POLAR COORDINATE SYSTEM

Initialization

Needs["Graphics`Graphics`"];

Plot of points on a polar coordinate system

Plot each of the following points on a polar coordinate system.

1.  P(3,60Ø)

To see the answer, double-click on the box.

Clear[r1,r2,r3,r4,r5,r6,l1,l2,l3,l4,pt1,t];
r1[t_] = 1;
r2[t_] = 2;
r3[t_] = 3;
r4[t_] = 4;
r5[t_] = 5;
r6[t_] = 6;
l1 = Line[{{-3,-3Sqrt[3]},{3,3Sqrt[3]}}];
l2 = Line[{{-3,3Sqrt[3]},{3,-3Sqrt[3]}}];
l3 = Line[{{-3Sqrt[3],-3},{3Sqrt[3],3}}];
l4 = Line[{{3Sqrt[3],-3},{-3Sqrt[3],3}}];
pt1 = {RGBColor[1,0,0],PointSize[.03],
            Point[{1.5,1.5Sqrt[3]}]};
Show[PolarPlot[{r1[t],r2[t],r3[t],r4[t],
                r5[t],r6[t]},{t,0,2Pi},
                DisplayFunction->Identity],
     Graphics[{l1,l2,l3,l4,pt1}],
     DisplayFunction->$DisplayFunction];

2.  P(-4,315Ø)

To see the answer, double-click on the box.

Clear[r1,r2,r3,r4,r5,r6,l1,l2,pt1,t];
r1[t_] = 1;
r2[t_] = 2;
r3[t_] = 3;
r4[t_] = 4;
r5[t_] = 5;
r6[t_] = 6;
l1 = Line[{{-6,-6},{6,6}}];
l2 = Line[{{-6,6},{6,-6}}];
pt1 = {RGBColor[1,0,0],PointSize[.03],
            Point[{-2Sqrt[2],2Sqrt[2]}]};
Show[PolarPlot[{r1[t],r2[t],r3[t],r4[t],
                r5[t],r6[t]},{t,0,2 Pi},
                DisplayFunction->Identity],
     Graphics[{l1,l2,pt1}],
     DisplayFunction->$DisplayFunction];

3.  P(4,™‹`)

To see the answer, double-click on the box.

Clear[r1,r2,r3,r4,r5,r6,pt1,t];
r1[t_] = 1;
r2[t_] = 2;
r3[t_] = 3;
r4[t_] = 4;
r5[t_] = 5;
r6[t_] = 6;
pt1 = {RGBColor[1,0,0],PointSize[.03],
            Point[{0,-4}]};
Show[PolarPlot[{r1[t],r2[t],r3[t],r4[t],
                r5[t],r6[t]},{t,0,2 Pi},
                DisplayFunction->Identity],
     Graphics[pt1],
     DisplayFunction->$DisplayFunction];

4.  P(5, -£›`)

To see the answer, double-click on the box.

Clear[r1,r2,r3,r4,r5,r6,l1,l2,l3,l4,pt1,t];
r1[t_] = 1;
r2[t_] = 2;
r3[t_] = 3;
r4[t_] = 4;
r5[t_] = 5;
r6[t_] = 6;
l1 = Line[{{-3,-3Sqrt[3]},{3,3Sqrt[3]}}];
l2 = Line[{{-3,3Sqrt[3]},{3,-3Sqrt[3]}}];
l3 = Line[{{-3Sqrt[3],-3},{3Sqrt[3],3}}];
l4 = Line[{{3Sqrt[3],-3},{-3Sqrt[3],3}}];
pt1 = {RGBColor[1,0,0],PointSize[.03],
            Point[{-2.5,2.5Sqrt[3]}]};
Show[PolarPlot[{r1[t],r2[t],r3[t],
                r4[t],r5[t],r6[t]},{t,0,2 Pi},
               DisplayFunction->Identity],
     Graphics[{l1,l2,l3,l4,pt1}],
     DisplayFunction->$DisplayFunction];

Names of points in polar coordinates

Give three other pairs of polar coordinates for each of the following.

1.  P(2,60Ø)

To see the answer, double-click on the box.

P(2,-300Ø)     P(-2,240Ø)     P(-2,-120Ø)

2.  P(4,-15Ø)

To see the answer, double-click on the box.

P(4,345Ø)      P(-4,165Ø)      P(-4,-195Ø)

3.  P(-3,`)

To see the answer, double-click on the box.

P(3,0)        P(3,2`)        P(-3,-`)

4.  P(1, -™‹`)

To see the answer, double-click on the box.

