How fast is boat B approaching point Q

Two boats, boat A and boat B, are attached to a pulley 5 feet under the surface of the water with a 47-feet rope. Point Q is a point on the surface of the water that is directly above the pulley. If boat A is pulled 3 ft/s away from point Q, how fast is boat B approaching Q?

Suppose that boat A is being held by x feet of rope. Then, boat B is being held by 47 - x feet of rope. Then, suppose that boat A is a feet from point Q and that boat B is b feet from point Q.

By right triangles, we see that:
(a) Boat A:
a^2 + 5^2 = x^2 ==> x^2 = a^2 + 25.
(b) Boat B:
b^2 + 5^2 = (47 - x)^2 ==> x^2 - 94x + 2209 = b^2 + 25.

We know that da/dt = 3 and we want to know db/dt when a = 10. To do this, we need to link da/dt and db/dt together by linking a and b together.

By solving x^2 = a^2 + 25 for x, we see that:
x = √(a^2 + 25).

Substituting this into the equation in (b) gives:
(a^2 + 25) - 94√(a^2 + 25) + 2209 = b^2 + 25.
==> a^2 - 94√(a^2 + 25) + 2209 = b^2.

By differentiating implicitly with respect to time:
(d/dt)[a^2 - 94√(a^2 + 25) + 2209] = (d/dt)(b^2).
==> 2a(da/dt) - 94(da/dt)/√(a^2 + 25) = 2b(db/dt)
==> db/dt = [2a(da/dt) - 94(da/dt)/√(a^2 + 25)]/(2b), by solving for db/dt.

Using a^2 - 94√(a^2 + 25) + 2209 = b^2, we see that, when a = 10:
10^2 - 94√(10^2 + 25) + 2209 = b^2 ==> b = √(2309 - 470√5)

Therefore, when boat A is 10 feet from Q:
db/dt = [2a(da/dt) - 94(da/dt)/√(a^2 + 25)]/(2b)
= [2(10)(3) - 94(3)/√(10^2 + 25)]/[2√(2309 - 470√5)].
= 0.49 ft/s.

I hope this helps!