(************** 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[ 51014, 1280]*) (*NotebookOutlinePosition[ 52039, 1312]*) (* CellTagsIndexPosition[ 51995, 1308]*) (*WindowFrame->Normal*) Notebook[{ Cell[TextData[{ StyleBox["Lab 3C - Tangent Planes", FontSize->24, FontWeight->"Bold", FontVariations->{"Underline"->True}], "\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[CellGroupData[{ Cell[TextData[StyleBox["Introduction", FontSize->16]], "Section"], Cell[TextData[{ "In the last few labs, we've investigated properties of continuity, \ differentiability, partial derivatives, directional derivatives, and \ gradients. In this lab, all of the functions we consider will be \ differentiable, which as we learned means that we can approximate the \ function at any given point with a linear approximation, or \"tangent \ plane.\"\n\nThere's an important point here: earlier we talked about whether \ a function is differentiable at the point ", Cell[BoxData[ \(TraditionalForm\`x\&\[RightVector] = a\&\[RightVector]\)]], "; if we say that a function is \"differentiable\" without any modifier, we \ mean that it is differentiable at ", StyleBox["every", FontSlant->"Italic"], " point, or at least every point where it is defined. Go to the following \ web page:\n\nhttp://www.math.umn.edu/~rogness/multivar/tanplane.shtml\n\n(Or, \ alternatively, expand the subsection below labeled \"Tangent Plane Example\" \ and evaluate the cell there.) This demo shows you part of a paraboloid with \ a tangent plane at a certain point; you can click and drag the point to move \ it around, and LiveGraphics3D will automatically show you the tangent plane \ at the new point.\n\nThis is intended to reinforce the point that a \ \"differentiable\" function doesn't have one single tangent plane (or linear \ approximation, if you prefer that language). Rather, it has a different \ linear approximation at each point!\n\nToday we'll learn various methods of \ finding the equation of the tangent plane / linear approximation of a \ differentiable function. In some sense this is one of the most important \ things you can learn in this course, and all of the material for the first \ month of the class has led to this point. In particular, we'll use a lot of \ the ideas from the previous two labs." }], "Text"], Cell[CellGroupData[{ Cell["Tangent Plane Example", "Subsection"], Cell["\<\ As with the code in the previous example, you don't have to worry \ about understanding the commands here. It's just intended for people who \ would like to view the example without opening up a web browser. This is a slightly modified version of an example on the LiveGraphics3D \ website, so I make no claim of originality.\ \>", "Text", CellFrame->True, Background->GrayLevel[0.833326]], Cell[BoxData[ \(\(\( (*\ plane, \ paraboloid, \ point, \ vector\ *) \)\(\[IndentingNewLine]\)\(\(independentVariables = {x \ \[Rule] .5, y \[Rule] .5};\)\n \(dependentVariables = {x \[Rule] If[x < \(-1\), \(-1\), If[x > 1, 1, x]], y \[Rule] If[y < \(-1\), \(-1\), If[y > 1, 1, y]], z \[Rule] x*x + y*y};\)\n minx = \(-1\); miny = \(-1\); maxx = 1; maxy = 1;\[IndentingNewLine] n = 10; dx = \((maxx - minx)\)/n; dy = \((maxy - miny)\)/n;\[IndentingNewLine] \(point = {PointSize[0.04], RGBColor[1, 0, 0], Point[{x, y, z}], Point[{1. *x, 1. *y, z + .2}], Point[{1. *x, 1. *y, z - .2}]};\)\[IndentingNewLine] \(paraboloid = Table[Polygon[{{i, j, i*i + j*j}, {i + dx, j, \((i + dx)\)*\((i + dx)\) + j*j}, {i + dx, j + dy, \((i + dx)\)*\((i + dx)\) + \((j + dy)\)*\((j + dy)\)}, {i, j + dy, i*i + \((j + dy)\)*\((j + dy)\)}}], {i, minx, maxx - dx/2, dx}, {j, miny, maxy - dy/2, dy}];\)\n \(plane = Table[Polygon[{{i, j, \((i - x)\)*2*x + \((j - y)\)*2*y + z}, {i + dx, j, \((i + dx - x)\)*2*x + \((j - y)\)*2*y + z}, {i + dx, j + dy, \((i + dx - x)\)*2*x + \((j + dy - y)\)*2*y + z}, {i, j + dy, \((i - x)\)*2*x + \((j + dy - y)\)*2*y + z}}], {i, minx, maxx - dx/2, dx}, {j, miny, maxy - dy/2, dy}];\)\[IndentingNewLine]\[IndentingNewLine] \(g = Graphics3D[{paraboloid, point, plane}, BoxRatios \[Rule] {1, 1, 1}, Axes \[Rule] True, AxesLabel \[Rule] {"\", "\", "\"}];\)\n\ \[IndentingNewLine] (*\ JLink\ loaded\ and\ Java\ installed\ in\ math2374 . nb\ *) \[IndentingNewLine] \(BeginJavaBlock[];\)\[IndentingNewLine] \(dirderApplet = JavaNew["\"];\)\n \(dirderFrame = JavaNew["\", dirderApplet, {"\" <> ToString[ InputForm[ N[g]]], "\", "\", \ "\", \ "\1,1,x]],y\[Rule]If[y<-1,-1,If[\ y>1,1,y]],z\[Rule]x*x+y*y}\>", "\", \ "\"}];\)\[IndentingNewLine] \(EndJavaBlock[];\)\[IndentingNewLine] \(Clear[g];\)\[IndentingNewLine] \)\)\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Plotting Planes", FontSize->16]], "Section"], Cell["\<\ Before we begin considering issues of tangency, let's review how to \ plot planes. It might help you to look in your textbook and review Cartesian \ and parametric equations for planes; you can also ask your TA for help. The nicest situation is when we can describe the plane as a function of x and \ y; in that case, we can write the equation of the plane as ``z = ax + by + \ c'' where a, b, and c are real numbers. We can just use Plot3D in those \ cases; for instance, if we had z = 3x - y, we could just use\ \>", "Text"], Cell[BoxData[ \(Plot3D[3 x\ - \ y, \ {x, \(-2\), 2}, {y, \(-2\), 2}]\)], "Input"], Cell["\<\ to plot the plane, adjusting the range of x and y if necessary. Even if we can describe a plane with a nice, easy equation, it is sometimes \ more convenient or enlightening to describe the plane with parametric \ equations. If we are given a point and two non-parallel vectors, there is a \ unique plane containing that point parallel to each of the vectors. Let's say \ our vectors are (2,3,5) and (-7,-11,13), and we wish to find the plane \ parallel to those two vectors passing through the point (17, 29, 0). In \ parametric equations, we could write\ \>", "Text"], Cell[BoxData[ \(\((17, 29, 0)\)\ + \ t\ \((2, 3, 5)\)\ + \ s \((\(-7\), \(-11\), 13)\)\)], "DisplayFormula"], Cell["\<\ where s and t range over all real numbers. To plot this plane, we \ use the properties of scalar multiplication and vector addition, and use \ ParametricPlot3D:\ \>", "Text"], Cell[BoxData[ \(ParametricPlot3D[{17 + 2*t - 7*s, \ 29 + 3*t - 11*\ s, \ 5*t + 13*s}, \ {t, \(-1\), 1}, {s, \(-1\), 1}]\)], "Input"], Cell[TextData[{ "You can keep the parametric equation of the plane in ``un-multiplied out'' \ form and use ParametricPlot3D, although earlier versions of ", StyleBox["Mathematica", FontSlant->"Italic"], " might complain a bit before showing you the graph. Either way works; \ mathematically they are exactly equivalent. The following form may be easier \ to write." }], "Text"], Cell[BoxData[ \(ParametricPlot3D[{17, \ 29, 0}\ + \ t*{2, 3, 5} + s*{\(-7\), \(-11\), 13}, \ {t, \(-1\), 1}, {s, \(-1\), 1}]\)], "Input"], Cell[TextData[{ "One important thing to note is that we've only plotted a small portion of \ the plane. Above we chose values of t and s between -1 and 1, but we can \ choose any range we like. The ranges on s and t don't have to be the same. \ You don't have to use s and t if you don't like; ", StyleBox["Mathematica", FontSlant->"Italic"], " will let you use pretty much any variable name you like. The world is \ your oyster.\n\nThere's one other point you should be aware of. In this \ class you've seen how difficult it can be sometimes to represent \ three-dimensional pictures on a two-dimensional computer screen or piece of \ paper. A picture of a plane can be particularly hard to decipher; it's \ sometimes difficult to tell which part is sloping uphill, which part is going \ downhill, or even whether a plane is horizontal or vertical. As an example, \ look at the previous picture and try to decide how the plane is situated in \ space. Then evaluate the following command, which will make the same picture \ pop up in an interactive window. Rotate the plane around and see how \ accurately you interpreted the two-dimensional picture." }], "Text"], Cell[BoxData[ \(ParametricPlot3D[{17, \ 29, 0}\ + \ t*{2, 3, 5} + s*{\(-7\), \(-11\), 13}, \ {t, \(-1\), 1}, {s, \(-1\), 1}]\)], "Input"], Cell[CellGroupData[{ Cell["Example", "Subsection"], Cell["Plot the plane that's described by", "Text"], Cell[BoxData[ \(s \((0, 0, 1)\)\ + \ t \((0.2, \ 0.2, \ 2)\)\)], "DisplayFormula"], Cell["\<\ when s and t range from 2 to 3. It won't look very \ good\[LongDash]it's ``too skinny''. Experiment with some different ranges on \ s and t until you get a nicer picture\[LongDash]one that's a little more \ ``square''.\ \>", "Text"], Cell[TextData[{ StyleBox["Exercise 1 - A Thought Experiment", FontSize->16, FontWeight->"Bold"], "\n\nAre there planes that are not the graph of a function of x and y? If \ so, which planes are they? If there are no such planes, explain why.\n\nAre \ there planes that ", StyleBox["cannot", FontSlant->"Italic"], " be described by a parametrization? If so, which planes are they? If there \ are no such planes (i.e., any plane whatsoever can be described by a \ parametrization), explain why.\n\nExplain your reasoning!" }], "Text", CellFrame->True, Background->RGBColor[1, 0.498039, 0.498039]] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Tangent Planes Via Tangent Vectors", FontSize->16]], "Section"], Cell[TextData[{ "We're going to start out this section with a crucial observation. Go back \ and think about the interpretation of a directional derivative -- it might \ help to reopen Lab 3B and run the example there, or go to:\n\n\ http://www.math.umn.edu/~rogness/multivar/dirderiv.shtml\n\nWe sliced through \ a surface with a vertical plane, and the intersection of the plane and \ surface gave us a ", StyleBox["cross-section", FontSlant->"Italic"], " of the surface. (This is represented by the curvy part of the blue line; \ the straight blue lines are just there to help you visualize the vertical \ plane that you're using as a meat cleaver.) The red arrow is part of the \ tangent line of the cross-section.\n\nHere's the crucial observation: because \ the arrow is tangent to a cross-section through that point, then the arrow \ must be tangent to the ", StyleBox["surface", FontSlant->"Italic"], " at that point! In other words, that arrow is actually part of the \ tangent plane at that point. As you rotate the cross-section, the arrow \ moves around a lot, so this might be hard to believe. There's another online \ example you can look at which might help convince you:\n\n\ http://www.math.umn.edu/~rogness/multivar/tanplane_withvectors.shtml\n\n\ (Alternatively, you can expand the subsection titled \"Interactive Example of \ Tangent Plane with Vectors\" down below and evaluate that cell to start the \ example from within ", StyleBox["Mathematica", FontSlant->"Italic"], ". In either case you should read on so that you know what you're looking \ at.)\n\nThe picture shows you the same surface as the previous directional \ derivative example, but now we've got ", StyleBox["two", FontSlant->"Italic"], " cross-sections of the surface, both of which go through the point ", Cell[BoxData[ \(TraditionalForm\`\((1, 0, 0)\)\)]], ". The red and purple vectors are tangent to the cross-sections at the \ point. The green vector is the cross product of the red and purple vectors.