How do you graph #r = 2cos2θ#?

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Answer 1 · 2018-07-18T13:15:18

See the lemniscate-like graph and details..

#0 <= r = 2 cos 2theta in [-2, 2 ]#.

As #r >= 0, 2theta in Q_1 or Q_4# and so

#theta in [ kpi - pi/4, kpi + pi/4 ], k = 0, +-1, +-2, +-3, ...#

The period is #(2pi)/2 = pi#.

In one period, r >= 0, for one half only.

For Socratic graph, use

# ( x, y ) = r ( cos theta, sin theta )# ,

#r = sqrt( x^2 + y^2) >= 0# and

#cos 2theta = cos^2theta - sin^2theta#

to get the Cartesian equation

#( x^2 + y^2 )^1.5-2( x^2- y^2 ) = 0#.

The graph is immediate. Two loops created, in two complete

rotations and the same repeated, for further rotations.
graph{(x^2+y^2)^1.5-2(x^2-y^2)=0[-4 4 -2 2]}.