(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.0' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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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[ 119721, 3276]*) (*NotebookOutlinePosition[ 120793, 3312]*) (* CellTagsIndexPosition[ 120749, 3308]*) (*WindowFrame->Normal*) Notebook[{ Cell[TextData[{ StyleBox["Lab 2A - Continuity and Differentiability", FontSize->24, FontWeight->"Bold"], "\nMath 2374 - University of Minnesota\nhttp://www.math.umn.edu/math2374\n \ questions to rogness@math.umn.edu, drake@math.umn.edu" }], "Text", CellFrame->True, TextAlignment->Center, FontColor->GrayLevel[1], Background->RGBColor[0, 0, 1]], Cell[BoxData[""], "Input"], Cell[CellGroupData[{ Cell[TextData[StyleBox["Introduction", FontSize->16]], "Section"], Cell[TextData[{ "This week we will investigate the ideas of ", StyleBox["continuity", FontSlant->"Italic"], " and ", StyleBox["differentiability", FontSlant->"Italic"], " for functions of several variables. The relationships between limits, \ continuity, partial derivatives, and differentiability are quite subtle and \ complicated and you might find that your intuition will often lead you \ astray. Most of the functions with which we'll concern ourselves in this \ course are very ``nice'', but it's important that you are aware of the sorts \ of pathological functions which do exist." }], "Text"], Cell[TextData[{ "Some of the functions and equations in this lab are complicated and \ difficult to read in a small font. If that bothers you, the magnification can \ be changed in the ", StyleBox["Format \[Rule] Magnification", FontFamily->"Courier"], " menu. Another way is ", StyleBox["Format \[Rule] Screen Style Environment \[Rule] Presentation", FontFamily->"Courier"], ", which changes the general look of your notebook and pretties things up a \ bit.\n\nAlso, many of the examples here will use ", StyleBox["ShowLive", FontWeight->"Bold"], " or ", StyleBox["Plot3DLive", FontWeight->"Bold"], ". If you're viewing this on a computer where those commands don't work, \ you can remove the ", StyleBox["Live", FontWeight->"Bold"], " from each command and still read the lab, although it will be less \ interactive." }], "Text", CellFrame->True, Background->GrayLevel[0.849989]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Continuity", FontSize->16]], "Section"], Cell[TextData[{ "From lecture and our textbooks, we know that the idea of continuity\ \[LongDash]for both single- and multi-variable functions\[LongDash]largely \ depends upon the idea of a limit. In both cases, a function ", StyleBox["f", FontSlant->"Italic"], " is continuous at a point ", StyleBox["a", FontSlant->"Italic"], " if" }], "Text"], Cell[BoxData[ StyleBox[ RowBox[{ UnderscriptBox[ StyleBox["lim", FontSlant->"Plain"], \(\(\ \)\(x\ \[Rule] \ a\)\)], " ", \(\(f\) \((x)\)\(\ \)\)}], FontFamily->"Times", FontSlant->"Italic"]], "DisplayFormula", TextAlignment->Center, FontSize->16], Cell["exists and", "Text"], Cell[BoxData[ StyleBox[ RowBox[{ RowBox[{ UnderscriptBox[ StyleBox["lim", FontSlant->"Plain"], \(\(\ \)\(x\ \[Rule] \ a\)\)], " ", \(f \((x)\)\)}], " ", "=", " ", \(f \(\((a)\)\(.\)\)\)}], FontFamily->"Times", FontSlant->"Italic"]], "DisplayFormula", TextAlignment->Center, FontSize->16], Cell["\<\ Intuitively, this means that the function has no holes or jumps; in \ the single-variable case, we often talk about being able to draw a function \ without lifting your pencil from the paper. Determining whether a function is continuous or not is therefore mostly a \ matter of determining the value of the limit (if it even exists, which it may \ not). For single-variable functions, we talk about right-hand and left-hand \ limits: one can approach a value on the real line from either the right or \ the left. But what about a function whose domain is two-dimensional? Not only \ can you approach from the left and right, but from many other directions as \ well. Say we have a function g(x) whose domain includes the point (2,2). 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An important \ observation is that ", StyleBox["the limit of a function as ", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" goes to ", FontSlant->"Italic"], StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" will exist if and only if the limits as you approach ", FontSlant->"Italic"], StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" along ANY path agree.", FontSlant->"Italic"], " As we just saw, there are infinitely many paths that approach a point, \ and we certainly can't test each one of them individually. This means that, \ in practice, we usually won't evaluate any \"two-dimensional\" limits. We \ will, however, use the contrapositive of the above statement: ", StyleBox["if the limits as you approach ", FontSlant->"Italic"], StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" along two or more different paths disagree, then the limit as ", FontSlant->"Italic"], StyleBox["x", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" goes to ", FontSlant->"Italic"], StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[" does not exist", FontSlant->"Italic"], ".\n\nNow carefully re-read the above paragraph once or twice. This is a \ tricky concept!" }], "Text"], Cell[CellGroupData[{ Cell["Example 1 - A continuous function", "Subsection"], Cell["\<\ We mentioned above that in the single variable case, you can \ intuitively read \"f is continuous\" as \"You can draw the graph of f without \ lifting your pencil from the paper.\" Unfortunately, it's impossible to \ generalize this sort of statement to multivariable functions. You're not \ drawing a line on a piece of paper -- you're trying to draw a surface in \ three-dimensional space! In practice, that means we have to work with the slightly less intuitive idea \ that \"a multivariable function f is continuous if its graph has no jumps or \ holes.