Examples

Converting to polar: pi-correction

Compute the polar coordinates of $(-\frac{1}{2},+\frac{\sqrt{3}}{2})$ and of $(+\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$.

Solution

For $(-\frac{1}{2},\frac{\sqrt{3}}{2})$ we observe first that it lies in Quadrant II.

Next compute:

$${\tan }^{-1}\left(\frac{\sqrt{3}/2}{-1/2}\right)\gg \gg {\tan }^{-1}(-\sqrt{3})\gg \gg -\pi /3$$

This angle is in Quadrant IV. We add $\pi$ to get the polar angle in Quadrant II:

$$\theta =\pi -\pi /3\gg \gg 2\pi /3$$

The radius is of course $1$ since this point lies on the unit circle. Therefore polar coordinates are $(r,\theta )=(\mathrm{1,2}\pi /3)$.

For $(+\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$ we observe first that it lies in Quadrant IV. (No extra $\pi$ needed.)

Next compute:

$${\tan }^{-1}\left(\frac{-\sqrt{2}/2}{+\sqrt{2}/2}\right)\gg \gg {\tan }^{-1}(-1)\gg \gg -\pi /4$$

So the point in polar is $(1,-\pi /4)$.

Shifted circle in polar

For example, let’s convert a shifted circle to polar. Say we have the Cartesian equation:

$$x^2+{(y-3)}^2=9$$

Then to find the polar we substitute $x=r\cos \theta$ and $y=r\sin \theta$ and simplify:

$$\begin{array}{c}{x}^2+{(y-3)}^2=9\\ \\ \gg \gg r^2{\cos }^2\theta +{(r\sin \theta -3)}^2=9\\ \\ \gg \gg r^2{\cos }^2\theta +r^2{\sin }^2\theta -6r\sin \theta +9=9\\ \\ \gg \gg r^2({\sin }^2\theta +{\cos }^2\theta )-6r\sin \theta =0\\ \\ \gg \gg r^2-6r\sin \theta =0\gg \gg r=6\sin \theta \end{array}$$

So this shifted circle is the polar graph of the polar function $r(\theta )=6\sin \theta$.