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Math Help - Polar to Rectangular and Vice Versa

  1. #1
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    Polar to Rectangular and Vice Versa

    Well I've figured out how to do most of my questions I need to get done, but there are still a few that I cannot quite figure out.

    r = 1 + 18cosѲ
    r^2 = r + 8rcosѲ
    x^2 + y^1 = r + 8y
    This is where I get stumped.


    8x^2 + 8y^2 - 88y = 0
    x^2 + Y^2 - 11y = 0
    r^2 - 11cosѲ = 0
    r(r - 11cosѲ) = 0
    r=11cosѲ

    Not sure if I did this one correctly.
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  2. #2
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    Hello, Crell!

    r \:=\: 1 + 18\cos\theta
    r^2 \:=\: r + 8r\cos\theta
    x^2 + y^2 \:=\: r + 8y
    This is where I get stumped.
    We must eliminate that last r.

    We have: . x^2+y^2-18x \:=\:r

    Square: . (x^2+y^2 - 18x)^2 \:=\:r^2

    Therefore: . (x^2+y^2 - 18x)^2 \:=\:x^2+y^2

    . . I'd leave it like that.


    8x^2 + 8y^2 - 88y \:=\: 0
    x^2 + y^2 - 11y \:=\: 0
    r^2 - 11r{\color{red}\cos}\theta \:=\: 0 ?
    No . . . y \:=\:r\sin\theta

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, Crell!

    We must eliminate that last r.

    We have: . x^2+y^2-18x \:=\:r

    Square: . (x^2+y^2 - 18x)^2 \:=\:r^2

    Therefore: . (x^2+y^2 - 18x)^2 \:=\:x^2+y^2

    . . I'd leave it like that.


    No . . . y \:=\:r\sin\theta

    Thanks a lot, guess I was pretty much there. Lets hope this online assignments takes that for an answer.
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