How do you convert #r = 2 sin theta# into cartesian form?

1 Answer
Ken C.
Jul 14, 2016

Make use of a few formulas and do some simplification. See below.

Explanation:

When dealing with transformations between polar and Cartesian coordinates, always remember these formulas:

  • #x=rcostheta#
  • #y=rsintheta#
  • #r^2=x^2+y^2#

From #y=rsintheta#, we can see that dividing both sides by #r# gives us #y/r=sintheta#. We can therefore replace #sintheta# in #r=2sintheta# with #y/r#:
#r=2sintheta#
#->r=2(y/r)#
#->r^2=2y#

We can also replace #r^2# with #x^2+y^2#, because #r^2=x^2+y^2#:
#r^2=2y#
#->x^2+y^2=2y#

We could leave it at that, but if you're interested...

Further Simplification
If we subtract #2y# from both sides we end up with this:
#x^2+y^2-2y=0#

Note that we can complete the square on #y^2-2y#:
#x^2+(y^2-2y)=0#
#->x^2+(y^2-2y+1)=0+1#
#->x^2+(y-1)^2=1#

And how about that! We end up with the equation of a circle with center #(h,k)->(0,1)# and radius #1#. We know that polar equations of the form #y=asintheta# form circles, and we just confirmed it using Cartesian coordinates.

Related questions
See all questions in Converting Equations from Polar to Rectangular
Impact of this question
61732 views around the world
You can reuse this answer
Creative Commons License