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Cell[TextData[{
StyleBox["Lab 08 - Arcs, Parametrizations, and Arclength",
FontSize->18,
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FontVariations->{"Underline"->True}],
"\nMath 2374 - University of Minnesota\nhttp://www.math.umn.edu/math2374\n\
Questions to: rogness@math.umn.edu"
}], "Text",
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Cell[TextData[StyleBox["Introduction",
FontSize->14,
FontWeight->"Bold"]], "Section"],
Cell["\<\
In Lab 4A we worked with parametric equations and curves, which are also \
called \"arcs.\" In this lab we're going to examine these objects a little \
deeper. We'll talk about arc length, as well as line integrals of scalar \
functions. For most of these purposes we're going to use the unit circle, \
because you're already familiar with it, and yet it's complicated enough to \
illustrate many different concepts.\
\>", "Text"],
Cell[TextData[{
StyleBox["FYI",
FontWeight->"Bold"],
": In many textbooks, \"curve\" and \"arc\" are defined to be the same \
thing. In practice most of us will generally refer to these things as \
curves. However when we deal with the length of a curve, we suddenly change \
terms and talk about ",
StyleBox["arc",
FontSlant->"Italic"],
" length as opposed to ",
StyleBox["curve",
FontSlant->"Italic"],
" length. Apparently the term \"arc length\" is so deeply ingrained in \
mathematicians' brains that we'll never be able to switch! Please don't get \
confused by this change in terminology.\n\nOne other reminder: a \"path\" is \
actually a function which is a parametrization for some curve. Sometimes in \
an abuse of language we'll say \"path\" when we really mean \"curve\" and \
vice versa, but you should remember that there is a (subtle) difference! \
(Ask your TA if you don't understand what the difference is.)\n\nOf course, \
we basically just have to throw our hands up in the air about the terminology \
when we get to line integrals, which are really integrals along a curve, not \
a straight line. Why on earth would they be called line integrals instead of \
curve integrals? Let's just agree not to go there..."
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Cell[TextData[{
"The rest of this lab assumes that you remember what the derivative of a \
parametrization ",
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" is. Remember, if ",
Cell[BoxData[
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" represents the position of the particle, then the tangent vector \
represents the velocity of the particle. You can review this at the end of \
Lab 4 if you'd like."
}], "Text"]
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Cell[CellGroupData[{
Cell[TextData[StyleBox["Estimating Arc Length",
FontSize->14]], "Section"],
Cell[TextData[{
"In the first part of this lab, we will explore how to estimate the length \
of a curve by approximating the curve with line segments. It turns out that \
the accuracy of these estimates can be tied in with the tangent vector, which \
is why we asked you to review that if necessary.\n\nTo begin, we're going to \
use larger and larger numbers of line segments to estimate the length of the \
upper half of the unit circle. We're using this particular curve because -- \
as you should definitely know by now -- the arc length of the upper half of \
the unit circle is \[Pi].\n\n\[Pi] is a very famous number which shows up \
everywhere. You probably know that it is irrational, i.e. it cannot be \
represented as a fraction. In fact, the decimal expansion of \[Pi] goes on \
forever and ever, with no apparent pattern that we've been able to discern. \
To 40 decimal places,\n\n\t\t\[Pi] = \
3.141592653589793238462643383279502884197.\n\nHistorically, people have used \
many different approximations for \[Pi]. The ancient Egyptians and \
Babylonians both realized that it is slightly bigger than 3. The Babylonians \
used an approximation of 3 ",
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"= 3.125, a bit low. The Egyptians' estimate of ",
Cell[BoxData[
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"\[TildeTilde] 3.16049 is too high. In the fifth century, a Chinese \
mathematician named Tsu Chung Chi estimated \[Pi] as ",
Cell[BoxData[
FormBox[
FractionBox["355", "113"], TraditionalForm]]],
"\[TildeTilde] 3.14159292, which agrees with the true value of \[Pi] for six \
decimal places!\n\nIn high school you may have used approximations such as \
3.14, or 3.14159, etc. Your calculator probably has the correct value of \
\[Pi] stored up to a few dozen digits or so, which is usually more than \
enough. In fact, knowing \[Pi] to just 39 decimal places is sufficient for \
calculating the circumference of the universe accurate to the radius of a \
hydrogen atom, but in recent years computer scientists have tried to \
calculate as many digits of \[Pi] as possible. By the late 1990s we knew the \
value of \[Pi] to more than 206 ",
StyleBox["billion",
FontSlant->"Italic"],
" digits! In 2002, some of the same researchers involved in that record \
computed an absolutely ridiculous 1.24 ",
StyleBox["trillion",
FontSlant->"Italic"],
" digits of Pi. (See http://pw1.netcom.com/~hjsmith/Pi.html. If you'd like \
to know more about \[Pi], you also could read a book called ",
StyleBox["A History of \[Pi]",
FontSlant->"Italic"],
", written by Petr Beckman.)\n\nFor over one thousand years the most common \
method of estimating \[Pi] was due to Archimedes, who used it to calculate \
that 3 ",
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"\[Precedes] \[Pi] \[Precedes] 3 ",
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". Essentially, part of his method was to do exactly what we're about to \
do: estimate the length of a circle using line segments. Let's get started!\n\
\nTo plot the upper half of the unit circle along with 4 approximating \
segments, evaluate this command:"
}], "Text"],
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Cell["\<\
There is another command which will add up the lengths of these segments and \
give us an estimate of the arclength.\
\>", "Text"],
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Cell["\<\
You can tell that we're in the right ballpark, but still not very close! In \
fact we need to use many more segments before we get anything close to an \
accurate estimate. For his lower bound on \[Pi], Archimedes would have used \
about 48 line segments:\
\>", "Text"],
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Cell[TextData[{
StyleBox["Exercise 1",
FontSize->14,
FontWeight->"Bold"],
"\n\nFind the minimum number n of segments required such that the estimate \
of \[Pi] is accurate to five decimal places, i.e. such that the estimate is \
3.14159. You should hand in what you think the number n is, as well as the \
estimates with n segments and (n-1) segments. (In other words, show that \
you've actually found the smallest such n.) You do not need to hand in \
graphs of the line segments; once you use more than about 10 segments, the \
graph of the circle is nearly indistinguishable from the line segments \
anyway!"
