## Calculus: Early Transcendentals 8th Edition

$8160 \pi \dfrac{in^3}{s}$
We need to use the chain rule. $\dfrac{dV}{dt}=(\dfrac{\partial V}{\partial r})(\dfrac{dr}{ dt})+(\dfrac{\partial V}{\partial h})(\dfrac{dh}{dt})$ ...(1) Here, we have $\dfrac{\partial V}{ \partial r} \approx \dfrac{\partial }{ \partial r} (\dfrac{1}{3} \pi r^2 h)$ or, $\dfrac{\partial V}{ \partial r} =\dfrac{2}{3}\pi r^2 h$ and $\dfrac{\partial V}{ \partial h} \approx \dfrac{\partial }{ \partial h} (\dfrac{1}{3} \pi r^2 h)$ or, $\dfrac{\partial V}{ \partial h} \approx \dfrac{1}{3}\pi r^2$ The equation (1), becomes: $\dfrac{dV}{dt}=(\dfrac{\partial V}{\partial r})(\dfrac{dr}{ dt})+(\dfrac{\partial V}{\partial h})(\dfrac{dh}{dt})=(11200 \pi) \times (1.8)+(4800 \pi) \times (-2.5)= 8160 \pi \dfrac{in^3}{s}$