How do you graph #r=-8cos2theta#?

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Answer 1 · 2018-07-30T14:10:13

See 8-like graph and details.

#r = -8 cos 2theta in [ - 8, 8 ], # with period #( 2pi )/2 = pi#

No pixels for # < 0#

#rArr 2theta in [ pi/2, 3/2pi ] U [ 5/2pi, 7/2pi]#

# rArr theta in [ pi/4, 3/4pi ] U [ 5/4pi, 7/4pi]#. So,

in two periods #theta in [ 0, 2pi ]#.

Use

0 <= r = sqrt ( x^2 + y^2 ) and ( x, y ) = r ( cos theta, sin theta )# d

convert to

#( x^2 + y^2 )^ 1.5 = - 8 (x^2 - y^2 )#, using

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

The Socratic 8-like graph is immediate.
graph{ ( x^2 + y^2 )^ 1.5 + 8 (x^2 - y^2 ) = 0[ -18 18 -9 9 ]}