MHB Points of intersection of polar equations

[Dethrone]

Find the points of intersection of $\rho=\cos\left({2\theta}\right)$ and $\rho=\cos\left({\theta}\right)$

By setting $\cos\left({2\theta}\right)=\cos\left({\theta}\right)$, we get the solutions $\theta=0,\frac{2\pi}{3},\frac{4\pi}{3}$.

My question is how come that doesn't give us all the points of intersection? There are some points of intersection at the poles that were not obtained above, such as $(0, \frac{\pi}{3})$ and $(0, \frac{\pi}{2})$

 

[I like Serena]

Find the points of intersection of $\rho=\cos\left({2\theta}\right)$ and $\rho=\cos\left({\theta}\right)$

By setting $\cos\left({2\theta}\right)=\cos\left({\theta}\right)$, we get the solutions $\theta=0,\frac{2\pi}{3},\frac{4\pi}{3}$.

That's not quite the solution!

Properly solving it gives:
$$2 \theta = \pm \theta + 2 k \pi
\quad\Rightarrow\quad \theta = 2 k \pi \vee \theta = \frac 2 3 k \pi
\quad\Rightarrow\quad \theta = \frac 2 3 k \pi$$
where $k$ is any integer.

My question is how come that doesn't give us all the points of intersection? There are some points of intersection at the poles that were not obtained above, such as $(0, \frac{\pi}{3})$ and $(0, \frac{\pi}{2})$

Huh?
How did you get vectors?

Do you mean that $\theta=\frac \pi 2$?
If so, that is not a solution!

Or do you mean $y=\frac \pi 2$?
That makes even less sense! (Wasntme)

 

[Dethrone]

I'm getting this straight from an example given by the textbook.

"Setting $\cos\left({2\theta}\right)=\cos\left({\theta}\right)$, then $\theta=0$, $\frac{2\pi}{3}$, $\frac{4\pi}{3}$, yielding points of intersection $(1,0)$, $(-\frac{1}{2}, \frac{2\pi}{3})$, and $(-\frac{1}{2}, \frac{4\pi}{3})$. But the pole is also an intersection point, appearing as $(0, \frac{\pi}{4})$ on $\rho=\cos\left({2\theta}\right)$ and as $(0, \frac{\pi}{2})$ on $\rho=\cos\left({\theta}\right)$"

I don't think they're using bracket notation to represent vectors, but rather $(\rho, \theta)$, where $\rho$ is the radius.

Do you mean that $\theta=\frac \pi 2$?
If so, that is not a solution!

My point and question exactly. $\theta=\frac \pi 2$ is indeed not a solution to $\cos\left({2\theta}\right)=\cos\left({\theta}\right)$, but it is a point of intersection, or at least the book says it is. My question again: Why are there points of intersection other than the ones that satisfy the above equation. Those points of intersections also happen to be at the origin/pole.

 

[Prove It]

Find the points of intersection of $\rho=\cos\left({2\theta}\right)$ and $\rho=\cos\left({\theta}\right)$

By setting $\cos\left({2\theta}\right)=\cos\left({\theta}\right)$, we get the solutions $\theta=0,\frac{2\pi}{3},\frac{4\pi}{3}$.

My question is how come that doesn't give us all the points of intersection? There are some points of intersection at the poles that were not obtained above, such as $(0, \frac{\pi}{3})$ and $(0, \frac{\pi}{2})$

Just putting in my two cents as I found your original solution to this equation very sloppy...

$\displaystyle \begin{align*} \cos{ \left( 2\theta \right) } &= \cos{ \left( \theta \right) } \\ 2\cos^2{ \left( \theta \right) } - 1 &= \cos{ \left( \theta \right) } \\ 2\cos^2{ \left( \theta \right) } - \cos{ \left( \theta \right) } - 1 &= 0 \\ 2\cos^2{\left( \theta \right) } - 2\cos{ \left( \theta \right) } + \cos{ \left( \theta \right) } - 1 &= 0 \\ 2\cos{ \left( \theta \right) } \left[ \cos{ \left( \theta \right) } - 1 \right] + 1 \left[ \cos{ \left( \theta \right) } - 1 \right] &= 0 \\ \left[ \cos{ \left( \theta \right) } - 1 \right] \left[ 2\cos{ \left( \theta \right) } + 1 \right] &= 0 \\ \cos{ \left( \theta \right) } - 1 &= 0 \textrm{ or } 2\cos{ \left( \theta \right) } + 1 = 0 \\ \cos{ \left( \theta \right) } &= 1 \textrm{ or } \cos{ \left( \theta \right) } = -\frac{1}{2} \\ \theta &= 2\pi\,n \textrm{ or }\theta = \left\{ \frac{2\pi}{3} , \frac{4\pi}{3} \right\} + 2\pi\,n \textrm{ where } n \in \mathbf{Z} \\ \theta &= \left\{ 0 , \frac{2\pi}{3} , \frac{4\pi}{3} \right\} + 2\pi n \end{align*}$

