Polar Graphs

Rose curves and lemniscates

Lesson Overview

Instructor Presentation

Introduction

Rose curves

We've already looked at circles and limaçons in the polar graph family. In both of these cases, the angle θ did not have a multiplier other than 1. In today's lesson we will examine what happens when this value is increased so that we are applying the sine and cosine functions to angles 2θ, 3θ, and so on.

Once again, here is an example of the same curves - one in polar form, and the other in rectangular form. It is obvious which equation form should be the preferred choice of graphers.

rose1 ={PolarPlot[3Sin[3θ],{θ,0,2π}, PolarTicks->{"Radians",Automatic},PolarAxes->Automatic,PolarGridLines->Automatic,PlotStyle->{Thick,Orange},ImageSize->400]}
rose2 = {ContourPlot[((((x^2)+(y^2)))^2)==3(3(x^2)y-(y^3)),{x,-4,4},{y,-4,4},GridLines->{Range[-6,6], Range[-6,6]},PlotRange->{-6,6},Axes->True, AspectRatio->Automatic, ContourStyle->{Thick,Green},PlotRange->Automatic,ImageSize->400]}
Grid[Join[{{Style["r = 3 sin (3θ)",FontFamily->"Arial",Bold,FontSize->20],Style["((x^2) + (y^2))^2 = 3(3(x^2)y - (y^3))",FontFamily->"Arial",Bold,FontSize->20]}},
Partition[Riffle[rose1,rose2],2]],Frame->All,FrameStyle->Thick]

The equations for rose curves are as follows, where n is an integer greater than 1, and |a| is the maximum petal length:

({{r=a sin (n θ)}, {r=a cos(n θ)}})

Because of the difficult nature of the trigonometric expansions and identities as n increases, it is unnecessarily difficult to give a general rectangular form for rose curves.  We'll just stick with polar form for this lesson!

Lemniscates

The equations for lemniscates are also much easier to handle in polar coordinates than in rectangular coordinates.

rose1 ={PolarPlot[(Sqrt[4Sin[2θ]]),{θ,0,2π}, PolarTicks->{"Radians",Automatic},PolarAxes->Automatic,PolarGridLines->Automatic,PlotStyle->{Thick,Orange},ImageSize->400]}
rose2 = {ContourPlot[((((x^2)+(y^2)))^2)==8x y,{x,-4,4},{y,-4,4},GridLines->{Range[-6,6], Range[-6,6]},PlotRange->{-6,6},Axes->True, AspectRatio->Automatic, ContourStyle->{Thick,Green},PlotRange->Automatic,ImageSize->400]}
Grid[Join[{{Style["(r^2) = 4 sin (2θ)",FontFamily->"Arial",Bold,FontSize->20],Style["((x^2) + (y^2))^2 = 8xy",FontFamily->"Arial",Bold,FontSize->20]}},
Partition[Riffle[rose1,rose2],2]],Frame->All,FrameStyle->Thick]

Lemniscate (rectangular):  ((((x^2)+(y^2)))^2)=2 (a^2) x y

Lemniscate (polar): (r^2)=(a^2) sin(2 θ) and (r^2)=(a^2) cos(2 θ)

Notice that the lemniscate has an "r-squared" rather than just "r."  This feature sets it apart from the rose curves, even though the lemniscate resembles a rose curve of 2 petals.

Goals

Students will be able to recognize from the equation whether the polar function is a rose curve, lemniscate, limaçon, or a circle.

Students will also be able to determine information about the polar graph, such as orientation, radius, diameter, number of petals, etc. from the equation.

Students will learn how to plot these graphs using Mathematica.

Concept Exploration

Rose curves

Look at the demonstration below. Try different functions by clicking on the buttons and then answer the questions below.

Manipulate[PolarPlot[3 trig[3θ],{θ,0,2π},PlotRange->5,PlotStyle->{Thick,Orange},PlotLabel->Style["r = 3 "<>ToString[trig]<>" 3θ ",Orange,FontSize->20]],{trig,{Sin,Cos}}]

What changes between the "sine" graph and the "cosine" graph?  What stays the same?

The graph seems rotated from one to the other.  However, the size and shape of the graph remains unchanged.

What about the symmetry of the graphs?

The "sine" graph is symmetric about the y-axis, while the "cosine" graph is symmetric about the line y=x.

Now look at a similar model and make some additional observations.

Manipulate[PolarPlot[3 Cos[a θ],{θ,0,2π},PlotRange->5,PlotStyle->{Thick,Red},PlotLabel ->Style[Row[{
Style["r",Italic]," = ",ToString[3]<>" cos "<>ToString[a]<>"θ"
}],{Red,20}]],{a,{2,3,4,5,6}}]

What is the correlation between the number of petals and the coefficient in front of the θ?

