Convert the rectangular coordinates (3sqrt(3), -3) into polar form. Express the ang...

1. Identify the rectangular coordinates: The given rectangular coordinates are $(x, y) = (3\sqrt{3}, -3)$.
2. Calculate the radial distance (r): The formula for the radial distance $r$ in polar coordinates is $r = \sqrt{x^2 + y^2}$.
Substitute the given values of $x$ and $y$ into the formula:
$$r = \sqrt{(3\sqrt{3})^2 + (-3)^2}$$
$$r = \sqrt{(9 \times 3) + 9}$$
$$r = \sqrt{27 + 9}$$
$$r = \sqrt{36}$$
$$r = 6$$
3. Calculate the angle (θ): The formula for the angle $\theta$ is $\tan(\theta) = \frac{y}{x}$.
Substitute the given values of $x$ and $y$ into the formula:
$$\tan(\theta) = \frac{-3}{3\sqrt{3}}$$
$$\tan(\theta) = -\frac{1}{\sqrt{3}}$$
To find $\theta$, we take the arctangent of $-\frac{1}{\sqrt{3}}$. The reference angle for $\tan(\theta) = \frac{1}{\sqrt{3}}$ is $\frac{\pi}{6}$.
Since $x$ is positive ($3\sqrt{3}$) and $y$ is negative ($-3$), the point lies in the fourth quadrant. The angle in the fourth quadrant with a reference angle of $\frac{\pi}{6}$ is $2\pi - \frac{\pi}{6} = \frac{12\pi - \pi}{6} = \frac{11\pi}{6}$.
Alternatively, using the atan2(y, x) function, $\theta = \text{atan2}(-3, 3\sqrt{3}) = -\frac{\pi}{6}$.
The problem requires the angle to be in the interval $0 \le \theta < 2\pi$. To convert $-\frac{\pi}{6}$ to this interval, we add $2\pi$:
$$ \theta = -\frac{\pi}{6} + 2\pi = -\frac{\pi}{6} + \frac{12\pi}{6} = \frac{11\pi}{6} $$
4. Express the polar coordinates: The polar coordinates are $(r, \theta)$.
So, the polar form is $(6, \frac{11\pi}{6})$.