\" That's actually not a bad way to think about it; the trick is that \ it's not always easy to tell if a function has a hole (or a jump) or not! Just for the record, here's a function which is continuous everywhere. \ Rotate the picture around and notice that it seems very smooth, without any \ jumps.\ \>", "Text"], Cell[BoxData[{ \(f[x_, y_] = Exp[\(-\((x^2 + y^2)\)\)]\), "\[IndentingNewLine]", \(Plot3DLive[f[x, y], {x, \(-1\), 1}, {y, \(-1\), 1}]\)}], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Example 2 - A function with a hole at (0,0) that can be fixed\ \>", \ "Subsection"], Cell[TextData[{ "In this course we're not going to deal with a lot of pathological \ functions. If a function ", Cell[BoxData[ \(TraditionalForm\`f(x, y)\)]], " is not continuous for all points ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\)\)]], ", then usually there will only be one bad point, and usually that bad \ point will be the origin. Here's an example where we've only changed the \ function in the previous example a little bit." }], "Text"], Cell[BoxData[{ \(f[x_, y_] = Exp[\(-1\)/\((x^2 + y^2)\)]\), "\[IndentingNewLine]", \(Plot3DLive[f[x, y], {x, \(-1\), 1}, {y, \(-1\), 1}]\)}], "Input"], Cell[TextData[{ "Notice that ", Cell[BoxData[ \(TraditionalForm\`f\)]], "has a hole at (0,0), because it's not even defined there!" }], "Text"], Cell[BoxData[ \(f[0, 0]\)], "Input"], Cell[TextData[{ "However, this is not at all evident from the picture, is it? This is \ another example of how you need to be very careful when using technology; ", StyleBox["Mathematica", FontSlant->"Italic"], " will help you visualize things, but it won't always tell you the whole \ story.\n\nNote that in this case, we could actually \"plug\" the hole. Look \ at the picture again and notice that as (x,y) get close to (0,0), the \ z-values all go to zero. In other words," }], "Text"], Cell[BoxData[ StyleBox[ RowBox[{ RowBox[{ RowBox[{ UnderscriptBox[ StyleBox["lim", FontSlant->"Plain"], \(\(\ \)\(\((x, y)\)\ \[Rule] \ \((0, 0)\)\)\)], " ", \(f \((x, y)\)\)}], " ", "=", " ", "0"}], ","}], FontFamily->"Times", FontSlant->"Italic"]], "DisplayFormula", TextAlignment->Center, FontSize->16], Cell[TextData[{ "although you should remember that graphical evidence isn't the same as \ actually calculating this limit! \n\nWhat this means is that we can make ", Cell[BoxData[ \(TraditionalForm\`f(x, y)\)]], "into a continuous function by simply defining ", Cell[BoxData[ \(TraditionalForm\`f(0, 0)\)]], "to be 0. In other words," }], "Text"], Cell[BoxData[ \(TraditionalForm\`f(x, y) = e\^\(\(-1\)/\((x\^2 + y\^2)\)\)\)], "DisplayFormula", TextAlignment->Center, FontSize->16], Cell["\<\ is not a continuous function at (0,0), because it doesn't even \ exist there, but\ \>", "Text"], Cell[BoxData[ RowBox[{\(f \((x, y)\)\), " ", "=", " ", TagBox[ RowBox[{ TagBox[ RowBox[{ StyleBox["{", ShowAutoStyles->False], StyleBox[GridBox[{ { RowBox[{ RowBox[{ SuperscriptBox[ StyleBox["e", FontSize->18], \(\(-1\)/\((x\^2 + y\^2)\)\)], ",", " ", RowBox[{ RowBox[{ StyleBox["if", FontFamily->"Times"], " ", \((x, y)\)}], "\[NotEqual]", \((0, 0)\)}]}], "\[IndentingNewLine]"}]}, { RowBox[{" ", RowBox[{"0", ",", StyleBox[ RowBox[{" ", StyleBox[" ", FontFamily->"Times"]}]], RowBox[{ RowBox[{ StyleBox["if", FontFamily->"Times"], StyleBox[" ", FontFamily->"Times"], \((x, y)\)}], "=", \((0, 0)\)}]}]}]} }], ShowAutoStyles->True]}], (#&)], "\[IndentingNewLine]"}], (#&)]}]], "DisplayFormula", TextAlignment->Center, FontSize->16], Cell[TextData[{ "is", StyleBox[" a continuous function at (0,0), because we've plugged the hole. \ Again, this is a tricky concept, but we won't be asking you to do any of \ this. Ask your TA if you've got questions, though.", FontSlant->"Plain"] }], "Text", FontSlant->"Italic"] }, Open ]], Cell[CellGroupData[{ Cell["Example 3 - A function which is not continuous at (0,0)", "Subsection"], Cell["\<\ Here's an example of a function that's not continuous at (0,0), and \ there's nothing we can do about it:\ \>", "Text"], Cell[BoxData[ RowBox[{\(f \((x, y)\)\), " ", "=", " ", TagBox[ RowBox[{ TagBox[ StyleBox[ RowBox[{"{", StyleBox[GridBox[{ { RowBox[{ StyleBox[\(xy\/\(x\^2 + \ y\^2\)\), FontSize->18], ",", " ", RowBox[{ RowBox[{ StyleBox["if", FontFamily->"Times"], " ", \((x, y)\)}], "\[NotEqual]", \((0, 0)\), "\[IndentingNewLine]"}]}]}, { RowBox[{" ", RowBox[{"0", ",", StyleBox[ RowBox[{" ", StyleBox[" ", FontFamily->"Times"]}]], RowBox[{ RowBox[{ StyleBox["if", FontFamily->"Times"], StyleBox[" ", FontFamily->"Times"], \((x, y)\)}], "=", \((0, 0)\)}]}]}]} }], ShowAutoStyles->True]}], ShowAutoStyles->False], (#&)], "\[IndentingNewLine]"}], (#&)]}]], "DisplayFormula", TextAlignment->Center, FontSize->14], Cell[TextData[{ "To work with this function in ", StyleBox["Mathematica", FontSlant->"Italic"], ", it's probably easiest just to use the following definition. If ", StyleBox["Mathematica", FontSlant->"Italic"], " ever complains that ", Cell[BoxData[ \(TraditionalForm\`f(0, 0)\)]], "is undefined, you'll just have to remember on your own that ", Cell[BoxData[ \(TraditionalForm\`f(0, 0) = 0\)]], "." }], "Text"], Cell[BoxData[ \(f[x_, y_] = x*y/\((x^2 + y^2)\)\)], "Input"], Cell["\<\ Let's take a look at the graph of f. The following commands will \ give you an interactive window with the picture, as well as a static image \ for later reference.\ \>", "Text"], Cell[BoxData[{ \(\(graph = Plot3D[f[x, y], {x, \(-1\), 1}, {y, \(-1\), 1}, AxesLabel \[Rule] {"\", "\", "\"}, PlotPoints \[Rule] 40];\)\), "\[IndentingNewLine]", \(\(ShowLive[graph];\)\)}], "Input"], Cell[TextData[{ "Notice that, even after using the ", StyleBox["PlotPoints", FontWeight->"Bold"], " option, the graph is still a little jagged in the middle. This is \ usually an indication that something is wrong; the graph is jumping around so \ much that ", StyleBox["Mathematica", FontSlant->"Italic"], " can't keep up!