Problem 34 - Precalculus

Question

In Exercises \(29-46,\) graph the functions over the indicated intervals. $$y=\frac{1}{2} \csc \left(\frac{\pi}{3} x\right),-6 \leq x \leq 6$$

Step-by-Step Solution

Answer

Sketch the csc curve with asymptotes at every 3 units, repeating every 6 units.

Step 1: Understanding the Function

The function given is \(y=\frac{1}{2} \csc \left(\frac{\pi}{3} x\right)\). The cosecant function is defined as \(\csc(x) = \frac{1}{\sin(x)}\), where \(\sin(x)\) cannot be zero. Thus, the function will have vertical asymptotes where the sine function is zero.

Step 2: Determine the Period

The period of the cosecant function \(\csc(x)\) is derived from the sine function, which is \(2\pi\). Given the argument \(\frac{\pi}{3} x\), the period becomes \(\frac{2\pi}{\frac{\pi}{3}} = 6\). Therefore, the function will repeat every 6 units along the x-axis.

Step 3: Identify the Asymptotes

Sine is zero at integer multiples of \(\pi\), i.e., \(x = n\pi\). Solving for \(x\) in \(\frac{\pi}{3} x = n \pi\) gives \(x = 3n\). Therefore, vertical asymptotes occur at \(x = 3n\), where \(n\) is an integer. Specifically, in the interval \(-6 \leq x \leq 6\), they occur at \(x = -6, -3, 0, 3, 6\).

Step 4: Calculate Key Points

Choose points between the asymptotes to sketch the graph. For example, to find values of \(y\), input \(x = -4.5, -1.5, 1.5, 4.5\): - At \(x = -4.5, y = \frac{1}{2\sin(-\frac{\pi}{2})} = -1/2\)- At \(x = -1.5, y = \frac{1}{2\sin(-\frac{\pi}{2})} = -1/2\)- At \(x = 1.5, y = \frac{1}{2\sin(\frac{\pi}{2})} = 1/2\)- At \(x = 4.5, y = \frac{1}{2\sin(\frac{\pi}{2})} = 1/2\)

Step 5: Sketch the Graph

Now that you have points and asymptotes, draw the graph:1. Plot vertical asymptotes at \(x = -6, -3, 0, 3, 6\).2. Plot the points calculated between the asymptotes.3. Sketch the curve approaching the vertical asymptotes and passing through the points determined, showing periodic cycles from 0 to 6 and likewise from -6 to 0.

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, particularly in understanding oscillations and waves. The cosecant function, denoted as \( \csc(x) \), is the reciprocal of the sine function. That is,

• \( \csc(x) = \frac{1}{\sin(x)} \)

The cosecant function is undefined at any point where the sine function equals zero because division by zero is undefined. In terms of graphing, this is represented by vertical asymptotes, making the graph appear as disconnected branches.
In the exercise problem, the modified cosecant function is \( y = \frac{1}{2} \csc \left(\frac{\pi}{3} x\right) \). The scaling factor (\( \frac{1}{2} \)) will affect the amplitude, shrinking it vertically, but not changing the period or the location of asymptotes. Remember, the basic shape of the cosecant graph resembles repeating arcs that branch outwards from its asymptotes.

Periodicity

Periodicity is a key feature of trigonometric functions. This property refers to the repeating nature of the function values over a specific interval, called the period. For the basic sine and cosecant functions, this period is \( 2\pi \).
In our exercise, due to the transformation applied by \( \frac{\pi}{3} x \), the period of the modified cosecant function becomes \( \frac{2\pi}{\frac{\pi}{3}} = 6 \).

• This calculation shows that the function repeats itself every 6 units along the x-axis.

Understanding periodicity helps anticipate where the function's values will repeat on the graph and ensures accurate plotting for extended intervals. Knowing that the graph will complete a full cycle from one period to the next allows you to draw it with the confidence that beyond this, the pattern simply continues.

Vertical Asymptotes

Vertical asymptotes are lines where the function approaches but never actually reaches a specific x-value. For the cosecant function, these occur at the zeros of the sine function, because wherever sine is zero, cosecant is undefined.
By setting the sine's argument to zero, we find these points more generally as multiples of \( \pi \). In the exercise’s context, we solve for where \( \frac{\pi}{3} x = n\pi \). Solving this gives:

• \( x = 3n \)

Within the interval \( -6 \leq x \leq 6 \), vertical asymptotes appear at \( x = -6, -3, 0, 3, 6 \). These lines effectively divide the x-axis into segments where the function is defined and can be sketched.
Graphically, these asymptotes serve as the boundaries within which the cosecant function can be plotted, making them critical for accurately representing the graph’s behavior.