A man on a wharf is pulling in a boat by means of a rope attached to the bow of the boat 1 meter above water level and passing through a simple pulley located on the dock 8 meters above water level. If he pulls in the rope at the rate of 2 meters per second, how fast is the boat approaching the wharf when the bow of the boat is 25 meters from a point that is directly below the pulley?
h height above bow (constant=7)
y hor distance to bow (25 at moment)
x = our rope length
x^2 = h^2 + y^2
2 x dx = 0 + 2ydy
dy/dx = x/y
dy/dt = dy/dx dx/dt = 2 (x/y)
x = sqrt (7^2 + 25^2) = 26
so
dy/dt = 2 (26/25) = 2.08 m/s
To solve this problem, we can use related rates, where we relate the distance between the boat and the wharf to the rate at which the rope is being pulled in.
Let's denote the distance between the boat and the wharf as x, and the height of the boat above the water level as h. We are given that the rate at which the rope is being pulled in is 2 m/s.
We can create a right triangle using the height of the boat, the distance between the boat and the wharf, and the distance between the pulley and the point directly below it. The height of the boat, h, can be represented as 25 - x, because the boat is 25 meters from a point directly below the pulley.
Using the Pythagorean theorem, we can express this relationship as follows:
x^2 + (25 - x)^2 = 8^2
Simplifying this equation gives us:
x^2 + 625 - 50x + x^2 = 64
Combining like terms:
2x^2 - 50x + 561 = 0
Now, to find the rate at which the boat approaches the wharf, we need to differentiate this equation with respect to time, t.
d/dt (2x^2 - 50x + 561) = d/dt(0)
4x(dx/dt) - 50(dx/dt) = 0
Factor out dx/dt:
(dx/dt)(4x - 50) = 0
Set each factor equal to zero:
dx/dt = 0 or 4x - 50 = 0
If dx/dt = 0, it means the boat is not moving and we can disregard this solution.
Setting 4x - 50 = 0, we can solve for x:
4x = 50
x = 50/4
x = 12.5
So, when the boat is 12.5 meters from the point directly below the pulley, the boat is approaching the wharf at the fastest rate.
To find this rate, we can substitute x = 12.5 back into the equation we obtained earlier:
dx/dt = 4x - 50
dx/dt = 4(12.5) - 50
dx/dt = 50 - 50
dx/dt = 0
Therefore, when the boat is 12.5 meters from the point directly below the pulley, the rate at which the boat is approaching the wharf is 0 m/s.
To solve this problem, we can use related rates. The key idea is to express the changing quantities in terms of each other. Let's break down the problem into smaller parts:
1. Draw a diagram of the situation to visualize it.
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B (boat)
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W (wharf)
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2. Identify the given information:
- The rate at which the man pulls the rope, given as 2 meters/sec.
- The distance between the bow of the boat and a point directly below the pulley, given as 25 meters.
3. Determine what needs to be found:
- The speed at which the boat is approaching the wharf.
4. Assign variables to the changing quantities:
- Let's call the distance between the boat and the wharf, x.
- Let's call the height of the boat above the water level, y.
- Let's call the height of the pulley above the water level, z.
5. Establish the relationship between the variables:
- Since the rope passes through a pulley, the distance the rope is pulled is the same as the distance the boat moves towards the wharf.
- Thus, we have an equation: x = y + z.
6. Differentiate both sides of the equation with respect to time t:
- d(x/dt) = d(y/dt) + d(z/dt).
7. Substitute known values and simplify:
- We know that d(y/dt) = 2 m/s (rate at which the rope is pulled).
- We are given that y = 1 m (height of the bow of the boat above the water level).
- We are given that z = 8 m (height of the pulley above the water level).
- Therefore, d(x/dt) = 2 + d(z/dt).
8. Determine the value of d(z/dt):
- We need to find how fast the height of the pulley is changing with time. Since there is no information given about this, we assume it to be constant.
- Thus, d(z/dt) = 0.
9. Substitute the values back into the equation:
- d(x/dt) = 2 + 0 = 2 m/s.
So, when the bow of the boat is 25 meters from a point directly below the pulley, the boat is approaching the wharf at a speed of 2 meters per second.