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.

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Aicharya · Accepted Answer

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

A ladder 5 meters long rests against a vertical wall. If its top slides downwards at the rate of $10…

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