Converting Cartesian coordinates to polar coordinates

To convert the given rectangular coordinates \((-3 \sqrt{2}, -3 \sqrt{2})\) into polar form, we need to find the radius \(r\) and the angle \(\theta\). 1. **Calculate the radius \(r\):** The radius \(r\) is given by the distance from the origin to the point, which can be found using the Pythagorean theorem: \[ r = \sqrt{x^2 + y^2} \] Given \(x = -3\sqrt{2}\) and \(y = -3\sqrt{2}\): \[ r = \sqrt{(-3\sqrt{2})^2 + (-3\sqrt{2})^2} \] Simplify the squares: \[ r = \sqrt{(9 \cdot 2) + (9 \cdot 2)} \] \[ r = \sqrt{18 + 18} \] \[ r = \sqrt{36} \] \[ r = 6 \] 2. **Calculate the angle \(\theta\):** The angle \(\theta\) is given by the arctangent of the ratio of \(y\) to \(x\): \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \] Substitute \(x = -3\sqrt{2}\) and \(y = -3\sqrt{2}\): \[ \theta = \tan^{-1}\left(\frac{-3\sqrt{2}}{-3\sqrt{2}}\right) \] Simplify the ratio: \[ \theta = \tan^{-1}(1) \] The arctangent of 1 is \(\frac{\pi}{4}\), but we need to consider the correct quadrant. Since both \(x\) and \(y\) are negative, the point is in the third quadrant. In the third quadrant, the angle is: \[ \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \] Therefore, the polar coordinates are: \[ (r, \theta) = (6, \frac{5\pi}{4}) \]