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
