A boat is pulled into a dock by a rope attached to the bow (front end) of the boat and passing through a pulley on the dock that is 4 m higher than the bow of the boat. If the rope is pulled in at a rate of 3 m/s, at what speed is the boat approaching the dock when it is 8 m from the dock?
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To solve this problem, we can use the concept of related rates. Since the boat is being pulled in, we are interested in finding the rate at which the boat is approaching the dock.
Let's assign variables to the quantities involved:
Let's call the distance between the boat and the dock at any given time "x" (measured in meters).
Let's call the distance between the boat and the dock when it is 8 m away "x₀" (measured in meters).
Let's call the height of the pulley "h" (measured in meters).
We are given that the rope is being pulled in at a rate of 3 m/s. This tells us that the rate of change of "x" with respect to time (dx/dt) is -3 m/s since the distance is decreasing.
We are also given that the height of the pulley is 4 m. This tells us that the rate of change of "h" with respect to time (dh/dt) is 0 m/s since the height remains constant.
Now, let's use the Pythagorean theorem to relate "x", "h", and the distance of the dock from the bow of the boat ("d"):
x² + h² = d²
Taking the derivative of both sides with respect to time t, we get:
2x(dx/dt) + 2h(dh/dt) = 2d(dd/dt)
Since we want to find the rate at which the boat is approaching the dock (dx/dt) when it is 8 m away (x₀ = 8 m), we can substitute the given values into the equation and solve for dx/dt.
Plugging in the known values:
x = 8 m
h = 4 m
dx/dt = -3 m/s
dh/dt = 0 m/s
8² + 4² = d²
64 + 16 = d²
80 = d²
d = √80 = 4√5
Now, substituting the values into the related rates equation:
2(8)(-3) + 2(4)(0) = 2(4√5)(dd/dt)
Simplifying:
-48 = 8√5(dd/dt)
dd/dt = -6/√5 m/s
Therefore, the boat is approaching the dock at a speed of -6/√5 m/s when it is 8 m away.
Note: The negative sign indicates that the boat is moving towards the dock.