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

In \(\triangle \mathrm{ABC}, \mathrm{AC}\) represents ladder
\(\mathrm{AB} \rightarrow\) vertical wall
Let \(\mathrm{AB}=x, \mathrm{BC}=y\)
\(\therefore \quad \angle \mathrm{ABC}=90^{\circ}\)
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
\(\therefore \quad\) Upper end is coming down at the rate of \(\frac{8}{15} \mathrm{~m} / \mathrm{s}\).