Polar Graphing
by 
Patrick McLoughlin
University High School
1212 West Springfield
Urbana, Illinois 61801

Description: This lesson shows how Mathematica plots a wide variety of polar graphs.  By varying the parameters in the equations, students can learn how the structure of the equation affects the shape of the graph.

Code for Generating the Title Graphic

Clear[r1,t,x1,y1,p1]
r1[t_] = ((Cos[t])^2 + Sin[3t]);
x1[t_] = r1[t] Cos[t];
y1[t_] = r1[t] Sin[t];

p1= ParametricPlot[{x1[t],y1[t]},{t,0, 2Pi},
     Axes->False,
     AspectRatio->Automatic];

Clear[r2,t,x2,y2,p2]
r2[t_] = Sin[3t];
x2[t_] = r2[t] Cos[t];
y2[t_] = r2[t] Sin[t];

p2= ParametricPlot[{x2[t],y2[t]},{t,0, 2Pi},
     Axes->False,
     AspectRatio->Automatic];

Clear[r3,t,x3,y3,p3]
r3[t_] = 1-2Sin[t];
x3[t_] = r3[t] Cos[t];
y3[t_] = r3[t] Sin[t];

p3= ParametricPlot[{x3[t],y3[t]},{t,0, 2Pi},
     Axes->False,
     AspectRatio->Automatic];

Show[Graphics[
	{Rectangle[Scaled[{0,0}],Scaled[{1,1}],p1],
		Rectangle[Scaled[{.075,.6}],Scaled[{.325,.85}],p2],
		Rectangle[Scaled[{.675,.6}],Scaled[{.925,.85}],p3]}]];

SECTION ONE: The Basics

Introduction

If a polar coordinate system and a rectangular coordinate system are superimposed in the standard manner any point P can be indicated either by rectangular coordinates, P(x,y), or by polar coordinates, P(r,q), where the relationship between (x,y) and (r,q) is given by:

x = rCosq

y = rSinq.

A polar equation is an equation in the variable r and q, where (r,q) represents coordinates of a point in a polar coordinate plane, for example
	
	r = Cos2q,
	r = 1 - 2Sinq,or
	r = 1 + Cosq.
	
	A graph of a polar equation in a polar coordinate plane is the set of all points in the plane having at least one pair of polar coordinates that satisfy the equation.

Graphing Polar Equations

Mathematica uses the equations x = rCosq  and y = rSinq  to graph a polar equation. For example, the equation r = 1 - 2Sinq  is graphed using the system
	
	x = (1 - 2Sinq )Cosq
	y = (1 - 2Sinq )Sinq .

Let's look at how Mathematica does this:

Clear[r,t,x,y,g]
r[t_] = 1 - 2Sin[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

Clear[lim]
lim = ParametricPlot[{x[t],y[t]},{t,0,2Pi},
    	PlotStyle->{Thickness[0.0053]},
    	PlotRange->{{-2,2},{-3.2,1.3}},
    	Axes->True,
     	AspectRatio->Automatic,
		DisplayFunction->Identity];

g = Graphics[Text["r = 1 - 2Sint",{1,.5}]];

Show[g,lim,PlotRange->{{-2,2},{-3.2,1.3}},
    	Axes->True,
     	AspectRatio->Automatic];

Now let's look at the same graph with a polar grid background.  This is just to give an idea of what the graph would look like on
polar graph paper.  It won't be used on subsequent graphs.

Clear[r,t,x,y,lim2]
r[t_] = 1 - 2Sin[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

lim2 = ParametricPlot[{x[t],y[t]},{t,0,2Pi},
    	PlotStyle->{Thickness[0.01]},
    	Axes->False,
     	AspectRatio->Automatic,
     	DisplayFunction->Identity];

plrgrd = Graphics[{
			Circle[{0,0},.5],Circle[{0,0},1],
			Circle[{0,0},1.5],Circle[{0,0},2],
			Line[{{0,0},{4,0}}],
			Text["r = 1 - 2Sint",{3,2}],
			Text["Polar Axis",{3,.2}],
			Text["0.5",{0.6,.1}],Text["1",{1.1,.1}],
				Text["1.5",{1.6,.1}],Text["2",{2.1,.1}]},
			AspectRatio->Automatic];

Show[plrgrd,lim2];

SECTION TWO: The Roses

Introduction

Plot the graph of r = Sin(2q)

Clear[r,t,x,y]
r[t_] = Sin[2t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p=ParametricPlot[{x[t],y[t]},{t,0,2Pi},
	PlotRange->{{-.8,.8},{-1,1}},
    AxesLabel->{"x","y"},
    AspectRatio->Automatic,
	DisplayFunction->Identity];

g = Graphics[Text["r = Sin[2t]",{.4,.85}]];

This type of polar graph is called a "rose."  Could you have use your knowledge of the sine function to predict its shape?  If you don't see how, maybe this will help.

