## Calculus (3rd Edition)

Let x be the distance from the bottom of the ladder to the wall and h be the height of its top. The goal is to find $\frac{dh}{dt}$ at $t=2s$. We are given that $\frac{dx}{dt}=0.8\, m/s$ and $x(0)=1.5\,m$ Using Pythagoras' theorem, we have $x^{2}+h^{2}=(5\,m)^{2}$ Differentiating both sides with respect to $t$, we have $2x\frac{dx}{dt}+2h\frac{dh}{dt}=0$ $\implies \frac{dh}{dt}=-\frac{x}{h}\frac{dx}{dt}=-\frac{x}{h}\times0.8\,m/s$ $x(2)=x(0)+dt\frac{dx}{dt}=1.5\,m+ 2\,s(0.8\,m/s)=3.1\,m$ $h(2)=\sqrt {(5\,m)^{2}-(3.1\,m)^{2}}\approx 3.923\,m$ $\frac{dh}{dt}|_{t=2}=-\frac{3.1\,m}{3.923\,m}\times0.8\,m/s=0.632\,m/s$