At 5 P.M. an oil tanker traveling west in the ocean at 16 kilometers per hour pa

Question

At 5 P.M. an oil tanker traveling west in the ocean at 16 kilometers per hour passes the same spot as a luxury liner that arrived at the same spot at 4 P.M. while traveling north at 19 kilometers per hour. If the "spot" is represented by the origin, find the location of the oil tanker and the location of the luxury liner t hours after 4 P.M. Then find the distance D between the oil tanker and the luxury liner at that time. D(t) = _______ At what time were the ships closest together? (Hint: Minimize the distance (or the square of the distance!) between them.) The time is ___:___ P.M. (Round down to the nearest minute.)

Explanation / Answer

At 9:00 PM the oil tanker is 16 km east of the origin. At time t after that, it is at (16 - 16t, 0). At 9:00 PM the luxury liner is at the origin. At time t after that, it is at (0, 24t) The distance d between them is D(t) = ((0 - 24t)^2 + ((16 - 16t) - 0)^2)^(1/2) D(t) = (24^2*t^2 + 16^2*(t^2 - 2t + 1))^(1/2) D(t) = (3^2*8^2*t^2 + 2^2*8^2*(t^2 - 2t + 1))^(1/2) D(t) = 8(9t^2 + 4(t^2 - 2t + 1))^(1/2) D(t) = 8(13t^2 - 8t + 4)^(1/2) D'(t) = (1/2)*8(13t^2 - 8t + 4)^(-1/2)(26t - 8) = 0 26t - 8 = 0 26t = 8 t = 0.3077 hr t = 18 mins They are closest together at 9:18 PM. ]

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