Calculus- Converting an equation from Cartesian to polar Cordinate

Question

I'm having difficulties with the following equation to simplify it and converting to a polar form

$$x^4 + y^4 + 2x^2y^2 + 2x^3 + 2xy^2 - y^2 = 0$$ its from my book thomas calculus and only the final answer is given which: $$r = 1-cos\theta$$

Answer 1

First neaten up the original equation \begin{eqnarray*} (x^2+y^2)^2+2x(x^2+y^2)-y^2 \end{eqnarray*} Now substitute $x=r \cos \theta$ & $y = r \sin \theta$ \begin{eqnarray*} r^4+2r^3 \cos \theta -r^2 \sin^2 \theta=0 \\ r^2+2r \cos \theta -1 +\cos^2 \theta=0 \\ (r+\cos \theta)^2 -1=0 \end{eqnarray*} So we have $\color{red}{r = \pm 1 -\cos \theta}$.

Answer 2

Recall that the standard conversion from Cartesian to polar is $x=r\cos\theta,$ $y=r\sin\theta.$ Do those substitutions in the equation until there are no $x$s or $y$s remaining.

You will then have something with only $r$ and trigonometric functions of $\theta.$ Try to group and factor the terms usefully. Putting all the $r^4$ terms together, all the $r^3$ terms together, and so forth will be helpful. Trigonometric identities also will help; actually you only really need one identity, namely, $\sin^2\theta + \cos^2\theta = 1.$

Eventually you may reduce the equation to one that you can solve for $r$; that is, you can derive a new equation with just $r$ on the left side and some expression involving $\theta$ and constants on the right. (In fact it will be $1-\cos\theta$ on the right, as the answer key says, so that is a way to check whether you did the other steps correctly.)

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Some other thoughts that may be useful: recall that when converting polar to Cartesian, $r = \sqrt{x^2 + y^2},$ that is, $r^2 = x^2 + y^2.$ What expression of $x$s and $y$s is equal to $r^4$?