How do you convert $x^{2} + y^{2} - 2 y = 0$ $x^2+y^2 - 2y=0$ to polar form?

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How do you convert $x^{2} + y^{2} - 2 y = 0$ $x^2+y^2 - 2y=0$ to polar form?

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Answer 1 · 2016-05-16T09:22:31

$r = 2 sin θ$ $r=2sintheta$

Explanation:

Using the formulae that link Cartesian to Polar coordinates.

$∙ x = r cos θ and y = r sin θ$ $• x = rcostheta" and "y=rsintheta$

and substituting into the given equation

$\Rightarrow {(r cos θ)}^{2} + {(r sin θ)}^{2} - 2 r sin θ = 0$ $rArr(rcostheta)^2+(rsintheta)^2-2rsintheta=0$

expanding brackets to obtain.

$r^{2} {cos}^{2} θ + r^{2} {sin}^{2} θ = 2 r sin θ$ $r^2cos^2theta+r^2sin^2theta=2rsintheta$

Take out a common factor of $r^{2}$ $r^2$

$\Rightarrow r^{2} ({cos}^{2} θ + {sin}^{2} θ) = 2 r sin θ$ $rArrr^2(cos^2theta+sin^2theta)=2rsintheta$

using the trig. identity $color(red)(|bar(ul(color(white)(a/a)color(black)(cos^2theta+sin^2theta=1)color(white)(a/a)|)))$

$rArrr^2=2rsintheta$

divide both sides by r

$rArrr=2sintheta$

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How do you convert $x^{2} + y^{2} - 2 y = 0$ $x^2+y^2 - 2y=0$ to polar form?

1 Answer

Explanation:

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How do you convert $x^{2} + y^{2} - 2 y = 0$ $x^2+y^2 - 2y=0$ to polar form?