A boat is being pulled into a dock by a rope attached to it, which is passing through a pulley on the dock, positioned 6 meters higher than the boat. If the rope is being pulled in at a rate of 3 meters/sec, at what rate is the boat approaching the dock when it is 8 meters from the dock?
Please enter your answer in decimal format with three significant digits after the decimal point.
Let's consider the situation using the concept of related rates.
We know that the rope is being pulled in at a rate of 3 meters/sec, which means the length of the rope is changing at that rate. Let's denote the length of the rope as L(t), where t is the time in seconds.
We are interested in finding the rate at which the boat is approaching the dock, which is the rate of change of the distance between the boat and the dock. Let's denote this distance as D(t) and the rate of change as dD/dt.
From the problem statement, we know that the rope is passing through a pulley on the dock, which means the length of the rope and the distance of the boat from the dock are related by the Pythagorean theorem:
L(t)^2 = D(t)^2 + 6^2
Differentiating both sides of this equation with respect to time t, we get:
2L(t) * dL/dt = 2D(t) * dD/dt
We are given dL/dt = 3 meters/sec, and we want to find dD/dt when D(t) = 8 meters.
Substituting the given values and solving for dD/dt:
2(8) * 3 = 2(8) * dD/dt
48 = 16 * dD/dt
dD/dt = 48/16 = 3 meters/sec
Therefore, the boat is approaching the dock at a rate of 3 meters/sec.
To find the rate at which the boat is approaching the dock, we can use related rates and the concept of similar triangles.
Let's denote the distance between the boat and the dock as x (in meters). We are given that the rope is being pulled in at a rate of 3 meters/sec, which means that dx/dt (rate at which x is changing) is 3 meters/sec. We need to find the rate at which the boat is approaching the dock, which is given by dy/dt (the rate at which the distance y is changing).
Notice that we have a right triangle formed by the boat, the dock, and the rope. The height of this triangle is 6 meters (the height difference between the boat and the dock), and the base is x meters. Since the triangle formed by the rope is similar to this right triangle, we can write the following proportion:
(x + y) / y = x / 6
Now, we can differentiate both sides of this equation implicitly with respect to time t:
d(x + y)/dt / dy/dt = dx/dt / 6
Substituting the known values:
(3 + dy/dt) / dy/dt = 3 / 6
Simplifying the equation:
(3 + dy/dt) / dy/dt = 1/2
Cross-multiplying:
2(3 + dy/dt) = dy/dt
6 + 2(dy/dt) = dy/dt
dy/dt - 2(dy/dt) = -6
-dy/dt = -6
dy/dt = 6
Therefore, the rate at which the boat is approaching the dock when it is 8 meters away is 6 meters/sec.
To solve this problem, we can use the concept of related rates. We need to find the rate at which the boat is approaching the dock, which is the rate of change of the distance between the boat and the dock.
Let's use the Pythagorean theorem to express the relationship between the distance x between the boat and the dock and the height y of the boat above the ground:
x^2 + (y+6)^2 = 8^2
Now, let's differentiate both sides of the equation with respect to time t:
2x(dx/dt) + 2(y+6)(dy/dt) = 0
We're given that dx/dt (the rate at which the distance is changing) is 3 meters/sec and we want to find dy/dt (the rate at which the height of the boat is changing) when x = 8.
Plugging in the given values into our equation:
2(8)(3) + 2(y+6)(dy/dt) = 0
Simplifying the equation:
16 + 2(y+6)(dy/dt) = 0
Now we can solve for dy/dt:
2(y+6)(dy/dt) = -16
(y+6)(dy/dt) = -8
dy/dt = -8/(y+6)
Substituting x = 8 into the original equation (x^2 + (y+6)^2 = 8^2):
8^2 + (y+6)^2 = 8^2
(y+6)^2 = 0
y+6 = 0
y = -6
Since the boat is 8 meters from the dock, we can see that y = -6 is not a valid solution, as it would mean the boat is underground. Therefore, we ignore it.
Now, plugging y = 0 into dy/dt = -8/(y+6):
dy/dt = -8/(0+6)
dy/dt = -8/6
Simplifying the expression:
dy/dt = -4/3
Therefore, when the boat is 8 meters from the dock, it is approaching the dock at a rate of -4/3 meters/sec.