question_answer

A ladder 5 m long, standing on a horizontal floor, leans against a vertical wall. If the two of the ladder slides downwards at the rate of 10 cm/s, then find the rate at which the angle between the floor and ladder is decreasing when the lower end of ladder is 2 m from the wall.

Answer:

Given, length of the ladder = 5 cm = 500 cm Let \[\theta \] be the angle between floor and ladder, then \[\sin \theta =\frac{y}{500}\] On differentiating w.r.t. ?t?, we get \[\cos \theta \frac{d\theta }{dt}=\frac{1}{500}\cdot \frac{dy}{dt}\] It is given that\[\frac{dy}{dt}=10\,\,cm/\sec \] \[\therefore \]      \[\cos \theta =\frac{x}{500},\]when \[x=200\] \[\Rightarrow \]   \[\cos \theta =\frac{200}{500}=\frac{2}{5}\] Substituting the respective values, we get \[\frac{2}{5}\frac{d\theta }{dt}=\frac{1}{500}\times 10\] \[\Rightarrow \]   \[\frac{d\theta }{dt}=\frac{1}{50}\times \frac{5}{20}=\frac{1}{20}rad/s\]