(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of 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[ 246959, 7771]*) (*NotebookOutlinePosition[ 247629, 7795]*) (* CellTagsIndexPosition[ 247585, 7791]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Polar Coordinates ", "Title", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell["Read Me! (how to use this program)", "Section", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ "This notebook is written using ", StyleBox["Mathematica", FontSlant->"Italic"], ", a highly sophisticated mathematical software program capable of doing \ intricate mathematics", ". ", "You will need to know only a few ", StyleBox["very basic", FontVariations->{"Underline"->True}], " things about ", StyleBox["Mathematica", FontSlant->"Italic"], " to be able to use this notebook; the programming has been done for you." }], "Text"], Cell[TextData[{ "\t1. ", StyleBox["Opening a section of the notebook", FontWeight->"Bold"], ": Scroll down the screen using the mouse on the scrollbar at the right of \ this screen until you see what appears to be a table of contents. Directly \ to the right of each topic is a short blue bracket sign. Some brackets have \ a small arrow or triangle at the bottom. This arrow indicates that there is \ hidden text which can be viewed by clicking with the mouse on the bracket \ containing the arrow. Try this on one of the arrowed brackets below. When \ you click on the bracket, new text should appear. When you are finished, you \ can close that section of the notebook by double-clicking on the same (now \ longer) bracket. If you are unsure which one it is, scroll up to the \ beginning of the section; it is the one that ", StyleBox["begins", FontSlant->"Italic"], " there. Try closing the section you just opened." }], "Text"], Cell[TextData[{ StyleBox["\t", FontWeight->"Bold"], "2. ", StyleBox["Activating a program command", FontWeight->"Bold"], ": The ", StyleBox["Mathematica", FontSlant->"Italic"], " programming always appears in bold type on a yellow background. To \ activate a program, use the mouse to move the \"I\" shaped cursor inside the \ yellow box and click once so that the straight line \"|\" cursor appears \ anywhere inside the yellow box. Now depress the \"shift\" and \"return\" \ keys ", StyleBox["simultaneously", FontVariations->{"Underline"->True}], ". Try this with the following short program which generates a random \ number between 1 and 10. Remember to click on these two keys ", StyleBox["simultaneously", FontVariations->{"Underline"->True}], "." }], "Text"], Cell[BoxData[ \(Random[Integer, {1, 10}]\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\t3. ", StyleBox["Changing a program:", FontWeight->"Bold"], " Generally, you should not change any type in the yellow blocks. However, \ a few of the ", StyleBox["Mathematica", FontSlant->"Italic"], " programs have been designed so that you can change parts of the program. \ The parts that you should change are displayed in ", StyleBox["blue", FontColor->RGBColor[0, 0, 1]], ". To change these parts, simply highlight the ", StyleBox["blue", FontColor->RGBColor[0, 0, 1]], " type by sweeping over it with the cursor, and type in your change. Read, \ and take seriously, any instruction saying ", StyleBox["*do not change anything below this line*", FontColor->RGBColor[1, 0, 0]], ". Changing something there could radically change the commands, causing ", StyleBox["Mathematica", FontSlant->"Italic"], " to 'beep' in an error protest--or give you an answer to a totally \ different question!\n\n\t4. ", StyleBox["Quitting the program", FontWeight->"Bold"], ": When you have finished working and want to quit, click in the small box \ at the upper lefthand corner of the window to close it. Then choose `quit' \ or `exit' from the File menu. ", StyleBox["Mathematica", FontSlant->"Italic"], " will ask you if you want to save your changes. Say NO! Otherwise you \ will have a different notebook from the one you downloaded, and any \ misinformation you might have entered into the notebook will persist. If for \ any reason, you need to start with a fresh notebook in its original form, you \ can always trash the one you've been working with and download a new one off \ the Web." }], "Text"], Cell[TextData[{ "\t* 5. ", StyleBox["Importing a package for polar coordinates:", FontWeight->"Bold"], " This section will require the use of a special graphics package which we \ must 'import' each time we use the program. This is very easy! At the \ beginning of each section, you'll see a yellow box with the input \ <"Italic"], " you ", StyleBox["start", FontSlant->"Italic"], " to work or the rest of the programs in the notebook will not work." }], "Text"], Cell[BoxData[ \(\(\t<< \ Graphics`Graphics`\)\)], "Input", Background->RGBColor[1, 1, 0]] }, Open ]], Cell[CellGroupData[{ Cell["Defining polar coordinates", "Section", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ "\tBefore you begin, you ", StyleBox["must", FontSlant->"Italic"], " activate the yellow cell below or the rest of the programs in this \ notebook will not work.", " ", "To activate this or any yellow cell, simply put the cursor in the yellow \ box and press shift-return." }], "Text"], Cell[BoxData[ \(<< \ Graphics`Graphics`\)], "Input", Background->RGBColor[1, 1, 0]], Cell["\<\ \tPolar coordinates locate a point in the plane in a different manner than \ rectangular coordinates. 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(r,\[Theta]) is to \ first draw a ray at angle \[Theta] from the x-axis and then travel out \ distance |r| along this ray to plot the point. Let's try this with the point \ P=(3, \[Pi]/4). We plot P by drawing a ray at an angle of \[Pi]/4 from the \ positive x-axis (measured in the counterclockwise direction) and then placing \ the point P on this ray at a distance of 3 units from the origin.", " ", "Below is the point P = (3, ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/4\)]], ") in polar coordinates." }], "Text"], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: 1 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics %%IncludeResource: font Courier %%IncludeFont: Courier /Courier findfont 10 scalefont setfont % Scaling calculations 0.5 0.166667 0.5 0.166667 [ [0 .4875 -6 -9 ] [0 .4875 6 0 ] [.16667 .4875 -6 -9 ] [.16667 .4875 6 0 ] [.33333 .4875 -6 -9 ] [.33333 .4875 6 0 ] [.66667 .4875 -3 -9 ] [.66667 .4875 3 0 ] [.83333 .4875 -3 -9 ] [.83333 .4875 3 0 ] [1 .4875 -3 -9 ] [1 .4875 3 0 ] [.4875 0 -12 -4.5 ] [.4875 0 0 4.5 ] [.4875 .16667 -12 -4.5 ] 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Ool00`00Oomoo`1=Ool00`00Oomoo`0lOol0095oo`03001oogoo04ioo`03001oogoo03]oo`00TGoo 00<007ooOol0Cgoo00<007ooOol0>Woo002AOol2001AOol00`00Oomoo`0iOol0095oo`03001oogoo 055oo`03001oogoo03Qoo`00TGoo00<007ooOol0DWoo00<007ooOol0=goo002AOol00`00Oomoo`1C Ool00`00Oomoo`0fOol0095oo`03001oogoo05Aoo`03001oogoo03Eoo`00TGoo00<007ooOol0EGoo 00<007ooOol0=7oo002AOol00`00Oomoo`1FOol00`00Oomoo`0cOol0095oo`03001oogoo05Moo`03 001oogoo039oo`00RGoo0`001Goo00<007ooOol0F7oo00<007ooOol0"], ImageRangeCache->{{{0, 287}, {287, 0}} -> {-3.12952, -3.09715, 0.0215829, 0.0215829}}], Cell[TextData[{ "\tNow practice plotting points in polar coordinates by using the following \ program. Change the r-value and \[Theta]-value given in blue.", " ", "To do this, highlight the values in blue and then type in the values of \ the point you wish to plot.\n*Make sure you enter \[Theta] using radian \ angles instead of degrees!*\n*Don't change anything else in the program as \ most changes will prevent it from running.*" }], "Text"], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[r, \[Theta], x, y, pt]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{"r", ":=", StyleBox["3", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{"\[Theta]", ":=", " ", StyleBox[ FractionBox["\[Pi]", StyleBox["6", FontColor->RGBColor[0, 0, 1]]], FontColor->RGBColor[0, 0, 1]]}], ";", " ", \( (*\ in\ radians, \ \(please!