# Thread: How to graph a lemniscate in polar coordinates?

1. ## How to graph a lemniscate in polar coordinates?

Hey guys,

The basic form of a lemniscate is $r^2=sin(theta)$ or $r^2=cos(theta)$
if taking root on both sides of the equation,than it will become
$r= sqrt(cost theta)$

now since square root only has positive values ,than how can a lemniscate be drawn if the value of cos theta becomes negative inside the square root??

2. Originally Posted by Maha211
Hey guys,

The basic form of a lemniscate is $r^2=sin(theta)$ or $r^2=cos(theta)$
if taking root on both sides of the equation,than it will become
$r= sqrt(cost theta)$

now since square root only has positive values ,than how can a lemniscate be drawn if the value of cos theta becomes negative inside the square root??
You should note that the basic form is $r^2 = a^2 cos 2 \theta$

This equation, as you mentioned, is only defined for certain angles.

$- \frac{ \pi }{4} \le \theta \le \frac{ \pi }{4}$ and $\frac{ 3 \pi }{4} \le \pi \le \frac{ 5 \pi }{4}$ this should be sufficient to draw the lemniscate.

Alternatively, you could turn back into x and y and get

$x = \frac{ acost } { 1 + sin^2 t }$

and

$y = \frac{ asintcost }{ 1 + sin^2 t }$

More can be found here: Lemniscate -- from Wolfram MathWorld

3. Hello, Maha211!

The basic form of a lemniscate is: . $r^2\:=\:\sin\theta\,\text{ or }\,r^2\:=\:\cos\theta$

If taking root on both sides of the equation, than it will become: . $r\:=\: \sqrt{\cos\theta}$ . . . . no

Now since square root only has positive values, then how can a lemniscate be drawn
if the value of $\cos\theta$ becomes negative inside the square root?

When we take the square root, we get: . $r \;=\:{\color{red}\pm}\sqrt{\cos\theta}$

You are right: . $\cos\theta \:\geq\:0$

If $\cos\theta$ is negative, $\tfrac{\pi}{2}<\theta <\tfrac{3\pi}{2}$, there are no points to plot.

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# graphing lemniscate

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