Extra Problems for 3.8

3.8.1: Modified Tank Model.  In class we discussed a tank model, as follows.  The tank has a capacity of 100 gallons, and initially is filled with water in which 5 pounds of salt have been dissolved.  Pure water flows into the tank at a rate of 10 gallons a minute, and the salt water mixture flows out of the tank at the same rate, so that the amount of liquid in the tank is constant.  Under these circumstances, we developed a model for S (t) = the amount of salt in the tank at time t, where S is in units of pounds and t in units of minutes.  According to the model, S satisfies a differential equation S ' (t) = -.1 S (t).  This leads to the equation S (t) = 5 e -.1t .
  1. Develop a modified version of this model based on the assumption that what flows into the tank is not pure water, but rather salt water at a fixed concentration of  .5 pounds per gallon.  Use your model to find a differential equation for S(t).
     
  2. Try to solve your differential equation.  That is, try to find a function S (t) whose derivative has the required form.  For a hint, click here.

  3. Based on the answer to b, will the amount of salt in the tank grow indefinitely, decrease to 0, or approach a constant equilibrium value over time?  Based on the physical situation we are modeling, is this prediction reasonable?
Solution: (a)  In 1 minute, 10 gallons of water are lost and replaced.  The lost water carries out 1/10th of the salt in the tank, and the gained water brings in 10 gallons at .5 pounds per gallon, for a total of 5 pound of salt.  So, if the amount of salt already in the tank is S, we have

Delta S = 5 - (1/10)S  for Delta t = 1 minute
and that means
(Delta S)/(Delta t) = 5 - (1/10)S.

Now let's consider how that changes if you go for 1/2 a minute.  Basically, the amount of salt coming in and going out are both cut in half.  So

Delta S = (1/2)(5 - (1/10)S)  for Delta t = 1/2 minute.
Again compute (Delta S) / (Delta t), finding
(Delta S)/(Delta t) = 5 - (1/10)S.

Similarly, for Delta t = 1/10 of a minute, we would find

Delta S = (1/10)(5 - (1/10)S)  for Delta t = 1/10 minute
and as before
(Delta S)/(Delta t) = 5 - (1/10)S.

Now as these examples show, for any time period Delta t , we will find the same result: (Delta S)/(Delta t) = 5 - (1/10)S.  So taking a limit as Delta t goes to 0 tells us that

S '(t) = 5 - .1 S(t)

This is a differential equation that completely determines what the function S (t) must be, assuming we know S (0) = 5.

(b) To solve this differential equation, rewrite it in the form

S '(t) / (5 - .1 S(t)) = 1.

On the left side, we almost have a fraction with the top equal to the derivative of the bottom.  In fact, if we multiply both sides of the equation by -.1 then we obtain

-.1S '(t) / (5 - .1 S(t)) = -.1

and now the left side is exactly in the form [5 - .1S]'/[5 - .1S], which we know is the derivative of   ln(5 - .1S).  That means we can rewrite the equation asform

[ ln (5 - .1 S(t))]' = -.1.

Now this tells us that ln(5-.1S) has a constant derivative, and so is a straight line with a slope of -.1.  That means it has to be a function of the form f(t) = -.1t + b.  So we now know

ln (5 - .1 S(t)) = -.1t + b.

We get rid of the ln on both sides by applying the function ex, and then we have

5 - .1 S(t) = e-.1t + b = A e-.1t,

where A = eb.  Now we can solve for S (t), finally obtaining

  S(t) = 50 - 10A e-.1t

We still have to find the constant A, but for that purpose we can use the fact that S = 5 when t = 0.  Substituting those values into the equation gives

  5 = 50 - 10A 

and that leads to A = 4.5.  In conclusion, this modeling exercise leads us a step at a time to a single possible equation for S:

  S(t) = 50 - 45 e-.1t

We can check that this does have the correct initial value of 5 pounds of salt.  And if you graph this function, you will see that it shows the amount of salt in the tank steadily increasing and approaching (without ever reaching) 50 pounds.  That makes sense because in the long run we expect the entire tank to be filled with water having the same concentration of salt as the inflow -- 1/2 a pound per gallon -- and in a 100 gallon tank, that would be a total of 50 pounds of salt.