Question

A tank originally contains 120 gal of fresh water. Then water containing 1/3 lb of salt per gallon is poured into the tank at a rate of 3 gal/min, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 3 gal/min, with the mixture again leaving at the same rate. Find the amount of salt Q(10) in the tank at the end of an additional 10 min.

Answer 1

ok sorry it took so long...i wanted to work through it first to make sure i was doing it right

this problem involves solving differential equations:
first you have to set up 2 equations for each situation
set them up using what is given about the rate the salt enters or leaves tank
--> dQ/dt = rate In - rate Out
Salt is coming in at rate of 1/3 lb/gal at 3gal/min --> 1 lb/min
Salt is leaving at rate of Q/120 lb/gal at 3gal/min ---> Q/40 lb/min
where Q is the amount of salt at any time "t"
\[\frac{dQ}{dt} = 1 - \frac{Q}{40}\]
solving for Q will give the function for Salt for first 10 minutes...use initial value Q(0) =0
next equation for when fresh water is poured in
-Salt is coming in at rate of 0 since its fresh water
- Salt is leaving at same rate of Q/40 lb/min
\[\frac{dQ}{dt} = -\frac{Q}{40}\]
solving for Q here gives function for additional minutes after first 10 ...use initial value Q(0) = Q*(10) , where Q* is first function

ok hope that helps, i can work through solving the DE's if you need me too