Answer :
If you have these conditions:
1. (r ,theta) where r > 0
2. (r, theta) where r < 0
The solution would be:
r = sqrt(x^2 + y^2)
t = arctan(y/x)
r = sqrt(12 + 4) = sqrt(16) = +/- 4
t = arctan(2 / -2sqrt(3)) = arctan(-1 / sqrt(3)) = 5pi/6 , 11pi/6
1)
r > 0
(4 , 11pi/6)
2)
r < 0
(-4 , 5pi/6)
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1. (r ,theta) where r > 0
2. (r, theta) where r < 0
The solution would be:
r = sqrt(x^2 + y^2)
t = arctan(y/x)
r = sqrt(12 + 4) = sqrt(16) = +/- 4
t = arctan(2 / -2sqrt(3)) = arctan(-1 / sqrt(3)) = 5pi/6 , 11pi/6
1)
r > 0
(4 , 11pi/6)
2)
r < 0
(-4 , 5pi/6)
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!