearest minute, the time of the highest tide height on the 4th January 2017.