How do you convert the polar equation #r=8csctheta# into rectangular form?

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Answer 1 · 2016-09-28T11:15:15

#y=8#

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

#x=rcostheta# and #y=rsintheta#

Hence #r=8csctheta=8/sintheta#

or #rsintheta=8#

or #y=8#

Answer 2 · 2016-09-28T11:45:07

#y = 8#

The conversion equation is #r(cos theta, sin theta)=(x, y)#, giving

#r = sqrt(x^2+y^2)# (principal square root) #>=0#,

#cos theta = x/sqrt(x^2+y^2) and sin theta =y/sqrt(x^2+y^2)#.

Here, #r>=#minimum #(8csc theta)=8#, for #theta in (0, pi)# .

Now, #r=8csc theta =8/sin theta=8r/y#.

Cancelling non-zero r,

#y=8#

Interestingly,

as #theta to 0_+, csc theta to oo# and,

as #theta to# 0#_, csc to -oo#.

Likewise, there is irremovable infinite discontinuity at #theta=pi#.

So, it is proper to state that

#y = 8, x in (-oo, oo)#, in cartesian form, is equivalent to

#r=8 csc theta, theta in (0. pi)#, in polar form.

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