\n\ \nNote that we've also drawn in a portion of the plane tangent to the surface \ at the point (1,0,0). Rotate the cross-sections (using the slider on the \ bottom) and the entire picture (by clicking and dragging on the picture) \ until you're convinced that the tangent vectors are ", StyleBox["always", FontSlant->"Italic"], " contained in the tangent plane. (This isn't always immediately obvious \ because LiveGraphics3D sometimes has trouble drawing lines and planes that \ overlap.)" }], "Text"], Cell[CellGroupData[{ Cell["Interactive Example of Tangent Plane with Vectors", "Subsection"], Cell["\<\ As always with these special examples, you don't have to worry \ about understanding the code here. Just evaluate the next cell to start the \ example. Then collapse this subsection (so you don't have to look at the \ messy commands) and continue reading the lab.\ \>", "Text", CellFrame->True, Background->GrayLevel[0.833326]], Cell[BoxData[{ \(\(f[x_, y_] = 2 y*Exp[\(-x^2\) - y^2];\)\), "\[IndentingNewLine]", \(\(gf[x_, y_] = {D[f[x, y], x], D[f[x, y], y]};\)\ (*\ Gradient\ of\ f\ *) \), "\n", \(\(\(Lf[x_, y_] = f[1, 0] + gf[1, 0] . {x - 1, y};\)\(\[IndentingNewLine]\) \)\), "\n", \(\(\(SliderX[xmin_, xmax_, yval_, zval_, ptlabel_] := \[IndentingNewLine]{{RGBColor[1, 0, 0], Thickness[0.01], Line[{{xmin, yval, zval}, {xmax, yval, zval}}]}, {RGBColor[0, 0, 1], PointSize[0.04], Point[{x, yval, zval}]}, Text[ptlabel, {x, yval, zval* .8}]};\)\(\n\) \)\), "\n", \(\(n = 16;\)\), "\n", \(xmin = \(-1\); xmax = 3.15; ymin = \(-2\); ymax = 2;\), "\n", \(dx = \((xmax - xmin)\)/n; dy = \((ymax - ymin)\)/n;\), "\n", \(\(mesh = {GrayLevel[0.7], Table[Line[{{i, j, f[i, j]}, {i + dx, j, f[i + dx, j]}, {i + dx, j + dy, f[i + dx, j + dy]}, {i, j + dy, f[i, j + dy]}, {i, j, f[i, j]}}], {i, xmin, xmax, dx}, {j, ymin, ymax, dy}]};\)\), "\n", \(\(\(point = {RGBColor[0, 1, 0], PointSize[0.03], Point[{1, 0, 0}]};\)\(\[IndentingNewLine]\) \)\), "\n", \(\(n = 22;\)\), "\n", \(rmin = \(-2\); rmax = 2;\), "\n", \(\(dr = \((rmax - rmin)\)/n;\)\), "\n", \(\(dt = .9*Pi/2;\)\), "\[IndentingNewLine]", \(\(section1 = {RGBColor[0, 0, 1], Thickness[0.005], Table[Line[{{1 + \((r)\) Cos[t], \((r)\) Sin[t], 2 \((r)\) Sin[t] Exp[\(-1\) - r^2 - 2 r*Cos[t]]}, {1 + \((r + dr)\) Cos[t], \((r + dr)\) Sin[t], 2 \((r + dr)\) Sin[t] Exp[\(-1\) - \((r + dr)\)^2 - 2 \((r + dr)\)*Cos[t]]}}], {r, rmin, rmax - dr, dr}]};\)\), "\[IndentingNewLine]", \(\(section2 = {RGBColor[0, 0, 1], Thickness[0.005], Table[Line[{{1 + \((r)\) Cos[t + dt], \((r)\) Sin[t + dt], 2 \((r)\) Sin[t + dt] Exp[\(-1\) - r^2 - 2 r*Cos[t + dt]]}, {1 + \((r + dr)\) Cos[t + dt], \((r + dr)\) Sin[t + dt], 2 \((r + dr)\) Sin[t + dt] Exp[\(-1\) - \((r + dr)\)^2 - 2 \((r + dr)\)*Cos[t + dt]]}}], {r, rmin, rmax - dr, dr}]};\)\[IndentingNewLine]\[IndentingNewLine]\), "\ \[IndentingNewLine]", \(dx = .1; dy = .1;\), "\[IndentingNewLine]", \(\(\(tanplane = {RGBColor[ .9, .8, .8], Table[Polygon[{{i, j, Lf[i, j]}, {i + dx, j, Lf[i + dx, j]}, {i + dx, j + dy, Lf[i + dx, j + dy]}, {i, j + dy, Lf[i, j + dy]}}], {i, .5, 1.4, dx}, {j, \(- .5\), .4, dy}]};\)\(\n\) \)\), "\[IndentingNewLine]", \(\(vector1 = {RGBColor[1, 0, 0], Thickness[0.01], Arrow3D[{1, 0, 0}, {1 + 1 Cos[t], 1 Sin[t], 2 Sin[t]/E}]};\)\), "\n", \(\(vector2 = {RGBColor[1, 0, 1], Thickness[0.01], Arrow3D[{1, 0, 0}, {1 + 1 Cos[t + dt], 1 Sin[t + dt], 2 Sin[t + dt]/E}]};\)\), "\[IndentingNewLine]", \(\(normalvector = {RGBColor[0, 1, 0], Thickness[0.01], Arrow3D[{1, 0, 0}, {1, \(-2\)/E, 1}]};\)\), "\[IndentingNewLine]", \(\(g = Graphics3D[{mesh, point, vector1, vector2, tanplane, normalvector, section1, \ section2, SliderX[0, Pi, \(-3\), \(-1\), "\<\>"], Text["\", {Pi/2, \(-3\), \(-1.5\)}]}, Boxed \[Rule] False, PlotRange \[Rule] {{\(-1\), 3.15}, {\(-3\), 2}, {\(-1.5\), .8}}, ViewPoint \[Rule] {2, \(-3\), 1.5}, Lighting \[Rule] False];\)\n\[IndentingNewLine] (*\ JLink\ loaded\ and\ Java\ installed\ in\ math2374 . nb\ *) \), "\[IndentingNewLine]", \(\(BeginJavaBlock[];\)\), "\[IndentingNewLine]", \(\(dirderApplet = JavaNew["\"];\)\), "\n", \(\(dirderFrame = JavaNew["\", dirderApplet, {"\" <> ToString[ InputForm[ N[g]]], "\", "\", \ "\ .785}\>", "\ If[x < \ 0, 0, If[x > 3.15, 3.15, x]],t -> 2x}\>", "\", \ "\"}];\)\), "\[IndentingNewLine]", \(\(EndJavaBlock[];\)\), "\[IndentingNewLine]", \(\(Clear[g];\)\)}], "Input"], Cell[BoxData[""], "Input"], Cell[BoxData[""], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["\<\ Using Tangent Vectors to find Parametric Equations of Tangent \ Planes\ \>", "Subsection"], Cell[TextData[{ "So now you've seen that tangent vectors of cross sections through a point \ are in the tangent plane at that point. But wait -- all we need to write \ down the parametric equation of a plane is a point on the plane and two \ vectors in the plane! So, in this situation, we've got all the information \ we need. Given two cross sections of a surface which intersect at a point ", Cell[BoxData[ \(TraditionalForm\`x\&\[RightVector] = a\&\[RightVector]\)]], ", the equation of the tangent plane is:" }], "Text"], Cell[BoxData[ \(TraditionalForm\`p(s, t) = a\&\[RightVector] + s\[CenterDot]\((tan . \ vector\ of\ first\ cross\ section\ at\ the\ point\ a\&\ \[RightVector])\)\ + \ t\[CenterDot]\(\((tan . \ vector\ of\ second\ cross\ section\ at\ the\ point\ a\&\ \[RightVector])\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center], Cell["\<\ Let's do a real-life example now, by finding the plane tangent to \ the surface\ \>", "Text"], Cell[BoxData[ \(z\ = \ x\^2\ + \ 3 xy\ + \ y\^2\)], "DisplayFormula", TextAlignment->Center, FontSize->14], Cell[TextData[{ "at the point (1,-2,-1). First let's plot the surface; we'll use the ", StyleBox["Mesh", FontWeight->"Bold"], " option to turn off the grid, so that later on we can see our cross \ sections a little better." }], "Text"], Cell[BoxData[{ \(f[x_, y_]\ = \ x^2\ + \ 3\ x*y\ + \ y^2\), "\[IndentingNewLine]", \(surf\ = \ Plot3D[f[x, y], \ {x, 0, 2}, {y, \(-3\), \(-1\)}, \ Mesh \[Rule] False]\)}], "Input"], Cell[TextData[{ "For our first cross section, let's just slice the surface with the plane \ x=1. [Note that this ", StyleBox["will", FontSlant->"Italic"], " give us a cross section through the point (1,-2,-1). If you're not sure \ why this is true, as your TA.]\n\nWe'll use a parametric description for the \ curve that represents the restriction of f(x,y) to x=1. The x-value is always \ 1; we'll let y vary -- let's say y=t in our parametric equation -- and the \ z-value is described by ", Cell[BoxData[ \(TraditionalForm\`f(1, t)\)]], ":" }], "Text"], Cell[BoxData[{ \(c1[t_]\ = \ {1, \ t, \ f[1, t]}\), "\[IndentingNewLine]", \(one\ = ParametricPlot3D[ Evaluate[c1[t]], \ {t, \(-3.1\), \(-0.9\)}]\)}], "Input"], Cell[TextData[{ "For our second cross section, let's use the vertical plane y=-2. The \ parametric description of this cross section is very similar: we fix y=-2, \ and let x vary -- let's say x=s in our parametric equation, and the z-value \ is just ", Cell[BoxData[ \(TraditionalForm\`f(s, \(-2\))\)]], "." }], "Text"], Cell[BoxData[{ \(c2[s_]\ = \ {s, \ \(-2\), \ f[s, \ \(-2\)]}\), "\[IndentingNewLine]", \(two\ = ParametricPlot3D[Evaluate[c2[s]], \ {s, \(-0.1\), 2.1}]\)}], "Input"], Cell["\<\ Now we'll plot the surface and these curves together. Remember, we \ used \"Mesh -> False\" back in the Plot3D command to make it easier to see \ the curves.\ \>", "Text"], Cell[BoxData[ \(Show[surf, \ one, two]\)], "Input"], Cell[TextData[{ "Note how the paths run right through the surface. The nice thing about \ going to the trouble of using these paths is that", StyleBox[" the tangent vectors of the paths are also tangent to the \ surface", FontSlant->"Italic"], ". We can use those two tangent vectors to describe the plane that is \ tangent to the surface at the point where the curves intersect. It's easy to \ find the tangent vectors for the paths; we differentiate with respect to t \ (or s) and plug in the appropriate values of t or s. We get" }], "Text"], Cell[BoxData[{ \(dc1[t_] = D[c1[t], t]\), "\[IndentingNewLine]", \(dc2[s_] = D[c2[s], s]\)}], "Input"], Cell[TextData[{ "We're interested in the tangent vectors at the point (1,-2,-1), so we have \ to do a little bit of work here. For what value of ", Cell[BoxData[ \(TraditionalForm\`t\)]], " does ", Cell[BoxData[ \(TraditionalForm\`c1(t) = \((1, \(-2\), \(-1\))\)\)]], "? Well, we can solve that easily enough:" }], "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`c1(t) = \((1, \(-2\), \(-1\))\)\)]], "\n\n", Cell[BoxData[ \(TraditionalForm\`\((1, t, 1 + 3 t + t\^2)\) = \((1, \(-2\), \(-1\))\)\)]] }], "Text", TextAlignment->Center, FontSize->14], Cell["\<\ Look at the y-component; it's pretty clear that c1 hits the \ intersection point (1,-2,-1) when t=-2. You can check that t=-2 gives you \ the right z-value, too. That means our first tangent vector is:\ \>", "Text"], Cell[BoxData[ \(dc1[\(-2\)]\)], "Input"], Cell["\<\ Similarly, the point of intersection occurs at s = 1 for c2 (check \ this!), yielding a tangent vector of (1,0,-4):\ \>", "Text"], Cell[BoxData[ \(dc2[1]\)], "Input"], Cell[TextData[{ "Now we know everything we need to describe the plane: we know a point it \ passes through, namely ", Cell[BoxData[ \(TraditionalForm\`\((1, 2, f(1, \(-2\))\)\)]], "), and two vectors parallel to the plane. In parametric terms, the plane \ is" }], "Text"], Cell[BoxData[ \(p[s_, t_]\ = \ {1, \(-2\), f[1, \(-2\)]}\ + \ t*{0, 1, \(-1\)}\ + \ s*{1, 0, \(-4\)}\)], "Input"], Cell["\<\ Let's plot the plane and the surface together in a LiveGraphics3D \ pop-up window:\ \>", "Text"], Cell[BoxData[{ \(\(tanplane\ = ParametricPlot3D[ p[s, t], \ {s, \(-1\), 1}, \ {t, \(-1\), 1}];\)\), "\[IndentingNewLine]", \(\(Show[surf, \ tanplane];\)\), "\[IndentingNewLine]", \(\)}], "Input"], Cell[TextData[{ "Rotate the picture and see if you can convince yourself that this plane \ really is tangent to the surface. (We've talked about how appearances can be \ deceiving, but in this case your eyes aren't lying to you.) Note how the \ plane is above the surface at some points and below it at others; that's \ perfectly fine. It's analogous to the line tangent to the curve y = ", Cell[BoxData[ \(TraditionalForm\`sin\ x\)]], " at x = 0, which goes above and beneath the curve:" }], "Text"], Cell[BoxData[ \(Plot[{Sin[x], x}, {x, \(-Pi\)/2, Pi/2}, AspectRatio \[Rule] Automatic]\)], "Input"], Cell["\<\ Experience has taught us that the most confusing part of this \ process is where we set t=-2 and s=1. Before you go on, make sure you \ understand what we did there. There's nothing magical going on; we're just trying to find the values of s \ and t so that c1(t) and c2(s) equal the point we're interested in. In \ particular, s and t don't have to be equal; they also don't have to be equal \ to 1, or 0, or any other \"nice\" value.