A $13$-foot ladder is leaning against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of $0.5$ feet per second. At what rate is the top of the ladder sliding down the wall when the bottom of the ladder is $5$ feet from the wall?

Core Concept

Correct. By the Pythagorean theorem, $x^2 + y^2 = 13^2$. Differentiating gives $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$. When $x=5$, $y=12$. Substituting $2(5)(0.5) + 2(12) \frac{dy}{dt} = 0$ yields $\frac{dy}{dt} = -\frac{5}{24}$. The speed is $5/24$ ft/sec.