\n\nRotate the picture around until you get a good feel for \ what the graph looks like. Along a diagonal on the bottom, the height seems \ to be ", Cell[BoxData[ FormBox[ RowBox[{"-", StyleBox[\(1\/2\), FontSize->14]}], TraditionalForm]]], ", while on the opposite diagonal, on the top, the height seems to be ", Cell[BoxData[ FormBox[ StyleBox[\(+\(1\/2\)\), FontSize->14], TraditionalForm]]], ". And that, in a nutshell, is why the function is discontinuous at the \ origin: if you walk along the ridge on the top, the value at (0,0) \"should\" \ be ", Cell[BoxData[ FormBox[ StyleBox[\(+\(1\/2\)\), FontSize->14], TraditionalForm]]], ", because otherwise there'd be a jump! But if you walk along the valley \ in the bottom, the value at (0,0) \"should\" be ", Cell[BoxData[ FormBox[ RowBox[{"-", StyleBox[\(1\/2\), FontSize->14]}], TraditionalForm]]], ", because otherwise there'd be a jump there too. There's no way to \ satisfy both conditions, so the function has a jump discontinuity at (0,0), \ even though it's defined there.\n\nWhat that last paragraph says is really \ that " }], "Text"], Cell[BoxData[ StyleBox[ RowBox[{ RowBox[{ RowBox[{ RowBox[{ UnderscriptBox[ StyleBox["lim", FontSlant->"Plain"], \(\(\ \)\(\((x, y)\)\ \[Rule] \ \((0, 0)\)\)\)], " ", \(f \((x, y)\)\)}], " ", "\[NotEqual]", \(f \((0, 0)\)\)}], ";", " ", \(in\ fact\)}], ",", " ", \(the\ limit\ \ does\ not\ \(exist!\)\)}], FontFamily->"Times", FontSlant->"Italic"]], "DisplayFormula", TextAlignment->Center, FontSize->16], Cell["\<\ How could we prove this? We definitely don't want to have to \ compute a limit with two variables. As we mentioned above, this is too hard. \ Instead, we could try to make the difference between the \"ridge\" and the \ \"valley\" a little more mathematical. Evaluate the following cell -- don't \ worry about understanding the commands -- and then continue reading. You \ should get a LiveGraphics3D window with a wireframe drawing of the surface, \ along with two lines.\ \>", "Text"], Cell[BoxData[{ \(\(dx = \(dy = 1.1/10\);\)\), "\[IndentingNewLine]", \(ShowLive[ Graphics3D[ Table[Line[{{i, j, f[i, j]}, {i + dx, j, f[i + dx, j]}, {i + dx, j + dy, f[i + dx, j + dy]}, {i, j + dy, f[i, j + dy]}}], {i, \(-1\), 1, dx}, {j, \(-1\), 1, dy}]], Graphics3D[{{RGBColor[1, 0, 0], Thickness[0.03], Line[{{\(-1\), \(-1\), .5}, {1, 1, .5}}]}, {RGBColor[0, 0, 1], Thickness[0.03], Line[{{\(-1\), 1, \(- .5\)}, {1, \(-1\), \(- .5\)}}]}}], Axes \[Rule] True, AxesLabel \[Rule] {"\", "\", "\"}]\)}], "Input"], Cell[TextData[{ "The ", StyleBox["red", FontWeight->"Bold"], " line show you all of the points on the surface which are above the line \ ", Cell[BoxData[ \(TraditionalForm\`y = x\)]], ". Look at the picture from above until you see this. Notice that the \ z-values for these points are always 1/2, with the possible exception of the \ point above the origin. We can prove this mathematically. If we want to \ know what the function values are equal to when ", Cell[BoxData[ \(TraditionalForm\`y = x\)]], ", simply do that substitution :" }], "Text"], Cell[BoxData[ RowBox[{\(f \((x, x)\)\), "=", StyleBox[\(\(x \((x)\)\)\/\(x\^2 + \ \((x)\)\^2\) = \ \(x\^2\/\(x\^2 + \ \ x\^2\) = \ \(x\^2\/\(2 x\^2\) = \ 1\/2\)\)\), FontSize->18]}]], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " will do all of this for you." }], "Text"], Cell[BoxData[ \(f[x, x]\)], "Input"], Cell[TextData[{ "Similarly, the ", StyleBox["blue", FontWeight->"Bold"], " line shows you all of the points on the graph which have (x,y) values on \ the line ", Cell[BoxData[ \(TraditionalForm\`y = \(-x\)\)]], ". Rotate the picture until you can see this. The function values for \ these points [again, with the possible exception of (0,0)] is always ", Cell[BoxData[ FormBox[ RowBox[{"-", StyleBox[\(1\/2\), FontSize->14]}], TraditionalForm]]], "." }], "Text"], Cell[BoxData[ \(f[x, \(-x\)]\)], "Input"], Cell[TextData[{ "You've now shown mathematically -- instead of just that if you approach \ the origin along these two different lines, you end up with different \ heights. Or, a little more mathematically, the limit as you approach the \ origin on the red line is ", StyleBox["+", FontSize->16], Cell[BoxData[ \(TraditionalForm\`1\/2\)], FontSize->14], ", which the limit as you approach on the blue line is ", StyleBox["-", FontSize->16], Cell[BoxData[ \(TraditionalForm\`1\/2\)], FontSize->14], ". Since they don't agree, we've shown that" }], "Text"], Cell[BoxData[ StyleBox[ RowBox[{ UnderscriptBox[ StyleBox["lim", FontSlant->"Plain"], \(\(\ \)\(\((x, y)\)\ \[Rule] \ \((0, 0)\)\)\)], " ", \(f \((x, y)\)\ \ \ does\ not\ exist\)}], FontFamily->"Times", FontSlant->"Italic"]], "DisplayFormula", TextAlignment->Center, FontSize->16], Cell[TextData[{ "without having to actually calculate limits. Woohoo!\n\nIncidentally, to \ show a function isn't continuous at the origin, you need to find two ways to \ approach (0,0) where you get different heights, but those approaches don't \ necessarily have to be along these lines -- or even on a line at all, as the \ next two exercises show. With our particular function in this example, we \ could have used just about any two lines. \n\n Remember that any non-vertical \ line through the origin has the form ", Cell[BoxData[ \(TraditionalForm\`y = m\ x\)]], ". If you look at the points on the surface above these lines (or below, \ as the case may be), their height will be:" }], "Text"], Cell[BoxData[{ \(f[x, m*x]\), "\[IndentingNewLine]", \(Simplify[%]\)}], "Input"], Cell[TextData[{ "In our example, we used the slopes ", Cell[BoxData[ \(TraditionalForm\`m = 1\)]], " and ", Cell[BoxData[ \(TraditionalForm\`m = \(-1\)\)]], "." }], "Text"], Cell[TextData[{ StyleBox["Exercise 1", FontSize->16, FontWeight->"Bold"], "\n\nShow that the function\n\n", Cell[BoxData[ RowBox[{\(g \((x, y)\)\), " ", "=", " ", TagBox[ RowBox[{ TagBox[ StyleBox[ RowBox[{"{", StyleBox[GridBox[{ { RowBox[{ RowBox[{ StyleBox[\(\(\(\ \)\(2 \( x\^2\) y\)\)\/\(x\^4 + \ y\^2\)\), FontSize->18], ",", " ", RowBox[{ RowBox[{ StyleBox["if", FontFamily->"Times"], " ", \((x, y)\)}], "\[NotEqual]", \((0, 0)\)}]}], "\[IndentingNewLine]"}]}, { RowBox[{" ", RowBox[{"0", ",", StyleBox[ RowBox[{" ", StyleBox[" ", FontFamily->"Times"]}]], RowBox[{ RowBox[{ StyleBox["if", FontFamily->"Times"], StyleBox[" ", FontFamily->"Times"], \((x, y)\)}], "=", \((0, 0)\)}]}]}]} }], ShowAutoStyles->True]}], ShowAutoStyles->False], (#&)], "\[IndentingNewLine]"}], (#&)]}]], "Text", CellFrame->False, SpanMaxSize->Infinity, FontSize->16], "\nis not continuous at the origin. You should use ", StyleBox["Mathematica", FontSlant->"Italic"], " to look at pictures of this surface, but the actual calculations might be \ easier done by hand. You can copy the following commands and evaluate them \ to see a picture, but remember that the picture \"breaks down\" a bit near \ the origin.\n\n", Cell[BoxData[{ \(\(Clear[g, x, y];\)\), "\[IndentingNewLine]", \(g[x_, y_] = \((2 x^2\ *y)\)/\((x^4 + y^2)\)\), "\n", \(\(graph\ = \ Plot3D[g[x, y], {x, \(-1\), 1}, {y, \(-1\), 1}, AxesLabel \[Rule] {"\", "\", "\"}, PlotPoints \[Rule] 30];\)\), "\n", \(ShowLive[graph]\)}], "Input"], "\n\nDo the following steps:\n\n(a) Show that the limits agree if you \ approach above/below any line ", Cell[BoxData[ \(TraditionalForm\`y = m\ x\)]], ". (Do this by figuring out what ", Cell[BoxData[ \(TraditionalForm\`g(x, m\ x)\)]], "is and then letting x\[Rule]0.)\n\n(b) Calculate the limit as you approach \ the origin above the parabola ", Cell[BoxData[ \(TraditionalForm\`y = x\^2\)]], ".\n\n(c) Calculate the limit as you approach the origin below the parabola \ ", Cell[BoxData[ \(TraditionalForm\`y = \(-x\^2\)\)]], ".\n\nYou should actually show your calculations here; don't just let ", StyleBox["Mathematica", FontSlant->"Italic"], " do them for you. Now interpret your results!\n\n", StyleBox["Exercise 2", FontSize->16, FontWeight->"Bold"], "\n\nRepeat Exercise 1 for the following function.\n\n", Cell[BoxData[ RowBox[{\(g \((x, y)\)\), " ", "=", " ", TagBox[ RowBox[{ TagBox[ StyleBox[ RowBox[{"{", StyleBox[GridBox[{ { RowBox[{ RowBox[{ StyleBox[\(\(x\^4\ y\^2\)\/\((x\^4 + y\^2)\)\^2\ \), FontSize->18], ",", " ", RowBox[{ RowBox[{ StyleBox["if", FontFamily->"Times"], " ", \((x, y)\)}], "\[NotEqual]", \((0, 0)\)}]}], "\[IndentingNewLine]"}]}, { RowBox[{" ", RowBox[{"0", ",", StyleBox[ RowBox[{" ", StyleBox[" ", FontFamily->"Times"]}]], RowBox[{ RowBox[{ StyleBox["if", FontFamily->"Times"], StyleBox[" ", FontFamily->"Times"], \((x, y)\)}], "=", \((0, 0)\)}]}]}]} }], ShowAutoStyles->True]}], ShowAutoStyles->False], (#&)], "\[IndentingNewLine]"}], (#&)]}]], "Text", CellFrame->False, SpanMaxSize->Infinity, FontSize->16], "\n\nYou can use the following commands to see a picture of its graph.\n\n\ ", Cell[BoxData[{ \(\(Clear[g, x, y];\)\), "\[IndentingNewLine]", \(g[x_, y_] = \((x^4\ *y^2)\)/\((x^4 + y^2)\)^2\), "\n", \(\(graph\ = \ Plot3D[g[x, y], {x, \(-1\), 1}, {y, \(-1\), 1}, AxesLabel \[Rule] {"\", "\", "\"}, PlotPoints \[Rule] 50];\)\), "\n", \(ShowLive[graph]\)}], "Input"] }], "Text", CellFrame->True, Background->RGBColor[0.996078, 0.627451, 0.627451]] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Differentiability", FontSize->16]], "Section"], Cell[TextData[{ "You've already been told (or will be told shortly) in lecture that the \ derivative of a multivariable function ", Cell[BoxData[ FormBox[ StyleBox[\(f : \[DoubleStruckCapitalR]\^n \[Rule] \ \ \[DoubleStruckCapitalR]\^m\), FontSize->14], TraditionalForm]]], " is a ", StyleBox["matrix", FontSlant->"Italic"], ", sometimes called ", Cell[BoxData[ \(TraditionalForm\`D\ f\)]], " (for \"Derivative matrix of ", Cell[BoxData[ \(TraditionalForm\`f\)]], "), and sometimes called ", Cell[BoxData[ \(TraditionalForm\`J\ f\)]], " (for \"Jacobian of ", Cell[BoxData[ \(TraditionalForm\`f\)]], "\"). Both ", StyleBox["Df", FontSlant->"Italic"], " and ", StyleBox["Jf", FontSlant->"Italic"], " mean the same thing: the matrix which contains all of the partial \ derivatives of the function:" }], "Text"], Cell[BoxData[ RowBox[{"(", "\[NegativeThinSpace]", GridBox[{ {\(\[PartialD]f\_1\/\[PartialD]x\_1\), \(\[PartialD]f\_1\/\ \[PartialD]x\_2\), "\[CenterEllipsis]", \(\[PartialD]f\_1\/\[PartialD]x\_n\)}, {\(\[PartialD]f\_2\/\[PartialD]x\_1\), \(\[PartialD]f\_2\/\ \[PartialD]x\_12\), "\[CenterEllipsis]", \(\[PartialD]f\_2\/\[PartialD]x\_n\)}, {"\[VerticalEllipsis]", "\[VerticalEllipsis]", "\[DescendingEllipsis]", "\[VerticalEllipsis]"}, {\(\[PartialD]f\_m\/\[PartialD]x\_1\), \(\[PartialD]f\_m\/\ \[PartialD]x\_2\), "\[CenterEllipsis]", \(\[PartialD]f\_m\/\[PartialD]x\_n\)} }], "\[NegativeThinSpace]", ")"}]], "DisplayFormula", TextAlignment->Center, FontSize->24], Cell[TextData[{ "Once we've found this matrix, we can evaluate it at a point ", StyleBox["x=a", FontWeight->"Bold", FontSlant->"Italic"], " by plugging in numbers for x, y, or whatever other variables we have.", " We write this as ", StyleBox["Df(", FontSlant->"Italic"], StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[")", FontSlant->"Italic"], ". ", "The technical definition of differentiability says that a function is \ differentiable at a point ", StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], " if" }], "Text"], Cell[BoxData[ RowBox[{" ", RowBox[{ RowBox[{ UnderscriptBox["lim", RowBox[{ StyleBox["x", FontWeight->"Bold"], " ", "\[Rule]", " ", StyleBox["a", FontWeight->"Bold"]}]], FractionBox[ RowBox[{"||", RowBox[{ RowBox[{"f", RowBox[{"(", StyleBox["x", FontWeight->"Bold"], ")"}]}], " ", "-", " ", RowBox[{"f", RowBox[{"(", StyleBox["a", FontWeight->"Bold"], ")"}]}], " ", "-", " ", RowBox[{"D", " ", "f", RowBox[{"(", StyleBox["a", FontWeight->"Bold"], ")"}], " ", RowBox[{"(", RowBox[{ StyleBox["x", FontWeight->"Bold"], "-", StyleBox["a", FontWeight->"Bold"]}], ")"}]}]}], "||"}], RowBox[{"||", RowBox[{ StyleBox["x", FontWeight->"Bold"], "-", StyleBox["a", FontWeight->"Bold"]}], "||"}]]}], " ", "=", " ", "0"}]}]], "DisplayFormula", TextAlignment->Center, FontFamily->"Times", FontSize->16], Cell[TextData[{ "Yikes! While this definition is correct, it's hard to work with, and it's \ not at all clear what it means. Let's take a step backwards and figure out \ what it means for a function to be differentiable.