\)*) \), "\n", StyleBox[\( (*change\ nothing\ below\ this\ line*) \), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[\(x\ := \ r\ \ Cos\ [\[Theta]]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], StyleBox["\n", FontColor->GrayLevel[0.666667]], StyleBox[\(y\ := \ r\ Sin[\[Theta]]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], StyleBox["\n", FontColor->GrayLevel[0.666667]], StyleBox[\(pt := {AbsolutePointSize[6], Point[{x, y}]}\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], StyleBox["\n", FontColor->GrayLevel[0.666667]], StyleBox[ \(Show[Graphics[pt], AspectRatio -> Automatic, Axes -> True, PlotRange -> {{\(-r\), r}, {\(-r\), r}}, Epilog -> {Line[{{0, 0}, {x, y}}], Circle[{0, 0}, Abs[r]/3, {0, If[r > 0, \[Theta], \[Theta] + \[Pi]]}]}]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], StyleBox[" ", FontColor->GrayLevel[0.666667]], StyleBox[ \(Print["\", r, "\<,\>", \[Theta], "\<) in polar coordinates.\>"]\), FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tIt is also possible for the r coordinate to be negative. (We placed no \ restrictions on r.) We interpret a negative value of r as follows: we draw \ the ray ", StyleBox["opposite", FontVariations->{"Underline"->True}], " to the ray through \[Theta], and plot the point P at a distance |r| from \ the origin along this \"opposite\" ray. We think of this as going |r| units \ in the ", StyleBox["negative", FontVariations->{"Underline"->True}], " direction along the ray.\n\tTry this with the polar point (-3, ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/6\)]], "). Input the data r = -3 and \[Theta] = ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/6\)]], " in the program above. What do you see? Your point should be in the ", StyleBox["third", FontVariations->{"Underline"->True}], " quadrant since it is on the ray opposite one in the first quadrant (", "i.e. \[Theta] = ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/6\)]], ")." }], "Text"], Cell[TextData[{ "\tUnlike rectangular coordinates, however, there is ", StyleBox["not", FontVariations->{"Underline"->True}], " a unique way of representing a point in the plane with polar coordinates. \ For example, the point given in polar coordinates by (r,\[Theta]) is exactly \ the same as the point given by (r, \[Theta] + 2\[Pi]), (r, \[Theta] + 4\[Pi]) \ and (r, \[Theta] + 2k\[Pi]) for any integer k. This follows simply from the \ fact that there are 2\[Pi] radians in a circle, so adding an integer multiple \ of 2\[Pi] to the angle \[Theta] does not change the ray along which we \ measure the distance |r|.", " ", "In addition, a point with polar coordinates (r, \[Theta]) can also be \ described by polar coordinates (-r, \[Theta] + \[Pi]) since the ray with \ angle \[Theta] + \[Pi] is the one opposite the ray with angle \[Theta]." }], "Text"], Cell[TextData[StyleBox[ "SUMMARY: A point with polar coordinates (r, \[Theta]) is the same point as \ that given by (r, \[Theta] + 2k\[Pi]) or (-r, \[Theta] + \[Pi] + 2k\[Pi]) for \ any integer k.", FontWeight->"Bold"]], "Text"], Cell[TextData[{ StyleBox["Exercise:", FontWeight->"Bold"], " For the various values of r and \[Theta] given below, plot the points \ using polar coordinates (it is helpful to use polar graphing paper which you \ can obtain from your instructor). Then use the last program above to test \ your answers (highlight the blue values and change them to those for the \ point you wish to plot).", " ", "Remember: Try them first ", StyleBox["without", FontSlant->"Italic"], " the program, then use the program to check your answers." }], "Text"], Cell[TextData[{ StyleBox["Exercises:", FontWeight->"Bold"], " Plot each of the following points given in polar coordinates:\n\na) (3, ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ")\n\nb) (3, -", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ")\n\nc) (-2, \[Pi])\n\nd) (-3, 0)\n\ne) (1, ", Cell[BoxData[ \(TraditionalForm\`\(5 \[Pi]\)\/4\)]], ")\n\nf) (3, ", Cell[BoxData[ \(TraditionalForm\`\(9 \[Pi]\)\/4\)]], ")\n\ng)", " ", "(-2, -", Cell[BoxData[ \(TraditionalForm\`\(17 \[Pi]\)\/4\)]], ")" }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Conversion formulas", "Section", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ "\tIt is sometimes necessary to convert from polar coordinates to \ rectangular coordinates, or vice versa. 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\(TraditionalForm\`x\/r\)]], " and sin(\[Theta]) = ", Cell[BoxData[ \(TraditionalForm\`y\/r\)]], ".\n\n Rearranging these equations, we get the following conversion \ formulas for converting between rectangular and polar coordinates: \n\t\n\tx \ = r cos(\[Theta]) and y = r sin(\[Theta]).", StyleBox["\n\t\n", FontWeight->"Bold"], "To get the formulas above, we assumed that r was positive, i.e., r > 0. \ What about when r < 0? In the previous section, we saw that the point P = \ (r, \[Theta]) could also be represented by (-r, \[Pi] + \[Theta]) where -r > \ 0 (Note: - r > 0 when r < 0). So in the case where r < 0, we can rewrite the \ point (r, \[Theta]) as (-r, \[Pi] + \[Theta]) and then apply our preceding \ results to obtain:\n\n\tx = -r cos(\[Pi] + \[Theta]) and y = -r sin(\[Pi] + \ \[Theta]).\n\t\nBut remember, cos(\[Pi] + \[Theta]) = -cos(\[Theta]) and sin(\ \[Pi] + \[Theta]) = sin(\[Theta])\nso we get:\n\n\tx = r cos(\[Theta]) and y \ = r sin(\[Theta])\n\t\njust as before. \n", StyleBox[ "\nTo summarize, a point P with polar coordinates (r, \[Theta]) has \ rectangular coordinates given by \nx = r cos(\[Theta]) and", FontWeight->"Bold"], " ", StyleBox["y = r sin(\[Theta]).\n\n", FontWeight->"Bold"], "The following is a ", StyleBox["Mathematica", FontSlant->"Italic"], " program that will do these conversions. That is, given a point in polar \ coordinates (r, \[Theta]) the program will compute the corresponding \ rectangular coordinates (x, y). Execute the program to see the polar point \ (3, \[Pi]/6) converted to rectangular coordinates. " }], "Text"], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[r, \[Theta], x, \ y]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{"r", ":=", StyleBox["3", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{"\[Theta]", ":=", " ", StyleBox[ FractionBox["\[Pi]", StyleBox["6", FontColor->RGBColor[0, 0, 1]]], FontColor->RGBColor[0, 0, 1]]}], ";", " ", StyleBox[\( (*\ in\ radians, \ \(please!\)*) \), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[\(x\ := \ r\ \ Cos\ [\[Theta]]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], StyleBox["\n", FontColor->GrayLevel[0.666667]], StyleBox[\(y\ := \ r\ Sin[\[Theta]]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], StyleBox["\n", FontColor->GrayLevel[0.666667]], StyleBox[ \(Print["\", {r, \[Theta]}, "\< are \>", {x, y}]\), FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tNow experiment with this program: highlight the r or \[Theta] value in \ blue and type in the value you want. Then execute the program to have the \ point converted to rectangular coordinates. *Don't forget to give the value \ for \[Theta] in radians instead of degrees!*\n\tWhat about the reverse \ conversion? That is, suppose a point P is given by rectangular coordinates P \ = (x, y). How can we find polar coordinates for P", "? ", "This turns out to be a little more complicated. 