Answer:
(a) [tex]A = 1500(1 - e^{-0.01t})[/tex]
(b) [tex]t = 0[/tex]
Step-by-step explanation:
Given
[tex]V= 300gal[/tex] --- Volume of tank
[tex]B = 5lb/gal[/tex] --- Brine solution
[tex]R_1 = 3gal/min[/tex] --- Rate in, in gallon/min
[tex]R_2 = 6gal/min[/tex] --- Rate out, in gallon/min
Required
Determine A(t)
First, calculate the rate in (R in) and (R out) in lb/min
[tex]R_{in} = B * R_1[/tex]
[tex]R_{in} = 5lb/gal * 3gal/min[/tex]
[tex]R_{in} = 15lb/min[/tex]
R(out) is calculated by multiply the rate at which brine leaves the tank (lb/gal) * R2
So, we have:
[tex]R_{out} = \frac{A(t)}{300} lb/gal * 3gal/min[/tex]
[tex]R_{out} = \frac{3*A(t)}{300} lb/min[/tex]
[tex]R_{out} = \frac{A(t)}{100} lb/min[/tex]
The change in the amount of brin e in the tank at time t is given as:
[tex]A'(t) = R_{in} - R_{out}[/tex]
[tex]A'(t) = 15 - \frac{A(t)}{100}[/tex]
Rewrite:
[tex]\frac{dA}{dt} = 15 - \frac{A}{100}[/tex]
Multiply through by dt
[tex]dA = [15 - \frac{A}{100}]* dt[/tex]
Make dt stand alone
[tex]\frac{dA}{15 - \frac{A}{100}} = dt[/tex]
[tex]ln|15 - \frac{A}{100}| = -0.01t + ln\ c[/tex]
Express as exponents
[tex]15 - \frac{A}{100} = ce^{-\frac{t}{100}}[/tex]
Multiply through by 100
[tex]1500 - A = 100ce^{-\frac{t}{100}}[/tex]
[tex]A = 1500 - 100ce^{-\frac{t}{100}}[/tex]
[tex]A = 1500 - 100ce^{-0.01t}[/tex]
Apply initial conditions
[tex]A(0) = 0[/tex]
Substitute 0 for t in: [tex]A = 1500 - 100ce^{-0.01t}[/tex]
[tex]0 = 1500 - 100ce^{-0.01*0}[/tex]
[tex]0 = 1500 - 100ce^{0}[/tex]
[tex]0 = 1500 - 100c[/tex]
[tex]100c = 1500[/tex]
[tex]c = 15[/tex]
Substitute 15 for c in [tex]A = 1500 - 100ce^{-0.01t}[/tex]
[tex]A = 1500 - 100*15e^{-0.01t}[/tex]
[tex]A = 1500 - 1500e^{-0.01t}[/tex]
Factorize:
[tex]A = 1500(1 - e^{-0.01t})[/tex]
How long to empty the tank
Set A to 0
[tex]1500(1 - e^{-0.01t}) = 0[/tex]
Divide through by 1500
[tex]1 - e^{-0.01t} = 0[/tex]
Collect Like Terms
[tex]e^{-0.01t} = 1[/tex]
Take ln of both sides
[tex]ln\ (e^{-0.01t}) = ln\ 1[/tex]
[tex]-0.0t = 0[/tex]
[tex]t = 0[/tex]
