Problem 34 - Precalculus : Building Concepts and Connections

Question

Graph at least two cycles of the given functions. $$r(x)=\sin (4 \pi x)-1$$

Step-by-Step Solution

Answer

The given function \(r(x) = \cot \left(\frac{\pi}{3} x\right) -1\) is a vertically shifted (down by 1 unit) cotangent function with a period of \(3\pi\). The graph of this function over at least two cycles will show repeating decreasing curves between vertical asymptotes each \(3\pi\) units apart, made by shifting down the basic cotangent function by 1.

Step 1: Identify the basic shape and characteristics of cotangent function

The cotangent function has key values at intervals of pi. At \(x = 0\) and \(x = \pi\), the function is undefined and has vertical asymptotes. Between these values, the function decreases from positive infinity at \(x = 0\) to negative infinity at \(x = \pi\) . This pattern repeats for every interval of pi.

Step 2: Calculate period of the modified cotangent function

The period of a cotangent function is normally \(\pi\). However, the cotangent function in the equation is multiplied by a factor of \(\pi/3\) inside the function, which changes the period. The period of \(r(x) = \cot\left(\frac{\pi}{3} x\right)\) is \(3\pi\).

Step 3: Add the vertical shift

The given function \(r(x) = \cot \left(\frac{\pi}{3} x\right) -1\) is a vertically shifted cotangent curve. The '-1' outside the function nips down the cotangent curve by 1 unit. This means every point on the previous graph \(r(x) = \cot \left(\frac{\pi}{3} x\right)\) will be moved 1 unit down.

Step 4: Draw the graph

To draw two cycles of the function, plot at least two periods of the function. Start by drawing vertical asymptotes at \(x=0\) and \(x = 3\pi\). Between these asymptotes, sketch a decreasing curve that starts from positive infinity and goes to negative infinity. Do the same for the next period. Then shift each point on the curve down by one unit to represent the '-1' shift. This is the graph of \(r(x)\) over two cycles.

Key Concepts

Period of Trigonometric Functions

Understanding the concept of the period is crucial when working with trigonometric functions like the cotangent. Generally, the period of a function is the interval after which it repeats its values. For the basic cotangent function, the period is \(\pi\), meaning it repeats its pattern and behavior every \(\pi\) units.
However, when transformations are applied, such as \(\cot\left(\frac{\pi}{3} x\right)\), the period changes. In this modified function, the \(x\)-variable is multiplied by a fraction, \(\frac{\pi}{3}\), which affects the period length. The period becomes \(3\pi\).
To determine the new period, use the formula for transformed trigonometric functions:\[ \text{New period} = \frac{\text{Original period}}{|b|} \]where \(b\) is the coefficient of \(x\) in the argument. Thus, calculating for the modified cotangent:\[ \text{New period} = \frac{\pi}{|\frac{\pi}{3}|} = 3\pi \]
This means that every \(3\pi\) units, the function repeats its values, creating a new repeating pattern.

Vertical Asymptotes

Vertical asymptotes are important features of the cotangent function. They are the \(x\)-values where the function does not exist and it tends to infinity. For the standard cotangent function \(\cot(x)\), vertical asymptotes occur at every integer multiple of \(\pi\), such as \(x = 0, \pi, 2\pi\), etc.
These asymptotes occur because the cotangent is the reciprocal of the tangent function, and wherever the tangent equals zero, the cotangent becomes undefined, shooting up to positive or negative infinity.
For the function \(\cot\left(\frac{\pi}{3} x\right)\), the asymptotes are moved by the transformation. Use this principle to locate them:

• Set the inner argument equal to each standard asymptote value: \(\frac{\pi}{3}x = \pi \times n\) (where \(n\) is an integer).
• Solve for \(x\): \(x = 3 \times n\).

This results in vertical asymptotes at \(x = 0, 3, 6, \ldots\)