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 equationdCdt = − FV C + 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.C(t) = (to enter C0 in your answer, type "C_0", that is, "capital C, underscore, zero")
(b) Find lim t→[infinity] C(t) = Incorrect: Your answer is incorrect.
(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. Give your answer in decimal form, rounded to the nearest year. years

(d) For Lake Superior, V = 12221 km3 and F = 65.2 km3/year. Answer the same question as in part (c) for Lake Superior. years

a.) [tex]\frac{dc}{dt}=-\frac{Fc+I}{v}\\\\\frac{dc}{dt}=-\frac{(-I+Fc)}{v}\\\\\frac{dc}{dt}=-\frac{-\frac{I}{F}+c}{\frac{v}{F}}=\frac{-F}{v}[c-\frac{I}{F}][/tex]

[tex]\frac{dc}{c-\frac{I}{F}}=\frac{-F}{v}dt[/tex]

Integrating;

ln[tex]c-\frac{I}{F}=\frac{-Ft}{v}+ k[/tex]

at t = 0; c = c0

ln[tex]c_0-\frac{I}{F}=k[/tex]

ln[tex]c-\frac{I}{F}=\frac{-Ft}{v}+ ln(c_0-\frac{I}{F}\\\\\frac{Ft}{v}=ln[\frac{c_0-\frac{I}{F}}{c-\frac{I}{F}}][/tex]

[tex]e^{\frac{Ft}{v}}=\frac{c_0-\frac{I}{F}}{c-\frac{I}{F}}\\\\c-\frac{I}{F}=(c_0-\frac{I}{F})e^{\frac{-Ft}{v}}\\\\c=(c_0-\frac{I}{F})e^{\frac{-Ft}{v}}+\frac{I}{F}[/tex]

B.) [tex]\lim_{t \to \infty} c(t)=(c_0 -\frac{I}{F})e^{-\infty}+\frac{I}{F}\\\\=0+\frac{I}{F}=\frac{I}{F}[/tex]

given [tex]I=0;c=c_0e^{\frac{-Ft}{v}}[/tex]

At [tex]c=\frac{c_0}{10};t=?[/tex]

[tex]\frac{c_0}{10}=c_0e^{\frac{-Ft}{v}}\\\\\frac{1}{10}=e^{\frac{-Ft}{v}}\\\\ln10=\frac{Ft}{v}\\\\t=\frac{vln10}{F}[/tex]

C.) [tex]t=\frac{458ln(10)}{175}=6.026[/tex]years≅6 years

D.) [tex]t=\frac{12221ln(10)}{65.2}=431.59[/tex]years≅432 years

xero099 xero099 · Answer 2 · 2022-03-16T04:42:00+01:00

(a) The concentration at any time [tex]t[/tex] is determined by [tex]C(t) = \left(C_{o}-\frac{I}{F} \right)\cdot e^{-\frac{F}{V}\cdot t } + \frac{I}{F}[/tex].

(b) The final concentration when [tex]t \to +\infty[/tex] is [tex]C_{\infty} =\frac{I}{V}[/tex].

(c) 6 years are needed to drop pollution to [tex]\frac{C_{o}}{10}[/tex] in the lake Eire.

(d) 432 years are needed to drop pollution to [tex]\frac{C_{o}}{10}[/tex] in the lake Superior.

How to determine a model for pollution effects in the Great Lakes

(a) The differential equation is equivalent to:

[tex]\frac{dC}{dt} +\frac{F}{V} \cdot C = \frac{I}{V}[/tex] (1)

Whose solution is [tex]C(t) = \left(C_{o}-\frac{I}{F} \right)\cdot e^{-\frac{F}{V}\cdot t }+\frac{I}{F}[/tex], where [tex]t[/tex] is the time, in minutes.

The concentration at any time [tex]t[/tex] is determined by [tex]C(t) = \left(C_{o}-\frac{I}{F} \right)\cdot e^{-\frac{F}{V}\cdot t } + \frac{I}{F}[/tex]. [tex]\blacksquare[/tex]

(b) [tex]e^{-\frac{F}{V}\cdot t } \to 0[/tex] inasmuch [tex]t \to + \infty[/tex], then we have that [tex]C_{\infty} = \frac{I}{V}[/tex].

The final concentration when [tex]t \to +\infty[/tex] is [tex]C_{\infty} =\frac{I}{V}[/tex]. [tex]\blacksquare[/tex]

(c) In this case, (1) is reduced into this form:

[tex]C(t) = C_{o}\cdot e^{-\frac{F}{V}\cdot t }[/tex] (2)

If we know that [tex]C(t) = \frac{C_{o}}{10}[/tex], then the time required is:

[tex]\frac{C_{o}}{10} = C_{o}\cdot e^{-\frac{F}{V}\cdot t }[/tex]

[tex]t = -\frac{V}{F}\cdot \ln \frac{1}{10}[/tex]

[tex]t \approx 6.026\,yr[/tex]

6 years are needed to drop pollution to [tex]\frac{C_{o}}{10}[/tex] in the lake Eire. [tex]\blacksquare[/tex]

(d) The final concentration in lake Superior is:

[tex]t = -\frac{V}{F}\cdot \ln \frac{1}{10}[/tex]

[tex]t\approx 431.593\,yr[/tex]

432 years are needed to drop pollution to [tex]\frac{C_{o}}{10}[/tex] in the lake Superior. [tex]\blacksquare[/tex]

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