25. rope on a boat a rope passing through a capstan on a dock is attached to a boat offshore. the rope is…

Explanation:

Step1: Set up the relationship

Let (x) be the horizontal distance of the boat from the dock and (y) be the length of the rope. By the Pythagorean theorem, (y^{2}=x^{2}+25).

Step2: Differentiate with respect to time (t)

Differentiating both sides of (y^{2}=x^{2}+25) with respect to (t), we get (2y\frac{dy}{dt}=2x\frac{dx}{dt}), which simplifies to (y\frac{dy}{dt}=x\frac{dx}{dt}).

Step3: Find (y) when (x = 10)

When (x = 10), (y=\sqrt{10^{2}+25}=\sqrt{125}=5\sqrt{5}).

Step4: Substitute values

We know that (\frac{dy}{dt}=- 3) (negative because (y) is decreasing). Substitute (x = 10), (y = 5\sqrt{5}) and (\frac{dy}{dt}=-3) into (y\frac{dy}{dt}=x\frac{dx}{dt}). [5\sqrt{5}\times(-3)=10\times\frac{dx}{dt}] [ \frac{dx}{dt}=\frac{-15\sqrt{5}}{10}=-\frac{3\sqrt{5}}{2}\approx - 3.35] The negative sign indicates the direction (the boat is moving towards the dock). The speed is (\vert\frac{dx}{dt}\vert).

Answer:

The boat is traveling at a speed of (\frac{3\sqrt{5}}{2}\approx3.35) ft/s.