Express given equation in polar coordinates

[hank]

x^2 + y^2 + 6y = 0

The answer is r = -6sin(theta)

Ok, here's what I am trying to do:

x^2 + y^2 = -6y

Then I reasoned that y = rsin(theta)

x^2 + y^2 = -6(rsin(theta))

That gets me in the ballpark, but I need to know how to justify r being one so I can get the right answer.

Am I even going about this the right way?

Originally I thought about completing the square, but that left me with r = 3 which was wrong as well.

Please help.

 

[galactus]

Divide through by r.

 

[stapel]

Try this:

. . . . .x² + y² + 6y = 0

. . . . .x² + y² + 6y + 9 = 9

. . . . .x² + (y + 3)² = 3²

...so you know that this is a circle of radius r = 3 and center (h, k) = (0, -3). (Why did you think this was wrong?)

You know that x = r cos(@), y = r sin(@), and r² = x² + y², so:

. . . . .x² + y² + 6y = 0

. . . . .r² + 6y = 0

. . . . .r² = -6y

. . . . .r² = -6r sin(@)

Since you know that r isn't zero, you know you can divide through.

Eliz.

 

[hank]

Ahh...ok.

I didn't realize that I needed both parts to solve the problem.

It makes sense now, tyvm.