Agh need help on visualizing r^2=4cso2thadum

[degreeplus]

All I need help with is visualizing r^2=4cso2thadum

 

[galactus]

What is 'thadum'?. Surely you don't mean 'theta'?. :roll:

 

[soroban]

Re: Agh need help on polar equations

Hello, degreeplus!

That Greek letter is "theta".
. . I'll take a guess at what you meant . . .

\(\displaystyle \begin{array}{cc}(1)\; & r^2\:=\:4\cos^2\theta \\ \\ \\ \\ & \text{or} \\ \\ \\ \\
(2)\; & r^2\:=\:4\cos2\theta\end{array}\)

$\displaystyle (1)\;r^2\:=\:4\cos^2\theta$

We have: $\displaystyle \:r\:=\:\pm2\cos\theta$

$\displaystyle r\:=\:2\cos\theta$ is a circle with radius 1
. . Its center is at $\displaystyle (1,\,0)$

$\displaystyle r\:=\:-2\cos\theta$ is a circle with radius 1
. . Its center is at $\displaystyle (1,\,\pi)$

We have two unit circles, horizontally tangent at the origin.


$\displaystyle (2)\;r^2 \:=\:4\cos2\theta$

This is a two-leaf rose curve.
. . The ends of its 'petals' are: \(\displaystyle \,(2,\,0),\2,\,\pi)\)