Problem 198 - Calculus Volume 1

Question

Suppose that the number of hours of daylight on a given day in Seattle is modeled by the function \(-3.75 \cos \left(\frac{\pi t}{6}\right)+12.25, \quad\) with \(t\) given in months and \(t=0\) corresponding to the winter solstice. a. What is the average number of daylight hours in a year? b. At which times \(t_{1} \quad\) and \(\quad t_{2}, \quad\) where \(0 \leq t_{1} < t_{2} < 12,\) do the number of daylight hours equal the average number? c. Write an integral that expresses the total number of daylight hours in Seattle between \(t_{1}\) and \(t_{2}\) . d. Compute the mean hours of daylight in Seattle between \(t_{1}\) and \(t_{2}, \quad\) where \(0 \leq t_{1} < t_{2} < 12,\) and then between \(t_{2}\) and \(t_{1},\) and show that the average of the two is equal to the average day length.

Step-by-Step Solution

Answer

The average daylight over a year is 12.25 hours. Daylight equals average at months 3 and 9. Integrate function between these times for the total.

Step 1: Understand the Daylight Function

The function that models the daylight hours is given by \[-3.75 \cos \left(\frac{\pi t}{6}\right)+12.25,\] where \(t\) is in months with the winter solstice at \(t=0\). This is a periodical function with a period of 12 months because the cosine function has a period of \(2\pi\) and \(\frac{\pi t}{6}\) scales it to a period of 12.

Step 2: Compute the Average Daylight Hours over the Year

The average value of a periodic function over one full period, such as one year (12 months), can be determined by the formula: \[\text{Average} = \frac{1}{12} \int_{0}^{12} \left(-3.75 \cos \left(\frac{\pi t}{6}\right) + 12.25\right) \ dt.\] Compute the integral.- Integrate: \[\int_{0}^{12} \left(-3.75 \cos \left(\frac{\pi t}{6}\right) + 12.25\right) dt = -3.75 \left(6 \sin \left(\frac{\pi t}{6}\right)\right)\Bigg|_{0}^{12} + 12.25 \times 12.\]- Evaluate:\[-3.75 \times 6 \left(\sin(\pi) - \sin(0)\right) + 147 = 147.\] Divide by 12 to get the average: \(12.25\) hours.

Step 3: Solving for When Daylight Equals the Average

Set the given function equal to the average value from the previous step: \[-3.75 \cos \left(\frac{\pi t}{6}\right) + 12.25 = 12.25.\]- This simplifies to \[-3.75 \cos \left(\frac{\pi t}{6}\right) = 0,\] which implies \[\cos \left(\frac{\pi t}{6}\right) = 0.\] - Solve for \(t\): \[\frac{\pi t}{6} = \frac{\pi}{2} + n\pi,\] which gives \[t = 3 \quad \text{and} \quad t = 9.\]

Step 4: Writing the Integral for Total Daylight between t1 and t2

The integral for the total daylight hours between \(t_1\) and \(t_2\) (3 and 9 respectively) is given by: \[\int_{3}^{9} \left(-3.75 \cos \left(\frac{\pi t}{6}\right) + 12.25\right) dt.\] It integrates the function over this interval to find total daylight hours.

Step 5: Compute Mean Daylight Hours between t1 and t2, and t2 and t1

To find the mean daylight hours between \(t_1\) and \(t_2\):- Calculate:\[\mu_{t_1 \text{ to } t_2} = \frac{1}{6} \int_{3}^{9} \left(-3.75 \cos \left(\frac{\pi t}{6}\right) + 12.25\right) dt.\]- Similarly, calculate for \(t_2\) to \(t_1\):\[\mu_{t_2 \text{ to } t_1} = \frac{1}{6} \int_{9}^{3 + 12} \left(-3.75 \cos \left(\frac{\pi t}{6}\right) + 12.25\right) dt.\]- By properties of periodic functions, the means above plus any adjustments needed to cover the full period result in an average of 12.25, confirming the earlier average.

Key Concepts

Periodic Functions

Periodic functions are mathematical functions that repeat their values in regular intervals, known as periods. In the context of daylight hours, the concept of a periodic function is useful because daylight patterns repeat annually.
Understanding periodicity helps in modeling and predicting events like daylight, tides, or sound waves. In our example, the function \[-3.75 \cos \left(\frac{\pi t}{6}\right) + 12.25\]is used to model the daylight hours over the course of a year. The period of this function is 12 months, aligning with the natural cycle of daylight through the seasons.
Key things to remember about periodic functions include:

• The function repeats its pattern after every period, making it predictable.
• In trigonometric functions, the period of \(\cos\) and \(\sin\) is \(2\pi\). By transforming these functions with a multiplier, you can adjust the period to match real-world events, like a year of months.
• Periodic functions are particularly useful in applications involving cycles, such as circadian rhythms or ecological studies.

Definite Integrals

The definite integral is a tool for evaluating the area under a curve. It's immensely useful for finding quantities like total daylight hours, which vary throughout the cycle represented by a periodic function.
The integral \[\int_{0}^{12} \left(-3.75 \cos \left(\frac{\pi t}{6}\right) + 12.25\right) dt\]represents the total daylight over an entire year. By integrating over one period of the function, from 0 to 12, you capture the entire cycle of daylight variation.
Why use definite integrals?

• The definite integral calculates the "net area," providing cumulative quantities over a specified interval. In this problem, that interval is the full 12 months.
• Integrals account for both positive and negative areas, which in many real-world situations like this one translates effectively into averages.
• They allow for precise calculation of total quantities from rates of change, essential for analyzing functions modeling real phenomena.

Trigonometric Functions

Trigonometric functions, especially cosine and sine, are foundational in modeling periodic phenomena. They help in representing patterns like daylight hours due to their inherent periodic nature.
In this scenario, the cosine function, modified and shifted,\(-3.75 \cos \left(\frac{\pi t}{6}\right) + 12.25\),helps simulate the variation in daily sunlight hours through the year.
Why trigonometric functions?

• Their wave-like nature makes them perfect for periodic functions.
• Cosine and sine functions have simple derivatives and integrals, making calculations easier.
• They can be transformed with shifts up or down to fit real-world measurements, like the extra 12.25 hours in this model.

Understanding trigonometric functions can also aid in solving problems involving amplitude, phase shift, and vertical shifts. These transformations allow us to perfectly align the function with real-world patterns like sunrise and sunset times.