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

**(a)** Given \[ V = \pi r^2 h, \] where both \( r \) and \( h \) depend on \( t \), we differentiate \( V \) with respect to \( t \) using the product rule. Notice that \[ V(t) = \pi (r(t))^2 h(t). \] Differentiate with respect to \( t \): \[ V'(t) = \pi \frac{d}{dt}\bigl[(r(t))^2 h(t)\bigr]. \] Apply the product rule: \[ \frac{d}{dt}\bigl[(r(t))^2 h(t)\bigr] = \frac{d}{dt}(r(t))^2 \cdot h(t) + (r(t))^2 \cdot \frac{d}{dt}h(t). \] Using the chain rule for \(\frac{d}{dt}(r(t))^2\): \[ \frac{d}{dt}(r(t))^2 = 2r(t)r'(t). \] Thus, we have: \[ V'(t) = \pi \left[2r(t)r'(t)h(t) + (r(t))^2 h'(t)\right]. \] So the answer to part (a) is: \[ V'(t) = 2\pi\,r(t)\,h(t)\,r'(t) + \pi\,(r(t))^2\,h'(t). \] **(b)** Now, let \[ r(t) = e^{6t} \quad \text{and} \quad h(t) = e^{-6t}. \] Compute the derivatives: \[ r'(t) = \frac{d}{dt}\,e^{6t} = 6e^{6t}, \] \[ h'(t) = \frac{d}{dt}\,e^{-6t} = -6e^{-6t}. \] Substitute these into the expression from part (a): \[ V'(t) = 2\pi\,r(t)\,h(t)\,r'(t) + \pi\,(r(t))^2\,h'(t). \] Substitute \( r(t) = e^{6t} \), \( h(t) = e^{-6t} \), \( r'(t) = 6e^{6t} \), and \( h'(t) = -6e^{-6t} \): \[ V'(t) = 2\pi\,(e^{6t})(e^{-6t})(6e^{6t}) + \pi\,(e^{6t})^2(-6e^{-6t}). \] Simplify the terms: - Notice that \( e^{6t}\,e^{-6t} = e^0 = 1 \). Thus, the first term simplifies to: \[ 2\pi \cdot 1 \cdot (6e^{6t}) = 12\pi \,e^{6t}. \] - For the second term, combine exponents: \[ (e^{6t})^2\,e^{-6t} = e^{12t}\,e^{-6t} = e^{6t}. \] Therefore, the second term becomes: \[ \pi \cdot (-6e^{6t}) = -6\pi\,e^{6t}. \] Now, combine both terms: \[ V'(t) = 12\pi\,e^{6t} - 6\pi\,e^{6t} = 6\pi\,e^{6t}. \] **(c)** Since \( e^{6t} > 0 \) for all \( t \) and \( 6\pi > 0 \), we have: \[ V'(t) = 6\pi\,e^{6t} > 0. \] Thus, the volume of the cylinder increases as \( t \) increases.