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?

Explanation

Let AB is a ladder of length 5 m which touches the wall at A and ground at B.
Let BC = x metre
and AC = y metre.
Given `(dx)/(dt) = 2 ` cm/sec
In ` Delta ABC`,
`x^(2) + y^(2) = 5^(2)" "(.:' angle ACB = 90^(@))`
`rArr y = sqrt(25 - x^(2))`
` rArr (dy)/(dt) = 1/(2sqrt(25-x^(2)))*(-2x) (dx)/(dt) = (-x)/(sqrt(25-x^(2))) (dx)/(dt)`
at x = 4 metre
`(dy)/(dt) = - 4/sqrt(25-4^(2))xx 2 =- 8/3` cm/sec
Therefore, the height of ladder on wall is decreasing at the rate of `8/3` cm/sec