\n\nThink back to single \ variable calculus. You know the limit definition of a derivative in the \ single variable case, but intuitively ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " is differentiable at ", Cell[BoxData[ \(TraditionalForm\`x = a\)]], " if the graph has a tangent line there, and the derivative is that slope. \ Another way of thinking about this is that the graph of ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " looks linear (\"like a line\") near ", Cell[BoxData[ \(TraditionalForm\`x = a\)]], ". Evaluate the next cell to see the graph of ", Cell[BoxData[ \(TraditionalForm\`y = x\^2\)]], "along with its tangent line for ", Cell[BoxData[ \(TraditionalForm\`x = 1\)]], ". (You haven't seen this before, but you can give ", StyleBox["Plot", FontWeight->"Bold"], " more than one function at a time, as long as you surround them with curly \ brackets.)" }], "Text"], Cell[BoxData[ \(Plot[{x^2, 2 x - 1}, {x, \(-1\), 3}]\)], "Input"], Cell[TextData[{ "Now go back and change the range for x to 0.5 and 1.5. Then .9 and 1.1, \ or even .999 and 1.001. Notice how the more you \"zoom in,\" the harder it \ gets to distinguish between the two graphs. In other words, the graph of ", Cell[BoxData[ \(TraditionalForm\`y = x\^2\)]], "looks linear when you zoom in far enough. This is also why the tangent \ line is called the ", StyleBox["linear approximation", FontSlant->"Italic"], " of the graph at ", Cell[BoxData[ \(TraditionalForm\`x = 1\)]], "; it's the line which most closely approximates the graph there.\n\nSo, a \ single variable function ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " is differentiable at ", Cell[BoxData[ \(TraditionalForm\`x = a\)]], " if it has a linear approximation there; ", Cell[BoxData[ \(TraditionalForm\`f' \((a)\)\)]], "is the slope of that line.\n\nHere's the quintessential function from \ single variable calculus which isn't differentiable at ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], ":" }], "Text"], Cell[BoxData[ \(Plot[Abs[x], {x, \(-1\), 1}]\)], "Input"], Cell[TextData[{ "The graph of ", Cell[BoxData[ \(TraditionalForm\`y = \(\(|\)\(x\)\(|\)\)\)]], " does ", StyleBox["not", FontSlant->"Italic"], " look like a line at ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], ". In fact it has a very sharp corner there, and no matter how far you \ zoom in --- try a range of -0.00001 to 0.00001 for x! --- you can't make the \ graph straighten out and look linear.\n\nThe same idea works for \ multivariable functions. If you zoom in far enough on a point on a graph, it \ should either look linear (which in higher dimensions could be replaced with \ \"planar\") or not. If it's linear when you zoom in, then you can find a \ linear approximation there. Make sense? \n\nSo, to draw a direct analog to \ the single variable case, a multivariable function is differentiable at ", Cell[BoxData[ \(TraditionalForm\`x\&\[RightVector] = a\&\[RightVector]\)]], "if it has a linear approximation -- i.e. a tangent plane -- there.\n\nThis \ can be very tricky to figure out! Evaluate the following cells and decide if \ you think they look linear at the origin. (Remember, you can zoom in/out \ with LiveGraphics3D by holding down the shift key, clicking on the picture, \ and dragging the mouse up and down.)" }], "Text"], Cell[TextData[{ "Note that the section function, ", Cell[BoxData[ \(TraditionalForm\`\((x\^3 - y\^3)\)/\((x\^2 + y\^2)\)\)]], ", isn't defined at (0,0), but you can \"plug the hole\" by defining the \ function to be equal to zero there. This makes it a continuous function, \ which you can assume without checking." }], "Text", CellFrame->True, Background->GrayLevel[0.833326]], Cell[BoxData[ \(\(surf1 = Plot3DLive[ Exp[\(-\((x^2 + y^2)\)\)] - 1, {x, \(-1\), 1}, {y, \(-1\), 1}, AxesLabel \[Rule] {"\", "\", "\"}, PlotPoints \[Rule] 40];\)\)], "Input"], Cell[BoxData[ \(\(surf2 = Plot3DLive[\((x^3 - y^3)\)/\((x^2 + y^2)\), {x, \(-1\), 1}, {y, \(-1\), 1}, AxesLabel \[Rule] {"\", "\", "\"}, PlotPoints \[Rule] 40];\)\)], "Input"], Cell[TextData[{ "It turns out that the first one ", StyleBox["is", FontSlant->"Italic"], " differentiable, and the second one ", StyleBox["isn't", FontSlant->"Italic"], ". That's hard to see from the second picture, but notice that there are \ \"ridges\" and \"valleys\" leaving the origin along the coordinate axes; the \ particular way that some of them go up, while some go down, make it \ impossible to approximate the graph with a tangent plane. The only real \ possibility is the plane" }], "Text"], Cell[BoxData[ \(z = x - y\)], "DisplayFormula", TextAlignment->Center, FontSize->14], Cell["\<\ (You're asked to show this in the exercises.) But look at the \ following picture of the graph and the plane together. Rotate and zoom \ in/out as needed. See if you can believe that the plane isn't actually \ tangent to the surface at (0,0). Basically, the surface has a few \"kinks\" \ at (0,0) which prevent the plane from being tangent.\ \>", "Text"], Cell[BoxData[{ \(plane2 = Plot3D[x - y, {x, \(-1\), 1}, {y, \(-1\), 1}, PlotPoints \[Rule] 40]\), "\[IndentingNewLine]", \(\(ShowLive[surf2, plane2, BoxRatios \[Rule] {1, 1, 2}];\)\)}], "Input"], Cell["\<\ Just for comparison, here's the first surface, along with its \ tangent plane.