a dock is 5 feet above water. suppose you stand on the edge of the dock and pull a rope to a boat at a constant rate of 2 ft/s. assume the boat remains at water level. at what speed is the boat approaching the dock when it is 4 feet from the dock

The boat is approaching the dock at a speed of 3.20 ft/s when it is 4 feet from the dock.

Step-by-step explanation:

The diagram of the situation described is shown in the attached image.

The distance of the boat to the dock along the water level at any time is x

The distance from the person on the dock to the boat at any time is y

The height of the dock is 5 ft.

These 3 dimensions form a right angle triangle at any time with y being the hypotenuse side.

According to Pythagoras' theorem

y² = x² + 5²

y² = x² + 25

(d/dt) y² = (d/dt) (x² + 5²)

2y (dy/dt) = 2x (dx/dt) + 0

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

When the boat is 4 ft from dock, that is x = 4 ft,

The boat is being pulled at a speed of 2 ft/s, that is, (dy/dt) = 2 ft/s

The speed with which the boat is approaching the dock = (dx/dt)

Since we are asked to find the speed with which the boat is approaching the dock when the boat is 4 ft from the dock

When the boat is 4 ft from the dock, x = 4 ft.

And we can obtain y at that point.

y² = x² + 5²

y² = 4² + 5² = 16 + 25 = 41

y = 6.40 ft.

So, to the differential equation relation

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

when x = 4 ft,

y = 6.40 ft

(dy/dt) = 2 ft/s

(dx/dt) = ?

2 × 6.40 × 2 = 2 × 4 × (dx/dt)

25.6 = 8 (dx/dt)

(dx/dt) = (25.6/8) = 3.20 ft/s.

Hope this Helps!!!

true? is that what you're asking or do you mean something else?