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Math Help - Polar to rectangular form.

  1. #1
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    Polar to rectangular form.

    I'm assuming this is a circle and will end up in x^2+y^2=r^2.

    r=-8cos0 (0 being Omega)
    Should I start by squaring each side knowing that r^2 will be in the rectangular form?? A little lost, not sure where to begin.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dashreeve View Post
    I'm assuming this is a circle and will end up in x^2+y^2=r^2.

    r=-8cos0 (0 being Omega)
    Should I start by squaring each side knowing that r^2 will be in the rectangular form?? A little lost, not sure where to begin.
    The thing that you need to do here is multiply both sides by r:

    r(r)=-8(r)\cos\varpi

    \implies r^2=-8r\cos\varpi

    Thus, in rectangular form, we get x^2+y^2=-8x

    \implies x^2+8x+y^2=0\implies x^2+8x+16+y^2=16\implies \color{red}\boxed{(x+4)^2+y^2=16}

    Does this make sense?

    --Chris
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  3. #3
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    That's where I was going wrong, by squaring each side. When you put the 16 in, is that like completing the square?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dashreeve View Post
    That's where I was going wrong, by squaring each side. When you put the 16 in, is that like completing the square?
    Yes, it appears as a result of completing the square. Does this seem to make more sense now? This polar stuff can get a little confusing.

    --Chris
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  5. #5
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    Yes it makes much more since now, thanks for the help.
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