What is the Cartesian form of #-2r^3 = -theta+sec(theta)+ 4cos(theta) #?

1 Answer
A. S. Adikesavan
Aug 8, 2018

See answer in explanation.

Explanation:

Use

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

#theta = phi = arctan ( y/x )#

#= pi + phi, (x, y ) in# the diagonally opposite quadrant.

#- 2 r^3 = - theta + sec theta + 4 cos theta # converts to

#- 2 ( x^2 + y^2 ) 1.5 =( phi or ( pi + phi ) ) + 1/x sqrt( x^2 + y^2 ) + x/sqrt ( x^2 + y^2 )#,

giving

# phi or ( pi + phi ) = sqrt ( x^2 + y^2 )( 2 ( x^2 + y^2 ) + 1/x ) + x/sqrt ( x^2 + y^2 )#

graph{sqrt ( x^2 + y^2 )( 2 ( x^2 + y^2 ) + 1/x ) + x/sqrt ( x^2 + y^2 )-arctan(y/x)=0[-1 0 -10 10]}

The graph is confined to a rectangle, as 'arctan' is used to value

#theta#

As cosine is present as secant and cosine, the general value of #theta = 2kpi +- phi#.

See this funnel graph, for k = 5 and + sign.
graph{sqrt ( x^2 + y^2 )( 2 ( x^2 + y^2 ) + 1/x ) + x/sqrt ( x^2 + y^2 )-arctan(y/x)-10pi=0[-1 0 -10 10]}

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