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Adding the individual measurements together, you get the \ approximate flow-across measurement\n\n ", Cell[BoxData[ RowBox[{"flowtime", " ", RowBox[{"Sum", "[", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{\(Field[x[t], y[t]]\), "\[CenterDot]", RowBox[{"{", RowBox[{ RowBox[{ SuperscriptBox["y", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], ",", RowBox[{"-", RowBox[{ SuperscriptBox["x", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}]}], "}"}]}], "dt"}], ")"}], ",", "\n", "\t\t ", \({t, tlow, thigh - dt, \ dt}\)}], "]"}]}]]], ".\n \nAs ", Cell[BoxData[ \(dt\)]], " closes in on ", Cell[BoxData[ \(0\)]], ", these approximate measurements close in on the exact measurement\n \ ", Cell[BoxData[ RowBox[{"flowtime", " ", RowBox[{\(\[Integral]\_tlow\%thigh\), " ", RowBox[{ RowBox[{"(", RowBox[{\(Field[x[t], y[t]]\), "\[CenterDot]", RowBox[{"{", RowBox[{ RowBox[{ SuperscriptBox["y", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], ",", RowBox[{"-", RowBox[{ SuperscriptBox["x", 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Math happens again.\ \>", "SmallText"] }, Closed]] }, Closed]], Cell["Tutorial Problem", "Subsubsection"], Cell[CellGroupData[{ Cell["T.5) Summary of main ideas", "Subsection", CellTags->"VC.05.T5"], Cell[TextData[{ "Vector Calculus&", StyleBox["Mathematica", FontSlant->"Italic"], " offers this summary to you \nfor your good use and enjoyment.\nIt comes \ from the home office to you.\n'93" }], "Special2"], Cell["T.5.a.i) Flow along", "Subsubsection"], Cell[TextData[{ "If a curve ", Cell[BoxData[ \(C\)]], " is parameterized by ", Cell[BoxData[ \({x[t], y[t]}\)]], " with ", Cell[BoxData[ \(tlow \[LessEqual] t \[LessEqual] thigh\)]], ", then\n ", Cell[BoxData[ \(\[Integral]\_\(\(\ \)\(C\)\)\ m[x, y] \[DifferentialD]x + \ n[x, y] \[DifferentialD]y\)]], "\n ", Cell[BoxData[ RowBox[{\(\[Integral]\_tlow\%thigh\), " ", RowBox[{ RowBox[{\(Field[x[t], y[t]]\), ".", RowBox[{"{", RowBox[{ RowBox[{ SuperscriptBox["x", "\[Prime]", MultilineFunction->None], "[", "t", "]"}], ",", RowBox[{ 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Here's a parameterization of ", Cell[BoxData[ \(C\_1\)]], " with ", Cell[BoxData[ \(t\)]], " running from ", Cell[BoxData[ \(0\)]], " to ", Cell[BoxData[ \(1\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x1, y1, t];\)\), "\n", \({x1[t_], y1[t_]} = {0, 0} + t\ {1, 2}\)}], "Input", AspectRatioFixed->True], Cell[BoxData[ \({t, 2\ t}\)], "Output"] }, Open ]], Cell[TextData[{ "The curve ", Cell[BoxData[ \(C\_2\)]], " also starts at ", Cell[BoxData[ \({0, 0}\)]], " and ends at ", Cell[BoxData[ \({1, 2}\)]], ", but ", Cell[BoxData[ \(C\_2\)]], " runs on the parabola \n ", Cell[BoxData[ \(y = 2 x\^2\)]], " \nHere's a parameterization of ", Cell[BoxData[ \(C\_2\)]], " with ", Cell[BoxData[ \(t\)]], " running from ", Cell[BoxData[ \(0\)]], " to ", Cell[BoxData[ \(1\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[x2, y2, t];\)\), "\n", \({x2[t_], y2[t_]} = {t, 2\ t\^2}\)}], "Input", AspectRatioFixed->True], Cell[BoxData[ \({t, 2\ t\^2}\)], "Output"] }, 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RowBox[{\(m[x3[t], y3[t]]\), " ", RowBox[{ SuperscriptBox["x3", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "+", RowBox[{\(n[x3[t], y3[t]]\), " ", RowBox[{ SuperscriptBox["y3", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}]}], ",", \({t, 0, 1}\)}], "]"}]], "Input", AspectRatioFixed->True], Cell[BoxData[ \(1.416146836547142`\)], "Output"] }, Open ]], Cell[TextData[{ "Here are calculations of the three path integrals\n ", Cell[BoxData[ \(\[Integral]\_\(\(\ \)\(C\_1\)\)\ \ x\ Sin[ x\ y] \[DifferentialD]x\ + \ y\ Sin[x\ y] \[DifferentialD]y\)]], ",\n ", Cell[BoxData[ \(\[Integral]\_\(\(\ \)\(C\_2\)\)\ \ x\ Sin[ x\ y] \[DifferentialD]x\ + \ y\ Sin[x\ y] \[DifferentialD]y\)]], ", and\n ", Cell[BoxData[ \(\[Integral]\_\(\(\ \)\(C\_3\)\)\ \ x\ Sin[ x\ y] \[DifferentialD]x\ + \ y\ Sin[x\ y] \[DifferentialD]y\)]], "\nin order:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{\(Clear[m, n, x, y]\), ";", "\n", \({m[x_, y_], n[x_, y_]} = {x\ Sin[x\ y], y\ Sin[x\ y]}\), 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\[DifferentialD]x\ + \ x\ \(e\^\(\(-x\)\ y\)\) \[DifferentialD]y\)]], "\nCome up with the substitute curve ", Cell[BoxData[ \(C\_1\)]], ", make the calculation, and explain how you know that your calculation is \ correct." }], "Text"], Cell["G.6.b.ii)", "Subsubsection"], Cell[TextData[{ "Go with the same curve ", Cell[BoxData[ \(C\)]], " as in part i) immediately above.\nIf you had been asked to calculate\n \ ", Cell[BoxData[ \(\[Integral]\_\(\(\ \)\(C\)\)\ \ x\ \(e\^\(\(-x\)\ y\)\) \ \[DifferentialD]x\ + \ y\ \(e\^\(\(-x\)\ y\)\) \[DifferentialD]y\)]], ",\nwould you have even considered the use of a substitute curve?\nWhy or \ why not?" }], "Text"], Cell["G.6.c.i)", "Subsubsection"], Cell[TextData[{ "Here's a function ", Cell[BoxData[ \(f[x, y]\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[f, x, y];\)\), "\n", \(f[x_, y_] = x\^2\ y\)}], "Input", AspectRatioFixed->True], Cell[BoxData[ \(x\^2\ y\)], "Output"] }, Open ]], Cell["Here's the gradient field of this 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", "Text"], Cell["G.6.d)", "Subsubsection"], Cell["Take:", "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(Clear[f, x, y];\)\), "\n", \(f[x_, y_] = Sin[x\^2\ y]\)}], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Sin[x\^2\ y]\)], "Output"] }, Open ]], Cell[TextData[{ "Look at ", Cell[BoxData[ \(gradf[x, y]\)]], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \({D[f[x, y], \ x], D[f[x, y], \ y]}\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \({2\ x\ y\ Cos[x\^2\ y], x\^2\ Cos[x\^2\ y]}\)], "Output"] }, Open ]], Cell[TextData[{ "If ", Cell[BoxData[ \(C\)]], " is any curve starting at ", Cell[BoxData[ \({0, 1.2}\)]], " and ending at ", Cell[BoxData[ \({2.7, \(-5.1\)}\)]], ", then\n ", Cell[BoxData[ \(\[Integral]\_\(\(\ \)\(C\)\)\ 2 x\ y\ Cos[\(x\^2\) y] \[DifferentialD]x\ + \ \(x\^2\) Cos[\(x\^2\) y] \[DifferentialD]y\)]], "\n ", Cell[BoxData[ \(\[Integral]\_\(\(\ \)\(C\)\)\ \ \[DifferentialD]f\)]], " \n ", Cell[BoxData[ \(\(\(=\)\(f[2.7, \(-5.1\)] - \ f[0, 1.2]\)\)\)]], ": " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(f[2.7, \(-5.1\)] - f[0, 1.2]\)], "Input", AspectRatioFixed->True], Cell[BoxData[ \(0.49697719430361215`\)], "Output"] }, Open ]], Cell["\<\ Do you agree or disagree? 