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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,
TextAlignment->Center,
FontColor->GrayLevel[1],
Background->RGBColor[0, 0, 1]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Introduction",
FontSize->14,
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 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 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.)\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..."
}], "Text",
CellFrame->True,
Background->GrayLevel[0.849989]],
Cell[TextData[{
"The rest of this lab assumes that you remember what the derivative of a \
parametrization ",
Cell[BoxData[
\(TraditionalForm\`f(t)\)]],
" is. Remember, if ",
Cell[BoxData[
\(TraditionalForm\`f(t)\)]],
" 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 2A if you'd like."
}], "Text"]
}, Open ]],
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\[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 estimated \[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 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 ",
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",
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Background->RGBColor[0.996109, 0.500008, 0.500008]]
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Cell[CellGroupData[{
Cell[TextData[StyleBox["Arc Length and the Derivative",
FontSize->14]], "Section"],
Cell["\<\
At the end of Lab 2A we looked at two different parametrizations \
for the unit circle:\
\>", "Text"],
Cell[BoxData[{
\(\(f[t_] = {Cos[t], Sin[t]};\)\), "\[IndentingNewLine]",
\(\(g[t_] = {Cos[t^2], Sin[t^2]};\)\)}], "Input"],
Cell[TextData[{
"You should go back to that lab and look at the discussion about the \
differences in the derivatives of ",
StyleBox["f",
FontSlant->"Italic"],
" and ",
StyleBox["g",
FontSlant->"Italic"],
". 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:"
}], "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 derivatives of ",
StyleBox["f",
FontSlant->"Italic"],
" and ",
StyleBox["g",
FontSlant->"Italic"],
". Remember, if ",
StyleBox["f",
FontSlant->"Italic"],
" and ",
StyleBox["g",
FontSlant->"Italic"],
" represent the position of a particle at time ",
StyleBox["t",
FontSlant->"Italic"],
", 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"
}], "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. 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 ",
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.\"\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."
}], "Text",
CellFrame->True,
Background->RGBColor[0.996109, 0.500008, 0.500008]]
}, Open ]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Dangerous Curve Ahead",
FontSize->14]], "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",
CellFrame->True,
Background->RGBColor[0.996109, 0.500008, 0.500008]]
}, Open ]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Line Integrals of Scalar Functions",
FontSize->14]], "Section"],
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 \
",
StyleBox["C",
FontSlant->"Italic"],
" is parametrized by ",
StyleBox["f(t)",
FontSlant->"Italic"],
", where ",
Cell[BoxData[
\(TraditionalForm\`a \[LessEqual] t \[LessEqual] b\)]],
". Then the length of ",
StyleBox["C",
FontSlant->"Italic"],
" is:"
}], "Text"],
Cell[BoxData[
\(Length = \[Integral]\_a\^b\(\(\[DoubleVerticalBar]\)\(f' \((t)\)\)\(\
\[DoubleVerticalBar]\)\(dt\)\)\)], "DisplayFormula",
TextAlignment->Center,
FontSize->14],
Cell[TextData[{
"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 ",
Cell[BoxData[
\(TraditionalForm\`g(x, y, z)\)]],
"on the same curve ",
StyleBox["C",
FontSlant->"Italic"],
". Then the line integral of ",
StyleBox["g",
FontSlant->"Italic"],
" on ",
StyleBox["C",
FontSlant->"Italic"],
" is:"
}], "Text"],
Cell[BoxData[
\(\[Integral]\_C
g\ dL = \[Integral]\_a\^b g \((f \((t)\))\) \[DoubleVerticalBar]
f' \((t)\) \[DoubleVerticalBar] dt\)], "DisplayFormula",
TextAlignment->Center,
FontSize->14],
Cell[TextData[{
"So the only difference is that we stick ",
Cell[BoxData[
\(TraditionalForm\`g(f(t))\)]],
" into the integral; essentially, we're trying to add up the values of ",
StyleBox["g",
FontSlant->"Italic"],
" along our curve. ",
"If we use the special function ",
Cell[BoxData[
\(TraditionalForm\`g(x, y, z) = 1\)]],
", then ",
Cell[BoxData[
\(TraditionalForm\`g(f(t)) = 1\)]],
", and then we get the integral for arc length. That's actually the reason \
for the \"d",
StyleBox["L",
FontWeight->"Bold"],
"\" -- if you integrate 1, you get the ",
StyleBox["L",
FontWeight->"Bold"],
"ength of ",
StyleBox["C",
FontSlant->"Italic"],
".\n\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 ",
StyleBox["straight",
FontSlant->"Italic"],
" wire, with a ",
StyleBox["constant",
FontSlant->"Italic"],
" density, then the mass is simply length times density. When the wire is \
curvy, and the density can vary, we need a line integral: "
}], "Text"],
Cell[BoxData[
\(Mass = \(\[Integral]\_C
g\ dL = \[Integral]\_a\^b g \((f \((t)\))\) \[DoubleVerticalBar]
f' \((t)\) \[DoubleVerticalBar] dt\)\)], "DisplayFormula",
