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:
    tutorial.math.lamar.edu

    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 x-axis, 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.

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