Question

What is the Cartesian form of `r^2-rtheta = 2costheta+3tantheta`?

Answer 1

k-specific
`(x^2 + y^2 )^1.5 - ( x^2 + y^2 )( kpi + arctan ( y/x ))`
`= 2 x + (y/x)sqrt( x^2 + y^2 )`. See graphs, in grandeur.

Explanation:

Like `r, theta >= 0`. This is important, for reading this answer,

wherein arctan values `in [ 0, pi ]` and not `[ -pi/2, pi/2 ]`.

And, you can choose any `theta` in any Q.

The chosen `theta = kpi + arctan( y/x), arctan ( y/x ) in [ 0, pi ]`,

for a specific k, from { 0, 1, 2, 3, ...}

Use

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

`theta = kpi + arctan (y/x), theta in Q_1, Q_2 or Q_4`, choosing

befitting k from 0, 1, 2, 3, .... Now,

`r^2 - rtheta = 2 cos theta + 3 tan theta` converts to k-specific

`(x^2 + y^2 )^1.5 - ( x^2 + y^2 )( kpi + arctan ( y/x ))`

`= 2 x + 3 (y/x) sqrt( x^2 + y^2 )`

See graph, with k = 0.
graph{(x^2 + y^2 )^1.5 - ( x^2 + y^2 )( arctan ( y/x )) - 2 x - 3(y/x)sqrt( x^2 + y^2 )=0}

Note that the graph is non-periodic, despite that the joint period of

`cos theta and tan theta` is `2pi`. .

Graph for `theta to theta + 2pi`, in the next round,

graph{(x^2 + y^2 )^1.5 - ( x^2 + y^2 )( 2pi + arctan ( y/x )) - 2 x - 3(y/x)sqrt( x^2 + y^2 )=0}

Graph for `theta to theta + 4pi`
graph{(x^2 + y^2 )^1.5 - ( x^2 + y^2 )( 4pi + arctan ( y/x )) - 2 x - 3(y/x)sqrt( x^2 + y^2 )=0}

Slide the graphs `uarr larr darr rarr` for graphs, in grandeur.