Give your answer accurate to 3 decimal places.
A man on a dock is pulling in a boat by means of a rope attached to the bow of the boat at a point that is 1 ft above water level. The rope goes from the bow of the boat to a pulley located at the edge of the dock 7 ft above water level. If he pulls in the rope at a rate of 2 ft/sec, how fast (in feet per second) is the boat approaching the dock when the point of attachment is 12 ft from the dock?
ft/sec
To solve this problem, we can use the concept of related rates. Let's call the distance between the boat and the dock x.
The rope forms a right triangle with the dock and the horizontal distance x. The length of the rope (r) is given by the Pythagorean Theorem: r^2 = x^2 + 7^2.
To find how fast the boat is approaching the dock, we need to find dx/dt when x = 12 ft.
Differentiating both sides of the equation with respect to time t, we get: 2r(dr/dt) = 2x(dx/dt).
Substituting the given values:
2(12)(dx/dt) = 2(12)(dr/dt).
Now, we need to find dr/dt. Since the rope is being pulled in at a rate of 2 ft/sec, dr/dt = 2 ft/sec.
Plugging this value back into the equation:
2(12)(dx/dt) = 2(12)(2).
Simplifying the equation:
24(dx/dt) = 48.
Dividing both sides by 24:
dx/dt = 48/24.
Now, solving the equation:
dx/dt = 2 ft/sec.
Therefore, the boat is approaching the dock at a rate of 2 ft/sec when the point of attachment is 12 ft from the dock.
To solve this problem, we can use the concept of related rates. Let's consider the triangle formed by the dock, the boat, and the rope attached to the bow of the boat.
First, let's label the sides of the triangle:
- The distance from the dock to the point of attachment will be represented by x (in ft).
- The height from the dock to the pulley (7 ft above water level) will be represented by h (in ft).
- The height from the pulley to the point of attachment (1 ft above water level) will be represented by y (in ft).
We want to find the rate at which the boat is approaching the dock, which is the rate of change of x with respect to time (dx/dt). We are given that dy/dt = 2 ft/sec.
Now, let's use the Pythagorean theorem to relate the sides of the triangle:
x^2 + (h - y)^2 = h^2
Differentiating both sides of the equation with respect to time (t), we get:
2x(dx/dt) + 2(h - y)(-dy/dt) = 2h(dh/dt)
We are given that h = 7 ft and dy/dt = 2 ft/sec, so we can substitute these values into the equation:
2x(dx/dt) + 2(7 - y)(-2) = 2(7)(dh/dt)
Simplifying the equation, we have:
2x(dx/dt) - 4(7 - y) = 14(dh/dt)
2x(dx/dt) - 28 + 4y = 14(dh/dt)
2x(dx/dt) + 4y = 14(dh/dt) + 28
We want to find dx/dt when x = 12 ft. To do this, we also need to find y when x = 12 ft.
Using the Pythagorean theorem and the values given, we can solve for y:
12^2 + (7 - y)^2 = 7^2
144 + (7 - y)^2 = 49
(7 - y)^2 = 49 - 144
(7 - y)^2 = -95
Since we have a negative value inside the square, this means that the point of attachment is beyond the dock, which is not physically possible. Therefore, there is an error in the problem statement. Please double-check the values given or provide additional information to correct the error.
To find the speed at which the boat is approaching the dock, we need to find the rate of change of the distance between the boat and the dock.
Let's label the distance between the boat and the dock as "x" (in feet). We are given that the rope is attached to the bow of the boat at a point 1 ft above water level and that the point of attachment is 12 ft from the dock.
Using the Pythagorean theorem, we can relate the distance x with the height of the pulley above the water level (7 ft) and the height of the attachment point above the water level (1 ft). The equation is:
x^2 + 7^2 = (x + 1)^2
Expanding and simplifying this equation, we get:
x^2 + 49 = x^2 + 2x + 1
Subtracting x^2 from both sides, we have:
49 = 2x + 1
Subtracting 1 from both sides, we get:
48 = 2x
Dividing both sides by 2, we obtain:
24 = x
Now we have the distance between the boat and the dock. To find how fast the boat is approaching the dock, we need to find the derivative of the distance with respect to time.
Let's differentiate both sides of the equation with respect to time (t):
d/dt (24) = d/dt (x)
The derivative of a constant is zero, so the left side simplifies to:
0 = d/dt (x)
Now we can find the derivative of the right side of the equation. Differentiating (x + 1)^2, we get:
0 = 2(x + 1) * dx/dt
We are given that the rope is being pulled in at a rate of 2 ft/sec. Thus, dx/dt = -2 ft/sec (negative since the distance x is decreasing).
Plugging in the value for dx/dt, we have:
0 = 2(24 + 1) * (-2)
Simplifying this, we get:
0 = 50 * (-2)
Therefore, the speed at which the boat is approaching the dock when the point of attachment is 12 ft from the dock is 0 ft/sec.