The radius of a cylinder is decreasing at a rate of 2 cm / s and the altitude is increasing at the rate of 3 cm / s. Find the rate of change of…

\[\begin{aligned} & \frac{d \mathrm{r}}{d t}=-2 \mathrm{~cm} / \mathrm{s}, \quad \frac{d \mathrm{h}}{d t}=3 \mathrm{~cm} / \mathrm{s},\left(\frac{d \mathrm{~V}}{d t}\right)_{\mathrm{r}=4, \mathrm{h}=6}=? \\ & \mathrm{~V}=\pi \mathrm{r}^{2} \mathrm{~h} \Rightarrow \frac{d V}{d t}=2 \pi \mathrm{r} . \frac{d \mathrm{r}}{d t} \cdot \mathrm{~h}+\pi \mathrm{r}^{2} \frac{d \mathrm{h}}{d t} \\ & \text { When } \mathrm{r}=4 \mathrm{~cm} \text { and } \mathrm{h}=6 \mathrm{~cm}, \\ & \frac{d V}{d t}=2 \pi(4)(-2)(6)+\pi(4)^{2}(3)=-48 \pi \mathrm{~cm}^{3} / \mathrm{s} \end{aligned} \] Volume is decreasing at the rate of $\displaystyle \mathbf{4 8} \boldsymbol{\pi} \mathbf{~ c m}^{\mathbf{3}} \boldsymbol{/} \mathbf{s}$