Polar to Cartesian conversion - how related?

[page13]

Homework Statement

Find the cartesian equation for the curve r=csctheta

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?

 

[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.

[edit] fixed two typos

 

[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?

 

[HallsofIvy]

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.

 

[LCKurtz]

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?

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