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Math Help - How do i convert this into a rectangular equation

  1. #1
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    How do i convert this into a rectangular equation

    y = cos(2t), y = sin(2t);

    t is greater than or equal to -pie and less than or equal to pie.
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  2. #2
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    Quote Originally Posted by cityismine View Post
    y = cos(2t), y = sin(2t);

    t is greater than or equal to -pie and less than or equal to pie.
    y^2 = \cos^2 (2t).
    x^2 = \sin^2 (2t).

    Therefore x^2 + y^2 = 1.

    Domain is [-1, 1] and range is [-1, 1] so it's the entire circle.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Just so you know, given

    x=a\cos(n\theta)

    and y=a\sin(n\theta)

    You will always get a circle of radius a

    x^2+y^2=a^2\cos^2(n\theta)+a^2\sin^2(n\theta)=a^2

    and if you have

    x=a\cos(n\theta)

    and

    y=b\sin(n\theta)

    You will get an ellipse

    \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{a^2\cos^2(n\  theta)}{a^2}+\frac{b^2\sin^2(n\theta)}{b^2}=1
    Last edited by Mathstud28; June 25th 2008 at 10:31 AM. Reason: I said "r" instead of a, for the radius
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  4. #4
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    -pie and less than or equal to pie.
    Come on...this is math, not a bakery.
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  5. #5
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    The answer in the back of the book is: y=(1/2)x^2-1

    Is the answer in the book wrong?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cityismine View Post
    The answer in the back of the book is: y=(1/2)x^2-1

    Is the answer in the book wrong?
    Yes

    For we have that our equation is

    x^2+y^2=1\Rightarrow{y=\pm\sqrt{{\color{red}1-x^2}}}
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  7. #7
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    Quote Originally Posted by cityismine View Post
    The answer in the back of the book is: y=(1/2)x^2-1

    Is the answer in the book wrong?
    If the parametric equations were actually x = \cos t and  y = \cos (2t) then the cartesian equation would be y = 2x^2 - 1 ........

    If the parametric equations were actually x = 2 \sin t and  y = \cos (2t) then the cartesian equation would be y = 1 - \frac{x^2}{2} ........

    Are the equations you posted correct?
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  8. #8
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    Yes, I posted them correctly, I just double checked. I guess it's a typo in the book.
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