Polar Functions and Graphing Guide (Section 3.14)

Introduction to Section 3.14 - Polar Functions

The lecture covers Section 3.14, focused on graphing polar functions. The speaker notes the prevalence of C0 in these equations, jokingly referencing the section number 3.14. Additionally, the speaker associates polar functions with the "Polar Express" character, humorously suggesting the character's enthusiasm for graphing circular coordinate systems.

Graphing Lines in a Polar Coordinate System

Although the system is inherently circular, it is possible to graph lines. For example, to graph the line represented by $\theta = \frac{\pi}{3}$, one must identify the angle $\frac{\pi}{3}$ on the polar grid. The resulting graph is a line passing through the pole that extends infinitely in both directions. The rule is that when $\theta$ is constant, the graph is indeed a line.

Graphing and Analyzing Circles

The polar coordinate system is constructed of concentric circles, making it particularly suitable for graphing circular functions. Consider the example $r = 4\cos(\theta)$. For point plotting, angles are plugged into a table where at $\theta = 0$, we have $r = 4\cos(0) = 4$, marking the starting point or the polar axis intercept. As we proceed with various angles, for instance, at $\theta = \frac{\pi}{6}$, $r$ is approximately 3.46, and at $\theta = \frac{\pi}{4}$, $r$ is approximately 2.83, culminating at $\theta = \frac{\pi}{3}$ with $r = 2$.
 As the angle approaches $\frac{2\pi}{3}$, the cosine becomes negative; for instance, at this angle, $r$ may be $-2$. Thus, instead of moving towards $\frac{2\pi}{3}$, the point would be plotted in the opposite direction, bringing the graph back onto itself. The complete circle cycles through $\pi$, and by $\theta = \pi$, one full trip around the circle is finished, with continuing to $2\pi$ simply retracing the same circle.

In another example, for $r = 4\sin(\theta)$ and point plotting at $\theta = 0$, $r = 0$ marking the origin as the starting point. As $\theta$ increases to $\frac{\pi}{6}$, we find that $r = 2$, and at $\theta = \frac{\pi}{2}$, $r = 4\sin(\frac{\pi}{2}) = 4$, indicating the maximum distance from the pole. As $\theta$ continues to rise from $\frac{\pi}{2}$ to $\pi$, the values gradually decrease back to the pole. Notably, when reaching $\theta = \frac{7\pi}{6}$, the value of $r$ becomes negative (for example, $-2$). In such cases, while facing towards $\frac{7\pi}{6}$, moving $-2$ results in landing back on previously graphed points in the first quadrant, completing the sine circle cycle within $\pi$.

General Formulas and Orientation for Circles

The standard forms for circular equations are $r = a\cos(\theta)$ and $r = a\sin(\theta)$. In these equations, the value $a$ indicates the maximum distance from the pole, essentially representing the diameter of the circle when originating from the pole. The orientation of these graphs is dictated by the sign of $a$: positive cosine graphs open to the right, negative cosine graphs to the left, positive sine graphs open up, and negative sine graphs open down.

Graphing and Analyzing Roses

The general formula for rose graphs is either $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$. Here, the amplitude $a$ determines how far the petals extend from the pole, effectively setting the radius for the boundary circle of the rose. Notably, if $n$ is odd, the graph will feature exactly $n$ petals; however, if $n$ is even, there will be $2n$ petals. Cosine roses initiate from the polar axis at length $a$ when $\theta = 0$, while sine roses start at the origin, with their petals extending away from the polar axis. The cycles for the roses vary: odd roses complete within a cycle of $\pi$, while even roses require a complete cycle of $2\pi$ due to their unique petal count characteristics.

Calculator Techniques and Settings

To ensure a smoother graph, it is essential to adjust calculator settings to prevent jagged outputs. The theta step ($\theta_{\text{step}}$) affects how many angles the calculator plots. A larger theta step (e.g., $\frac{\pi}{3}$) may yield a poorly defined graph, while a smaller step (e.g., $0.005$) creates a more rounded graph. Many calculators also allow for a “pin” mode that enables users to visualize the graph being formed step-by-step, revealing the loops and cycles of the graph. Standard window settings range from $-10$ to $10$ for both $x$ and $y$ coordinates, with $0$ to $2\pi$ for $\theta$; utilizing “Zoom Square” is recommended to maintain proportional graphing.

Practice: Describing Polar Functions

Regarding various practice examples, the function $r = 2\cos(7\theta)$ is identified as an odd rose with 7 petals and a maximum distance of 2, completing a cycle from 0 to $\pi$. In contrast, the function $r = 9\sin(\theta)$ reveals circle properties, given that $n=1$ and opens upwards, hitting a maximum distance of 9 and completing its cycle from 0 to $\pi$. Additionally, $r = 8\cos(6\theta)$ indicates an even rose with 12 petals due to the $2n$ petal rule, also cycling from 0 to $2\pi$.

Practice: Writing Equations from Graphs

Different graph representations provide varied equations; for Graph A, a circle that opens to the right and extends to 2 on the polar axis yields the equation $r = 2\cos(\theta)$. For Graph B, with 4 petals and a maximum distance of 4 (not starting on the polar axis), the equation becomes $r = 4\sin(2\theta)$ because it follows the sine petal characteristics. Lastly, Graph C depicts a circle opening down with an extent of 5, corresponding to the equation $r = -5\sin(\theta)$.

Radius-Only Functions and Endpoints

Constant radius functions, such as $r = 3$, produce a circle centered at the pole, consistently equidistant from it at all angles. Finding specific regions may involve restricted endpoints; for instance, during calculations for $r = 3$ from $\theta = \frac{\pi}{6}$ to $\theta = \frac{\pi}{3}$, the points yield $(3, \frac{\pi}{6})$ and $(3, \frac{\pi}{3})$. Analyzing $r = 4\cos(3\theta)$ from the same angles results in $(0, \frac{\pi}{6})$ and $(4, \frac{\pi}{3})$, demonstrating how negative radius impacts point plotting visually to reveal overlaps with the petal. A practice problem involving $r = 5\sin(\theta)$ from $\theta = 0$ to $\theta = \frac{\pi}{2}$ illustrates a circle that opens upwards, yielding points at $(0, 0)$ and $(5, \frac{\pi}{2})$, highlighting the first quadrant half of the full circle.

Questions & Discussion

Students have engaged in questions, noting, for example, that the sine function begins at the origin due to $\sin(0) = 0$, with the sign indicating whether it opens upwards or downwards, peaking at $\frac{\pi}{2}$. The inquiry about the function $5\sin(4\theta)$ leads to anticipations of 8 petals and the need for a full cycle of $2\pi$ due to its even nature. The speaker encourages students to explore and use their calculators creatively to visualize different cycles and patterns.