A pulley is on the edge of a dock, 6 feet above the water level. A rope is being used to pull In a boat. The rope is attached to the boat at water level. The rope is being pulled in at the rate of 2 ft./s find the rate at which the boat is approaching the dock at the instant the boat is 3 feet from the dock.
Let:
- y be the distance between the boat and the dock
- x be the distance between the boat and the pulley
- z be the length of the rope
Since the pulley is 6 feet above the water level and the rope is being pulled in at a rate of 2 ft/s, we have:
x^2 + 6^2 = z^2
2x(dx/dt) = 2(dz/dt)
When the boat is 3 feet from the dock (y = 3), we want to find dy/dt when y = 3. At this point, we have x = 5 (6-3-1) and z = 8 (from Pythagoras' theorem).
From the first equation:
5^2 + 6^2 = 8^2
25 + 36 = 64
61 = 64
Differentiating implicitly with respect to time t:
10x(dx/dt) = 16(dz/dt)
Solve for (dz/dt):
(dz/dt) = 5/4(dx/dt)
Substitute x = 5 and (dx/dt) = -2 into the equation:
(dz/dt) = 5/4(-2) = -5/2 ft/s
Since (dy/dt) = (dx/dt), substitute (dx/dt) = -2 and x = 5 into the equation:
(dy/dt) = -2 ft/s
Therefore, the boat is approaching the dock at a rate of 2 ft/s at the instant when the boat is 3 feet from the dock.