A 20' ladder leans against a building. The bottom of the ladder is sliding away horizontally at a rate of 2 feet per second. How fast is the top of the ladder sliding down the building when it is 12' above the ground?
If you draw a diagram, let x be the distance of the base of the ladder from the wall, and h the height of the ladder against the wall:
x^2 + h^2 = 400
When h = 12,
x^2 = 400 - 144 = 256
x = 16
2x dx/dt + 2h dh/dt = 0
32 * 2 + 24 dh/dt = 0
dh/dt = -64/24 = -8/3
To find the rate at which the top of the ladder is sliding down the building, we can use the concept of related rates.
Let's assign some variables to the given information:
- Let x be the distance from the bottom of the ladder to the building.
- Let y be the height of the ladder above the ground.
- Let ẋ represent the rate at which x is changing (2 ft/s in this case).
- Let ẏ represent the rate at which y is changing, which is what we are trying to find.
We know that the ladder has a constant length of 20 feet, so we can write the Pythagorean theorem:
x^2 + y^2 = 20^2
Differentiating both sides of this equation with respect to time t, we get:
2xẋ + 2yẏ = 0
Since we are interested in finding ẏ when y = 12 ft, we can substitute the known values into the equation:
2x(2) + 2(12)ẏ = 0
Simplifying the equation:
4x + 24ẏ = 0
Now we need to find the value of x when y = 12 ft. Using the Pythagorean theorem:
x^2 + 12^2 = 20^2
x^2 + 144 = 400
x^2 = 256
x = 16 ft
Substituting the values back into the equation:
4(16) + 24ẏ = 0
64 + 24ẏ = 0
24ẏ = -64
ẏ = -64/24
ẏ = -8/3 ft/s
So, when the ladder is 12 feet above the ground, the top of the ladder is sliding down the building at a rate of 8/3 ft/s, or approximately 2.67 ft/s.