A boat is approaching a straight shoreline at a constant speed of 12 meters per second. A person on the boat is shining a spotlight on a point on the shoreline. If the boat is 9 meters from the shore, find the rate at which the angle of the spotlight's beam is changing when the boat is 12 meters from the point on the shore closest to it.
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Let $P$ be the point on the shoreline closest to the boat, and let $Q$ be the location of the boat, with $Q$ closer to the shoreline. Let $X$ be the point on the beam of the spotlight that hits the water. Finally, let $Y$ be the point on the shoreline that is directly opposite $X$.
[asy]
unitsize (0.5 cm);
pair P, Q, X, Y;
P = (0,0);
Q = (12,0);
X = (9,0);
Y = (9,12);
draw((0,0)--(20,0));
draw(rightanglemark((0,0),X,Y,50));
label("$P$", P, SW);
label("$Q$", Q, S);
label("$X$", X, S);
label("$Y$", Y, N);
[/asy]
We want to find $d\theta/dt$ when $Q=(9,0)$. Let $s$ be the distance between $Q$ and $P$ at any time $t$. Note that $PQ=s$, so $dPQ/dt=ds/dt$. Also note that $\triangle XQP$ and $\triangle XYQ$ are similar, so $XY/PQ=QX/QP$. This gives \[\frac{12}{s}=\frac{9}{PQ}.\] Then $9s=12\cdot PQ\implies ds=4dPQ/3$. Taking $d/dt$ of both sides gives \[ds/dt=4(dPQ/dt)/3,\] from which $dPQ/dt=3ds/dt/4$. Substituting our previous value gives \[dPQ/dt=3(ds/dt)/4=3(12)/4=\boxed{9}.\]
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