How do you find the polar coordinate for (-4, -4)?

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Answer 1 · 2018-05-05T18:50:08

#(4sqrt2, (5pi)/4)# (radians) or #(4sqrt2, 225^@)# (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((-4)^2 + (-4)^2)#

#r = sqrt((16)+16)#

#r = sqrt(32)#

#r = sqrt(16*2)#

#r = 4sqrt2#

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

#tantheta = (-4)/(-4)#

#tantheta = 1#

#theta = tan^-1(1)#

#theta = (pi)/4# or #(5pi)/4#

To determine which one it is, we have to look at our coordinate #(-4, -4)#. 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 = (5pi)/4#.

From #r# and #theta#, we can write our polar coordinate:
#(4sqrt2, (5pi)/4)# (radians) or #(4sqrt2, 225^@)# (degrees)

Hope this helps!