(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 4.0, MathReader 4.0, or any compatible application. 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). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. 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[ 16071, 564]*) (*NotebookOutlinePosition[ 17029, 595]*) (* CellTagsIndexPosition[ 16985, 591]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["\<\ 11 - How High the Moon or How Low Can You Go?\ \>", "Title"], Cell["Maxima and Minima in Three Dimensions", "Subtitle"], Cell[CellGroupData[{ Cell["\<\ William D. Emerson\t Metropolitan State College of Denver\ \>", "Section"], Cell[CellGroupData[{ Cell["First and Second Order Partial Derivatives", "Subsection"], Cell[TextData[{ "Given a function we can compute the partial derivatives using ", StyleBox["Mathematica", FontSlant->"Italic"], ". Here's an example." }], "Text"], Cell[BoxData[{ \(Clear[f, x, y]\), "\[IndentingNewLine]", \(f[x_, y_]\ = \ x^3\ \ + \ Sqrt[y] + \ x^2\ y^2\)}], "Input"], Cell["\<\ We can compute the partial derivative of f with respect to x \ :\ \>", "Text"], Cell[BoxData[ \(D[f[x, y], x]\)], "Input"], Cell["\<\ Similarly, we can compute the partial derivative with respect to y:\ \ \>", "Text"], Cell[BoxData[ \(D[f[x, y], y]\)], "Input"], Cell["\<\ The syntax for second partial derivatives is similar. Here is the \ second partial with respect to x twice and the second partial with respect to \ y twice.\ \>", "Text"], Cell[BoxData[ \(D[f[x, y], x, x]\)], "Input"], Cell[BoxData[ \(D[f[x, y], y, y]\)], "Input"], Cell["\<\ Finally, here is the mixed second partial derivative, first with \ respect to x and then with respect to y.\ \>", "Text"], Cell[BoxData[ \(D[f[x, y], x, y]\)], "Input"], Cell["\<\ If we want the mixed second partial, first with respect to y and \ then with respect to x we should get the same answer.\ \>", "Text"], Cell[BoxData[ \(D[f[x, y], y, x]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["Tangent Planes to Surfaces", "Subsection"], Cell[TextData[{ "Given a surface ", Cell[BoxData[ \(TraditionalForm\`z\ = \ f[x, y]\)]], " and a point ", Cell[BoxData[ \(TraditionalForm\`\((a, b, f[a, b])\)\)]], " the equation of the tangent plane at this point is ", Cell[BoxData[ \(TraditionalForm\`z\ = \ f[a, b]\ + \ f\_x[a, b] \((x - a)\)\ + \ f\_y[a, b] \(\((y - b)\)\(.\)\)\)]], " Here is an example." }], "Text"], Cell[BoxData[{ \(Clear[f, x, y, surface]\), "\[IndentingNewLine]", \(f[x_, y_]\ = \ 16\ - \ x^2\ - \ y^2\), "\[IndentingNewLine]", \(surface\ = \ Plot3D[f[x, y], {x, \(-3\), 3}, {y, \(-3\), 3}]\)}], "Input"], Cell["\<\ We will find the tangent plane at the point (2, -2, f[2,-1]). \ First we need to find the partial derivatives.\ \>", "Text"], Cell[BoxData[{ \(Clear[fx, fy]\), "\[IndentingNewLine]", \(fx[x_, y_]\ = \ D[f[x, y], x]\), "\[IndentingNewLine]", \(fy[x_, y_]\ = \ D[f[x, y], y]\)}], "Input"], Cell["Now we can define and plot the tangent plane.", "Text"], Cell[BoxData[{ \(Clear[tangent, tanPlane]\), "\[IndentingNewLine]", \(tangent[x_, y_]\ = \ f[2, \(-2\)]\ + \ fx[2, \(-2\)] \((x - 2)\)\ + \ fy[2, \(-2\)] \((y + 2)\)\), "\[IndentingNewLine]", \(tanPlane\ = \ Plot3D[tangent[x, y], {x, \(-3\), 3}, {y, \(-3\), 3}]\)}], "Input"], Cell["Let's look at point the surface and the plane together.", "Text"], Cell[BoxData[ \(Show[surface, tanPlane]\)], "Input"], Cell["Here's a view from a different angle.", "Text"], Cell[BoxData[ \(Show[surface, tanPlane, ViewPoint \[Rule] {\(-3\), \(-3\), 0}]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Max/Min and Saddle Points and their Horizontal Tangent \ Planes.