You may consider the following rate problem as an example that uses a multivariate function. If a liquid is poured into a cylinder shape recipient, with circular cross section radius of `1.5 m` , at a rate of `1 m^3/min` , at what rate is the height of liquid increasing?

You need to remember the equation of volume such that:

`V = pi*r^2*h`

r represents the radius of circular cross section of cylinder

h represents the height of cylinder

You need to solve for h the equation of volume such that:

`h = V/(pi*r^2) => h = V/(pi*2.25)`

You need to differentiate the equation of height with respect to t such that:

`(dh)/(dt) = (dh)/(dV)*(dV)/(dt) => (dh)/(dt) = (d(V/(pi*2.25))/(dV)`

Since the problem provides the information that the liquid is poured at a rate of `1 m^3/min` , hence `(dV)/(dt) = 1` .

`(dh)/(dt) = 1/pi*2.25 ~~ 0.318`

**Hence, the height of liquid increases at a rate of about `0.318 m/min.` **

Possibly you may recognize the expression

`v=pi r^2 h`

as the formula for the volume of a cylinder. Thus, we could come up with a situation involving cylinders, and try to view the scenario as a function. Here is an example:

A soup company makes cans of soup. They are deciding what the dimensions of the soup can will be. The volume of the cans depends on the radius of the cans, and the height of the cans. The volume of a soup can can be expressed as:

volume = pi * (radius of can)^2 * (height of can)

Thus, v depends on both r, and h, and is a multivariable function