Answer:
0.65 ft/s
Explanation:
Let l be the length of the rope and x the distance from the boat to the dock, this forms a right triangle, hence:
l² = x² + 5²
substituting l = 13 ft, to get x gives:
13² = x² + 5²
x² = 13² - 5²
x² = 144
x = √144 = 12 ft
But, l² = x² + 5²; differentiating both sides with respect to time (t):
2l(dl/dt) = 2x(dx/dt)
dx/dt = l(dl / dt)/x
but dl / dt = 0.6 ft/s, hence:
dx/dt = 13 ft * 0.6 ft/s / 12 ft
dx/dt = 0.65 ft/s
The boat is approaching the dock at 0.65 ft/s
When 13 ft of rope is out, the boat will be approaching the dock at the rate of; 0.65 ft/sec.
Let x be the horizontal distance of the boat from the dock at time t
Let y be the length of rope at time t
We are given;
Rate at which rope is being pulled through the ring; dy/dt = -0.6 ft/s
We want to find dx/dt when y = 13
We are told that a ring attached to the dock at a point 5 feet higher than the front of the boat. Thus by Pythagorean Theorem, we can say;
x² + 5² = y²
Differentiating each term with respect to t gives;
2x(dx/dt) = 2y(dy/dt)
Now, When y = 13, it means that;
x² + 5² = 13²
x² + 25 = 169
x² = 169 - 25
x = √144 x = 12
We have 2x(dx/dt) = 2y(dy/dt)
Thus;
2(12)(dx/dt) = 2(13)(-0.6)
dx/dt = -(7.8/12)
dx/dt = -0.65 ft/s
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