# A ladder is 5 m long is leaning against the 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?

**Solution:**

Derivatives are used to find the rate of changes of a quantity with respect to the other quantity. By using the application of derivatives we can find the approximate change in one quantity with respect to the

change in the other quantity.

Let the height of the wall at which the ladder is touching it be y meter and the distance of its foot from the wall on the ground be x meter.

Hence, by using pythagoras theorem

⇒x^{2} + y^{2} = 5^{2}

⇒ y^{2} = 25 - x^{2}

⇒ y = 25 - x^{2}

Therefore,

dy/dx

= d/dt √25 - x²

= d/dx √25 - x² dx/dt

= - x/(√25 - x²) dx/dt

We have,

dx/dt = 2 cm/s

Thus,

dy/dt = - 2x/(√25 - x²)

When x = 4 cm

Then,

dy/dt = (- 2 × 4)/(√25 - 16)

= - 8/3 cm/s

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.1 Question 10

## A ladder is 5 m long is leaning against the 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?

**Summary:**

Given that the bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. The height on the wall decreasing when the foot of the ladder is 4 m away from the wall is - 8/3 cm/s