(* Content-type: application/mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 6.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 145, 7] NotebookDataLength[ 37407, 1097] NotebookOptionsPosition[ 34699, 1014] NotebookOutlinePosition[ 35069, 1030] CellTagsIndexPosition[ 35026, 1027] WindowFrame->Normal ContainsDynamic->False*) (* Beginning of Notebook Content *) 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[ FormBox[ RowBox[{"f", "(", "t", ")"}], TraditionalForm]]], " is. Remember, if ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", "t", ")"}], TraditionalForm]]], " 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[ FormBox[ RowBox[{ FractionBox["1", "8"], " "}], TraditionalForm]]], "= 3.125, a bit low. The Egyptians' estimate of ", Cell[BoxData[ FormBox[ FractionBox["256", "81"], TraditionalForm]]], "\[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 ", Cell[BoxData[ FormBox[ FractionBox["10", "71"], TraditionalForm]]], "\[Precedes] \[Pi] \[Precedes] 3 ", Cell[BoxData[ FormBox[ FractionBox["1", "7"], TraditionalForm]]], ". 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[ RowBox[{"Segments", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"Cos", "[", "t", "]"}], ",", RowBox[{"Sin", "[", "t", "]"}]}], "}"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", "Pi"}], "}"}], ",", "4", ",", RowBox[{"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[ RowBox[{"Estimate", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"Cos", "[", "t", "]"}], ",", RowBox[{"Sin", "[", "t", "]"}]}], "}"}], ",", RowBox[{"{", RowBox[{"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[{ RowBox[{"Segments", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"Cos", "[", "t", "]"}], ",", RowBox[{"Sin", "[", "t", "]"}]}], "}"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", "Pi"}], "}"}], ",", "48", ",", RowBox[{"AspectRatio", "\[Rule]", "Automatic"}]}], "]"}], "\[IndentingNewLine]", RowBox[{"Estimate", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{"Cos", "[", "t", "]"}], ",", RowBox[{"Sin", "[", "t", "]"}]}], "}"}], ",", RowBox[{"{", RowBox[{"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[{ RowBox[{ RowBox[{ RowBox[{"f", "[", "t_", "]"}], "=", RowBox[{"{", RowBox[{ RowBox[{"Cos", "[", "t", "]"}], ",", RowBox[{"Sin", "[", "t", "]"}]}], "}"}]}], ";"}], "\[IndentingNewLine]", RowBox[{ RowBox[{ RowBox[{"g", "[", "t_", "]"}], "=", RowBox[{"{", RowBox[{ RowBox[{"Cos", "[", RowBox[{"t", "^", "2"}], "]"}], ",", RowBox[{"Sin", "[", RowBox[{"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[{ RowBox[{"Segments", "[", RowBox[{ RowBox[{"f", "[", "t", "]"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", RowBox[{"2", "Pi"}]}], "}"}], ",", "12", ",", RowBox[{"AspectRatio", "\[Rule]", "Automatic"}]}], "]"}], "\[IndentingNewLine]", RowBox[{"Estimate", "[", RowBox[{ RowBox[{"f", "[", "t", "]"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", RowBox[{"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[ FormBox[ SqrtBox[ RowBox[{"2", "\[Pi]"}]], TraditionalForm]]], " when we use