I try to find a manner to convert cartesian rectangular equation like $y = f(x)$ to it's polar equation $r = r(\theta)$.

Given the function $f(x) = y = 5x^4$, lets try to get the polar equation:

$$x = r\cdot \cos(\theta), \quad y = r\cdot \sin(\theta)$$


$$r\cdot \sin(\theta) = 5\cdot r^4 \cdot \cos^4(\theta)$$ $$ \Rightarrow r^3= \frac{\sin(\theta)}{5\cdot \cos^4(\theta)} $$ $$ \Rightarrow r(\theta) = \sqrt[3] {\frac {\sin(\theta)}{5\cdot \cos^4(\theta)}}$$

My question is, That's really the correct answer? if so, how we can convert it back to rectangular equation? I tried to convert the polar equation we received to a rectangular equation using the following equations

$$\theta = \arctan\Bigr(\frac yx\Bigr), \quad r = \sqrt{x^2+y^2}$$

The result I got was everything except $y = 5x^4$.

Can someone point me the right direction to get the correct solution?

Thanks for Help!!

  • $\begingroup$ In the line 2 you divided by zero... I mean, you can not divide by r, because the point $(0,0)$ lies in the graph of the function, put $r$ in evidence instead... $\endgroup$
    – DiegoMath
    Jul 9, 2018 at 19:31
  • $\begingroup$ "was everything" ? $\endgroup$
    – user65203
    Jul 9, 2018 at 19:32
  • $\begingroup$ Why the $\pm$ symbol? If the root is a cubic one this is not necessary. $\endgroup$
    – DiegoMath
    Jul 9, 2018 at 19:35
  • $\begingroup$ The transformation equations $\theta = arctan\Bigr(\frac yx\Bigr), r = \sqrt{x^2+y^2}$ are not the only ones. You can use $x = r\cdot cos(\theta),y = r\cdot sin(\theta)$ as well. $\endgroup$
    – DiegoMath
    Jul 9, 2018 at 19:37

1 Answer 1


Using the inverse relations, the Cartesian equation reads

$$\sqrt{x^2+y^2}=\sqrt[3]{\frac{\sin\left(\arctan\dfrac yx\right)}{5\cos^4\left(\arctan\dfrac yx\right)}}=\sqrt[3]{\frac{\tan\left(\arctan\dfrac yx\right)}{5\cos^3\left(\arctan\dfrac yx\right)}}.$$

Now notice that $$\frac1{\cos^2\theta}=\frac{\sin^2\theta+\cos^2\theta}{\cos^2\theta}=\tan^2\theta+1$$ and $$\frac1{\cos^3\theta}=(\tan^2\theta+1)^{3/2}.$$

Substituting, we have $$\sqrt{x^2+y^2}=\sqrt[3]{\frac{\tan\left(\arctan\dfrac yx\right)}{5\cos^3\left(\arctan\dfrac yx\right)}} =\sqrt[3]{\dfrac y{5x}\left(\tan^2\left(\arctan\dfrac yx\right)+1\right)^{3/2}} =\sqrt[3]{\dfrac y{5x}\left(\dfrac{y^2}{x^2}+1\right)^{3/2}}=\sqrt[3]{\dfrac y{5x^4}}\sqrt{x^2+y^2}.$$

A final simplification yields

$$\frac y{5x^4}=1.$$

Easy as that :-)


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