How do you graph #r^2 = sin2(t)#?

1 Answer
Jul 23, 2018

See graphs galore.

Explanation:

The power-scaling of r is done,

when #r# is changed to #r^n#, n > 1, in #r = f( theta )#.

It is r-more, when #r in ( 0, 1 )#. Otherwise, jt is r-less

Here, this is an example.

Use #r = sqrt( x^2 + y^2 ) >= 0, r ( cos theta, sin theta )#

and #sin 2theta = 2 sintheta cos theta#,

to get the Cartesian form of

#r^2 = sin 2theta # as

#( x^2 + y^2 ) ^2 ) - 2xy = 0#.

The Socratic graph is immediate.
graph{( x^2 + y^2 ) ^2 - 2xy =0[-2 2 -1 1]}
Graph of #r = sin 2theta#, for contrast in r-scaling:
graph{( x^2 + y^2 ) ^1.5 - 2xy =0[-2 2 -1 1]}

Easy to see which is which, jn the combined graph, along with the

third graph of #r^5 = sin 2theta#:

graph{(( x^2 + y^2 ) ^2 - 2xy)( ( x^2 + y^2 ) ^1.5 - 2xy) ( ( x^2 + y^2 ) ^3.5 - 2xy) =0[-2 2 -1 1]}