how fast in ft/s is the boat approaching the dock when it is 33 ft from the dock?

Question

A boat is pulled toward a dock by means of a rope wound on a drum that is located 6 ft above the bow of the boat. If the rope is being pulled in at the rate of 5 ft/s, how fast (in ft/s) is the boat approaching the dock when it is 33 ft from the dock?

Answer 1

The Pythagorean theorem says:

a² + b² = c²

Taking the time derivative of both sides, we have:

2a(da/dt) + 2b(db/dt) = 2c(dc/dt)

The location of the drum on the boat is constant, so da/dt = 0. The rope is the hypotenuse, and it is being pulled at a rate of 5 ft/s, so dc/dt = 5. The length of the rope is c = sqrt(33² + 6²) = 15sqrt(5), b = 33, and we need to solve for db/dt. Therefore, we have:

66(db/dt) = 30sqrt(5)(5)

db/dt = (150sqrt(5))/66 = 5.0820

Therefore, the boat is approaching the dock at a rate of 5.0820 ft/s.