The girl at C stands near the edge of the pier and pulls in the rope horizontally at constant speed 6 ft/s. Determine how fast the boat approaches the pier at the instant the rope length AB is 50 ft. (h = 8 ft)

Answer

0 ft/s

Determine the value of AC (distance between girl and pier). AC = sqrt(BC^2 + h^2) A:AC = sqrt(50^2 + 8^2) = 50.8ft
Determine the rate at which the length of rope AB is changing. AB^2 = AC^2 + BC^2 (by Pythagoras theorem) Differentiating w.r.t time, 2AB(dAB/dt) = 2AC(dAC/dt) + 2BC(dBC/dt) Since the rope is being pulled horizontally, dBC/dt = 0 Also, at the instant when AB = 50 ft, AC = 50.8 ft and dAC/dt = 0 (because girl is standing still) Substituting these values, we get 2(50)(dAB/dt) = 2(50.8)(0) + 2(50)(0) So, dAB/dt = 0 A: The boat is approaching the pier at 0 ft/s.