A ladder 13 m long is leaning against a vertical wall. The bottom of the ladder is dragged away from the wall along the ground at the rate of 2 cm/sec. How fast is the height on the wall decreasing when the foot of the ladder is 5 m away from the wall.

#### Solution

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:**

x^{2} + y^{2 }= 169 ......................[Length of the ladder = 13 m]

⇒ `"y" = sqrt(169-"x"^2)`

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

`"dy"/"dx" = (-"x")/sqrt(169-"x"^2) "dx"/"dt"`

It is given that `"dx"/"dt" = 2 "cm"//"s".`

`"dy"/"dt" = (-2"x")/sqrt(169-"x"^2)`

**Now, when x = 5 m, we have:**

`"dy"/"dt" = (-2xx5)/sqrt(169-5^2) = (-10)/sqrt144 = -5/6`

Hence, the height of the ladder on the wall is decreasing at the rate of `5/6`cm/sec.