Problem 21 - Precalculus

Question

For the following exercises, graph one full period of each function, starting at \(x=0 .\) For each function, state the amplitude, period, and midine. State the maximum and minimum \(y\) -values and their corresponding \(x\) -values on one period for \(x>0\) . State the phase shift and vertical translation, if applicable. Round answers to two decimal places if necessary. $$ f(t)=-\sin \left(\frac{1}{2} t+\frac{5 \pi}{3}\right) $$

Step-by-Step Solution

Answer

Amplitude: 1; Period: \(4\pi\); Midline: \(y=0\). Max at \((\frac{4\pi}{3}, 1)\), Min at \((\frac{10\pi}{3}, -1)\), Phase shift: \(-\frac{5\pi}{3}\).

Step 1: Identify Amplitude and Period

The function is given by \( f(t) = -\sin \left( \frac{1}{2} t + \frac{5\pi}{3} \right) \). To determine the amplitude, note the coefficient of \( \sin \) function. The amplitude is the absolute value of the coefficient in front of \( \sin \), which is 1. The period is calculated using the formula \( \frac{2\pi}{|B|} \) where \( B = \frac{1}{2} \). This gives a period of \( 4\pi \).

Step 2: Determine the Midline

Since there is no vertical translation in the function \( f(t) = -\sin \left( \frac{1}{2} t + \frac{5\pi}{3} \right) \), the midline is \( y = 0 \).

Step 3: Identify Phase Shift

The phase shift is calculated by solving for \( t \) in the equation \( \frac{1}{2} t + \frac{5\pi}{3} = 0 \). This results in \( t = -\frac{5\pi}{3} \). Therefore, the phase shift is left \( \frac{5\pi}{3} \).

Step 4: Vertical Translation Assessment

The function has no constant term added outside the sine function, indicating that there is no vertical translation.

Step 5: Graph the Function

To graph the function, start at the phase shift \( x = 0 - \frac{5\pi}{3} \), and plot over one period (\( 4\pi \)). The function \( f(t) \) oscillates with maximum and minimum values at the respective midline crossings.

Step 6: Determine Max and Min Values

The maximum \( y \) value occurs at \( \frac{1}{2}(3\pi - 5\pi/3) = \frac{4\pi}{3} \) with \( y = 1 \). The minimum \( y \) value occurs at \( \frac{1}{2}(7\pi - 5\pi/3) = \frac{10\pi}{3} \) with \( y = -1 \).

Key Concepts

Amplitude

When talking about graphs of trigonometric functions, the amplitude is an essential concept. Amplitude refers to the height of the wave from its central axis, which is often the midline of the function. In the equation, \( f(t) = -\sin \left( \frac{1}{2} t + \frac{5\pi}{3} \right) \), the amplitude can be found by looking at the coefficient of the sine function. Here, the coefficient of \( \sin \) is -1, but since amplitude is always a positive value, we take the absolute value which gives us an amplitude of 1.

• Amplitude measures the wave's height.
• It's noted as the absolute value of the coefficient in front of the trigonometric function.
• For our function, the amplitude is 1.

Understanding the amplitude helps in determining how 'tall' or 'short' the wave will be compared to the midline, which in this case is simply the x-axis.

Period

The period of a trigonometric function indicates how long it takes for the function to complete one full cycle. It informs us about the horizontal length of one full wave on the graph. This is calculated using the formula \( \frac{2\pi}{|B|} \), where \( B \) is the coefficient of the variable \( t \) inside the function. In this instance, \( B \) is \( \frac{1}{2} \), so the period is \( \frac{2\pi}{\frac{1}{2}} = 4\pi \).

• The period determines the cycle length on the x-axis.
• It is calculated as \( \frac{2\pi}{|B|} \).
• The function here has a period of \( 4\pi \).

This means the function repeats every \( 4\pi \) units along the x-axis, so we see the same wave pattern every \( 4\pi \) interval.

Phase Shift

Phase shift refers to how much the function is moved left or right along the x-axis from its usual position. It arises from the angle part of the trigonometric function. To determine the phase shift, we solve the equation \( \frac{1}{2} t + \frac{5\pi}{3} = 0 \) for \( t \). Solving this, we find \( t = -\frac{5\pi}{3} \). This means our phase shift is to the left by \( \frac{5\pi}{3} \).

• Phase shift is the horizontal movement of the graph.
• Calculated by solving for \( t \) when the angle part equals zero.
• This function shifts to the left by \( \frac{5\pi}{3} \).

Recognizing the phase shift is crucial because it sets the starting point of the cycle on the graph along the x-axis.

Midline

The midline of a trigonometric function is an imaginary horizontal line that runs through the middle of the wave. It indicates the average value of the function. If there is no vertical translation (a number added or subtracted outside the sine or cosine function), then the midline is simply the x-axis, or \( y = 0 \).

• Midline is the horizontal center of the wave.
• It is \( y = 0 \) when no vertical translation exists.
• This function's midline is at \( y = 0 \).

The midline helps us understand the oscillation of the wave above and below this line. In graphing, it's where the wave would naturally rest if it were not oscillating.