## Calculus (3rd Edition)

The goal is to find $\frac{dx}{dt}$ at $h=3\,m$. We are given that $\frac{dh}{dt}= -1.2\,m/s$ Using Pythagoras' theorem, we have $h^{2}+x^{2}=5^{2}$ Differentiating both sides with respect to $t$, we have $2h\frac{dh}{dt}+2x\frac{dx}{dt}=0$ $\implies \frac{dx}{dt}=-\frac{h}{x}\frac{dh}{dt}$ When $h=3\,m$, $x=\sqrt {5^{2}-3^{2}}=4\,m$ $\implies \frac{dx}{dt}|_{h=3\,m}=-\frac{3}{4}\times-1.2\,m/s$ $=0.9\,m/s$