Question

How do you convert `y=x-2y+x^2y^2` into a polar equation?

Answer 1

`r = root(3)((3sin(t) - cos(t))/(cos(t)^2sin(t)^2))`

Explanation:

Converting a rectangular equation to a polar equation is fairly simple, it is accomplished using:

`x = rcos(t)`
`y = rsin(t)`

Another useful rule is that since `cos(x)^2 + sin(x)^2 = 1`:

`x^2 + y^2 = r^2cos(t)^2 + r^2sin(t)^2 = r^2`

But we won't need that for this problem. We also want to rewrite the equation as:
`0 = x - 3y + x^2y^2`

And we perform substitution:

`0= rcos(t) - 3rsin(t) + r^4cos(t)^2sin(t)^2`
`0 = cos(t) - 3sin(t) + r^3cos(t)^2sin(t)^2`

Now we can solve for `r`:

`-r^3cos(t)^2sin(t)^2 = cos(t) - 3sin(t)`

`r^3cos(t)^2sin(t)^2 = 3sin(t) - cos(t)`

`r^3 = (3sin(t) - cos(t))/(cos(t)^2sin(t)^2)`

`r = root(3)((3sin(t) - cos(t))/(cos(t)^2sin(t)^2))`