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" }], "Text"], Cell["G.6.f)", "Subsubsection"], Cell[TextData[{ "Suppose you know for sure that ", Cell[BoxData[ \(Field[x, y] = {m[x, y], n[x, y]}\)]], " is a gradient field. \nExplain how you know that the flow of ", Cell[BoxData[ \(Field[x, y]\)]], " along any one curve starting at ", Cell[BoxData[ \({x\_0, y\_0}\)]], " and ending at ", Cell[BoxData[ \({x\_1, y\_1}\)]], " is the same as the flow of ", Cell[BoxData[ \(Field[x, y]\)]], " along any other curve starting at ", Cell[BoxData[ \({x\_0, y\_0}\)]], " and ending at ", Cell[BoxData[ \({x\_1, y\_1}\)]], "." }], "Text"] }, Closed]], Cell["Literacy Problem", "Subsubsection"], Cell["L.1)", "Subsubsection"], Cell[TextData[{ "You have a given vector field \n ", Cell[BoxData[ \(Field[x, y] = {m[x, y], n[x, y]}\)]], ".\nYou have parameterized the ellipse \n ", Cell[BoxData[ RowBox[{ RowBox[{ SuperscriptBox[ RowBox[{"(", StyleBox[\(x\/2\), FontSize->16], ")"}], "2"], "+", SuperscriptBox[ RowBox[{"(", StyleBox[\(y\/3\), FontSize->16], ")"}], "2"]}], "=", "1"}]]], " \nin the counterclockwise way with \n ", Cell[BoxData[ \({x[t], y[t]} = {2 Cos[t], 3 Sin[t]}\)]], " \nwith ", Cell[BoxData[ \(0 \[LessEqual] t \[LessEqual] 2 \[Pi]\)]], ".\nYou calculate\n ", Cell[BoxData[ \(\[ContourIntegral]\_\(\(\ \)\(C\)\)\ m[x, y] \[DifferentialD]x\ + \ n[x, y] \[DifferentialD]y\)]], "\n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(\[Integral]\_0\%\(2 \[Pi]\)\), RowBox[{ RowBox[{"(", RowBox[{ RowBox[{\(m[x[t], y[t]]\), " ", RowBox[{ SuperscriptBox["x", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "+", " ", RowBox[{\(n[x[t], y[t]]\), " ", RowBox[{ SuperscriptBox["y", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}]}], ")"}], \(\[DifferentialD]t\)}]}]}]]], ",\nand learn that \n ", Cell[BoxData[ \(\[ContourIntegral]\_\(\(\ \)\(C\)\)\ m[x, y] \[DifferentialD]x\ + \ n[x, y] \[DifferentialD]y = 9.01\)]], ".\nThen you calculate\n ", Cell[BoxData[ \(\[ContourIntegral]\_\(\(\ \)\(C\)\)\ \(-n[x, y]\) \[DifferentialD]x\ + \ m[x, y] \[DifferentialD]y\)]], " \n ", Cell[BoxData[ RowBox[{"=", RowBox[{\(\[Integral]\_0\%\(2 \[Pi]\)\), RowBox[{ RowBox[{"(", RowBox[{ RowBox[{\(-n[x[t], y[t]]\), " ", RowBox[{ SuperscriptBox["x", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}], "+", " ", RowBox[{\(m[x[t], y[t]]\), " ", RowBox[{ SuperscriptBox["y", "\[Prime]", MultilineFunction->None], "[", "t", "]"}]}]}], ")"}], \(\[DifferentialD]t\)}]}]}]]], ",\nand learn that \n ", Cell[BoxData[ \(\[ContourIntegral]\_\(\(\ \)\(C\)\)\ \(-n[x, y]\) \[DifferentialD]x\ + \ m[x, y] \[DifferentialD]y = \(-3.52\)\)]], ".\nIs the net flow of ", Cell[BoxData[ \(Field[x, y]\)]], " along ", Cell[BoxData[ \(C\)]], " clockwise or counterclockwise?\nIs the net flow of ", Cell[BoxData[ \(Field[x, y]\)]], " across ", Cell[BoxData[ \(C\)]], " from outside to inside, or is it from inside to outside?" }], "Text"] }, FrontEndVersion->"4.0 for Macintosh", ScreenRectangle->{{0, 640}, {0, 460}}, AutoGeneratedPackage->None, WindowToolbars->{}, 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