The radius 
of a right circular cylinder is decreasing at a rate of 
, while its height 
is decreasing at a rate of 
. How is the volume changing when 
and 
? Is the volume increasing or decreasing?
Solution.
The volume of a right circular cylinder is calculated by the formula
.
The volume 
(a dependent variable) of a cylinder varies with its radius and height (both independent variables). The change of volume, simply put it, is a derivative of volume with respect to time 
:
.
Differentiate 
wrt. time :
![\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(r,h,t) & = \frac{\mathrm{d}}{\mathrm{d}t} (\pi r^2 h) \\ & = \pi \frac{\mathrm{d}}{\mathrm{d}t} (r^2h) \\ & = \pi \bigg[ r^2 \frac{\mathrm{d}}{\mathrm{d}t}(h) + h \frac{\mathrm{d}}{\mathrm{d}t} (r^2) \bigg] \\ & = \pi \bigg[ \big( r(t) \big)^2 \frac{\mathrm{d}}{\mathrm{d}t}\big( h(t) \big) + \big( h(t)\big) \bigg( \frac{\mathrm{d}}{\mathrm{d}t}\big( r(t) \big)^2 \bigg) \bigg] \\ & = \pi (r^2h' +2hrr' ) \\ & = \pi r^2h' + 2\pi hrr' \\ \end{aligned}](https://physicspupil.com/wp-content/ql-cache/quicklatex.com-6799ee3438e9e88d3a6b70a1e4e814ca_l3.svg "Rendered by QuickLaTeX.com")
Given , , 
, and 
, you would have it.
In some formalism of partial derivatives,
you could have it also.
Afterthought.
It just so happens that there are two lines of attack, by taking total/ordinary derivatives and by taking partial derivatives. Is here anyhow the difference? Is there anything the matter?