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Math Help - Where do these two polar equations meet

  1. #1
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    Where do these two polar equations meet

    I recently came across this problem and would like some clarification as to why one of the answers to theta cannot be \frac{\pi }{2}

    The problem asks to sketch two polar curves and find where they intersect, and then present those points in rectangular coordinates. I can easily see from the graph that the points are (1,0) and (-1,0). The two polar functions are,

    r= \cos ^{2}\theta and r = -1.

    I set them equal to each other

    \cos ^{2}\theta= 1

    \frac{1}{2}\left ( 1+\cos 2\theta  \right )= 1

    1+\cos 2\theta = 2

    \cos 2\theta = 1

    So theta should be... 0,\pi,-\pi,\frac{-\pi}{2},\frac{\pi}{2}

    Yet the graph tells me only pi and -pi can be the answer. I'm confused...
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  2. #2
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    Quote Originally Posted by Jukodan View Post
    I recently came across this problem and would like some clarification as to why one of the answers to theta cannot be \frac{\pi }{2}

    The problem asks to sketch two polar curves and find where they intersect, and then present those points in rectangular coordinates. I can easily see from the graph that the points are (1,0) and (-1,0). The two polar functions are,

    r= \cos ^{2}\theta and r = -1.

    I set them equal to each other

    \cos ^{2}\theta= 1


    Ok, is it r = 1 or r = -1??


    \frac{1}{2}\left ( 1+\cos 2\theta \right )= 1

    1+\cos 2\theta = 2

    \cos 2\theta = 1

    So theta should be... 0,\pi,-\pi,\frac{-\pi}{2},\frac{\pi}{2}

    This is wrong: \cos 2\theta=1\Longrightarrow 2\theta=2k\pi\,,\,\,k\in\mathbb{Z} \Longrightarrow \theta = k\pi = 0,\,\pm \pi,\,\pm 2\pi,\ldots


    Yet the graph tells me only pi and -pi can be the answer. I'm confused...

    Easier: \cos^2\theta=1\Longrightarrow \cos\theta=\pm 1 \Longrightarrow \theta=k\pi ...etc

    Tonio
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  3. #3
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    Quote Originally Posted by Jukodan View Post
    I recently came across this problem and would like some clarification as to why one of the answers to theta cannot be \frac{\pi }{2}

    The problem asks to sketch two polar curves and find where they intersect, and then present those points in rectangular coordinates. I can easily see from the graph that the points are (1,0) and (-1,0). The two polar functions are,

    r= \cos ^{2}\theta and r = -1.

    I set them equal to each other

    \cos ^{2}\theta= 1

    \frac{1}{2}\left ( 1+\cos 2\theta  \right )= 1

    1+\cos 2\theta = 2

    \cos 2\theta = 1

    So theta should be... 0,\pi,-\pi,\frac{-\pi}{2},\frac{\pi}{2}

    Yet the graph tells me only pi and -pi can be the answer. I'm confused...
    i am saying this very empirically. perhaps u should consider only the principal values because
    -1<cos(theta)<1
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  4. #4
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    Thank you gentlemen, it seems I made this problem harder for myself than it really is. Much much thanks.

    And also yes R = -1 but since it forms a circle I figured the radius is simply 1, hence why I set cos2theta equal to 1
    Last edited by Jukodan; January 24th 2010 at 04:27 AM.
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  5. #5
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    Quote Originally Posted by Jukodan View Post
    I recently came across this problem and would like some clarification as to why one of the answers to theta cannot be \frac{\pi }{2}

    The problem asks to sketch two polar curves and find where they intersect, and then present those points in rectangular coordinates. I can easily see from the graph that the points are (1,0) and (-1,0). The two polar functions are,

    r= \cos ^{2}\theta and r = -1.

    I set them equal to each other

    \cos ^{2}\theta= 1

    \frac{1}{2}\left ( 1+\cos 2\theta \right )= 1

    1+\cos 2\theta = 2

    \cos 2\theta = 1

    So theta should be... 0,\pi,-\pi,\frac{-\pi}{2},\frac{\pi}{2}

    Yet the graph tells me only pi and -pi can be the answer. I'm confused...
    Thread of related interest: http://www.mathhelpforum.com/math-he...ar-graphs.html
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