5.) a 10-foot ladder is placed against a vertical wall such that the bottom of the ladder is sliding away from the wall at the rate of 3 feet per second. how far is the top of the ladder sliding down the wall when the bottom of the ladder is 6 feet away from the wall. use the pythagorean theorem to set up the equation.

If x is the distance the base of the ladder is from the wall, and y is the distance to top of the ladder is up the wall, then the Pythagorean theorem tells you

... x² + y² = 10²

Differentiating with respect to time, we have

... 2x·dx/dt + 2y·dy/dt = 0

... dy/dt = (-x/y)·dx/dt

We are given x = 6 ft and dx/dt = 3 ft/s. Using the first equation, we can find y as

... y = √(10² -6²) = 8

Then

... dy/dt =(-6/8)·(3 ft/s)

... dy/dt = -2.25 ft/s

The top of the ladder is sliding down the wall at -2.25 ft/s.

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

12

step-by-step explanation: