# 1st order DE - word problem

Gold Member
A 120gal tank initially contains 90lbs of salt dissolved in 90gal of water. Saltwater containing 2lb/gal of salt flows into the tank at the rate of 4gal/min, and the well stirred mixture flows out of the tank at the rate of 3gal/min. How much salt does the tank contain when it is full?

Well here is my train of thought. Since they are asking for the amount of salt when the tank is full, it seems logical to model the differential equation with the amount of salt and time, since we know the rate of entering and rate of leaving, we can find how much time 30gals will take to accumulate.

So I will let Q be the quantity of salt. It seems that the rate of change of salt in the tank should be the difference between the rate of entering and rate of leaving, correct? Now If I'm wrong about this, then dont bother reading the rest, because I entire approach is based on this assumption.

I get the equation is this form:
$$\frac{dQ}{dt} = rate_{entering} - rate_{leaving}$$

It seems logical to have the rate of entering and leaving as lb/min since dQ/dt is also lb/min.

Rate of entering: 2lb/gal * 4gal/min = 8lb/min

Rate of leaving: this one is giving me some problems.

--thanks.

## Answers and Replies

AKG
Science Advisor
Homework Helper
To find the rate entering, you multiplied the concentration of salt in the entering water by the rate at which the entering water is entering. So the rate of leaving is calculated the same way. You know that the rate at which the leaving water is leaving, all you need to know is the concentration of the leaving water. Since the tank is well-stirred, the concentration of the leaving water at time t is precisely the concenctration of the tank at time t, which would be Q(t)/A(t) where A(t) is the amount of water at time t. It's easy to find an expression for A(t).

Gold Member
I think I figured it out.
Rate of leaving = 3Q/(90+t)

correct?