Calculus question on rates

Question

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 4 ft/sec, how fast is the ladder sliding down the house when the top of the ladder is 8 feet from the ground.

Very confused on this

Answer 1

dy/dt = dy/dx dx/dt

The problem is asking for dy/dt or, rather, - dy/dt, as it asks for the rate that the ladder is sliding down the wall.

We are given dx/dt, so we only need to figure dy/dx

Using the Pythagorean theorem, x² + y² = 20², or y = (400 - x²)^1/2

dy/dx = 1/2 * -2x * (400 - x²)^-1/2 or -x/√(400 - x²)

Note that we can rewrite this as -x/y (this will come in handy later).

Now, in the problem, we are given that the top of the ladder is 8 feet from the top of the ground, so y = 8, and x = (400 - y²)^1/2 = (400 - y²)^1/2 = (400 - 64)^1/2 = √336 = 4√21

Then, dy/dx = -x/y = -4√21/8 = -√21/2

Thus, dy/dt = dy/dx dx/dt = -√21/2 * 4 = -2√21

-dy/dt, the rate the ladder is sliding down, is - -2√21 = 2√21