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Notebook[{
Cell[TextData[{
StyleBox["Lab 4A - Arcs, Parametrizations, and Arclength",
FontSize->18,
FontWeight->"Bold",
FontVariations->{"Underline"->True}],
"\nMath 2374 - University of Minnesota\nhttp://www.math.umn.edu/math2374\n\
Questions to: rogness@math.umn.edu"
}], "Text",
CellFrame->True,
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Cell[CellGroupData[{
Cell[TextData[StyleBox["Introduction",
FontSize->16,
FontWeight->"Bold"]], "Section"],
Cell["\<\
In Lab 2A 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 take a look at what the derivative of a parametrization \
is, and we'll also talk about arc length. 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 your textbook, \"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.)"
}], "Text",
CellFrame->True,
Background->GrayLevel[0.849989]],
Cell[TextData[{
"We're going to use some commands in this lab which are not normally part \
of ",
StyleBox["Mathematica",
FontSlant->"Italic"],
". Two of the commands are part of a package from Lafayette College, and \
the other commands were written here at the University of Minnesota. In \
order to use them, you must download the package \"Arctools.m\" from the main \
course webpage, http://www.math.umn.edu/math2374. Once you've saved this \
file in your home directory, you can load the commands by evaluating the \
following command:"
}], "Text"],
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"You ",
StyleBox["must",
FontSlant->"Italic"],
" load the package before you run any of the commands below! We're going \
to use ",
StyleBox["ParametricPlot",
FontWeight->"Bold"],
" in this lab, so let's also turn off those blue error messages about \
compiling functions:"
}], "Text"],
Cell[BoxData[
\(Off[ParametricPlot::ppcom]\)], "Input"],
Cell["Now we can begin the lab.", "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell["The Derivative of a Parametrization", "Section"],
Cell["\<\
Suppose we have a parametrization f(t) = (x(t), y(t)), where t \
ranges from 0 to 10. If we say that t represents time in seconds, then we \
can say f(t) represents the position of an object at time t. For example, \
consider the following function:\
\>", "Text"],
Cell[BoxData[
\(\(f[t_] = {Cos[t], Sin[t]};\)\)], "Input"],
Cell[TextData[{
"You should know by now that this is a parametrization for the unit circle. \
As you learned in Lab 2A, you can plot this function using ",
StyleBox["ParametricPlot",
FontWeight->"Bold"],
":"
}], "Text"],
Cell[BoxData[
\(ParametricPlot[f[t], {t, 0, 2 Pi},
AspectRatio \[Rule] Automatic]\)], "Input"],
Cell["\<\
This doesn't give you a feel for what I'm telling you, however: \
that f(t) can be viewed as the location function for a particle moving around \
the unit circle, where the particle finishes its trip when t=2\[Pi] seconds \
(or minutes, or hours, or whatever our unit is).
Try executing the following command. It will produce an animation of the \
particle moving around, which is actually just a number of different pictures \
in rapid succession. Your TA will demonstrate how to view the \
animation.\
\>", "Text"],
Cell[BoxData[
\(PathAnimate[f[t], {t, 0, 2 Pi}, 40]\)], "Input"],
Cell[TextData[{
"Now you can see the particle moving around the curve. It looks as if the \
speed of the particle is constant; let's see if we can prove that somehow.\n\n\
Recall from section 1.10 in your textbook that if f(t) represents the \
location of a particle at time t, then v(t)=",
Cell[BoxData[
\(TraditionalForm\`f\^\[Prime]\)]],
"(t), the derivative of f, represents the ",
StyleBox["velocity",
FontSlant->"Italic"],
" of the particle at time t. For example, in our case we have:"
}], "Text"],
Cell[BoxData[
\(v[t_] = D[f[t], t]\)], "Input"],
Cell[TextData[{
"Velocity is a little different than speed. Speed is just a number \
representing how fast a particle is moving. Velocity is a vector whose ",
StyleBox["length",
FontSlant->"Italic"],
" represents speed, and whose ",
StyleBox["direction",
FontSlant->"Italic"],
" represents the direction the particle is moving at that specific instance \
of time.\n\nTo see an animation of the particle moving around the circle \
along with its tangent vectors, run this command:"
}], "Text"],
Cell[BoxData[
\(PathTangentAnimate[f[t], {t, 0, 2 Pi}, 40]\)], "Input"],
Cell[TextData[{
"It certainly looks as if the length of the velocity vector is constant, \
which would confirm that the speed of the particle is constant. In fact, we \
can show this algebraically. You should find the length of the tangent \
vector v(t) on paper and show that it is constant -- it does not depend on t.\
\n\nLet's look at a different parametrization of the unit circle,\n\ng(t) = \
(",
Cell[BoxData[
\(TraditionalForm\`\(\(Cos(t\^2)\)\(,\)\(\ \)\(Sin(t\^2)\)\(\ \)\)\)]],
"),\t0\[LessEqual]t\[LessEqual]",
Cell[BoxData[
\(TraditionalForm\`\@\(2 \[Pi]\)\)]],
".\n\nThis is a parametrization of the unit circle because as t ranges from \
0 to ",
Cell[BoxData[
\(TraditionalForm\`\@\(2 \[Pi]\)\)]],
", ",
Cell[BoxData[
\(TraditionalForm\`t\^2\)]],
" ranges from 0 to 2\[Pi]:"
}], "Text"],
Cell[BoxData[{
\(\(g[t_] = {Cos[t^2], Sin[t^2]};\)\), "\[IndentingNewLine]",
\(ParametricPlot[g[t], {t, 0, Sqrt[2 Pi]},
AspectRatio \[Rule] Automatic]\)}], "Input"],
Cell["\<\
Now let's watch the particle whose motion is represented by g[t]:\
\
\>", "Text"],
Cell[BoxData[
\(PathTangentAnimate[g[t], {t, 0, Sqrt[2 Pi]}, 40]\)], "Input"],
Cell["\<\
As you can see, the particle starts out very slowly and picks up \
speed as time goes on. You can see this either by watching the particle \
itself, or by watching the length of the tangent vector grow. (Remember, the \
length of the tangent vector represents the speed of the particle!)
