6.3 Example

From [1, p. 423, 29]

Given: A circular cylinder of changing radius and height . At a given time, inch, $\dot r = 0.2$ in/sec, in, $\dot h = -0.4$ in/sec.

Asked: $\dot V$ and $\dot A$ at that time.

$\begin{displaymath} V= \pi r^2 h \qquad A = 2 \pi r h + 2 \pi r^2 \end{displaymath}$

The total differential gives

$\begin{displaymath} {\rm d}V = \frac{\partial V}{\partial h} {\rm d}h + \frac{\partial V}{\partial r} {\rm d}r \end{displaymath}$

where differential changes become time derivatives if you divide by ${\rm d}t$ . So identifying the partial derivatives gives:

$\begin{displaymath} \dot V = \pi r^2 \dot h + \pi 2 r h \dot r = 15.08 \mbox{ in$^3$/sec } \end{displaymath}$

Similarly:

$\begin{displaymath} \dot A = 2 \pi r \dot h + \left(2 \pi h + 4 \pi r\right) \dot r = 10.05 \mbox{ in$^2$/sec } \end{displaymath}$