Let's solve each part step-by-step. Recall the conversion formulas: - \( r = \sqrt{x^2 + y^2} \) - \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \) (Adjust \(\theta\) based on the quadrant.) --- ### (a) \( (3\sqrt{2},\ 3\sqrt{2}) \) **Step 1:** Find \( r \) \[ r = \sqrt{(3\sqrt{2})^2 + (3\sqrt{2})^2} = \sqrt{18 + 18} = \sqrt{36} = 6 \] **Step 2:** Find \( \theta \) \[ \theta = \tan^{-1}\left(\frac{3\sqrt{2}}{3\sqrt{2}}\right) = \tan^{-1}(1) = \frac{\pi}{4} \] (Both coordinates are positive, so it's in the first quadrant.) **Final Answer:** \[ \boxed{(6,\ \frac{\pi}{4})} \] --- ### (b) \( (-4\sqrt{2},\ 4\sqrt{2}) \) **Step 1:** Find \( r \) \[ r = \sqrt{(-4\sqrt{2})^2 + (4\sqrt{2})^2} = \sqrt{32 + 32} = \sqrt{64} = 8 \] **Step 2:** Find \( \theta \) \[ \theta = \tan^{-1}\left(\frac{4\sqrt{2}}{-4\sqrt{2}}\right) = \tan^{-1}(-1) = -\frac{\pi}{4} \] But the point is in the **second quadrant** (x negative, y positive), so add \(\pi\): \[ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \] **Final Answer:** \[ \boxed{(8,\ \frac{3\pi}{4})} \] --- ### (c) \( (-3,\ -3\sqrt{3}) \) **Step 1:** Find \( r \) \[ r = \sqrt{(-3)^2 + (-3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6 \] **Step 2:** Find \( \theta \) \[ \theta = \tan^{-1}\left(\frac{-3\sqrt{3}}{-3}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \] But the point is in the **third quadrant** (x negative, y negative), so add \(\pi\): \[ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \] **Final Answer:** \[ \boxed{(6,\ \frac{4\pi}{3})} \] --- ## **Summary of Answers** - (a) \(\boxed{(6,\ \frac{\pi}{4})}\) - (b) \(\boxed{(8,\ \frac{3\pi}{4})}\) - (c) \(\boxed{(6,\ \frac{4\pi}{3})}\)