Question

Water flows from the bottom of a storage tank at a rate of r(t) = 210 - 5t liters per minute?

Water flows from the bottom of a storage tank at a rate of `r(t) = 210 - 5t` liters per minute, where `0 <= t <= 40`. Find the amount of water that flows from the tank during the first 45 minutes?

Answer 1

If we remove the constraint `0 le t le 40` for `r(t)` then total flow after `45` minutes is `4387.5 \ l`

Explanation:

The rate of water flow (litres) wrt to time `t` (minutes) is given by:

`r(t) = 210-5t`

We are given that `0 le t le 40`, and we seek the total litres that have flowed when `t=45`, thus there is insufficient information to answer the question as we have no information whatsoever regarding `r(t)` in the interval `40 le t le 45`.

If, however, we remove this constraint and If we denote the number of litres that have flowed in total by `n(t)`, then we can write:

`(dn)/(dt) = r(t) => n(t) = int \ 210-5t \ dt`

And integrating we get:

`n = 210t - 5/2t^2 + c`

Initially there is zero water, so `n=0` when `t=0`, giving:

`0 = 0 -0 + c => c = 0`

Thus we have:

`n = 210t - 5/2t^2`

So after `45` minutes, we have:

`n = 210(45) - 5/2(45^2)`
`\ \ = 9450 - 5/2(2025)`
`\ \ = 9450 - 5062.5`
`\ \ = 4387.5`