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
A man pulls a rope over a simple pulley attached to the bow of a rowboat. The man’s hands are 8 feet above the water and the bow of the boat is 1 foot above the water. How fast must the man pull the rope to tow the boat in at 3 feet per second when it is 25 feet from the dock? How fast must he pull the rope to tow the boat at that same speed when it is 10 feet from the dock? If he pulls the rope at 3 feet per second when the boat is 18 feet from the dock, how fast is the boat moving?
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