\ \>", "Text"], Cell[BoxData[{ \(plane1 = Plot3D[0, {x, \(-1\), 1}, {y, \(-1\), 1}, PlotPoints \[Rule] 40]\), "\[IndentingNewLine]", \(\(ShowLive[surf1, plane1];\)\)}], "Input"], Cell["\<\ You can see that the linear approximation isn't very good once you \ move away from the origin, but the surface certainly doesn't have any kinks \ in it, or \"jab through\" the plane near the origin. So, back to the $64,000 question: why does the big, ugly derivative matrix \ (or \"Jacobian\" matrix, which is the same thing) matter? As it turns out, \ it matters because you use it to find the linear approximation --- or \ \"tangent plane\" --- for your function. If you'd like to understand the technical definition a bit more, expand the \ next subsection; otherwise go on to the summary of what differentiable \ functions are.\ \>", "Text"], Cell[CellGroupData[{ Cell["\<\ (Expand this Subsection to read an explanation of the technical \ definition)\ \>", "Subsection"], Cell[TextData[{ "Here again, for your reading pleasure, is the technical definition of \ differentiability: a function is differentiable at a point ", StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], " if there's a linear transformation ", StyleBox["T", FontSlant->"Italic"], " such that" }], "Text"], Cell[BoxData[ RowBox[{" ", RowBox[{ RowBox[{ UnderscriptBox["lim", RowBox[{ StyleBox["x", FontWeight->"Bold"], " ", "\[Rule]", " ", StyleBox["a", FontWeight->"Bold"]}]], FractionBox[ RowBox[{"||", RowBox[{ RowBox[{"f", RowBox[{"(", StyleBox["x", FontWeight->"Bold"], ")"}]}], " ", "-", " ", RowBox[{"f", RowBox[{"(", StyleBox["a", FontWeight->"Bold"], ")"}]}], " ", "-", " ", RowBox[{"T", RowBox[{"(", RowBox[{ StyleBox["x", FontWeight->"Bold"], "-", StyleBox["a", FontWeight->"Bold"]}], ")"}]}]}], "||"}], RowBox[{"||", RowBox[{ StyleBox["x", FontWeight->"Bold"], "-", StyleBox["a", FontWeight->"Bold"]}], "||"}]]}], " ", "=", " ", "0"}]}]], "DisplayFormula", TextAlignment->Center, FontFamily->"Times", FontSize->16], Cell[TextData[{ "(The double bars in that equation mean ``length of a vector'.') Let's \ think about what that equation means: the limit is zero, so as ", StyleBox["x", FontWeight->"Bold"], " approaches ", StyleBox["a", FontWeight->"Bold"], ", the numerator becomes much smaller than the denominator. The denominator \ is just the distance from ", StyleBox["x", FontWeight->"Bold"], " to ", StyleBox["a", FontWeight->"Bold"], ". Since T is linear, T(x-a) = T(x) - T(a), and the part of the numerator \ inside the bars can be written in a somewhat more illuminating way:" }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"f", RowBox[{"(", StyleBox["x", FontWeight->"Bold"], ")"}]}], " ", "-", " ", RowBox[{"f", RowBox[{"(", StyleBox["a", FontWeight->"Bold"], ")"}]}], " ", "-", " ", RowBox[{"(", RowBox[{ RowBox[{"T", RowBox[{"(", StyleBox["x", FontWeight->"Bold"], ")"}]}], " ", "-", " ", RowBox[{"T", RowBox[{"(", StyleBox["a", FontWeight->"Bold"], ")"}]}]}], ")"}]}], " ", "=", " ", RowBox[{ RowBox[{"f", RowBox[{"(", StyleBox["x", FontWeight->"Bold"], ")"}]}], " ", "-", " ", RowBox[{"f", RowBox[{"(", StyleBox["a", FontWeight->"Bold"], ")"}]}], " ", "-", " ", RowBox[{"T", RowBox[{"(", StyleBox["x", FontWeight->"Bold"], ")"}]}], " ", "+", " ", RowBox[{"T", RowBox[{ RowBox[{"(", StyleBox["a", FontWeight->"Bold"], ")"}], "."}]}]}]}]], "DisplayFormula", FontFamily->"Times"], Cell[TextData[{ "But wait...T is a tangent plane (or ``tangent 3-space'', etc) so ", StyleBox["f(", FontSlant->"Italic"], StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[")", FontSlant->"Italic"], " must equal ", StyleBox["T(", FontSlant->"Italic"], StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], StyleBox[")", FontSlant->"Italic"], ", and in the above equation they'll cancel. So the numerator becomes even \ simpler\[LongDash]" }], "Text"], Cell[BoxData[ RowBox[{ StyleBox[ RowBox[{"f", RowBox[{"(", StyleBox["x", FontWeight->"Bold"], ")"}]}], FontFamily->"Times"], StyleBox[" ", FontFamily->"Times"], StyleBox["-", FontFamily->"Times"], StyleBox[" ", FontFamily->"Times"], RowBox[{ StyleBox["T", FontFamily->"Times"], RowBox[{ StyleBox["(", FontFamily->"Times"], StyleBox["x", FontFamily->"Times", FontWeight->"Bold"], StyleBox[")", FontFamily->"Times"]}], StyleBox[" ", FontFamily->"Times"]}]}]], "DisplayFormula"], Cell["\<\ \[LongDash]and we can rewrite the above limit as\ \>", "Text"], Cell[BoxData[ RowBox[{" ", RowBox[{ RowBox[{ UnderscriptBox["lim", RowBox[{ StyleBox["x", FontWeight->"Bold"], " ", "\[Rule]", " ", StyleBox["a", FontWeight->"Bold"]}]], FractionBox[ RowBox[{"|", RowBox[{ RowBox[{"f", RowBox[{"(", StyleBox["x", FontWeight->"Bold"], ")"}]}], " ", "-", " ", RowBox[{"T", RowBox[{"(", StyleBox["x", FontWeight->"Bold"], ")"}]}]}], "|"}], RowBox[{"|", RowBox[{ StyleBox["x", FontWeight->"Bold"], "-", StyleBox["a", FontWeight->"Bold"]}], "|"}]]}], " ", "=", " ", "0."}]}]], "DisplayFormula", FontFamily->"Times", FontSize->16], Cell[TextData[{ "If that limit is zero, then as ", StyleBox["x", FontWeight->"Bold"], " approaches ", StyleBox["a", FontWeight->"Bold"], " (along any path, of course) the numerator is much smaller than the \ denominator, which means that the linear transformation approximates f(", StyleBox["x", FontWeight->"Bold"], ") very well near ", StyleBox["a", FontWeight->"Bold"], ". You might say that in a very small neighborhood of ", StyleBox["a", FontWeight->"Bold"], ", f(x) is ``just about'' linear. This is what we were trying to visualize \ earlier in this section.\n\nThis definition of differentiability is, \ unfortunately, not very practical. If we're interested in deciding whether a \ function is differentiable or not (and also in finding the particular linear \ transformation that approximates it), we have to use a test which will be \ described below." