A 6 m ladder leans against a vertical wall. The foot of the ladder is sliding away from the wall at a speed of $0.8\text{ m/s}$. Find the rate at which the top of the ladder is descending when the foot is $4\text{ m}$ 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 above ground. The ladder satisfies $x^2+y^2=6^2=36$.
2. Differentiate implicitly: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$.
3. Solve for $\frac{dy}{dt}$: $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$.
4. When $x=4$, find $y$ from $4^2+y^2=36Rightarrowy=\sqrt{20}=2sqrt5$.
5. Substitute $\frac{dx}{dt}=0.8$ and $x=4$ into $\frac{dy}{dt}=-\frac{4}{2\sqrt{5}}(0.8)=-\frac{2}{\sqrt{5}}\cdot 0.8=-\frac{1.6}{\sqrt{5}}$.
6. Simplify $\frac{1.6}{\sqrt{5}}=\frac{1.6\sqrt{5}}{5}=\frac{0.32\sqrt{5}}{1}\approx 0.4$. Therefore $\frac{dy}{dt}\approx -0.4\text{ m/s}$. For exact answer, $\frac{dy}{dt}=-\frac{2}{5}\text{ m/s}$.
7. So the top of the ladder is descending at $-\frac{2}{5}\text{ m/s}$ when the foot is 4 m from the wall.

Concept tested: Related rates – ladder problem

Published 8 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked