Question

How do you convert `y=4` to polar form?

Answer 1

`r= 4 csc theta, theta in Q_1 or Q_2` only.., ,

Explanation:

`y = r sin theta`

So, the polar form is `r = 4 csc theta`

This is the polar equation of the straight line y = 4, parallel to the x-

axis at a height 4 units.

Only for those who are interested in "`+ooto-oo` jumps for r":.

Interestingly, as `thetato0_(+-), rto(+-)oo`.

At `theta=pi/2, r=4`.

As `thetato pi_ -, rto+oo`, and `thetato pi_ +, rto-oo`.

Can you Imagine r jumping from `+ooto-oo`?

For retracing y = 4?..

No problem. #r = sqrt( x^2 + y^2 ) >= 0.

For `theta in ( pi, 2pi ), r < 0`.

`csc theta` is discontinuous at `theta=0, pi, 2pi, ..`.

Yet, the endless line is created iteratively ( `larr` ),

for `theta in .( 2 k pi, ( 2 k + 1 ) pi)`,

`k = 0, 1, 2, 3, ...`

This complication is avoided, by using the Cartesian form, y = 4.

graph{y - 4 +0x=0}