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- Apr 21st 2010, 02:40 PMyoman360the rate at which a purification process can remove contaminants from a tank of water
- Apr 21st 2010, 03:32 PMione
Let A = the amount of contaminant in the water

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

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

$\displaystyle ln|A|=kt+C$

Let the amount of contaminant when t = 0 be 100

$\displaystyle ln100=C$

$\displaystyle ln|A|=kt+ln100$

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

A=80 when t = 1

$\displaystyle ln80=k(1)+ln100$

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

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

You want to find t when A = 2