Question

A ladder, 5 metre long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides down wards at the rate of 10 cm/sec, then find the rate at which the angle between the floor and ladder is decreasing when lower end of ladder is 2 metres from the wall.

Answer 1

$$\begin{gathered} \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{ladder}=500cm \\ \mathrm{Let}\mathrm{the}\mathrm{horizontal}\mathrm{length}\mathrm{covered}\mathrm{between}\mathrm{the}\mathrm{wall}\mathrm{and}\mathrm{the}\mathrm{ladder}\mathrm{be}x\mathrm{and}\mathrm{vertical}\mathrm{length}\mathrm{covered}\mathrm{between}\mathrm{the}\mathrm{wall}\mathrm{and}\mathrm{the}\mathrm{ladder}\mathrm{be}y. \\ \mathrm{And}\mathrm{let}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{floor}\mathrm{and}\mathrm{ladder}\mathrm{be}\theta . \\ \mathrm{Then},\sin \theta =\frac{y}{500} \\ \mathrm{On}\mathrm{differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}t,\mathrm{we}\mathrm{get} \\ \cos \theta \frac{d\theta }{dt}=\frac{1}{500}\frac{dy}{dt}...\left(1\right) \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\frac{dy}{dt}=-10cm/sec....\left(2\right) \\ \mathrm{Also}, \\ \cos \theta =\frac{x}{500} \\ \mathrm{When}x=200cm,\cos \theta =\frac{200}{500}=\frac{2}{5}...\left(3\right) \\ \mathrm{Substituting}\left(2\right)\mathrm{and}\left(3\right)\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{2}{5}\frac{d\theta }{dt}=\frac{1}{500}\left(-10\right) \\ \frac{d\theta }{dt}=-\frac{1}{20}\mathrm{radian}/\mathrm{second} \\ \mathrm{Hence},\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{floor}\mathrm{and}\mathrm{the}\mathrm{ladder}\mathrm{is}\mathrm{decreasing}\mathrm{at}\mathrm{the}\mathrm{rate}\mathrm{of}\frac{1}{20}\mathrm{radian}/\mathrm{second}. \end{gathered}$$