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?
Potential Answer: got 1.89m/s, but not entirely sure if I'm doing this correctly.
B) How fast is the rope is pulled as the distance from the dock grows large? Justify your answer
Potential Answer: If the rope gets larger, the speed of the rope increases as well?
as we saw earlier, the distance z is
z^2 = 1+x^2
z dz/dt = x dx/dt
plugging in the numbers,
z(3) = ?10, so
?10 dz/dt = 3 * 2
dz/dt = 6/?10 = 1.897
So, I guess you were right on that one.
As for dz/dt when x gets large, consider that
z^2 = 1+x^2
as x gets large, that is just
z^2 ? x^2
z dz/dt ? x dx/dt
But z ? x, so, when the boat is far away, the rope's speed approaches the boat's speed.
http://www.wolframalpha.com/input/?i=derivative+sqrt(1%2Bx%5E2)
To calculate the speed at which the rope is being pulled out when the boat is 3 m away from the dock, you can use the concept of similar triangles.
Let's denote the distance between the boat and the dock as x, and the height above the dock where the rope goes through the pulley as h.
Since the boat is moving at a constant speed of 2 m/s, the rate at which the boat is moving away from the dock is also 2 m/s. Therefore, the rate at which the boat is stretching the rope and pulling it through the pulley is also 2 m/s.
To find the speed at which the rope is being pulled out, we need to find the rate at which the length of the rope between the boat and the pulley is changing. This can be calculated using similar triangles.
Let's consider a right-angled triangle formed by the boat, the pulley, and the dock. The sides of this triangle can be labeled as follows:
- The horizontal distance from the boat to the pulley is x.
- The vertical distance from the pulley to the dock is h + 1 m (since the pulley is attached to the dock 1 m higher than the attachment point on the boat).
- The hypotenuse of this triangle represents the length of the rope between the boat and the pulley.
We can find the length of the hypotenuse using the Pythagorean theorem:
hypotenuse^2 = x^2 + (h + 1)^2
Differentiating both sides of this equation with respect to time gives us:
2 * hypotenuse * rate of change of hypotenuse = 2 * x * rate of change of x + 2 * (h + 1) * rate of change of h
Simplifying this equation by dividing both sides by 2 gives us:
hypotenuse * rate of change of hypotenuse = x * rate of change of x + (h + 1) * rate of change of h
Since the rate of change of x is given as 2 m/s (the boat's speed), we can substitute that into the equation:
hypotenuse * rate of change of hypotenuse = 2x + (h + 1) * rate of change of h
We want to find the rate of change of the hypotenuse (the rope) when the boat is 3 m away from the dock, so we substitute x = 3 into the equation:
hypotenuse * rate of change of hypotenuse = 2(3) + (h + 1) * rate of change of h
Now, the rate of change of h can be calculated using the fact that the rope is being pulled at a constant rate of 2 m/s. As the boat moves away from the dock, the height of the pulley above the dock will decrease. Therefore, the rate of change of h is -2 m/s.
Substituting these values into the equation:
hypotenuse * rate of change of hypotenuse = 2(3) + (h + 1)(-2)
Simplifying:
hypotenuse * rate of change of hypotenuse = 6 - 2h - 2
hypotenuse * rate of change of hypotenuse = 4 - 2h
Finally, substituting the value of hypotenuse as x + (h + 1), we get:
(x + h + 1) * rate of change of (x + h + 1) = 4 - 2h
(x + h + 1) * rate of change of (x + h + 1) = 4 - 2h
Now, substituting x = 3 and h = 1 into the equation:
(3 + 1 + 1) * rate of change of (3 + 1 + 1) = 4 - 2(1)
5 * rate of change of 5 = 4 - 2
5 * rate of change of 5 = 2
rate of change of 5 = 2 / 5
rate of change of 5 = 0.4 m/s
Therefore, when the boat is 3 m away from the dock, the rope is being pulled out at a speed of 0.4 m/s.
Regarding part B of your question, as the distance between the boat and the dock grows large, the speed at which the rope is being pulled out will also increase. This is because the rate at which the rope is being pulled out is directly influenced by the rate at which the boat is moving away from the dock, which is a constant 2 m/s. As the boat moves farther away, the length of rope between the boat and the pulley increases, resulting in a larger hypotenuse. Since the rate of change of the hypotenuse remains constant at 0.4 m/s (as calculated in part A), the speed at which the rope is being pulled out will continue to increase.