If the coefficient is even, there are twice as many petals as that number.  
If the coefficient is odd, there are exactly that many petals.

Now, open up the cell that contains the previous example and replace the "cosine" function with the "sine" function.  Do you still get the same outcome?

Examples: Rose curves 1

Given the equations below, describe the petal length, number of petals and symmetry of the rose curves.  Then check your answer by making a polar plot.

r=5 sin(4 θ)

Petal length:

Number of petals:

Symmetry:

PolarPlot[

r = 2 sin (5θ)

Petal length:

Number of petals:

Symmetry:

PolarPlot[

r = 3.5 cos (2θ)

Petal length:

Number of petals:

Symmetry:

PolarPlot[

Examples: Rose curves 2

Given the descriptions of the rose curves below, create a rose curve that matches those characteristics.  Then check your answer by making a polar plot.

A rose curve with y-axis symmetry that has 3 petals, each of length (Sqrt[7]).

PolarPlot[

A rose curve with 8 petals, each of length 2.  The rose curve has y-axis symmetry.

PolarPlot[

A rose curve with petals of length 4.  This 11-petaled curve must also have line y=x symmetry.

PolarPlot[

Concept Exploration 2

Lemniscates

As we have seen already, lemniscates are a bit different in the structure of their equations.  Let's look at the following demonstration:

Manipulate[PolarPlot[{(Sqrt[9trig[2θ]]),3},{θ,0,2π},PlotRange->5,PlotStyle->{{Thick,Blue},Yellow},PlotLabel ->Style[Row[{
Style["(r^2)",Italic]," = ",ToString[9  trig  ]<>ToString[ 2]<>"θ"
}],{Blue,20}]],{trig,{Sin,Cos}}]

What do you notice about the symmetry of the graph?

The "sine" graph appears to be symmetric about either the line y=x or the origin.  The "cosine" graph appears to be symmetric about either the x-axis or y-axis.

What do you notice about the length of the petals in comparison with the equation?

The coefficient in front of the trig function is the square of the petal-length.

What generalities can be made about lemniscates now?

Lemniscates must "start" with an (r^2).

All lemniscates must have a 2 in front of the θ.

The coefficient in front of the "sine" or "cosine" is the square of the petal length.

We can generalize that the lemniscate behaves, in the way of symmetry, that the rectangular plot of sine and cosine do—sine is symmetric about the 
origin (the sine lemniscate is as well) and cosine is symmetric about the y-axis (similarly with the cosine lemniscate).

Examples - Lemniscates 1

Given the descriptions of the lemniscates below, create an equation that matches those characteristics. Then check your answer by making a polar plot.

A lemniscate with petals of length 4 and with x-axis symmetry.

PolarPlot[

A lemniscate with origin symmetry and whose petals have length 2.5.

PolarPlot[

Examples - Lemniscates 2

Convert the following polar equations of lemniscates to rectangular equations.  Make sure in the "equation conversion" cell below, that you hit Ctrl+Enter to move to the next line.  This will keep the equal signs lined up.  Then, type in the polar equation in the first entry box and the rectangular equation in the second box below.  Compare the two results.

(r^2)= 9cos(2θ)

({{({{(r^2) = 9 cos 2θ}, {□}})}})

Panel[DynamicModule[{f=Sin[θ]},Column[{InputField[Dynamic[f]],Dynamic[PolarPlot[f,{θ,0,2π},PolarTicks->{"Radians",Automatic},PolarAxes->Automatic,PlotStyle->{Thick,Blue},PolarGridLines->Automatic,ImageSize->400,PlotLabel->Style["\!\(\*SuperscriptBox[\(r\), \(\(2\)\(\\ \)\)]\)= \!\(\*SqrtBox[\(9\\ cos\\ \((2  θ)\)\)]\)",Blue,18]]]}]]]

Panel[DynamicModule[{f=(((x^2+y^2))^2)==9(x^2-y^2)},Column[{InputField[Dynamic[f]],Dynamic[ContourPlot[Evaluate@Apply[Subtract,f],{x,-5,5},{y,-5,5},GridLines->{Range[-6,6],Range[-6,6]},PlotRange->{-6,6},Axes->True,AspectRatio->Automatic,ContourStyle->{Thick,Black},PlotRange->Automatic,ImageSize->400,PlotLabel->Style["((x^2) + (y^2))^2 = 9((x^2) - (y^2))",Blue,18]]]}]]]

(r^2)= 16sin 2θ

({{({{(r^2) = 16 sin 2θ}, {□}})}})

