What the is the polar form of #y = x^2y-x/y^2 +xy^2 #?

Question

Carter Anderson · Accepted Answer

#color(blue)[r^2=(-costheta)/[sin^3theta-r^2*sin^3theta*cos^2theta-r^2costheta*sin^4theta]]#

Note that

#color(red)[y=r*sintheta]#

#color(red)[x=r*costheta]#

#y = x^2y-x/y^2 +xy^2#

#(r*sintheta)=(r*costheta)^2*(r*sintheta)-(r*costheta)/(r*sintheta)^2+(r*costheta)(r*sintheta)^2#

#rsintheta=r^3sintheta*cos^2theta-(costheta)/(rsin^2theta)+r^3costheta*sin^2theta#

#[rsintheta-r^3sintheta*cos^2theta-r^3costheta*sin^2theta]/1=-(costheta)/(rsin^2theta)#

#r^2sin^3theta-r^4*sin^3theta*cos^2theta-r^4costheta*sin^4theta=-costheta#

#color(blue)[r^2=(-costheta)/[sin^3theta-r^2*sin^3theta*cos^2theta-r^2costheta*sin^4theta]]#

Delilah Aguilar · Answer

undefined To find the polar form of the given Cartesian equation $y = x^2y - \frac{x}{y^2} + xy^2$, we need to express $x$ and $y$ in terms of $r$ and $\theta$, where $r$ is the distance from the origin and $\theta$ is the angle from the positive x-axis.

Using the relationships $x = r \cos(\theta)$ and $y = r \sin(\theta)$, we substitute these into the given equation:

$$y = (r\cos(\theta))^2(r\sin(\theta)) - \frac{r\cos(\theta)}{(r\sin(\theta))^2} + (r\cos(\theta))(r\sin(\theta))^2$$

Now, simplify this expression.