### 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 coordinates to polar coordinates, we use the formulas $r=\sqrt{x^2+y^2}$ and $\theta=\arctan(\frac{y}{x})$
Where $\arctan$ is the inverse tangent function.
Right. So in this case, $r=\sqrt{(\sqrt{2})^2+(-\sqrt{2})^2}=\sqrt{2+2}=\sqrt{4}=2$.
And $\theta=\arctan(\frac{-\sqrt{2}}{\sqrt{2}})=\arctan(-1)$.
Now we just need to find the angle $\theta$ that has a tangent of $-1$.
That would be $\frac{3\pi}{4}$.