Associated Topics || Dr. Math Home || Search Dr. Math

### 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.

```

```
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

(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

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

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search