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 m/s, how fast is the boat approaching the dock when it is 4 m from the dock?
i don't get how the dy/dt is -1 and dx/dt -y/x??
You haven't defined your x and y, so I can't answer the last part of your question.
On my diagram, I have a right-angled triangle with a height of 1 m, a horizontal of x m and a hypotenuse of y m
given: dy/dt = -1 m/s
( I made it negative, because the value of y will be decreasing as the rope is being pulled in )
Since we are involving variables x and y, I will need some kind of equation that contains x and y.
Heh, how about Pythagoras
x^2 + 1^2 = y^2
2x dx/dt = 2y dy/dt
x dx/dt = y dy/dt
(notice I have not worried about the x=4, now I will)
4^2 + 1^2 = y^2
so y = √17
sub in our stuff into
x dx/dt = y dy/dt
4 dx/dt = √17(-1)
dx/dt = -√17/4 m/s or appr -1.03 m/s
again, my answer is consistent with our concept that if x is decreasing, then dx/dt will be negative
So the boat it approaching the dock at appr 1.03 m/s
To solve this problem, we can use related rates, which involves finding the rates at which different quantities are changing. In this case, we want to find the rate at which the boat is approaching the dock, which is given by the derivative dy/dt (change in y with respect to time).
Let's consider a right triangle formed by the dock, the boat, and the rope. The horizontal distance between the boat and the dock is x, and the vertical distance between the boat and the dock (the height of the pulley) is y.
We are given that the rope is being pulled in at a rate of 1 m/s, so dx/dt = -1 (the negative sign indicates that the distance x is decreasing).
Now, we need to find an equation relating x, y, and the rate dy/dt. From the right triangle, we can use the Pythagorean theorem:
x^2 + y^2 = (4)^2
Differentiating with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Plugging in the values we know: x = 4, dx/dt = -1, and solving for dy/dt, we have:
2(4)(-1) + 2y(dy/dt) = 0
-8 + 2y(dy/dt) = 0
2y(dy/dt) = 8
dy/dt = 8/(2y)
Since the height of the pulley is 1 meter higher than the boat, y = 1. Plugging this in, we have:
dy/dt = 8/(2(1)) = 8/2 = 4 m/s
Therefore, the boat is approaching the dock at a rate of 4 m/s when it is 4 m from the dock.
To understand why dy/dt is -1 and dx/dt is -y/x in this scenario, we can apply the concepts of related rates.
First, let's define some variables:
- y: the height of the rope above the waterline (measured from the bow of the boat)
- x: the horizontal distance between the boat and the dock
From the problem description, we know that the rope is being pulled in at a rate of 1 m/s, which means that dy/dt (the rate at which y is changing) is -1 m/s. The negative sign indicates that the height is decreasing as the rope is being pulled in.
Now, let's consider how the rope's movement influences the boat's position. As the rope is being pulled in, the boat is also being pulled towards the dock. Since the rope is attached to the bow of the boat, the boat moves in the direction opposite to the rope's movement. This means that as y decreases (rope height), x also decreases (distance from the dock).
The relationship between y and x can be defined using similar triangles. Since the pulley is 1 m higher than the bow of the boat, and the rope is directly above the bow, we have a right triangle with legs x and y and a hypotenuse of x + 1. So we can write the equation:
(x + 1)^2 = x^2 + y^2
Now, we can differentiate both sides of this equation with respect to time (t), since we are interested in the rates of change. Using the chain rule, we get:
2(x + 1)(dx/dt) = 2x(dx/dt) + 2y(dy/dt)
Simplifying this equation, we can cancel out the 2's and rearrange:
(x + 1)dx/dt = x(dx/dt) + y(dy/dt)
Since we're interested in finding the rate at which the boat approaches the dock (dx/dt) when it is 4 m away (x = 4), we can substitute the known values into the equation:
(4 + 1)dx/dt = 4(dx/dt) + y(-1)
Simplifying further:
5dx/dt = 4dx/dt - y
Finally, rearrange the terms:
dx/dt = -y/5
So, we have dx/dt = -y/5. This shows that the rate at which the boat approaches the dock (dx/dt) is negatively proportional to the height of the rope (y). As the rope is pulled in (y decreases), the boat approaches the dock at a faster rate (positive dx/dt value).