### 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 $r$ from the origin and the angle $\theta$ with the positive $x$-axis.
To find $r$, I can use the Pythagorean theorem: $r^2 = x^2 + y^2 = 6^2 + (2 \sqrt{3})^2 = 36 + 12 = 48.$
To find $\theta$, I can use the inverse tangent function: $\tan \theta = \frac{y}{x} = \frac{2 \sqrt{3}}{6} = \frac{\sqrt{3}}{3}.$
However, I have to be careful with the inverse tangent function, because it only gives me an angle in the first or fourth quadrant, and I need to check which quadrant the point is in.
Since $x$ and $y$ are both positive, the point is in the first quadrant, so the inverse tangent function gives me the correct angle.
Therefore, $\theta = \tan^{-1} \left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}.$
So, the polar coordinates of the point are $(\sqrt{48}, \frac{\pi}{6}).$