Question

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

Answer 1

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]}

Sailing in the ocean of Mathematics needs patience.