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

\[\text { Length of the ladder }= 500cm\]

\[\text { Let the horizontal length covered between the wall and the ladder be x and vertical length covered between the wall and the ladder be y } . \]

\[\text { And let the angle between the floor and ladder be } \theta . \]

\[\text { Then,} \sin\theta = \frac{y}{500}\]

\[\text { On differentiating with respect to t, we get }\]

\[\cos\theta\frac{d \theta}{d t} = \frac{1}{500}\frac{d y}{d t} . . . (1)\]

\[\text {It is given that } \frac{d y}{d t} = text{- 10 cm} /sec . . . . (2)\]

\[\text { Also,} \]

\[\cos\theta = \frac{x}{500}\]

\[\text { When }x = \text{200 cm}, \cos\theta = \frac{200}{500} = \frac{2}{5} . . . (3)\]

\[\text { Substituting } (2) \text { and } (3)\text { in } (1), \text { we get }\]

\[\frac{2}{5}\frac{d \theta}{d t} = \frac{1}{500}\left( - 10 \right)\]

\[\frac{d \theta}{d t} = - \frac{1}{20} \text { radian/second }\]

\[\text { Hence, the angle between the floor and the ladder is decreasing at the rate of }\frac{1}{20}\text { radian/second } .\]