# How fast is the volume of a cylinder changing with respect to the radius when the radius is mm and the height is a constant 5mm?

Dec 11, 2017

The change will be proportional to the square of the radius.

#### Explanation:

Given two cylinders with the same height and radii r1 and r2 their volume will be:

${V}_{1} = h \pi {r}_{1}^{2}$ and ${V}_{2} = h \pi {r}_{2}^{2}$

The ratio of volumes between them will be:

${V}_{2} / \left({V}_{1}\right) = \frac{h \pi {r}_{2}^{2}}{h \pi {r}_{1}^{2}} = \frac{{r}_{2}^{2}}{{r}_{1}^{2}}$

This means that these two cylinders are correlated by the square of radius.

Dec 11, 2017

$\frac{\mathrm{dV}}{\mathrm{dr}} = 10 \pi r$

#### Explanation:

The general formula for the volume of a cylinder is $V = \pi {r}^{2} h$

We are told that the height is constant, so this cylinder has volume $V = 5 \pi {r}^{2}$

The rate of change of volume with respect to radius is $\frac{\mathrm{dV}}{\mathrm{dr}}$

$\frac{d}{\mathrm{dr}} \left(V\right) = \frac{d}{\mathrm{dr}} \left(5 \pi {r}^{2}\right) = 5 \pi \frac{d}{\mathrm{dr}} \left({r}^{2}\right) = 5 \pi \left(2 r\right) = 10 \pi r$

The volume is changing at a rate of

$10 \pi r$ $m {m}^{3} \text{(of Volume)}$ / $m m \text{(of radius)}$