Question

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 4 m from the dock

Answer 1

-1.031 m/s or
`(-√(17) )/(4)`

Explanation:

We take the length of the rope from the dock to the bow of the boat as y.

We take x be the horizontal distance from the dock to the boat.

We know that the rate of change of the rope length is
`(dy)/(dt)` = -1 m/s

We need to find the rate of change of the horizontal distance from the dock to the boat =
`(dx)/(dt)` = ?

for x = 4

Applying Pythagorean Theorem we have

`1^(2) +x^(2) =y^(2)` .... equ 1

solving, where x = 4, we have

`1^(2) +4^(2) =y^(2)`

`y^(2) = 17`

`y = √(17)`

Differentiating equ 1 implicitly with respect to t, we have

`2x(dx)/(dt) = 2y(dy)/(dt)`

substituting values of

x = 4

y =
`√(17)`

`(dy)/(dt)` = -1

into the equation, we get

`2(4)(dx)/(dt) = 2(√(17) )(-1)`

`(dx)/(dt) = (-√(17) )/(4)` = -1.031 m/s