Question

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

Answer 1

See graph and details, particularly on `r >= 0`.

Explanation:

The period is `(2pi)/((3pi)/2) = (4pi)/3`. So, three loops are created

for `theta in [ -2pi, 2pi]`.

As `r >= 0, (3theta)/2 in [ - pi/2, pi/2 ] rArr theta in [ -pi/3, pi/3 ]`,

for the first loop.

`r = 2 cos ((3theta)/2) = 2 sqrt(1/2(1+ cos 3theta)) >= 0`

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

Converting to Cartesian ( x, y ) = `r ( cos theta, sin theta )`,

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

Graph of `r = 2 cos ((3theta)/2)`, 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 [pi/3, pi ], 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 ((5/2)theta).`
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]}