1. ## Diff. Eq./Related rates

Hey guys, can you help me with this problem?

A tank initially contains 120 liters of pure water. A mixture containing a concentration of x g/liter of salt enters the tank at a rate of 2 liters/min, and the well-stirred mixture is allowed to leave the tank at the same rate. Find an expression in terms of x for the amount of salt in the tank at any time t. Also find the limiting amount of salt in the tank as t-->infinity

Thanks guys

2. Originally Posted by Jhevon
Hey guys, can you help me with this problem?

A tank initially contains 120 liters of pure water. A mixture containing a concentration of x g/liter of salt enters the tank at a rate of 2 liters/min, and the well-stirred mixture is allowed to leave the tank at the same rate. Find an expression in terms of x for the amount of salt in the tank at any time t. Also find the limiting amount of salt in the tank as t-->infinity

Thanks guys
Let Q(t) represent the amount of salt after t minutes.

Then,
dQ/dt = Rate in - Rate out

The rate is is (x g/liter)*(2 liters/min) = 2x g/min.

The rate out is (Concentration)*(2 liters/min)
Now the concentration = Amount/Volume = Q/120
Thus, rate out is (Q/120 g/liter)(2 liters/min)=Q/60 g/min.

Thus, the differencial equation is,
dQ/dt = 2x - Q/60
Thus,
dQ/dt + Q/60 = 2x
With initial conditions Q(0)=0.

This is a first order linear differencial equation, it solves easily. I am sure you can do it from there.

3. Yes, I can take it from there, thanks.