Introduction to Finding the Intersection of Polar Equations

By Debra Woods

Initialization

Off[General::spell1];
Off[Plot::plnr];
Needs["Graphics`Colors`"]
Needs["Miscellaneous`Units`"]
Needs["Graphics`Graphics`"]

Statement of Problem

The following is a classroom example of solving systems of polar equations.

Example

Problem:  Find the point(s) of intersection of two polar graphs.

Sometimes, when trying to find the point(s) of intersection algebraically, one or more of the roots may become "lost" due to the parameterization of the equations.  For example, consider the equations
		r = 2 sin θ
		r = 1 + 2 cos θ.
Trying to find the roots algebraically 
		2 sin θ = 1 + 2 cos θ
		4 sin\(TraditionalForm\`\^2\)θ = 1 + 4 cos θ + cos\(TraditionalForm\`\^2\)θ
		4 - 4 cos\(TraditionalForm\`\^2\)θ = 1 + 4 cos θ + cos\(TraditionalForm\`\^2\)θ				
		5 cos\(TraditionalForm\`\^2\)θ + 4 cos θ - 3 = 0
leads to an equation that is clearly of quadratic form, and thus will only yield two solutions.  Let's use Mathematica to find these  roots.

Do[
Print[FindRoot[2Sin[theta]==1+2Cos[theta],
				{theta,k}]],{k,0,2Pi,Pi}]

These may look like three solutions, but note that 2π times the first solution is the third solution, and we're talking polar coordinates after all.

N[1.14677 + 2 π]

Now, let's see what the graph of the intersection's looks like:

Clear[r1,r2,theta];
r1[theta_] := 2Sin[theta];
r2[theta_] := 1 + 2 Cos[theta];

PolarPlot[{r1[theta],r2[theta]},
		{theta, 0, 2 π},
		  PlotStyle->{{Red},{Blue}}];

You may notice that there are three points of intersection!  Why didn't our algebra, or Mathematica for that matter, find the root at (0,0)?  Let's see what happens if we trace a point around the graphs as θ changes from 0 to 2π.

Do[Show[PolarPlot[{r1[theta],r2[theta]},
				{theta, 0, 2π},
		  		PlotStyle->{{Red},{Blue}},
	      		DisplayFunction->Identity],
				Graphics[{{RGBColor[0,1,0],
	        			PointSize[0.03],
	        			Point[{r1[a] Cos[a],r1[a] Sin[a]}]},
	       				{RGBColor[1,0,1],
	        			PointSize[0.03],
	        			Point[{r2[a] Cos[a],r2[a] Sin[a]}]}}],
		DisplayFunction->$DisplayFunction,
		Epilog->Text[a,{2.5,1.75}]]
	,
{a,0,2 π,π/8}];

Select all of the above graphs and press apple-y to see the animation.

Notice that as the points move around the curves, the two points are never at the point (0,0) simultaneously, but they did meet at the other two points of intersection, sometimes.

Example (Reparameterized)

Now, let's reparameterize r = 1 + 2 Cos θ by plotting the curve using the points 
(r(θ + π/3)cos(θ + π/3), r(θ + π/3)sin(θ + π/3))
As you can see, the curves are the same.

ParametricPlot[{{r1[theta] Cos[theta],r1[theta] Sin[theta]},
				{r2[theta + π/3] Cos[theta + π/3],
				r2[theta + π/3] Sin[theta + π/3]}},
				{theta, 0, 2π},
			AspectRatio->Automatic,
		  	PlotStyle->{{Red},{Blue}}];

It is clear that these plots yield the same curves.  Next, observe the movement of the points using the new parameterization.

Do[Show[PolarPlot[{r1[theta],r2[theta]},
				{theta, 0, 2π},
		  		PlotStyle->{{Red},{Blue}},
	      		DisplayFunction->Identity],
				Graphics[{{RGBColor[0,1,0],
	        			PointSize[0.04],
	        			Point[{r1[a] Cos[a],r1[a] Sin[a]}]},
	       				{RGBColor[1,0,1],
	        			PointSize[0.02],
	        			Point[{r2[a + π/3] Cos[a + π/3],
	        				   r2[a + π/3] Sin[a + π/3]}]}}],
		DisplayFunction->$DisplayFunction,
		Epilog->Text[a,{2.5,1.75}]]
	,
{a,0,2 π,π/8}];

Select all of the graphs and press apple-y to see the animation.

Notice that in the new parameterization,the points arrive at the origin simultaneously when θ = π, but are never at the other two points simultaneously as before.