# How do you convert r^2 = 9cos5(theta) into cartesian form?

Dec 7, 2016

$\frac{1}{9} {\left({x}^{2} + {y}^{2}\right)}^{3.5} - {x}^{5} + 10 {x}^{3} {y}^{2} - 5 x {y}^{4} = 0$. Graph is inserted

#### Explanation:

Here, $9 \cos 5 \theta = {r}^{2} \ge 0$. So, $5 \theta$ is in Q1 or Q4.

And so, $\theta$ is in $2 k \pi \pm \left(\frac{1}{5}\right) \frac{\pi}{2}$, k = 0, 1, 2, 3...

The period for the graph is $\frac{2 \pi}{5} = {72}^{o}$.

For one half (${36}^{o}$) ${r}^{2}$ is positive. For the other, ${r}^{2} < 0$.

Only for half period = $\frac{\pi}{5} = {36}^{o}$, the graph appears.

I expect five equal loops like the ones that are marked red in the

second graph, contributed by the other author.

The conversion formula is $r \left(\cos \theta , \sin \theta\right) = \left(x , y\right)$, with

$r = \sqrt{{x}^{2} + {y}^{2}} \ge 0$.

Substitution gives the form in the answer.

graph{(x^2+y^2)^3.5-9x^5+90x^3y^2-45xy^4=0}

Debugging was elusive for me, for so long. I was able to do it now.