# Math Help - A couple things: Polar coordinate problems

1. ## A couple things: Polar coordinate problems

Hello everyone:

I'm reviewing for a test, and I got some questions regarding polar coordinates:

Question one:

Show that the point $(2, 3\pi/4)$ lies on the curve $r = 2 sin2\theta$

Specifically, when you simply "plug in" the $r$ and $\theta$, it gives you a false equality. I know that there are multiple ways to express the point $(2, 3\pi/4)$, and that apparently you must replace this point with an equivalent point in order to make the equality work. What exactly must I do to make this work?

Question two:

Find the intersections of this pair of curves:
$r = 1 + cos\theta$ and $r = 1 - cos\theta$

Thanks!

2. Originally Posted by Skinner
Hello everyone:

I'm reviewing for a test, and I got some questions regarding polar coordinates:

Question one:

Show that the point $(2, 3\pi/4)$ lies on the curve $r = 2 sin2\theta$

Specifically, when you simply "plug in" the $r$ and $\theta$, it gives you a false equality. I know that there are multiple ways to express the point $(2, 3\pi/4)$, and that apparently you must replace this point with an equivalent point in order to make the equality work. What exactly must I do to make this work?

Question two:

Find the intersections of this pair of curves:
$r = 1 + cos\theta$ and $r = 1 - cos\theta$

Thanks!
Q1 The point specified by [2, 3pi/4] is also specified by [2, -5pi/4].

Q2 You should first draw both curves.

Algebra tells you that the curves intersect at [1, pi/2] and [1, 3pi/2], that is, at (0, 1) and (0, -1):

$1 + \cos \theta = 1 - \cos \theta \Rightarrow \cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2}, \, \frac{3 \pi}{2}$.

But drawing both curves will tell you that the curves also intersect at the origin. The reason algebra doesn't tell you this is that this intersection point has polar coordinates [0, pi] for the first curve and [0, 0] for the second curve .... that is, this point has no simultaneous solution for $\theta$.