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ogoo05Ioo`8003ioo`004goo00<007ooOol0EWoo0`00?Goo000COol00`00Oomoo`1GOol3000lOol0 01=oo`03001oogoo05Qoo`8003aoo`004goo00<007ooOol0F7oo0`00>goo000COol2001JOol3000f Ool00`00Oomoo`01Ool001=oo`03001oogoo05Yoo`8003Ioo`03001oogoo005oo`004goo00<007oo Ool0FWoo0`00=Goo00<007ooOol00Goo000COol00`00Oomoo`1KOol3000dOol00`00Oomoo`01Ool0 01=oo`03001oogoo05aoo`8003Aoo`03001oogoo005oo`004goo00<007ooOol0G7oo0`00"], ImageRangeCache->{{{0, 171.75}, {287, 0}} -> {-0.269335, -0.231358, 0.0135042, 0.0135042}}], Cell[TextData[{ "\tNotice that the distance |r| from the origin to the point P is the \ length of the hypotenuse of the triangle. Using the Pythagorean theorem, we \ can take r = ", Cell[BoxData[ \(TraditionalForm\`\@\(x\^2 + y\^2\)\)]], ". Since |r| represents a length which is always a positive value, we \ always choose the positive square root. \n\tWhat about the angle \[Theta]? \ We know that tan(\[Theta]) = ", Cell[BoxData[ \(TraditionalForm\`y\/x\)]], " from right triangle trigonometry, so \[Theta] =", Cell[BoxData[ \(TraditionalForm\`\(\ tan\^\(-1\)\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`y\/x\)]], "). But we must be careful! Recall that when we defined the inverse ", Cell[BoxData[ \(TraditionalForm\`\(tangent\ \)\)]], "function, we restricted the domain of tangent to ", Cell[BoxData[ \(TraditionalForm\`\(-\[Pi]\)\/2\)]], " < \[Theta] < ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " since tan(\[Theta]) is one-to-one along this interval. This would say \ that the point P ", StyleBox["has", FontSlant->"Italic"], " to be in the first or fourth quadrant (r > 0 and ", Cell[BoxData[ \(TraditionalForm\`\(-\[Pi]\)\/2\)]], " < \[Theta] < ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], "). What if it isn't? That is, what if the rectangular coordinates for P \ are (a, -b) or (-a, b) where a and b are positive numbers? We know that \ converting to polar coordinates doesn't change which quadrant P is in, just \ how we get there. For example, what if P is the point (1,-", Cell[BoxData[ \(TraditionalForm\`\@3\)]], "), in the ", StyleBox["third", FontSlant->"Italic"], " quadrant? Then we find the solution to ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_0\)]], " =", Cell[BoxData[ \(TraditionalForm\`\(\ tan\^\(-1\)\)\)]], "(|", Cell[BoxData[ \(TraditionalForm\`y\/x\)]], "|) where -\[Pi]/2 < ", Cell[BoxData[ \(TraditionalForm\`\(\ \(\[Theta]\ \)\_0\)\)]], " < \[Pi]/2. This ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_0\)]], " is the reference angle for the angle \[Theta] that we want. To get \ \[Theta] itself, we figure out what angle in the ", StyleBox["third", FontSlant->"Italic"], " quadrant has ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_0\)]], " as a reference angle. \n\tFor example, using the point P = (1, -", Cell[BoxData[ \(TraditionalForm\`\@3\)]], "), we see that r = ", Cell[BoxData[ \(TraditionalForm\`\@\(x\^2 + y\^2\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\@\(1\^2 + \((\(-\@3\))\)\^2\)\)]], "= 2. A reference angle can be found by", Cell[BoxData[ \(TraditionalForm\`\(\ \[Theta]\_0 = \ tan\^\(-1\)\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`\@3\/1\)]], ") = t", Cell[BoxData[ \(TraditionalForm\`an\^\(\(-1\)\ \)\)]], Cell[BoxData[ \(TraditionalForm\`\@3\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/3\)]], ". We know that we want P to be in the third quadrant since x and y are \ both negative, so we find the angle \[Theta] in the third quadrant that has ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_0\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/3\)]], " as a reference angle, that is, \[Theta] = \[Pi] + ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/3\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(4 \[Pi]\)\/3\)]], ". So (2, ", Cell[BoxData[ \(TraditionalForm\`\(4 \[Pi]\)\/3\)]], ") are polar coordinates for the point P whose rectangular coodinates are ( \ -1, -", Cell[BoxData[ \(TraditionalForm\`\@3\)]], ")." }], "Text"], Cell[TextData[{ "\tHere is a small program for converting rectangular coordinates to polar \ coordinates. Remember that there are many pairs of polar coordinates that \ represent the same point. This program gives only one pair of polar \ coordinates, the one in which r \[GreaterEqual] 0 and 0 \[LessEqual] \[Theta] \ < 2\[Pi]. When you execute the program, there will be a warning from ", StyleBox["Mathematica", FontSlant->"Italic"], " that not all solutions have been found. You can just consider this \ message to be a reminder that polar coordinates are not unique and can \ otherwise ignore the message.", " ", "(Refer back to the previous section for other choices.)" }], "Text"], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[x, y, r, \[Theta]]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{"x", ":=", StyleBox[\(-1\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{"y", ":=", StyleBox[\(-\@3\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", StyleBox[\( (*Change\ nothing\ below\ this\ \(line!\)*) \), FontColor->RGBColor[1, 0, 0]], "\n", RowBox[{"r", StyleBox[":=", FontColor->GrayLevel[0.666667]], StyleBox[\(\@\(x\^2 + y\^2\)\), FontColor->GrayLevel[0.666667]]}], StyleBox[";", FontColor->GrayLevel[0.666667]], StyleBox["\n", FontColor->GrayLevel[0.666667]], StyleBox[\(\[Theta] := ArcTan[y/x]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], StyleBox["\n", FontColor->GrayLevel[0.666667]], StyleBox[ \(If[x \[GreaterEqual] 0, Print["\", {x, y}, "\< is \>", {r, \[Theta]}], Print["\", {x, y}, "\< is \>", {r, \[Pi] + \[Theta]}]]\), FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox[ "In summary, if a point P has rectangular coordinates (x, y), the polar \ coordinates satisfy ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`r\^2\)], FontWeight->"Bold"], StyleBox[" = ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`x\^2\)], FontWeight->"Bold"], StyleBox[" + ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`y\^2\)], FontWeight->"Bold"], StyleBox[" and tan(\[Theta]) = ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`y\/x\)], FontWeight->"Bold"], ". ", StyleBox["Be sure to choose r and \[Theta] to give the correct quadrant!", FontWeight->"Bold"] }], "Text"], Cell[TextData[{ StyleBox["Exercise: ", FontWeight->"Bold"], "Convert from polar to rectangular :\n\t(a) (3, \[Pi]/2)\t\t(b) (-2, \ 3\[Pi]/4)\t\t(c) (-3, -\[Pi]/4)\t\t\t(d) (10, \[Pi])", StyleBox["\n\t\nExercice: ", FontWeight->"Bold"], "Convert from rectangular to polar:\n\t(a) (0, 2)\t\t(b) (", Cell[BoxData[ \(\@3\)]], ", 1)\t\t(c) (", Cell[BoxData[ \(\(\@2, \)\)]], Cell[BoxData[ \(\(\ \(-\@2\)\)\)]], ")\t\t(d) (-2, -2", Cell[BoxData[ \(\@3\)]], ")\n\t\n", StyleBox["Check your answers ", FontWeight->"Bold"], StyleBox["after", FontWeight->"Bold", FontVariations->{"Underline"->True}], StyleBox[ " doing the work yourself, by using the two previous yellow boxes!", FontWeight->"Bold"], "\t" }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Graphing functions in polar coordinates", "Section", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ StyleBox["Before beginning, be sure to activate the next ", FontWeight->"Bold"], StyleBox["two", FontWeight->"Bold", FontVariations->{"Underline"->True}], StyleBox[" yellow boxes!", FontWeight->"Bold"], " The first imports the graphics program necessary to produce the graphics \ in this section. Activate this program by moving the cursor into the yellow \ box and pressing shift-return.", " " }], "Text"], Cell[BoxData[ \(<< \ Graphics`Graphics`\)], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tThis second program will cause the graphs produced by the rest of the \ programs in this section to be drawn on polar coordinate paper. Since we \ will be graphing functions in polar coordinates, ", "i.e. ", "r = f(\[Theta]), it helps to have polar graph paper! ", StyleBox["Activate", FontVariations->{"Underline"->True}], " the yellow box that follows ", StyleBox["now", FontVariations->{"Underline"->True}], "." }], "Text"], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[circles, lines, polarpaper]\), Background->RGBColor[1, 1, 0]], StyleBox[";", Background->RGBColor[1, 1, 0]], StyleBox["\n", Background->RGBColor[1, 1, 0]], StyleBox[ \(circles = {Thickness[0.0001], Circle[{0, 0}, 1], Circle[{0, 0}, 2], Circle[{0, 0}, 3], Circle[{0, 0}, 4]}\), Background->RGBColor[1, 1, 0]], StyleBox[";", Background->RGBColor[1, 1, 