How do you convert $2 y = - x^{2} + 3 x$ $2y= -x^2+3x$ into a polar equation?

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How do you convert $2 y = - x^{2} + 3 x$ $2y= -x^2+3x$ into a polar equation?

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Answer 1 · 2016-05-28T00:50:05

The polar equation is
$r = \frac{3 \cdot cos θ - 2 \cdot sin θ}{{cos}^{2} θ}$ $color(blue)(r=(3*cos theta-2*sin theta)/cos^2 theta)$

Explanation:

To convert $2 y = - x^{2} + 3 x$ $2y=-x^2+3x$

Use $x = r \cdot cos θ$ $x=r*cos theta$ and $y = r \cdot sin θ$ $y=r*sin theta$

Let's do it

$2 y = - x^{2} + 3 x$ $2y=-x^2+3x$

$2 (r \cdot sin θ) = - {(r \cdot cos θ)}^{2} + 3 (r \cdot cos θ)$ $2(r*sin theta)=-(r*cos theta)^2+3(r*cos theta)$

$2 \cdot r \cdot sin θ = - r^{2} \cdot {cos}^{2} θ + 3 \cdot r \cdot cos θ$ $2*r*sin theta=-r^2*cos^2 theta+3*r*cos theta$

divide both sides of the equation by $r$ $r$

$\frac{2 \cdot r \cdot sin θ}{r} = \frac{- r^{2} \cdot {cos}^{2} θ}{r} + \frac{3 \cdot r \cdot cos θ}{r}$ $(2*r*sin theta)/r=(-r^2*cos^2 theta)/r+(3*r*cos theta)/r$

$(2*cancelr*sin theta)/cancelr=(-cancelr^2*cos^2 theta)/cancelr+(3*r*cos theta)/cancelr$

$2*sin theta=-r*cos^2 theta+3*cos theta$

Transposition

$r*cos^2 theta=3*cos theta-2*sin theta$

divide both sides by $cos^2 theta$

$(r*cos^2 theta)/cos^2 theta=(3*cos theta-2*sin theta)/cos^2 theta$

$(r*cancelcos^2 theta)/cancel(cos^2 theta)=(3*cos theta-2*sin theta)/cos^2 theta$

$color(red)(r=(3*cos theta-2*sin theta)/cos^2 theta)$

God bless ....I hope the explanation is useful.

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How do you convert $2 y = - x^{2} + 3 x$ $2y= -x^2+3x$ into a polar equation?

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Explanation:

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How do you convert $2 y = - x^{2} + 3 x$ $2y= -x^2+3x$ into a polar equation?