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.`