6 Fig. 6 shows the triangle OAP , where O is the origin and A is the point $( 0,3 )$. The point $\mathrm { P } ( x , 0 )$ moves on the positive $x$-axis. The point $\mathrm { Q } ( 0 , y )$ moves between O and A in such a way that $\mathrm { AQ } + \mathrm { AP } = 6$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{666dc19e-f293-4738-8530-fce90df23d17-3_490_839_438_612}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}

(i) Write down the length AQ in terms of $y$. Hence find AP in terms of $y$, and show that

$$( y + 3 ) ^ { 2 } = x ^ { 2 } + 9 .$$

(ii) Use this result to show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { y + 3 }$.\\
(iii) When $x = 4$ and $y = 2 , \frac { \mathrm {~d} x } { \mathrm {~d} t } = 2$. Calculate $\frac { \mathrm { d } y } { \mathrm {~d} t }$ at this time.

Section B (36 marks)\\

\hfill \mbox{\textit{OCR MEI C3 2007 Q6 [8]}}