Exploring Pollution in the Great Lakes-Solution
Problem:
Most of the water flowing into Lake Erie comes from Lake Huron, and most of the water flowing
into Lake Ontario is from Lake Erie. Each year, 11% of the water in Lake Huron flows into Lake
Erie, while 36% of the water in Lake Erie flows into Lake Ontario, and 12% of the water in Lake
Ontario flows out to the sea.
For generations, factories on the lakes had been dumping a pollutant into the water. Presently,
there are 4000 units of pollutant in Lake Huron, 2000 units in Lake Erie, and 3000 units in
Lake Ontario. For the most part this form of pollution has stopped. Only two such factories
remain. One, on Lake Huron, is dumping 25 units of pollutant into the water each year; the
other on Lake Ontario is dumping 20 units of the pollutant into the water each year.
How long will it be before the amount of pollutant in the three lakes is reduced to 10% of its
present level? What is the long-term level of pollutant in the lakes?
Source: Intermath: Four Sample Problems, COMAP, Inc., Lexington, MA, 1992.
Solution:
Let H n , En , and On represent the level of pollutants in year n for Lakes Huron, Erie and
Ontario, respectively. We can set up a set of recursive equations as follows:
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H n  H n 1  0.11H n 1  25
En  En 1  0.36 En 1  0.11H n 1
On  On 1  0.12On 1  0.36 En 1  20
With the original pollutant levels: H 0  4000 , E0  2000 , and O0  3000 .
Combining like terms in the recursive equations, we have:
H n  0.89 H n 1  25
En  0.64 En 1  0.11H n 1
On  0.88On 1  0.36 En 1  20
Using the Table feature, we see the value of the iterates below, where un , and vn represent H n ,
and En , respectively. Unfortunately we are not able to see all three lists at once. You can arrow
over to see the values for wn .
The graph below shows the level of pollutants in each lake:
Window is Xmin = 0, Xmax = 30, Xscl = 5, Ymin = 0, Ymax = 4500, Yscl = 500
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Where the level of pollutants in Lake Huron is represented by the solid trace type, Lake Erie is
the thick solid trace type and Lake Ontario is the dotted trace type.
To answer the question “How long will it be before the amount of pollutant in the three lakes is
reduced to 10% of its present level?”, we will find what 10% of the original pollutant levels are
for each lake. These values are: 400, 200 and 300 units for Lakes Huron, Erie and Ontario,
respectively. Now we can trace on the graph above to find when these levels are reached.
For Lake Huron, we see that the levels of pollutant is approximately 410 units after 26 years and
390 units after 27 years, so the level of pollutants is 400 units between 26 and 27 years. For
Lake Erie, we see that the level of pollutants is approximately 219 units after 21 years and 197
units after 22 years, so the level of pollutants is 200 units between 21 and 22 years. For Lake
Ontario, we see that the level of pollutants is approximately 961 units when n is 30, so we need a
bigger X window. Doubling the X window to 60 and using the trace key, we see that the level of
pollutants is 403 after 60 years. Experimenting with bigger windows, we should see that the
level of pollutants in Lake Ontario seems to level out to 375 units; therefore, the level of
pollutants in Lake Ontario will never reach 300 units.
In fact all of the pollutant levels seem to level off to an equilibrium value. For Lake Huron, we
notice that the equilibrium value is 227.27 units and for Lake Erie, the equilibrium is
69.4 units. We have just answered the second question posed: “What is the long-term level of
pollutant in the lakes?”
We can find these values analytically by thinking about what it means for the level of pollutants
to stabilize. The pollutant level in any of the lakes will stabilize when the amount of pollutants
flowing out of the lake is equal to the amount of pollutants flowing into the lake. Let’s consider
our original recursive equation below for Lake Huron:
H n  H n1  0.11H n1  25
The amount of pollutants flowing out of Lake Huron is equal to the amount of pollutants flowing
into Lake Huron when 0.11H n 1  25 .
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Solving for H n 1 , we have H n 1 
25
or H n1  227.27 .
0.11
Another way to think about finding the equilibrium value is to find the value for which the
iterates do not change. That is, finding the level of pollutants for which H n  H n 1 .
In our recursive equation, we can set H n  H n 1 , and solve for H n as follows:
H n  H n 1  0.11H n 1  25
H n  0.89 H n 1  25
H n  0.89 H n  25
0.11H n  25
Hn 
25
0.11
Now we can use this equilibrium value to find the equilibrium values for Lakes Erie and Ontario
using analytic methods.
We will set En  En 1 and H n  H n 1 .
En  0.64 En  0.11H n
0.36 En  0.11H n
En 
0.11
Hn
0.36
In the long run, we know H n  227.27 . Substituting this value in for H n , we have En  69.4 .
Similarly we can find the equilibrium value for the pollutant level in Lake Ontario as shown
below:
On  0.88On  0.36 En  20
0.12On  0.36 En  20
On 
0.36 En  20
0.12
Substituting the equilibrium value for the pollutant level in Lake Erie, we get On  375 .
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This problem offers an interesting look at these dependent recursive equations and allows us to
solve for the equilibrium values analytically. The students have an opportunity to consider
numerical solutions and graphical solutions using the calculator and then confirm those
solutions using analytical methods.
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