How the Angle Variable Is Used To Build
a Rose Graph.

Show how the graph of r = Sin(2q) is traced out as q varies from 0 to 2¹. To see a "movie" of how the graph is traced, select all 8 graphics and press "apple-y."

Clear[r,t,x,y]
r[t_] = Sin[2t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

Do[ParametricPlot[{x[t],y[t]},{t,0,k Pi/4},
     PlotRange->{{-1,1},{-1,1}},
     AxesOrigin->{0,0},
     AxesLabel->{"x","y"},
     AspectRatio->Automatic],
    {k,1,8}];

A New Rose.

Let's try a new value for "a" in r = Sin(aq).  Be really creative;  try a = 3.  Graph r = Sin(3q).

Clear[r,t,x,y,p,g]
r[t_] = Sin[3t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p=ParametricPlot[{x[t],y[t]},{t,0,2Pi},
				AxesLabel->{"x","y"},
				PlotRange->{{-.9,.9},{-1,.75}},
				DisplayFunction->Identity,
     			AspectRatio->Automatic];

g = Graphics[Text["r = Sin[3t]",{.5,.65}]];

Show[p,g,DisplayFunction->$DisplayFunction];

We'll look at how this one "grows" as q varies from 0 to 2¹.

Clear[r,t,x,y]
r[t_] = Sin[3t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

Do[ParametricPlot[{x[t],y[t]},{t,0,k Pi/6},
     PlotRange->{{-1,1},{-1,1}},
     AxesOrigin->{0,0},
     AxesLabel->{"x","y"},
     AspectRatio->Automatic],
    {k,1,12}];

That was interesting.  Did you see what happened as q  varied from ¹ to 2¹?  Let's see what eliminating that part of the range from will do.

Clear[r,t,x,y]
r[t_] = Sin[3t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

Do[ParametricPlot[{x[t],y[t]},{t,0,k Pi/6},
     PlotRange->{{-1,1},{-1,1}},
     AxesOrigin->{0,0},
     AxesLabel->{"x","y"},
     AspectRatio->Automatic],
    {k,1,6}];

Can you see why changing a = 2 to a = 3 in r = Sin(aq) changes the effect of the size of the domain?  Think about it.

Changing the value of "a" in r = Sin(aq)

Now try several values for "a" in r = Sin(aq).  Try to find a pattern.

i.		a = 4.  Graph r = Sin(4q).

Clear[r,t,x,y,r4,g4]
r4[t_] = Sin[4t];
x[t_] = r4[t] Cos[t];
y[t_] = r4[t] Sin[t];

r4:= ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     				AxesLabel->{"x","y"},
     				AspectRatio->Automatic,
					DisplayFunction -> Identity];

g4:= Graphics[Text["r = Sin[4t]",{.75,1}]];

Show[r4,g4,DisplayFunction -> $DisplayFunction];

ii.		a = 5.  Graph r = Sin(5q).

Clear[r,t,x,y,r5,g5]
r5[t_] = Sin[5t];
x[t_] = r5[t] Cos[t];
y[t_] = r5[t] Sin[t];

r5:= ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     				AxesLabel->{"x","y"},
     				AspectRatio->Automatic,
					DisplayFunction -> Identity];

g5:= Graphics[Text["r = Sin[5t]",{.75,1}]];

Show[r5,g5,DisplayFunction -> $DisplayFunction];

iii.		a = 6.  Graph r = Sin(6q).

Clear[r,t,x,y,r6,g6]
r6[t_] = Sin[6t];
x[t_] = r6[t] Cos[t];
y[t_] = r6[t] Sin[t];

r6:= ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     				AxesLabel->{"x","y"},
     				AspectRatio->Automatic,
					DisplayFunction -> Identity];

g6:= Graphics[Text["r = Sin[6t]",{.75,1}]];

Show[r6,g6,DisplayFunction -> $DisplayFunction];

iv.		a = 7.  Graph r = Sin(7q).

