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Allisa [31]
3 years ago
14

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 500 liters per hour. Lake Alpha contains 400 thousand liters of water, and Lake Beta contains 100 thousand liters of water. A truck with 500 kilograms of Kool-Aid drink mix crashes into Lake Alpha. Assume that the water is being continually mixed perfectly by the stream.
A. 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 x: dx/dt = ?
B. find a formula for the amount of Kook-Aid in kilograms, in Lake Alpha t hours after the crash
x(t) = ?
500 - 500/400000 * 500t is incorrect
C. 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.
D.Find a formula for the amount of Kool-Aid in Lake Beta t hours after the crash.
y(t)=
Mathematics
1 answer:
Gala2k [10]3 years ago
6 0

Answer:

a) dx / dt = - x / 800

b) x = 500*e^(-0.00125*t)

c) dy/dt = x / 800 - y / 200

d) y(t) = 0.625*e^(-0.00125*t)*( 1  - e^(-4*t) )

Step-by-step explanation:

Given:

- Out-flow water after crash from Lake Alpha = 500 liters/h

- Inflow water after crash into lake beta = 500 liters/h

- Initial amount of Kool-Aid in lake Alpha is = 500 kg

- Initial amount of water in Lake Alpha is = 400,000 L

- Initial amount of water in Lake Beta is = 100,000 L

Find:

a) 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 x:

b) find a formula for the amount of Kook-Aid in kilograms, in Lake Alpha t hours after the crash

c) 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.

d) Find a formula for the amount of Kool-Aid in Lake Beta t hours after the crash.

Solution:

- We will investigate Lake Alpha first. The rate of flow in after crash in lake alpha is zero. The flow out can be determined:

                              dx / dt = concentration*flow

                              dx / dt = - ( x / 400,000)*( 500 L / hr )

                              dx / dt = - x / 800

- Now we will solve the differential Eq formed:

Separate variables:

                              dx / x = -dt / 800

Integrate:

                             Ln | x | = - t / 800 + C

- We know that at t = 0, truck crashed hence, x(0) = 500.

                             Ln | 500 | = - 0 / 800 + C

                                  C = Ln | 500 |

- The solution to the differential equation is:

                             Ln | x | = -t/800 + Ln | 500 |

                                x = 500*e^(-0.00125*t)

- Now for Lake Beta. We will consider the rate of flow in which is equivalent to rate of flow out of Lake Alpha. We can set up the ODE as:

                  conc. Flow in = x / 800

                  conc. Flow out = (y / 100,000)*( 500 L / hr ) = y / 200

                  dy/dt = con.Flow_in - conc.Flow_out

                  dy/dt = x / 800 - y / 200

- Now replace x with the solution of ODE for Lake Alpha:

                  dy/dt = 500*e^(-0.00125*t)/ 800 - y / 200

                  dy/dt = 0.625*e^(-0.00125*t)- y / 200

- Express the form:

                               y' + P(t)*y = Q(t)

                      y' + 0.005*y = 0.625*e^(-0.00125*t)

- Find the integrating factor:

                     u(t) = e^(P(t)) = e^(0.005*t)

- Use the form:

                    ( u(t) . y(t) )' = u(t) . Q(t)

- Plug in the terms:

                     e^(0.005*t) * y(t) = 0.625*e^(0.00375*t) + C

                               y(t) = 0.625*e^(-0.00125*t) + C*e^(-0.005*t)

- Initial conditions are: t = 0, y = 0:

                              0 = 0.625 + C

                              C = - 0.625

Hence,

                              y(t) = 0.625*( e^(-0.00125*t)  - e^(-0.005*t) )

                             y(t) = 0.625*e^(-0.00125*t)*( 1  - e^(-4*t) )

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