# Polar to Cartesian conversion - how related?

1. Jan 25, 2010

### page13

1. The problem statement, all variables and given/known data
Find the cartesian equation for the curve r=csctheta

3. The attempt at a solution
I understand how to get the answer, by changing it to r=1/sin, and then rsin=1, and then since y=rsin, then y=1.

What I'm not understanding is the relationship between y=1 an r=csc. I thought that when you convert to a cartesian equation, it's supposed to either look the same when graphed or be able to be used to plot the r and theta coordinates onto a polar plane. But y=1 translates to r=1 on a polar plane, and r=csc is nothing like that?? What am I not understanding here?

2. Jan 25, 2010

### LCKurtz

Draw a cartesian coordinate system and draw the horizontal line y = 1. Then draw a line from the origin to the point (3,1) on your horizontal line and make that a radius vector r. Label the polar coordinate angle $\theta$ in the proper position. Drop a line from your point (3,1) to the x axis at (3,0) forming a little triangle with r and $\theta$ as part of it. The sides of that triangle are r, 3, and 1 units long. Notice that for that triangle that

$$\csc\theta = \frac r 1$$

Then notice if you do the same thing using (x,1) instead of (3,1) you will still get

$$\csc\theta = \frac r 1$$

So that equation $r =\csc\theta$ literally gives the same graph as y = 1.

 fixed two typos

Last edited: Jan 26, 2010
3. Jan 26, 2010

### page13

Excellent! That helps a lot. But the last bit about using (x,1) - not sure how you're getting csc=r/x. Wouldn't it still be r/1?

4. Jan 26, 2010

### HallsofIvy

Staff Emeritus
Yes, that was probably a typo. It is precisely because $csc(\theta)= r/1$ that [/itex]r= (1)csc(\theta)[/itex].

"r= 1" is NOT "y= 1". It's graph is the circle with center at the origin and radius 1.

5. Jan 26, 2010

### LCKurtz

Yes. That was a typo as Halls surmised. There was another one too. I'll fix both.