A 25-ft ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.18 ft/sec, how fast, in ft/sec, is the top of the ladder sliding down the wall, at the instant when the bottom of the ladder is 20 ft from the wall?

If the base of the ladder is x away from the wall, and the top of the ladder is at height y, then

x^2+y^2 = 25^2

x dx/dt + y dy/dt = 0

So, figure y when x=20, and plug in dx/dt=0.18 to find dy/dt.

To find the rate at which the top of the ladder is sliding down the wall, we can use the related rates method.

Let's denote:
- h as the distance between the bottom of the ladder and the ground,
- x as the distance between the bottom of the ladder and the wall, and
- y as the distance between the top of the ladder and the ground.

We are given that dx/dt (the rate at which x is changing) is -0.18 ft/sec.

We need to find dy/dt (the rate at which y is changing) when x = 20 ft.

Using the Pythagorean theorem, we have the relation:

x^2 + y^2 = h^2

Differentiating this equation with respect to time (t), we get:

2x(dx/dt) + 2y(dy/dt) = 2h(dh/dt)

Substituting the given values dx/dt = -0.18 ft/sec and x = 20 ft, and solving for dy/dt:

2(20)(-0.18) + 2y(dy/dt) = 2h(dh/dt)

-7.2 + 2y(dy/dt) = 0

2y(dy/dt) = 7.2

dy/dt = 7.2 / (2y)

To find y, we can use the Pythagorean theorem:

x^2 + y^2 = h^2

Plugging in x = 20 ft and h = 25 ft:

20^2 + y^2 = 25^2

400 + y^2 = 625

y^2 = 625 - 400

y^2 = 225

y = √225

y = 15 ft

Now that we have the value of y, we can substitute it into the equation dy/dt = 7.2 / (2y):

dy/dt = 7.2 / (2 * 15)

dy/dt = 7.2 / 30

dy/dt = 0.24 ft/sec

Therefore, at the instant when the bottom of the ladder is 20 ft from the wall, the top of the ladder is sliding down the wall at a rate of 0.24 ft/sec.