A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat.
If the rope is pulled in at a rate of 1.4 m/s, how fast is the boat approaching the dock when it is 9 m from the dock?
To answer this question, we can use the concept of related rates. First, let's assign some variables to the given information:
Let:
- x represent the distance between the boat and the dock.
- y represent the distance between the pulley and the dock.
- z represent the length of the rope.
We are given:
- The rate at which the rope is being pulled in, which is 1.4 m/s.
Our goal is to find the rate at which the boat is approaching the dock, which we'll call dx/dt.
To relate these variables, we can use the Pythagorean theorem. Since the rope is attached to the bow of the boat and passes through a pulley on the dock that is higher than the bow, we can form a right triangle with x as the horizontal distance, y as the vertical distance, and z as the hypotenuse:
z^2 = x^2 + y^2
Now, let's differentiate this equation with respect to time (t):
2z * dz/dt = 2x * dx/dt
Simplifying, we can cancel out the 2's:
z * dz/dt = x * dx/dt
We need to find dx/dt when x = 9 m. To do this, we need to find the values of z, y, and dz/dt.
Given that the pulley is 1 m higher than the bow of the boat and x = 9 m, we can use the Pythagorean theorem to find the value of y:
y^2 + 9^2 = (y + 1)^2
y^2 + 81 = y^2 + 2y + 1
80 = 2y
y = 40 m
Next, using the Pythagorean theorem again, we can find the value of z:
z^2 = 9^2 + 40^2
z^2 = 81 + 1600
z^2 = 1681
z = √1681
z = 41 m
Now, we have the values of z, x, and y. We also know that dz/dt is given as 1.4 m/s.
Plugging these values into the differentiated equation, we can solve for dx/dt:
41 * 1.4 = 9 * dx/dt
dx/dt = (41 * 1.4) / 9
dx/dt = 6.38 m/s
Therefore, when the boat is 9 m from the dock, it is approaching the dock at a rate of approximately 6.38 m/s.