Answer:
Thus, the pollutant concentration in lake will be reduced to 0.2% after 6.931471 days
Explanation:
From the information given:
A(t) = amount of pollutant for time (t)
A 4 billion cubic ft is the same as 4000 million cubic ft:
∴
The Initial amount of pollutant is [tex]A_o = (\dfrac{0.4}{100})\times 4000[/tex]
[tex]A_o = 16 \ million \ cubic \ feet[/tex]
However;
the pollutant rate (input) = 400 × 0 = 0
the pollutant rate (output) = [tex]400( \dfrac{A(t)} {4000})[/tex]
[tex]= 1( \dfrac{A(t)} {10})[/tex]
The net rate = [tex]A'(t) = 0 - 1( \dfrac{A(t)} {10})[/tex]
[tex]\implies A'(t) = - 1( \dfrac{A(t)} {10})[/tex]
[tex]\implies \dfrac{1}{A(t)}A'(t) = -(\dfrac{1}{10})[/tex]
[tex]\implies \int (\dfrac{1}{A(t)}A'(t) ) dt = \int -(\dfrac{1}{10}) dt[/tex]
[tex]\implies In (A(t)) = -(\dfrac{1}{10})t + c[/tex]
[tex]\implies A(t) = e^{-(\dfrac{1}{10})t+c}[/tex]
[tex]\implies A(t) = Ce^{-\dfrac{1}{10}^t}[/tex]
A(0) = 16
[tex]\implies Ce^{ -(1/20)^0} = 16 \\ \\ C = 16[/tex]
[tex]\implies A(t) = 16e^{(-1/10)t}[/tex]
[tex]0.2\% \ pollutant = (\dfrac{0.2}{100})*4000 =8 \ million \ cubic \ feet[/tex]
A(t) = 8
[tex]\implies 16e^{(-1/10)t}= 8 \\ \\ \implies e^{1/10)t} = 2 \\ \\ (\dfrac{1}{10} )^t = In(2) \\ \\ t = 10\ In(2) \\ \\ \mathbf{ t = 6.931471}[/tex]
