How do you convert #r =4sin(2θ)# to rectangular form?

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Answer 1 · 2016-09-25T05:17:27

#(x^2+y^2)^3=64x^2y^2#

Use the conversion equation #r(cos theta, sin theta)=(x, y)# that

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

#sin 2theta = 2 sin theta cos theta = (2xy)/r^2=(2xy)/(x^2+y^2)#.

So, after rationalization, the given polar equation becomes

#(x^2+y^2)^3=64x^2y^2#

Some nuances for the interested readers:

If r is allowed to be negative, there are four petals in the four

quadrants, from the center r = 0, for #theta in #[0, pi]#.

For strictly #r >=0,# there are just two petals, in #Q_1 and Q_3#..

Interestingly, as the period of r is #pi#, the second is

contributed by this periodicity. .A Table for #theta in [0, 2pi]# would

reveal all this.

I welcome a graphical depiction of these aspects.