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

1 Answer
Apr 10, 2016

r= 4 csc theta, theta in Q_1 or Q_2r=4cscθ,θQ1orQ2 only.., ,

Explanation:

y = r sin thetay=rsinθ

So, the polar form is r = 4 csc thetar=4cscθ

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(+-)ooθ0±,r(±).

At theta=pi/2, r=4θ=π2,r=4.

As thetato pi_ -, rto+ooθπ,r+, and thetato pi_ +, rto-ooθπ+,r.

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 < 0θ(π,2π),r<0.

csc thetacscθ is discontinuous at theta=0, pi, 2pi, ..θ=0,π,2π,...

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

for theta in .( 2 k pi, ( 2 k + 1 ) pi)θ.(2kπ,(2k+1)π),

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

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

graph{y - 4 +0x=0}