### 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 use the formulas $r = \sqrt{x^2 + y^2}$ and $\theta = \tan^{-1}(y/x),$ where $\theta$ is the angle measured counterclockwise from the positive $x$-axis.
In this problem, $x = 6$ and $y = 2 \sqrt{3},$ so I can plug these values into the formulas and simplify.
For $r,$ I get $r = \sqrt{6^2 + (2 \sqrt{3})^2} = \sqrt{36 + 12} = \sqrt{48} = 4 \sqrt{3}.$
For $\theta,$ I get $\theta = \tan^{-1}(2 \sqrt{3} / 6) = \tan^{-1}(\sqrt{3} / 3).$
I know that $\tan^{-1}(\sqrt{3} / 3)$ is the same as the angle whose tangent is $\sqrt{3} / 3,$ which is one of the special angles that I have memorized from the unit circle.
It is equal to $\pi / 6.$
However, I also need to check the quadrant of the point $(6,2 \sqrt{3})$ to make sure that $\theta$ is in the correct range.
Since $x$ and $y$ are both positive, the point is in the first quadrant, so $\theta$ is between $0$ and $\pi / 2.$
Therefore, $\theta = \pi / 6$ is a valid choice and I don't need to adjust it by adding or subtracting $\pi.$