degreeplus
New member
- Joined
- Oct 7, 2006
- Messages
- 24
All I need help with is visualizing r^2=4cso2thadum
Agh need help on visualizing r^2=4cso2thadum
- Thread starter degreeplus
- Start date
Re: Agh need help on polar equations
Hello, degreeplus!
That Greek letter is "theta".
. . I'll take a guess at what you meant . . .
\(\displaystyle (1)\;r^2\:=\:4\cos^2\theta\)
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: \(\displaystyle \,(2,\,0),\
2,\,\pi)\)
Hello, degreeplus!
That Greek letter is "theta".
. . I'll take a guess at what you meant . . .
\(\displaystyle (1)\;r^2\:=\:4\cos^2\theta\)
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: \(\displaystyle \,(2,\,0),\
2,\,\pi)\)