Problem 7 - Contemporary Precalculus

Question

State the amplitude, period, and phase shift of the function. \(p(t)=6 \cos (3 \pi t+1)\)

Step-by-Step Solution

Answer

Answer: The amplitude is 6, the period is \(\frac{2}{3}\), and the phase shift is \(\frac{1}{3\pi}\).

Step 1: Identify the amplitude

The amplitude is the coefficient on the cosine function, which in this function is 6. Therefore, the amplitude is: A = 6

Step 2: Identify B and find the period

The value inside the cosine function is \(3 \pi t + 1\). We must identify B and find the period from the expression \(3 \pi t\). In this case, the value of B is \(3 \pi\), so we can use \(\frac{2 \pi}{B}\) to find the period. Period = \(\frac{2\pi}{3\pi}\)

Step 3: Simplify the period expression

Now we simplify the period expression: Period = \(\frac{2\pi}{3\pi}\) • \(\frac{1}{\pi}\) (multiply both the numerator and denominator by the reciprocal of \(\pi\)) Period = \(\frac{2}{3}\)

Step 4: Identify the phase shift

To find the phase shift, we need to express the given function in the form \(f(t)=A \cos (B (t - C))\). To do so, we can rewrite the given function as follows: \(p(t)=6 \cos (3\pi (t + \frac{1}{3\pi}))\) Now, the phase shift C can be identified as \(\frac{1}{3\pi}\). So, the amplitude is 6, the period is \(\frac{2}{3}\), and the phase shift is \(\frac{1}{3\pi}\).

Key Concepts

Amplitude

Amplitude in trigonometric functions like cosine and sine dictates how far the peaks and valleys of the wave go from the center line, also known as the horizontal axis. The amplitude tells us how "tall" the wave is from its middle point. In the function, \(p(t) = 6 \cos(3 \pi t + 1)\), the amplitude is easy to spot. It's simply the number that is multiplying the cosine function. Here, this number is 6.

This means the highest point of the wave will reach 6 units above the center line, and the lowest will go 6 units below it. Therefore:

• Amplitude = 6

Remember, amplitude is always a positive number that represents the wave's height, regardless of whether there's a negative sign in front of the function's coefficient. This property ensures the amount of energy conveyed by the wave's oscillations is correctly interpreted.

Period

The period of a trigonometric function measures the distance over which the function repeats itself. For the cosine function, the standard period is \(2\pi\). However, when we have a coefficient multiplying the variable \(t\), like in \(3\pi t\) in \(p(t) = 6 \cos(3\pi t + 1)\), this changes the period.

The general formula to find the period of a cosine or sine function given \(B\) (the coefficient of \(t\)) is \(\frac{2\pi}{B}\). In this function:

• \(B = 3\pi\)
• Period = \(\frac{2\pi}{3\pi}\)

When you calculate \(\frac{2\pi}{3\pi}\), you simplify it to \(\frac{2}{3}\). This means the cosine wave completes one full cycle every \(\frac{2}{3}\) of a unit in the horizontal direction.

A shortened period implies the cosine wave oscillates more frequently, leading to a more "squished" appearance horizontally.

Phase Shift

Phase shift tells us how much and in which direction a wave is shifted from its usual starting point on a graph, often compared to the basic form of the function such as \(\cos(t)\). To find the phase shift for the function \(p(t) = 6 \cos(3\pi t + 1)\), we aim to rewrite this function in the form \(A \cos(B(t - C))\), where \(C\) represents the phase shift.

In this specific function, we rewrite it as follows:

• \(p(t) = 6 \cos(3\pi (t + \frac{1}{3\pi}))\)

Here, the term \(+ \frac{1}{3\pi}\) inside the parentheses indicates a leftward phase shift. This occurs because adding to \(t\) shifts the graph left, whereas subtracting would shift it right. Hence, the phase shift is:

• Phase Shift = \(\frac{1}{3\pi}\) units left

Phase shifts are essential as they determine the starting point of the wave cycle, affecting how functions represent various real-world periodic phenomena.