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Summary of what \"differentiable\" means", "Subsection"], Cell[TextData[{ "To recap what we've gone over (and add a few more details from lecture \ that you can look up in your notes or your book):\n\n", StyleBox["Single Variable Calculus", FontWeight->"Bold"], "\n\n", "A function ", Cell[BoxData[ FormBox[ StyleBox[\(f : \[DoubleStruckCapitalR]\^1 \[Rule] \ \ \[DoubleStruckCapitalR]\^1\), FontSize->14], TraditionalForm]]], "is differentiable at ", Cell[BoxData[ \(TraditionalForm\`x = a\)]], " if has linear approximation (that is, a tangent line) at that point. \ Moreover, the derivative ", Cell[BoxData[ \(TraditionalForm\`f' \((a)\)\)]], " is the slope of the tangent line. A little less intuitively, the \ derivative ", Cell[BoxData[ \(TraditionalForm\`f' \((a)\)\)]], " is the number which makes the following equation be the linear \ approximation to ", Cell[BoxData[ \(TraditionalForm\`f\)]], " at ", Cell[BoxData[ \(TraditionalForm\`x = a\)]], "." }], "Text"], Cell[BoxData[ \(L \((x)\) = f \((a)\) + f' \((a)\) \((x - a)\)\)], "DisplayFormula", TextAlignment->Center, FontSize->16], Cell[TextData[{ "We know that some things, such as polynomials and trig functions, are \ differentiable. If you're ever in doubt about a given function, you can \ always check it using the limit definition you learned in Calc I.\n\n", StyleBox["Multivariable Calculus", FontWeight->"Bold"], "\n\nA function ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["f", FontSize->14, FontWeight->"Bold"], StyleBox[":", FontSize->14], StyleBox[\(\[DoubleStruckCapitalR]\^n\), FontSize->14]}], StyleBox["\[Rule]", FontSize->14], StyleBox[" ", FontSize->14], StyleBox[\(\[DoubleStruckCapitalR]\^m\), FontSize->14]}], TraditionalForm]]], "is differentiable at ", Cell[BoxData[ \(TraditionalForm\`x\&\[RightVector] = a\&\[RightVector]\)]], " if has linear approximation (that is, a tangent ", StyleBox["plane", FontSlant->"Italic"], ") at that point. Moreover, the derivative ", StyleBox["Df(", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm\`a\&\[RightVector]\)]], StyleBox[")", FontSlant->"Italic"], " is a matrix which, in some sense, tells you the \"slope\" of the tangent \ plane. More accurately, if a little less intuitively, the derivative ", Cell[BoxData[ FormBox[ StyleBox[\(Df(a\&\[RightVector])\), FontSlant->"Italic"], TraditionalForm]]], " is the matrix which makes the following equation be the linear \ approximation to ", Cell[BoxData[ \(TraditionalForm\`f\)]], " at ", Cell[BoxData[ \(TraditionalForm\`x = a\)]], "." }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{ StyleBox["L", FontWeight->"Bold"], \((x\&\[RightVector])\)}], "=", RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], \((a\&\[RightVector])\)}], "+", \(D\ f \((a\&\[RightVector])\) \((x\&\[RightVector] - a\&\[RightVector])\)\)}]}]], "DisplayFormula", TextAlignment->Center, FontSize->16], Cell["\<\ How can you tell if a multivariable function is differentiable? We \ have the following handy-dandy (if somewhat complicated) theorem.\ \>", "Text"], Cell[TextData[{ StyleBox["Differentiability Theorem", FontSlant->"Italic"], ": ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["f", FontSize->14, FontWeight->"Bold"], StyleBox[":", FontSize->14], StyleBox[\(\[DoubleStruckCapitalR]\^n\), FontSize->14]}], StyleBox["\[Rule]", FontSize->14], StyleBox[" ", FontSize->14], StyleBox[\(\[DoubleStruckCapitalR]\^m\), FontSize->14]}], TraditionalForm]]], " is differentiable at a point ", StyleBox["x=a", FontWeight->"Bold", FontSlant->"Italic"], " if each of the entries of the matrix ", Cell[BoxData[ \(TraditionalForm\`D\ \(f(a)\)\)]], " exists at ", StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], " and is continuous at ", StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], "; in other words, all of the partial derivatives which make up ", Cell[BoxData[ \(TraditionalForm\`D\ \(f(a)\)\)]], " exist and are continuous functions at ", StyleBox["a", FontWeight->"Bold", FontSlant->"Italic"], ". " }], "Text", CellFrame->True, FontSize->14], Cell["\<\ The important feature of this theorem is that it requires the \ partial derivatives to be continuous, and not to merely be defined at some \ point. That doesn't sound too bad, but you know from the first section of \ this lab that it can be tricky just to tell if a given function (i.e. one of \ the partial derivatives) is continuous or not!\ \>", "Text"], Cell[TextData[{ StyleBox["Exercise 3", FontSize->16, FontWeight->"Bold"], "\n\nConsider the following function, which we examined above. You may \ wish to go back and look at the graph and the \"alleged\" tangent plane at \ (0,0), which we've already told you isn't really a tangent plane. In this \ exercise you're going to prove that.\n\n", Cell[BoxData[ RowBox[{\(f \((x, y)\)\), " ", "=", StyleBox[" ", SpanMaxSize->Infinity], RowBox[{ StyleBox[ TagBox[ StyleBox["{", ShowAutoStyles->False], (#&)], SpanMaxSize->Infinity], GridBox[{ {\(\(x\^3\ - \ y\^3\)\/\(x\^2\ + \ y\^2\), \ \(if\ \((x, y)\)\ \[NotEqual] \ \((0, 0)\);\)\)}, {\(0, \ \ \ \ \ \ \ \ \ \ \ \ if\ \((x, y)\)\ = \ \(\((0, 0)\)\(.\)\)\)} }]}]}]], FontFamily->"Times", FontSize->16], "\n\nYou may assume that ", Cell[BoxData[ \(TraditionalForm\`f(x, y)\)]], " is a continuous function without proving it.\n \n(a) Find the partial \ derivatives of ", Cell[BoxData[ \(TraditionalForm\`f\)]], " at the origin. Don't use ", StyleBox["Mathematica", FontSlant->"Italic"], " to do this; do it by hand. Remember that ", Cell[BoxData[ \(TraditionalForm\`f(0, 0) = 0\)]], ".\n\nNow let's see why ", Cell[BoxData[ \(TraditionalForm\`f\)]], " is not differentiable at the origin.