How do you convert #(-2, -2sqrt{3})#into polar coordinates?

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Answer 1 · 2018-05-05T18:42:15

#(4, (4pi)/3)# (radians) or #(4, 240^@)# (degrees)

Rectangular #-># Polar: #(x, y) -> (r, theta)#

Find #r# (radius) using #r = sqrt(x^2 + y^2)#
Find #theta# by finding the reference angle: #tantheta = y/x# and use this to find the angle in the correct quadrant

#r = sqrt((-2)^2 + (-2sqrt3)^2)#

#r = sqrt(4+(4*3))#

#r = sqrt(4+12)#

#r = sqrt(16)#

#r = 4#

Now we find the value of #theta# using #tantheta = y/x#.

#tantheta = (-2sqrt3)/-2#

#tantheta = sqrt3#

#theta = tan^-1(sqrt3)#

#theta = (pi)/3# or #(4pi)/3#

To determine which one it is, we have to look at our coordinate #(-2, -2sqrt3)#. First, let's graph it:
enter image source here

As you can see, it is in the third quadrant. Our #theta# has to match that quadrant, meaning that #theta = (4pi)/3#.

From #r# and #theta#, we can write our polar coordinate:
#(4, (4pi)/3)# (radians) or #(4, 240^@)# (degrees)

Hope this helps!