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# **Title:** Converting Rectangular Coordinates to Polar Form

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Question: Convert the rectangular coordinates left parenthesis, minus, 3, square root of, 2, end square root, comma, minus, 3, square root of, 2, end square root, right parenthesis(−3 2 ​ ,−3 2 ​ ) into polar form. Express the angle using radians in terms of piπ over the interval 0, is less than or equal to, theta, is less than, 2, pi0≤θ<2π, with a positive value of rr.

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To convert rectangular coordinates $$(-3\sqrt{2}, -3\sqrt{2})$$ into polar coordinates $$(r, \theta)$$, you need to determine both the radius $$r$$ and the angle $$\theta$$. 1. **Calculate the radius $$r$$:** The radius $$r$$ is the distance from the origin to the point $$(-3\sqrt{2}, -3\sqrt{2})$$. It can be calculated using the Pythagorean theorem: $r = \sqrt{(x^2 + y^2)}$ Substitute $$x = -3\sqrt{2}$$ and $$y = -3\sqrt{2}$$: $r = \sqrt{(-3\sqrt{2})^2 + (-3\sqrt{2})^2}$ $r = \sqrt{(9 \cdot 2) + (9 \cdot 2)}$ $r = \sqrt{18 + 18}$ $r = \sqrt{36}$ $r = 6$ So, $$r = 6$$. 2. **Calculate the angle $$\theta$$:** The angle $$\theta$$ can be found using the arctangent function. For Cartesian coordinates $$(x, y)$$: $\theta = \tan^{-1}\left(\frac{y}{x}\right)$ Substitute $$x = -3\sqrt{2}$$ and $$y = -3\sqrt{2}$$: $\theta = \tan^{-1}\left(\frac{-3\sqrt{2}}{-3\sqrt{2}}\right)$ $\theta = \tan^{-1}(1)$ Since both coordinates are negative, the point is located in the third quadrant. Thus, the reference angle is $$\frac{3\pi}{4}$$, but the actual angle in the third quadrant should be: $\theta = \pi + \frac{3\pi}{4}$ $\theta = \frac{4\pi}{4} + \frac{3\pi}{4}$ $\theta = \frac{7\pi}{4}$ So, the angle $$\theta = \frac{5\pi}{4}$$ (since we should not exceed $$2\pi$$). Therefore, the polar coordinates are: $\boxed{(6, \frac{5\pi}{4})}$

May 30, 2024, 6:56 a.m.

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