Thread: Converting Polar to Rectangular Equations

Was curious if anyone new how to find an equivalant equation in cartesian coordinates for the problem r=2sin@+2cos@ I have to turn in my trigonometry final tomorrow and am having problems with this one. I also am not sure how to make a theta so @ is theta. thank you...

Originally Posted by Thalstead

Was curious if anyone new how to find an equivalant equation in cartesian coordinates for the problem r=2sin@+2cos@ I have to turn in my trigonometry final tomorrow and am having problems with this one. I also am not sure how to make a theta so @ is theta. thank you...

recall that , also recall that, and

so, start with

multiply through by ,



Now what?

Originally Posted by Jhevon

recall that , also recall that, and

so, start with

multiply through by ,



Now what?

The only next step that I can think would be to then have r^2=2y + 2x and possibly change the r^2 to x^2 + y^2

Originally Posted by Thalstead

The only next step that I can think would be to then have r^2=2y + 2x and possibly change the r^2 to x^2 + y^2

yes. we want to get rid of the r's and the theta's and get x's and y's, so that's exactly what we do. so we get x^2 + y^2 = 2x + 2y. but what kind of curve is that?

I'm not really sure.... I was looking through the chapters in the book and didn't find anything similar. It wouldn't be a polar graph because we replaced the r's and theta's

Originally Posted by Thalstead

I'm not really sure.... I was looking through the chapters in the book and didn't find anything similar. It wouldn't be a polar graph because we replaced the r's and theta's

of course it's not a polar graph anymore. it is a circle. now that you know that, could you say, find it's center and radius?

Oh I understand now! yes I know how to work with circles! thank you so much for you're help. I really appreciate it =)