Problem 8 - Contemporary Precalculus

Question

(a) What is the period of \(f(t)=\sin 2 \pi t ?\) (b) For what values of \(t\) (with \(0 \leq t \leq 2 \pi\) ) is \(f(t)=0 ?\) (c) For what values of \(t\) (with \(0 \leq t \leq 2 \pi\) ) is \(f(t)=1 ?\) or \(f(t)=-1 ?\)

Step-by-Step Solution

Answer

Question: Determine the period of the function \(f(t) = \sin 2\pi t\), and find the values of \(t\) in the interval \([0, 2\pi]\) for which \(f(t) = 0\), \(f(t) = 1\), or \(f(t) = -1\). Answer: The period of the function \(f(t) = \sin 2\pi t\) is \(1\). In the interval \([0, 2\pi]\), the values of \(t\) for which \(f(t) = 0\) are \(0, \frac{1}{2}, 1, \cdots, 2\pi - \frac{1}{2}\). The values of \(t\) for which \(f(t) = 1\) are \(\frac{1}{4}, \frac{5}{4}, \ldots , 2\pi - \frac{3}{4}\), and for \(f(t) = -1\), they are: \(\frac{3}{4}, \frac{7}{4}, \ldots , 2\pi - \frac{1}{4}\).

Step 1: (a) Finding the period of the function

The period of a function describes how often the function repeats its values. For a sinusoidal function like \(f(t) = \sin 2\pi t\), the period is given by \(\frac{2\pi}{\omega}\), where \(\omega\) is the angular frequency. In this case, \(\omega = 2\pi\), so the period of the function is: $$ \text{Period} = \frac{2\pi}{2\pi} = 1 $$

Step 2: (b) Finding the values of \(t\) for which \(f(t) = 0\)

From part (a), we know that the function repeats its values every \(1\) unit of \(t\). To find the values of \(t\) in the interval \([0, 2\pi]\) for which \(f(t) = 0\), we'll set the function equal to zero and solve for \(t\): $$ \sin 2\pi t = 0 $$ The sine function equals zero whenever \(2\pi t\) is an integer multiple of the \(\pi\). So, we have: $$ 2\pi t = k\pi, \quad k \in \mathbb{Z} $$ Dividing both sides by \(2\pi\), we get: $$ t = \frac{k}{2}, \quad k \in \mathbb{Z} $$ In the interval \([0, 2\pi]\), the values of \(k\) are \(0, 1, 2, \cdots , 4\pi - 1\), and the corresponding values of \(t\) are \(0, \frac{1}{2}, 1, \cdots , 2\pi - \frac{1}{2}\).

Step 3: (c) Finding the values of \(t\) for which \(f(t) = 1\) or \(f(t) = -1\)

To find the values of \(t\) in the interval \([0, 2\pi]\) for which \(f(t) = 1\) or \(f(t) = -1\), we'll first solve for \(f(t) = 1\) and then for \(f(t) = -1\): $$ \sin 2\pi t = 1 $$ The sine function equals one whenever \(2\pi t\) is \(\frac{\pi}{2}\) plus an integer multiple of \(2\pi\). So, we have: $$ 2\pi t = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z} $$ Dividing both sides by \(2\pi\), we get: $$ t = \frac{1}{4} + k, \quad k \in \mathbb{Z} $$ In the interval \([0, 2\pi]\), the values of \(k\) are \(0, 1, 2, \cdots , 4\pi - 2\), and the corresponding values of \(t\) for which \(f(t) = 1\) are \(\frac{1}{4}, \frac{5}{4}, \ldots , 2\pi - \frac{3}{4}\). Now we'll solve for \(f(t) = -1\): $$ \sin 2\pi t = -1 $$ The sine function equals negative one whenever \(2\pi t\) is \(-\frac{\pi}{2}\) plus an integer multiple of \(2\pi\). So, we have: $$ 2\pi t = -\frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z} $$ Dividing both sides by \(2\pi\), we get: $$ t = -\frac{1}{4} + k, \quad k \in \mathbb{Z} $$ In the interval \([0, 2\pi]\), the values of \(k\) are \(1, 2, 3, \cdots , 4\pi - 1\), and the corresponding values of \(t\) for which \(f(t) = -1\) are \(\frac{3}{4}, \frac{7}{4}, \ldots , 2\pi - \frac{1}{4}\).

Key Concepts

Periodicity

In mathematics, periodicity refers to the property of a function to repeat its values over regular intervals. In trigonometry, this is particularly important when talking about sine and cosine functions, which are inherently periodic. The period of a function is the smallest interval over which it repeats itself completely. For the function \(f(t) = \sin 2\pi t\), the period can be found using the formula \( \frac{2\pi}{\omega} \), where \( \omega \) represents the angular frequency. In our example, \( \omega = 2\pi \). Therefore, the period is \( \frac{2\pi}{2\pi} = 1 \). This means that the function repeats its values every 1 unit along the \(t\)-axis.

Understanding periodicity is crucial since it allows us to predict the behavior of the function over any required stretch. Here are a few properties of periodic functions that are handy:

• The graph of a periodic function forms a repeating pattern.
• Adding the period to the variable of a periodic function results in the same function value.
• Periodic functions can be shifted horizontally if needed, but the period remains the same.

These properties are widely used in solving trigonometric problems.

Sine Function

The sine function is one of the essential functions in trigonometry. It is defined for all real numbers and has a repeating wave-like pattern, which makes it a periodic function. The sine function helps describe phenomena like sound waves, light waves, and other cyclic occurrences in nature.

For the function \( f(t) = \sin 2 \pi t \), the sine value indicates the vertical position on the unit circle as it moves around, starting from the origin. The function value lies between -1 and 1, indicating that the sine wave oscillates between these limits. This makes it useful for modeling fluctuations or any scenario where values rise and fall in a regular pattern.

Here are a few key points about the sine function:

• Sine has a period of \(2\pi\), but it can be altered with the angular frequency, as seen in \(f(t) = \sin 2\pi t\).
• The sine function has zeros at integer multiples of \(\pi\), thus repeating every \(\pi\) for standard sine but it repeats every 1 unit for our example due to the angular frequency.
• It reaches a maximum value of 1 at \(\frac{\pi}{2} + 2k\pi\) and a minimum value of -1 at \(-\frac{\pi}{2} + 2k\pi\), where \(k\) is an integer.

Understanding the sine function's actions and behaviors is key to mastering trigonometry.

Angular Frequency

Angular frequency, denoted by \( \omega \), is a crucial concept when dealing with periodic phenomena, especially in trigonometry and physics. It refers to the rate of change of the function's angle with respect to time and is often used in equations of motion involving circular or oscillatory paths.

For the function \(f(t) = \sin 2\pi t\), the angular frequency is \(2\pi\), which means this function completes one full cycle every 1 unit of \(t\). The formula for determining the period \( T \) of the function from the angular frequency is \( T = \frac{2\pi}{\omega} \). Higher angular frequencies result in faster oscillations, which means shorter periods, and vice versa.

Here are a few points to understand angular frequency:

• Angular frequency is directly proportional to the number of cycles a function completes in a given period.
• It is expressed in radians per unit of time, which refers to the distance moved along the unit circle per each time unit.
• In the context of waves, it is linked to the frequency with the relation \(\omega = 2\pi f \), where \(f\) is the standard frequency in cycles per unit of time.

Exploring angular frequency gives insight into how frequency and period relate in the realm of periodic functions.