\ \>", "Subsection"], Cell[CellGroupData[{ Cell["First Example", "Subsubsection"], Cell["\<\ Let's look at an example of a function that clearly has a \ maximum.\ \>", "Text"], Cell[BoxData[{ \(Clear[f, x, y, surface]\), "\[IndentingNewLine]", \(f[x_, y_]\ = \ 9\ - \ x^2\ - y^2\), "\[IndentingNewLine]", \(surface\ = \ Plot3D[f[x, y], {x, \(-2\), 2}, {y, \(-2\), 2}]\)}], "Input"], Cell["\<\ This function has a maximum at (0,0). Let's find the tangent plane \ at that point.\ \>", "Text"], Cell[BoxData[{ \(Clear[fx, fy, tangent, tanPlane]\), "\[IndentingNewLine]", \(Clear[tangent, tanPlane]\), "\[IndentingNewLine]", \(fx[x_, y_]\ = \ D[f[x, y], x]\), "\[IndentingNewLine]", \(fy[x_, y_]\ = \ D[f[x, y], y]\), "\n", \(tangent[x_, y_]\ = \ f[0, 0]\ + \ fx[0, 0] \((x - 0)\)\ + \ fy[0, 0] \((y - 0)\)\), "\n", \(tanPlane\ = \ Plot3D[tangent[x, y], {x, \(-2\), 2}, {y, \(-2\), 2}]\[IndentingNewLine]\)}], "Input"], Cell["\<\ Now we can look at the suface and the tangent plane at the same \ time.\ \>", "Text"], Cell[BoxData[ \(Show[surface, tanPlane]\)], "Input"], Cell["\<\ Notice that since the tangent plane is horizontal its equation is \ just a constant.\ \>", "Text"], Cell[BoxData[ \(tangent[x, y]\)], "Input"], Cell[TextData[{ "This means that the partial derivatives must be 0, otherwise the plane \ would not be a constant. That means that the gradient of ", Cell[BoxData[ \(TraditionalForm\`f\)]], " must be the zero vector. Let's check" }], "Text"], Cell[BoxData[ \(gradf[x_, y_]\ = \ {fx[x, y], fy[x, y]}\)], "Input"], Cell[BoxData[ \(gradf[0, 0]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["Second Example", "Subsubsection"], Cell["We can also look at a saddle point. Here is an example.", "Text"], Cell[BoxData[{ \(Clear[f, x, y, surface]\), "\[IndentingNewLine]", \(f[x_, y_]\ = \ x^2 - y^2\), "\[IndentingNewLine]", \(surface\ = \ Plot3D[f[x, y], {x, \(-2\), 2}, {y, \(-2\), 2}]\)}], "Input"], Cell["Let's find and plot the tangent plane at (0,0).", "Text"], Cell[BoxData[{ \(Clear[fx, fy, tangent, tanPlane]\), "\[IndentingNewLine]", \(Clear[tangent, tanPlane]\), "\[IndentingNewLine]", \(fx[x_, y_]\ = \ D[f[x, y], x]\), "\[IndentingNewLine]", \(fy[x_, y_]\ = \ D[f[x, y], y]\), "\n", \(tangent[x_, y_]\ = \ f[0, 0]\ + \ fx[0, 0] \((x - 0)\)\ + \ fy[0, 0] \((y - 0)\)\), "\n", \(tanPlane\ = \ Plot3D[tangent[x, y], {x, \(-2\), 2}, {y, \(-2\), 2}]\)}], "Input"], Cell["Let's look at both the surface and the plane.", "Text"], Cell[BoxData[ \(Show[surface, tanPlane]\)], "Input"], Cell[TextData[{ "In this case we also have a horizontal tangent plane given by ", Cell[BoxData[ \(TraditionalForm\`z\ = \ k\)]], " (here ", Cell[BoxData[ \(TraditionalForm\`k\ = \ 0\)]], "). Again, this means that the partial derivatives must be 0 and that \ means that the gradient of ", Cell[BoxData[ \(TraditionalForm\`f\)]], " mst be the zero vector. Let's check." }], "Text"], Cell[BoxData[ \(tangent[x, y]\)], "Input"], Cell[BoxData[ \(gradf[x_, y_]\ = \ {fx[x, y], fy[x, y]}\)], "Input"], Cell[BoxData[ \(gradf[0, 0]\)], "Input"], Cell["\<\ We could continue by looking at a surface with a minimum, \ but....well, you get the point by now. Surfaces that have maximums, minimums \ or saddle points have horizontal (level) tangent planes. A point, \ {a,b,f[a,g]}, with a horizontal tangent plane corresponds to a point,{a,b}, \ where the gradient of the function is {0,0}.\ \>", "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Finding Critical Points and Testing with the Discriminant", "Subsection"], Cell["Let's start with an example.", "Text"], Cell[BoxData[{ \(Clear[f, x, y]\), "\[IndentingNewLine]", \(f[x_, y_]\ = \((4 x\ - \ x^2)\) \((4 y - y^2)\)\)}], "Input"], Cell["Here's a plot of this function.", "Text"], Cell[BoxData[ \(Plot3D[f[x, y], {x, \(-5\), 5}, {y, \(-5\), 5}]\)], "Input"], Cell["\<\ To find the critical points we find where the gradient is \ {0,0}.\ \>", "Text"], Cell[BoxData[ \(gradf[x_, y_]\ = {D[f[x, y], x], D[f[x, y], y]}\)], "Input"], Cell[BoxData[ \(Solve[gradf[x, y] \[Equal] {0, 0}, {x, y}]\)], "Input"], Cell["\<\ To test the critical points for maxima, minima, and saddle points \ we need to define the discriminant and that requires that we have a few first \ and second order partial derivatives.\ \>", "Text"], Cell[BoxData[{ \(fx[x_, y_]\ \ = \ D[f[x, y], x]\), "\[IndentingNewLine]", \(fy[x_, y_]\ \ = \ D[f[x, y], y]\), "\[IndentingNewLine]", \(fxx[x_, y_]\ \ = \ D[f[x, y], x, x]\), "\[IndentingNewLine]", \(fyy[x_, y_]\ \ = \ D[f[x, y], y, y]\), "\[IndentingNewLine]", \(fxy[x_, y_]\ \ = \ D[f[x, y], x, y]\), "\[IndentingNewLine]", \(discriminant[x_, y_]\ = \ fxx[x, y] fyy[x, y] - \((fxy[x, y])\)^2\)}], "Input"], Cell["Now we can test our critical points.", "Text"], Cell[BoxData[ \(discriminant[0, 0]\)], "Input"], Cell["This is a saddle point.", "Text"], Cell[BoxData[ \(discriminant[0, 4]\)], "Input"], Cell["This is a saddle point.", "Text"], Cell[BoxData[ \(discriminant[4, 0]\)], "Input"], Cell["This is a saddle point.", "Text"], Cell[BoxData[ \(discriminant[4, 4]\)], "Input"], Cell["This is a saddle point.", "Text"], Cell[BoxData[ \(discriminant[2, 2]\)], "Input"], Cell["\<\ Ah! We have finally located a maximum or minimum. We'll use the \ second partial with respect to x to determine which.\ \>", "Text"], Cell[BoxData[ \(fxx[2, 2]\)], "Input"], Cell["\<\ Since the second partial with respect to x is negative the surface \ is concave down in the x direction and this indicates that the critical \ points is a maximum.\ \>", "Text"], Cell["\<\ Another way to determine whether we have a maximum, minimum or \ saddle point is to graph the function in a small neighborhood of the point. \ First, we look at the point (4,4).\ \>", "Text"], Cell[BoxData[ \(Plot3D[f[x, y], {x, 3.9, 4.1}, {y, 3.9, 4.1}]\)], "Input"], Cell["\<\ Yep, that's a saddle point. Now we look closely at the point at \ (2,2).\ \>", "Text"], Cell[BoxData[ \(Plot3D[f[x, y], {x, 1.9, 2.1}, {y, 1.9, 2.1}]\)], "Input"], Cell["Clearly, we have a maximum at (2,2).", "Text"], Cell[TextData[{ "Sometimes, we are able to find a portion of the surface which clearly \ depicts the behavior. Check out the following. We are able to avoid some of \ the very large values of ", Cell[BoxData[ \(TraditionalForm\`f\)]], " for negative values of ", Cell[BoxData[ \(TraditionalForm\`x\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y\)]], ". As a result we can see more detail of the surface and the maximum and \ saddle points become much more visible." }], "Text"], Cell[BoxData[ \(Plot3D[f[x, y], {x, \(-1\), 5}, {y, \(-1\), 5}]\)], "Input"], Cell["\<\ Another way to examine our surface for critical points is to look \ at a contour map. Here is one for our function.\ \>", "Text"], Cell[BoxData[ \(ContourPlot[f[x, y], {x, \(-1\), 5}, {y, \(-1\), 5}, Contours \[Rule] 30]\)], "Input"], Cell["\<\ We can see that we have saddle points at (0,0), (4,0),(0,4) and \ (4,4) and a maximum at (0,0). The lighter shading at (0,0) shows that we \ have a maximum there and not a minimum. The darker shading indicates a lower \ position on the surface.