\n\n(b) Earlier you were told that if \ there were a tangent plane at the origin, it would have to be the plane ", Cell[BoxData[ \(TraditionalForm\`z = x - y\)]], ". Prove this. Remember that the equation for the tangent plane -- that \ is, the linear approximation -- looks like this:\n\n\t\t\t\t\t\t", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["L", FontWeight->"Bold"], \((x\&\[RightVector])\)}], "=", RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], \((a\&\[RightVector])\)}], "+", \(D\ \(f( a\&\[RightVector])\) \((x\&\[RightVector] - a\&\[RightVector])\)\)}]}], TraditionalForm]]], "\n\nIn particular, if we have a function of two variables, the equation \ for the linear approximation at a point (a,b) becomes\n\n\t\t\t\t\t\t", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ StyleBox["L", FontWeight->"Bold"], "(", \(x, y\), ")"}], "=", RowBox[{ RowBox[{ StyleBox["f", FontWeight->"Bold"], "(", \(a, b\), ")"}], "+", \(D\ \(f(a, b)\)\[CenterDot]\((x - a, y - b)\)\)}]}], TraditionalForm]]], "\n\n(c) Is the plane tangent to the graph when restricted to the line ", StyleBox["x = ", FontSlant->"Italic"], "0? ", StyleBox["y", FontSlant->"Italic"], " = 0? ", StyleBox["y = -x", FontSlant->"Italic"], "? Check this by making two-dimensional plots of L and ", Cell[BoxData[ \(TraditionalForm\`f\)]], " when restricted to those lines. Note that tangency is somewhat like \ continuity: if there's some line along which L fails to be tangent to ", Cell[BoxData[ \(TraditionalForm\`f\)]], ", then L (in all of 3-space) is not tangent to ", Cell[BoxData[ \(TraditionalForm\`f\)]], ".\n\n(d) Now you've shown that L is not tangent to ", Cell[BoxData[ \(TraditionalForm\`f\)]], ". But L is the only possible linear approximation for ", Cell[BoxData[ \(TraditionalForm\`f\)]], " at the origin; hence there ", StyleBox["is", FontSlant->"Italic"], " no linear approximation for ", Cell[BoxData[ \(TraditionalForm\`f\)]], " there. In other words, ", Cell[BoxData[ \(TraditionalForm\`f\)]], " is not differentiable at the origin. Explain why this does not \ contradict the \"Differentiability Theorem\" above.\n\nNote that we've split \ this problem up into pieces to help you work through it, but your written \ solution should not be split up in this way; it should be a seamless essay \ which investigates this function and hits the important points mentioned \ here.\n\n", StyleBox["Exercise 4", FontSize->16, FontWeight->"Bold"], "\n\nConsider the following function:\n\n", Cell[BoxData[ RowBox[{\(f \((x, y)\)\), " ", "=", StyleBox[" ", SpanMaxSize->Infinity], RowBox[{ StyleBox[ TagBox[ StyleBox["{", ShowAutoStyles->False], (#&)], SpanMaxSize->Infinity], GridBox[{ {\(1\/2\ y\ Log[ x\^2 + y\^2], \ \ \ \ \ \ \(if\ \((x, y)\)\ \[NotEqual] \ \((0, 0)\);\)\)}, {\(0, \ \ \ \ \ \ \ \ \ \ \ \ \ if\ \((x, y)\)\ = \ \(\((0, 0)\)\(.\)\)\)} }]}]}]], FontFamily->"Times", FontSize->16], "\n\nYou may assume that this is a continuous function (although that's not \ easy to show!) Prove that ", Cell[BoxData[ \(TraditionalForm\`f\)]], " does not satisfy the conditions of the \"Differentiability Theorem\" \ above at the point (0,0)." }], "Text", CellFrame->True, Background->RGBColor[1, 0.501961, 0.501961]], Cell[CellGroupData[{ Cell["Credits", "Subsubsection"], Cell[TextData[{ "This lab has a long history. Originally we had a lab 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"}]]]], " by Cindy Kaus which described all of the pathological examples of things \ that can go wrong -- everything from discontinuous functions to \ differentiable functions whose mixed partials are not equal. Students were \ asked to prove these facts about the functions and were given some guidance \ about how to do it in ", StyleBox["Mathematica", FontSlant->"Italic"], ". In Spring 2002 Dan Drake completely rewrote the entire thing as a ", StyleBox["Mathematica", FontSlant->"Italic"], " notebook, which had two advantages for us: (1) consistency, because now \ all of the labs were in notebook form, and (2) he could include graphing \ commands and other pictures to help make his point. Dan wrote pages and \ pages of terrific explanations which helped a lot; the only overlap with \ Cindy's original lab were the four exercises, although he restated split \ those into steps to help students along.\n\nIt was still a very hard \ assignment. People weren't learning the subtle details in the labs, and we \ weren't using the really pathological examples later on in the course anyway. \ In January 2004 I revamped the lab to be more interactive and \ example-driven. The exercises have been scaled back and are based on the \ examples. Hopefully this version will go over well.\n\nEverything up to \ example 1 was written by Dan, as well as the excellent explanation of the \ \"technical definition\" of differentiability. Example 3 is an expansion of \ an example he added to the lab, and exercise 3 is a slight reworking of ", StyleBox["his", FontSlant->"Italic"], " version of one of Cindy's exercises. There's not much else left of \ Cindy's original lab, I guess. Example 3 is based on her \"Example (i).\"\n\n\ This version is copyright 2004 by Jonathan Rogness (rogness@math.umn.edu), \ with the exception of those parts described above, which are copying 2002 by \ Dan Drake and 2000 by Cindy Kaus. All of us have 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!" }], "Text"] }, Closed]] }, Open ]] }, Closed]] }, FrontEndVersion->"5.0 for X", ScreenRectangle->{{0, 1280}, {0, 1024}}, ScreenStyleEnvironment->"Working", PrintingStyleEnvironment->"Printout", WindowToolbars->{}, WindowSize->{754, 744}, WindowMargins->{{Automatic, 109}, {Automatic, 29}}, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"PrintingMargins"->{{54, 54}, {72, 72}}, "PaperSize"->{612, 792}, "PaperOrientation"->"Portrait", "PrintCellBrackets"->False, "PrintRegistrationMarks"->True, "PrintMultipleHorizontalPages"->False, "PostScriptOutputFile":>"", "Magnification"->1}, SpanMaxSize->Infinity, Magnification->1 ] (******************************************************************* Cached data follows. 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