A ladder 5 m long rests against a vertical wall. If its top slides downwards at the rate of $10 \mathrm{~cm} / \mathrm{sec}$., then the foot of the ladder is sliding at the rate of _______ $\mathrm{m} / \mathrm{sec}$, when it is 4 m away from the wall.

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Aicharya · Accepted Answer

According to the figure, $x^2+y^2=25$ At $x=4, y=3$ Differentiating (i) with respect to ' $t$ ', we get $\begin{aligned} & 2 x \frac{\mathrm{~d} x}{\mathrm{dt}}+2 y \cdot \frac{\mathrm{~d} y}{\mathrm{dt}}=0 \\ & x \frac{\mathrm{~d} x}{\mathrm{dt}}=-y \frac{\mathrm{~d} y}{\mathrm{dt}} \\ & \frac{\mathrm{~d} x}{\mathrm{dt}}=\frac{-y}{x} \frac{\mathrm{~d} y}{\mathrm{dt}}=\frac{-3}{4} \times(-0.1)=0.075 \end{aligned}$

A ladder 5 m long rests against a vertical wall. If its top slides downwards at the rate of $10 \mathrm{~cm}…

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