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

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Answer 1 · 2016-07-29T17:10:43

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

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))#