Respuesta :
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) )