g!" }], "Text"], Cell[BoxData[{ RowBox[{"Segments", "[", RowBox[{ RowBox[{"g", "[", "t", "]"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", RowBox[{"Sqrt", "[", RowBox[{"2", "Pi"}], "]"}]}], "}"}], ",", "12", ",", RowBox[{"AspectRatio", "\[Rule]", "Automatic"}]}], "]"}], "\[IndentingNewLine]", RowBox[{"Estimate", "[", RowBox[{ RowBox[{"g", "[", "t", "]"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", RowBox[{"Sqrt", "[", RowBox[{"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[ FormBox[ RowBox[{ RowBox[{"Cos", "(", SuperscriptBox["t", "2"], ")"}], ",", " ", RowBox[{"Sin", "(", SuperscriptBox["t", "2"], ")"}], " "}], TraditionalForm]]], "),\t0\[LessEqual]t\[LessEqual]", Cell[BoxData[ FormBox[ SqrtBox[ RowBox[{"2", "\[Pi]"}]], TraditionalForm]]], ".\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[ FormBox[ FractionBox["\[Pi]", "2"], TraditionalForm]]], ", 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[ RowBox[{ RowBox[{"h", "[", "t_", "]"}], "=", RowBox[{"{", RowBox[{ RowBox[{"Cos", "[", "t", "]"}], ",", " ", RowBox[{ RowBox[{"Sin", "[", "t", "]"}], "+", RowBox[{"0.01", " ", RowBox[{"Sin", "[", RowBox[{"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[{ RowBox[{"Segments", "[", RowBox[{ RowBox[{"h", "[", "t", "]"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", "Pi"}], "}"}], ",", "100"}], "]"}], "\[IndentingNewLine]", RowBox[{"Estimate", "[", RowBox[{ RowBox[{"h", "[", "t", "]"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", "Pi"}], "}"}], ",", "100"}], "]"}]}], "Input",\ CellChangeTimes->{3.412350938446646*^9}], 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[ 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.)" }], "Text", CellChangeTimes->{{3.412351225072183*^9, 3.412351275129219*^9}}], Cell[BoxData[ RowBox[{"Segments", "[", RowBox[{ RowBox[{"h", "[", "t", "]"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", RowBox[{"Pi", "/", "50"}]}], "}"}], ",", RowBox[{"1000", "/", "50"}], ",", RowBox[{"AspectRatio", "\[Rule]", "1"}]}], "]"}]], "Input", CellChangeTimes->{{3.412350955687699*^9, 3.412350961534317*^9}, { 3.41235103607021*^9, 3.412351113848474*^9}, {3.412351173464161*^9, 3.412351215350271*^9}, {3.41235128415637*^9, 3.412351289853468*^9}}], 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[ FormBox[ FractionBox["\[Pi]", "50"], TraditionalForm]]], ") 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[ FormBox[ RowBox[{"a", "\[LessEqual]", "t", "\[LessEqual]", "b"}], TraditionalForm]]], ". Then the length of ", StyleBox["C", FontSlant->"Italic"], " is:" }], "Text"], Cell[BoxData[ RowBox[{"Length", "=", RowBox[{ SuperscriptBox[ SubscriptBox["\[Integral]", "a"], "b"], RowBox[{"\[DoubleVerticalBar]", RowBox[{ RowBox[{"f", "'"}], RowBox[{"(", "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[ FormBox[ RowBox[{"g", "(", RowBox[{"x", ",", "y", ",", "z"}], ")"}], TraditionalForm]]], "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[ RowBox[{ RowBox[{ SubscriptBox["\[Integral]", "C"], RowBox[{"g", " ", "ds"}]}], "=", RowBox[{ RowBox[{ SuperscriptBox[ SubscriptBox["\[Integral]", "a"], "b"], RowBox[{"g", RowBox[{"(", RowBox[{"f", RowBox[{"(", "t", ")"}]}], ")"}]}]}], "\[DoubleVerticalBar]", RowBox[{ RowBox[{"f", "'"}], RowBox[{"(", "t", ")"}]}], "\[DoubleVerticalBar]", "dt"}]}]], "DisplayFormula", TextAlignment->Center, FontSize->14], Cell[TextData[{ "So the only difference is that we stick ", Cell[BoxData[ FormBox[ RowBox[{"g", "(", RowBox[{"f", "(", "t", ")"}], ")"}], TraditionalForm]]], " 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[ FormBox[ RowBox[{ RowBox[{"g", "(", RowBox[{"x", ",", "y", ",", "z"}], ")"}], "=", "1"}], TraditionalForm]]], ", then ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"g", "(", RowBox[{"f", "(", "t", ")"}], ")"}], "=", "1"}], TraditionalForm]]], ", 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[ FormBox[ RowBox[{"s", "(", "C", ")"}], TraditionalForm]]], ", 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[ RowBox[{"Mass", "=", RowBox[{ RowBox[{ SubscriptBox["\[Integral]", "C"], RowBox[{"g", " ", "ds"}]}], "=", RowBox[{ RowBox[{ SuperscriptBox[ SubscriptBox["\[Integral]", "a"], "b"], RowBox[{"g", RowBox[{"(", RowBox[{"f", RowBox[{"(", "t", ")"}]}], ")"}]}]}], "\[DoubleVerticalBar]", RowBox[{ RowBox[{"f", "'"}], RowBox[{"(", "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[{ RowBox[{ RowBox[{"f", "[", "t_", "]"}], "=", RowBox[{"{", RowBox[{ RowBox[{"Cos", "[", RowBox[{"2", "*", "Pi", "*", "t"}], "]"}], ",", RowBox[{"Sin", "[", RowBox[{"2", "*", "Pi", "*", "t"}], "]"}], ",", RowBox[{"1", "+", RowBox[{"t", "/", "5"}]}]}], "}"}]}], "\[IndentingNewLine]", RowBox[{"ParametricPlot3D", "[", RowBox[{ RowBox[{"f", "[", "t", "]"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", "10"}], "}"}], ",", RowBox[{"ViewPoint", "\[Rule]", RowBox[{"{", RowBox[{"0", ",", RowBox[{"-", "10"}], ",", "1"}], "}"}]}]}], "]"}]}], "Input", CellChangeTimes->{{3.412351307208001*^9, 3.41235130765128*^9}}], 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[ RowBox[{ RowBox[{"g", "[", RowBox[{"x_", ",", "y_", ",", "z_"}], "]"}], "=", "z"}]], "Input"], Cell["Then the total mass of the wire is:", "Text"], Cell[BoxData[ RowBox[{"Mass", "=", RowBox[{ RowBox[{ SubscriptBox["\[Integral]", "C"], RowBox[{"g", " ", "ds"}]}], "=", RowBox[{ RowBox[{ SuperscriptBox[ SubscriptBox["\[Integral]", "0"], "10"], RowBox[{"g", RowBox[{"(", RowBox[{"f", RowBox[{"(", "t", ")"}]}], ")"}]}]}], "\[DoubleVerticalBar]", RowBox[{ RowBox[{"f", "'"}], RowBox[{"(", "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[{ RowBox[{"D", "[", RowBox[{ RowBox[{"f", "[", "t", "]"}], ",", "t"}], "]"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"Norm", "[", "%", "]"}], "//", "Simplify"}]}], "Input"], Cell[TextData[{ "Next we have to plug ", Cell[BoxData[ FormBox[ RowBox[{"f", "(", "t", ")"}], TraditionalForm]]], " into ", Cell[BoxData[ FormBox[ RowBox[{"g", "(", RowBox[{"x", ",", "y", ",", "z"}], ")"}], TraditionalForm]]], ":" }], "Text"], Cell[BoxData[ RowBox[{"Apply", "[", RowBox[{"g", ",", RowBox[{"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[ RowBox[{ RowBox[{"f", "[", "t", "]"}], "=", RowBox[{"{", RowBox[{ RowBox[{"t", " ", RowBox[{"Cos", "[", RowBox[{"2", " ", "\[Pi]", " ", "t"}], "]"}]}], ",", RowBox[{"t", " ", RowBox[{"Sin", "[", RowBox[{"2", " ", "\[Pi]", " ", "t"}], "]"}]}], ",", RowBox[{"2", " ", "t"}]}], "}"}]}]], "Input"], "\n\nFor our integral, we want to evaluate ", Cell[BoxData[ RowBox[{"g", "[", RowBox[{ RowBox[{"t", " ", RowBox[{"Cos", "[", RowBox[{"2", " ", "\[Pi]", " ", "t"}], "]"}]}], ",", RowBox[{"t", " ", RowBox[{"Sin", "[", RowBox[{"2", " ", "\[Pi]", " ", "t"}], "]"}]}], ",", RowBox[{"2", " ", "t"}]}], "]"}]], "Input"], ",\nbut ", StyleBox["Mathematica", FontSlant->"Italic"], " reads ", StyleBox["g[f[t]]", FontFamily->"Courier", FontWeight->"Bold"], " as ", Cell[BoxData[ RowBox[{"g", "[", RowBox[{"{", RowBox[{ RowBox[{"t", " ", RowBox[{"Cos", "[", RowBox[{"2", " ", "\[Pi]", " ", "t"}], "]"}]}], ",", RowBox[{"t", " ", RowBox[{"Sin", "[", RowBox[{"2", " ", "\[Pi]", " ", "t"}], "]"}]}], ",", RowBox[{"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[{ RowBox[{"Integrate", "[", RowBox[{ RowBox[{ RowBox[{"Apply", "[", RowBox[{"g", ",", RowBox[{"f", "[", "t", "]"}]}], "]"}], "*", RowBox[{"Norm", "[", RowBox[{"D", "[", RowBox[{ RowBox[{"f", "[", "t", "]"}], ",", "t"}], "]"}], "]"}]}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", "10"}], "}"}]}], "]"}], "\[IndentingNewLine]", RowBox[{"N", "[", "%", "]"}]}], "Input"], Cell[TextData[{ StyleBox["Exercise 5", FontSize->14, FontWeight->"Bold"], "\n\nConsider the curve ", StyleBox["C", FontSlant->"Italic"], " parametrized by ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"f", "(", "t", ")"}], "=", RowBox[{"(", RowBox[{"t", ",", SuperscriptBox["t", "3"], ",", SuperscriptBox["t", "2"]}], ")"}]}], TraditionalForm]]], ", where ", Cell[BoxData[ FormBox[ RowBox[{"0", "\[LessEqual]", "t", "\[LessEqual]", "1"}], TraditionalForm]]], ". 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[ FormBox[ RowBox[{ RowBox[{"g", "(", RowBox[{"x", ",", "y", ",", "z"}], ")"}], "=", RowBox[{ RowBox[{"2", "x"}], "+", RowBox[{"4", "y"}]}]}], TraditionalForm]]], " 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[ FormBox[ RowBox[{ RowBox[{"f", "(", "t", ")"}], "=", RowBox[{"(", RowBox[{"t", ",", RowBox[{"Cos", " ", "t"}], ",", " ", RowBox[{"Sin", " ", "t"}]}], ")"}]}], TraditionalForm]]], "where ", Cell[BoxData[ FormBox[ RowBox[{"0", "\[LessEqual]", "t", "\[LessEqual]", RowBox[{"2", "\[Pi]"}]}], TraditionalForm]]], ",\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"g", "(", RowBox[{"x", ",", "y", ",", "z"}], ")"}], "=", RowBox[{ RowBox[{"-", "x"}], " ", "y", " ", "z"}]}], TraditionalForm]]], "." }], "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", BoxBaselineShift->-0.2, BoxMargins->{{-0.36, -0.1}, {0, 0}}], FontSize->Smaller], "T", AdjustmentBox["E", BoxBaselineShift->0.5, BoxMargins->{{-0.075, -0.085}, {0, 0}}], "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\nExtremely minor changes made in Spring 2008 for ", StyleBox["Mathematica", FontSlant->"Italic"], " 6.0.