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Cell[CellGroupData[{
Cell["Calculus: The Language of Change", "Title"],
Cell[TextData[{
"\n",
StyleBox["Exercise Set 3.1\n",
FontVariations->{"Underline"->True}],
"Linear Approximation of Ox-bows"
}], "Subtitle"],
Cell["by Sharon Portz", "Subsubtitle"],
Cell[CellGroupData[{
Cell["Exercise 1", "Subsection"],
Cell["\<\
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?\
\>", "Text",
TextJustification->0],
Cell[CellGroupData[{
Cell["Solution ", "Subsubsection"],
Cell["\<\
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.\
\>", "Text"],
Cell[TextData[{
"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 (",
Cell[BoxData[
\(TraditionalForm\`mi\/yr\)]],
") . We use the values Twain gave us (it shrunk 242 miles in 176 years) to \
get:\n\t\n\t\t\t\t",
Cell[BoxData[
\(TraditionalForm\`\(242\ miles\)\/\(176\ years\)\)]],
"= 1.375 ",
Cell[BoxData[
\(TraditionalForm\`mi\/yr\)]]
}], "Text",
TextAlignment->Left],
Cell[TextData[{
"\nUsing the above calculation we can write the function like this:\n\n\t\t\
\t\tf[t+\[CapitalDelta]t] = f[t] - 1.375\[CapitalDelta]t\n\t\t\t\t\nThis says \
that to get the length, f[t + t\[CapitalDelta]], of the river in some \
amount of time, \[CapitalDelta]t, we take the current length of the \
river (f[t]) and subtract the length it will shrink in that amount of time \
(1.375\[CapitalDelta]t). \n\nWe can rewrite the function by dividing by \
\[CapitalDelta]t to get:\n\n\t\t\t\t",
Cell[BoxData[
\(TraditionalForm
\`\(f[t + \[CapitalDelta]t]\ - \ f[t]\)\/\[CapitalDelta]t\)]],
" = -1.375\n\t\t\t\t"
}], "Text",
TextAlignment->Left],
Cell[TextData[{
"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:\n\t\t\t\tf[t] = f[t + \[CapitalDelta]t] \
+ 1.375\[CapitalDelta]t\n\nLet f[t + \[CapitalDelta]t] be equal to 1.75 miles \
using Twain's statement. \[CapitalDelta]t 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 :\n\n\t\t\t\tf[t] = \
1.75 mi+ (1.375",
Cell[BoxData[
\(TraditionalForm\`mi\/yr\)]],
")(742 yr)\n\t\t\t\tf[t] = 1.75 mi + 1020.5 mi\n\t\t\t\tf[t] = 1022 mi\n\t\t\
\t\t\nSo the Lower Mississippi was 1022 miles long at the time of Twain's \
comments."
}], "Text"],
Cell["\<\
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.\
\>", "Text"]
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