A ladder 20 ft long is leaning against an embankment inclined 60 degrees to the horizontal. If the bottom of the ladder is being moved horizontally toward the embankment at 1 ft/sec, how fast is the top of the ladder moving when the bottom is 4 ft from the embankment?
from the triangle:
20^2=h^2 + x^2
h is vertical height, and x is distance base.
take the derivative...
0=2h dh/dt + 2x dx/dt
dh/dt=-dx/dt x/h
= 1 * 4/h where h= sqrt(400-16)
To solve this problem, we can use trigonometry and related rates.
Let's define the variables:
- Let x be the horizontal distance between the bottom of the ladder and the embankment.
- Let y be the vertical distance between the top of the ladder and the ground.
- Let θ be the angle between the ladder and the ground.
Now, we are given:
- The ladder is 20 ft long, so we have y = 20 ft.
- The ladder is leaning 60 degrees to the horizontal, so we have θ = 60 degrees.
- The bottom of the ladder is being moved horizontally toward the embankment at 1 ft/sec, so we have dx/dt = 1 ft/sec.
Our goal is to find dy/dt, the rate at which the top of the ladder is moving.
Using trigonometry, we can relate x, y, and θ:
y = x * tan(θ)
Differentiating both sides of this equation with respect to time t:
dy/dt = dx/dt * tan(θ)
Now, let's substitute the known values:
dy/dt = 1 ft/sec * tan(60 degrees)
Since tan(60 degrees) = √3, we have:
dy/dt = 1 ft/sec * √3
Therefore, when the bottom is 4 ft from the embankment, the top of the ladder is moving at a rate of √3 ft/sec.
To solve this problem, we can use related rates. We are given that the ladder is 20 ft long and the bottom of the ladder is being moved horizontally towards the embankment at a rate of 1 ft/sec. We need to find the rate at which the top of the ladder is moving when the bottom is 4 ft from the embankment.
Let's break down the problem using a right triangle. Let's call the height of the embankment h, the distance between the bottom of the ladder and the embankment x, and the length of the ladder L.
We are given that L = 20 ft and x = 4 ft. We need to find dx/dt, the rate at which x is changing, and we want to find dy/dt, the rate at which y (the height of the ladder) is changing when x = 4 ft.
By using the properties of a right triangle, we can find a relation between h, x, and L:
sin(60°) = h / L
h = L * sin(60°)
Now, we can differentiate both sides of this equation with respect to t:
dh/dt = d/dt (L * sin(60°))
dh/dt = dL/dt * sin(60°)
Since we know dL/dt = 0, because the length of the ladder is not changing, dh/dt = 0.
Now, let's find dx/dt. We have x = 4 ft, and we know that dx/dt = 1 ft/sec. Therefore, dx/dt = 1 ft/sec.
Finally, let's find dy/dt. We can use the Pythagorean theorem to relate x, y, and L:
x^2 + y^2 = L^2
Taking the derivative of both sides with respect to t gives us:
2x * dx/dt + 2y * dy/dt = 2L * dL/dt
Since we know dx/dt = 1 ft/sec, x = 4 ft, L = 20 ft, and dL/dt = 0, we can solve for dy/dt:
2 * 4 * 1 + 2y * dy/dt = 2 * 20 * 0
8 + 2y * dy/dt = 0
2y * dy/dt = -8
dy/dt = -8 / (2y)
Now, let's substitute the value of y. Using the relation h = L * sin(60°):
h = 20 * sin(60°)
h = 20 * (√3/2)
h = 10√3 ft
Therefore, when x = 4 ft and h = 10√3 ft, we can find y using the Pythagorean theorem:
x^2 + y^2 = L^2
4^2 + y^2 = 20^2
16 + y^2 = 400
y^2 = 384
y ≈ 19.6 ft
Substituting y = 19.6 ft into dy/dt = -8 / (2y):
dy/dt = -8 / (2 * 19.6)
dy/dt ≈ -0.2041 ft/sec
Therefore, when the bottom of the ladder is 4 ft from the embankment, the top of the ladder is moving at a rate of approximately -0.2041 ft/sec.