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Math Help - Polar equations, limited information given

  1. #1
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    Polar equations, limited information given

    Hi, for polar equations, you usually need r and theta, in this case, I am only given one bit of information, how do I do it?

    1) Convert these polar equations to Cartesian:
    a] r = 4
    b] r = 2cos\theta

    2) Convert from Cartesian to polar:
    a] x^2 + (y - 1)^2 = 1
    b] y = x + 1

    Thanks for any help
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  2. #2
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    There are 2 equations that define the relationship between cartesian and polar coordinates:

    x=r\cos{\theta}
    y=r\sin{\theta}

    or, equivalently
    x^2 + y^2 = r^2
    \theta = \tan{\frac{y}{x}}


    To do your conversions, just make the substitutions, eg:

    1a
    r=4
    r^2=16
    x^2 + y^2=16
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    Okay, I've got that, but what about the others. For example, 2b) surely can't be:

    y = r sin\theta (formula)
    y = x + 1 (given)

    x + 1 = r sin\theta

    \therefore r = \frac{x + 1}{sin\theta}

    ...could it?
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  4. #4
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    Quote Originally Posted by SpringFan25 View Post
    There are 2 equations that define the relationship between cartesian and polar coordinates:

    x=r\cos{\theta}
    y=r\sin{\theta}

    or, equivalently
    x^2 + y^2 = r^2
    \theta = \tan{\frac{y}{x}}
    Actually, it's

    \tan{\theta} = \frac{y}{x}

    or \theta = \arctan{\frac{y}{x}}.

    It's also important to take into account which quadrant you are working in.
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  5. #5
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    *ahem* yes i rushed it
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    Quote Originally Posted by BG5965 View Post
    Okay, I've got that, but what about the others. For example, 2b) surely can't be:

    y = r sin\theta (formula)
    y = x + 1 (given)

    x + 1 = r sin\theta

    \therefore r = \frac{x + 1}{sin\theta}

    ...could it?
    You need to substitute for the x as well.

    so, to start you off:
    y = x + 1
    r\sin{\theta} = r\cos{\theta} + 1
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  7. #7
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    Is it:
     r = \frac{1}{sin\theta - cos\theta} ?
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  8. #8
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    i think so, but i fear the wrath of prove it if im wrong
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  9. #9
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    Quote Originally Posted by SpringFan25 View Post
    i think so, but i fear the wrath of prove it if im wrong
    Yes it's correct.

    And yes, I'm always watching
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