Question

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

Answer 1

`y=8`

Explanation:

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

`y = 8`

Explanation:

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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