Converting Between Systems
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Key Questions

To convert from polar to rectangular:
#x=rcos theta #
#y=rsin theta# To convert from rectangular to polar:
#r^2=x^2+y^2#
#tan theta= y/x# This is where these equations come from:
Basically, if you are given an
#(r,theta)# a polar coordinate , you can plug your#r# and#theta# into your equation for#x=rcos theta # and#y=rsin theta# to get your#(x,y)# .The same holds true for if you are given an
#(x,y)# a rectangular coordinate instead. You can solve for#r# in#r^2=x^2+y^2# to get#r=sqrt(x^2+y^2)# and solve for#theta# in#tan theta= y/x# to get#theta=arctan (y/x)# (arctan is just tan inverse, or#tan^1# ). Note that there can be infinitely many polar coordinates that mean the same thing. For example,#(5, pi/3)=(5,5pi/3)=(5,4pi/3)=(5,2pi/3)# ...However, by convention, we are always measuring positive#theta# COUNTERCLOCKWISE from the xaxis, even if our#r# is negative.Let's look at a couple examples.
( 1)Convert
#(4,2pi/3)# into Cartesian coordinates.So we just plug in our
#r=4# and#theta= 2pi/3# into#x=4cos 2pi/3=2#
#y=4sin 2pi/3=2sqrt3# The cartersian coordinate is
#(2,2sqrt3)# (2) Convert
#(1,1)# into polar coordinates. ( since there are many posibilites of this, the restriction here is that#r# must be positive and#theta# must be between 0 and#pi# )So,
#x=1# and#y=1# . We can find# r# and#theta# from:
#r=sqrt(1^2+1^2)=sqrt2#
#theta=arctan (y/x)=arctan(1)=pi/4# The polar coordinate is
#(sqrt2,pi/4)# 
It is usually suitable to use polar coordinates when you deal with round objects like circles, and to use rectangular coordinates when you deal with more straight edges like rectangles.
I hope that this was helpful.

All you have to do is a simple change of variables:
# x=rcostheta #
# y=rsintheta # from polar to rectangular and:
#r=sqrt(x^2+y^2)#
#theta=arctan(y/x)# from rectangular to polar.
Example: what is the polar equation of the parabola
#y=x^2# ?By using the first to relations we have:
#rsintheta=r^2cos^2theta=>r=sintheta/(cos^2theta)#