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

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Answer 1 · 2016-04-10T17:05:13

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

#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}