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 6 feet below the level of the pulley. If the rope is pulled through the pulley at a rate of 18 ft/min, at what rate will the boat be approaching the dock when 90 ft of rope is out?
The answer is x' = 16.07 ft/min and here is the procedure to get this one: <span>y = 8 ft, the distance the boat is below the pulley. x =the distance from the bottom of the dock to the front of the boat. c= the amount of rope out. c is the hypotenuse of a right triangle formed between the height of the dock and the distance from the dock t the boat.
y = constant, while x and c are all changing with time, so:
[x(t)]² + y² = [c(t)]²
Dropping the (t) notation and differentiating implicitly with respect to time,
x² + 64 = c² 2xx' = 2cc' xx' = cc' From the problem statement, c = 90 ft c' = 16 ft/min We know that when c = 90 ft, y = 8 ft, so we can solve for x at this time. x² + 64 = 8100 x = 89.64 ft x' = cc'/x = (90 ft)(16 ft/min)/89.64 ft x' = 16.07 ft/min</span>
