How do you convert $r^{2} = 9 cos 5 (θ)$ $r^2 = 9cos5(theta)$ into cartesian form?

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Answer 1 · 2016-12-07T09:36:50

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

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

And so, $θ$ $theta$ is in $2 k π \pm (\frac{1}{5}) \frac{π}{2}$ $2kpi+-(1/5)pi/2$ , k = 0, 1, 2, 3...

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

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

Only for half period = $\frac{π}{5} = 36^{o}$ $pi/5 = 36^o$ , 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 $r (cos θ, sin θ) = (x, y)$ $r(cos theta, sin theta)=(x, y)$ , with

$r = \sqrt{x^{2} + y^{2}} \geq 0$ $r=sqrt(x^2+y^2)>=0$ .

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.

Adikesavan's revised answer ends here.

Here is a graph where the red is the positive value for r and the green is the negative values for r:
![Desmos.com](useruploads.socratic.org)

How do you convert r^2 = 9cos5(theta)r2=9cos5(θ)r^2 = 9cos5(theta) into cartesian form?