Question

How do you convert `r=2sin(3theta)` to rectangular form?

Answer 1

`(x^2+y^2)^2=2y(3x^2-y^2)`
I have inserted graph for this cartesian frame, using Socratic graphic facility, for the purpose of making some remarks.

Explanation:

The conversion formula is

`r(cos theta, sin theta)=( x, y ) to cos theta=x/r and sin theta =y/r`

`and r=sqrt(x^2+y^2)`.

Here, `r = 2 sin 3theta = 2(3cos^2theta sin theta-sin^3theta)`

`=2((3/r)^2(y/r)-(y/r)^3)`

`=2y(3x^2-y^2)/r^3`. So,

`r^4=(x^2+y^2)^2=2y(3x^2-y^2)`

The graph is a 3-petal rose. .

graph{(x^2+y^2)^2=2y(3x^2-y^2) [-4 4 -2 2]]}

This is to inform the interested readers that

`r^n = 2 sin 3theta, n = 1, 2, 3, ...`

graphs are similar. with radial scale factors

`1, 2^(1/2), 2^(1/3), ...`.

See the illustrative graph.

graph{((x^2+y^2)^2-2y(3x^2-y^2))((x^2+y^2)^2.5-2y(3x^2-y^2))((x^2+y^2)^3-2y(3x^2-y^2))=0[-4 4 -2 2]}

Note on polar scaling: In `k r^m = sin ntheta`, m is for power

scaling of r, k is for scalar multiplication of r and n is for scalar

multiplication of `theta`. In `r = 2 sin 3theta`, k = 0.5, n = 3 and m

= 1. See graph for mixed scaling.

graph{((x^2+y^2)^2-2y(3x^2-y^2))(.33(x^2+y^2)^2.5-2y(3x^2-y^2))(1.4(x^2+y^2)^3-2y(3x^2-y^2))=0[-4 4 -2 2]}