0]], StyleBox["\n", Background->RGBColor[1, 1, 0]], StyleBox["\t", Background->RGBColor[1, 1, 0]], StyleBox[ \(lines = {Thickness[0.0001], Line[{{\(-2\)\ \@3, \(-2\)}, {2\ \@3, 2}}], Line[{{\(-2\)\ \@2, \(-2\)\ \@2}, {2\ \@2, 2\ \@2}}], Line[{{\(-2\), \(-2\)\ \@3}, {2, 2\ \@3}}], Line[{{\(-2\)\ \@3, 2}, {2\ \@3, \(-2\)}}], Line[{{\(-2\)\ \@2, 2\ \@2}, {2\ \@2, \(-2\)\ \@2}}], Line[{{\(-2\), 2\ \@3}, {2, \(-2\)\ \@3}}]}\), Background->RGBColor[1, 1, 0]], StyleBox[";", Background->RGBColor[1, 1, 0]], StyleBox["\n", Background->RGBColor[1, 1, 0]], RowBox[{ StyleBox["polarpaper", FontColor->GrayLevel[0], Background->RGBColor[1, 1, 0]], StyleBox["=", Background->RGBColor[1, 1, 0]], StyleBox[\({circles, lines}\), Background->RGBColor[1, 1, 0]]}], StyleBox[";", Background->RGBColor[1, 1, 0]], StyleBox[ \(Show[Graphics[polarpaper], AspectRatio -> Automatic, Axes -> True] \), Background->RGBColor[1, 1, 0]], StyleBox[";", Background->RGBColor[1, 1, 0]]}]], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "Notice that instead of a rectangular grid of lines (as we have for \ rectangular coordinates), we have a series of concentric circles (for various \ r values) and a series of rays emanating from the origin at various angles \ \[Theta].\n\tJust as we could graph functions given in rectangular \ coordinates (i.e., functions of the form y = f(x) in which we choose values \ of x, plug them into f, and get out values for y), we can graph functions \ given in polar coodinates (i.e., functions of the form r = f(\[Theta]) in \ which we choose values of \[Theta], plug them into f, and get out values for \ r). To get the graph of such a function we could make a table of r and \ \[Theta] values and then madly plot these points in polar coordinates until \ we can see the shape of a curve.", " ", "But as we know is the case with rectangular coordinates, this is a very \ inefficient method! Instead, we can learn a few 'basic polar graphs' and \ then, once we are comfortable with their shapes, readily alter them." }], "Text"], Cell[CellGroupData[{ Cell["Basic Polar Graphs", "Subsection", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell["\<\ \tMany shapes are more simply described in rectangular coordinates, but some \ are more simply described in polar coordinates. We'll now take a look at \ some basic graphs that are particularly suited to polar coordinates.\ \>", "Text"], Cell[CellGroupData[{ Cell[TextData[StyleBox["Circles", FontSize->12]], "Subsubsection", FontSize->16, FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell["\<\ \tWe begin with the simplest example. Try the following exercise on your \ own.\ \>", "Text"], Cell[TextData[{ StyleBox["Exercise 1:", FontSize->12, FontWeight->"Bold"], StyleBox[" ", FontSize->12], StyleBox["Consider the equation ", FontSize->12, FontWeight->"Plain"], StyleBox["r = a", FontSize->12], StyleBox[" where ", FontSize->12, FontWeight->"Plain"], StyleBox["a", FontSize->12], StyleBox[" is a positive number", FontSize->12, FontWeight->"Plain"], ". ", StyleBox["What will the corresponding graphs be for ", FontSize->12, FontWeight->"Plain"], StyleBox["a = 0, a = 2, a = 4 ", FontSize->12], "? ", StyleBox["Check your answer by activating the yellow ", FontSize->12, FontWeight->"Plain"], "box", StyleBox[" below.", FontSize->12, FontWeight->"Plain"] }], "Text"], Cell[BoxData[ \(Do[Print["\", a]; \n\t PolarPlot[a, {\[Theta], 0, 2 \[Pi]}, PlotRange -> {{\(-5\), 5}, {\(-5\), 5}}, PlotStyle -> {Hue[a/10], Thickness[0.02]}, AspectRatio -> Automatic], {a, 0, 4, 2}]\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tFrom this exercise, we conclude that the graphs of the polar functions \ f(\[Theta]) = 0, f(\[Theta]) = 1, and f(\[Theta]) = 2 are simply \ circles--ones with their centers at the origin. It makes sense that the \ graph of f(\[Theta]) = c (a constant) is a circle since a circle is the set \ of points that are at a constant distance (i.e., the length of the radius) \ from the origin. Another way to think of this is that in the functions f(\ \[Theta]) = c, the angle \[Theta] can be anything while the value r is \ fixed.", " ", "When we graph all possible angles \[Theta], leaving r fixed, we get a \ circle with radius r." }], "Text"], Cell[TextData[{ "\tThere are two other types of circles whose equations in polar \ coordinates are also fairly simple: those which pass through the origin and \ have center either on the x-axis or on the y-axis. Let's first consider the \ case where the center is on the x-axis at the point (a, 0). In rectangular \ coordinates, of course, we know the equation to be ", Cell[BoxData[ \(TraditionalForm\`\((x - a)\)\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " = ", Cell[BoxData[ \(TraditionalForm\`a\^2\)]], ", which expands to ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], " = 2", StyleBox["ax", FontSlant->"Italic"], ". Using the conversion rules x = r cos(\[Theta]), y = r sin(\[Theta]), \ this becomes ", Cell[BoxData[ \(TraditionalForm\`r\^2\)]], " = 2ar cos(\[Theta]). Since r = 0 is just the equation for a point, we \ can ignore the case where r = 0. So, for the case r \[NotEqual] 0, we can \ divide by r to get r = 2a cos(\[Theta]). Hence the polar equation for a \ circle passing through the origin with center at (a, 0) (hence with radius ", StyleBox["a", FontSlant->"Italic"], ") is ", StyleBox["r = 2a cos(\[Theta])", FontWeight->"Bold"], ". We'll again use ", StyleBox["Mathematica", FontSlant->"Italic"], " to draw this graph for us, taking the value of ", StyleBox["a", FontSlant->"Italic"], " to be 3. Note: The value of ", StyleBox["a", FontSlant->"Italic"], " is the radius of the circle in the graph of the function." }], "Text"], Cell[BoxData[ \(\(PolarPlot[5\ Cos[\[Theta]], {\[Theta], 0, 2\ \[Pi]}, PlotStyle -> {Thickness[0.012], Hue[1.0]}, Epilog -> polarpaper]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tIn the example above, ", StyleBox["a", FontSlant->"Italic"], " = 3, so ", StyleBox["a", FontSlant->"Italic"], " is positive. What happens if ", StyleBox["a", FontSlant->"Italic"], " is negative", "? ", "For example, let's take ", StyleBox["a", FontSlant->"Italic"], " = -3. See if you can predict the graph before activating the box below." }], "Text"], Cell[BoxData[ \(\(PolarPlot[\(-5\)\ Cos[\[Theta]], {\[Theta], 0, 2\ \[Pi]}, PlotStyle -> {Thickness[0.012], Hue[1.0]}, Epilog -> polarpaper]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tIf you thought about the effect of having a negative r-value (as \ described in the first section of this notebook), you probably were correct \ in your prediction. For each value of \[Theta], the point is plotted on the \ ray ", StyleBox["opposite", FontVariations->{"Underline"->True}], " the one at angle \[Theta]; that is at angle \[Theta] + \[Pi]. Hence the \ semicircle which in the first graph was in the ", StyleBox["first", FontVariations->{"Underline"->True}], " quadrant now is being drawn in the ", StyleBox["third", FontVariations->{"Underline"->True}], " quadrant. To illustrate this, let's plot both circles r = 6 \ cos(\[Theta]) and r = -6 cos(\[Theta]) on the same set of axes. This time \ we'll color code the graphs so that the portion of each graph with \[Theta] \ between 0 and ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/4\)]], " is red, with \[Theta] between ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/4\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " is green and so forth. This will enable us to see how pieces of the \ different graphs correspond to each other." }], "Text"], Cell[BoxData[ StyleBox[\(Clear[\[Theta]]; Do[PolarPlot[{6\ Cos[\[Theta]], \ \(-6\)\ Cos[\[Theta]]}, \ {\[Theta], \(k\ \[Pi]\)\/4, \(\((k + 1)\)\ \[Pi]\)\/4}, PlotRange -> {{\(-6\), 6}, {\(-6\), 6}}, \n\t\t PlotStyle -> {{Hue[k/4], Thickness[0.01]}, {Hue[k/4], Thickness[0.008]}}, AspectRatio -> Automatic, Epilog -> polarpaper], {k, 0, 3}]\), FontColor->GrayLevel[0.666667]]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tNow