A $7$m ladder leans against a vertical wall. The foot of the ladder slides away from the wall at $0.3$m/s. Find the speed at which the top of the ladder descends when the foot is $4$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; the ladder gives $x^2+y^2=7^2=49$.
2. Differentiate with respect to time $t$: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$.
3. Solve for $\frac{dy}{dt}$: $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$.
4. When $x=4$m, compute $y=\sqrt{49-4^2}=\sqrt{33}\approx 5.7446$m.
5. Substitute $x$, $y$ and $\frac{dx}{dt}=0.3$m/s: $\frac{dy}{dt}=-\frac{4}{5.7446}\times 0.3\approx -0.209$m/s.
6. The negative sign indicates the top is descending; round to two decimal places: $0.21$m/s downwards.

Concept tested: Related rates – ladder problem

Published 17 September 2026 · How this solution was prepared and checked