Question

In Exercises 24–31 find all polar coordinate representations for the point given in rectangular coordinates.

$\left(6,2\sqrt{3}\right)$

Short answer

The polar coordinates are $\left(4\sqrt{3},\frac{\mathrm{\pi }}{6}+2\mathrm{k\pi }\right)\mathrm{and}\left(-4\sqrt{3},\frac{7\mathrm{\pi }}{6}+2\mathrm{k\pi }\right)$, for any integer k.

Step-by-step solution

Step 1. Given information.

The given rectangular coordinate is:

$\left(6,2\sqrt{3}\right)$

Here$x=6\mathrm{and}y=2\sqrt{3}$.

Step 2. Find the value of r.

To find the value of r, use the formula $r=\sqrt{x^2+y^2}$:

$$\begin{gathered} r=\sqrt{{\left(6\right)}^2+{\left(2\sqrt{3}\right)}^2} \\ r=\sqrt{36+12} \\ r=\sqrt{48} \\ r=\pm 4\sqrt{3} \end{gathered}$$

Step 3. Find the value of θ.

Use the formula $\tan \theta =\frac{y}{x}$,

$$\begin{gathered} \tan \theta =\frac{y}{x};\mathrm{then}\mathrm{\theta }={\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right) \\ \mathrm{\theta }={\tan }^{-1}\left(\frac{2\sqrt{3}}{6}\right) \\ \mathrm{\theta }={\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right);\mathrm{\theta }={\tan }^{-1}\left(\frac{\sqrt{3}}{\sqrt{3}·\sqrt{3}}\right) \\ \mathrm{\theta }={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ \mathrm{\theta }=\frac{\mathrm{\pi }}{6},\frac{7\mathrm{\pi }}{6} \end{gathered}$$

Step 4. Find the polar coordinates.

Take $r=4\sqrt{3},\mathrm{\theta }=\frac{\mathrm{\pi }}{6};\mathrm{then}\left(\mathrm{r},\mathrm{\theta }\right)=\left(4\sqrt{3},\frac{\mathrm{\pi }}{6}\right)$,

The coordinates of $\left(\mathrm{r},\mathrm{\theta }\right)\mathrm{are}\left(4\sqrt{3},\frac{\mathrm{\pi }}{6}+2\mathrm{k\pi }\right)$, for any integer k.

Now take $\mathrm{r}=-4\sqrt{3},\mathrm{\theta }=\frac{7\mathrm{\pi }}{6};\mathrm{then}\left(\mathrm{r},\mathrm{\theta }\right)=\left(-4\sqrt{3},\frac{7\mathrm{\pi }}{6}\right)$,

The coordinates of $\left(\mathrm{r},\mathrm{\theta }\right)\mathrm{are}\left(-4\sqrt{3},\frac{7\mathrm{\pi }}{6}+2\mathrm{k\pi }\right)$, for any integer k.

Therefore, all the polar coordinates are $\left(4\sqrt{3},\frac{\mathrm{\pi }}{6}+2\mathrm{k\pi }\right)\mathrm{and}\left(-4\sqrt{3},\frac{7\mathrm{\pi }}{6}+2\mathrm{k\pi }\right)$, for any integer k.