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/sec. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

The correct answer is $-\dfrac{8}{3}$

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

By Pythagoras theorem, we have: $x^2 + y^2 = 25$

⇒ $y = \sqrt{25−x^2}$


Then, the rate of change of height ($y$) with respect to time ($t$) =
$\dfrac{dy}{dt}$ = $\dfrac{-x}{\sqrt{25−x^2}}$ $\dfrac{dx}{dt}$


​It is given that $\dfrac{dx}{dt}$ = 2 cm/sec

∴ $\dfrac{dy}{dt}$ = $\dfrac{-2x}{\sqrt{25−x^2}}$



When $x=4m$,
$\dfrac{dy}{dt}$ = $\dfrac{-2 * 4}{\sqrt{25−4^2}}$ = $-\dfrac{8}{3}$