A $2 \mathrm{~m}$ ladder leans against a vertical wall. If the top of the ladder begins to slide dow

Question

A $2 \mathrm{~m}$ ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate $25 \mathrm{~cm} / \mathrm{sec}$., then the rate (in $\mathrm{cm} / \mathrm{sec}$.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is $1 \mathrm{~m}$ above the ground is\

Professor PC Thomas Classes · Answer

(a) : Let $A B$ be the ladder. Given, $\frac{d y}{d t}=-25$\ We have, $x^{2}+y^{2}=4$\ Differentiating w.r.t. ' $t$ ' we get $$ \begin{aligned} & 2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0 \ & \Rightarrow \frac{d x}{d t}=-\frac{y}{x} \frac{d y}{d t} \end{aligned} $$ When top of ladder is 1 m above the ground, i.e., $y=1$, then $x=\sqrt{4-1}=\sqrt{3}$\ $	herefore \frac{d x}{d t}=\frac{-1}{\sqrt{3}}(-25)=\frac{25}{\sqrt{3}} \mathrm{~cm} / \mathrm{sec}$\