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

##### 1 Answer

See graph and details, particularly on

#### Explanation:

The period is

for

As

for the first loop.

Converting to Cartesian ( x, y ) =

Graph of

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

period,

Zooming might reveal this happening.

In addition, I add here 5-loop graph of

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]}