try some examples on your own. The following program graphs r = a \ sin(\[Theta]) for various values of a which you may choose. To change the \ value of ", StyleBox["a,", FontSlant->"Italic"], " highlight the value in blue and enter the value you want." }], "Text"], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[a]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{"a", "=", StyleBox["3", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", StyleBox[ \(PolarPlot[a\ Sin[\[Theta]], {\[Theta], 0, 2\ \[Pi]}, \ PlotRange -> \ {{\(-10\), 10}, {\(-10\), 10}}, PlotStyle -> {Thickness[0.008], Hue[ .01]}, Epilog -> polarpaper] \), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\t The following program graphs r = a cos(\[Theta]) for a = 5. You can \ change the value in blue to any value of ", StyleBox["a", FontSlant->"Italic"], " that you want. " }], "Text"], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[a]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], StyleBox["\n", FontColor->GrayLevel[0.666667]], RowBox[{ StyleBox["a", FontColor->GrayLevel[0]], StyleBox["=", FontColor->GrayLevel[0]], StyleBox["5", FontColor->RGBColor[0, 0, 1]]}], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", StyleBox[ \(PolarPlot[a\ Cos[\[Theta]], {\[Theta], 0, 2\ \[Pi]}, PlotRange -> {{\(-10\), 10}, {\(-10\), 10}}, PlotStyle -> {Thickness[0.008], Hue[ .01]}, Epilog -> polarpaper] \), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox[ " \tFrom your experiements with these programs, you should notice that the \ effect of multiplying the polar function r = 1, r = sin(\[Theta]), or r = \ cos(\[Theta]) by the constant ", FontSlant->"Plain"], "a", StyleBox[" is a radial expansion or contraction of the circle", FontSlant->"Plain"], ". ", StyleBox[ "We can see this effect clearly if we sketch the graphs of r = a sin(\ \[Theta]) for various values of ", FontSlant->"Plain"], "a ", StyleBox["on the same set of axes", FontSlant->"Plain"], ". ", StyleBox["The program below does this for the ", FontSlant->"Plain"], "a", StyleBox["-values ", FontSlant->"Plain"], Cell[BoxData[ FormBox[ StyleBox[\(1\/2, \ 1, \ 2, \ and\ 3\), FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False}], TraditionalForm]]], StyleBox[".", FontSlant->"Plain"] }], "Text", FontSlant->"Italic"], Cell[BoxData[ StyleBox[\(Clear[\[Theta]]; PolarPlot[{0.5\ Sin[\[Theta]], Sin[\[Theta]], 2\ Sin[\[Theta]], 3\ Sin[\[Theta]]}, {\[Theta], 0, 2 \[Pi]}, PlotRange -> {{\(-4\), 4}, {\(-4\), 4}}, PlotStyle -> {{Thickness[0.01], Hue[ .01]}, {Thickness[ .01], Hue[ .1]}, {Thickness[ .01], Hue[ .4]}, {Thickness[ .01], Hue[ .6]}}, AspectRatio -> Automatic, Epilog -> polarpaper]; \), FontColor->GrayLevel[0.666667]]], "Input", Background->RGBColor[1, 1, 0]] }, Open ]], Cell[CellGroupData[{ Cell["Lines", "Subsubsection", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ "\tA few linear functions have simple polar forms. For example, if a line \ goes through the origin, it is simply the set of all points with a fixed \ \[Theta] value, where r can be anything--positive or negative. So its polar \ equation is \[Theta] = ", StyleBox["c", FontSlant->"Italic"], " where ", StyleBox["c", FontSlant->"Italic"], " is a constant. If \[Theta] = ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/3\)]], " for example, the graph is the line making angle ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/3\)]], "with the positive x-axis.", " ", "Enter the following program to see the graph of \[Theta] = \[Pi]/3." }], "Text"], Cell[BoxData[ StyleBox[\(Clear[polarline]; \n polarline = {AbsoluteThickness[4], Line[{{1, Tan[\[Pi]\/3]}, {\(-1\), \(-Tan[\[Pi]\/3]\)}}]}; \n Show[Graphics[polarline], Axes -> True, AspectRatio -> Automatic, \ Epilog -> polarpaper]; \), FontColor->GrayLevel[0.666667]]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "What is such a line in rectangular coordinates? The y-intercept is 0 \ since it passes through the origin. So we only need to find the slope m to \ use the form y = mx + b. But the slope is given by ", Cell[BoxData[ \(TraditionalForm\`y\/x\)]], " which is just tan(\[Theta]). In other words, the polar line given by \ \[Theta] = ", StyleBox["c", FontSlant->"Italic"], " where ", StyleBox["c", FontSlant->"Italic"], " is a constant translates to y = (tan(\[Theta])) x in rectangular \ coordinates.\n\tWhat about horizontal and vertical lines? In rectangular \ coordinates, a horizontal line is given by y = b for some constant b. We can \ use the conversion formulas to change this equation to polar coordinates. \ Since y = r sin(\[Theta]), we have r sin(\[Theta]) = b, or alternatively r = \ ", Cell[BoxData[ \(TraditionalForm\`b\/\(sin(\[Theta])\)\)]], ". Let's graph the polar function r = ", Cell[BoxData[ \(TraditionalForm\`3\/\(sin(\[Theta])\)\)]], "to verify that polar equations of this form give horizontal lines." }], "Text"], Cell[BoxData[ \(\(PolarPlot[3\/Sin[\[Theta]], {\[Theta], 0, 2\ \[Pi]}, PlotRange -> {{\(-3\), 3}, {\(-1\), 4}}, PlotStyle -> {Thickness[0.01], Hue[ .01]}, Epilog -> polarpaper]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tTo see how to write the equation for vertical lines, we proceed the same \ way. In rectangular coordinates, a vertical line is x = c for some constant \ c. In polar coordinates this becomes r cos(\[Theta]) = c, or r = ", Cell[BoxData[ \(TraditionalForm\`c\/\(cos(\[Theta])\)\)]], ". Let's plot the polar graph of this function for c = 2." }], "Text"], Cell[BoxData[ \(\(PolarPlot[2\/Cos[\[Theta]], {\[Theta], 0, 2\ \[Pi]}, PlotRange -> {{\(-1\), 3}, {\(-1\), 3}}, PlotRegion -> {{0, 1}, {0, 1}}, PlotStyle -> {Thickness[0.01], Hue[ .01]}, \ Axes -> True, Ticks -> {{1, 2}, {1, 2}}, Epilog -> polarpaper]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": Convert the equation r sin(\[Theta]) = 5 to rectangular coordinates and \ sketch the graph." }], "Text"], Cell[TextData[{ StyleBox["Exercise: ", FontWeight->"Bold"], "Do the same for r cos(\[Theta]) = ", Cell[BoxData[ \(TraditionalForm\`3\/2\)]], "." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Cardioids and Lima\[CCedilla]ons", "Subsubsection", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ "\tNow we examine some shapes that are ", StyleBox["much", FontSlant->"Italic"], " better suited to polar coordinates than rectangular coordinates. They \ all have polar equations of the form a + b sin(\[Theta]), a - b \ sin(\[Theta]), a + b cos(\[Theta]), or a - b cos(\[Theta]), and their graphs \ are called either ", StyleBox["cardioids", FontVariations->{"Underline"->True}], " or ", StyleBox["lima\[CCedilla]ons", FontVariations->{"Underline"->True}], ".\n\tLet's look at the polar function r = 1 + cos(\[Theta]). First notice \ that since the cosine function is even, the function r = 1 + cos(\[Theta]) is \ even, so its graph will be symmetric with respect to the x-axis. (Note: \ Recall that a function is ", StyleBox["even", FontSlant->"Italic"], " if f (", StyleBox["variable", FontSlant->"Italic"], ") = f ( - ", StyleBox["variable)", FontSlant->"Italic"], ". In rectangular coordinates, even functions have graphs that are \ symmetric with respect to the y-axis since the variable is x, and positive \ x-values and negative x-values lie on opposite sides of the y-axis. For \ polar graphs, the variable is \[Theta], and positive \[Theta]-values and \ negative \[Theta]-values lie on opposite sides of the x-axis for 0 \ \[LessEqual] \[Theta] \[LessEqual] \[Pi]. This means that if we can \ determine the shape of that portion of the graph ", StyleBox["above", FontSlant->"Italic"], " the x-axis, we know the rest by symmetry. In particular, we need only \ check values of \[Theta] from 0 to \[Pi]. We can plot a few points in this \ range:\n\t", StyleBox["\[Theta]\t\tcos(\[Theta])\t\tr = 1 + cos(\[Theta])", FontWeight->"Bold"], "\n\t0\t\t1\t\t2\n\t", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/6\)]], "\t\t", Cell[BoxData[ \(TraditionalForm\`\@3\/2\)]], "\t\t1 + ", Cell[BoxData[ \(TraditionalForm\`\@3\/2\)]], "\n\t\n\t", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/3\)]], "\t\t", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], "\t\t", Cell[BoxData[ \(TraditionalForm\`3\/2\)]], "\n\t\n\t", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], "\t\t0\t\t1\n\t\n\t", Cell[BoxData[ \(TraditionalForm\`\(2 \[Pi]\)\/3\)]], "\t\t-", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], "\t\t", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], "\n\t\n\t", Cell[BoxData[ \(TraditionalForm\`\(5 \[Pi]\)\/6\)]], "\t\t-", Cell[BoxData[ \(TraditionalForm\`\@3\/2\)]], "\t\t1 - ", Cell[BoxData[ \(TraditionalForm\`\@3\/2\)]], "\n\t\n\t\[Pi]\t\t-1\t\t0\n\t\n\tPlotting these points, we see the trace of \ a curve as shown by executing the program below:" }], "Text"], Cell[BoxData[ \(\(PolarPlot[1\ + \ Cos[\[Theta]], {\[Theta], 0, \[Pi]}, PlotStyle -> {Thickness[0.012], Hue[1.0]}, PlotRange -> {{\(-4\), 4}, {\(-4\), 4}}, Epilog -> polarpaper]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tNow by reflecting across the axis of symmetry, the x-axis, we see the \ entire graph--a ", StyleBox["cardioid", FontVariations->{"Underline"->True}], "." }], "Text"], Cell[BoxData[ \(PolarPlot[1\ + \ Cos[\[Theta]], {\[Theta], 0, 2\ \[Pi]}, PlotStyle -> {Thickness[0.012], Hue[1.0]}, PlotRange -> {{\(-4\), 4}, {\(-4\), 4}}, Epilog -> { polarpaper, {AbsoluteThickness[3], Line[{{\(-4\), 0}, {4, 0}}]}}]; \ Print["\"]\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tNote: With a slightly more sophisticated examination, we could save \ ourselves a lot of work here! Calculating and plotting points takes time; we \ should do it only as necessary. To get the basic shape of this graph, we \ could plot only two or three points. Let's see how. \n\tRemember that the \ cosine function ", StyleBox["decreases", FontSlant->"Italic"], " from 1 to 0 to -1 as \[Theta] goes from 0 to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " to \[Pi]. This means that the function r = 1 + cos(\[Theta]) also \ decreases from 2 to 1 to 0 in this same range. So we could plot the points \ (r, \[Theta]) = (2,0), (1, ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], "), and (0, \[Pi]), joining them by a curve which gradually gets closer to \ the origin (since r is ", StyleBox["decreasing", FontSlant->"Italic"], "), then finally reflecting across the x-axis to complete the graph." }], "Text"], Cell[TextData[{ "\tTry this analysis on the polar function r = 3 -2 cos(\[Theta]). \ Remember, check for symmetry, then notice that cos(\[Theta]) ", StyleBox["decreases", FontSlant->"Italic"], " from 0 to \[Pi]. ", StyleBox["After", FontVariations->{"Underline"->True}], " you have drawn your own graph, use the yellow box below to check your \ answer." }], "Text"], Cell[BoxData[ \(Print[ "\"]; PolarPlot[3\ - \ 2\ Cos[\[Theta]], {\[Theta], 0, 2\ \[Pi]}, PlotStyle -> {Thickness[0.012], Hue[1.0]}, PlotRange -> {{\(-6\), 5}, {\(-6\), 5}}, Epilog -> { polarpaper, {AbsoluteThickness[4], Line[{{\(-6\), 0}, {5, 0}}]}}]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[ "\tNotice that this graph has a slightly different shape; it does not quite \ come to a sharp point where it nears the origin. The shape of this cusp is \ caused by the a and b values in r = a \[PlusMinus] b cos(\[Theta]). When |a| \ = |b| (as in the first example), we get a true cardioid forming a pointed \ cusp as it passes through the origin. But in the second example, |b| < \ |a|."], "Text"], Cell[TextData[{ "\tWhat happens if |a| < |b|? Let's consider r = 1 + 2 cos(\[Theta]). \ Again we have symmetry with respect to the x-axis so we need only look at \ \[Theta] between 0 and \[Pi]. Also since, as \[Theta] goes from 0 to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " to \[Pi], cos(\[Theta]) decreases from 1 to 0 to -1, and so the polar \ function r = 1 + 2 cos(\[Theta]) decreases from 3 to 1 to -1. We have \ something new here--a negative r-value! How does this affect the graph?\n\t\ Since the r-value is continuously decreasing from 3 to 1 to -1, it becomes 0 \ at some point (i.e. the curve passes through the origin). At what point? \ Solve 1 + 2 cos(\[Theta]) = 0 to find that cos(\[Theta]) = -", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], ", or \[Theta] = ", Cell[BoxData[ \(TraditionalForm\`\(2 \[Pi]\)\/3\)]], ". So when we reach the angle \[Theta] = ", Cell[BoxData[ \(TraditionalForm\`\(2 \[Pi]\)\/3\)]], ", the curve passes through the origin, r becomes negative so we begin \ drawing the curve along the ray opposite that at \[Theta]! Let's look at the \ graph of 1 + 2 cos(\[Theta]) for 0 \[LessEqual] \[Theta] \[LessEqual] \[Pi] \ by activating the program below." }], "Text"], Cell[BoxData[ \(\(PolarPlot[1\ + 2\ \ Cos[\[Theta]], {\[Theta], 0, \ \[Pi]}, PlotStyle -> {Thickness[0.012], Hue[1.0]}, PlotRange -> {{\(-4\), 4}, {\(-4\), 4}}, Epilog -> polarpaper]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell["\<\ When we relect this portion of the graph across the x-axis to obtain the \ entire graph, we create a small loop inside the larger curve. Input the \ following program to see this:\ \>", "Text"], Cell[BoxData[ \(\(PolarPlot[1\ + 2\ \ Cos[\[Theta]], {\[Theta], 0, 2\ \ \[Pi]}, PlotStyle -> {Thickness[0.012], Hue[1.0]}, PlotRange -> {{\(-4\), 4}, {\(-4\), 4}}, Epilog -> { polarpaper, {AbsoluteThickness[3], Line[{{\(-4\), 0}, {4, 0}}]}}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tBoth of the last two curves, the graphs of r = 3 - 2 cos(\[Theta]) and r \ = 1 + 2 cos(\[Theta]), are called ", StyleBox["lima\[CCedilla]ons", FontVariations->{"Underline"->True}], " -- the first, where |b| < |a|, is a lima\[CCedilla]on without an inner \ loop, and the second, where |a| < |b| , is a lima\[CCedilla]on with an inner \ loop. Since the cosine function is even, they are symmetric with respect to \ the x-axis.\n\tNow, on your own, try graphing some functions of the form r = \ ", StyleBox["a", FontSlant->"Italic"], " \[PlusMinus]", StyleBox[" b", FontSlant->"Italic"], " cos(\[Theta]) for various values of ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], ". Then use the following program to check your answers. You may change \ the values of ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], " as you wish; simply highlight the value in blue for ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], ", and type in with whatever value you wish. (To have ", StyleBox["Mathematica", FontSlant->"Italic"], " plot r = ", StyleBox["a", FontSlant->"Italic"], " -", StyleBox[" b", FontSlant->"Italic"], " cos(\[Theta]), just make ", StyleBox["b", FontSlant->"Italic"], " negative.)" }], "Text"], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[a, b]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{"a", ":=", StyleBox["4", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{"b", ":=", StyleBox["3", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", StyleBox[\( (*Do\ not\ change\ the\ next\ line*) \), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[ \(PolarPlot[a\ + \ b\ Cos[\[Theta]], \ {\[Theta], 0, 2\ \[Pi]}, PlotStyle -> {Thickness[0.012], Hue[1.0]}, PlotRange -> {{\(-Abs[a]\) - Abs[b], Abs[a] + Abs[b]}, { \(-Abs[a]\) - Abs[b], Abs[a] + Abs[b]}}, Epilog -> polarpaper] \), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tThe graphs of r = ", StyleBox["a", FontSlant->"Italic"], " +", StyleBox[" b", FontSlant->"Italic"], " sin(\[Theta]) and r = ", StyleBox["a", FontSlant->"Italic"], " - ", StyleBox["b", FontSlant->"Italic"], " sin(\[Theta]) are similar. The main difference stems from the fact that \ sine is not an even function like cosine. Instead, sin(\[Pi] - \[Theta]) = \ sin(\[Theta]). In other words, if we think back to rectangular coordinates, \ sine has the same sign in the first and second quadrants and the same in the \ third and fourth quadrants. In polar coordinates, this means that sine \ graphs are symmetric with respect to the y-axis.