P(1,™∕`)       P(-1,™‹`)        P(-1,-™∕`)

Graphs of polar-coordinate equations

We will sketch the graph of several polar coordinate equations by plotting points.  To see each of the graphs, double-click on the boxes below.

r = 3 + sin t

To construct the graph, we will make a table of values for t and r.  The table below shows the value for sin t and r when t = 30Ø, 60Ø, . . . , 360Ø.

Clear[values,chart];
values = Table[{180t/Pi,N[Sin[t],2],N[3+Sin[t],2]},
               {t,0,2Pi,Pi/6}];
chart = TableForm[values,
                  TableHeadings->{Automatic,
                                 {"t",
                                  "sin t","r"}}]

The graph below shows the points (r,t) above plotted.

Clear[x,y,r,t]
r[t_] = 3 + Sin[t];
x[t_] = r[t] Cos[t];
y[t_] = r[t] Sin[t];
ListPlot[Table[{x[t],y[t]},{t,0,2Pi,Pi/6}],
           PlotStyle->{PointSize[0.03],
                       RGBColor[1,0,0]},
           AspectRatio->1];

If we connect these points into a smooth curve, we get the following figure.

Clear[r,t];
r[t_] = 3 + Sin[t];
PolarPlot[r[t],{t,0,2Pi},
  PlotStyle->{RGBColor[1,0,0]}];

r = 4 cos t

Clear[r,t];
r[t_] = 4 Cos[t];
PolarPlot[r[t],{t,0,2Pi},
  PlotStyle->{RGBColor[1,0,0]}];

r = 1 / (1 + sin t)

When you make your table of values,what is the problem with this graph?

Answer:

The function is undefined at 3`/2 since sin(3`/2) = -1

Clear[p1,p2,t];
p1 = PolarPlot[1 / (1 + Sin[t]),{t,0,5Pi/4},
  PlotStyle->{RGBColor[1,0,0]},
  AspectRatio->1,
  DisplayFunction->Identity];
p2 = PolarPlot[1 / (1 + Sin[t]),{t,7Pi/4,2Pi},
  PlotStyle->{RGBColor[1,0,0]},
  AspectRatio->1,
  DisplayFunction->Identity];
Show[p1,p2,DisplayFunction->$DisplayFunction];

Special types of equations

Limacons

The graph of an equation of the form r = a + b sin t or 
r = a + b cos t is called a limacon.  Let's look at the graph of a limacon.  The following is the graph of the equation 
r = 1 + 3 cos t.

Clear[r,t];
r[t_] = 1 + 3Cos[t];
PolarPlot[r[t],{t,0,2Pi},
  PlotStyle->{RGBColor[1,0,0]}];

You will notice that some limacons look a little different than others.  The graph below is the sketch of the equation 
r = 3 + cos t.  No, this is not a circle!!!

Clear[r,t];
r[t_] = 3 + Cos[t];
PolarPlot[r[t],{t,0,2Pi},
  PlotStyle->{RGBColor[1,0,0]}];

If a = b in the equation of the limacon, you get a special case called a cardioid.  The graph below is the graph of 
r = -2 - 2 sin t.

Clear[r,t];
r[t_] = -2 - 2Sin[t];
PolarPlot[r[t],{t,0,2Pi},
  PlotStyle->{RGBColor[1,0,0]}];

Cardiac means "of or near the heart".  Can you see why the graph is called a carioid?

Remember:
     1)  r = a + b sin t or r = a + b cos t is a limacon. 
     2)  The shape of a limacon is one of the following:
         --similar to a circle
         --heart shaped (called a cardioid), when a = b  
         --heart shaped with a loop in the middle

Rose curves

If a polar equation has the form r = a sin n(t) or 
r = a cos n(t), the graph is a rose curve.  The graph below is the graph of r = 6 sin 2t

Clear[r,t];
r[t_] = 6 Sin[2t];
PolarPlot[r[t],{t,0,2Pi},
  PlotStyle->{RGBColor[1,0,0]}];

Rose curves all have the same general shape, but the number of leaves will change.  The graphs below are graphs of r = 4 cos n(t) for n = 1, 2, 3, 4, 5, and 6.
     See if you can determine how many leaves a rose curve will have from the following graphs.

Clear[r,t,n,chart,setup];
r[t_] = 4 Cos[n t];
chart = Table[PolarPlot[r[t],{t,0,2Pi},
                PlotLabel->n" = n",
                PlotStyle->{RGBColor[1,0,0]},
                DisplayFunction->Identity],
             {n,1.,6}];
setup = Partition[chart,2];
Show[GraphicsArray[setup,
                  DisplayFunction->$DisplayFunction]];

To see the answer, double-click on the box.

