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!!!

hey there!

if she's saving for six months, let's divide the total number we need for the trip by six, dividing it into the amount that needs to be saved for each month.

500 ÷ 6 =

she needs to save at least this amount each month, so we should use this symbol: ≥

let x be the amount she saves each month.

your final answer is x ≥

hope this !

this shows a reflection over the x-axis and is translated one unit down and 6 units to the right (x + 6, y - 1)

hope this !

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