You will be asked to explain what's going on in this parametrization down \
below in Exercise 2.\
\>", "Text"],
Cell[TextData[{
"If you'd like to play around with these two animation commands, their \
syntax is:\n\n",
StyleBox["PathAnimate[f[t],{t,tmin,tmax},n]\n\
PathTangentAnimate[f[t],{t,tmin,tmax},n]",
FontFamily->"Courier",
FontWeight->"Bold"],
StyleBox["\n",
FontWeight->"Bold"],
"\nwhere n represents the number of frames to draw. (So the bigger n is, \
the longer it takes to create the animation, but the smoother it looks. 40 \
seems to be a reasonable number for most parametrizations. Here are a couple \
of parametrizations to try; you can copy these and paste them into commands:\n\
\n",
StyleBox["f[t_]={Cos[t],Sin[t]^3}, {t, 0, 2Pi}\t\t(at least 40 frames)\n\
f[t_]={Cos[2t],Sin[3t]}, {t, 0, 2Pi}\t\t(at least 50 frames)\n\
f[t_]={Cos[2t],Sin[4t]}, {t, 0, 2Pi}\t\t(at least 50 frames)\n\
f[t_]={Cos[5t],Sin[3t]}, {t, 0, 2Pi}\t\t(at least 100 frames)\nf[t_]={t,t^2}, \
{t, -1,1}\t\t\t\t\t(at least 30 frames)\nf[t_]={t^3,t^2}, {t, -1,1}\t\t\t\t\t\
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Cell[CellGroupData[{
Cell["Estimating Arc Length", "Section"],
Cell[TextData[{
"During the rest of this lab, we will explore how to estimate the length of \
a curve by approximating the curve with line segments. If this seems like a \
total subject change, it will all make sense in a bit; the accuracy of these \
estimates can be tied in with the tangent vector, which is why we examined \
that above.\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\[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 ",
Cell[BoxData[
\(TraditionalForm\`\(\(1\/8\)\(\ \)\)\)]],
"= 3.125, a bit low. The Egyptians' estimate of ",
Cell[BoxData[
\(TraditionalForm\`256\/81\)]],
"\[TildeTilde] 3.16049 is too high. In the fifth century, a Chinese \
mathematician named Tsu Chung Chi estimate \[Pi] as ",
Cell[BoxData[
\(TraditionalForm\`355\/113\)]],
"\[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 100 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 mid 1990s we knew the value of \[Pi] to \
more than 6 ",
StyleBox["billion",
FontSlant->"Italic"],
" digits! (If you'd like to know more about \[Pi], you 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 ",
Cell[BoxData[
\(TraditionalForm\`10\/71\)]],
"\[Precedes] \[Pi] \[Precedes] 3 ",
Cell[BoxData[
\(TraditionalForm\`1\/7\)]],
". 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"],
Cell[BoxData[
\(Segments[{Cos[t], Sin[t]}, {t, 0, Pi}, 4,
AspectRatio \[Rule] Automatic]\)], "Input"],
Cell["\<\
There is another command which will add up the lengths of these \
segments and give us an estimate of the arclength.\
\>", "Text"],
Cell[BoxData[
\(Estimate[{Cos[t], Sin[t]}, {t, 0, Pi}, 4]\)], "Input"],
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"],
Cell[BoxData[{
\(Segments[{Cos[t], Sin[t]}, {t, 0, Pi}, 48,
AspectRatio \[Rule] Automatic]\), "\[IndentingNewLine]",
\(Estimate[{Cos[t], Sin[t]}, {t, 0, Pi}, 48]\)}], "Input"],
Cell[TextData[{
StyleBox["Exercise 1",
FontSize->16,
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!"
}], "Text",
CellFrame->True,
Background->RGBColor[0.996109, 0.500008, 0.500008]]
}, Closed]],
Cell[CellGroupData[{
Cell["Arc Length and the Derivative", "Section"],
Cell["\<\
Earlier in this lab we used two different parametrizations for the \
unit circle. You can redefine them here, just in case you've used the \
letters f or g for other functions since that part of the lab:\
\>", "Text"],
Cell[BoxData[{
\(\(f[t_] = {Cos[t], Sin[t]};\)\), "\[IndentingNewLine]",
\(\(g[t_] = {Cos[t^2], Sin[t^2]};\)\)}], "Input"],
Cell["\<\
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.
