5.3 Example

From [1, p. 423, 29]

Given: A circular cylinder of changing radius $r$ and height $h$. At a given time, $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 6 inch, $\dot r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.2 in/sec, $h$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8 in, $\dot h$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-0.4$ in/sec.

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

5.3.1 Solution

$$V= \pi r^2 h \qquad A = 2 \pi r h + 2 \pi r^2$$

The total differential gives

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

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

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

Similarly:

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