How do you convert #y = (pi/6)x# into polar form?

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How do you convert #y = (pi/6)x# into polar form?

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Answer 1 · 2016-04-09T05:51:54

#theta = tan^(-1)(pi/6) and theta = pi+tan^(-1)(pi/6)#, representing radial lines, in opposite directions for the complete line #y=(pi/6)x#, through the origin, with slope #pi/6#.

Explanation:

#x=r cos theta and y=r sin theta#.
The polar equation is #tan theta = pi/6#.

Explicitly, #theta# is as stated in the answer.

The half lines reach out, from r = 0, in opposite directions..

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How do you convert #y = (pi/6)x# into polar form?

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