A boat is pulled into a dock by a rope attached to the bow of the boat an passing through the pulley on the deck that is 1m high than the bow of the boat ,if the rope is pulled im at the rate of 0.8m/s how fast does the boat approach the deck when it is 10m from the dock?
I hope you made a diagram.
Let the distance of the boat to the dock be x m
let the length of the rope be y m
given: dy/dt = -.8 m/s
find dx/dt when x = 10
you will have the right-angled triangle such that
x^2 + 1 = y^2
2x dx/dt = 2y dy/dt
dx/dt = y/x (dy/dt)
when x=10
y^2 = 101
y = √101
dx/dt = √101/10(-.8) = -.804
the distance is decreasing at a rate of .804 m/second
Yea i did made a diagraam and thanks .
To find how fast the boat approaches the deck when it is 10m from the dock, we need to use the concept of related rates.
Let's denote the distance from the dock to the boat as x (measured in meters), and the height of the boat above the water as y (measured in meters). Since the rope passes through a pulley on the deck that is 1m higher than the bow of the boat, the height of the rope from the water is y + 1.
We are given that the rope is being pulled in at a rate of 0.8m/s. This means that the rate at which the distance from the dock to the boat is changing, dx/dt, is also 0.8m/s.
Now, we need to find the rate at which the boat is moving vertically, dy/dt, when x = 10m.
Using the Pythagorean theorem, we have:
x^2 + (y + 1)^2 = (10)^2
Differentiating both sides of this equation with respect to time t, we get:
2x dx/dt + 2(y + 1) dy/dt = 0
Substituting the values we know:
2(10)(0.8) + 2(y + 1) dy/dt = 0
20(0.8) + 2(y + 1) dy/dt = 0
16 + 2(y + 1) dy/dt = 0
Dividing both sides by 2(y + 1), we obtain:
dy/dt = -16/(2(y + 1))
Now, we can substitute the value of x = 10m into the equation to find dy/dt when x = 10m:
dy/dt = -16/(2(y + 1))
When x = 10m, we need to solve for y:
10^2 + (y + 1)^2 = (10)^2
y^2 + 2y + 1 = 0
(y+1)(y+1) = 0
y = -1
Substituting y = -1 into the equation for dy/dt, we find:
dy/dt = -16/(2(-1 + 1))
dy/dt = -16/0
Since the denominator is 0, this means that dy/dt is undefined. In other words, the boat doesn't approach the deck when it is 10m from the dock.