Panel[DynamicModule[{f=(Sqrt[16Sin[2θ]])},Column[{InputField[Dynamic[f]],Dynamic[PolarPlot[f,{θ,0,2π},PolarTicks->{"Radians",Automatic},PolarAxes->Automatic,PlotStyle->{Thick,Blue},PolarGridLines->Automatic,ImageSize->400,PlotLabel->Style["\!\(\*SuperscriptBox[\(r\), \(\(2\)\(\\ \)\)]\)= 16 sin (2θ)",Blue,18]]]}]]]

Panel[DynamicModule[{f=(((x^2+y^2))^2)==32x y},Column[{InputField[Dynamic[f]],Dynamic[ContourPlot[Evaluate@Apply[Subtract,f],{x,-5,5},{y,-5,5},GridLines->{Range[-6,6],Range[-6,6]},PlotRange->{-6,6},Axes->True,AspectRatio->Automatic,ContourStyle->{Thick,Black},PlotRange->Automatic,ImageSize->400,PlotLabel->Style["((x^2) + (y^2))^2 = 32xy",Blue,18]]]}]]]

Examples - Rose Curve and Lemniscate Identification

Look at the diagram of polar graphs below.  Label each with the type of polar graph it is and its equation.

To label these easily, go to Graphics ▶ Drawing Tools.  Click on the button in the lower right-hand corner with an "A."  This is the button that allows text boxes to be inserted in images.

Grid[{{PolarPlot[(Sqrt[16Cos[2x]]),{x,0,2π},PlotStyle->{Thick,Purple},PlotRange->5],PolarPlot[16Sin[3x],{x,0,2π},PlotStyle->{Thick,Purple}],PolarPlot[2Sin[6x],{x,0,2π},PlotStyle->{Thick,Purple}]},{PolarPlot[{(Sqrt[4Sin[2x]]),2},{x,0,2π},PlotStyle->{{Thick,Purple},Yellow}],PolarPlot[6Cos[7x],{x,0,2π},PlotStyle->{Thick,Purple}],PolarPlot[(Sqrt[25Cos[2x]]),{x,0,2π},PlotStyle->{Thick,Purple},PlotRange->6]}},Spacings->{5,5}]

Examples Answer Key

Examples - Rose curves 1

Given the equations below, describe the petal length, number of petals and symmetry of the rose curves.  Then check your answer by making a polar plot.

r = 5 sin (4θ)

Petal length:  5

Number of petals: 8

Symmetry:  y-axis, x-axis, origin

PolarPlot[5Sin[4θ],{θ,0,2π},PlotStyle->Thick,ColorFunction->"Rainbow"]

r=2 sin(5θ)

Petal length:  2

Number of petals: 5

Symmetry: y-axis, origin

PolarPlot[2Sin[5θ],{θ,0,2π},PlotStyle->{Thick,Green}]

r = 3.5 cos (2θ)

Petal length: 3.5

Number of petals: 4

Symmetry:  x-axis, y-axis, origin

PolarPlot[3.5Cos[2θ],{θ,0,2π}, PlotStyle->Thick,ColorFunction->Function[{θ},Hue[θ]],Mesh->16,MeshShading->{Black,Automatic}]

Examples - Rose curves 2

Given the equation of the rose curves below, create a rose curve that matches those characteristics.  Then check your answer by making a polar plot.

A rose curve with y-axis symmetry that has 3 petals, each of length (Sqrt[7]).

r = (Sqrt[7])sin (3θ)

PolarPlot[(Sqrt[7])Sin[3θ],{θ,0,2π},PlotRange->4,PlotStyle->{Thickness[0.01],Orange}]

A rose curve with 8 petals, each of length 2.  The rose curve has y-axis symmetry.

r = 2sin (4θ) or r = 2 cos (4θ)

Manipulate[PolarPlot[2trig[4θ],{θ,0,2π},PlotStyle->{Thick,Red}],{trig,{Sin,Cos}}]

A rose curve with petals of length 4.  This 11-petaled curve must also have line y=x symmetry.

r = 4 cos (11θ)

PolarPlot[4Cos[11θ],{θ,0,2π},PlotStyle->{Thick,Blue}]

Examples - Lemniscates 1

Given the descriptions of the lemniscates below, create an equation that matches those characteristics.  Then check your answer by making a polar plot.

A lemniscate with petals of length 4 and with x-axis symmetry.

(r^2) = 16 cos (2θ)

PolarPlot[(Sqrt[16Cos[2θ]]),{θ,0,2π},PlotRange->5,PlotStyle->Thick,Mesh->18,MeshShading->{Red, Orange,Yellow}]

A lemniscate with origin symmetry and whose petals have length 2.5.

(r^2) = 6.25 sin(2θ)

PolarPlot[(Sqrt[6.25Sin[2θ]]),{θ,0,2π},PlotStyle->Thick,ColorFunction->"DeepSeaColors",PlotRange->3]

Examples - Lemniscates 2

Convert the following polar equations of lemniscates to rectangular equations.  Make sure in the "equation conversion" cell below, that you hit Ctrl+Enter to move to the next line.  This will keep the equal signs lined up.  Then, type in the polar equation in the first entry box and the rectangular equation in the second box below.  Compare the two results.

