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]
\includegraphics[alt={},max width=\textwidth]{666dc19e-f293-4738-8530-fce90df23d17-3_490_839_438_612}
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\caption{Fig. 6}
\end{figure}
- Write down the length AQ in terms of \(y\). Hence find AP in terms of \(y\), and show that
$$( y + 3 ) ^ { 2 } = x ^ { 2 } + 9 .$$
- Use this result to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { y + 3 }\).
- 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)