The problem states that a tank originally contains 100 gal of fresh water. Then water containg 1/2 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poure dinto the tank at a rate of 2 gal/min, with the mixture agian leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 min.
So I drew 2 tanks, showed the rate in and rate out... etc. I stated that Q(t) = amount of salt in tank at time t.
dQ/dt = rate in - rate out
= (1/2)*2 - (Q(t)/100)*2
= 1 - (1/50)Q(t)
It is seperable so...
dQ/dt + (1/50)*Q(t) = 1
u(t) = e^[INT(1/50)dt) = e^(t/50)
After multiplying both sides by u(t) and integrating...
e^(t/50)*Q = 50*e^(t/50) + C
Solving for Q...
Q = 50 + C/e^(t/50)
Solving for C...
C = (Q - 50)*e^(t/50)
C = (Q(10) - 50)*e^(1/5) ... right?
This is where I am lost. I'm not sure where to go from here.. I have a feeling I messed up on that last part. Any thoughts? And apologies... I haven't figured out LaTeX yet -_-
This is a two staged problem. You want to use the first stage as the initial data for the second. You seem to be on the right track for most of the way though. Here it is
Originally Posted by pakman
MHF tells me my file is too large, so i will post it in two pieces, stage1 and stage2, at the end of stage2 is the answer