A $6\text{m}$ ladder is leaning against a vertical wall. The foot of the ladder slides away from the wall at $0.4\text{m/s}$. Find the speed at which the top of the ladder descends when the foot is $5\text{m}$ 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 length is constant: $x^2+y^2=6^2=36$.
2. Differentiate implicitly with respect to time $t$: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$.
3. Simplify: $x\frac{dx}{dt}+y\frac{dy}{dt}=0$.
4. Given $\frac{dx}{dt}=0.4\text{m/s}$ and $x=5\text{m}$, first find $y$ using the Pythagorean relation: $y=\sqrt{36-5^2}=\sqrt{11}\approx 3.3166\text{m}$.
5. Substitute the known values: $5(0.4)+\sqrt{11}\frac{dy}{dt}=0$.
6. Solve for $\frac{dy}{dt}$: $\frac{dy}{dt}=-\frac{5\times 0.4}{\sqrt{11}}=-\frac{2}{\sqrt{11}}\approx -0.48\text{m/s}$.
7. The negative sign indicates the top is moving downward, so the speed of descent is $0.48\text{m/s}$.

Concept tested: Related rates for a ladder

Published 15 September 2026 · How this solution was prepared and checked