## Calculus (3rd Edition)

$x=h$
Using Pythagoras' theorem, we have $x^{2}+h^{2}=5^{2}$ Differentiating with respect to $t$, we have $2x\frac{dx}{dt}+2h\frac{dh}{dt}=0$ But we are given that the top and bottom of the ladder move at the same speed. That is, $\frac{dh}{dt}=-\frac{dx}{dt}$ which gives $2x\frac{dx}{dt}+2h(-\frac{dx}{dt})=0$ Or $2x\frac{dx}{dt}=2h\frac{dx}{dt}$ $\implies x=h$