A pulley system on a pier is used to pull boats towards the pier for docking. Suppose that a boat is being docked and has a rope attaching the boat's bow on one end and feeding through the pier's pulley system on the other end. The pulley system is 1 meter higher than the height of the boat's bow. The rope is being pulled in at a rate of 1 m/s. When the boat is 5 meters from the pier, how fast is the boat approaching the pier?
Draw a diagram. If the boat is x meters from the pier, the rope haas length r, with
r^2 = x^2+1
So,
2r dr/dt = 2x dx/dt
when x=5, r=√26, so
2√26(-1) = 2(5) dx/dt
dx/dt = -√26/5
so, the boat is approaching the dock at about 1.0 m/s
Makes sense, since x and r are almost the same for x relatively large compared to 1.
To find how fast the boat is approaching the pier, we need to find the rate at which the boat's distance from the pier is changing with respect to time.
Let's denote the distance between the boat and the pier as "x" and the height of the pulley system above the boat's bow as "h".
Given:
Rate of pulling the rope, dx/dt = 1 m/s
Distance between the boat and the pier, x = 5 meters
Height of the pulley system, h = 1 meter
We need to find:
The rate at which the boat is approaching the pier, d(Area)/dt
To solve this problem, we can use similar triangles. Consider a right triangle formed by the boat, the pulley system, and a vertical line from the pulley system to the water:
------
/ /
/ /
/ h /
/ /
/ /
------------
Let's label the distance from the pulley system to the water as "d" and the distance from the boat's bow to the water as "y". We have two similar triangles - the larger one formed by the boat, the smaller one formed by the pulley system. Thus, we can write the following ratios:
h / y = x / (x + d) ----(1)
Differentiating both sides of equation (1) with respect to time "t":
d(h / y) / dt = d(x / (x + d)) / dt
We need to find d(y) / dt, the rate at which the boat's distance from the pier is changing with respect to time. Differentiating the right side of the equation above requires the chain rule:
d(h / y) / dt = (1/(x + d) - x/(x + d)^2) * dx / dt ----(2)
d(h / y) / dt = (1/(x + 1) - x/(x + 1)^2) * 1 m/s (Substituting values)
Now, let's substitute the given values:
x = 5 meters
h = 1 meter
dx/dt = 1 m/s
Plug these values into equation (2) to find the rate at which the boat is approaching the pier:
d(h / y) / dt = (1/(5 + 1) - 5/(5 + 1)^2) * 1 m/s
d(h / y) / dt = (1/6 - 5/36) * 1 m/s
d(h / y) / dt = (6/36 - 5/36) * 1 m/s
d(h / y) / dt = 1/36 * 1 m/s
d(h / y) / dt = 1/36 m/s
Therefore, the boat is approaching the pier at a rate of 1/36 m/s.