Let a = amount of water (gallons) in the tank at time t (minutes).
The drainage rate is proportional to the amount of water left in the tank, therefore
[tex] \frac{da}{dt} = ka[/tex]
where k is the proportionality constant.
Therefore, obtain
[tex] \frac{da}{a} =k\, dt[/tex]
Integrate to obtain
ln(a) = kt + c.
where c = the constant of integration.
When t = 0, a = 1000 gal, therefore
ln(1000) = c
When t = 10 min, then a = 1000 - 200 = 800 gal.
Therefore
ln(800) = 10k + ln(1000)
10k = ln(800) - ln(1000) = -0.2231
k = -0.02231
Therefore
ln(a) = -0.02231t + ln(1000)
ln(a/1000) = -0.02231t
[tex]a(t) = 1000\,e^{-0.02231t}[/tex]
A graph of the solution is shown below.
Answer: [tex]a(t) = 1000 \, e^{-0.02231t}[/tex]
