Aboat 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: 
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

C.9

Step-by-step explanation: