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I stumped at one of the exercise in my multivariable calculus textbook.
I try to search online but I can't seem to search
on how answer no 3 and 4 below is derived.
I also plot both of polar coordinates in mathematica but I can't seem to get around on
how to solve them algebraically.
The question and answer as appear in textbook is as below:

The polar equation for the curve $C_{1}$ is $r=\cos 2\theta$,while the curve
$C_{2}$ is described by the polar equation $r=1+\cos\theta$. Find all points
at which $C_{1}$ and $C_{2}$ intersect.

Solution below
1) r=0 (the origin)
2) ($\frac{5-\sqrt{17}}{4}$ , $\pm\arccos \frac{1-\sqrt{17}}{4}$)
(I manage this solve this, just put both equation equal to each other)

I'm at loss what the formula to derive answer for 3 and 4 below.
3) ($-1$ , $\pm 90 ^\circ$)
4) ($\frac{-1}{2}$ , $\pm 60 ^\circ$)

Can anyone point out to me what's the formula used ?

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First, note that $r=\cos(\theta)+1$ and $r=\cos(\theta)-1$ are the same curve... –  J. M. Jul 15 '12 at 16:44
(3)-(4) look fishy...is the first coordinate supposed to be $\,r\,$? Because that must be non-negative, then... –  DonAntonio Jul 15 '12 at 16:45
@DonAntonio yes, the first coordinate for (3) and (4) supposed to be r according to my textbook.But that's what the solution is printed. I guess there's error on the textbook. –  kypronite Jul 15 '12 at 17:00
What book is that? I guess they must be using the convention Andre's mentions below. –  DonAntonio Jul 15 '12 at 17:13
It's not actually a textbook, more like online handout.It's Calculus revisited: Multivariable calculus by Prof. Herbert Gross.It's question no (3) under pretest section.The url to pdf is ocw.mit.edu/resources/… –  kypronite Jul 15 '12 at 17:18
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4 Answers

up vote 3 down vote accepted

The issues come from a lack of uniqueness: $(r,\theta),(-r,\theta+\pi),(r,\theta+2\pi)$ are all the same. Also, all $(0,\theta)$ are the same for any $\theta$. Fortunately these equations are periodic (with period $2\pi$), which simplifies the problem. So you have to do this as well:


i.e, we need to see where/if one is negative and halfway across the circle, since this will also count as an intersection. $$-\cos(2\theta)=\cos(\theta)+1$$ $$\sin^2(\theta)-\cos^2(\theta)=\cos(\theta)+1$$ $$2\cos^2(\theta)+\cos(\theta)=\cos(\theta)(2\cos(\theta)+1)=0$$ $$\cos(\theta)=0\implies\theta=\pi/2,3\pi/2$$ $$2\cos(\theta)+1=0\implies\theta=2\pi/3, 4\pi/3$$

graphing both should convince you that these, plus the ones you found, give the full set of solutions.

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There are two common conventions about polar equations.

Convention (i): The curve $r=f(\theta)$ is undefined for values of $\theta$ such that $f(\theta)\lt 0$;

Convention (ii): $r \lt 0$ is allowed, and the point with coordinates $(-|r|,\theta)$ is obtained by taking the point with coordinates $(|r|,\theta)$ and reflecting in the origin, or equivalently rotating by $180^\circ$.

Convention (ii) tends to yield nicer pictures than Convention (i).

The book is using Convention (ii). So in addition to $\cos 2\theta=1+\cos\theta$, we need to consider $\cos 2\theta=-(1+\cos\theta)$. Using $\cos 2\theta=2\cos^2\theta-1$, we arrive at $2\cos^2\theta+\cos\theta=0$. That gives $\cos\theta=0$ or $\cos\theta=-1/2$, and leads to the solutions described in answers $(3)$ and $(4)$.

Let's check directly the correctness of the answer $r=-1$, $\theta=90^\circ$. When we substitute $90^\circ$ for $\theta$ in $\cos 2\theta$, we get $-1$. So the point $(-1, 90^\circ)$ is on $r=\cos2\theta$. And the same point, this time called $(1,-90^\circ)$, is on the curve $r=1+\cos\theta$.

Also, $r=0$ is clearly on both curves, as any graph will reveal. It cannot be obtained by setting $\cos 2\theta=\pm (1+\cos 2\theta)$ because the origin has infinitely many addresses.

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I need to redeem my earlier egregious errors:

The thing to remember with a polar plot of the form $r=f(\theta)$ is that it represents the set $\{f(\theta)(\cos \theta, \sin \theta)\}_{\theta \in \mathbb{R}}$. Thus to solve the problem, we need to determine $\Omega = \{\cos 2\theta(\cos \theta, \sin \theta)\}_{\theta \in \mathbb{R}} \cap \{(1+\cos\theta)(\cos \theta, \sin \theta)\}_{\theta \in \mathbb{R}}$.

So, we look for $\theta, \phi$ that solve the equation: $$\cos 2\theta(\cos \theta, \sin \theta) = (1+\cos\phi)(\cos \phi, \sin \phi).$$

To simplify life, first look for solutions to $\cos 2\theta = 0$, $(1+\cos\phi)=0$. Setting $\theta = \pi, \phi = \frac{\pi}{2}$ shows that $(0,0) \in \Omega$. This gives solution 1).

Now suppose that either $\theta$ and $\phi$ are such that $\cos 2\theta \neq 0$, $(1+\cos\phi) \neq 0$. Then, since the points are the same, we must have $\phi = \theta + n \pi$, for some integer $n$ (since both $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$ lie on the same line through the origin). In our case, we need only consider $n=0,1$.

Taking $n=0$, we look for solutions to $1+ \cos \theta = \cos 2 \theta$. Since $\cos 2 \theta = 2 (\cos \theta)^2-1$, we have $1+ \cos \theta = 2 (\cos \theta)^2-1$. We can solve this by looking for solutions to $1+x = 2 x^2-1$. Solving the quadratic yields $x = \frac{1 \pm \sqrt{17}}{4}$. Since only one of these points lies in $[-1,1]$ we have $\theta = \pm \arccos \frac{1 - \sqrt{17}}{4}$. This gives solution 2).

Now take $n=1$. Look for solutions to $1+ \cos \theta = -\cos 2 \theta$. Using the previous expansion for $\cos 2 \theta$ we end up with the equation $\cos \theta = -2 (\cos \theta)^2$. As above, we look for solutions to $x = -2x^2$ that lie in $[-1,1]$. The solutions are $x=0$, and $x=-\frac{1}{2}$ which both lie in $[-1,1]$. This gives $\theta = \pm \frac{\pi}{2}$ and $\theta = \pm \frac{\pi}{3}$, which gives solutions 3) and 4).

You can use Octave (or Matlab) to see the plot with:

t = 0:0.1:10*pi;
plot((1+cos(t)).*cos(t), (1+cos(t)).*sin(t), cos(2*t).*cos(t), cos(2*t).*sin(t))

enter image description here

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I know this is a pretty late reply, but I found your post while investigating polar intersection for my teaching, and ended up with the following strategy:

The graphs for two polar functions $r=f(\theta)$ and $r=g(\theta)$ have possible intersections in 3 cases:

  1. In the origin if the equations $f(\theta)=0$ and $g(\theta)=0$ have at least one solution each.
  2. All the points $[g(\theta_i),\theta_i]$ where $\theta_i$ are the solutions to the equation $f(\theta)=g(\theta)$.
  3. All the points $[g(\theta_i),\theta_i]$ where $\theta_i$ are the solutions to the equation $f(\theta+(2k+1)\pi)=-g(\theta)$.

A comment on this: In case 2 you just as well could calculate the points with $[f(\theta_i),\theta_i]$ but in case 3 you have to use $g(\theta_i)$ so it is better just to get used to calculating the points in both case 2 and 3 using $g$.

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