related rates calculus

Question

1 point) A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the front of the boat, which is 7 feet below the level of the pulley. If the rope is pulled through the pulley at a rate of 12 ft/min, at what rate will the boat be approaching the dock when 100 ft of rope is out?

The boat will be approaching the dock at  ft/min.

Answer 1

The height of the dock, distance from the boat to the dock and the length of the rope from the pulley to the

boat form a right triangle. Let x be the distance from the dock to the boat and s be the length of the rope, which is the hypotenuse of the triangle.

This relationship is:

s² = 7²+x²

ds/dt is given as 12 ft/min.

Taking the derivative of both sides of the equation above,

2ss' = 2xx'

We want the value at x' when x = 100 ft. The length of the rope then is, s= √(49+10000) =√(10049) ≈100.24 ft. Solving for x',

x' = ss'/x = (100.24)(12)/100 ≈ 12.029 ft/min

The boat will be approaching the dock at 12.029 ft/min.