Worksheet for plotting polar coordinates with four exercises.

Here are the step-by-step solutions for each problem on the worksheet.

1. Give the polar coordinate of the point with $r > 0$.
* Find $r$ (radius): Look at the concentric circles. The red dot is on the first circle out from the center. So, $r = 1$.
* Find $\theta$ (angle): The angle lines are spaced out by $\frac{\pi}{6}$ (30 degrees). The positive y-axis is $\frac{\pi}{2}$. The line just before it (clockwise) is $\frac{\pi}{3}$. The line before that is $\frac{\pi}{6}$. The red dot is on the negative y-axis, which corresponds to the angle $\frac{3\pi}{2}$.
* Coordinate: $(1, \frac{3\pi}{2})$

2. Give the polar coordinate of the point with $r > 0$.
* Find $r$ (radius): Count the rings from the center. The dot is on the 5th ring. So, $r = 5$.
* Find $\theta$ (angle): The point is in the second quadrant (top-left). The vertical axis is $\frac{\pi}{2}$ and the horizontal axis is $\pi$. There are two grid lines between them, dividing that 90-degree section into three parts. Each part is $\frac{\pi}{6}$.
* Start at $\frac{\pi}{2}$ and add $\frac{\pi}{6}$: $\frac{3\pi}{6} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$.
* Add another $\frac{\pi}{6}$: $\frac{4\pi}{6} + \frac{\pi}{6} = \frac{5\pi}{6}$.
* The dot lies on this second line. So, $\theta = \frac{5\pi}{6}$.
* Coordinate: $(5, \frac{5\pi}{6})$

3. Give the polar coordinate of the point with $r > 0$.
* Find $r$ (radius): Count the rings from the center. The dot is on the 4th ring. So, $r = 4$.
* Find $\theta$ (angle): The point is in the first quadrant (top-right). It lies on the grid line exactly halfway between the x-axis ($0$) and the y-axis ($\frac{\pi}{2}$).
* Half of $\frac{\pi}{2}$ is $\frac{\pi}{4}$.
* Coordinate: $(4, \frac{\pi}{4})$

4. Identify the correct polar point.
* Analyze the graph:
* Radius ($r$): The red dot is on the 5th ring. This means the magnitude of the radius is 5.
* Angle ($\theta$): The dot is in the second quadrant. As calculated in Problem 2, this angle is $\frac{5\pi}{6}$.
* So, a standard positive representation is $(5, \frac{5\pi}{6})$. However, looking at the options, they all involve negative numbers or different angles. We need to find an equivalent coordinate.
* Check the options:
* Recall that a point $(-r, \theta)$ is the same as $(r, \theta + \pi)$. Essentially, if you have a negative radius, you go in the opposite direction through the origin.
* Let's test the option $(-5, \frac{11\pi}{6})$.
* Angle $\frac{11\pi}{6}$ is in the 4th quadrant (just above the positive x-axis).
* A negative radius of $-5$ means we go backwards through the origin, into the opposite quadrant (2nd quadrant).
* Does this match our point? Yes. The point is in the 2nd quadrant at distance 5.
* Let's double-check the math:
* Point $(-5, \frac{11\pi}{6})$ is equivalent to $(5, \frac{11\pi}{6} - \pi)$.
* $\frac{11\pi}{6} - \frac{6\pi}{6} = \frac{5\pi}{6}$.
* This matches our visual analysis perfectly.

Final Answer:
1. $(1, \frac{3\pi}{2})$
2. $(5, \frac{5\pi}{6})$
3. $(4, \frac{\pi}{4})$
4. $(-5, \frac{11\pi}{6})$