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Cell["\<\
At the end of Lab 4 we looked at two different parametrizations for the unit \
circle:\
\>", "Text"],
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Cell[TextData[{
"You should go back to that lab and look at the discussion about the \
differences in the derivatives of ",
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" and ",
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". Let's look at what happens when we try to use these parametrizations to \
find the circumference (or arc length) of the unit circle. Initially we'll \
use 12 line segments for picture.\n\nWith f(t), we'll get a picture similar \
to the one above when we used 4 segments to estimate the length of the upper \
half circle:"
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"(Remember, since we're using the whole circle, the length is really 2\[Pi] \
\[TildeTilde] 6.2831853.)\n\nWe get a nice symmetric picture, as we would \
expect. But things are different if we use g(t). Remember that t only goes \
to ",
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" when we use g!"
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"As you can see, the picture is all out of whack, and the estimate is much \
worse than what we obtained using f(t). The reason for this has to do with \
the derivatives of ",
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" and ",
StyleBox["g",
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". Remember, if ",
StyleBox["f",
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" and ",
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FontSlant->"Italic"],
" represent the position of a particle at time ",
StyleBox["t",
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", then the derivatives represent the velocity of the particle.",
"\n\nThese pictures were obtained by dividing the time interval into 12 \
equal pieces; each blue segment represents the movement of the particle \
during one of those subintervals. The first picture is symmetric because a \
particle moving according to f(t) has constant speed. The second picture is \
uglier because a particle moving according to g(t) picks up speed as time \
passes, so it moves further during each successive subinterval.\n\nThat last \
paragraph is a little dense, but you should re-read it until you understand \
it because it holds the key to understanding how the derivative relates to \
the accuracy of given estimations. Once you understand what's happening for \
these pictures, you are ready for problem 2.\n"
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StyleBox["Exercise 2",
FontSize->14,
FontWeight->"Bold"],
"\n\nIn this problem we'll work with the same two parametrizations for the \
unit circle,\n\nf(t) = (Cos(t), Sin(t)),\t0\[LessEqual]t\[LessEqual]2\[Pi]\n\
g(t) = (",
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"),\t0\[LessEqual]t\[LessEqual]",
Cell[BoxData[
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".\n\n(a) Calculate the length of the tangent vector using the first \
parametrization. Then do the same for the second tangent vector. Simplify \
your answer!\n\n(b) Look at the pictures above where the circumference is \
estimated using 12 segments with each parametrization. Although the estimate \
using g(t) is clearly the worse of the two, if you look in the first quadrant \
it's another story. If you were to add up the lengths of the line segments \
in the first quadrant of each picture, thereby estimating ",
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", it appears that you'd get a better estimate from the ",
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" picture, which was made using g(t)!\n\nConfirm this and explain why it is \
so! \n\nHint: to confirm it you can use ",
StyleBox["Segments",
FontWeight->"Bold"],
" and ",
StyleBox["Estimate",
FontWeight->"Bold"],
" to analyze the quarter circle, but you need the correct number of segments \
and the correct domain for for each parametrization. To explain why it is so \
will take some thought, and we expect you to present a clear explanation of \
what's going on -- something beyond \"the Estimate command shows that it is \
more accurate.\"\n\nYour writeup to this problem does not need to be split \
into parts (a) and (b); rather, it should be one seamless essay which uses \
the information in (a) and (b) to describe why g(t) gives you a better \
approximation for \[Pi]/2."