\ \>", "Text", CellFrame->True, Background->GrayLevel[0.833326]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Curves Not Parallel to Coordinate Planes", FontVariations->{"CompatibilityType"->0}]], "Subsection"], Cell[TextData[{ StyleBox["In the previous example, we used the cross sections determined by \ x=1 and y=-2, but it's not necessary for us to choose such nice equations -- \ in fact, it's not even necessary that the values for x and y follow a \ straight line; we just need to make sure the x- and y-values follow some \ curve which goes through our point -- (1,-2), in this case. For example we \ could have used a cross section where the x- and y-values are on the parabola \ ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`y = \(-x\^2\) - 1\)]], ". [Check that this parabola ", StyleBox["does", FontSlant->"Italic"], " in fact go through (1,-2).] ", "It would look like this:" }], "Text"], Cell[BoxData[{ \(\(f[x_, y_]\ = \ x^2\ + \ 3\ x*y\ + \ y^2;\)\), "\[IndentingNewLine]", \(\(cs[t_] = {t, \(-t^2\) - 1, f[t, \(-t^2\) - 1]};\)\), "\[IndentingNewLine]", \(\(parabcs = ParametricPlot3D[cs[t], {t, 0, 1.4}];\)\), "\[IndentingNewLine]", \(\(Show[surf, parabcs, AxesLabel \[Rule] {"\", "\", "\"}];\)\)}], "Input"], Cell[TextData[{ StyleBox["You should look at the above commands until you're sure you \ understand the definition for ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`cs1[t]\)]], ". Ask your TA if you need help. However, in practice we usually don't \ use complicated cross sections like this because it's simply easier to work \ with \"straight ones.\"", StyleBox["\n\nIt's certainly not necessary for the vertical planes to be as \ simple as x=1 or y=-2, however. Let's work through the previous example \ again, but this time, instead of restricting f(x,y) to the planes x=1 and \ y=-2, let's restrict to the planes ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`y = 2 x - 4\)]], ".", StyleBox[" and ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`y = \(-x\) - 1\)]], ".", StyleBox[" [You should check that these do in fact slice through the \ surface at the point ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`\((x, y)\) = \((1, \(-2\))\)\)]], ".]", StyleBox["\n\nLet's define c3 and c4 to represent the restrictions to y = \ 2x - 4 and y = -x-1 respectively. Again, make sure you understand the \ following definitions; ask for help if you're not sure about them.", FontVariations->{"CompatibilityType"->0}] }], "Text"], Cell[BoxData[{ \(c3[s_]\ = \ {s, \ 2 s\ - 4, \ f[s, 2 s - 4]}\), "\[IndentingNewLine]", \(c4[t_]\ = \ {t, \ \(-t\) - 1, \ f[t, \(-t\) - 1]}\)}], "Input"], Cell[TextData[{ "Now let's look at the original surface with these two curves; you could \ replace the ", StyleBox["Show", FontWeight->"Bold"], " command with ", StyleBox["Show", FontWeight->"Bold"], " if you prefer." }], "Text"], Cell[BoxData[{ \(three\ = ParametricPlot3D[ Evaluate[c3[s]], \ {s, 0.5, 1.5}]\), "\[IndentingNewLine]", \(four\ = \ ParametricPlot3D[ c4[t], \ {t, \(-0.1\), 2.1}]\), "\[IndentingNewLine]", \(Show[surf, three, four]\)}], "Input"], Cell[TextData[{ "The tangent vectors are again found by differentiation: the tangent \ vectors are (1, 2, 22s-28) and (1, -1, -2t-1) and we are interested in the \ point when s and t both equal one, so we get (1,2,-6) and (1,-1,-3). Just \ like above, ", StyleBox["the vectors tangent to the paths at the point of intersection are \ both tangent to the surface at that point", FontSlant->"Italic"], ". The resulting plane is then" }], "Text"], Cell[BoxData[{ \(p2[s_, t_]\ = \ {1, \(-2\), f[1, \(-2\)]}\ + \ t*{1, 2, \(-6\)}\ + \ s*{1, \(-1\), \(-3\)}\), "\[IndentingNewLine]", \(tanplane2\ = \ ParametricPlot3D[ p2[s, t], {t, \(- .5\), .5}, {s, \(- .5\), .5}]\), "\ \[IndentingNewLine]", \(Show[surf, \ tanplane2]\)}], "Input"], Cell["For comparison, here's the first tangent plane we found:", "Text"], Cell[BoxData[ \(Show[surf, tanplane]\)], "Input"], Cell["\<\ Are these planes the same? They had better be, and in this case \ they are. The pictures don't look exactly the same, but that's because we \ used different cross sections, and therefore we have different vectors in our \ parametric equations for the plane. You'll get a chance to verify that these \ two different methods produce identical planes.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Using Tangent Vectors to find Cartesian Equations of Tangent Planes\ \ \>", "Subsection"], Cell["\<\ Sometimes we need to find a Cartesian equation for a tangent plane. \ Recall that the Cartesian equation of a plane looks like\ \>", "Text"], Cell[BoxData[ \(A \((x - x\_0)\) + B \((y - y\_0)\) + C \((z - z\_0)\) = 0\)], "DisplayFormula", TextAlignment->Center, FontSize->14], Cell[TextData[{ "where ", Cell[BoxData[ \(TraditionalForm\`\((x\_0, y\_0, z\_0)\)\)]], " is a point on the plane, and the vector ", Cell[BoxData[ \(TraditionalForm\`\((A, B, C)\)\)]], " is normal (i.e. perpendicular) to the plane. Now think back to what \ we've been doing so far: given a surface, we choose a point; then we find two \ curves through that point; then we find the tangent vectors at that point.\n\n\ So we've got a point, and now we need a normal vector. Why not just use the \ cross product of the two tangent vectors? They're both in the tangent plane, \ so their cross product is perpendicular to the plane.