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

Question

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

1 Answer

Impact of this question

1435 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-15T09:12:53

$5 r^{3} sin θ {cos}^{2} θ - r^{2} {sin}^{2} θ + r cos θ + 2 + cot θ = 0$ $5r^3sinthetacos^2theta-r^2sin^2theta+rcostheta+2+cottheta=0$

Explanation:

The relation between polar coordinates $(r, θ)$ $(r,theta)$ and Cartesian or rectangular coordinates $(x, y)$ $(x,y)$ is given by

$x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$

Hence, $2 = - x - 5 x^{2} y - \frac{x}{y} + y^{2}$ $2=-x-5x^2y-x/y+y^2$ can be written as

$2 = - r cos θ - 5 r^{2} {cos}^{2} θ r sin θ - \frac{r cos θ}{r sin θ} + r^{2} {sin}^{2} θ$ $2=-rcostheta-5r^2cos^2thetarsintheta-(rcostheta)/(rsintheta)+r^2sin^2theta$

or $2 = - r cos θ - 5 r^{3} sin θ {cos}^{2} θ - cot θ + r^{2} {sin}^{2} θ$ $2=-rcostheta-5r^3sinthetacos^2theta-cottheta+r^2sin^2theta$

or $5 r^{3} sin θ {cos}^{2} θ - r^{2} {sin}^{2} θ + r cos θ + 2 + cot θ = 0$ $5r^3sinthetacos^2theta-r^2sin^2theta+rcostheta+2+cottheta=0$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What the is the polar form of 2 = -x-5x^2y-x/y +y^2 2=−x−5x2y−xy+y22 = -x-5x^2y-x/y +y^2 ?

1 Answer

Explanation:

Related questions

Impact of this question

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