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Thread: [SOLVED] Conversion of Polar and Rectangular Equations

  1. #1
    Junior Member Nerdfighter's Avatar
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    [SOLVED] Conversion of Polar and Rectangular Equations

    I need to convert this rectangular equation into polar form:
    $\displaystyle x^4 + 2x^2y^2 + y^4 = 2xy$

    I know that $\displaystyle x = rcos\theta$ and $\displaystyle y = rsin\theta$. I've tried a little substitution but have not had a lot of success in simplifying it.


    I also need to convert this polar equation into rectangular form:
    $\displaystyle r^2 = \frac{36}{9cos^2\theta - 4sin^2\theta}$

    I have a suspicion that $\displaystyle r^2 = x^2 + y^2$ is a useful tool, but I am having trouble with the conversion still. I used it to create this simplified equation in rectangular form:

    $\displaystyle y = \frac{6}{3cos\theta - 2sin\theta} -x$ Is this right? I simply substituted $\displaystyle r^2$, subtracted $\displaystyle x^2$ and took the square root of everything. Is this correct?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Nerdfighter View Post
    I need to convert this rectangular equation into polar form:
    $\displaystyle x^4 + 2x^2y^2 + y^4 = 2xy$

    I know that $\displaystyle x = rcos\theta$ and $\displaystyle y = rsin\theta$. I've tried a little substitution but have not had a lot of success in simplifying it.


    I also need to convert this polar equation into rectangular form:
    $\displaystyle r^2 = \frac{36}{9cos^2\theta - 4sin^2\theta} $

    I have a suspicion that $\displaystyle r^2 = x^2 + y^2$ is a useful tool, but I am having trouble with the conversion still. I used it to create this simplified equation in rectangular form:

    $\displaystyle y = \frac{6}{3cos\theta - 2sin\theta} -x$ Is this right? I simply substituted $\displaystyle r^2$, subtracted $\displaystyle x^2$ and took the square root of everything. Is this correct?
    For the first one note that
    $\displaystyle x^4+2x^2y^2+y^4=(x^2+y^2)^2$

    Then use $\displaystyle x^2+y^2=r^2$

    For the next one clear the fractions to get

    $\displaystyle r^2(9cos^2\theta - 4sin^2\theta) = 36 \iff 9(r\cos(\theta))^2-4(r\sin(\theta))^2=36 $

    See if this can get you started
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  3. #3
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    Question 1:

    $\displaystyle x^4 + 2x^2y^2 + y^4 = 2xy$

    Make the substitutions you suggested

    $\displaystyle r^4cos^4(\theta) +2r^4cos^2(\theta)\sin^2(\theta) + r^4sin^4(\theta) = 2r^2\sin(\theta)\cos(\theta)$

    Divide by r^4

    $\displaystyle cos^4(\theta) + sin^4(\theta) + 2cos^2(\theta)\sin^2(\theta)= 2r^{-2}\sin(\theta)\cos(\theta)$


    You can factorise the left hand side

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

    But you know sin^2 + cos^2 = 1

    $\displaystyle 1^2 = 2r^{-2}\sin(\theta)\cos(\theta)$

    Tidy up to finish.


    editToo slow...again
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  4. #4
    Junior Member Nerdfighter's Avatar
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    Thank you for the tips and help everyone, this is really useful. I can't believe I didn't see that factoring from the first problem, it's so obvious now!
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  5. #5
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    I have the same questions.
    For the first conversion I ended up with:

    $\displaystyle r^3 = 2cosOsinO$
    Is this a valid polar equation, or do I need to simplify it further?

    For the second conversion my answer is:
    $\displaystyle 3x - 2y = 6 $ OR $\displaystyle y=2/3x-3$


    Do these look accurate?
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