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

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Answer 1 · 2016-04-20T03:00:58

Called Lemniscate, the graph looks like #oo#, symmetrical about the initial line #theta=0# and the pole r = 0.

In the 1st and 4th quadrants, #|theta|<=pi/4# In the 2nd and 3rd,

#|pi-theta|<=pi/4#.

I used infinity symbol #oo# to depict the shape of this double-

loop, looking like a fallen 8. For getting 8-erect, the equation is

#r^2=-cos 2theta#

Use a table

#{(r, theta)}.= {(0, -pi/4) (1/sqrt2, -pi/8) (1, 0) (1/sqrt2, pi/8) (0,#

#pi/4)}#, for one loop. Its mirror image with respect to #theta=pi/2#

is the other loop.

Strictly, #r =sqrt(x^2 + y^2)>=0#.

Graphs of both #r^2 =+- cos 2theta# are combined.
graph{((x^2+y^2)^2-x^2+y^2)((x^2+y^2)^2+x^2-y^2)=0[-2 2 -1 1]}.

Interestingly, an easy rotation of this graph through #pi/4#

produces a grand 8-petal flower.

graph{((x^2+y^2)^2-x^2+y^2)((x^2+y^2)^2+x^2-y^2)((x^2+y^2)^2-2xy)((x^2+y^2)^2+2xy)=0[-2 2 -1 1]}.