Originally Posted by

**Aryth** I just want to verify my work... Hopefully I was right:

A tank initially contains 100 gallons of brine in which 50 lbs. of salt are dissolved. A brine containing 2 lbs/gal of salt runs into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring and flows out of the tank at 4 gal/min.

Initial Information:

IN: 2 lbs/gal AT: 5 gal/min

OUT: 4 gal/min

$\displaystyle V(t) = Volume$

$\displaystyle y(t) = Concentration$

$\displaystyle \frac{dy}{dt} = \text{RATE IN} - \text{RATE OUT}$

a) At what rate (lbs/min) does the salt enter the container at time t?

$\displaystyle \text{RATE IN} = 2\frac{lbs}{gal}*5\frac{gal}{min} = 10\frac{lbs}{min}$

b) What is the volume of brine in the tank at time t?

$\displaystyle V(t) = 100 + t$

c) At what rate (lbs/min) does salt leave the tank at time t?

$\displaystyle \text{RATE OUT} = 4*\left(\frac{y(t)}{100 + t}\right)$

d) Write and solve initial value problem.

$\displaystyle \frac{dy}{dt} = 10 - \left(\frac{4}{100 + t}\right)y$

$\displaystyle \frac{dy}{dt} + \left(\frac{4}{100 + t}\right)y = 10$

$\displaystyle IF = (100 + t)^4$

$\displaystyle [(100 + t)^4y]' = 10(100 + t)^4$

$\displaystyle \int [(100 + t)^4y]' = \int 10(100 + t)^4 dt$

$\displaystyle (100 + t)^4y = \frac{(100 + t)^5}{2} + C$ *

Mr F says: This line should be $\displaystyle (100 + t)^4y = {\color{blue}2} (100 + t)^5 + C$. But that's the least of your worries ......

$\displaystyle y = \frac{100 + t}{2} + C$ Mr F says: No. It's $\displaystyle y = \frac{100 + t}{2} + \frac{C}{(100 + t)^4}$. So I'm afraid that everything that follows is no good.

$\displaystyle y(0) = 50 + C = 0$

$\displaystyle C = -50$

$\displaystyle y = \frac{100 + t}{2} - 50$

$\displaystyle y = \frac{t}{2}$

e) Find the concentration of salt in the tank at 25 minutes after process starts.

$\displaystyle y(25) = \frac{25}{2}$

$\displaystyle y = 12.5 lbs.$