Home

About

Advanced Search

Browse by Discipline

Scientific Societies

E-print Alerts

Add E-prints

E-print Network
FAQHELPSITE MAPCONTACT US


  Advanced Search  

 
Differential Equation Example: Salt in a Solution
 

Summary: Differential Equation Example:
Salt in a Solution
February 8, 2012
Problem. Suppose that a full 100-gallon tank contains 40 pounds of salt in a solution at
time t = 0, that a salt solution with a concentration of 1.5 pounds per gallon is being added
at a rate of 4 gallons per minute, and that thoroughly mixed salt solution is owing out of the
tank at a rate of 4 gallons per minute. Find S(t), the amount of chemical in the tank at time t.
Solution. Reviewing the information in the problem statement, we see that we are given
an initial condition; that is, we know that there is 40 pounds of salt initially in the tank.
We may write this information as an equation:
S(t = 0) = 40.
This initial condition will be useful for determining the unknown constant that we will nd
in our equation for S(t). Now, in order to calculate the amount of salt in the tank at later
times, we must take into account the ow of salt in and out of the tank. This ow will
determine the rate of change for the amount of salt in the tank, which is the derivative of
S(t) with respect to the time t. So,
dS
dt
= (Flow of salt in) - (Flow of salt out).
We know that a salt solution with a concentration of 1.5 pounds per gallon is being added

  

Source: Anderson, Greg W. - School of Mathematics, University of Minnesota

 

Collections: Mathematics