Find stationary/turning points

7 A curve has parametric equations $$x = 2 \sin t , \quad y = \cos 2 t + 2 \sin t$$ for \(- \frac { 1 } { 2 } \pi \leqslant t \leqslant \frac { 1 } { 2 } \pi\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 - 2 \sin t\) and hence find the coordinates of the stationary point.
2. Find the cartesian equation of the curve.
3. State the set of values that \(x\) can take and hence sketch the curve.

Fig. 7a shows the curve with the parametric equations $$x = 2\cos\theta, \quad y = \sin 2\theta, \quad -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}.$$ The curve meets the \(x\)-axis at O and P. Q and R are turning points on the curve. The scales on the axes are the same. \includegraphics{figure_7a}

1. State, with their coordinates, the points on the curve for which \(\theta = -\frac{\pi}{2}\), \(\theta = 0\) and \(\theta = \frac{\pi}{2}\). [3]
2. Find \(\frac{dy}{dx}\) in terms of \(\theta\). Hence find the gradient of the curve when \(\theta = \frac{\pi}{2}\), and verify that the two tangents to the curve at the origin meet at right angles. [5]
3. Find the exact coordinates of the turning point Q. [3]

When the curve is rotated about the \(x\)-axis, it forms a paperweight shape, as shown in Fig. 7b. \includegraphics{figure_7b}

1. Express \(\sin^2\theta\) in terms of \(x\). Hence show that the cartesian equation of the curve is \(y^2 = x^2(1 - \frac{1}{4}x^2)\). [4]
2. Find the volume of the paperweight shape. [4]

A curve has parametric equations $$x = t + a \sin t, \quad y = 1 - a \cos t,$$ where \(a\) is a positive constant.

1. Draw, on separate diagrams, sketches of the curve for \(-2\pi < t < 2\pi\) in the cases \(a = 1\), \(a = 2\) and \(a = 0.5\). By investigating other cases, state the value(s) of \(a\) for which the curve has
   1. loops,
   2. cusps. [7]
2. Suppose that the point P\((x, y)\) lies on the curve. Show that the point P\('(-x, y)\) also lies on the curve. What does this indicate about the symmetry of the curve? [3]
3. Find an expression in terms of \(a\) and \(t\) for the gradient of the curve. Hence find, in terms of \(a\), the coordinates of the turning points on the curve for \(-2\pi < t < 2\pi\) and \(a \neq 1\). [5]
4. In the case \(a = \frac{1}{2}\pi\), show that \(t = \frac{1}{3}\pi\) and \(t = \frac{5}{3}\pi\) give the same point. Find the angle at which the curve crosses itself at this point. [3]

A design for a surfboard is shown in Figure 1. Figure 1 \includegraphics{figure_6_1} The curve of the top half of the surfboard can be modelled by the parametric equations $$x = -2t^2$$ $$y = 9t - 0.7t^2$$ for \(0 \leq t \leq 9.5\) as shown in Figure 2, where \(x\) and \(y\) are measured in centimetres. Figure 2 \includegraphics{figure_6_2}

1. Find the length of the surfboard. [2 marks]
2. 1. Find an expression for \(\frac{dy}{dx}\) in terms of \(t\). [3 marks]
   2. Hence, show that the width of the surfboard is approximately one third of its length. [4 marks]