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A boat floating several feet away from a dock is pulled in by a rope being wound up by a windlass at the rate of [tex]3 \text{ ft/sec}[/tex]. If the windlass is [tex]4[/tex] feet above the level of the boat, how fast is the boat moving through the water when it is [tex]12[/tex] feet from the dock?

Answer :

Final answer:

At the moment the boat is 12 feet away from the dock and being pulled in by a rope winding at 3 ft/s, the boat is moving towards the dock at a speed of 0.75 ft/s.

Explanation:

This is a problem aligned with calculus and specifically the branch known as related rates. Here the rate of change of one quantity (distance of the boat from the dock), is related to another quantity (the speed the boat is moving through the water).

Let's denote the horizontal distance of the boat from the dock as x(t) and the length of the rope as y(t). We know that x(t) and y(t) are linked by the Pythagorean theorem as y(t) = sqrt[x(t)² + 16].

We're given dy/dt = -3 ft/s (the windlass lets in the rope). We're asked to find dx/dt when x(t) is 12 feet. For that, we can use the related rates formula: dx/dt = -x/y * dy/dt. We plug our values: dx/dt = -12/sqrt[144+16] * -3 = 0.75 ft/sec.

So, when the boat is 12 feet away from the dock, it is moving towards the dock at a speed of 0.75 ft/sec.

Learn more about Related Rates here:

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