Problem 11 - Applied Calculus

Question

Values of a function are given in the following table. Explain why this function appears to be periodic. Approximately what are the period and amplitude of the function? Assuming that the function is periodic, estimate its value at \(t=15,\) at \(t=75,\) and at \(t=135\). $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline t & 20 & 25 & 30 & 35 & 40 & 45 & 50 & 55 & 60 \\\\\hline f(t) & 1.8 & 1.4 & 1.7 & 2.3 & 2.0 & 1.8 & 1.4 & 1.7 & 2.3 \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

Periodic with period 25 and amplitude 0.45; \(f(15) = 2.0\), \(f(75) = 1.4\), \(f(135) = 1.8\).

Step 1: Identify Periodicity

We start by observing the given function values in the table: \(f(t) = 1.8, 1.4, 1.7, 2.3, 2.0, 1.8, 1.4, 1.7, 2.3\). Notice that the sequence begins to repeat itself starting from \(t = 45\). Since the values from \(t = 20\) to \(t = 60\) repeat from \(t = 45\) to \(t = 60\), the function appears to be periodic.

Step 2: Calculate the Period

To determine the period of the function, we observe where the repetition begins. The function repeats every five steps, which are \(t = 20, 25, 30, 35, 40\) and then starts over at \(t = 45, 50, 55, 60\). Thus, the period of the function is \(40 - 20 = 25\).

Step 3: Determine the Amplitude

The amplitude is calculated as the half of the difference between the maximum and minimum values of the function. Here, the maximum value is 2.3 and the minimum value is 1.4. Therefore, the amplitude is \(\frac{2.3 - 1.4}{2} = 0.45\).

Step 4: Estimate f(t) at t=15

Since the function is periodic with a period of 25, \(t = 15\) is equivalent to \(t = 15 + 25 = 40\). The value of the function at \(t = 40\) is 2.0. Therefore, \(f(15) = 2.0\).

Step 5: Estimate f(t) at t=75

Similarly, for \(t = 75\), since each period is 25, \(t = 75\) is equivalent to \(t = 75 - 25 \times 2 = 25\). The value of the function at \(t = 25\) is 1.4. Therefore, \(f(75) = 1.4\).

Step 6: Estimate f(t) at t=135

For \(t = 135\), similarly applying the periodicity, \(t = 135 - 25 \times 5 = 10\), it is in the same phase as \(t=20\). We examine the first value in the sequence at \(t=20\) and find it is 1.8. Therefore, \(f(135) = 1.8\).

Key Concepts

Function Periodicity

Periodicity in functions refers to the property where the function's values repeat at regular intervals over its domain. Because of this pattern, these functions are often predictable and can be easier to work with once their periodic nature is identified. The given exercise showcases a function where values like 1.8, 1.4, 1.7, and 2.3 repeat.

By inspecting these values, it becomes apparent that starting at a specific point, the pattern repeats every time interval. This repetition suggests a rhythmic or cyclic nature inherent to the function. If you spot this kind of repetition, then you are likely dealing with a periodic function. This is essential when making predictions or extending the function beyond the given values, as we can do in this exercise. Once confirmed, this characteristic allows the function to be extended indefinitely while maintaining its pattern.

In practical terms, periodic functions are everywhere – they help model waves, seasonal phenomena, and even economic cycles. Recognizing periodicity is the first step in utilizing mathematical tools and techniques designed to capitalize on these repetitions.

Function Period

The period of a periodic function is the smallest positive interval after which the function repeats its values. Think of it like the "cycle time" for one full repeat of the pattern. A function's period can help simplify complex problems, as recognizing it allows us to infer values beyond initial observations.

• In our exercise, observe from the table that the function repeats from time intervals 20 to 60. Identifying where the first cycle ends gives you its period.
• Here, the sequence begins to repeat at every five time steps, such as at 20, 25, 30, continuing to 45, 50, etc. This means the "function period" is 25 units of time.

If you understand this concept, you can take the period and apply it to further calculations beyond the data provided. If asked to find the function's value at a time not in the table, like at 15, 75 or 135, you shift your timeline based on the period. It's a powerful tool for making predictions in a variety of fields, from engineering to finance!

Function Amplitude

Amplitude in a periodic function measures the height of the wave from its central axis or equilibrium position. For something like sound waves, this would tell us how "loud" or "soft" the sound is, while for a tide, it shows the maximum rise and fall from sea level. Simply put, it is half the distance between the maximum and minimum values in one complete cycle of the function.

In the given exercise, you can find the amplitude by taking the highest value (2.3) and the lowest value (1.4) of the function. Compute the amplitude as \( \frac{2.3 - 1.4}{2} = 0.45 \).

Understanding the amplitude helps in discerning the magnitude of the oscillations or fluctuations of the function. Not only does it give an idea about the height of the wave or the strength of the signal, but also how drastic the changes are over one complete cycle. Mathematical analysis of function amplitude is integral in various fields such as physics, engineering, and even finance, to analyze trends and patterns effectively.