A boat pulls away from a dock at 2 m/s, but the operator has neglected to remove the tow rope used to pull the boat up to the dock. This rope runs thru a pulley which is attached to the dock at a point 1 m higher than the point at which the rope is attached to the boat.
A) How fast is the rope being pulled out when the boat is 3 m away from the dock
B) How fast is the rope is pulled as the distance from the dock grows large? Justify your answer
at a distance x, the length of rope (z) is
z^2 = x^2+1
now just find dz/dx at x=3
it should be clear how z changes when x is large.
oops. I meant find dz/dt.
To solve this problem, we can use the concept of related rates. We need to find the rate at which the rope is being pulled out when the boat is a certain distance away from the dock.
Let's start by defining some variables:
- Let x be the horizontal distance between the boat and the dock.
- Let h be the vertical distance between the rope and the dock.
A) To find the rate at which the rope is being pulled out when the boat is 3 m away from the dock, we need to find dx/dt (the rate of change of x) when x = 3.
To go about solving this, we can use the Pythagorean Theorem to relate x, h, and the length of the rope. Notice that the rope and the horizontal line between the rope and the dock form a right triangle.
Using the Pythagorean Theorem, we have:
x^2 + h^2 = (3 + 1)^2 = 16
Differentiating both sides with respect to time (t), we get:
2x(dx/dt) + 2h(dh/dt) = 0
Since we are interested in finding dx/dt when x = 3, and we know that h = 1, we can substitute these values into the equation:
2(3)(dx/dt) + 2(1)(dh/dt) = 0
6(dx/dt) + 2(dh/dt) = 0
Simplifying the equation, we have:
6(dx/dt) = -2(dh/dt)
Dividing by 2, we get:
3(dx/dt) = -dh/dt
Now, we want to find dx/dt when x = 3. Plugging the given values into the equation, we have:
3(dx/dt) = -dh/dt
3(dx/dt) = -v
where v represents the rate at which the rope is being pulled out.
Therefore, when the boat is 3 m away from the dock, the rope is being pulled out at a rate of v = -3(dx/dt) m/s.
B) To determine how fast the rope is being pulled as the distance from the dock grows large, we observe that the right triangle formed by the boat, the dock, and the rope approaches a right triangle similar to a 3-4-5 triangle.
As x becomes very large, the ratio of h to x closely approximates 3/4.
Let's say that as distance increases, the rope is being pulled out at a rate of w m/s. Then we have:
w = -3(dx/dt)
Since dx/dt is the rate at which the boat is moving horizontally, it remains constant regardless of the distance from the dock. This means that as x becomes larger, dx/dt does not change.
Therefore, as the distance from the dock grows large, the rope is being pulled out at a constant rate equal to 3 times the rate at which the boat is moving horizontally, regardless of the specific value of dx/dt.
In summary:
A) When the boat is 3 m away from the dock, the rope is being pulled out at a rate of v = -3(dx/dt) m/s.
B) As the distance from the dock grows large, the rope is being pulled out at a constant rate of 3 times the rate at which the boat is moving horizontally, regardless of the specific value of dx/dt.