\ \>", "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problems for Notebook 11 - MaxMin Problems", "Section"], Cell["\<\ Name: Name: Name:\ \>", "Text"], Cell[TextData[{ "You may want to cut and paste the following definitions to save yourself \ some typing. Of course, you will need to modify the definition of the \ function ", Cell[BoxData[ \(TraditionalForm\`f\)]], " depending on which problem you are working." }], "Text"], Cell[BoxData[{ \(Clear[f, x, y, fx, fy, fxx, fyy, fxy, gradf, discriminant]\), "\[IndentingNewLine]", \(f[x_, y_]\ = \ x^4\ + y^4\ - \ 2 x\ y\ + \ 3\), "\[IndentingNewLine]", \(fx[x_, y_]\ \ = \ D[f[x, y], x]\), "\n", \(fy[x_, y_]\ \ = \ D[f[x, y], y]\), "\n", \(fxx[x_, y_]\ \ = \ D[f[x, y], x, x]\), "\n", \(fyy[x_, y_]\ \ = \ D[f[x, y], y, y]\), "\n", \(fxy[x_, y_]\ \ = \ D[f[x, y], x, y]\), "\[IndentingNewLine]", \(gradf[x_, y_]\ = \ {fx[x, y], fy[x, y]}\), "\n", \(discriminant[x_, y_]\ = \ fxx[x, y] fyy[x, y] - \((fxy[x, y])\)^2\)}], "Input"], Cell[CellGroupData[{ Cell["Problem 1", "Subsubsection"], Cell[TextData[{ "For the function ", Cell[BoxData[ \(TraditionalForm\`f[x, y]\ = \ x\^4\ + \ y\^4\ - \ 2 xy\ + \ 3\)]], " find the critical points and use the discriminant to test them for max, \ min., and saddle points. (You can cut and paste the definitions from above \ here.)" }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Problem 2", "Subsubsection"], Cell[TextData[{ "For the function ", Cell[BoxData[ \(TraditionalForm\`f[x, y]\ = \ 3 x\^3\ + \ xy\^2\ + \ 4 x\^2\ + \ y\^2\)]], " find the critical points and use the discriminant to test them for max, \ min and saddle points." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Problem 3", "Subsubsection"], Cell[TextData[{ "For the function ", Cell[BoxData[ \(TraditionalForm\`f[x, y]\ = \ Cos[x\ + \ y]\)]], " use a contour plot to find the critical points in the region ", Cell[BoxData[ \(TraditionalForm\`R\ = \ \([\(-4\), 4]\)\[Times]\(\([\(-4\), 4]\)\(.\)\)\)]], " Try to determine the nature of the critical point from the contour plot. \ Verify your results by \"zooming in\" on the surface near the point." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Problem 4", "Subsubsection"], Cell[TextData[{ "For the function ", Cell[BoxData[ \(TraditionalForm\`f[x . y]\ = \ Sin[x]\ + \ Sin[y]\ + \ Sin[x + y]\)]], " use a contour plot to find the critical points in the region ", Cell[BoxData[ \(TraditionalForm\`R\ = \ \([0, 2 \[Pi]]\)\[Times]\([0, 2 \[Pi]]\)\)]], ". Try to determine the nature of the critical point from the contour \ plot. Verify your results by \"zooming in\" on the surface near the point." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Problem 5", "Subsubsection"], Cell[TextData[{ "Show that the function ", Cell[BoxData[ \(TraditionalForm\`f[x, y]\ = \ \((y\^2 - 1)\)\^\(\(2\)\(\ \)\) + \ \((xy\^2 - y - 1)\)\^2\ \)]], " has two critical points which are both minimums. Plot this function so \ that this phenomena can be clearly seen." }], "Text"] }, Open ]] }, Open ]] }, Open ]] }, FrontEndVersion->"4.0 for Macintosh", ScreenRectangle->{{0, 1152}, {0, 748}}, PrintingStyleEnvironment->"Presentation", WindowSize->{520, 624}, WindowMargins->{{5, Automatic}, {Automatic, 31}}, PrintingCopies->1, PrintingPageRange->{1, Automatic}, Magnification->1, MacintoshSystemPageSetup->"\<\ 00<0004/0B`000002mT8o?mooh<" ] (*********************************************************************** Cached data follows. 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