A ladder $5$ m long leans against a wall. The foot is pulled away along the ground at $2$ cm/s. When the foot is $4$ m from the wall, the top is sliding down at the rate of

1. Let $x$=distance of foot, $y$=height; $x^2+y^2=25$. 2. At $x=4$: $y=\sqrt{25-16}=3$. 3. Differentiate: $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$. 4. $\dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}=-\dfrac{4}{3}(2)=-\dfrac{8}{3}$. 5. The height decreases at $\dfrac{8}{3}$ cm/s. The trap $\dfrac{4}{3}$ drops the factor $2$. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.4_