Problem 10.4.40

I made a big mistake in my first section and a small mistake in my second section.

We could solve problems like number 40 as follows.

Problem: When do two curves $r_1(\theta)$ and $r_2(\theta)$ intersect?

They will intersect when $(r_1(x),x)=(r_2(y),y)$, where the pair $(r,\theta)$ is the polar representation of the point $(r\cos\theta,r\sin\theta)$ in Cartesian coordinates.

This implies either
1) $r_1(x)=r_2(y)$ and that $x=y+2\pi k$, for some integer $k$.
2) $r_1(x)=-r_2(y)$ and that $x=y+\pi+2\pi k$, for some integer $k$.
or
3) $r_1(x)=0$ for some $x$ and $r_2(y)=0$ for some $y$.

Solution:
1) implies $r_1(y+2\pi k)=r_2 (y)$. Solve for $y$. Then solve for $x$. Check answers.
2) implies $r_1(y+\pi+2\pi k)=r_2 (y)$. Solve for $y$. Then solve for $x$. Check answers.
3) is easy to determine.

Example:
Let's do problem 40.

$r_1(\theta)=\cos(3\theta)$
$r_2(\theta)=\sin(3\theta)$

At first, we note that they have period $\pi$. We can check this without graphing, but I'll omit showing that. Because they both have period $\pi$, we can ignore the second case. I'll also omit explaining that. Feel free to ask why below, but try to give it some thought on your own first.

Following the steps above we get $\cos(3(y+2\pi k))=\sin(3y)$ which simplifies to $\cos(3y)=\sin(3y)$.

For the first equation we get that $3y=\pi/4 +\pi k $ for any integer $k$. We should further take advantage of the fact that the curve $r_2$ has period $\pi$. Thus, we only need values of $y$ between $0$ and $\pi$. This gives $\pi/12$, $5\pi/12$, and $9\pi/12$.

Now $x=y+2\pi k$, but the period of $r_1$ is also $\pi$ and we only need values of $x$ between $0$ and $\pi$. Thus, in this case, and probably in most cases you'll encounter, $x=y$.

We should plug these three values in and check they are correct. Then because both curves pass through the origin, that is also a point of intersection.