Question

what is the polar graph of r=2cos3(theta) and a characteristic?

Answer 1

\[r =2cos(3\theta)\]
\[r^2 =x^2 +y^2 \]
\[x =rcos3\theta\]
\[\frac{x}{r} = cos3\theta\]
\[y =rsin3\theta\]
\[r = 2\frac{x}{r}\]
\[r^2 = 2x\]
\[x^2 +y^2 -2x = 0\]
Looks like a circle, It is circle

Answer 2

what a nice picture. not a circle though
http://www.wolframalpha.com/input/?i=r+%3D+2+cos%283+theta%29

Answer 3

I must have done something stupid. am I to consider 3theta and then break it into theta ?

Answer 4

or just 3theta =a then solve the whole function

Answer 5

this one is the circle
http://www.wolframalpha.com/input/?i=r%3D+2+cos%28theta%29

Answer 6

ah break into theta got it

then the thing gets complicated

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

Answer 7

\[x = r cos\theta \]
\[\frac{x}{r}=cos\theta\]
\[r = 8\frac{x^3}{r^3} -3\frac{x}{r}\]
\[r^4 = 8x^3 - 3xr^2\]

Answer 8

\[(x^2 +y^2)^2 = 8x^3 - 3x(x^2+y^2)\]

Answer 9

\[x^4 +y^4 +2x^2y^2 = 8x^3 -3x^3 -3xy^2\]
\[x^4 +y^4 +2x^2y^2 -5x^3 +3xy^2=0\]

Answer 10

something's wrong again

http://www.wolframalpha.com/input/?i=x^4+%2By^4%2B2x^2y^2-5x^3+%2B3xy^2%3D0

Answer 11

\[(x,y)=(rcos(\theta),rsin(\theta))\]
\[r^2=x^2+y^2\]
----------------------------
\[r=2\cos(3\theta)=2(\cos(2\theta+\theta))=2(\cos(2\theta)\cos(\theta)-\sin(2\theta)\sin(\theta))\]
\[r=2([\cos^2(\theta)-\sin^2(\theta)]\cos(\theta)-2\sin(\theta)\cos(\theta)\sin(\theta))\]
\[r=2([\frac{x^2}{r^2}-\frac{y^2}{r^2}]\frac{x}{r}-2\frac{y}{r}\frac{x}{r}\frac{y}{r})\]
\[r=2\frac{x^2-y^2}{r^3}x-4\frac{xy^2}{r^3}\]
\[r^4=2(x^3-y^2x)-4xy^2=>(x^2+y^2)^2=2x^3-6xy^2\]

Answer 12

\[x^4+2x^2y^2+y^4=2x^3-6xy^2\]
\[x^4+2x^2y^2+y^4-2x^3+6xy^2=0\]

Answer 13

Just a quick question: is the polar equation \[r=2cos^3(\theta)\] or is it\[r=2cos(3\theta)\] ??

Answer 14

I did a mistake before typo
yeah the equation myininaya got is correct


http://www.wolframalpha.com/input/?i=x^4+%2By^4%2B2x^2y^2-2x^3+%2B6xy^2%3D0

I never thought I could make such complicated graphs through polars : )