The polar coordinates of the point whose Cart | vedclass

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Question

(A) We know that $r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$.Since the point $(-2, -2)$ lies in the $III$ q

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$(2 \sqrt{2}, \frac{5 \pi}{4})$

Vedclass · Answer

(A) We know that $r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$.Since the point $(-2, -2)$ lies in the $III$ quadrant,the angle $	heta$ is given by $	an 	heta = \frac{y}{x} = \frac{-2}{-2} = 1$.Since the point is in the $III$ quadrant,$	heta = \pi + 	an^{-1}(1) = \pi + \frac{\pi}{4} = \frac{5 \pi}{4}$.Therefore,the polar coordinates $(r, 	heta)$ are $(2 \sqrt{2}, \frac{5 \pi}{4})$.

The polar coordinates of the point whose Cartesian coordinates are $(-2, -2)$ are given by

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