dV}{dt}$. This is a
      first-semester calculus derivative; we'll use it to check our
      work below.</li>

      <li>We know $DV(r,h)=\begin{bmatrix} 2\pi rh &amp; \pi r^2
      \end{bmatrix}$ and $Dg(t)= \begin{bmatrix} 3\\ 2t
      \end{bmatrix}.$ In first semester calculus, the chain rule
      was the product of derivatives. Multiply these matrices
      together to get $$\dfrac{dV}{dt}=DV(g(t))\, D(r,h)(t).$$ Did
      you get the same answer as the first part?</li>

      <li>To get the correct a