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Thread: How to graph a lemniscate in polar coordinates?

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

    Hey guys,

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

    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??
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Maha211 View Post
    Hey guys,

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

    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 $\displaystyle r^2 = a^2 cos 2 \theta $

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

    $\displaystyle - \frac{ \pi }{4} \le \theta \le \frac{ \pi }{4} $ and $\displaystyle \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

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

    and

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

    More can be found here: Lemniscate -- from Wolfram MathWorld
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  3. #3
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    Hello, Maha211!

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

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

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

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


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

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

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