# How do you graph r=2cos(3theta/2)?

Jul 5, 2018

See graph and details, particularly on $r \ge 0$.

#### Explanation:

The period is $\frac{2 \pi}{\frac{3 \pi}{2}} = \frac{4 \pi}{3}$. So, three loops are created

for $\theta \in \left[- 2 \pi , 2 \pi\right]$.

As $r \ge 0 , \frac{3 \theta}{2} \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] \Rightarrow \theta \in \left[- \frac{\pi}{3} , \frac{\pi}{3}\right]$,

for the first loop.

$r = 2 \cos \left(\frac{3 \theta}{2}\right) = 2 \sqrt{\frac{1}{2} \left(1 + \cos 3 \theta\right)} \ge 0$

= 2sqrt(1/2(1+(cos^3theta - 3 cos theta sin^2theta )) .

Converting to Cartesian ( x, y ) = $r \left(\cos \theta , \sin \theta\right)$,

( x^2 + y^2 )^2 = sqrt 2 sqrt( ( x^2 + y^2 )^1.5+ x^3 - 3xy^2))

Graph of $r = 2 \cos \left(\frac{3 \theta}{2}\right)$, on uniform scale:
graph{( x^2 + y^2 )^2 - (sqrt 2 sqrt( ( x^2 + y^2 )^1.5 + x^3 - 3xy^2)) = 0[-4 4 -2 2]}

Observe the three nodes, all at r = 0, where the tangent turns.

For $\theta \in \left[\frac{\pi}{3} , \pi\right] , r < 0$, and so on. For every second-half

period, $r < 0$. All loops have r = 0 as the common point.

Zooming might reveal this happening.

In addition, I add here 5-loop graph of $r = 2 \cos \left(\left(\frac{5}{2}\right) \theta\right) .$
graph{(x^2 + y^2 )^3 - (sqrt 2 sqrt( ( x^2 + y^2 )^2.5+ x^5 - 10 x^3y^2 + 5xy^4)) = 0[-4 4 -2 2]}