\n\nHere's the \ parametric equation of the tangent plane we just found, which touches the \ surface at ", Cell[BoxData[ \(TraditionalForm\`\((1, \(-2\), \(-1\))\)\)]], ". " }], "Text"], Cell[BoxData[ \(p2[s_, t_]\ = \ {1, \(-2\), f[1, \(-2\)]}\ + \ t*{1, 2, \(-6\)}\ + \ s*{1, \(-1\), \(-3\)}\)], "Input"], Cell[TextData[{ "The two tangent vectors are ", Cell[BoxData[ \(TraditionalForm\`\((1, \(-1\), \(-3\))\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\((1, 2, \(-6\))\)\)]], ", so our normal vector is their cross product. ", StyleBox["Mathematica", FontSlant->"Italic"], " can compute cross products with a command named, not surprisingly, ", StyleBox["Cross", FontWeight->"Bold"], ":" }], "Text"], Cell[BoxData[ \(Cross[{1, \(-1\), \(-3\)}, {1, 2, \(-6\)}]\)], "Input"], Cell["Therefore our Cartesian equation for this same plane is", "Text"], Cell[BoxData[{ \(\(\(12 \((x - 1)\) + 3 \((y + 2)\) + 3 \((z + 1)\) = 0\)\(\[IndentingNewLine]\) \)\), "\[IndentingNewLine]", \(z = \(-4\) x - y + 1\)}], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "Let's plot this using ", StyleBox["Plot3D", FontWeight->"Bold"], " and show it with the original surface:" }], "Text"], Cell[BoxData[{ \(\(tanplane3 = Plot3D[\(-4\) x - y + 1, {x, 0, 2}, {y, \(-1\), \(-3\)}];\)\), "\[IndentingNewLine]", \(\(Show[surf, tanplane3];\)\)}], "Input"], Cell["\<\ Are you convinced that this is the same tangent plane? It \ certainly looks that way, and in fact it is!\ \>", "Text"], Cell[TextData[{ StyleBox["Exercise 2", FontSize->16, FontWeight->"Bold"], "\n\nIn this exercise you're going to find the linear approximation (or \ tangent plane, if you prefer) of the function ", Cell[BoxData[ \(TraditionalForm\`g(x, y) = x\^2 - y\^2\)]], "at the point ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\) = \((1, 2)\)\)]], ". Your work should include the following steps:\n\n(a) Find two curves on \ the surface which go through the point ", Cell[BoxData[ \(TraditionalForm\`\((1, 2)\)\)]], "; one of them can be the cross section given by ", Cell[BoxData[ \(TraditionalForm\`x = 1\)]], ", but the other should be something different than the cross section given \ by ", Cell[BoxData[ \(TraditionalForm\`y = 2\)]], ".\n\n(b) Find the derivatives of each of these curves, and find their \ tangent vectors at the point ", Cell[BoxData[ \(TraditionalForm\`\((1, 2)\)\)]], ". \n\n(c) Give simplified parametric and cartesian equations for the \ plane.\n\nYour writeup should include a good picture of both the surface and \ the tangent plane. You should put some thought into the ranges of the x- and \ y-values in your picture; if you're too close, you won't be able to tell the \ difference between the plane and the surface, but if you're too far away, the \ resulting picture may not be very useful. In particular, if your picture \ clearly shows that your plane is ", StyleBox["not", FontSlant->"Italic"], " tangent, or if it's impossible to tell, you will most likely lose \ points.", "\n\n", StyleBox["Exercise 3", FontSize->16, FontWeight->"Bold"], "\n\nIn this exercise you're going to find the linear approximation (or \ tangent plane, if you prefer) of the function ", Cell[BoxData[ \(TraditionalForm\`g(x, y) = \(sin(x)\) \(cos(y)\)\)]], "at the point ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\) = \((\[Pi]/4, \(-\[Pi]\)/4)\)\)]], ". Your work should include the following steps:\n\n(a) Find two curves on \ the surface which go through the point (\[Pi]/4,-\[Pi]/4); one of them can be \ the cross section given by ", Cell[BoxData[ \(TraditionalForm\`y = \(-\[Pi]\)/4\)]], ", but the other should be something different than the cross section given \ by ", Cell[BoxData[ \(TraditionalForm\`x = \[Pi]/4\)]], ".\n\n(b) Find the derivatives of each of these curves, and find their \ tangent vectors at the point ", Cell[BoxData[ \(TraditionalForm\`\((\[Pi]/4, \(-\[Pi]\)/4)\)\)]], ". \n\n(c) Give simplified parametric and cartesian equations for the \ plane.\n\nYour writeup should include a good picture of both the surface and \ the tangent plane. You should put some thought into the ranges of the x- and \ y-values in your picture; if you're too close, you won't be able to tell the \ difference between the plane and the surface, but if you're too far away, the \ resulting picture may not be very useful. In particular, if your picture \ clearly shows that your plane is ", StyleBox["not", FontSlant->"Italic"], " tangent, or if it's impossible to tell, you will most likely lose \ points.", "\n" }], "Text", CellFrame->True, Background->RGBColor[1, 0.498039, 0.498039]] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Tangent Planes Via Gradients", FontSize->16]], "Section"], Cell["\<\ Another way to find the tangent plane to a surface is to use the \ gradient vector. Recall that the gradient vector of a function f(x,y) is \ defined to be\ \>", "Text"], Cell[BoxData[ \(\[Del]\&\[RightVector] f \((x, y)\) = \(\((\[PartialD]f\/\[PartialD]x, \ \ \[PartialD]f\/\[PartialD]y)\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "Hmmm. This is a two-dimensional vector, and to define a plane in 3-space, \ we'll definitely need vectors with three components. We get around this by \ doing some minor rearranging. Our surface is defined by ", Cell[BoxData[ \(TraditionalForm\`z = f(x, y)\)]], ". We could move everything over to the same side and say that our surface \ is defined by the equation" }], "Text"], Cell[BoxData[ \(TraditionalForm\`z - f(x, y) = 0\)], "DisplayFormula", TextAlignment->Center, FontSize->14], Cell[TextData[{ "You might be thinking that this is an odd step; if anything, it's a more \ complicated way to write it. That's true, and it's going to get a little \ worse. Rather than saying our surface is defined by this equation, let's \ defined a ", StyleBox["new", FontSlant->"Italic"], " function:" }], "Text"], Cell[BoxData[ \(g \((x, y, z)\)\ = \ z\ - \ f \((x, y)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "And now we say that our function is the ", StyleBox["level set", FontSlant->"Italic"], " defined by ", Cell[BoxData[ \(TraditionalForm\`g(x, y, z) = 0\)]], ". So now instead of the nice, simple statement ", Cell[BoxData[ \(TraditionalForm\`z = f(x, y)\)]], ", we're suddenly talking about level sets. Yikes! There's a reason for \ this, however...