Clear[r,t,x,y,r7,g7]
r7[t_] = Sin[7t];
x[t_] = r7[t] Cos[t];
y[t_] = r7[t] Sin[t];

r7:= ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     				AxesLabel->{"x","y"},
     				AspectRatio->Automatic,
					DisplayFunction -> Identity];

g7:= Graphics[Text["r = Sin[7t]",{.75,1}]];

Show[r7,g7,DisplayFunction -> $DisplayFunction];

You may see the pattern by now.  If so,state it explicitly and try to explain why the properties of the sine function and of plotting points in polar coordinates would have allowed you to predict it.  If you don't see what is happening and why, try a few more examples and think about it some more.

A Rose By Another Name

Now let's see what happens if we use the cosine function instead of the sine.  Try r = Cos(2q).

Clear[r,t,x,y,rc2,gc2]
rc2[t_] = Cos[2t];
x[t_] = rc2[t] Cos[t];
y[t_] = rc2[t] Sin[t];

rc2:= ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     AxesLabel->{"x","y"},
     AspectRatio->Automatic,DisplayFunction -> Identity];

gc2:= Graphics[Text["r = Cos[2t]",{.75,1}]];

Show[rc2,gc2,DisplayFunction -> $DisplayFunction];

You probably could have used your knowledge of the sine and cosine functions to predict how the graph of  r = Cos(2q) would have to relate to the graph of  r = Sin(2q).  Let's look at r = Cos(aq) for a few more values of "a".

Clear[t,x,y,r3c,p3l]
r3c[t_] = Cos[3t];
x[t_] = r3c[t] Cos[t];
y[t_] = r3c[t] Sin[t];

p3l = ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     	AxesLabel->{"x","y"},
     	AspectRatio->Automatic,
     	PlotLabel->"r = Cos[3t]",
     	DisplayFunction -> Identity];

Clear[t,x,y,r4c,p4l]
r4c[t_] = Cos[4t];
x[t_] = r4c[t] Cos[t];
y[t_] = r4c[t] Sin[t];

p4l = ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     	AxesLabel->{"x","y"},
     	AspectRatio->Automatic,
     	PlotLabel->"r = Cos[4t]",
     	DisplayFunction -> Identity];

Clear[t,x,y,r5c,p5l]
r5c[t_] = Cos[5t];
x[t_] = r5c[t] Cos[t];
y[t_] = r5c[t] Sin[t];

p5l = ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     	AxesLabel->{"x","y"},
     	AspectRatio->Automatic,
     	PlotLabel->"r = Cos[5t]",
     	DisplayFunction -> Identity];

Clear[t,x,y,r6c,p6l]
r6c[t_] = Cos[6t];
x[t_] = r6c[t] Cos[t];
y[t_] = r6c[t] Sin[t];

p6l = ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     	AxesLabel->{"x","y"},
     	AspectRatio->Automatic,
     	PlotLabel->"r = Cos[6t]",
     	DisplayFunction -> Identity];

Clear[t,x,y,r7c,p7l]
r7c[t_] = Cos[7t];
x[t_] = r7c[t] Cos[t];
y[t_] = r7c[t] Sin[t];

p7l = ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     	AxesLabel->{"x","y"},
     	AspectRatio->Automatic,
     	PlotLabel->"r = Cos[7t]",
     	DisplayFunction -> Identity];

Clear[t,x,y,r8c,p8l]
r8c[t_] = Cos[8t];
x[t_] = r8c[t] Cos[t];
y[t_] = r8c[t] Sin[t];

p8l = ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     	AxesLabel->{"x","y"},
     	AspectRatio->Automatic,
     	PlotLabel->"r = Cos[8t]",
     	DisplayFunction -> Identity];

Show[GraphicsArray[{{p3l,p4l,p5l},{p6l,p7l,p8l}}]
		/.(Ticks->Automatic)->(Ticks->None),
		DisplayFunction->$DisplayFunction];

You probably weren't suprised by any of these.  Previous experience would tend to lead you to know what to expect.

Further Exploration

This is pretty much the basic story for roses.  If you want you could explore what happens for fractional values of "a", for example r =  Sin(1.5q):

Clear[t,x,y,r]
r[t_] = Sin[1.5t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     	AxesLabel->{"x","y"},
     	AspectRatio->Automatic,
     	PlotLabel->"r = Sin[1.5t]"];

The asymetry is surprising;  perhaps we need a larger domain?

Clear[t,x,y,r]
r[t_] = Sin[1.5t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

ParametricPlot[{x[t],y[t]},{t,0,4Pi},
     	AxesLabel->{"x","y"},
     	AspectRatio->Automatic,
     	PlotLabel->"r = Sin[1.5t]"];

That looks more reasonable.  This line of investigation can get complicated.  Its pursuit is entirely optional.