With 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:\
\>", "Text"],
Cell[BoxData[{
\(Segments[f[t], {t, 0, 2 Pi}, 12,
AspectRatio \[Rule] Automatic]\), "\[IndentingNewLine]",
\(Estimate[f[t], {t, 0, 2 Pi}, 12]\)}], "Input"],
Cell[TextData[{
"(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 ",
Cell[BoxData[
\(TraditionalForm\`\@\(2 \[Pi]\)\)]],
" when we use g!"
}], "Text"],
Cell[BoxData[{
\(Segments[g[t], {t, 0, Sqrt[2 Pi]}, 12,
AspectRatio \[Rule] Automatic]\), "\[IndentingNewLine]",
\(Estimate[g[t], {t, 0, Sqrt[2 Pi]}, 12]\)}], "Input"],
Cell[TextData[{
"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 ",
StyleBox["first",
FontSlant->"Italic"],
" part of this lab, where we looked at particles whose position was \
described by f(t) and g(t). \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"
}], "Text"],
Cell[TextData[{
StyleBox["Exercise 2",
FontSize->16,
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) = (",
Cell[BoxData[
\(TraditionalForm\`\(\(Cos(t\^2)\)\(,\)\(\ \)\(Sin(t\^2)\)\(\ \)\)\)]],
"),\t0\[LessEqual]t\[LessEqual]",
Cell[BoxData[
\(TraditionalForm\`\@\(2 \[Pi]\)\)]],
".\n\n(a) Calculate the length of the tangent vector using the first \
parametrization. (You did this earlier.)\n\n(b) Do the same for the second \
parametrization. (Simplify your answer!)\n\n(c) 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 ",
Cell[BoxData[
\(TraditionalForm\`\[Pi]\/2\)]],
", it appears that you'd get a better estimate from the ",
StyleBox["second",
FontSlant->"Italic"],
" 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.\""
}], "Text",
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Cell[CellGroupData[{
Cell["Dangerous Curve Ahead", "Section"],
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"],
Cell[BoxData[
\(h[t_] = {Cos[t], \ Sin[t] + 0.01\ Sin[1000 t]}\)], "Input"],
Cell[TextData[{
"(For our purposes we'll assume 0\[LessEqual]t\[LessEqual]\[Pi].)\n\nFirst \
use the command ",
StyleBox["Segments",
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.)"
}], "Text"],
Cell[BoxData[{
\(\(Segments[h[t], {t, 0, Pi}, 100];\)\), "\[IndentingNewLine]",
\(Estimate[h[t], {t, 0, Pi}, 100]\)}], "Input"],
Cell[TextData[{
"You should also try ",
StyleBox["Estimate",
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[
\(TraditionalForm\`\[Pi]\/50\)]],
". 1000 segments over the whole curve corresponds to 1000/50=20 segments \
on this tiny piece of the curve."
}], "Text"],
Cell[BoxData[
\(Segments[h[t], {t, 0, Pi/50}, 1000/50]\)], "Input"],
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->16,
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[
\(TraditionalForm\`\[Pi]\/50\)]],
") so we can see how good the fit is.\n\n",
StyleBox["Exercise 4",
FontSize->16,
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?)"
}], "Text",
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Cell[CellGroupData[{
Cell["Credits", "Subsection"],
Cell[TextData[{
"This lab is a substantial rewrite of the old Arc Length lab, which was \
actually written in ",
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RowBox[{"L",
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". Exercises 1 and 3 are the same, but the text surrounding them has been \
completely changed. The introduction and the section with the animations are \
new as well. In other words, this lab was basically written from scratch, \
except for two of the exercises.\n \nThe old exercise 2 went something like \
this: start with two completely different arcs, and estimates of their \
lengths. If you evaluate the derivatives at the point t=\[Pi]/2 for each \
parametrization, you find that the second arc's tangent vector is longer than \
the other. From this the students were asked to extrapolate that the arc \
length estimate of the second curve is less accurate. At best this wasn't \
the whole story; at worst it's an incorrect claim. In any case, I hope the \
current exercise 2 does a better job.\n\n\"Segments\" and \"Estimate\" are \
part of a package used in the old lab. Evidently this package was from \
Lafayette, although I haven't found any references to it at Lafayette's web \
site. I wrote the animation commands in January 2002, and combined them into \
a new Arctools packages with the other two commands. To avoid licensing \
issues I may go back and write new versions of these commands.\n\nAs always, \
please send me any questions or comments!\n\nCopyright 2002,2003 by Jonathan \
Rogness (apart from the Lafayette Package)\nLicensed under GNU GPL to the \
University of Minnesota School of Mathematics.\nrogness@math.umn.edu\n"
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*)
(*******************************************************************
End of Mathematica Notebook file.
*******************************************************************)