A $5\text{m}$ ladder is leaning against a vertical wall. The foot of the ladder is sliding away from the wall at a speed of $0.6\text{m/s}$. Find the rate at which the top of the ladder is descending when the foot is $3\text{m}$ from the wall.

Step-by-step solution

1. Let $x$ be the distance of the foot from the wall and $y$ the height of the top on the wall. The ladder gives $x^2+y^2=5^2$.
2. Differentiate implicitly with respect to time $t$: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$.
3. Substitute $\frac{dx}{dt}=0.6\text{m/s}$ and the values $x=3\text{m}$, $y=\sqrt{5^2-3^2}=4\text{m}$ into the differentiated equation.
4. Solve for $\frac{dy}{dt}$: $3(0.6)+4\frac{dy}{dt}=0\Rightarrow \frac{dy}{dt}=-\frac{3\times 0.6}{4}=-0.45\text{m/s}$.
5. Check the arithmetic: $3\times 0.6=1.8$, $1.8/4=0.45$. Since the ladder length is $5\text{m}$, the correct $y$ is $4\text{m}$, giving $\frac{dy}{dt}=-0.45\text{m/s}$. However the ladder length is $5\text{m}$, not $5\text{m}$? Re‑evaluate: $x=3$, $y=\sqrt{25-9}=4$, correct. Hence $\frac{dy}{dt}=-0.45\text{m/s}$.
6. But the standard answer for a $5\text{m}$ ladder sliding with $0.6\text{m/s}$ at $x=3$ is $\frac{dy}{dt}=-0.8\text{m/s}$. Check the differentiation again: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\Rightarrow x\frac{dx}{dt}+y\frac{dy}{dt}=0\Rightarrow \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-\frac{3}{4}\times 0.6=-0.45\text{m/s}$. The correct rate is $-0.45\text{m/s}$, so the answer is $-0.45\text{m/s}$.
7. Therefore the top descends at $0.45\text{m/s}$ downward (negative sign indicates descent).

Concept tested: Related rates – sliding ladder

Published 6 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked