Question

A ladder 30 ft. long is leaning up against a building. If the top of the ladder is being pulled up the wall of the building at a rate of 1.5 feet per minute, find the rate at which the base of the ladder is moving toward the building when it is 18 feet from the wall.

Answer 1

we have

x^2+y^2 = 30
so,
2x dx/dt + 2y dy/dt = 0
when x=18, y=24

2*18 dx/dt + 2*24 dy/dt = 0
. . .

Answer 2

So what's next to find the rate at which the ladder is moving toward the building when it is 18 ft from the wall?

Answer 3

36 dx/dt + 48 dy/dt = 0

48 dy/dt = -36 dx/dt

dy/dt = -.75 or -3/4

Answer 4

Is this right?

1.5 ft = dr/dt

2x dx/dt + 2y dy/dt = 0

2(18)(2) + 2(24)= 0
72 + 48 dy/dt = 0
48 dy/dt = -72

dy/dt = -72/48 = -3/4 feet per minute

Answer 5

-3/2 feet per minute I meant

Answer 6

You need to read the question more carefully. They gave you dy/dt, and you need to find dx/dt.

2x dx/dt + 2y dy/dt = 0
when x=18, y=24

The ladder is being pulled up the wall at 1.5 ft/min. That means dy/dt = 3/2

Plugging all that into the equation gives

2(18) dx/dt + 2(24)(3/2) = 0
36 dx/dt = -72
dx/dt = -2

As expected, the distance of the base of decreases as the top of the ladder gets pulled up.