TextAlignment->Center,
FontSize->14],
Cell[CellGroupData[{
Cell["Example", "Subsubsection"],
Cell[TextData[{
StyleBox["Consider the following curve, a helix. It could also represent, \
say, a wire which has been made into a slinky. Notice that we're using the \
bounds 0\[LessEqual]",
FontWeight->"Plain"],
StyleBox["t",
FontWeight->"Plain",
FontSlant->"Italic"],
StyleBox["\[LessEqual]10:",
FontWeight->"Plain"]
}], "Text",
FontWeight->"Bold"],
Cell[BoxData[{
\(f[t_] = {Cos[2*Pi*t], Sin[2*Pi*t], 1 + t/5}\), "\[IndentingNewLine]",
\(ParametricPlot3D[f[t], {t, 0, 10},
ViewPoint \[Rule] {0, \(-10\), 1}, \[IndentingNewLine]\t
PlotPoints \[Rule] 120]\)}], "Input"],
Cell["\<\
Suppose we know that the density of the wire (in grams per \
centimeter) at any given point is equal to its distance from the xy-plane, \
i.e.\
\>", "Text"],
Cell[BoxData[
\(g[x_, y_, z_] = z\)], "Input"],
Cell["Then the total mass of the wire is:", "Text"],
Cell[BoxData[
\(Mass = \(\[Integral]\_C
g\ dL = \[Integral]\_0\^10 g \((f \((t)\))\) \[DoubleVerticalBar]
f' \((t)\) \[DoubleVerticalBar] dt\)\)], "DisplayFormula",
TextAlignment->Center,
FontSize->14],
Cell[TextData[{
"We need to learn how to evaluate this integral with the computer. First \
of all, we can have ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" find the length of the derivative of ",
StyleBox["f",
FontSlant->"Italic"],
". The derivative will be a vector, and the ",
StyleBox["Norm",
FontWeight->"Bold"],
" function will find its length:"
}], "Text"],
Cell[BoxData[{
\(D[f[t], t]\), "\[IndentingNewLine]",
\(Norm[%] // Simplify\)}], "Input"],
Cell[TextData[{
"Next we have to plug ",
Cell[BoxData[
\(TraditionalForm\`f(t)\)]],
" into ",
Cell[BoxData[
\(TraditionalForm\`g(x, y, z)\)]],
":"
}], "Text"],
Cell[BoxData[
\(Apply[g, f[t]]\)], "Input"],
Cell[TextData[{
"Huh? Why did we use ",
StyleBox["Apply[g,f[t]]",
FontFamily->"Courier",
FontWeight->"Bold"],
" when we want g",
StyleBox["[f[t]]",
FontFamily->"Courier",
FontWeight->"Bold"],
"? We had to do this to avoid an annoying problem. Remember, we have\n\t\t\
\t",
Cell[BoxData[
\(f[t] = {t\ Cos[2\ \[Pi]\ t], t\ Sin[2\ \[Pi]\ t], 2\ t}\)], "Input"],
"\n\nFor our integral, we want to evaluate ",
Cell[BoxData[
\(g[t\ Cos[2\ \[Pi]\ t], t\ Sin[2\ \[Pi]\ t], 2\ t]\)], "Input"],
",\nbut ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" reads ",
StyleBox["g[f[t]]",
FontFamily->"Courier",
FontWeight->"Bold"],
" as ",
Cell[BoxData[
\(g[{t\ Cos[2\ \[Pi]\ t], t\ Sin[2\ \[Pi]\ t], 2\ t}]\)], "Input"],
"\n\nSee the difference? The ",
StyleBox["Apply",
FontFamily->"Courier",
FontWeight->"Bold"],
" command makes this all work correctly."
}], "Text",
CellFrame->True,
Background->GrayLevel[0.849989]],
Cell["\<\
Now we can put them all together to find the mass of the \
wire:\
\>", "Text"],
Cell[BoxData[{
\(Integrate[
Apply[g, f[t]]*Norm[D[f[t], t]], {t, 0, 10}]\), "\[IndentingNewLine]",
\(N[%]\)}], "Input"],
Cell[TextData[{
StyleBox["Exercise 5",
FontSize->16,
FontWeight->"Bold"],
"\n\nConsider the curve ",
StyleBox["C",
FontSlant->"Italic"],
" parametrized by ",
Cell[BoxData[
\(TraditionalForm\`f(t) = \((t, t\^3, t\^2)\)\)]],
", where ",
Cell[BoxData[
\(TraditionalForm\`0 \[LessEqual] t \[LessEqual] 1\)]],
". The engineers building a new airplane realize that they'll have to have \
a wire in the shape of ",
StyleBox["C",
FontSlant->"Italic"],
" running between two systems; because of the different stresses placed on \
different parts of the hull, the wire has to be denser in some areas. If the \
density of the wire is given by ",
Cell[BoxData[
\(TraditionalForm\`g(x, y, z) = 2 x + 4 y\)]],
" grams/centimeter, find the mass of the wire.\n\nWhile we'd like to see \
the exact answer, a decimal answer might be more useful in the real world.\n\n\
",
StyleBox["Exercise 6",
FontSize->16,
FontWeight->"Bold"],
"\n\nFollow the instructions of exercise 5 using the following functions:\n\
\n",
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"The ArcLength section of this lab is a substantial rewrite of the old \
ArcLength lab written by Cindy Kaus, which was actually written in ",
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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 about line integrals of \
scalar functions are both brand new. In other words, this lab was basically \
written from scratch, except for two of the exercises. The section on line \
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 \
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. (Maybe I was misinterpreting the problem?) 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. 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 \
this lab used to include the animation commands which were moved to Lab 2A \
(Parametrizing Curves). To fill the space I added the section about line \
integrals.\n\nOther than the exercises from Cindy Kaus, this lab is copyright \
2002, 2004 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\n\
Although 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 any questions or comments!\n\n"
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