If n is an odd number, the rose curve will have n leaves.  If n is an even number, the rose curve will have 2n leaves.

Lemniscates

The graph of an equation of the form r¤ = a¤ sin 2t or 
r¤ = a¤ cos 2t is called a lemniscate.  
    One thing that you have to be very careful about when graphing a lemniscate is that the graph will have values of t that make r undefined.  Let us use r¤ = 9 cos 2t for an example. If we solve the equation for r we get

r = + Square Root (9 cos 2t)

Now, you know that you can only take the square root of a non-negative number so r is undefined whenever cos 2t is negative.  Where is cos 2t negative?

To see the answer, double-click on the box.

The cosine function is negative in the second and third quadrants which is between `/2 and 3`/2 and also between 5`/2 and 7`/2.  
    If we write these as the following inequalities
       `/2 < 2t < 3`/2         5`/2 < 2t < 7`/2
and solve for t, we get the following:
       `/4 < t < 3`/4         5`/4 < 2t < 7`/4
    This tells us that the function is undefined for the following values:  `/4 < t < 3`/4 and 5`/4 < 2t < 7`/4

Remembering that the function is undefined for these values, let's look at the graph of a lemniscate.  The following is the graph of the equation r¤ = 9 cos 2t.

Clear[r,t];
r[t_] = Sqrt[9 Cos[2t]];
ParametricPlot[{{r[t]Cos[t],r[t]Sin[t]},
                {r[t]Cos[Pi-t],r[t]Sin[Pi-t]},
                {r[t]Cos[Pi+t],r[t]Sin[Pi+t]},
                {r[t]Cos[2Pi-t],r[t]Sin[2Pi-t]}},
                {t,0,9999Pi/40000},
                AspectRatio->Automatic,
                PlotStyle->RGBColor[1,0,0]];

The following is the graph of r¤ = 9 sin 2t

Clear[r,t];
r[t_] = Sqrt[9 Sin[2t]];
ParametricPlot[{{r[t]Cos[t],r[t]Sin[t]},
                {r[t]Cos[Pi+t],r[t]Sin[Pi+t]}},
                {t,0,19999Pi/40000},
                AspectRatio->Automatic,
                PlotStyle->RGBColor[1,0,0]];

The graph of a lemniscate looks like a figure-8.

Lines

The equation of a line in polar form will have one of the following general forms:
   1)  r cos t = p
          This is a vertical line through (p,0).
   2)  r sin t = p
          This is a horizontal line through (p,`/2).
   3)  t = a
          This is a line through the origin with inclination a.
   4)  r = C / (A cos t + B sin t) 
          This is a line through (C/A,0) and (C/B,`).
Let us look at an example of each of the above.

r cos t = 3

To see the graph, click on the box.

Clear[r,t];
r[t_] = 3 / Cos[t];
PolarPlot[r[t],{t,-Pi/4,Pi/4},
  PlotStyle->{RGBColor[1,0,0]}];

This a vertical line through (3,0).

r sin t = -5

To see the graph, click on the box.

Clear[r1,r2,r3,r4,l1,t,caxis,graph];
r1[t_] = 1;
r2[t_] = 2;
r3[t_] = 3;
r4[t_] = 4;
r5[t_] = 5;
r6[t_] = 6;
l1[t_] = -5 / Sin[t];
caxis = PolarPlot[{r1[t],r2[t],r3[t],r4[t],
                   r5[t],r6[t]},
             {t,0,2Pi},DisplayFunction->Identity];
graph = PolarPlot[l1[t],{t,Pi/4,3Pi/4},
           PlotStyle->RGBColor[1,0,0],
           DisplayFunction->Identity];
Show[{caxis,graph},DisplayFunction->$DisplayFunction];

This a horizontal line through (-5,`/2).

t = 3Pi/4

To see the graph, click on the box.

Clear[r1,r2,r3,r4,r5,r6,l1,t];
r1[t_] = 1;
r2[t_] = 2;
r3[t_] = 3;
r4[t_] = 4;
r5[t_] = 5;
r6[t_] = 6;
l1 = Line[{{-3Sqrt[2],3Sqrt[2]},{3Sqrt[2],-3Sqrt[2]}}];
Show[PolarPlot[{r1[t],r2[t],r3[t],r4[t],
                r5[t],r6[t]},{t,0,2Pi},
                DisplayFunction->Identity],
     Graphics[l1],
     DisplayFunction->$DisplayFunction];

This a line with inclination 3`/4.

r = 6 / (2 cos t + 3 sin t)

To see the graph, click on the box.

