A boat leaves a dock at 2:00 PM and travels due south at a speed of….?

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A boat leaves a dock at 2:00 PM and travels due south at a speed of 20 km/h. Another boat departs from the same dock at 3:00 PM, traveling due east at a speed of 15 km/h. At what time were the two boats closest together? This is a calculus optimization problem, and I am unsure how to approach it.

Anonymous · Accepted Answer

make sure to sketch this out...make the dock the origin, with east being +x and north being + y (you could just as easily make west +x and south +y, the distance will work out the same)let t be the time that the south-bound boat leaves the dockit's position can be described as (0 , -20t)the east-bound boat can be described as (15(t - 1) , 0)note that t = 0 is 2:00, and at t = 0, the east-bound boat is at (-15 , 0)the distance between them squared will be a minimum when the distance between them is a minimum, thus we'll work with distance squaredd^2 = [15(t - 1) - 0]^2 + (0 + 20t)^2d^2 = [15(t - 1)]^2 + (20t)^2d^2 = 225(t^2 - 2t + 1) + 400t^2d^2 = 625t^2 - 450t + 225d^2 ' = 1250t - 450d^2 " = 1250, confirming that this distance will be a minimumd^2 ' = 0 when 1250t = 450t = 450 / 1250 = 45 / 125 = 9 / 25at 9/25 of an hour past 2:00, the boats were closest together(9/25) * 60 = (9 * 12) / 5 = 108 / 5 = 21.6 minutesclosest approach: 2:21.6 pmthis vertical leg will be [-(9/25)(20)]^2 = (-36/5)^2the horizontal leg will be [15(9/25 - 1)]^2 = [15(-16/25)]^2 = (-48/5)^2the distance squared will be (36/5)^2 + (48/5)^2 = (60/5)^2the minimum distance at this time will be 12 km apartnote that this is a 3-4-5 triangle..., which makes sense since the speeds are in a ratio of 3:4, so the hypotenuse will complete the 3:4:5...

Anonymous

A boat leaves a dock at 2:00 PM and travels due south at a speed of….?

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