Problem 35 - Algebra and Trigonometry

Question

In Exercises 29–44, graph two periods of the given cosecant or secant function. $$ y=\sec \frac{x}{3} $$

Step-by-Step Solution

Answer

The function \(y=\sec\frac{x}{3}\) is graphed over the range \(x = 0\) to \(x = 12\pi\) showing two periods of the function. There are seven points at \(x = 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi, 12\pi\), and six asymptotes between each pair of consecutive points.

Step 1: Understanding the Secant function

Secant is the reciprocal of the cosine function which means \(\sec(x) = 1 / \cos(x)\). Therefore, wherever \(\cos(x)\) is zero, \(\sec(x)\) has asymptotes, vertical lines that the graph approaches but never crosses.

Step 2: Understanding the effect of \(x/3\)

The argument of the secant function is \(x/3\) rather than \(x\). This stretches the function horizontally by a factor of 3. It means the regular period of \(2\pi\) of the secant function will become \(6\pi\). Thus, for this exercise, two periods of the function will cover the range from \(x = 0\) to \(x = 12\pi\).

Step 3: Determining the points and asymptotes

To graph the function, we also need the points where \(\sec(x/3)\) intersects the \(x\)-axis. This happens when \(\cos(x/3) = \pm1\). Solving for \(x\), we find \(x = 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi, 12\pi\). Between every two of these consecutive points, there is a range where \(\cos(x/3) = 0\). This will be an asymptote for our \(\sec(x/3)\) function.

Step 4: Plotting the graph

Plot the points and asymptotes on the graph. Then sketch the graph, keeping in mind that it has to approach but never cross the asymptote. The graph will have a 'U' shape between each pair of asymptotes.

Key Concepts

Graphing Trigonometric Functions

When graphing trigonometric functions like the secant function, it's helpful to remember that these graphs are based on the simpler sine and cosine functions. For the secant function, remember that it is the reciprocal of the cosine function. Therefore, places where the cosine function equals zero are crucial because the secant function approaches infinity there, creating vertical asymptotes.

To graph the secant function, start by identifying the cosine function graph. Then, find the points where the cosine graph touches its maximum and minimum, as these will correspond to the maximum and minimum points of the secant graph. Between these points, draw U-shaped curves open upwards, which follow the peaks and valleys of the cosine.

• The cosine graph’s crests correspond to the secant graph’s minimum points.
• Similarly, the troughs of the cosine graph correspond to the secant’s maximum points.

This reciprocal relationship is key in understanding how to create the graph for the secant function from the cosine graph, drawing lines that approach the asymptotes without ever crossing them.

Period of Trigonometric Functions

Trigonometric functions like the secant function have periods that define how often the function repeats its basic shape. The basic secant function, derived from the cosine function, has a standard period of \(2\pi\). This means the shape repeats itself every \(2\pi\) units along the x-axis.

In this exercise, the secant function is modified as \(y=\sec \frac{x}{3}\). Here, the \(\frac{x}{3}\) inside the function changes its period from \(2\pi\) to \(6\pi\). This is due to the horizontal scaling introduced by the division: the x-values take longer to impact the y-values, effectively stretching the graph horizontally.

• The original period of \(2\pi\) is multiplied by the factor inside the argument (which is \(3\) in this case).
• This results in a new period of \(6\pi\).
• For this exercise, two complete periods will cover from \(x = 0\) to \(x = 12\pi\).

Understanding how changing the argument inside the function affects the period is crucial for correctly graphing trigonometric functions with adjustments.

Trigonometric Asymptotes

Asymptotes are crucial when dealing with trigonometric functions like the secant and cosecant since they denote the values x where the function goes to infinity. For the secant function, vertical asymptotes occur where the cosine function equals zero because the secant is defined as \(1/\cos(x)\), and division by zero is undefined.

In the case of the function \(y=\sec \frac{x}{3}\), these asymptotes occur at points where \(\cos(\frac{x}{3}) = 0\). The values of x that satisfy this are solutions to the equation \(\frac{x}{3} = (2n+1)\cdot\frac{\pi}{2}\), where \(n\) is an integer. Solving this gives the specific x-values:

• \(x = \frac{3\pi}{2}, \frac{9\pi}{2}, \frac{15\pi}{2}, \ldots\)

Each of these values marks a vertical asymptote, where the secant function reaches infinity. Between these asymptotes, the secant function will pass through maxima and minima, creating the typical wave pattern. Recognizing these asymptotes is essential for plotting the secant function accurately, giving it its distinctive repeated U and inverted U shapes.