A ladder 15 meters long leans against a vertical wall. If the foot of the ladder is being pulled away from the wall at the rate of 2 meters per minute. How fast is the top of the ladder sliding down the wall at the instant when the foot of the ladder is 12 meters from the wall.

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

Let's denote the distance between the foot of the ladder and the wall as x (in meters), and the height of the ladder on the wall as y (also in meters). We are given that dx/dt (the rate at which x is changing) is 2 m/min.

By applying the Pythagorean theorem, we know that x^2 + y^2 = 15^2 (from the given 15-meter ladder). Now, we can differentiate both sides of the equation with respect to time (t):

d/dt(x^2 + y^2) = d/dt(15^2)

Differentiating each term separately, we get:

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

Now, we need to find dy/dt (the rate at which y is changing) when x = 12. We already know dx/dt = 2 m/min.

Substituting x = 12 and dx/dt = 2 into the equation, we can solve for dy/dt:

2(12)(2) + 2y(dy/dt) = 0

24 + 24y(dy/dt) = 0

24y(dy/dt) = -24

dy/dt = -1

Therefore, the top of the ladder is sliding down the wall at a rate of 1 m/min at the instant when the foot of the ladder is 12 meters from the wall.

x^2+y^2 = 225

x dx/dt + y dy/dt = 0

now just plug in your numbers and solve for dy/dt