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

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Ken C. Share
Jul 14, 2016

Answer:

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.

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