\n\nNotice that ", Cell[BoxData[ \(TraditionalForm\`g(x, y, z)\)]], " is a function of three variables, and its gradient ", Cell[BoxData[ \(TraditionalForm\`is\)]] }], "Text"], Cell[BoxData[ \(\[Del]\&\[RightVector] g \((x, y, z)\) = \(\((\(-\(\[PartialD]f\/\[PartialD]x\)\), \ \ \(-\(\[PartialD]f\/\[PartialD]y\)\), \ 1)\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "Now recall from lecture (or the previous lab) that ", StyleBox["the gradient of a function g is perpendicular to the level sets \ of g", FontSlant->"Italic"], ". In other words:\n\n-- We can start with a point (x,y,z) which is on the \ level set ", Cell[BoxData[ \(TraditionalForm\`g(x, y, z) = 0\)]], ".\n-- Because of the way we defined ", Cell[BoxData[ \(TraditionalForm\`g(x, y, z)\)]], ", this is the same as saying the point is on the surface ", Cell[BoxData[ \(TraditionalForm\`z = f(x, y)\)]], ".\n-- Furthermore, ", Cell[BoxData[ \(\[Del]\&\[RightVector] g \((x, y, z)\)\)], TextAlignment->Center], "will be perpendicular (or normal) to our surface at that point.\n-- In \ particular, we can use ", Cell[BoxData[ \(\[Del]\&\[RightVector] g \((x, y, z)\)\)], TextAlignment->Center], "as a normal vector to define the equation for our tangent plane!\n\nLet's \ use this method quickly to find the same tangent plane we've been working \ with this whole time. Remember that the function is given by:" }], "Text"], Cell[BoxData[ \(f[x_, y_]\ = \ x^2\ + \ 3\ x*y\ + \ y^2\)], "Input"], Cell["So we can do our magic \"new\" function like this:", "Text"], Cell[BoxData[ \(g[x_, y_, z_] = z - f[x, y]\)], "Input"], Cell[TextData[{ "We're interested in the gradient at the point ", Cell[BoxData[ \(TraditionalForm\`\((1, \(-2\), \(-1\))\)\)]], ":" }], "Text"], Cell[BoxData[{ \(gradg[x_, y_, z_] = Grad[g[x, y, z]]\), "\[IndentingNewLine]", \(gradg[1, \(-2\), \(-1\)]\)}], "Input"], Cell["Thus the Cartesian equation of the tangent plane is:", "Text"], Cell[BoxData[{ \(\(\(4 \((x - 1)\) + \((y + 2)\) + \((z + 1)\) = 0\)\(\[IndentingNewLine]\) \)\), "\[IndentingNewLine]", \(z = \(-4\) x - y + 1\)}], "DisplayFormula", TextAlignment->Center], Cell["\<\ Which is exactly what we found before. Good! Although the \"gradient method\" of finding a tangent plane might seem more \ complicated -- particularly if you don't like level sets -- it's very, very \ useful. We didn't have to find any cross sections, find tangent vectors, \ solve for values of s and t, and so on. We just rewrote the original \ equation, computed one gradient, and we were basically done!\ \>", "Text"], Cell[TextData[{ StyleBox["Exercise 4", FontSize->16, FontWeight->"Bold"], "\n\nUse the gradient method to find the cartesian equation of the plane \ tangent to\n\n", Cell[BoxData[ \(TraditionalForm\`f(x, y) = \(Exp(\(-x\^2\) - y\^2)\)*x\)]], "\n\nat the point ", Cell[BoxData[ \(TraditionalForm\`\(\((0, 0, f(0, 0))\)\(.\)\)\)]], " Show the graph of f(x,y) and this tangent plane together on the same \ plot. Be sure to show your work and explain your reasoning.\n\n", StyleBox["Exercise 5", FontSize->16, FontWeight->"Bold"], "\n\nUse the gradient method to find the cartesian equation of the plane \ tangent to\n\n", Cell[BoxData[ \(TraditionalForm\`f(x, y) = y*\(Cos(x)\)\)]], "\n\nat the point ", Cell[BoxData[ \(TraditionalForm\`\(\((0, \[Pi]/4, f(0, \[Pi]/4))\)\(.\)\)\)]], " Show the graph of f(x,y) and this tangent plane together on the same \ plot; use the ranges -\[Pi] to \[Pi] for both x and y. Be sure to show your \ work and explain your reasoning. In particular, with the suggested ranges \ the \"true\" tangent plane looks like it sticks through part of the surface. \ You may want to explain why that's the case." }], "Text", CellFrame->True, Background->RGBColor[1, 0.498039, 0.498039]] }, Closed]], Cell[CellGroupData[{ Cell[TextData[StyleBox["Tangent Plane Conspiracy Theory", FontSize->16]], "Section"], Cell["\<\ You are not responsible for the material in this section, but we've \ included it for those people who are interested in the mathematics behind all \ of this; it turns out that all of the different methods of finding tangent \ planes aren't really all that different!\ \>", "Text", CellFrame->True, Background->GrayLevel[0.833326]], Cell[TextData[{ "In this section, we'll learn that, despite all appearances to the \ contrary, the ``tangent vectors'' and ``gradient vector'' methods are \ surreptitiously working together and are engaged in a ", StyleBox["secret government conspiracy!", FontSlant->"Italic"], "\n\nOkay, there's no government conspiracy (well...none that ", StyleBox["I'm", FontSlant->"Italic"], " aware of), but it is true that the two methods are working together\ \[LongDash]in fact, they are just different ways of looking at the same \ thing. Let's talk about directional derivatives for a bit before explaining \ why they're the same.\n\nAbove we were talking about partial derivatives and \ directions, which should remind you of directional derivatives. In the \ Tangent Vectors section, when specifying ``the planes y = 2x - 4 and y = -x - \ 1'', we were really talking about direction vectors. In the above cases, the \ corresponding direction vectors would be" }], "Text"], Cell[BoxData[ \(\(\(\(1\/\@5\) \((1, 2)\)\ \ \ \ and\ \ \ \ \ \(1\/\@2\) \((1, \(-1\))\)\)\(,\)\)\)], \ "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "respectively. We are basically using the slope of a line: y = 2x - 4 is a \ line (in two dimensions) with slope 2. The vector (1,2) is parallel to that \ line, and above we just normalized it. The other direction vector was \ obtained in a similar way. Once we know direction vectors, we just need to \ take the dot product with the gradient vector to find the directional \ derivative. The conspiracy is already unravelling.