What is the polar form of #(-3,3)#?

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Answer 1 · 2016-10-16T19:00:25

From Cartesian form to polar form:

#(-3,3)-> (3sqrt(2), (3pi)/4)#

Because the x coordinate is negative, one must add #pi# to the inverse tangent function to obtain the correct angular component:

#theta = tan^-1(3/-3) + pi#

#theta = (3pi)/4#

The radial component is:

#r = sqrt((-3)^2 + 3^2)#

#r = sqrt(18) = 3sqrt(2)#