A rectangle's width is $w(t)=3t$ meters and its height is $h(t)=t^2+1$ meters, where $t$ is in seconds. How fast is its area increasing at $t=2$ seconds?

Core Concept

Using the product rule, $\dfrac{dA}{dt}=w'\,h+w\,h'$. At $t=2$, $w=6$, $h=5$, $w'=3$, and $h'=4$, so $\dfrac{dA}{dt}=3(5)+6(4)=39$ m$^2$/s.