# Thread: the rate at which a purification process can remove contaminants from a tank of water

1. ## the rate at which a purification process can remove contaminants from a tank of water

2. Originally Posted by yoman360
Let A = the amount of contaminant in the water

$\frac{dA}{dt}=kA$

$\frac{dA}{A}=k\,dt$

$ln|A|=kt+C$

Let the amount of contaminant when t = 0 be 100

$ln100=C$

$ln|A|=kt+ln100$

Since 20% of the contaminants can be removed during the first minute

A=80 when t = 1

$ln80=k(1)+ln100$

$k=ln\frac{4}{5}$

$ln|A|=(ln\frac{4}{5})t+ln100$

You want to find t when A = 2