How do you convert $r = 1 + 2 sin θ$ $r = 1 + 2 sin theta$ to rectangular form?

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Answer 1 · 2018-03-12T08:48:49

${(x^{2} + y^{2} - 2 y)}^{2} = x^{2} + y^{2}$ $(x^2+y^2-2y)^2=x^2+y^2$

Multiply each term by $r$ $r$ to get $r^{2} = r + 2 r sin θ$ $r^2=r+2rsintheta$

$r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$
$r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$
$2 r sin θ = 2 y$ $2rsintheta=2y$

$x^{2} + y^{2} = \sqrt{x^{2} + y^{2}} + 2 y$ $x^2+y^2=sqrt(x^2+y^2)+2y$

$x^{2} + y^{2} - 2 y = \sqrt{x^{2} + y^{2}}$ $x^2+y^2-2y=sqrt(x^2+y^2)$

${(x^{2} + y^{2} - 2 y)}^{2} = x^{2} + y^{2}$ $(x^2+y^2-2y)^2=x^2+y^2$

How do you convert r = 1 + 2 sin thetar=1+2sinθr = 1 + 2 sin theta to rectangular form?