# Converting cartesian equation to polar issue

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.

• 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. Jan 11 at 17:50
• Start with $y=-x^3$. The result will be the same starting from the second line, since $(-x^3)^2=(x^3)^2$ Jan 11 at 17:57

$$y=x^3$$

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

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 $$0\le \theta<\pi/2\lor \pi\le \theta<3\pi/2$$

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

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

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

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

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