A ladder 5 meters long rests against a vertical wall. If its top slides downwards at the rate of \(10 \mathrm{~cm} / \mathrm{s}\), then the angle between the ladder and the floor is decreasing at the rate of rad./s when it's lower end is \(4 \mathrm{~m}\) away from the wall.

According to the figure, \(x^2+y^2=25...(i)\)
Note that \(\cos \theta=\frac{\mathrm{OB}}{\mathrm{AB}}=\frac{x}{5}\)
\(\therefore \quad x=5 \cos \theta\)
\(\therefore \quad\) (i) \(\Rightarrow 25 \cos ^2 \theta+y^2=25\)
Differentiating w.r.t. ' \(t\) ', we get
\(
\begin{aligned}
& \quad-50 \cos \theta \sin \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}+2 y \frac{\mathrm{d} y}{\mathrm{dt}}=0 \\
& 25 \sin \theta \cos \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}=y \frac{\mathrm{d} y}{\mathrm{dt}} \\
& \therefore \quad 25 \sin \theta \cos \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}=y(-0.1) \\
& \therefore \quad \ldots\left[\because \frac{\mathrm{d} y}{\mathrm{~d} x}=-10 \mathrm{~cm} / \mathrm{s}=-0.1 \mathrm{~m} / \mathrm{s}\right] \\
& \quad 25 \sin \theta \cos \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}=-(0.1) y \quad, .(\mathrm{ii}) \\
& \quad \text { at } x=4, \cos \theta=\frac{4}{5}, \sin \theta=\frac{3}{5} \text { and } y=3 \\
& \quad \text { (ii) } \Rightarrow 25 \times \frac{3}{5} \times \frac{4}{5} \times \frac{\mathrm{d} \theta}{\mathrm{dt}}=-0.3 \\
& \Rightarrow \frac{\mathrm{d} \theta}{\mathrm{dt}}=-0.025
\end{aligned}
\)
i.e., the angle is decreasing at the rate of \(0.025 \mathrm{rad} / \mathrm{s}\)