SECTION THREE: Limacons and Cardioids

Introduction

The graph of r = 2 - Cosq  is one type of limacon.

Clear[r,t,x,y,p,g]
r[t_] = 2 - Cos[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p=ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     				AxesLabel->{"x","y"},
					PlotRange->All,
     				AspectRatio->Automatic,
					DisplayFunction->Identity];

g=Graphics[Text["r = 2 - Cos[t]",{1,2.1}]];

Show[p,g,DisplayFunction->$DisplayFunction];

Could you have used your knowledge of the cosine function to predict the shape of r = 2 - Cosq?

Now let's try a similar function;  the graph of r = 1 - Cosq  is a cardioid.

Clear[r,t,x,y,p,g]
r[t_] = 1 - Cos[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p=ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     				AxesLabel->{"x","y"},
					PlotRange->{{-2.1,1.5},{-1.5,1.5}},
     				AspectRatio->Automatic,
					DisplayFunction->Identity];

g=Graphics[Text["r = 1 - Cos[t]",{.77,1.1}]];

Show[p,g,DisplayFunction->$DisplayFunction];

Could you have used your knowledge of the cosine function to predict the shape of r = 1 - Cosq?

Now another variation; the graph of r = 1 -2Cosq  is another type of limacon.

Clear[r,t,x,y,p,g]
r[t_] = 1 - 2 Cos[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p=ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     				AxesLabel->{"x","y"},
					PlotRange->All,
     				AspectRatio->Automatic,
					DisplayFunction->Identity];

g=Graphics[Text["r = 1 - 2Cos[t]",{.9,1.3}]];

Show[p,g,DisplayFunction->$DisplayFunction];

Could you have used your knowledge of the cosine function to predict the shape of r = 1 - 2Cosq?

The graphs of  r = 2 - Cosq, r = 1 - Cosq, and r = 1 - 2Cosq,are examples of the three basic types of limacons and cardioids. Let's look at each type in more detail.

Limacons of the Type r = a - bCosq, for a > b.

The graph of  r = a - bCosq, for a > b is one type of limacon.  First let's look at how the graph of r = 2 - Cosq is traced out as q varies from 0 to 2¹.  Select and animate the following graphics.

Clear[r,t,x,y]
r[t_] = 2 - Cos[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

Do[ParametricPlot[{x[t],y[t]},{t,0,k Pi/6},           
     PlotRange->{{-3.5,1.5},{-3,3}},                  
     AxesOrigin->{0,0},AxesLabel->{"x","y"},          
     AspectRatio->Automatic],                         
     {k,1,12}];

Now show how varying a affects the graph of r = a - bCosq, for b = 1,a > b.  	
	Plot		r = a - Cosq for a = 1.25, 1.5, 1.75,and 2.

Clear[r,t,x,y]
r[t_] = (.25 k) - Cos[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

Do[ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     PlotRange->{{-3,1},{-2.5,2.5}},
     AxesOrigin->{0,0},AxesLabel->{"x","y"},
     AspectRatio->Automatic],
     {k,5,8}];

To get a better idea of what happens to  r = a - Cosq as a varies from 1.25 to 2, select and animate the above graphics.

Next show how varying b affects the graph of r = a - bCosq, for a = 2, a > b.  	
Plot		r = 2 - bCosq for b = 1, 1.25, 1.5, and 1.75.

Clear[r,t,x,y]
r[t_] = 2 - (.25 k) Cos[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

Do[ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     PlotRange->{{-4.5,1.5},{-3,3}},
     AxesOrigin->{0,0},AxesLabel->{"x","y"},
     AspectRatio->Automatic],
     {k,4,7}];

To get a better idea of what happens to  r = 2 - bCosq as b varies from 1 to 1.75, select and animate the above graphics.

By now you have a pretty good idea of how varying b affects the graph of r = a - bCosq, for a > b.  Now let's look at what happens if a = b.

Cardiods:  Functions of the Type r = a - bCosq, for a = b.

Let's look at r = 1 - Cosq, r = 2 - 2Cosq, and r = 3 - 3Cosq.

