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}

\begin{enumerate}[label=(\roman*)]
\item State, with their coordinates, the points on the curve for which $\theta = -\frac{\pi}{2}$, $\theta = 0$ and $\theta = \frac{\pi}{2}$. [3]

\item 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]

\item Find the exact coordinates of the turning point Q. [3]
\end{enumerate}

When the curve is rotated about the $x$-axis, it forms a paperweight shape, as shown in Fig. 7b.

\includegraphics{figure_7b}

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{3}
\item 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]

\item Find the volume of the paperweight shape. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C4 2012 Q7 [19]}}