Posted by jonh on Tuesday, February 19, 2013 at 2:10am.
Your situation can be described by a right-angled triangle, where the base is along the water.
let the boat be x ft from the dock, let the length of rope from the man to the boat be y ft
then y^2 = x^2 + 7^2
2y dy/dt = 2x dx/dt
y dy/dt = x dx/dt
case #1, dx/dt = -3 ft/sec, x = 25
y^2 = 25^2 + 7^ = 674 , ----> y = √674
√674 dy/dt = 25(-3)
dy/dt = -75/√674 = appr -2.89 ft/sec
case 2, when x = 10, dx/dt = -3
y^2 = 100+49 = 149 , ---> y = √149
√149 dy/dt = 10(-3)
dy/dt = --2.46 ft/s
case3 , when x = 28 , dy/dt = -3
y^2 = 784+49 = 833 , -----> y = √833
√833(-3 = 28 dx/dt
dx/dt = √833(-3)/28 = -3.09 ft/s
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