Clear[r,t,x,y,p1l]
r1c[t_] = 1 - Cos[t];
r2c[t_] = 2 - 2Cos[t];
r3c[t_] = 3 - 3Cos[t];
x1[t_] = r1c[t] Cos[t];
y1[t_] = r1c[t] Sin[t];
x2[t_] = r2c[t] Cos[t];
y2[t_] = r2c[t] Sin[t];
x3[t_] = r3c[t] Cos[t];
y3[t_] = r3c[t] Sin[t];

p1l=ParametricPlot[{x1[t],y1[t]},{t,0,2Pi},
     PlotRange->{{-6.5,3.5},{-5,5}},
     AxesOrigin->{0,0},AxesLabel->{"x","y"},
     PlotLabel->"r = 1 - Cos[t]",
     AspectRatio->Automatic,
     DisplayFunction->Identity];

p2l=ParametricPlot[{x2[t],y2[t]},{t,0,2Pi},
     PlotRange->{{-6.5,3.5},{-5,5}},
     AxesOrigin->{0,0},AxesLabel->{"x","y"},
     PlotLabel->"r = 2 - 2Cos[t]",
     AspectRatio->Automatic,
     DisplayFunction->Identity];

p3l=ParametricPlot[{x3[t],y3[t]},{t,0,2Pi},
     PlotRange->{{-6.5,3.5},{-5,5}},
     AxesOrigin->{0,0},AxesLabel->{"x","y"},
     PlotLabel->"r = 3 - 3Cos[t]",
     AspectRatio->Automatic,
     DisplayFunction->Identity];

Show[GraphicsArray[{p1l,p2l,p3l},
					DisplayFunction->$DisplayFunction]];

The graph of r = a - bCosq, for a = b, is a cardioid.  As you can see, other than size, cardioids don't exhibit much variety.

Limacons of the Type r = a - bCosq, for a < b.

The graph of r = a - bCosq, for a < b, is a different type of limacon.  
Recall our earlier example, r = 1 - 2Cosq.

Clear[r,t,x,y]
r[t_] = 1 - 2 Cos[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     AxesLabel->{"x","y"},
     PlotLabel->"r = 1 - 2Cos[t]",
     AspectRatio->Automatic];

Now look at how this graph is traced out as q varies from 0 to 2¹.

Clear[r,t,x,y]
r[t_] = 1 - 2Cos[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

Do[ParametricPlot[{x[t],y[t]},{t,0,k Pi/6},        
     PlotRange->{{-3.5,1},{-3,3}},                 
     AxesOrigin->{0,0},AxesLabel->{"x","y"},       
     PlotLabel->"r = 1 - 2Cos[t]",                 
     AspectRatio->Automatic],                      
     {k,1,12}];

Let's look at how varying a affects the graph of r = a - bCosq, for b = 2, a < b.  	
Plot		r = a - 2Cosq for a = 1, 1.25, 1.5, and 1.75.

Clear[r,t,x,y]
r[t_] = (.25 k) - 2 Cos[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

Do[ParametricPlot[{x[t],y[t]},{t,0,2Pi},												
     PlotRange->{{-4,1.5},{-3.5,3.5}},              
     AxesOrigin->{0,0},AxesLabel->{"x","y"},        
     AspectRatio->Automatic],                       
     {k,4,7}];

For more help in understanding what is happening, animate the above graphics.

Next show how varying b affects the graph of r = a - bCosq, fora = 1, a < b.  	
Plot		r =1 -bCosq for b =  1.25, 1.5, 1.75, and 2.

Clear[r,t,x,y]
r[t_] = 1 - .25k Cos[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

Do[ParametricPlot[{x[t],y[t]},{t,0,2Pi},												
     	PlotRange->All,              
     	AxesOrigin->{0,0},
		AxesLabel->{"x","y"},        
     	AspectRatio->Automatic],                       
     	{k,5,8}];

Again,animation may help you to see what is going on.

Limacons and Cardioids

What you have seen so far probably leads you to believe that there is a strong relatioship between the two types of limacons and cardioids.  The following may help to reinforce the relationship.  Select and animate the graphics below.  They show what happens to the the graph of r = a - bCosq, first for a varying from 2 to 1 with b = 1, then for b varying from 1 to 2 with a = 1.

Explain why the two types of limacons and the cardioids are all special cases of the same concept.

Changing Cosine to Sine

Only one major issue remains in our investigation of limacons and the cardioids:  suppose we change the cosine function to the sine function in r = a - bCosq.  How will this effect the graphs?
		Let's make that change in our original three graphs.

i.	Graph of r = 2 - Sinq .

Clear[r,t,x,y,p,g]
r[t_] = 2 - Sin[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p=ParametricPlot[{x[t],y[t]},{t,0,2Pi},
				PlotRange->All,
     			AxesLabel->{"x","y"},
				DisplayFunction->Identity,
     			AspectRatio->Automatic];

g=Graphics[Text["r = 2 - Sin[t]",{1.3,1.2}]];

Show[p,g,DisplayFunction->$DisplayFunction];

ii.	Graph r = 1 - Sinq.

Clear[r,t,x,y,p,g]
r[t_] = 1 - Sin[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p=ParametricPlot[{x[t],y[t]},{t,0,2Pi},
				PlotRange->All,
     			AxesLabel->{"x","y"},
				DisplayFunction->Identity,
     			AspectRatio->Automatic];

g=Graphics[Text["r = 1 - Sin[t]",{.8,.4}]];

Show[p,g,DisplayFunction->$DisplayFunction];

iii.	Graph r = 1 -2Sinq.

Clear[r,t,x,y,p,g]
r[t_] = 1 - 2Sin[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p=ParametricPlot[{x[t],y[t]},{t,0,2Pi},
				PlotRange->All,
     			AxesLabel->{"x","y"},
				DisplayFunction->Identity,
     			AspectRatio->Automatic];

g=Graphics[Text["r = 1 - 2Sin[t]",{1.2,.35}]];

Show[p,g,DisplayFunction->$DisplayFunction];

Explain why you could have used your knowledge of the cosine function and the sine function to predict the above changes.

SECTION FOUR: Other Polar Graphs

Introduction

Roses, limacons, and cardioids are examples of graphs whose equations look particularly nice in polar coordinates.  We will now look at some other graphs whose equations also have this property.  We will specifically not consider lines, parabolas, ellipses, and hyperbolas at this point since they tend to be handled more easily in rectangular coordinates.

Circles

Certain types of circles have nice-looking equations in polar coordinates.  The most obvious type is a circle with center at the pole (or "Origin" in rectangular coordinates).  Consider the graph of r = 3.

Clear[r,t,x,y]
r[t_] = 3;
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];
p = ParametricPlot[{x[t],y[t]},{t,0,2Pi},
     AxesLabel->{"x","y"},
     PlotRange->All,
     AspectRatio->Automatic,
     DisplayFunction->Identity];
t = Graphics[Text["r = 3",{3,3}]];

Show[p,t, DisplayFunction->$DisplayFunction];

You should also be able to write the equation of this circle in rectangular coordinates.

Other types of circles with simple polar representation are those of form r = aSinq and r = aCosq.  Consider two examples.

Clear[r1,r2,t,x1,y1,x2,y2,a,b]

r1[t_] = Sin[t];
x1[t_] = r1[t] Cos[t];
y1[t_] = r1[t] Sin[t];
r2[t_] = Cos[t];
x2[t_] = r2[t] Cos[t];
y2[t_] = r2[t] Sin[t];

p1 = ParametricPlot[{x1[t],y1[t]},{t,0,2Pi},
     AxesLabel->{"x","y"},
     PlotRange->All,
     AspectRatio->Automatic,
     DisplayFunction->Identity];

t1 = Graphics[Text["r = Sin[t]",{.8,.9}]];

a=Show[p1,t1];

p2 = ParametricPlot[{x2[t],y2[t]},{t,0,2Pi},
     AxesLabel->{"x","y"},
     PlotRange->All,
     AspectRatio->Automatic,
     DisplayFunction->Identity];

t2 = Graphics[Text["r = Cos[t]",{1.05,.5}]];

b=Show[p2,t2];

Show[GraphicsArray[{a,b},
	DisplayFunction->$DisplayFunction]];

Given what you know about the sine and cosine functions, these results are probably not surprising.  Try a few more examples if you want; there is not really too much more to the story on r = aSinq and r = aCosq.  Circles other than the types already mentioned have much more complicated equations in polar coordinates.  For most purposes they are more easily handled in rectangular coordinates.

Lemniscates

The graphs of the polar equations rÛ = aÛSin2q and rÛ = aÛCos2q
are lemniscates. Consider two examples.

First, rÛ = 9Sin2q.

Clear[t,r1,r2,x1,y1,x2,y2,t1]
r1[t_] = 3Sqrt[Sin[2t]];
x1[t_] = r1[t] Cos[t];
y1[t_] = r1[t] Sin[t];
r2[t_] = -3Sqrt[Sin[2t]];
x2[t_] = r2[t] Cos[t];
y2[t_] = r2[t] Sin[t];
r3[t_] = 3Sqrt[Cos[2t]];
x3[t_] = r3[t] Cos[t];
y3[t_] = r3[t] Sin[t];
r4[t_] = -3Sqrt[Cos[2t]];
x4[t_] = r4[t] Cos[t];
y4[t_] = r4[t] Sin[t];

p1 = ParametricPlot[{x1[t],y1[t]},{t,0.0001,Pi/2},
     AxesLabel->{"x","y"},
     PlotRange->All,
     AspectRatio->Automatic,
     DisplayFunction->Identity];
p2 = ParametricPlot[{x2[t],y2[t]},{t,0.0001,Pi/2},
     AxesLabel->{"x","y"},
     PlotRange->All,
     AspectRatio->Automatic,
     DisplayFunction->Identity];
t1 = Graphics[Text["r^2 = 9Sin[2t]",{3,2.5}]];

Show[p1,p2,t1,DisplayFunction->$DisplayFunction];

Now, rÛ = 9Cos2q.

