: A 20m ladder is placed against a wall. The bottom of the ladder is slipping at the rate of 2m/sec. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 12m from the wall

To solve this problem, we can use related rates, as we are given the rate at which the bottom of the ladder is sliding and we need to find the rate at which the top of the ladder is sliding.

Let's define the following variables:
- x: the distance between the bottom of the ladder and the wall
- y: the distance between the top of the ladder and the ground
- z: the length of the ladder

We are given that dx/dt (the rate at which x is changing) is -2 m/sec, as the bottom of the ladder is slipping at a rate of 2 m/sec towards the wall. We need to find dy/dt (the rate at which y is changing) when x = 12 m.

Using the Pythagorean theorem, we know that x^2 + y^2 = z^2. To find dy/dt, we need to differentiate this equation implicitly with respect to time (t):

2x * (dx/dt) + 2y * (dy/dt) = 2z * (dz/dt)

Substituting the given values and differentiating:
2(12) * (-2) + 2y * (dy/dt) = 2(20) * (dz/dt)
-48 + 2y * (dy/dt) = 40 * (dz/dt)

Since we are trying to find dy/dt, let's solve for it:
2y * (dy/dt) = 40 * (dz/dt) + 48
dy/dt = (40 * (dz/dt) + 48) / (2y)

We know that z = 20 and x = 12, so we can find y using the Pythagorean theorem again:
12^2 + y^2 = 20^2
144 + y^2 = 400
y^2 = 256
y = 16

Now we can substitute the known values into the equation for dy/dt:
dy/dt = (40 * (dz/dt) + 48) / (2 * 16)
dy/dt = (20 * (dz/dt) + 24) / 16
dy/dt = (20 * (dz/dt) + 3) / 2

Therefore, the rate at which the top of the ladder is sliding down the wall is (20 * (dz/dt) + 3) / 2.