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Math Help - need help to sketch the curve

  1. #1
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    need help to sketch the curve

    would you sketch the curve with the polar equation r = 2cos(2theta), and express the equation in rectangular coordinates, that is what must be represented.
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  2. #2
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    Quote Originally Posted by rcmango View Post
    would you sketch the curve with the polar equation r = 2cos(2theta), and express the equation in rectangular coordinates, that is what must be represented.
    I've attached the graph of the curve.

    As you easily can see you'll need 4 equations to describe this curve by the usual x,y-coordinates.
    Attached Thumbnails Attached Thumbnails need help to sketch the curve-kleeblatt_polar.gif  
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  3. #3
    Super Member wingless's Avatar
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    r = 2 \cos2\theta


    r^2 = x^2 + y^2
    ...for r>0~,~~r = + \sqrt{x^2+y^2}
    ...for r<0~,~~r = - \sqrt{x^2+y^2}


    \tan\theta = \frac{y}{x}
    ...for -\frac{\pi}{2}<\theta<\frac{\pi}{2}~,~~\theta = \arctan(\frac{y}{x})
    ...for \frac{\pi}{2}<\theta<\frac{3\pi}{2}~,~~\theta = \arctan(\frac{y}{x})+\pi


    Well, if we look carefully at \theta..
    \theta = \arctan(\frac{y}{x})
    \theta = \arctan(\frac{y}{x})+\pi
    r = 2 \cos(2\theta)
    r = 2 \cos(2(\arctan(\frac{y}{x})+\pi))
    r = 2 \cos(2\arctan(\frac{y}{x})+2\pi))
    which is equal to r = 2 \cos2(\arctan(\frac{y}{x}))

    That means, \theta = \arctan(\frac{y}{x}) and \theta = \arctan(\frac{y}{x})+\pi are the same. Then, using one of them is enough.



    So, we have
    r = 2 \cos2\theta
    ...for r>0~,~~r = + \sqrt{x^2+y^2}
    ...for r<0~,~~r = - \sqrt{x^2+y^2}
    \theta = \arctan(\frac{y}{x})

    Placing them will give us two functions:
    \sqrt{x^2+y^2} = 2\cos(2\arctan(\frac{y}{x}))
    -\sqrt{x^2+y^2} = 2\cos(2\arctan(\frac{y}{x}))





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