Both ways

This diagram is the same as the previous except there is an additional grey rectangle of width U and height delta U on top of the white rectangle

This white rectangle has width \(u\) and height \(v\), both of which vary as functions of time, \(t\). We can write the rates of change as \(\dot{u}\) and \(\dot{v}\).

Our rectangle has \(v=\quantity{\sqrt{t+1}}{cm}\) and \(u=\quantity{\left(\frac{1}{2}t^2+1\right)}{cm}\) at time \(\quantity{t}{s}\).

What is the rate of change of area, \(\dfrac{dA}{dt}\), at

  1. \(t=0\)?

  2. \(t=3\)?