Clear[t,r3,r4,r5,r6,x3,y3,x4,y4,x5,y5,x6,y6,t3]
r3[t_] = 3Sqrt[Cos[2t]];
x3[t_] = r3[t] Cos[t];
y3[t_] = r3[t] Sin[t];
r4[t_] = -3Sqrt[Cos[2t]];
x4[t_] = r4[t] Cos[t];
y4[t_] = r4[t] Sin[t];
r5[t_] = 3Sqrt[Cos[2t]];
x5[t_] = r5[t] Cos[t];
y5[t_] = r5[t] Sin[t];
r6[t_] = -3Sqrt[Cos[2t]];
x6[t_] = r6[t] Cos[t];
y6[t_] = r6[t] Sin[t];

p3 = ParametricPlot[{x3[t],y3[t]},{t,0,.999 Pi/4},
     AxesLabel->{"x","y"},
     PlotRange->All,
     AspectRatio->Automatic,
     DisplayFunction->Identity];
p4 = ParametricPlot[{x4[t],y4[t]},{t,0,.999 Pi/4},
     AxesLabel->{"x","y"},
     PlotRange->All,
     AspectRatio->Automatic,
     DisplayFunction->Identity];
p5 = ParametricPlot[{x5[t],y5[t]},
			{t,1.0001 (3 Pi/4),Pi},
     AxesLabel->{"x","y"},
     PlotRange->All,
     AspectRatio->Automatic,
     DisplayFunction->Identity];
p6 = ParametricPlot[{x6[t],y6[t]},
			{t,1.0001 (3 Pi/4),Pi},
     AxesLabel->{"x","y"},
     PlotRange->All,
     AspectRatio->Automatic,
     DisplayFunction->Identity];
t3 = Graphics[Text["r^2 = 9Cos[2t]",{1.5,1.8}]];

Show[p3,p4,p5,p6,t3,DisplayFunction->$DisplayFunction];

You should consider the following questions about these two graphs:
	i.	Why do the graphs need to be generated in seperate sections?
	ii.	Why is it necessary to be very careful in choosing ranges for q
		in the various sections?

Spirals

Spirals tend to have nice equations in polar coordinates.  One famous spiral is the Spiral of Archimedes,
 r = 2q.

Clear[x,y,r,t,p,tex]
r[t_] = 2 t;
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p = ParametricPlot[{x[t],y[t]},{t,-10 Pi, 10Pi},
		AspectRatio->Automatic,
		DisplayFunction->Identity];

tex = Graphics[Text["r = 2t",Scaled[{.9,.9}]]];

Show[p,tex,DisplayFunction->$DisplayFunction];

This is the correct graph of the Spiral of Archimedes, but it is visually somewhat overwhelming.  Splitting the q range into two parts may help to clarify things.

First let's look at positive values of q:

Clear[x,y,r,t,p1,tex1]
r[t_] = 2 t;
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p1 = ParametricPlot[{x[t],y[t]},{t,0, 10Pi},
		AspectRatio->Automatic,
		DisplayFunction->Identity];

tex1 = Graphics[Text["r = 2t, t > 0",Scaled[{.85,.9}]]];

Show[p1,tex1,DisplayFunction->$DisplayFunction];

Now consider negative values of q:

Clear[x,y,r,t,p2,tex2]

r[t_] = 2 t;
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p2 = ParametricPlot[{x[t],y[t]},{t,-10Pi,0},
		AspectRatio->Automatic,
		DisplayFunction->Identity];

tex2 = Graphics[Text["r = 2t, t < 0",Scaled[{.15,.9}]]];

Show[p2,tex2,DisplayFunction->$DisplayFunction];

Another interesting type of spiral is the Logarithmic Spiral, r = e .

Clear[x,y,r,t,p,tex]
r[t_] = E^((.01) t);
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];
p = ParametricPlot[{x[t],y[t]},{t,0, 10Pi},
		AspectRatio->Automatic,
		PlotRange->All,
		DisplayFunction->Identity];
tex = Graphics[Text["r = E^((.01) t)",{1.4,1.3}]];

Show[p,tex,DisplayFunction->$DisplayFunction];

The interesting things to experiment with on this one include:
	i.	Varying the value of the parameter "a".
	ii.	Varying the range for q.

