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Polar Equations

Date: 05/05/2003 at 11:21:18
From: Derek
Subject: Polar equations

If you have r^2 = 9*cos(theta) and r = 1 + cos(theta), how many points 
of intersection are there?

If I attempt to use simultaneous equations, by splitting the first 
equation into two:

r = -(9*cos(theta))^0.5 and r = (9*cos(theta))^0.5, I then have the 
following:

1 + cos(theta) = -(9*cos(theta))^0.5   and

1 + cos(theta) = (9*cos(theta))^0.5

I cannot square both sides to get rid of the root, because I end up 
with two identical equations and only two intercepts, 1.424 and 
(2pi - 1.424).

When I draw the two graphs using a software package called Autograph 
it clearly shows 4 intercepts.

Please help.


Date: 05/05/2003 at 12:40:54
From: Doctor Luis
Subject: Re: Polar equations

Hi Derek,

You weren't doing anything wrong by ending up with a single equation. 
You want to solve for values of t that satisfy the original two 
equations.

As you noted, by squaring both sides you end up with the same equation 
as if you had substituted for r in the first equation.

   (1 + cos(t))^2 = 9*cos(t)

For the purposes of solving this problem, let x = cos(t). We must 
satisfy the quadratic equation

       (1 + x)^2 = 9*x
   1 + 2*x + x^2 = 9x

or

   x^2 - 7*x + 1 = 0

This gives you two values of x using the quadratic equation (i.e. two 
distinct real roots x_1, x_2). However, one of roots lies outside the 
range [-1, 1]. If you recall, the cosine of a real number cannnot 
exceed 1, so the arccos is not defined (well, technically, it's an 
imaginary number). So one of the roots (say x_2 was the one with 
x_2 > 1) is bad.

The good root x_1 that's in the range [-1, 1] satisfies

   x_1 = cos(t)

or

   t = arccos(x_1)

There are two values of t that satisfy that equation. You can see this 
from the identity cos(2pi - t) = cos(t).

So, the only two values of t are

   t = arccos(x_1), 2pi - arccos(x_1)

And that's where they intersect (to find r, find r(t) for these t).

I'm not sure how you did your graph, but here's mine, done in Maple 8:



As you can see, there are only two intersections. The origin looks 
like a solution, but it doesn't satisfy the equations 0^2 = 9*cos(t),
0 = 1 + cos(t). (One says cos(t) = 0, the other cos(t) = -1.)

I hope this helped. Let us know if you have any other questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 05/05/2003 at 15:43:02
From: Derek
Subject: Polar equations

Many thanks for your prompt reply. I believe I was misled by a lesser 
software package than Maple. My software showed a dashed shape on the 
left-hand side of the graph, symmetrical to the right-hand side of 
your graph (r^2 = 9cos(theta)). I assume, now, the dashed lines meant 
that the values for r were negative(?) and in reality lay on top of 
the right-hand side. 

In that case am I right in saying that the two equations:

1 + cos (theta) = -(9*cos(theta))^0.5 and
1 + cos (theta) = +(9*cos(theta))^0.5

have the same real solutions and no others?

You mentioned that values outside -1 to 1 are not real. Can I do 
anything with these using complex numbers?

Again many thanks for your help.
Derek


Date: 05/06/2003 at 18:49:52
From: Doctor Luis
Subject: Re: Polar equations

Hi Derek,

When the value of r becomes negative you add pi radians to the angle 
theta before graphing. That's why you saw the dashed lines in your 
program.

In general, you want to avoid taking square roots. You'll be 
introducing extraneous solutions by doing that. It's better to square 
and then just use the quadratic formula.

   r^2 = 9*cos(theta)
   r   = 1 + cos(theta)

The REAL problem (pun intended, heh) with r^2 = 9*cos(theta) is that 
theta can only range from -pi to pi. It is forbidden from ever being 
in the range from pi to 3pi/2. As soon as theta enters that range, 
cos(theta) becomes negative, which would mean that r^2 is negative, or 
that r is imaginary. By taking the square root, you are making a 
reflected image on the (until now) forbidden left half of the plane.

If you want those "solutions" too, simply add or subtract pi radians
from the two original solutions. You can see in this diagram
(or from your program) that the solutions are indeed diametrically
opposite (180 degrees apart):



To find the respective radii, just use r = 1 + cos(theta).

As for the complex numbers, I don't think they'll be of much use in 
your diagram. They usually represent a *lack* of intersections. For 
example, when graphing the function y = x^2 + 1, you can see that it 
doesn't cross the x-axis at all. When you try to find an x-intercept 
by setting y = 0, you'll find imaginary roots x = +i, -i. Therefore, 
imaginary roots represent the non-existence of an intersection.

I hope this helped you understand the problem better. Let us know if 
you have any other questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Trigonometry

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