The volume of a right circular cylinder of radius \( r \) and height \( h \) is \( V=\pi r^{2} h \). \( \begin{array}{ll}\text { (a) Assume that } r \text { and } h \text { are functions of } t \text {. Find } V^{\prime}(t) \text {. } \\ \text { (b) Suppose that } r=e^{6 t} \text { and } h=e^{-6 t} \text {. Use part (a) to find } V^{\prime}(t) \text {. } \\ \begin{array}{ll}\text { (c) Does the volume of the cylinder of part (b) increase or decrease as } t \text { increases? } \\ \text { (a) Find } V^{\prime}(t) \text {. Choose the correct answer below. } \\ \text { A. } V^{\prime}(t)=2 \pi r(t) h(t) r^{\prime}(t) & \text { B. } V^{\prime}(t)=2 \pi r(t) h(t) r^{\prime}(t)+\pi(r(t))^{2} h^{\prime}(t)\end{array} \\ \text { C. } V^{\prime}(t)=\pi(r(t))^{2} h^{\prime}(t) & \text { D. } V^{\prime}(t)=2 \pi r(t) h(t) h^{\prime}(t)+\pi(r(t))^{2} r^{\prime}(t)\end{array} \)

To find \( V^{\prime}(t) \), we can apply the product rule since the volume \( V = \pi r^2 h \) is a product of the functions \( r(t) \) and \( h(t) \). This leads us to the formula: \[ V^{\prime}(t) = \pi \left(2r(t) h(t) r^{\prime}(t) + r(t)^2 h^{\prime}(t)\right) \] So, the correct answer is B: \( V^{\prime}(t) = 2 \pi r(t) h(t) r^{\prime}(t) + \pi (r(t))^2 h^{\prime}(t) \). Next, using the given functions \( r = e^{6t} \) and \( h = e^{-6t} \), we find their derivatives: \( r^{\prime}(t) = 6e^{6t} \) and \( h^{\prime}(t) = -6e^{-6t} \). Plugging these into our expression for \( V^{\prime}(t) \), we have: \[ V^{\prime}(t) = 2\pi e^{6t} e^{-6t} (6e^{6t}) + \pi (e^{6t})^2 (-6e^{-6t}) \] This simplifies to: \[ V^{\prime}(t) = 12\pi (6) + \pi (36e^{6t} e^{-6t})(-6) \] The overall behavior of the volume changes based on the values of \( \frac{6}{36} \) and \( -6 \), revealing that it increases when combining the growth of the base and the height. Ultimately, as \( t \) increases, the volume indeed increases!