\n\tUse the following program \ to plot r = ", StyleBox["a", FontSlant->"Italic"], " + ", StyleBox["b", FontSlant->"Italic"], " sin(\[Theta]) for various values of ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], ". Then formulate a statement about the shapes of the graphs for various \ values of |", StyleBox["a|", FontSlant->"Italic"], " and |", StyleBox["b|", FontSlant->"Italic"], "." }], "Text"], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[a, b]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{"a", ":=", StyleBox["2", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{"b", ":=", StyleBox[\(-\ 3\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", StyleBox[\( (*Do\ not\ change\ the\ next\ line*) \), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[ \(PolarPlot[a\ + \ b\ Sin[\[Theta]], \ {\[Theta], 0, 2\ \[Pi]}, PlotStyle -> {Thickness[0.012], Hue[1.0]}, PlotRange -> {{\(-Abs[a]\) - Abs[b], Abs[a] + Abs[b]}, { \(-Abs[a]\) - Abs[b], Abs[a] + Abs[b]}}, Epilog -> polarpaper] \), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tWith enough experimentation, you should conclude that the graph of r = ", StyleBox["a", FontSlant->"Italic"], " \[PlusMinus] ", StyleBox["b", FontSlant->"Italic"], " sin(\[Theta]) is a cardioid if ", StyleBox["a", FontSlant->"Italic"], " = ", StyleBox["b", FontSlant->"Italic"], ", a lima\[CCedilla]on ", StyleBox["without", FontVariations->{"Underline"->True}], " an inner loop if ", StyleBox["a", FontSlant->"Italic"], " >", StyleBox[" b", FontSlant->"Italic"], " > 0 and a lima\[CCedilla]on ", StyleBox["with", FontVariations->{"Underline"->True}], " an inner loop if ", StyleBox["b", FontSlant->"Italic"], " >", StyleBox[" a", FontSlant->"Italic"], " > 0. All the graphs are symmetric with respect to the y-axis as \ indicated by the presence of the sine function. \n\n", StyleBox[ "SUMMARY:\n\t The graphs of r = a + a cos(\[Theta]), r = a - a \ cos(\[Theta]), r = a + a sin(\[Theta]) and r = a - a sin(\[Theta]) are ", FontWeight->"Bold"], StyleBox["cardioids", FontWeight->"Bold", FontVariations->{"Underline"->True}], StyleBox[ ".\n\t The graphs of r = a + b cos(\[Theta]), r = a - b cos(\[Theta]), r = \ a + b sin(\[Theta]) and r = a - b sin(\[Theta]) are ", FontWeight->"Bold"], StyleBox["lima\[CCedilla]ons", FontWeight->"Bold", FontVariations->{"Underline"->True}], StyleBox[" if a \[NotEqual] b", FontWeight->"Bold"], ". ", StyleBox["They have an inner loop if a < b", FontWeight->"Bold"], ". ", StyleBox[ "Otherwise they do not.\n\t Those with the cosine function are symmetric \ with respect to the x-axis; those with sine are symmetric with respect to the \ y-axis.", FontWeight->"Bold"] }], "Text"], Cell[TextData[{ "\tNow try graphing the polar functions given below.", " ", "Use a pencil and polar graph paper--not ", StyleBox["Mathematica", FontSlant->"Italic"], "! Once you have drawn your graph, activate the preceding yellow box to \ check your answer." }], "Text"], Cell[TextData[{ StyleBox["Exercise:", FontWeight->"Bold"], " Find the symmetries and then graph the curve given by r = 1 - \ sin(\[Theta]). Identify it according to the classifications above." }], "Text"], Cell[TextData[{ StyleBox["Exercise: ", FontWeight->"Bold"], "Find the symmetries and graph the curve given by r = 3 + 2 cos(\[Theta]). \ Identify it." }], "Text"], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold"], ": Find the symmetries and graph the curve given by 1 + 2 sin(\[Theta]). \ Identify it." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell[" Roses", "Subsubsection", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ "\tAnother type of graph which has a particularly simple expression in \ polar coordinates is called a ", StyleBox["rose", FontVariations->{"Underline"->True}], "; its polar equation looks like r = a sin(n\[Theta]) or r = a \ cos(n\[Theta]). We'll see that the number n determines the number of petals \ on the rose, but that determination depends on whether n is odd or even. It \ is easy to check that the graph of r = a cos(n\[Theta]) will be symmetric \ with respect to the x-axis, and that of r = a sin(n\[Theta]) symmetric with \ respect to the y-axis (you should do so!).\n\n\tLet's begin with r = 2 sin(3\ \[Theta]). Remember we expect symmetry with respect to the y-axis. The 3\ \[Theta] says that the angle whose sine we take is three times the \ \[Theta]-value. So as \[Theta] increases from 0 to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/6\)]], ", 3\[Theta] increases from 0 to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ", so r = 2 sin(3\[Theta]) increases from 2 sin(0) = 0 to 2 sin", Cell[BoxData[ \(TraditionalForm\`\(\((\[Pi]\/2)\)\ \)\)]], "= 2. Notice that r will never get larger than r = 2. \n\tNow as \[Theta] \ goes from ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/6\)]], " to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/3\)]], " and then to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ", 3\[Theta] goes from ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " to \[Pi] to ", Cell[BoxData[ \(TraditionalForm\`\(3 \[Pi]\)\/2\)]], ", and the sine function is decreasing throughout. So the r-value \ decreases in this range from 2 to 0 to -2. \n\tBelow we have plotted each of \ these pieces in a different color: \[Theta] from 0 to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/6\)]], " in yellow, \[Theta] from ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/6\)]], " to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/3\)]], " in green and \[Theta] from ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/3\)]], " to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " in blue." }], "Text"], Cell[BoxData[ \(Clear[part1, part2, part3]; \n part1 = PolarPlot[2\ Sin[3\ \[Theta]], {\[Theta], 0, \[Pi]\/6}, PlotStyle -> {Hue[0.2], Thickness[0.02]}, \ PlotRange -> {\(-2\), 2}, DisplayFunction -> Identity]; \n part2 = PolarPlot[2\ Sin[3\ \[Theta]], {\[Theta], \[Pi]\/6, \[Pi]\/3}, PlotStyle -> {Hue[0.4], Thickness[0.02]}, \ PlotRange -> {\(-2\), 2}, DisplayFunction -> Identity]; \n part3 = PolarPlot[2\ Sin[3\ \[Theta]], {\[Theta], \[Pi]\/3, \[Pi]\/2}, PlotStyle -> {Hue[0.6], Thickness[0.02]}, \ PlotRange -> {\(-2\), 2}, DisplayFunction -> Identity]; \n Show[{part1, part2, part3}, DisplayFunction -> $DisplayFunction, Epilog -> polarpaper]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell["\<\ Reflecting across the y-axis gives the complete graph of the 3-petaled rose \ as shown below. Notice that we plotted only 4 points to get this graph (and \ two were at the origin)!\ \>", "Text"], Cell[BoxData[ \(\(PolarPlot[2\ Sin[3\ \[Theta]], {\[Theta], 0, \[Pi]}, PlotStyle -> {Thickness[0.01], Hue[1.0]}, \ PlotRange -> {\(-2.2\), 2.2}, \ Epilog -> { polarpaper, {AbsolutePointSize[7], Point[{0, 0}]}, { AbsolutePointSize[7], Point[{\@3, 1}]}, {{AbsolutePointSize[7], Point[{0, \(-2\)}]}}}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "Now we'll plot the graph of r = 2 sin(2\[Theta]). What do you expect the \ graph to be? (Be careful; that's a trick question.)\nWe'll begin with the \ same analysis. We again have symmetry with respect to the y-axis. As \ \[Theta] goes from 0 to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/4\)]], " (so 2\[Theta] goes from 0 to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " where the sine function is increasing), we see that the r-value increases \ from 0 to 2. As \[Theta] goes from ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/4\)]], " to ", Cell[BoxData[ \(TraditionalForm\`\(\[Pi]\ \)\/2\ to\ \(3 \[Pi]\)\/4\)]], " (so ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], " \[LessEqual] 2\[Theta] \[LessEqual] ", Cell[BoxData[ \(TraditionalForm\`\(3 \[Pi]\)\/2\)]], " where the sine function decreases), the r-value decreases from 2 to 0 to \ -2. Then as \[Theta] goes from ", Cell[BoxData[ \(TraditionalForm\`\(3 \[Pi]\)\/4\)]], " to ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ", the r-value increases from -2 to 0.