(r^2)= 9cos (2θ)

({{({{(r^2) = 9 cos 2θ}, {(r^2) = 9 ((cos^2)θ - (sin^2)θ)}, {(r^2) = 9 (((((x/r)))^2) - ((((y/r)))^2))}, {(r^2) = 9 ((((x^2) - (y^2))/(r^2)))}, {(r^2) * (r^2) = 9 ((x^2) - (y^2))}, {((x^2) + (y^2)) * ((x^2) + (y^2)) = 9 ((x^2) - (y^2))}, {((((x^2) + (y^2)))^2) = 9 ((x^2) - (y^2))}})}})

Panel[DynamicModule[{f=Sin[θ]},Column[{InputField[Dynamic[f]],Dynamic[PolarPlot[f,{θ,0,2π},PolarTicks->{"Radians",Automatic},PolarAxes->Automatic,PlotStyle->{Thick,Blue},PolarGridLines->Automatic,ImageSize->400,PlotLabel->Style["\!\(\*SuperscriptBox[\(r\), \(\(2\)\(\\ \)\)]\)= \!\(\*SqrtBox[\(9\\ cos\\ \((2  θ)\)\)]\)",Blue,18]]]}]]]

Panel[DynamicModule[{f=(((x^2+y^2))^2)==9(x^2-y^2)},Column[{InputField[Dynamic[f]],Dynamic[ContourPlot[Evaluate@Apply[Subtract,f],{x,-5,5},{y,-5,5},GridLines->{Range[-6,6],Range[-6,6]},PlotRange->{-6,6},Axes->True,AspectRatio->Automatic,ContourStyle->{Thick,Black},PlotRange->Automatic,ImageSize->400,PlotLabel->Style["((x^2) + (y^2))^2 = 9((x^2) - (y^2))",Blue,18]]]}]]]

(r^2)= 16sin (2θ)

({{({{(r^2) = 16 sin 2θ}, {(r^2) = 16 (2 sin θ cos θ)}, {(r^2) = 32 * (x/r) * (y/r)}, {(r^2) = 32 * (xy/(r^2))}, {(r^2) * (r^2) = 32xy}, {((x^2) + (y^2)) * ((x^2) + (y^2)) = 32xy}, {((((x^2) + (y^2)))^2) = 32xy}})}})

Panel[DynamicModule[{f=(Sqrt[16Sin[2θ]])},Column[{InputField[Dynamic[f]],Dynamic[PolarPlot[f,{θ,0,2π},PolarTicks->{"Radians",Automatic},PolarAxes->Automatic,PlotStyle->{Thick,Blue},PolarGridLines->Automatic,ImageSize->400,PlotLabel->Style["\!\(\*SuperscriptBox[\(r\), \(\(2\)\(\\ \)\)]\)= 16 sin (2θ)",Blue,18]]]}]]]

Panel[DynamicModule[{f=(((x^2+y^2))^2)==32x y},Column[{InputField[Dynamic[f]],Dynamic[ContourPlot[Evaluate@Apply[Subtract,f],{x,-5,5},{y,-5,5},GridLines->{Range[-6,6],Range[-6,6]},PlotRange->{-6,6},Axes->True,AspectRatio->Automatic,ContourStyle->{Thick,Black},PlotRange->Automatic,ImageSize->400,PlotLabel->Style["((x^2) + (y^2))^2 = 32xy",Blue,18]]]}]]]

Examples - Rose Curve and Lemniscate Identification

Look at the diagram of polar graphs below.  Label each with the type of polar graph it is and its equation.

To label these easily, go to Graphics ▶ Drawing Tools.  Click on the button in the lower right-hand corner with an "A."  This is the button that allows text boxes to be inserted in images.

Grid[{
{PolarPlot[(Sqrt[16Cos[2x]]),{x,0,2π},PlotStyle->{Thick,Purple},PlotRange->5],PolarPlot[16Sin[3x],{x,0,2π},PlotStyle->{Thick,Purple}],PolarPlot[2Sin[6x],{x,0,2π},PlotStyle->{Thick,Purple}]},{PolarPlot[{(Sqrt[4Sin[2x]]),2},{x,0,2π},PlotStyle->{{Thick,Purple},Yellow}],PolarPlot[6Cos[7x],{x,0,2π},PlotStyle->{Thick,Purple}],PolarPlot[(Sqrt[25Cos[2x]]),{x,0,2π},PlotStyle->{Thick,Purple},PlotRange->6]}},Spacings->{5,5}]

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