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I have tried converting the equation $ y = x^3 $ to a polar equation using the following steps but the graph of the resulting polar equation also appears to include $ y=-x^3 $.

$ r^2 = x^2 + y^2 $

$ r^2 = x^2 + (x^3)^2$

$ r^2 = r^2\times cos(\theta)^2 + r^6cos(\theta)^6 $

$ 1 = cos(\theta)^2+r^4\times cos(\theta)^6$

$ \frac{sin(\theta)^2}{cos(\theta)^6}=r^4$

$ r = (\frac{sin(\theta)^2}{cos(\theta)^6})^{1/4}$

However, the graph of this is not $ y = x^3 $.

I am wondering why converting a cartesian equation to a polar equation does not appear to work this way. I understand that $ y=x^3$ can be easily converted to polar by substituting $ y=rsin(\theta)$ and $x=rcos(\theta)$ into $y=x^3$ but I don't understand why I do not get the same answer when substituting $y = x^3$ for $x$ in $r^2 = x^2+y^2$.

Thanks.

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  • $\begingroup$ You get the extra solutions from the squaring operations, since $r^2=(-r)^2$. This has the effect of "mirroring" the graph across the $x,y$ axes, or in the polar case, across the center. $\endgroup$
    – abiessu
    Jan 11 at 17:50
  • $\begingroup$ Start with $y=-x^3$. The result will be the same starting from the second line, since $(-x^3)^2=(x^3)^2$ $\endgroup$ Jan 11 at 17:57
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$y=x^3$

$r\sin\theta=r^3\cos^3\theta$

cancel $r$ and solve

$r^2=\frac{\sin\theta}{\cos^3\theta}$

to solve for $r$ we must have $\frac{\sin\theta}{\cos\theta}\ge 0\to \tan \theta \ge 0$ that is $0\le \theta<\pi/2\lor \pi\le \theta<3\pi/2$

$r=\sqrt{\frac{\sin\theta}{\cos^3\theta}};\;\theta\in [0,\pi/2)\cup[\pi,3\pi/2)$

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$$r=\sqrt{x^2+x^6}$$ is fine, as the RHS is positive. Then

$$\tan\theta=-\frac{x^3}x=-x^2$$ and

$$r=\sqrt{-\tan^3\theta-\tan\theta}=\sqrt{-\dfrac{\sin\theta}{\cos^3\theta}}.$$

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  • $\begingroup$ $\cos^3\theta r^3=-r\sin\theta$ gives $y=-x^3$. You should take $r=\frac{\sin\theta}{\cos^3\theta}$ in the interval $[0,\pi/2)\cup [\pi,3\pi/2)$ $\endgroup$
    – Raffaele
    Jan 11 at 18:13
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    $\begingroup$ @Raffaele: why don't you enter this as an answer ? It is independent of mine. $\endgroup$
    – user65203
    Jan 11 at 18:29

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