A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?              

Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x m away from the wall.

Then, by Pythagoras theorem, we have:

x² + y² = 25 [Length of the ladder = 5 m]

x₁ < x₂

Then, the rate of change of height (y) with respect to time (t) is given by,

⇒ (x₁) > f(x₂)

It is given that x₁, x₂ ϵ l.

f' (x) > 0

Now, when x = 4 m, we have:

x ϵ (a, b)

Hence, the height of the ladder on the wall is decreasing at the rate of f' (x) < 0.