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Cell["\<\
In this last part of the lab we'll examine a curve whose length can be tricky \
to estimate. The following is a parametrization of the curve:\
\>", "Text"],
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Cell[TextData[{
"(For our purposes we'll assume 0\[LessEqual]t\[LessEqual]\[Pi].)\n\nFirst \
use the command ",
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FontWeight->"Bold"],
" with 100 segments to sketch the curve and the approximating segments. \
Then use ",
StyleBox["Estimate",
FontWeight->"Bold"],
" with 100 segments to calculate the approximate length. Do you think this \
is accurate? (You can use the following commands.)"
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"You should also try ",
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FontWeight->"Bold"],
" with 250, 500, and 1000 line segments. You should be building a large \
amount of evidence that the arc length of this curve is \[Pi]. Again, do you \
think this is accurate?\n\nNow that you're comfortable with this estimate, \
try using ",
StyleBox["Estimate",
FontWeight->"Bold"],
" with 950 segments. If you've done everything correctly, you should be \
thinking, \"Huh?\" when you see the answer. Now which estimate do you think \
is the most accurate?\n\nTo find out, let's examine a tiny little piece of \
the curve. Instead of letting t range from 0 to \[Pi], let's look at the \
section where 0\[LessEqual]t\[LessEqual]",
Cell[BoxData[
FormBox[
FractionBox["\[Pi]", "50"], TraditionalForm]]],
". 1000 segments over the whole curve corresponds to 1000/50=20 segments on \
this tiny piece of the curve. (The ",
StyleBox["AspectRatio",
FontWeight->"Bold"],
" option ensures that ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" streches out the x-axis enough to show us the whole graph; you could \
remove it to see what happens otherwise.)"
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Cell["\<\
Remember, the curve is red, while the approximating line segments are blue. \
Now do you understand what's going on? Look at this tiny piece with \
950/50=19 segments instead of 20 segments. Which estimate is more accurate?\
\>", "Text"],
Cell[TextData[{
StyleBox["Exercise 3",
FontSize->14,
FontWeight->"Bold"],
"\n\nExperiment with the number of line segments on the small piece of the \
curve until you think you have an accurate estimation of the arc length. \
Then multiply by 50 to find the corresponding number of line segments on the \
entire curve, and use this to find an estimation of the total arc length.\n\n\
You should hand in your estimate, the number of segments used, and a picture \
of this estimate at the \"tiny\" level (where 0\[LessEqual]t\[LessEqual]",
Cell[BoxData[
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") so we can see how good the fit is.\n\n",
StyleBox["Exercise 4",
FontSize->14,
FontWeight->"Bold"],
"\n\n(a) Suppose you have a parametrization f(t) (with some bounds on t) \
for a given curve.\nNow suppose you find the following estimates of the \
curve's length:\n\n100 segments: length \[TildeTilde] \[Pi].\n1000 \
segments: length \[TildeTilde] \[Pi].\n850 segments: length \[TildeTilde] \
400.\n\nCan you tell which number of segments gives the most accurate \
estimate? Why?\n\n(b) Suppose now your estimates look like this:\n\n100 \
segments: length \[TildeTilde] \[Pi].\n1000 segments: length \[TildeTilde] \
\[Pi].\n850 segments: length \[TildeTilde] 1.\n\nNow can you tell which \
number of segments gives the most accurate estimate? Why? (Has your answer \
changed?)"
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"In some textbooks, such as ",
StyleBox["Marsden and Tromba",
FontSlant->"Italic"],
", a line integral of a scalar function is referred to as a \"path \
integral.\""
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Cell[TextData[{
"So far we've only been estimating arc length, but you know from lecture \
(and your textbook) that we don't have to be satisfied with estimates. We \
can use an integral to find the exact length of a curve. Suppose our curve \
",
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"This is really just a special case of a \"Line Integral of a Scalar \
Function,\" which you've also learned about in lecture. Suppose we want to \
integrate a real-valued function ",
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"So the only difference is that we stick ",
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" along our curve. If we use the special function ",
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for the \"d",
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"\" -- if you integrate 1, you get ",
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", which is a common notation for the arclength of ",
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". ",
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"\nThere are a few different physical interpretations of line integrals of \
scalar functions. Perhaps the most common ones involves mass. Suppose our \
curve represents a wire in three-dimensional space. The wire is made up of \
different metals and might have a different density at each point. If we had \
a ",
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" wire, with a ",
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curvy, and the density can vary, we need a line integral: "
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integrals was added in March 2004; the rest of it was written in February \
2002.\n \nThe old exercise 2 went something like this: start with two \
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students were asked to extrapolate that the arc length estimate of the second \
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Lafayette's web site. To avoid licensing issues I may go back and write new \
versions of these commands, but the documentation with the commands says they \
can be freely used. You can find this in the math2374.nb file.\n\nNote that \
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2002, 2004, 2008 by Jonathan Rogness (rogness@math.umn.edu). We've both \
agreed to use the same license, so this lab is protected by the Creative \
Commons Attribution-NonCommercial-ShareAlike License. You can find more \
information on this license at \
http://creativecommons.org/licenses/by-nc-sa/1.0/. \n\nAlthough it's not \
specifically required by the license, I'd appreciate it if you let me know if \
you use parts of our labs, just so I can keep track of it. Please send me \
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