Problem 39 - Precalculus with Calculus Previews

Question

Find the period and the vertical asymptotes of the given function. Sketch at least one cycle of the graph. $$ y=3 \csc \pi x $$

Step-by-Step Solution

Answer

Period: 2; Vertical asymptotes: x = n for all integers n.

Step 1: Recognize the Basic Function

The given function is related to the cosecant function, which is defined as the reciprocal of the sine function. Therefore, the given function is related to \(y=\frac{1}{\sin(\pi x)}\). The csc function has vertical asymptotes wherever the sine function is zero.

Step 2: Determine the Period of the Function

The period of the cosecant function \(\csc(kx)\) is determined by the expression \(\frac{2\pi}{k}\). In this case, \(k=\pi\), so:\[ T = \frac{2\pi}{\pi} = 2 \] The period of the function is 2.

Step 3: Identify Vertical Asymptotes

Vertical asymptotes occur where the sine function is zero because division by zero is undefined. The sine function, \(\sin(\pi x)\), is zero when \(\pi x = n\pi\), where \(n\) is any integer. Therefore, solving for \(x\), the vertical asymptotes are at \(x = n\) for all integers \(n\).

Step 4: Sketch One Cycle of the Graph

Sketching the graph involves choosing the period interval from 0 to 2 (since the period is 2) and marking asymptotes at integer values of \(x\). The graph of \(y=3\csc(\pi x)\) will have curves opening towards the asymptotes between these intervals, reflecting the usual cosecant wave pattern but vertically stretched by a factor of 3.

Key Concepts

Period of Trigonometric Functions

The period of a trigonometric function is the distance over which the function's values repeat. This is a key characteristic because it tells us how often the wave-like pattern of the function occurs. For a function like \(y = 3 \csc(\pi x)\), we start by understanding that it is related to the basic cosecant function \(\csc(kx)\). The period formula for the cosecant function is determined as \(\frac{2\pi}{k}\).
In this problem, \(k = \pi\), leading to a period of:
\[ T = \frac{2\pi}{\pi} = 2 \] This means the pattern of the graph repeats every 2 units along the x-axis. Remember, recognizing the period helps in sketching just one cycle of the graph accurately. It's crucial for understanding the timing of peaks and troughs in trigonometric graphs.

Vertical Asymptotes

Vertical asymptotes are lines where the function grows infinitely in the positive or negative direction. They are very important in trigonometric functions like the cosecant because they indicate the points not included in the function's domain.
For the function \(\csc(\pi x)\), the vertical asymptotes occur at the zero points of the sine function, because division by zero is undefined. The sine function \(\sin(\pi x)\) equals zero when \(\pi x = n\pi\), where \(n\) is any integer.
For \(y = 3 \csc(\pi x)\), solving \(\pi x = n\pi\) gives us vertical asymptotes at \(x = n\) for every integer \(n\). Recognizing these lines is crucial as it shapes the graph by creating boundaries where the function cannot exist.

Graphing Trigonometric Functions

Graphing trigonometric functions starts with understanding their basic shape and key characteristics like period and asymptotes. The function \(y = 3 \csc(\pi x)\) involves sketching one complete cycle between 0 and 2, the period of the function.
To graph this function, follow these steps:

• Identify the period interval, which in this case is from 0 to 2.
• Mark vertical asymptotes at integer values of \(x\), such as \(x = 0, 1,\) and \(2\), where the function is undefined.
• Sketch the arcs of the cosecant function between these asymptotes. Each arc opens toward the asymptote from both sides, creating a leaf shape typical of the cosecant function.
• Apply the vertical stretch. Since there is a coefficient 3, stretch the typical "leaf" of the cosecant vertically by a factor of 3.

This approach will create a clear and accurate graph, illustrating how the function behaves over one full cycle.