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Thread: Converting Polar to Cartesian Equations

  1. #1
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    Converting Polar to Cartesian Equations

    Hi there!

    Im a little stuck on one of my questions and was hoping for a little bit of help.. So first, the question

    Convert from polar to cartesian form, using double angle formula:

    $\displaystyle r=\sqrt{1+cos2\theta}$

    Ok, so things that I know:

    $\displaystyle r^2=\sqrt{x^2+y^2}$
    $\displaystyle x = r cos\theta$
    $\displaystyle y = r sin\theta$

    The double angle formula which is reffered to, i believe is
    $\displaystyle cos2\theta = cos^2 \theta sin^2 \theta$
    But i could be wrong

    The furthest i have got to in the problem so far is

    $\displaystyle r^2 = 1+cos^2 \theta sin^2 \theta$

    And now im greatful if anyone is able to help me out please?

    Thanks in advance!
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  2. #2
    Moo
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    Hello,

    $\displaystyle \begin{aligned} \cos 2x &=\cos^2x-\sin^2x \\
    &=1-2\sin^2x \\
    &=2\cos^2x-1 \end{aligned}$

    this is the double angle formula for cos
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  3. #3
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    Ah yeah, thanks.. I knew that but forgot the - sign in my original post.

    I thought that the cos identity i chose was most appropriate as i could then relate x and y once i had moved things around further.

    Still not sure on how things are to be done?
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  4. #4
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by farso View Post
    Still not sure on how things are to be done?
    $\displaystyle r = \sqrt{1 + \cos2\theta}$

    $\displaystyle \Rightarrow r = \sqrt{1 + \left(2\cos^2\theta - 1\right)}$

    $\displaystyle \Rightarrow r = \sqrt{2\cos^2\theta} = \sqrt2\sqrt{\cos^2\theta} = \sqrt2\left\lvert\cos\theta\right\rvert$

    $\displaystyle \Rightarrow r^2 = r\sqrt2\,\lvert\cos\theta\rvert\qquad\qquad\color{ red}\text{Multiplying both sides by }r$

    $\displaystyle \Rightarrow r^2 = \sqrt2\,\lvert r\cos\theta\rvert\qquad\qquad\color{red}\text{Sinc e }r\text{ is nonnegative, }r\,\lvert\cos\theta\rvert = \lvert r\rvert\lvert\cos\theta\rvert = \lvert r\cos\theta\rvert$

    $\displaystyle \Rightarrow x^2 + y^2 = \sqrt2\,\lvert x\rvert\qquad\qquad\color{red}\text{Substitution}$

    $\displaystyle \Rightarrow y^2 = \sqrt2\,\lvert x\rvert - x^2$
    Last edited by Reckoner; Jun 10th 2008 at 07:22 PM. Reason: Fixed spacing on absolute value bars; I don't like ugly-looking LaTeX!
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  5. #5
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    Quote Originally Posted by Reckoner View Post
    $\displaystyle r = \sqrt{1 + \cos2\theta}$

    $\displaystyle \Rightarrow r = \sqrt{1 + \left(2\cos^2\theta - 1\right)}$

    $\displaystyle \Rightarrow r = \sqrt{2\cos^2\theta} = \sqrt2\sqrt{\cos^2\theta} = \sqrt2\left\lvert\cos\theta\right\rvert$

    $\displaystyle \Rightarrow r^2 = r\sqrt2|\cos\theta|\qquad\qquad\color{red}\text{Mu ltiplying both sides by }r$

    $\displaystyle \Rightarrow r^2 = \sqrt2|r\cos\theta|\qquad\qquad\color{red}\text{Si nce }r\text{ is nonnegative, }r|\cos\theta| = |r||\cos\theta| = |r\cos\theta|$

    $\displaystyle \Rightarrow x^2 + y^2 = \sqrt2|x|\qquad\qquad\color{red}\text{Substitution }$

    $\displaystyle \Rightarrow y^2 = \sqrt2|x| - x^2$

    That seems like a really good explanation, however, I thought $\displaystyle r^2 =\sqrt{x^2+y^2}$ Am I not correct in this assumption?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by farso View Post
    That seems like a really good explanation, however, I thought $\displaystyle r^2 =\sqrt{x^2+y^2}$ Am I not correct in this assumption?
    No it is $\displaystyle r^2 = x^2 + y^2$. This is easy to remember since it is merely the Pythagorean Theorem. r is the hypotenuse and x and y are the legs.

    -Dan
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  7. #7
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by farso View Post
    That seems like a really good explanation, however, I thought $\displaystyle r^2 =\sqrt{x^2+y^2}$ Am I not correct in this assumption?
    As topsquark explained, you may use the Pythagorean theorem to get $\displaystyle r^2 = x^2 + y^2$ (you can also use this triangle to see where $\displaystyle \tan\theta = \frac yx$ comes from):
    Attached Thumbnails Attached Thumbnails Converting Polar to Cartesian Equations-polar_coords.png  
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