\n\nOther than the exercises from Cindy Kaus, this lab is copyright \ 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 \ any questions or comments!\n\n" }], "Text", CellChangeTimes->{{3.412351344309447*^9, 3.412351363401014*^9}}] }, Closed]] }, Closed]] }, ScreenStyleEnvironment->"Working", WindowSize->{751, 836}, WindowMargins->{{135, Automatic}, {Automatic, 49}}, FrontEndVersion->"6.0 for Linux x86 (32-bit) (April 20, 2007)", StyleDefinitions->"Default.nb" ] (* End of Notebook Content *) (* Internal cache information *) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[568, 21, 378, 11, 107, "Text"], Cell[CellGroupData[{ Cell[971, 36, 87, 2, 65, "Section"], Cell[1061, 40, 443, 7, 83, "Text"], Cell[1507, 49, 1317, 25, 243, "Text"], Cell[2827, 76, 474, 13, 65, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[3338, 94, 75, 1, 35, "Section"], Cell[3416, 97, 3278, 63, 607, "Text"], Cell[6697, 162, 327, 9, 31, "Input"], Cell[7027, 173, 140, 3, 29, "Text"], Cell[7170, 178, 269, 8, 31, "Input"], Cell[7442, 188, 281, 5, 65, "Text"], Cell[7726, 195, 604, 18, 52, "Input"], Cell[8333, 215, 704, 14, 156, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[9074, 234, 83, 1, 35, "Section"], Cell[9160, 237, 110, 3, 29, "Text"], Cell[9273, 242, 484, 17, 52, "Input"], Cell[9760, 261, 552, 13, 119, "Text"], Cell[10315, 276, 474, 14, 52, "Input"], Cell[10792, 292, 373, 10, 86, "Text"], Cell[11168, 304, 544, 16, 52, "Input"], Cell[11715, 322, 1302, 29, 245, "Text"], Cell[13020, 353, 2154, 49, 486, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[15211, 407, 75, 1, 35, "Section"], Cell[15289, 410, 166, 3, 47, "Text"], Cell[15458, 415, 309, 10, 31, "Input"], Cell[15770, 427, 434, 11, 101, "Text"], Cell[16207, 440, 420, 13, 52, "Input"], Cell[16630, 455, 1241, 28, 212, "Text"], Cell[17874, 485, 496, 11, 31, "Input"], Cell[18373, 498, 250, 4, 47, "Text"], Cell[18626, 504, 1503, 29, 522, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[20166, 538, 88, 1, 35, "Section"], Cell[20257, 541, 251, 8, 63, "Text"], Cell[20511, 551, 564, 19, 65, "Text"], Cell[21078, 572, 319, 11, 45, "DisplayFormula"], Cell[21400, 585, 518, 18, 47, "Text"], Cell[21921, 605, 489, 18, 45, "DisplayFormula"], Cell[22413, 625, 1599, 48, 191, "Text"], Cell[24015, 675, 530, 19, 45, "DisplayFormula"], Cell[CellGroupData[{ Cell[24570, 698, 32, 0, 30, "Subsubsection"], Cell[24605, 700, 365, 11, 47, "Text"], Cell[24973, 713, 705, 20, 52, "Input"], Cell[25681, 735, 164, 3, 47, "Text"], Cell[25848, 740, 114, 3, 31, "Input"], Cell[25965, 745, 51, 0, 29, "Text"], Cell[26019, 747, 531, 19, 45, "DisplayFormula"], Cell[26553, 768, 383, 12, 47, "Text"], Cell[26939, 782, 197, 5, 52, "Input"], Cell[27139, 789, 262, 11, 29, "Text"], Cell[27404, 802, 110, 3, 31, "Input"], Cell[27517, 807, 1731, 60, 194, "Text"], Cell[29251, 869, 86, 2, 29, "Text"], Cell[29340, 873, 445, 14, 52, "Input"], Cell[29788, 889, 2018, 67, 303, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[31843, 961, 29, 0, 38, "Subsection"], Cell[31875, 963, 2796, 47, 570, "Text"] }, Closed]] }, Closed]] } ] *) (* End of internal cache information *)