\n\nFor the function f(x,y) \ = ", Cell[BoxData[ \(TraditionalForm\`x\^2\ + \ 3 xy + \ y\^2\)]], " at the point (1,-2), and the first direction vector\[LongDash]call it u\ \[LongDash]we have" }], "Text"], Cell[BoxData[ \(Df\_u = \ \(\((\[PartialD]f\/\[PartialD]x, \ \ \[PartialD]f\/\[PartialD]y)\)\[CenterDot]\((1\/\@5, 2\/\@5)\)\ = \ \(\((\(-4\), \ \(-1\))\)\[CenterDot]\((1\/\@5, 2\/\@5)\)\ = \ \(-\(6\/\@5\)\)\)\)\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "We interpret this by saying ``in the direction of u, f is increasing at a \ rate of ", Cell[BoxData[ \(TraditionalForm\`\(-6\)/\@5\)]], ".'' The upshot of this is that (", Cell[BoxData[ \(TraditionalForm\`1/\@5\)]], ", ", Cell[BoxData[ \(TraditionalForm\`2/\@5\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(-6\)/\@5\)]], ") is a tangent vector to f(x,y), and it's parallel to \ (1,2,-6)\[LongDash]the tangent vector we found by using the path we described \ in the first section. We get this vector by taking the direction vector\ \[LongDash]a two-dimensional vector\[LongDash]and adding a third component, \ which is given by the directional derivative of f(x,y). In general, this \ means that" }], "Text"], Cell[BoxData[ \(\((u1, \ u2, \ \(D\_u\) f)\)\)], "DisplayFormula"], Cell[TextData[{ "is a vector tangent to the surface z = f(x,y). You should do this same \ process with the other direction vector above and make sure that you end up \ with a tangent vector parallel to (1,-1,-3).\n\nOf course, there's nothing \ special about those two directional vectors above. We can use ", StyleBox["any", FontSlant->"Italic"], " two non-parallel direction vectors and we'll get the same \ plane\[LongDash]which is exactly what you would hope for, since a function \ can have only one tangent plane (or no tangent plane, if the function isn't \ differentiable). The following exercise will help you expose the conspiracy\ \[LongDash]er, show that the tangent planes from the ``tangent vectors \ method'' and from the ``gradient vectors method'' are in fact the same. We'll \ do that by comparing the normal vectors of the resulting planes." }], "Text"], Cell[TextData[{ StyleBox["Exercise ", FontSize->16, FontWeight->"Bold"], "\n\n(Notice that this isn't in a red box; you don't have to hand this \ exercise in. You might find it interesting, however.)\n\nAs discussed in the \ Gradient Vector section, the gradient of z - f(x,y) represents a normal \ vector to the tangent plane, so our main objective is to find the normal \ vector of the plane we get from the tangent vector method. We'll use an \ arbitrary differentiable function f(x,y) and two non-parallel direction \ vectors u = (u1, u2) and v = (v1, v2).\n\n\t(i) Find the vectors tangent to \ f(x,y) in the directions of u and v. Describe them in terms of u1, u2, v1, \ v2, and the partials of f (\[PartialD]f / \[PartialD]x and \[PartialD]f / \ \[PartialD]y).\n\n\t(ii) Find the normal vector of the plane spanned by those \ two vectors.\n\t\n\t(iii) Show that the normal vector you found in (ii) is \ parallel to the gradient vector of the function g(x,y,z) = z - f(x,y).\n\t\n\t\ (iii) What does this imply about the tangent planes found using the two \ different methods?\n " }], "Text", CellFrame->True, Background->GrayLevel[0.833326]], Cell[CellGroupData[{ Cell["Credits", "Subsubsection"], Cell[TextData[{ "This lab is a descendent of an earlier lab on Tangent Planes written by \ Cindy Kaus. Dan Drake did a major rewrite in 2002, and then I overhauled it \ again in February 2004. We changed the focus a little bit, and added a few \ things here and there; having taught this lab for a few years now, we've \ learned what the sticking points are for students, so I've tried to add extra \ explanations in these parts. I also changed the exercises, except for #1, \ which Dan wrote, and the pseudo-exercise in the \"Conspiracy Theory\" \ section, which is also Dan's. I changed the rest of them so we'd have a \ little variety; the previous exercises have been used for the last 4 years or \ so.\n\nIt's a little difficult to completely describe who did what, but \ here's an attempt:\n\n-- I don't believe any of the writing is Cindy's, but \ the particular surface and cross sections used in the \"Tangent Vectors\" \ sections (and exercises 2 and 3) are hers.\n\n-- The \"Plotting Planes\" and \ \"Tangent Plane Conspiracy Theory\" sections are all Dan's; at most I changed \ a few words here or there, or added a ", StyleBox["Live", FontWeight->"Bold"], " command.\n\n-- The other sections are fairly major rewrites of Dan's \ original sections. A lot of his original text remains. Some of it remains \ but has been modified a bit, because of reorganization, etc. Some of it is \ new text added by me. To make things more complicated, it's all intertwined; \ many paragraphs combine Dan's writing with mine.\n\nThis would be a major \ mess, but fortunately we've all agreed to use the same license, and it works \ out. The current version of the lab is copyright 2004 by Jonathan Rogness \ (rogness@math.umn.edu) and 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/. (As \ mentioned, parts of this are copyright 2000 by Cindy Kaus and 2002 by Dan \ Drake and are protected under the same license.)\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. 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