Jake throws a dart that hits (Its 8) the square
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Answer:
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!!!
