Question

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.

Answer 1

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Answer:

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

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

(c) T = 6.02619 years

(d) T = 431.593 years

Explanation:

(a)

`(dC)/(dt) = -(F)/(v)C + (I)/(v) \\ \\ \\ (dC)/(dt) + (F)/(v)C = (I)/(v)`

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

`= e \int\limits (F)/(v)t \\ \\ \\ = e (Ft)/(v)`

`C* e (Ft)/(v)= \int\limits (I)/(v)*e (Ft)/(v) dt`

`C* e (Ft)/(v)= (I)/(v)* (e (Ft)/(v) )/(F/v)+ K`

`C* e (Ft)/(v)= (I)/(v)* (V)/(F) e (Ft)/(v) + K`

`Ce (Ft)/(v) = (I)/(F) \ * \ e (Ft)/(v) + k \ \ \ \ (at \ t = 0 \ ; C = C_o)`

`C_o = (I)/(F) e^o + K`

`K = C_o - (I)/(F)`

`C_((t)) = [ (I)/(F)e (Ft)/(v)+ C_o - (I)/(F)] e(-Ft)/(v)`

`C_((t)) =(I)/(F) [ e (Ft)/(v) * e (-Ft)/(v)+ C_oe (-Ft)/(v) - 1* e (-Ft)/(v)]`

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

(b)

`\lim_(t \to \infty) C_t = (I)/(F)[1-e^(- \infty) + C_o e^(- \infty) ]`

since
`(e^(- \infty) = 0)`

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

(c)

V = 458 km³ and F = 175 km³ , I = 0

`(dC)/(dt) = - (-175)/(458)C`

`= \int\limits \ (dC)/(C) = - (175)/(458)\int\limits dt`

`In (C_((t))) = - (175)/(458) t + K`

`C_((t)) = e (-175)/(458)t + K`

Let at time t = 0
`C_((t))} = C_o \to C_o = e^(0+k) = e^K`

`C_((t)) = e (-175)/(458)t`

Now at time t = T ;
`C_(9t)) = (C_o)/(10)`

`(C_o)/(10) = C_o e (-175)/(458)T \to (1)/(10) = e (-175)/(458)T`

`In ( (1)/(10)) = (-175)/(459)T`

`- In (10) = (175)/(458)T`

`T = (458)/(175) In (10)`

T = 6.02619 years

(d) V = 12221 km³

F = 65.2 km³/ year

`\mathbf{T = (v)/(F)In (10)}`

`T = (12221)/(65.2)In (10)`

T = 431.593 years