### Instruction: Convert the point $(\sqrt{2},-\sqrt{2})$ 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 $\theta = \tan^{-1}(y/x).$
Plugging in the given values of $x$ and $y$, I get $r = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2.$
To find $\theta$, I need to be careful about the quadrant of the point.
Since $x > 0$ and $y < 0$, the point is in the fourth quadrant, so $\theta$ should be between $\frac{3}{2}\pi$ and $2\pi.$
Using the formula, I get $\theta = \tan^{-1}(-\sqrt{2}/\sqrt{2}) = \tan^{-1}(-1).$
Now, I know that $\tan^{-1}(-1)$ has two possible values, one in the second quadrant and one in the fourth quadrant.
The second quadrant value is $\frac{3}{4}\pi$, but that does not match the quadrant of the point, so I reject it.
The fourth quadrant value is $\frac{7}{4}\pi$, and that is consistent with the quadrant of the point, so I accept it.
Therefore, the polar coordinates of the point are $(2,\frac{7}{4}\pi).$