 

[Dethrone]

Yup, that's what ILS said too and it makes sense. The solution I have here was given by my workbook, "Schaum's Outlines". I think the author was assuming the domain $[0, 2\pi]$ which is consistent with the other questions in the section. Though, I still don't understand why solving the equation doesn't give us all the points of intersection, such as those at the poles.

 

[I like Serena]

Yup, that's what ILS said too and it makes sense. The solution I have here was given by my workbook, "Schaum's Outlines". I think the author was assuming the domain $[0, 2\pi]$ which is consistent with the other questions in the section.

With the domain $[0, 2\pi]$, $\theta=2\pi$ is also a solution. (Angel)
But yeah, a domain of $[0, 2\pi)$ is assumed somewhere.
Note that this should be specified somewhere in the book. Typically such assumptions are written at the beginning of the chapter or some such.

Though, I still don't understand why solving the equation doesn't give us all the points of intersection, such as those at the poles.

Nor do I.
However, it does look a bit odd to have $\rho = \cos \theta$.

It appears we're talking about polar coordinates.
This should be something like $x = \rho \cos \theta$ and $x = \rho \cos 2 \theta$.
Then we have a solution if $\cos\theta=\cos 2\theta$ or if $\rho = 0$.

 

[Dethrone]

I'm working on a chapter dedicated to polar coordinates, and they such formulas:

$$\tan\left({\psi}\right)=\frac{\rho}{\rho '}$$

any ideas what they call this? $\psi$ is the angle between the radius vector OP and the tangent at P.

also,

$$\tan\left({\phi}\right)=\frac{\tan\left({\psi_1}\right)-\tan\left({\psi_2}\right)}{1+\tan\left({\psi_1}\right)\tan\left({\varPsi_2}\right)}$$

which is the angle of intersection.

So they use $\rho$ like that a lot.

In this case, we have solutions $\cos\theta=\cos 2\theta$ and when $\rho = 0$ too. Why doesn't $\cos\theta=\cos 2\theta$ cover all the points of intersection? Why do we also have to consider when $\rho=0$?

 

[I like Serena]

I'm working on a chapter dedicated to polar coordinates, and they such formulas:

$$\tan\left({\psi}\right)=\frac{\rho}{\rho '}$$

any ideas what they call this? $\psi$ is the angle between the radius vector OP and the tangent at P.

What kind of name are you looking for? (Wondering)

Edit: I guess polar tangential angle will do.
Or the angle between the radial line and the tangent.

also,

$$\tan\left({\phi}\right)=\frac{\tan\left({\psi_1}\right)-\tan\left({\psi_2}\right)}{1+\tan\left({\psi_1}\right)\tan\left({\varPsi_2}\right)}$$

which is the angle of intersection.

That's the difference formula for the tangent:
$$\tan(\alpha \pm \beta) = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}$$

So basically it says:
$$\phi = \psi_1 - \psi_2$$

So they use $\rho$ like that a lot.

In this case, we have solutions $\cos\theta=\cos 2\theta$ and when $\rho = 0$ too. Why doesn't $\cos\theta=\cos 2\theta$ cover all the points of intersection? Why do we also have to consider when $\rho=0$?

Any point with $\rho=0$ is the origin.
This is independent of the angle.

In polar coordinates we can write a point as $(\rho,\theta) = (4,\frac\pi 3)$, meaning it is the cartesian point $(x,y)=(2,2\sqrt 3)$.

Note that the points $(\rho,\theta)=(0,\frac\pi 3)$ and $(\rho,\theta)=(0,\frac\pi 6)$ are the same point.
They are simply the origin $(x,y)=(0,0)$. (Nerd)

 

[Dethrone]

That makes sense now. They're are both the origin point... (Rofl)...but disguised (Smoking)