Team Homework 2 Solutions
Assignment: pp 20-22, problems 8-14
Mark Twains Mississippi
The Lower Mississippi River meanders over its flat valley, forming broad loops called oxbows. In a flood, the river can jump its banks and cut off one of these loops, getting shorter in
the process. In his book Life on the Mississippi (1884), Mark Twain suggests, with tongue in
cheek, that some day the river might even vanish! Here is a passage that shows us some of the
pitfalls in using rates to predict the future and the past.
In the space of one hundred and seventy six years the Lower Mississippi has shortened itself
two hundred and forty-two miles. That is an average of a trifle over a mile and a third per
year. Therefore, any calm person, who is not blind or idiotic, can see that in the Old Oolitic
Silurian Period, just a million years ago next November, the Lower Mississippi was upwards
of one million three hundred thousand miles long, and stuck out over the Gulf of Mexico like
a fishing-pole. And by the same token any person can see that seven hundred and forty-two
years from now the Lower Mississippi will be only a mile and three-quarters long, and Cairo
[Illinois] and New Orleans will have joined their streets together and be plodding comfortably
along under a single mayor and a mutual board of aldermen. There is something fascinating
about science. One gets such wholesome returns of conjecture out of such a trifling investment
of fact.
Let L be the length of the Lower Mississippi River. Then L is a variable quantity we shall
analyze.
Problems and Solutions:
8. According to Twain’s data, what is the exact rate at which L is changing, in miles per
year? What approximation does he use for this rate? Is this a reasonable approximation?
Is this rate positive or negative? Explain. In what follows, use Twain’s approximation.
Solution: Mark Twain said that in 176 years, L shortened itself 240 miles. Thus we can
calculate the rate at which L is changing by using the data that he provides:
L0 = −
240 miles
miles
= −1.36
176 years
year
Mark Twain uses the approximation of “a trifle over a mile and a third per year”, so
roughly 1 13 per year. While this may be a reasonable approximation during the 176 year
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time period given, it doesn’t appear to be reasonable if we extend it through time. As
Twain calculates using L, a million years ago the river was upwards of 1,300,000 miles
long and in 742 years, it will be only 1.75 miles long. These are absurd statements, and
so it is not a very reasonable approximation.
We note that the rate is negative since the length of the river has shortened over time.
9. Twain wrote his book in 1884. Suppose the Mississippi that Twain wrote about had been
1100 miles long; how long would it have become in 1990?
Solution: Since L0 = −1.33
miles
year ,
we know that:
L1990 = L1884 + (L0 ∗ 4t).
(1)
We know that 4t = 1990 − 1884 = 106 years, L1884 = 1100 miles and L0 = −1.33. Thus,
L1990 = 1100 miles + (−1.33
miles
)(106 years) = 958.67 miles.
year
Hence the river was 955.8 miles long in 1990 according to Twain’s approximation.
10. Twain does not tell us how long the Lower Mississippi was in 1884 when he wrote the
book, but he does say that 742 years later it will be only 1 34 miles long. How long must
the river have been when he wrote the book?
Solution: We can use the same ideas from question 9 by modifying equation (1) and
solving for L1884 :
L1884 = L2626 − (L0 ∗ 4t)
(2)
In this problem, we know that 4t = 2626 − 1884 = 742 years, L2626 = 1 34 miles and
L0 = −1.33. Thus,
L1884 = 1.75 miles − (−1.33
miles
)(742 years) = 988.61 miles.
year
So the river must have been 988.61 miles long when Twain wrote the book in 1884.
11. Suppose t is the number of years since 1884. Write a formula that describes how much L
has changed in t years. Your formula should complete the equation:
the change in L in t years = . . .
Solution:
Note that t is now the number of years since 1884, so from here on 4t = t and L1884 = L0 .
Now we can begin to solve the problem by modifying equation (1). So the change in L in
t years looks like:
the change in L in t years = Lt − L0
= (L0 + (L0 ∗ t)) − L0
= − 1.33 ∗ t
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12. From your answer to question 10, you know how long the river was in 1884. From question
11, you know how much the length has changed t years after 1884. Now write a formula
that describes how long the river is t years later.
Solution: We know that the river was 988.61 in 1884, and we know that
the change in L in t years = −1.36 ∗ t. Now we want to write a formula that describes
how long the river is t years later. To do this, we can modify equation (2) and plug in
some values.
Ft = F0 + (L0 ∗ t)
= 988.61 − 1.33 ∗ t
(3)
(4)
13. Use your formula to find what L was a million years ago. Does your answer confirm
Twain’s assertion that the river was “upwards of 1,300,000 miles long” then?
Solution: We want to find L at t = −1, 000, 000, so we can use equation (4):
Ft = 988.61 − 1.33 ∗ t
Now we plug in:
F1,000,000 = 988.61 − 1.33 ∗ −1, 000, 000
= 988.61 + 1, 330, 000
= 1, 330, 988.61
So Twain’s assertion that the river was “upwards of 1,300,000 miles long” in in fact correct
since our equation calculated that the river was 1,330,988.61 miles long a million years ago.
14. Was the river ever 1,300,000 miles long; will it ever be 1 34 miles long? (This is called a
reality check.) What, if anything, is wrong with the “trifling investment of fact” which
led to such “wholesale returns of conjecture” that Twain has given us?
Solution: Certainly the river was never 1,300,000 miles long nor 1 43 miles long, so something went wrong with Twain’s “trifling investment of fact” that produced the above
equations. Typical of models, our equations work well during a set time period, likely a
short time period around the time Twain wrote his book in 1884, but fail as time extends
back into the past and into the future. Perhaps a “s-curve” would model this behavior
better then our linear model, since L0 would vary with time.
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