i need to solve this using related rates.

Question

A boat is pulled in to a dock by a rope with one end attached to the front of the boat and the other end passing through a ring attached to the dock at a point 6 ft higher than the front of the boat. The rope is being pulled through the ring at the rate of 0.2 ft/sec. How fast is the boat approaching the dock when 10 ft of rope is out?

Answer 1

Let x = horizontal distance of the boat from the dock at time t

     y = length of rope at time t

 

Given:  dy/dt = -0.2

Find :   dx/dt when y = 10

 

Solution:  By the Pythagorean Theorem, x² + 6² = y²

 

              So, 2x(dx/dt) = 2y(dy/dt)   When y = 10, x² + 36 = 100

 

                                                                               x = 8

 

              We have 2(8)(dx/dt) = 2(10)(-0.2)

 

                                   dx/dt = -0.25

 

When 10 ft of rope is out, the boat is approaching the dock at

the rate of 0.25 ft/sec.