Problem 170 - Calculus Volume 1

Question

For the following exercises, find a. the amplitude, b. the period, and c. the phase shift with direction for each function. $$y=2 \cos \left(x-\frac{\pi}{3}\right)$$

Step-by-Step Solution

Answer

Amplitude is 2, period is \(2\pi\), phase shift is \(\frac{\pi}{3}\) right.

Step 1: Identify the Standard Form

The standard form for the cosine function is \( a \cos(bx - c) + d \). For the given function, \( y = 2 \cos \left( x - \frac{\pi}{3} \right) \), we can observe that it matches the standard form with \( a = 2 \), \( b = 1 \), and \( c = \frac{\pi}{3} \).

Step 2: Find the Amplitude

The amplitude of a cosine function \( a \cos(bx - c) + d \) is the absolute value of \( a \). In this case, \( a = 2 \). Therefore, the amplitude is \( |2| = 2 \).

Step 3: Calculate the Period

The period of a cosine function \( a \cos(bx - c) + d \) is given by \( \frac{2\pi}{b} \). Here, \( b = 1 \), so the period is \( \frac{2\pi}{1} = 2\pi \).

Step 4: Determine the Phase Shift

The phase shift of a cosine function is calculated using \( \frac{c}{b} \). From \( c = \frac{\pi}{3} \) and \( b = 1 \), we find the phase shift is \( \frac{\pi}{3} \). Since it is \( x - \frac{\pi}{3} \), the shift is to the right by \( \frac{\pi}{3} \).

Key Concepts

Amplitude

One of the first characteristics you encounter when studying trigonometric functions is amplitude. Amplitude corresponds to the peak value of the function, measuring how far the graph deviates from its middle line, both upwards and downwards. In simpler terms, it tells you the height of the wave from the center to its peak.For the function in our exercise, \( y = 2 \cos \left( x - \frac{\pi}{3} \right) \), the amplitude is determined by the coefficient \( a \) in the standard cosine form \( a \cos(bx - c) + d \). Here, \( a = 2 \). Since amplitude is always a positive distance, we take the absolute value: \(|a| = |2| = 2 \).Thus, the amplitude of this function is 2, which means the graph oscillates 2 units above and below its midline (typically the x-axis if no vertical shift is involved). Understanding amplitude is crucial because it helps you visualize how "tall" or "short" the wave forms are, enabling you to gauge the intensity of oscillations quickly.

Period

The period of a trigonometric function is the distance over which the function starts repeating its pattern. Essentially, it tells you how long it takes for the wave to complete one full cycle. For sine and cosine functions, the regular period is \( 2\pi \), which means every \( 2\pi \) units along the x-axis, the function returns to its starting position.In our cosine example \( y = 2 \cos \left( x - \frac{\pi}{3} \right) \), we calculate the period using the formula \( \frac{2\pi}{b} \). Here, \( b = 1 \), so the period remains \( \frac{2\pi}{1} = 2\pi \). This indicates that the graph completes one cycle every \( 2\pi \) units.Having knowledge of the period allows you to predict when the wave-like patterns will repeat, which is pivotal for applications ranging from sound waves to oscillating mechanical systems.

Phase Shift

Phase shift refers to the horizontal movement of the wave along the x-axis. It represents how much the graph is shifted from its usual position. In the function \( y = 2 \cos \left( x - \frac{\pi}{3} \right) \), the phase shift helps us determine where the cycle of the wave effectively starts. It is calculated using \( \frac{c}{b} \) in the context of the standard form \( a \cos(bx - c) + d \).For this particular function, we have \( c = \frac{\pi}{3} \) and \( b = 1 \), giving us a phase shift of \( \frac{\pi}{3} \). Since the function's argument is \( x - \frac{\pi}{3} \), it indicates a shift to the right by \( \frac{\pi}{3} \) units. If the expression were \( x + \frac{\pi}{3} \), the shift would be to the left.Understanding phase shift is useful in determining the new starting point of the cosine or sine wave, which is critical in timing and synchronization tasks in electronics, signal processing, and other applications.