Suppose the clean water of a stream flows into Lake Alpha, then into Lake Beta,

Question

Suppose the clean water of a stream flows into Lake Alpha, then into Lake Beta, and then further downstream. The in and out flow for each lake is 400 liters per hour. Lake Alpha contains 200 thousand liters of water, and Lake Beta contains 100 thousand liters of water. A truck with 100 kilograms of Kool-Aid drink mix crashes into Lake Alpha. Assume that the water is being continually mixed perfectly by the stream. Let x be the amount of Kool-Aid, in kilograms, in Lake Alpha t hours after the crash. Find a formula for the rate of change in the amount of Kool-Aid, dx/dt, in terms of the amount of Kool-Aid in the lake.

dxdt=

Find a formula for the amount of Kool-Aid, in kilograms, in Lake Alpha t hours after the crash.

x(t)=

Let y be the amount of Kool-Aid, in kilograms, in Lake Beta t hours after the crash. Find a formula for the rate of change in the amount of Kool-Aid, dy/dt, in terms of the amounts x,y. dydt= kg/hour Find a formula for the amount of Kool-Aid in Lake Beta t hours after the crash.

y(t)=

Explanation / Answer

Let x be amount of Koolaid in Lake Alpha, and let V be the volume. Water flowing out of Lake Alpha has (x/V) kg/liter The flow rates into and out of Lake Alpha are r = 100 liters/hour Lake Alpha looses Koolaid at (x/V)r kg/hour, so in Deltat hours Deltax = -(x/V)r Deltat is the change in the amount of Koolaid in Lake Alpha. Dividing by Deltat, using (x/V)r = (r/V)x and taking the limit Deltax to 0, dx/dt =-(r/V)x and x(t) = x(0)exp(-(r/V)t) For the given data, x(0)=400 and r/V=100/200000=1/2000 so x(t)=400 exp(-t/2000) Let y be amount of Koolaid in Lake Beta, and let W be the volume. Water flowing into Lake Beta has (x/V) kg/liter Water flowing out of Lake Beta has (y/W) kg/liter In Deltat hours, Lake Beta gains (x/V)r Deltat kg due to inflow, and looses (y/W)r Deltat kg due to outflow, so Deltay = (x/V)r Deltat - (y/W)r Deltat is the change in the amount of Koolaid in Lake Beta. Taking the limit Deltat to 0, dy/dt=(x/V)r - (y/W)r This is first order linear. standard form, dy/dt+(r/W)y = (r/V)x where we know x=x(t) = (r/V)x(0)exp(-(r/V)t) from above Integrating factor is exp((r/W)t) d/dt( y(t) exp((r/W)t)) = (r/V)x(0)exp( ((r/W)-(r/V)) t) Integrating with y(0) = 0 (no Koolaid in Lake Beta at time 0) y(t) exp((r/W)t) = (r/V)x(0) (1/((r/W)-(r/V))) (exp((r/W)-(r/V)t -1) Multiplying by exp((-r/W)t), y(t) = x(0)(r/V)(1/((r/W)-(r/V))(exp(-(r/V)t) - exp(-(r/W)t)) = x(0)(1/((V/W)-1)(exp(-(r/V)t) - exp(-(r/W)t)) where for the particular data, x(0) = 400, V/W=2/5, r/V=100/200000= 1/2000, r/W=100/500000=1/5000, y(t)=(3/2000)(exp(-t/5000)-exp(-t/2000))

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