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Math Help - Changing coordinated

  1. #1
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    Changing coordinated

    How do I determine the shape polar of the equations as:

    x-3y=0
    x^2+y^2+4x-2y-5=0
    .
    .
    .
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  2. #2
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    Quote Originally Posted by Apprentice123 View Post
    How do I determine the shape polar of the equations as:

    x-3y=0
    x^2+y^2+4x-2y-5=0
    .
    .
    .
    Substitute x = r \cos \theta and y = r \sin \theta and simplify where required.
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    thanks
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  4. #4
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    1) y^2+5x=0

    My solution:

    y=psen\theta => y^2=p^2sen^2\theta
    x=pcos\theta => 5x=5pcos\theta
    p^2+sen^2\theta+5pcos\theta=0

    But the answer is:
    psen^2\theta+5cos\theta=0

    2) xy=2

    My solution:

    (pcos\theta)(psen\theta)=2
    p^2cos(\theta).sen(\theta)=2

    The answer is:
    p^2sen2\theta=4

    3) x^2-4y-4=0

    My solution:

    (pcos\theta)^2-4psen\theta=4
    p(pcos^2\theta-4sen\theta)=4

    The answer is:
    p=\frac{2}{(1-sen\theta)}

    4) (x^2+y^2)^2-2a^2xy=0

    My solution:

    p^4=2a^2pcos(\theta).psen(\theta)

    The answer is:
    P^2=a^2sen2\theta
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  5. #5
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    Quote Originally Posted by Apprentice123 View Post
    1) y^2+5x=0

    My solution:

    y=psen\theta => y^2=p^2sen^2\theta
    x=pcos\theta => 5x=5pcos\theta
    p^2{\color{red}+}sen^2\theta+5pcos\theta=0

    Mr F says: Why the + (in red)?? It should be  {\color{red}\rho^2 \sin^2 \theta + 5 \rho \cos \theta = 0}.

    Then  {\color{red}\rho (\rho \sin^2 \theta + 5 \cos \theta ) = 0}. Therefore p = 0 or ......

    But the answer is:
    psen^2\theta+5cos\theta=0

    2) xy=2

    My solution:

    (pcos\theta)(psen\theta)=2
    p^2cos(\theta).sen(\theta)=2

    Mr F says: Do you know the double angle formula for {\color{red}\sin (2 \theta)} .....?

    The answer is:
    p^2sen2\theta=4

    [snip]

    4) (x^2+y^2)^2-2a^2xy=0

    My solution:

    p^4=2a^2pcos(\theta).psen(\theta)

    Mr F says: See my remarks for 1) and 2) .......

    The answer is:

    P^2=a^2sen2\theta
    ..
    Last edited by mr fantastic; September 2nd 2008 at 04:13 PM.
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  6. #6
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    Quote Originally Posted by Apprentice123 View Post
    [snip]
    3) x^2-4y-4=0

    My solution:

    (pcos\theta)^2-4psen\theta=4
    p(pcos^2\theta-4sen\theta)=4

    The answer is:
    p=\frac{2}{(1-sen\theta)}
    [snip]
    \rho^2 \cos^2 \theta - 4 \rho \sin \theta - 4 = 0

    \Rightarrow (\cos^2 \theta ) \rho^2 - (4 \sin \theta ) \rho - 4 = 0

    is a quadratic equation in \rho. Use the quadratic formula to get (after appropriate simplification) that \rho = \frac{2}{1 - \sin \theta} or \rho = \frac{-2}{1 + \sin \theta}.

    Note: \cos^2 \theta = 1 - \sin^2 \theta = (1 - \sin \theta) (1 + \sin \theta) will help in the simplifying .....


    You should convince yourself that the two solutions above for \rho give the same curve .....
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