Math  /  Geometry

Question4. Convert the rectangular coordinates to polar coordinates with r>0r>0 and 0θ<2π0 \leq \theta<2 \pi. a. (1,1)(-1,1) b. (33,3)(3 \sqrt{3},-3)

Studdy Solution
Convert the rectangular coordinates (33,3)(3\sqrt{3}, -3) to polar coordinates.
Calculate the radius rr:
r=(33)2+(3)2=27+9=36=6 r = \sqrt{(3\sqrt{3})^2 + (-3)^2} = \sqrt{27 + 9} = \sqrt{36} = 6
Calculate the angle θ\theta:
θ=tan1(333)=tan1(13)=π6 \theta = \tan^{-1}\left(\frac{-3}{3\sqrt{3}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}
Since the point (33,3)(3\sqrt{3}, -3) is in the fourth quadrant, adjust θ\theta:
θ=2ππ6=11π6 \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}
Polar coordinates: (6,11π6)(6, \frac{11\pi}{6})
The polar coordinates are: a. (2,3π4)(\sqrt{2}, \frac{3\pi}{4}) b. (6,11π6)(6, \frac{11\pi}{6})

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