Calculus: The Language of Change

Exercise Set 3.1
Linear Approximation of Ox-bows

by Sharon Portz

Exercise   1

1. Linear Ox-bows
How wholesome are Twain's returns?Express the length of the  river (in miles) as a function of time in years using the implicit  mathematical  assumption of Twain's statement.What   is this assumption? We are not given the length of the Lower Mississippi at the time of Twain's statement,but  we can find it from the information.What   is it?What  is the moral  of this whole exercise in terms of long-range  prediction?

Solution

Mark Twain's returns are not mathematically   wholesome.  His mathematical satire makes the point that  sometimes our answers are  not always reasonabe.  It is not possible for the Mississippi River to decrease so much that  it is only 1.75 miles long from Cairo to New Orleans since we know it cannot be less than the straight line distance between the two cities.  This minimum or lower "limit" is approximately 490 miles.  Twain's assumption, that the  river will steadily decrease at a linear rate for all time periods, is not realistic.

To express the length of the river (in miles) as a function of time (in years) using the implicit mathematical   assumption of Twain's statement  we need to determine how much the river shrinks in one  year.  Twain said it was "a trifle  over a mile and a third" a year, but what is the  exact number?  We need to find out how many miles the river decreased in a year ((mi/yr)) .  We use the values Twain gave us (it shrunk 242 miles in 176 years) to get:
	
				((242 miles)/(176 years))= 1.375 (mi/yr)

Using the above calculation we can write the function like this:

				f[t+Δt] = f[t] - 1.375Δt
				
This says that  to get the  length, f[t + tΔ], of the  river in some amount  of time, Δt, we take  the  current  length  of the  river (f[t]) and subtract  the length it will shrink in that  amount  of time (1.375Δt).  

We  can rewrite  the function by dividing by Δt to get:

				\(TraditionalForm
      \`\(f[t + Δt]\  - \ f[t]\)\/Δt\) = -1.375

To find how long the river was at the time of Twain's statement , f[t], we solve for f[t] in the above equation:
				f[t] = f[t + Δt] + 1.375Δt

Let f[t + Δt] be equal to 1.75 miles using Twain's statement.  Δt equals the change in time we are  calculating  which is 742 years.  At the time of Twain's statements  t=0.  Plug these values in, cancel the units, and calculate :

				f[t] = 1.75 mi+ (1.375(mi/yr))(742 yr)
				f[t] = 1.75 mi + 1020.5 mi
				f[t] = 1022 mi
				
So the Lower Mississippi was 1022 miles long at the time of Twain's  comments.

The predictions that Twain is making are linear because he is assuming the river changes at a constant rate over the entire period of time.  This type of long range  prediction has limitations.  It is not always accurate  and can sometimes, as in this case, lead to invalid answers.  So beware.  In general, linearity assumptions are more likely to be valid the shorter the time period involved.