A ladder 5 m in length is leaning against a wall. The bottom of the ladder is pulled along the ground away…
A ladder 5 m in length is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall, at the rate of $2 \mathrm{~m} / \mathrm{sec}$. How fast is the height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
$\frac{4}{3} \mathrm{~m} / \mathrm{sec}$
$\frac{2}{3} \mathrm{~m} / \mathrm{sec}$
$\frac{5}{3} \mathrm{~m} / \mathrm{sec}$
$\frac{8}{3} \mathrm{~m} / \mathrm{sec}$
Solution
In right angled $\triangle \mathrm{AOC}$,
$\begin{aligned}
& x^2+y^2=(5)^2 \\
& \Rightarrow y^2=5^2-x^2
\end{aligned}$
Differentiating w.r.t. t, we get
$\begin{aligned} 2 y \frac{\mathrm{~d} y}{\mathrm{dt}} & =-2 x \frac{\mathrm{~d} x}{\mathrm{dt}} \\ \Rightarrow \frac{\mathrm{d} y}{\mathrm{dt}} & =-\frac{x}{y} \cdot \frac{\mathrm{~d} x}{\mathrm{dt}} \\ & =\frac{-x}{\sqrt{5^2-x^2}} \cdot \frac{\mathrm{~d} x}{\mathrm{dt}}\end{aligned}$
$\begin{aligned}
\therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{dt}}\right)_{x=4} & =\frac{-4}{\sqrt{25-16}} \cdot(2) \\
& =\frac{-8}{3} \mathrm{~m} / \mathrm{sec}
\end{aligned}$ Thus; the height on the wall is decreasing at the rate of $\frac{8}{3} \mathrm{~m} / \mathrm{sec}$
