How do you convert # r= theta - 4sintheta +cos^2theta# to Cartesian form?

1 Answer
Aug 11, 2018

See explanation, with answer.

Explanation:

I use both #r and theta >= 0# only.

So, #arctan (y/x) in [ 0, pi ], sans pi/2#

#r = theta - 4 sin theta + cos^2theta = theta# + a periodic function

of period #2pi#. So, r is non-periodic.

Convert using

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

#0 <= theta = kpi + arctan ( y/x ), l = 0, 1, 2, 3, ...#.

#sqrt ( x^2 + y^2 ) = kpi + arctan ( y/x )#

#- 4 x/sqrt ( x^2 + y^2 ) + y^2/(x^2 + y^2 )#, giving

#0 <= (tan)^(-1) ( y/x ) = kpi + arctan ( y/x ) #

#= 4 x/sqrt ( x^2 + y^2 ) - y^2/(x^2 + y^2 )#

#+ sqrt ( x^2 + y^2 )#, k = 0, 1, 2, 3, ..#, piecewise, for successive

half rotations, about O.

Graph, for k = 0 using conventional #arctan ( y/x ) in ( - pi/2, pi/2 )# )#.graph{arctan ( y/x )-4 x/sqrt ( x^2 + y^2 ) + y^2/(x^2 + y^2 )=0[-1000 1000 -500 500]} Graph, for the assumed #theta to theta + pi/2# non- negative

#arctan ( y/x) in ( 0, pi )#, sans #pi/2 # .
graph{pi/2+arctan ( y/x )+4y/sqrt ( x^2 + y^2 ) + x^2/(x^2 + y^2 )=0[-1000 1000 -500 500]}

Observe that this is confined to #Q_3 and Q_4#, whereas, the first

that was expected to be in #Q_1 and Q_4# is not so.