Question

Convert polynomial to rectangular coordinates (use variables x and y as needed.)

r=8+cos(theta)

Answer 1

@Astrophysics ? :)

Answer 2

@zepdrix guess who!

Answer 3

@jim_thompson5910

Answer 4

so I multiplied both sides by r to ge
r^2=8r+rcos(theta)

Answer 5

and then
(r^2-rcos(theta)^2=8r^2

Answer 6

so
\[(x^2+y^2-x)^2=(8r)^2\]

Answer 7

I see how you multiply both sides by r, but I don't see how you got to step 2

Answer 8

I just moved rcos(theta) to the left hand side

Answer 9

but you have 8r in step 1, then you have 8r^2 in step 2

Answer 10

\[r^2=8r+rcos \theta\]
\[(r^2-rcos \theta) = 8r\]
\[(r^2-rcos \theta)^2=(8r)^2\]

Answer 11

I see. So in step 2 you meant to say (8r)^2 and not 8r^2

Answer 12

Yeah, sorry about that :)

Answer 13

that's ok

Answer 14

\[(x^2+y^2-x)^2=(8r)^2\] em..I'm not sure where to go from here

Answer 15

well r = sqrt(x^2 + y^2)

so r^2 = x^2 +y^2

(8r)^2 = 64r^2 = 64*(x^2 +y^2)

Answer 16

What's strange is that I'm getting a circle but there are going to be x^4 and y^4 terms

Answer 17

hm..I don't know o.0

Answer 18

Well but
(x^2+y^2-x)^2=64(x^2+y^2)
would be the final thing yeah? I'm not supposed to graph it, phew

Answer 19

yes that works. We can't solve for y because of the y^2 and y^4 terms

Answer 20

so it's best to leave it in that implicit form

Answer 21

Yay! it worked, fabulous :)
I had been putting 16 instead of 64 *facepalm. Thank youu

Answer 22

you're welcome