Clear[r1,r2,r3,r4,l1,t,caxis,graph];
r1[t_] = 1;
r2[t_] = 2;
r3[t_] = 3;
r4[t_] = 4;
l1[t_] = 6 / (2 Cos[t] + 3Sin[t]);
caxis = PolarPlot[{r1[t],r2[t],r3[t],r4[t]},
             {t,0,2Pi},DisplayFunction->Identity];
graph = PolarPlot[l1[t],{t,-Pi/16,11Pi/16},
           PlotStyle->RGBColor[1,0,0],
           DisplayFunction->Identity];
Show[{caxis,graph},DisplayFunction->$DisplayFunction];

This a line through (3,0) and (2,`/2).

Circles

The equation of a line in polar form will have one of the following general forms:
   1)  r = 2a cos t 
          This has center at (a,0) and radius of a.
   2)  r = 2a sin t 
          This has center at (a,`/2) and radius of a.
   3)  r = a
          This has center at origin and radius of a.
   4)  r¤ + r¡¤- 2rr¡ cos(t - t¡ ) = a¤
          This has center at (r¡ , t¡ ) and radius of a. Let us look at an example of each of the above.

r = 6 cos(t)

Clear[r,t];
r[t_] = 6 Cos[t];
PolarPlot[r[t],{t,0,2Pi},
               PlotStyle->RGBColor[1,0,0]];

You can tell from the equation that the center of the circle is (3,0) and the radius is 3.

r = 8 sin(t)

Clear[r,t];
r[t_] = 8 Sin[t];
PolarPlot[r[t],{t,0,2Pi},
               PlotStyle->RGBColor[1,0,0]];

You can tell from the equation that the center of the circle is at (4,`/2) and the radius is 4.

r = 4

Clear[r,t];
r[t_] = 4;
PolarPlot[r[t],{t,0,2Pi},
          PlotStyle->RGBColor[1,0,0]];

From the equation you can tell this is a circle with center at the origin and a radius of 4.

r¤ + 25 - 10cos(t - `/3) = 9

Clear[r,t,cir,cen];
cir = Circle[{5/2,5Sqrt[3]/2},3];
cen = Point[{5/2,5Sqrt[3]/2}];
Show[Graphics[{RGBColor[1,0,0],
             {cir,cen}}],Axes->True,
                 AspectRatio->1,
                 Epilog->Text["(5,π/3)",{3,5Sqrt[3]/2}]];

You can tell from the equation that this is a circle with a center at (5,`/3) and a radius of 3.

Intersectons of polar coordinate graphs

A simultaneous real solution of two equations in rectangular coordinates represents a point of intersection of their graphs.

Let us see what happens in polar coordinates.  If we solve simultaneously the two equations
    r = 6 sin t    and    r = 6 cos t
we get the following:

6 sin t = 6 cos t
tan t = 1
t = `/4, 5`/4
r = 3 Sqrt(2), -3 Sqrt(2)

The coordinates (3 Sqrt(2), `/4) and (-3 Sqrt(2), 5`/4) define the same point as you will see from the graph.

Clear[r1,r2,t];
r1[t_] = 6 Sin[t];
r2[t_] = 6 Cos[t];
PolarPlot[{r1[t],r2[t]},{t,0,2Pi},
   PlotStyle->{RGBColor[0,0,1]}];

The graph shows that the graphs both pass through the points (3 Sqrt(2), `/4) and (-3 Sqrt(2), 5`/4), but they also both pass through the origin.  We did not get the origin when we solved this equation algebraically.  Why not?  
    Let us look at these graphs again but this time we will look at what happens at t increases from 0 to 2`.

Clear[r1,r2,s,t];
r1[t_] = 6 Sin[t];
r2[t_] = 6 Cos[t];
Do[Show[PolarPlot[{r1[t],r2[t]},{t,0,2 Pi},
          PlotStyle->{RGBColor[1,0,0],
                      RGBColor[0,0,1]},
          DisplayFunction->Identity],
        Graphics[{{RGBColor[1,0,0],PointSize[0.03],
		          Point[{r1[s] Cos[s], r1[s] Sin[s]}]},
                  {RGBColor[0,0,1],PointSize[0.03],
                  Point[{r2[s] Cos[s], r2[s] Sin[s]}]}}],
  DisplayFunction->$DisplayFunction],
  {s,0,Pi,Pi/8}];

To animate this, click on the box and press Apple-Y.

You can see from this graph that the point (0,0) satisfies the first equation and (0,`/2) satisfies the second equation.  The origin has no pair of points that satisfy both equations.  You have to graph the equations to determine this.