A third type of spiral is the Hyperbolic Spiral, r = a/q, a > 0.

For example, consider  r = 2/q.

Clear[x,y,r,t,p,tex]
r[t_] = 2/t;
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p = ParametricPlot[{x[t],y[t]},{t,Pi/6, 5Pi},
		PlotRange->{{-3.6,3.6},{-2,2}},
		AspectRatio->Automatic,
		DisplayFunction->Identity];

tex = Graphics[Text["r = 2/t",Scaled[{.9,.9}]]];

Show[p,tex,DisplayFunction->$DisplayFunction];

To get a different view, we can try changing the range for q.

First try 6¹ < q < 12¹.

Clear[x,y,r,t,p,tex,tex2]
r[t_] = 2/t;
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];
p = ParametricPlot[{x[t],y[t]},{t,6Pi, 12Pi},
		AspectRatio->Automatic,
		DisplayFunction->Identity];
tex = Graphics[Text["r = 2/t",
				Scaled[{.9,.95}]]];
tex2 = Graphics[Text["6Pi<t<12Pi",
				Scaled[{.9,.92}]]];

Show[p,tex,tex2,DisplayFunction->$DisplayFunction];

This shows a part of the graph nearer the pole.

Now try ¹/12 < q < 4¹.

Clear[x,y,r,t,p,tex,tex2]
r[t_] = 2/t;
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];
p = ParametricPlot[{x[t],y[t]},{t,Pi/12, 4Pi},
		AspectRatio->Automatic,
		PlotRange->{{-2,6},{-2,2}},
		DisplayFunction->Identity];
tex = Graphics[Text["r = 2/t",Scaled[{.9,.9}]]];
tex2 = Graphics[Text["Pi/12 <t< 4Pi",
				Scaled[{.9,.8}]]];

Show[p,tex,tex2,DisplayFunction->$DisplayFunction];

This seems to indicate that y = 2 might be a horizontal asymptote.  If you think about the behaviour of (cosq)/q and (sinq)/q near q = 0, you might be able to figure out what is going on.  What would be the effect of changing the value of "a"
in r = a/q?

SECTION FIVE: Experimenting WithPolar Graphs

Mathematica allows you to experiment easily with a wide variety of polar graphs.  Here are a few examples to help you get started.

Graph r = 2Sinq + Cos2q.

Clear[r,t,x,y,p, tex]
r[t_] = (Sin[t] + Cos[2t]);
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p = ParametricPlot[{x[t],y[t]},{t,0, 2Pi},
     AxesOrigin->{0,0},AxesLabel->{"x","y"},
     AspectRatio->Automatic,
     DisplayFunction->Identity];

tex = Graphics[Text["r = Sin[t] + Cos[2t]",
								Scaled[{.85,.97}]]];

Show[p,tex,
	PlotRange->All,
	DisplayFunction->$DisplayFunction];

Graph r = Sinq + Sin2q.

Clear[r,t,x,y,p, tex]
r[t_] = (Sin[t] + Sin[2t]);
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p = ParametricPlot[{x[t],y[t]},{t,0, 2Pi},
     AxesOrigin->{0,0},AxesLabel->{"x","y"},
     PlotRange->{{-1.5,1.8},{-.4,2}},
     AspectRatio->Automatic,
     DisplayFunction->Identity];

tex = Graphics[Text["r = Sin[t] + Sin[2t]",
								Scaled[{.7,.97}]]];

Show[p,tex,
	DisplayFunction->$DisplayFunction];

Graph r = SinÛq + Cos3q.

Clear[r,t,x,y,p, tex]
r[t_] = (Sin[t])^2 + Cos[3t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];

p = ParametricPlot[{x[t],y[t]},{t,0, 2Pi},
     AxesOrigin->{0,0},AxesLabel->{"x","y"},
     PlotRange->All,
     AspectRatio->Automatic,
     DisplayFunction->Identity];

tex = Graphics[Text["r = (Sin[t])^2 + Cos[3t]",
								Scaled[{.85,.97}]]];

Show[p,tex,DisplayFunction->$DisplayFunction];

Try some of your own.  One caution, however:  many people find  once they get started creating these graphs that it is very difficult to stop.  Another note:  to say you are really doing mathematics you should constantly be asking yourself why  the graphs look the way they do.  It might help to think about the same function graphed in rectangular coordinates.