A ladder of length $17 \mathrm{~m}$ rests with one end against a vertical wall and the other on the level ground. If the lower end slips away at the rate of $1 \mathrm{~m} / \mathrm{sec}$., then when it is $8 \mathrm{~m}$ away from the wall, its upper end is coming down at the rate of

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Aicharya · Accepted Answer

In $	riangle \mathrm{ABC}, \mathrm{AC}$ represents ladder $\mathrm{AB} ightarrow$ vertical wall Let $\mathrm{AB}=x, \mathrm{BC}=y$ $	herefore \quad \angle \mathrm{ABC}=90^{\circ}$ By Pythagoras theorem, $\begin{aligned} & \mathrm{AB}^2+\mathrm{BC}^2=\mathrm{AC}^2 \ & \Rightarrow x^2+y^2=17^2 \ & \Rightarrow x^2=289-y^2 ... (i)\ & \Rightarrow x^2=289-64 \ & \Rightarrow x^2=225 \ & \Rightarrow x=15 \mathrm{~m} \end{aligned}$ Consider equation (i), $x^2=289-y^2$ Differentiating w.r.t. t, we get $\begin{aligned} & 2 x \frac{\mathrm{d} x}{\mathrm{dt}}=-2 y \frac{\mathrm{d} y}{\mathrm{dt}} \ & \Rightarrow 15 \frac{\mathrm{d} x}{\mathrm{dt}}=-8(1) \ & \Rightarrow \frac{\mathrm{d} x}{\mathrm{dt}}=\frac{-8}{15} \mathrm{~m} / \mathrm{s} \end{aligned}$ Negative sign shows that the ladder is moving down. i.e., vertical length is decreasing $	herefore \quad$ Upper end is coming down at the rate of $\frac{8}{15} \mathrm{~m} / \mathrm{s}$.

A ladder of length $17 \mathrm{~m}$ rests with one end against a vertical wall and the other on the level…

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