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This problem models pollution effects in the Great Lakes. We assume pollutants are flowing into a lake at a constant rate of I kg/year, and that water is flowing out at a constant rate of F km3/year. We also assume that the pollutants are uniformly distributed throughout the lake. If C(t) denotes the concentration (in kg/km3) of pollutants at time t (in years), then C(t) satisfies the differential equationdC dt = −FVC + IVwhere V is the volume of the lake (in km3). We assume that (pollutant-free) rain and streams flowing into the lake keep the volume of water in the lake constant.A) Suppose that the concentration at time t = 0 is C0. Determine the concentration at any time t by solving the differential equation.B) Find lim t→[infinity] C(t) =C) For Lake Erie, V = 458 km3 and F = 175 km3/year. Suppose that one day its pollutant concentration is C0 and that all incoming pollution suddenly stopped (so I = 0). Determine the number of years it would then take for pollution levels to drop to C0/10.D) For Lake Superior, V = 12221 km3 and F = 65.2 km3/year.

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Answer:

(a) [tex]\mathbf{C_{(t)} =\dfrac{I}{F} [ 1- e \dfrac{-Ft}{v}+ C_oe \dfrac{-Ft}{v} ]}[/tex]

(b) [tex]\mathbf{\lim_{t \to \infty} C_t = \dfrac{I}{F}[1-0+ 0 ] \ = \dfrac{I}{F}}[/tex]

(c) T = 6.02619 years

(d) T = 431.593 years

Step-by-step explanation:

(a)

[tex]\dfrac{dC}{dt} = -\dfrac{F}{v}C + \dfrac{I}{v} \\ \\ \\ \dfrac{dC}{dt} + \dfrac{F}{v}C = \dfrac{I}{v}[/tex]

By integrating the factor of this linear differential equation ; we have :

[tex]= e \int\limits \dfrac{F}{v}t \\ \\ \\ = e \dfrac{Ft}{v}[/tex]

[tex]C* e \frac{Ft}{v}= \int\limits \dfrac{I}{v}*e \dfrac{Ft}{v} dt[/tex]

[tex]C* e \frac{Ft}{v}= \dfrac{I}{v}* \dfrac{e \frac{Ft}{v} }{F/v}+ K[/tex]

[tex]C* e \frac{Ft}{v}= \dfrac{I}{v}* \dfrac{V}{F} e \dfrac{Ft}{v} + K[/tex]

[tex]Ce \dfrac{Ft}{v} = \dfrac{I}{F} \ * \ e \dfrac{Ft}{v} + k \ \ \ \ (at \ t = 0 \ ; C = C_o)[/tex]

[tex]C_o = \dfrac{I}{F} e^o + K[/tex]

[tex]K = C_o - \dfrac{I}{F}[/tex]

[tex]C_{(t)} = [ \dfrac{I}{F}e \dfrac{Ft}{v}+ C_o - \dfrac{I}{F}] e\dfrac{-Ft}{v}[/tex]

[tex]C_{(t)} =\dfrac{I}{F} [ e \dfrac{Ft}{v} * e \dfrac{-Ft}{v}+ C_oe \dfrac{-Ft}{v} - 1* e \dfrac{-Ft}{v}][/tex]

[tex]\mathbf{C_{(t)} =\dfrac{I}{F} [ 1- e \dfrac{-Ft}{v}+ C_oe \dfrac{-Ft}{v} ]}[/tex]

(b)

[tex]\lim_{t \to \infty} C_t = \dfrac{I}{F}[1-e^{- \infty} + C_o e^{- \infty} ][/tex]

since  [tex](e^{- \infty} = 0)[/tex]

[tex]\mathbf{\lim_{t \to \infty} C_t = \dfrac{I}{F}[1-0+ 0 ] \ = \dfrac{I}{F}}[/tex]

(c)

V = 458 km³   and F = 175 km³  , I = 0

[tex]\dfrac{dC}{dt} = - \dfrac{-175}{458}C[/tex]

[tex]= \int\limits \ \dfrac{dC}{C} = - \dfrac{175}{458}\int\limits dt[/tex]

[tex]In (C_{(t)}) = - \dfrac{175}{458} t + K[/tex]

[tex]C_{(t)} = e \dfrac{-175}{458}t + K[/tex]

Let at time t = 0 [tex]C_{(t)}} = C_o \to C_o = e^{0+k} = e^K[/tex]

[tex]C_{(t)} = e \dfrac{-175}{458}t[/tex]

Now at time t = T ; [tex]C_{9t)} = \dfrac{C_o}{10}[/tex]

[tex]\dfrac{C_o}{10} = C_o e \dfrac{-175}{458}T \to \dfrac{1}{10} = e \dfrac{-175}{458}T[/tex]

[tex]In ( \dfrac{1}{10}) = \dfrac{-175}{459}T[/tex]

[tex]- In (10) = \dfrac{175}{458}T[/tex]

[tex]T = \dfrac{458}{175} In (10)[/tex]

T = 6.02619 years

(d) V = 12221 km³

F = 65.2 km³/ year

[tex]\mathbf{T = \dfrac{v}{F}In (10)}[/tex]

[tex]T = \dfrac{12221}{65.2}In (10)[/tex]

T = 431.593 years

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