Question

A ladder 13 m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5 m/sec. How fast is the angle θ between the ladder and the ground is changing when the foot of the ladder is 12 m away from the wall.

Answer 1

Let the bottom of the ladder be at a distance of x m from the wall and its top be at a height of y m from the ground.


Then,

$$\begin{gathered} \tan \theta =\frac{y}{x}\mathrm{and}\mathit{} \\ {x}^2+y^2={\left(13\right)}^2 \\ \Rightarrow x^2\left(1+{\tan }^2\theta \right)=169 \\ \Rightarrow {\sec }^2\theta =\frac{169}{x^2} \\ \Rightarrow 2{\sec }^2\theta \tan \theta \frac{d\theta }{dt}=169\left(\frac{-2}{x^3}\right)\frac{dx}{dt} \\ \Rightarrow \frac{d\theta }{dt}=\frac{-338\times 1.5}{{\left(12\right)}^{3}2{\sec }^2\theta \tan \theta }...\left(1\right) \\ \mathrm{When}\mathit{}x=12,y=\sqrt{169-144}=5\mathrm{m} \\ \mathrm{So}, \\ \sec \theta =\frac{13}{12}\mathrm{and}\tan \theta =\frac{12}{5} \\ \mathrm{From}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{d\theta }{dt}=\frac{-338\times 1.5}{{\left(12\right)}^3\times 2\times {\left({\displaystyle \frac{13}{12}}\right)}^2\times {\displaystyle \frac{5}{12}}}=\frac{-338\times 1.5}{10\times 169}=-0.3\mathrm{rad}/\sec \end{gathered}$$