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

1 Answer
A. S. Adikesavan
Jul 5, 2018

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]}

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