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

Let \[ V=\pi r(t)^2 h(t). \] **(a) Differentiate \(V\) with respect to \(t\):** Using the product and chain rules, \[ V'(t)=\pi\frac{d}{dt}\bigl[r(t)^2\,h(t)\bigr] =\pi\Bigl(2r(t)r'(t)h(t)+r(t)^2h'(t)\Bigr). \] Thus, the correct answer is \[ \boxed{V'(t)=2\pi r(t)h(t)r'(t)+\pi r(t)^2h'(t)}. \] **(b) Substitute \(r(t)=e^{6t}\) and \(h(t)=e^{-6t}\):** First, compute the derivatives: \[ r'(t)=\frac{d}{dt}(e^{6t})=6e^{6t},\qquad h'(t)=\frac{d}{dt}(e^{-6t})=-6e^{-6t}. \] Substitute into the formula: \[ \begin{aligned} V'(t)&=\pi\Bigl[2e^{6t}\cdot e^{-6t}\cdot (6e^{6t})+ \Bigl(e^{6t}\Bigr)^2\cdot (-6e^{-6t})\Bigr]\\[1mm] &=\pi\Bigl[2\cdot (e^{6t}e^{-6t})\cdot 6e^{6t} -6e^{12t}e^{-6t}\Bigr]\\[1mm] &=\pi\Bigl[2\cdot 1\cdot 6e^{6t}-6e^{6t}\Bigr]\\[1mm] &=\pi\Bigl[12e^{6t}-6e^{6t}\Bigr] =\pi\Bigl[6e^{6t}\Bigr]. \end{aligned} \] Thus, \[ \boxed{V'(t)=6\pi e^{6t}}. \] **(c) Determine whether the volume increases or decreases as \(t\) increases:** Since \[ V'(t)=6\pi e^{6t}, \] and the exponential function \(e^{6t}\) is positive for all \(t\), it follows that \(V'(t)>0\) for all \(t\). Therefore, the volume of the cylinder is increasing as \(t\) increases.