Question

When converting polar equations in Cartesian equations, how do you deal with `theta` when it is on its own?

I know `theta= arctan(y/x)` , but if I use this substitution I end up with it in the final equation. I just don't get how to deal with `theta`.

Example.

`r^2=-theta + rsin(theta)`

Answer 1

We can only use the relation `theta=arctan(y/x)`.

Explanation:

The relation between Cartesian or rectangular coordinates `(x,y)` and polar coordinates `(r,theta)` is given by

`x=rcostheta` and `y=rsintheta`

and though we can have a simpler function for `r` given by `r^2=x^2+y^2`, for `theta` as `y/x=tantheta`, we have `theta=arctan(y/x)`, as you too have correctly mentioned.

Many times we do get trigonometric ratios of `theta` and it turns out to be easier as `sintheta=y/r=y/sqrt(x^2+y^2)`, `costheta=x/r=x/sqrt(x^2+y^2)` and `tantheta=y/x` (and for others we can just use their reciprocals),

however, if we just have `theta` there is no alternative to using `theta=arctan(y/x)`

and hence `r^2=-theta+rsintheta` in Cartesian or rectangular coordinates is just `x^2+y^2=-arctan(y/x)+y`