What the is the polar form of $y = \frac{y}{x^{2}} + (2 x - 2) (y - 5)$ $y = y/x^2+(2x-2)(y-5)$ ?

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What the is the polar form of $y = \frac{y}{x^{2}} + (2 x - 2) (y - 5)$ $y = y/x^2+(2x-2)(y-5)$ ?

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Answer 1 · 2016-04-15T18:45:36

$r^{4} {cos}^{2} θ sin 2 θ - \frac{3}{2} r^{3} cos θ sin 2 θ - 10 r^{3} {cos}^{3} θ + 10 r^{2} {cos}^{2} θ + r sin θ = 0$ $r^4cos^2thetasin2theta-3/2 r^3costhetasin2theta-10r^3cos^3theta+10r^2cos^2theta+rsintheta =0$

Explanation:

$y = \frac{y}{x^{2}} + (2 x - 2) (y - 5)$ $y=y/x^2 +(2x-2)(y-5)$

$y = \frac{y + x^{2} (2 x - 2) (y - 5)}{x^{2}}$ $y=(y+x^2(2x-2)(y-5))/x^2$

$x^{2} y = y + x^{2} (2 x - 2) (y - 5)$ $x^2y=y+x^2(2x-2)(y-5)$

$x^{2} y = y + x^{2} (2 x y - 10 x - 2 y + 10)$ $x^2y=y+x^2(2xy-10x-2y+10)$

$x^{2} y = y + 2 x^{3} y - 10 x^{3} - 2 x^{2} y + 10 x^{2}$ $x^2y=y+2x^3y-10x^3-2x^2y+10x^2$

$0 = y + 2 x^{3} y - 10 x^{3} - 2 x^{2} y + 10 x^{2} - x^{2} y$ $0=y+2x^3y-10x^3-2x^2y+10x^2-x^2y$

$0 = 2 x^{3} y - 3 x^{2} y - 10 x^{3} + 10 x^{2} + y$ $0=2x^3y-3x^2y-10x^3+10x^2+y$

Now use the formula
$x = r cos θ and y = r sin θ$ $x=rcostheta and y=rsin theta$

$0 = 2 r^{3} {cos}^{3} θ \times r sin θ - 3 r^{2} {cos}^{2} θ \times r sin θ - 10 r^{3} {cos}^{3} θ + 10 r^{2} {cos}^{2} θ + r sin θ$ $0=2r^3cos^3thetaxxrsintheta-3r^2cos^2thetaxxrsintheta-10r^3cos^3theta+10r^2cos^2theta+rsintheta$

$0 = 2 r^{4} {cos}^{3} θ sin θ - 3 r^{3} {cos}^{2} θ sin θ - 10 r^{3} {cos}^{3} θ + 10 r^{2} {cos}^{2} θ + r sin θ$ $0=2r^4cos^3thetasintheta-3r^3cos^2thetasintheta-10r^3cos^3theta+10r^2cos^2theta+rsintheta$

$0 = r^{4} {cos}^{2} θ (2 sin θ cos θ) - \frac{3}{2} r^{3} cos θ (2 sin θ cos θ) - 10 r^{3} {cos}^{3} θ + 10 r^{2} {cos}^{2} θ + r sin θ$ $0=r^4cos^2theta(2sinthetacostheta)-3/2 r^3costheta(2sinthetacostheta)-10r^3cos^3theta+10r^2cos^2theta+rsintheta$

$0 = r^{4} {cos}^{2} θ sin 2 θ - \frac{3}{2} r^{3} cos θ sin 2 θ - 10 r^{3} {cos}^{3} θ + 10 r^{2} {cos}^{2} θ + r sin θ$ $0=r^4cos^2thetasin2theta-3/2 r^3costhetasin2theta-10r^3cos^3theta+10r^2cos^2theta+rsintheta$

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What the is the polar form of $y = \frac{y}{x^{2}} + (2 x - 2) (y - 5)$ $y = y/x^2+(2x-2)(y-5)$ ?

1 Answer

Explanation:

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What the is the polar form of y = y/x^2+(2x-2)(y-5) y=yx2+(2x−2)(y−5)y = y/x^2+(2x-2)(y-5) ?

1 Answer

Explanation:

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What the is the polar form of $y = \frac{y}{x^{2}} + (2 x - 2) (y - 5)$ $y = y/x^2+(2x-2)(y-5)$ ?