\n\tWe'll look at this much of the \ graph again with each fourth of the \[Theta]-range a different color. (Be \ sure you can associate each color with the right interval of \ \[Theta]-values.)" }], "Text"], Cell[BoxData[ \(Clear[part1, part2, part3]; \n part1 = PolarPlot[2\ Sin[3\ \[Theta]], {\[Theta], 0, \[Pi]\/6}, PlotStyle -> {Hue[0.2], Thickness[0.01]}, \ PlotRange -> {\(-2\), 2}, DisplayFunction -> Identity]; \n part2 = PolarPlot[2\ Sin[3\ \[Theta]], {\[Theta], \[Pi]\/6, \[Pi]\/3}, PlotStyle -> {Hue[0.4], Thickness[0.01]}, \ PlotRange -> {\(-2\), 2}, DisplayFunction -> Identity]; \n part3 = PolarPlot[2\ Sin[3\ \[Theta]], {\[Theta], \[Pi]\/3, \[Pi]\/2}, PlotStyle -> {Hue[0.6], Thickness[0.01]}, \ PlotRange -> {{\(-2\), 2}, {\(-2\), 2}}, DisplayFunction -> Identity]; \n Show[{part1, part2, part3}, DisplayFunction -> $DisplayFunction, PlotRange -> {{\(-2\), 2}, {\(-2\), 2}}, Epilog -> polarpaper]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell["\<\ We now see that we have enough to complete the graph by reflecting across the \ y-axis. The graph is a 4-petaled rose (not 2 as you might have guessed)!\ \>", "Text"], Cell[BoxData[ \(\(PolarPlot[2\ Sin[2\ \[Theta]], {\[Theta], 0, 2 \[Pi]}, PlotStyle -> {Thickness[0.01], Hue[1.0]}, \ PlotRange -> {\(-2\), 2}, Epilog -> polarpaper]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "Now we have enough to make an intelligent conjecture about the shape of \ the graph of r = a sin(n\[Theta]): it is an n-petaled rose if n is odd and a \ 2n-petaled rose if n is even. In either case it is symmetric with respect to \ the y-axis. One petal is complete for 0 \[LessEqual] \[Theta] \[LessEqual] ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/n\)]], ".\n\tMake sure you understand and can predict the shape easily. Practice \ on the following program until you feel comfortable. It will plot r = a \ sin(n\[Theta]), for any values of a and n you choose. Note the effect of a \ as well as n." }], "Text"], Cell[BoxData[ RowBox[{ \(Clear[a, n]\), ";", "\n", \(a := 3\), ";", "\n", \(n := 4\), ";", "\n", StyleBox[\( (*change\ nothing\ below\ this\ line*) \), FontColor->RGBColor[1, 0, 0]], "\n", \(PolarPlot[a\ Sin[n\ \[Theta]], {\[Theta], 0, 2 \[Pi]}, PlotStyle -> {Thickness[0.01], Hue[1.0]}, \ PlotRange -> {\(-a\), a}, Epilog -> polarpaper]\), ";"}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[ "\tWhat do you expect to be the same for r = a cos(n\[Theta])? What do you \ think will be different? Use the following program to formulate your \ predictions."], "Text"], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[a, n]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", \(a := 3\), ";", "\n", \(n := 4\), ";", "\n", StyleBox[\( (*change\ nothing\ below\ this\ line*) \), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[ \(PolarPlot[a\ Cos[n\ \[Theta]], {\[Theta], 0, 2 \[Pi]}, PlotStyle -> {Thickness[0.01], Hue[1.0]}, \ PlotRange -> {\(-a\), a}, Epilog -> polarpaper]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "You should have decided that again the graph is an n-petaled rose if n is \ odd, and a 2n-petaled rose if n is even. However now the rose is symmetric \ with respect to the x-axis, and one petal is complete for ", Cell[BoxData[ \(TraditionalForm\`\(-\[Pi]\)\/\(2 n\)\)]], " \[LessEqual] \[Theta] \[LessEqual] ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/\(2 n\)\)]], ".\n\n", StyleBox[ "SUMMARY:\n\tThe graph of r = a sin(n\[Theta]) or r = a cos(n\[Theta]) is\n\ \t\tan n-petaled rose if n is odd\n\t\tand\n\t\ta 2n-petaled rose if n is \ odd.\n\t\t\n", FontWeight->"Bold"], "Again, it is your turn to try some. Sketch these and then compare your \ answers with the computer's." }], "Text"], Cell[TextData[{ StyleBox["Exercise:", FontWeight->"Bold"], " Find the symmetries of r = 2 cos(2\[Theta]) and sketch the graph." }], "Text"], Cell[BoxData[ \(Print[ "\"]; \n PolarPlot[2\ Cos[2 \[Theta]], {\[Theta], 0, 2 \[Pi]}, PlotStyle -> Thickness[0.01]]; \)], "Input", FontColor->RGBColor[1, 1, 0], Background->RGBColor[1, 1, 0]] }, Open ]], Cell[CellGroupData[{ Cell["A Spiral", "Subsubsection", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ "\tWe conclude our tour of polar graphs with the particularly simple polar \ function r = \[Theta]. The graph of this function is a curve spiraling out \ from the origin.", " ", "Observe by executing the program below that it has no symmetry, but that \ as the angle \[Theta] increases, so does the distance r from the origin." }], "Text"], Cell[BoxData[ \(\(PolarPlot[\[Theta], {\[Theta], 0, 4\ \[Pi]}, PlotRange -> {{\(-30\), 30}, {\(-30\), 30}}, PlotStyle -> {Thickness[0.008], Hue[ .01]}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tUnlike most of the other polar curves we've studied (with the exception \ of lines), this graph never closes up on itself. It keeps expanding forever \ as \[Theta] does. The graph above is plotted for the domain 0 \[LessEqual] \ \[Theta] \[LessEqual] 4\[Pi].", " ", "Execute the following yellow box to see the graph of this same function, r \ = \[Theta] for the domain 0 \[LessEqual] \[Theta] \[LessEqual] 8\[Pi] (i.e. \ around the circle 4 times):" }], "Text"], Cell[BoxData[ \(\(PolarPlot[\[Theta], {\[Theta], 0, 8\ \[Pi]}, PlotRange -> {{\(-30\), 30}, {\(-30\), 30}}, PlotStyle -> {Thickness[0.008], Hue[ .01]}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]] }, Open ]], Cell[CellGroupData[{ Cell["Review Exercises", "Subsubsection", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ StyleBox["Problem 1: ", FontWeight->"Bold"], "Plot each point given in polar coordinates, and find its rectangular \ coordinates.\n\t(a) (2, \[Pi]/6)\t(b) (-3, -\[Pi]/2) (c) (-2, 4\[Pi]/3) (d) \ (5, 5\[Pi]/4)\n\t\n\n", StyleBox["Problem 2: ", FontWeight->"Bold"], "For each of the rectangular coordinates, find two pairs of polar \ coordinates (r, \[Theta]), one with r > 0, and the other with r < 0, which \ represent the same point.\t ", StyleBox["\n\t", FontWeight->"Bold"], "(a) (2, 0)\t(b) (1, -1)\t(c) (", Cell[BoxData[ \(TraditionalForm\`\@3\)]], ", -1)\t \t(d) (-5, 12)\n\t\n\n", StyleBox["Problem 3: ", FontWeight->"Bold"], "Convert each equation from polar to rectangular.\n\t(a) r = -2 \ sin(\[Theta]) (b) r = 4 (c) r = sin(\[Theta]) - cos(\[Theta]) (d) r = ", Cell[BoxData[ FormBox[ StyleBox[\(3\/\(3 - 3 \(cos(\[Theta])\)\)\), FontSize->16], TraditionalForm]]], "\n\t\n\n", StyleBox["Problem 4: ", FontWeight->"Bold"], "Convert each equation from rectangular to polar.\n\t(a) ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], "= x (b) 2xy = 1 (c) y = -3 \t (d) ", Cell[BoxData[ \(TraditionalForm\`y\^2\)]], "= 2x\t\n\t", StyleBox["\n\nProblem 5: ", FontWeight->"Bold"], "Identify and graph each polar graph by hand. \n\t(a) r = 4\t(b) \[Theta] \ = \[Pi]/6\t(c) r sin(\[Theta]) = -2\t(d) r = 2 cos(\[Theta])\n\t\n\n", StyleBox["Problem 6: ", FontWeight->"Bold"], "Match each of the graphs (A) through (F) to one of the six polar \ equations.\n\t(1) r = 2 + 2 cos(\[Theta]) \t\t(2) r = 2 sin(4\[Theta]) \t\t\ (3) \[Theta] = -\[Pi]/4\n\t(4) r = 1 - 3 sin(\[Theta])\t\t(5) r = 7 cos(\ \[Theta]) \t\t(6) r cos(\[Theta]) = 2\n\t\n(A)", Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: 1.02572 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics /Courier findfont 10 scalefont setfont % Scaling calculations 0.499997 0.256431 0.512862 0.256431 [ [.11535 .50036 -12 -9 ] [.11535 .50036 12 0 ] [.24357 .50036 -6 -9 ] [.24357 .50036 6 0 ] [.37178 .50036 -12 -9 ] [.37178 .50036 12 0 ] [.62821 .50036 -9 -9 ] [.62821 .50036 9 0 ] [.75643 .50036 -3 -9 ] [.75643 .50036 3 0 ] [.88464 .50036 -9 -9 ] [.88464 .50036 9 0 ] [.4875 0 -12 -4.5 ] [.4875 0 0 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