Let's work through the problem step by step.

Part (a) - Cartesian coordinates (9, -9)

We are tasked with converting the Cartesian coordinates $(9, -9)$ to polar coordinates $(r, \theta)$, where $r > 0$ and $0 \leq \theta < 2\pi$.

The polar coordinate transformation is given by:

$r^2 = x^2 + y^2$ $\tan(\theta) = \frac{y}{x}$

Here, $x = 9$ and $y = -9$.

Step 1: Finding $r$

$r^2 = 9^2 + (-9)^2 = 81 + 81 = 162$ $r = \sqrt{162} = 9\sqrt{2}$

So, $r = 9\sqrt{2}$.

Step 2: Finding $\theta$

$\tan(\theta) = \frac{-9}{9} = -1$

Now, we need to find $\theta$ such that $\tan(\theta) = -1$. The point $(9, -9)$ is in the fourth quadrant (since $x > 0$ and $y < 0$), and $\tan(\theta) = -1$ corresponds to an angle of $\frac{7\pi}{4}$ radians in the fourth quadrant.

So, $\theta = \frac{7\pi}{4}$.

Final Answer for Part (a)(i):

The polar coordinates for the point $(9, -9)$ are:

$r = 9\sqrt{2}, \quad \theta = \frac{7\pi}{4}$

Would you like me to proceed with part (ii) and the rest of the problem, or do you have any questions about this part?

Here are some questions for further understanding:

1. What is the geometric interpretation of polar coordinates?
2. How can you determine the quadrant of an angle in polar coordinates?
3. What are some common applications of polar coordinates in physics or engineering?
4. How would you calculate the distance between two points in polar coordinates?
5. What happens when $r < 0$ in polar coordinates?

Tip: When converting from Cartesian to polar coordinates, always consider the quadrant of the point to determine the correct angle $\theta$.