### Instruction: Convert the point $(6,2 \sqrt{3})$ in rectangular coordinates to polar coordinates.  Enter your answer in the form $(r,\theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$.### Response: To convert from rectangular to polar coordinates, I need to find the distance from the origin and the angle with the positive x-axis.
The distance from the origin is the hypotenuse of a right triangle with legs $6$ and $2 \sqrt{3},$ so by the Pythagorean theorem, it is $\sqrt{6^2 + (2 \sqrt{3})^2} = \sqrt{36 + 12} = \sqrt{48} = 4 \sqrt{3}.$
The angle with the positive x-axis is given by the inverse tangent of the ratio of the y-coordinate to the x-coordinate, so it is $\tan^{-1} \left( \frac{2 \sqrt{3}}{6} \right).$
However, I need to be careful about the quadrant of the point, since the inverse tangent function only returns values between $-\frac{\pi}{2}$ and $\frac{\pi}{2}.$
Since the point is in the first quadrant, where both x and y are positive, the angle is positive and in the correct range, so I don't need to adjust it.
Therefore, the polar coordinates are $(4 \sqrt{3}, \tan^{-1} \left( \frac{2 \sqrt{3}}{6} \right)).$