Let's go through the solution based on the given polar coordinates $(3, \frac{5\pi}{4})$ and the instructions.

1. Plotting the Point

The point $(3, \frac{5\pi}{4})$ in polar coordinates is given by:

• $r = 3$: This is the radial distance from the origin.
• $\theta = \frac{5\pi}{4}$: This is the angle measured counterclockwise from the positive $x$-axis.

To plot this point:

• Start at the origin and move 3 units outward.
• Locate the angle $\frac{5\pi}{4}$, which is in the third quadrant (it’s slightly past $\pi$, or 180°).
• Place the point 3 units out in this direction.

2. Finding Additional Polar Coordinates

Condition (a): $r > 0$ and $-2\pi \leq \theta < 0$

For a positive radius ($r = 3$) and a negative angle, we can find an equivalent angle by subtracting $2\pi$ from $\frac{5\pi}{4}$: $\theta = \frac{5\pi}{4} - 2\pi = \frac{5\pi}{4} - \frac{8\pi}{4} = -\frac{3\pi}{4}$ Thus, one equivalent set of coordinates is: $\left( 3, -\frac{3\pi}{4} \right)$

Condition (b): $r < 0$ and $0 \leq \theta < 2\pi$

For a negative radius, we need to reverse the direction. So, $r = -3$. To keep the point in the same location, add $\pi$ to the angle $\frac{5\pi}{4}$: $\theta = \frac{5\pi}{4} + \pi = \frac{5\pi}{4} + \frac{4\pi}{4} = \frac{9\pi}{4}$ Since $\frac{9\pi}{4}$ is greater than $2\pi$, subtract $2\pi$ to bring it back within the range $[0, 2\pi)$: $\frac{9\pi}{4} - 2\pi = \frac{9\pi}{4} - \frac{8\pi}{4} = \frac{\pi}{4}$ Thus, the second set of coordinates is: $\left( -3, \frac{\pi}{4} \right)$

Summary of Solutions:

1. Original point: $(3, \frac{5\pi}{4})$
2. For condition (a): $(3, -\frac{3\pi}{4})$
3. For condition (b): $(-3, \frac{\pi}{4})$

Would you like further clarification on these steps?

Here are five related questions to deepen your understanding:

1. How can we convert any polar coordinate with a positive radius to one with a negative radius?
2. What are the steps for converting angles in polar coordinates between positive and negative values?
3. How can you determine the quadrant of a point using polar coordinates?
4. Why is subtracting $2\pi$ effective in finding equivalent angles?
5. What is the significance of using $\theta = \theta + \pi$ when switching the sign of $r$?

Tip: In polar coordinates, adding or subtracting $2\pi$ to an angle gives an equivalent point with the same position, due to the circular nature of polar coordinates.