In general, the volume
[tex]V=\pi r^2h[/tex]
has total derivative
[tex]\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)[/tex]
If the cylinder's height is kept constant, then [tex]\dfrac{\mathrm dh}{\mathrm dt}=0[/tex] and we have
[tex]\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}[/tex]
which is to say, [tex]\dfrac{\mathrm dV}{\mathrm dt}[/tex] and [tex]\dfrac{\mathrm dr}{\mathrm dt}[/tex] are directly proportional by a factor equivalent to the lateral surface area of the cylinder ([tex]2\pi r h[/tex]).
Meanwhile, if the cylinder's radius is kept fixed, then
[tex]\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}[/tex]
since [tex]\dfrac{\mathrm dr}{\mathrm dt}=0[/tex]. In other words, [tex]\dfrac{\mathrm dV}{\mathrm dt}[/tex] and [tex]\dfrac{\mathrm dh}{\mathrm dt}[/tex] are directly proportional by a factor of the surface area of the cylinder's circular face ([tex]\pi r^2[/tex]).
Finally, the general case ([tex]r[/tex] and [tex]h[/tex] not constant), you can see from the total derivative that [tex]\dfrac{\mathrm dV}{\mathrm dt}[/tex] is affected by both [tex]\dfrac{\mathrm dh}{\mathrm dt}[/tex] and [tex]\dfrac{\mathrm dr}{\mathrm dt}[/tex] in combination.
