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 0.4 ft/sec. How fast is the boat approaching the dock when 10 ft of rope is out?

Answer 1

The speed of the boat when is approaching the dock is

v=0.5
`(ft)/(s)`

Step-by-step explanation:

When the boat is 10ft os the rope using trigonometry can find the length to know the speed the boat is

`h^(2) =x^(2) +y^(2) \\h=10ft\\y=6ft\\10^(2)=x^(2)+6^(2)\\x^(2)=10^(2)-6^(2) \\ x=√(100-36) =√(64)\\  x=8 ft`

Now the velocity final is just at x axis so iis derivate the function can find the velocity

`(dh)/(dt)*h=(dx)/(dt)*x+(dy)/(dt)*y\\(dy)/(dt)=0\\0.4(ft)/(s)*10=(dx)/(dt)*8+0(ft)/(s)*6\\(dx)/(dt)*8ft=0.4(ft)/(s)*10ft\\(dx)/(dt)=0.5(ft)/(s)`