A ladder 20 feet long leans against a building. If the bottom of the ladder slides away from the building
horizontally at a rate of 2 ft./ sec. How fast is the ladder sliding down the house when the top of the ladder
is 8 feet from the ground?
you have
x^2 + y^2 = 20^2
x dx/dt + y dy/dt = 0
when y=8, x = √(400-64) = 4√21
so now just plug in your numbers:
4√21 * 2 + 8 dy/dt = 0
dy/dt = -√21 ft/s
To find the rate at which the ladder is sliding down the house, we can use the concept of related rates. Let's denote the distance from the top of the ladder to the ground as "x" and the distance from the bottom of the ladder to the building as "y".
We know that the ladder is 20 feet long, so we can use the Pythagorean theorem to establish a relationship between x and y:
x^2 + y^2 = 20^2
Now, let's differentiate both sides of the equation with respect to time (t), since we are dealing with rates:
2x(dx/dt) + 2y(dy/dt) = 0
Since we are interested in finding the rate at which the ladder is sliding down the house (dy/dt) when the top of the ladder is 8 feet from the ground (x = 8), we can substitute these values into our equation:
2(8)(0) + 2y(dy/dt) = 0
16(0) + 2y(dy/dt) = 0
2y(dy/dt) = 0
(dy/dt) = 0
Therefore, when the top of the ladder is 8 feet from the ground, the ladder does not slide down the house. Hence, the rate at which the ladder is sliding down the house is 0 ft/s.