Question

If a cylindrical tank with radius 5 meters is being filled with water at a rate of 3 cubic meters per minute, how fast is the height of the water increasing?

Answer 1

The answer is `(dh)/(dt)=3/(25 pi)m/(min)`.

With related rates, we need a function to relate the 2 variables, in this case it is clearly volume and height. The formula is:
`V=pi r^2 h`

There is radius in the formula, but in this problem, radius is constant so it is not a variable. We can substitute the value in:
`V=pi (5m)^2 h`

Since the rate in this problem is time related, we need to implicitly differentiate wrt (with respect to) time:
`(dV)/(dt)=(25 m^2) pi (dh)/(dt)`

In the problem, we are given `3(m^3)/min` which is `(dV)/(dt)`. So we substitute this in:
`(dh)/(dt)=(3m^3)/(min (25m^2) pi)=3/(25 pi)m/(min)`

In general
- find a formula to relate the 2 variables
- substitute values to remove the constant variables
- implicitly differentiate wrt time (most often the case)
- substitute the given rate
- and solve for the desired rate.