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Math Help - Polar Equations

  1. #1
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    Polar Equations

    Directions are simple either convert from polar to Cartesian or vice versa...

    1. r\sin(\theta - \frac{\pi}{4}) = 2

    Then....

    \sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{2}}{2}(sin\theta - cos\theta)

    x - y = 2\sqrt{2}

    y = x - 2\sqrt{2}

    Now, my book has y = x + 2\sqrt{2} and I just can't see how they got it, maybe I'm skipping something or making a mistake, but I can't see it as of now.

    And...

    2. r^2\sin2\theta = 2

    I got it to...

    r = \frac{1}{xy}

    But that doesn't seem right, so there's probably something obvious I'm missing.

    Also, as intuitive as it seems

    \theta = \frac{\pi}{2} and \theta = \frac{3\pi}{2} are both x = 0

    \theta = \pi and \theta = 0 are both y = 0

    Thanks for the help.
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  2. #2
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    Quote Originally Posted by JSB1917 View Post
    Directions are simple either convert from polar to Cartesian or vice versa...

    1. r\sin(\theta - \frac{\pi}{4}) = 2

    Then....

    \sin(\theta - \frac{\pi}{4}) = \frac{\sqrt{2}}{2}(sin\theta - cos\theta)

    x - y = 2\sqrt{2}

    y = x - 2\sqrt{2}

    Now, my book has y = x + 2\sqrt{2} and I just can't see how they got it, maybe I'm skipping something or making a mistake, but I can't see it as of now.

    [snip]
    r \frac{\sqrt{2}}{2}(\sin\theta - \cos\theta) = 2 \Rightarrow r (\sin\theta - \cos\theta) = \frac{4}{\sqrt{2}} \Rightarrow y - x = 2 \sqrt{2}.
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  3. #3
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    Quote Originally Posted by JSB1917 View Post
    Directions are simple either convert from polar to Cartesian or vice versa...

    [snip]
    2.

    I got it to...



    But that doesn't seem right, so there's probably something obvious I'm missing.

    Also, as intuitive as it seems

    \theta = \frac{\pi}{2} and \theta = \frac{3\pi}{2} are both x = 0

    \theta = \pi and \theta = 0 are both y = 0

    Thanks for the help.
    r^2 2 \sin \theta \cos \theta = 2 \Rightarrow 2xy = 2.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    r \frac{\sqrt{2}}{2}(\sin\theta - \cos\theta) = 2 \Rightarrow r (\sin\theta - \cos\theta) = \frac{4}{\sqrt{2}} \Rightarrow y - x = 2 \sqrt{2}.
    LOL. I see now. I guess it helps if I don't mix up sin and cos.

    Quote Originally Posted by mr fantastic View Post
    r^2 2 \sin \theta \cos \theta = 2 \Rightarrow 2xy = 2.
    Oh, ok. So it's more like thinking one r for each sin and cos in 2sincos. Oh yeah, because if it was only one r than I would still be left with a cos or sin.
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