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:
- Exit flow of water following the accident from Lake Alpha = 500 liters/h
- Entry flow of water post-accident into Lake Beta = 500 liters/h
- Initial Kool-Aid quantity in Lake Alpha = 500 kg
- Initial water volume in Lake Alpha = 400,000 L
- Initial water volume in Lake Beta = 100,000 L
Find:
a) Let x denote the mass of Kool-Aid, in kilograms, present in Lake Alpha t hours post-accident. Derive an equation for the changing rate of Kool-Aid, dx/dt, based on the Kool-Aid present in the lake x:
b) Derive an equation for the amount of Kool-Aid in kilograms, in Lake Alpha t hours after the accident
c) Let y represent the quantity of Kool-Aid, in kilograms, located in Lake Beta t hours post-accident. Derive an equation for the changing rate of Kool-Aid, dy/dt, in terms of quantities x and y.
d) Derive an equation for the quantity of Kool-Aid in Lake Beta t hours following the accident.
Solution:
- First, we focus on Lake Alpha. The inflow post-accident into Lake Alpha is nonexistent. The outflow can be expressed as:
dx / dt = concentration*flow
dx / dt = - ( x / 400,000)*( 500 L / hr )
dx / dt = - x / 800
- We'll now solve the resulting differential equation:
Separate variables:
dx / x = -dt / 800
Integrate:
Ln | x | = - t / 800 + C
- We recognize that at t = 0, the truck has crashed, so x(0) = 500.
Ln | 500 | = - 0 / 800 + C
C = Ln | 500 |
- The solution to the differential equation yields:
Ln | x | = -t/800 + Ln | 500 |
x = 500*e^(-0.00125*t)
- Next, for Lake Beta. We consider the inflow rate on par with the outflow rate from Lake Alpha. We can formulate 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 substituting x with the solution from Lake Alpha ODE:
dy/dt = 500*e^(-0.00125*t)/ 800 - y / 200
dy/dt = 0.625*e^(-0.00125*t)- y / 200
- Rearranging gives us:
y' + P(t)*y = Q(t)
y' + 0.005*y = 0.625*e^(-0.00125*t)
- Next, we determine the integrating factor:
u(t) = e^(P(t)) = e^(0.005*t)
- Finally, we apply the form:
( u(t). y(t) )' = u(t). Q(t)
- Inserting the values:
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)
- Given the initial conditions: t = 0, y = 0:
0 = 0.625 + C
C = - 0.625
Therefore,
y(t) = 0.625*( e^(-0.00125*t) - e^(-0.005*t) )
y(t) = 0.625*e^(-0.00125*t)*( 1 - e^(-4*t) )
