Question

How do I convert r=2-cos(3theta) to a cartesian equation?

Answer 1

@agent0smith

Answer 2

Also, the question was originally given in parametric form of (r,theta)=(2-cos(t3pi),tpi) Did I eliminate the parameters correctly?

Answer 3

@sourwing

Answer 4

looks good,
multiply "r" both sides

Answer 5

and use the trig identiy : cos(3x) = 4cos^3x - 3cosx

Answer 6

\( r=2-\cos(3\theta) \)

\(r^2 = 2r - r[4\cos^3\theta - 3\cos \theta]\)

\(x^2+y^2 = 2\sqrt{x^2+y^2} - r[4\frac{x^3}{r^3} - 3\frac{x}{r}]\)

\(x^2+y^2 = 2\sqrt{x^2+y^2} - 4\frac{x^3}{x^2+y^2} - 3x\)

Answer 7

Wow, that looks complicated. You're sure I got rid of the parameters correctly?

Answer 8

looks okay to me
good to have checked it wid multiple eyes @agent0smith

Answer 9

Ok, well thank you so much!

Answer 10

\[\Large r=2- 4\cos^3 \theta - 3\cos \theta\]
there's no simple way to do it... multiply both sides by r^3 and i'm not gonna explain all steps
\[\Large (r^2)^2=2r^2r- 4 (r\cos \theta)^3 - 3r^2 r\cos \theta\]
\[\Large (x^2 + y^2)^2=2 (x^2+y^2) \sqrt{ x^2+y^2 } - 4 x^3 - 3x(x^2+y^2) \]

Answer 11

\[\Large (x^2 + y^2)^2=2 (x^2+y^2)^{3/2} - 4 x^3 - 3x(x^2+y^2) \]

Answer 12

Is there anyway to arrive at an equation in cartesian form just using the given parametric equation?

Answer 13

Maybe. Idk tho, it's weird lookin http://www.wolframalpha.com/input/?i=convert+to+rectangular+r%3D2-cos%283theta%29+

Answer 14

Yup certainly is. Well thanks everyone for your help.

Answer 15

And based on how it looks in rectangular form, either mine or @ganeshie8's i doubt there's a simple representation (hence the usefulness of polar or parametric form)

Answer 16

Oh and @ganeshie8's is the same as mine if you multiply both sides by (x^2+y^2)