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Math Help - polar to cartesian

  1. #1
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    Talking polar to cartesian

    im having problems figuring out how my teacher came up with this solution.

    the problem is to convert the following polar to cartesian
    r=2-cos(2theta)

    we are in agreement for the following steps
    r=2-[(cos^2 (theta))-(sin^2(theta))] using the double angle identity

    then he has
    √(x^2 + y^2) = 2 - [(x/r)^2] + [(y/r)^2]
    where did the (x/r) and (y/r) come from and why are they squared??

    he goes on to further solve the problem as follows:
    √(x^2 + y^2) * (x^2 + y^2) = 2(x^2 + y^2) - (x^2) + (y^2)

    final step
    √(x^2 + y^2) * (x^2 + y^2) = (x^2) + 3(y^2)

    this guy is know to make some mistake ... OFTEN. but i dont want to write this problem off. can someone explain to me what is happening here and a better way to arrive at his answer .. or in the least tell me if the solution is correct.


    thanks. its exam review
    Last edited by osolage; December 9th 2009 at 12:05 PM.
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  2. #2
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    Quote Originally Posted by osolage View Post
    im having problems figuring out how my teacher came up with this solution.

    the problem is to convert the following polar to cartesian
    r=2-cos(2theta)

    we are in agreement for the following steps
    r=2-[(cos^2 (theta))-(sin^2(theta))] using the double angle identity

    then he has
    √(x^2 + y^2) = 2 - [(x/r)^2] + [(y/r)^2]
    where did the (x/r) and (y/r) come from and why are they squared??

    he goes on to further solve the problem as follows:
    √(x^2 + y^2) * (x^2 + y^2) = 2(x^2 + y^2) - (x^2) + (y^2)

    final step
    √(x^2 + y^2) * (x^2 + y^2) = (x^2) + 3(y^2)

    this guy is know to make some mistake ... OFTEN. but i dont want to write this problem off. can someone explain to me what is happening here and a better way to arrive at his answer .. or in the least tell me if the solution is correct.


    thanks. its exam review
    I don't really understand your question...

    Are you asking to change this equation

    r = 2 - \cos{(2\theta)}

    into Cartesians?


    If so, recall that r = \sqrt{x^2 + y^2}.

    Also, remember that \cos{(2\theta)} = \cos^2{\theta} - \sin^2{\theta}


    Now since x = r\cos{\theta} that means that \cos{\theta} = \frac{x}{r}

     = \frac{x}{\sqrt{x^2 + y^2}}.


    Also, y = r\sin{\theta}, which means that \sin{\theta} = \frac{y}{r}

     = \frac{y}{\sqrt{x^2 + y^2}}.


    So the equation becomes

    r = 2 - \cos{(2\theta)}

    r = 2 - (\cos^2{\theta} - \sin^2{\theta})

    r = 2 - \cos^2{\theta} + \sin^2{\theta}

    \sqrt{x^2 + y^2} = 2 - \left(\frac{x}{\sqrt{x^2 + y^2}}\right)^2 + \left(\frac{y}{\sqrt{x^2 + y^2}}\right)^2

    \sqrt{x^2 + y^2} = 2 - \frac{x^2}{x^2 + y^2} + \frac{y^2}{x^2 + y^2}

    \sqrt{x^2 + y^2} = 2 + \frac{y^2 - x^2}{x^2 + y^2}

    \sqrt{x^2 + y^2} = \frac{2x^2 + 2y^2 + y^2 - x^2}{x^2 + y^2}

    \sqrt{x^2 + y^2} = \frac{x^2 + 3y^2}{x^2 + y^2}...



    You may wish to simplify it further...
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