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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 28785, 765]*) (*NotebookOutlinePosition[ 29450, 788]*) (* CellTagsIndexPosition[ 29406, 784]*) (*WindowFrame->Normal*) Notebook[{ Cell[TextData[{ StyleBox["Lab 5A - 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 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..." }], "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 4 if you'd like." }], "Text"] }, Closed]], 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 ", 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->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!" }], "Text", CellFrame->True, Background->RGBColor[0.996109, 0.500008, 0.500008]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Arc Length and the Derivative", FontSize->14]], "Section"], Cell["\<\ At the end of Lab 4 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->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) = (", 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]] }, Closed]], 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->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[ \(TraditionalForm\`\[Pi]\/50\)]], ") 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?)" }], "Text", CellFrame->True, Background->RGBColor[0.996109, 0.500008, 0.500008]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Line Integrals of Scalar Functions", FontSize->14]], "Section"], Cell[TextData[{ "In some textbooks, such as ", StyleBox["Marsden and Tromba", FontSlant->"Italic"], ", a line integragl of a scalar function is referred to as a \"path \ integral.\"" }], "Text", CellFrame->True, Background->GrayLevel[0.833326]], 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\ ds = \[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["s", FontWeight->"Bold"], "\" -- if you integrate 1, you get ", Cell[BoxData[ \(TraditionalForm\`s(C)\)]], ", which is a common notation for the arclength of ", StyleBox["C", FontSlant->"Italic"], ". ", StyleBox["\n", FontWeight->"Bold"], "\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\ ds = \[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\ ds = \[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\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->14, 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->14, FontWeight->"Bold"], "\n\nFollow the instructions of exercise 5 using the following functions:\n\ \n", Cell[BoxData[ \(TraditionalForm\`f(t) = \((t, Cos\ t, \ Sin\ t)\)\)]], "where ", Cell[BoxData[ \(TraditionalForm\`0 \[LessEqual] t \[LessEqual] 2 \[Pi]\)]], ",\n", Cell[BoxData[ \(TraditionalForm\`g(x, y, z) = \(-x\)\ y\ z\)]], "." }], "Text", CellFrame->True, Background->RGBColor[0.996109, 0.500008, 0.500008]] }, Open ]], Cell[CellGroupData[{ Cell["Credits", "Subsection"], Cell[TextData[{ "The ArcLength section of this lab is a substantial rewrite of the old \ ArcLength lab written by Cindy Kaus, which was actually written in ", Cell[BoxData[ StyleBox[ RowBox[{"L", StyleBox[ AdjustmentBox["A", BoxMargins->{{-0.36, -0.1}, {0, 0}}, BoxBaselineShift->-0.2], FontSize->Smaller], "T", AdjustmentBox["E", BoxMargins->{{-0.075, -0.085}, {0, 0}}, BoxBaselineShift->0.5], "X"}]]]], ". 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" }], "Text"] }, Closed]] }, Closed]] }, FrontEndVersion->"5.1 for X", ScreenRectangle->{{0, 1280}, {0, 1024}}, ScreenStyleEnvironment->"Working", WindowSize->{741, 836}, WindowMargins->{{93, Automatic}, {48, Automatic}} ] (******************************************************************* Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. 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