Question

a boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. if the rope is pulled in at a rate of 1 m/s, how fast (in m/s) is the boat approaching the dock when it is 7 m from the dock? (round your answer to two decimal places.) m/s

Explanation:

Step1: Set up the Pythagorean theorem

Let \(x\) be the distance of the boat from the dock, and \(y\) be the length of the rope. By the Pythagorean theorem, \(y^{2}=x^{2}+1^{2}=x^{2} + 1\).

Step2: Differentiate both sides with respect to time \(t\)

Differentiating \(y^{2}=x^{2}+1\) with respect to \(t\) gives \(2y\frac{dy}{dt}=2x\frac{dx}{dt}\). Then \(\frac{dx}{dt}=\frac{y}{x}\cdot\frac{dy}{dt}\).

Step3: Find \(y\) when \(x = 7\)

When \(x = 7\), \(y=\sqrt{x^{2}+1}=\sqrt{7^{2}+1}=\sqrt{49 + 1}=\sqrt{50}=5\sqrt{2}\).

Step4: Substitute values into \(\frac{dx}{dt}\) formula

We know that \(\frac{dy}{dt}=- 1\) (negative because \(y\) is decreasing). Substituting \(x = 7\), \(y = 5\sqrt{2}\), and \(\frac{dy}{dt}=-1\) into \(\frac{dx}{dt}=\frac{y}{x}\cdot\frac{dy}{dt}\), we get \(\frac{dx}{dt}=\frac{5\sqrt{2}}{7}\times(-1)\approx - 1.01\). The negative sign indicates the boat is approaching the dock.

Answer:

\(1.01\)