Originally Posted by
pakman 
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)
or...
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 -_-
