Sketch the polar curve

[never_lose]

I'm trying to convert a particular polar curve to rectangular coordinates.

When I have something like r = 2cos(?) I can multiply r by both sides and get r[sup:2oylokpx]2[/sup:2oylokpx] = 2 * r cos(?) and just substitute in to get x[sup:2oylokpx]2[/sup:2oylokpx] + y[sup:2oylokpx]2[/sup:2oylokpx] = 2x.

I tried the same approach with r = 1 - sin(?), but I couldn't get rid of the r. How can I convert equations in this particular form to rectangular coordinates?

 

[tkhunny]

The same approach would be fine. You just gave up too soon. You should not expect lovely Cartesian forms.

\(\displaystyle r^{2} = r - r\cdot\sin(\theta)\)

\(\displaystyle x^{2}+y^{2} = \sqrt{x^{2}+y^{2}} - y\)

\(\displaystyle x^{2}+y^{2} + y = \sqrt{x^{2}+y^{2}}\)

\(\displaystyle x^{4}+y^{4} + y^{2} + 2x^{2}y^{2} + 2x^{2}y + 2y^{3} = {x^{2}+y^{2}\)

A little more can be done. I'll leave it to you. It is a worthwhile exercise.

 

[Deleted member 4993]

I'm trying to convert a particular polar curve to rectangular coordinates.

When I have something like r = 2cos(?) I can multiply r by both sides and get r[sup:1h2jn80a]2[/sup:1h2jn80a] = 2 * r cos(?) and just substitute in to get x[sup:1h2jn80a]2[/sup:1h2jn80a] + y[sup:1h2jn80a]2[/sup:1h2jn80a] = 2x.

I tried the same approach with r = 1 - sin(?), but I couldn't get rid of the r. How can I convert equations in this particular form to rectangular coordinates?

You can simply substitute:

\(\displaystyle r \ = \ \sqrt{x^2+y^2}\)

and

\(\displaystyle sin(\theta} \ = \ \frac{x}{\sqrt{x^2+y^2}}\)

and simplify as needed.