Problem 199 - Calculus Volume 1

Question

Suppose the rate of gasoline consumption in the United States can be modeled by a sinusoidal function of the form \(\left(11.21-\cos \left(\frac{\pi t}{6}\right)\right) \times 10^{9} \mathrm{gal} / \mathrm{mo}\). a. What is the average monthly consumption, and for which values of \(t\) is the rate at time \(t\) equal to the average rate? b. What is the number of gallons of gasoline consumed in the United States in a year? c. Write an integral that expresses the average monthly U.S. gas consumption during the part of the year between the beginning of April \((t=3)\) and the end of September \((t=9)\).

Step-by-Step Solution

Answer

The average monthly consumption is 11.21 billion gallons. The rate equals this average in March and September. Total yearly consumption is 134.52 billion gallons.

Step 1: Understand the Sinusoidal Function

The function given is modeling the rate of gasoline consumption: \( \left(11.21-\cos \left(\frac{\pi t}{6}\right)\right) \times 10^{9} \mathrm{gal} / \mathrm{mo} \). This is a sinusoidal function where \( t \) represents the month starting from January (\( t=0 \)).

Step 2: Calculate the Average Monthly Consumption Over One Year

The average value of the function over one period is obtained by integrating the function over one period and then dividing by the period length.1. Compute the integral of \( \left(11.21-\cos \left(\frac{\pi t}{6}\right)\right) \) from 0 to 12. \[ \int_0^{12} \left(11.21-\cos \left(\frac{\pi t}{6}\right)\right) \, dt \]2. Integrate: \[ \int \left(11.21-\cos \left(\frac{\pi t}{6}\right)\right) \, dt = 11.21t + 6\sin \left(\frac{\pi t}{6}\right) \]3. Evaluate: \[ \left[11.21t + 6\sin \left(\frac{\pi t}{6}\right)\right]_0^{12} = 11.21 \times 12 + 6 \sin \left(2\pi\right) - 0 = 134.52 \]4. Average: \[ \frac{134.52}{12} = 11.21 \, \text{billion gallons per month} \]

Step 3: Determine When Rate Equals Average

Set the function equal to the average rate:\[11.21 - \cos \left(\frac{\pi t}{6}\right) = 11.21\]This simplifies to:\[\cos \left(\frac{\pi t}{6}\right) = 0\]Solving \( \cos \left(\frac{\pi t}{6}\right) = 0 \) gives:\[\frac{\pi t}{6} = \frac{\pi}{2}, \frac{3\pi}{2}\]Thus, \( t = 3, 9 \).

Step 4: Compute Total Consumption in a Year

The total consumption over a year is the integral of the rate over 12 months:\[\int_0^{12} \left(11.21-\cos \left(\frac{\pi t}{6}\right)\right) \, dt\]This integral was previously calculated as 134.52 billion gallons.

Step 5: Express Average Consumption from April to September

Consider integration from \( t = 3 \) to \( t = 9 \):\[\int_3^9 \left(11.21-\cos \left(\frac{\pi t}{6}\right)\right) \, dt\]First, calculate the integral:\[11.21t + 6\sin \left(\frac{\pi t}{6}\right) \bigg|_3^9 = \left(11.21 \times 9 + 6\sin \frac{3\pi}{2}\right) - \left(11.21 \times 3 + 6\sin \frac{\pi}{2}\right)\]This simplifies to:\[= 100.89 - 33.63 = 67.26 \, \text{billion gallons}\]Then, divide by 6 for the average:\[\frac{67.26}{6} \approx 11.21 \, \text{billion gallons per month}\]

Key Concepts

Sinusoidal Functions

Sinusoidal functions are mathematical functions that depict a smooth oscillating curve, much like a sine or cosine wave. These functions are pivotal in modeling periodic phenomena and can describe various real-world situations, from sound waves to seasonal changes. In the context of the problem, the function \( \left(11.21 - \cos \left(\frac{\pi t}{6}\right)\right) \times 10^9 \mathrm{gal} / \mathrm{mo} \) is a sinusoidal function representing gasoline consumption over time in the United States.
Here, the term \(-\cos\left(\frac{\pi t}{6}\right)\) signifies the oscillatory nature of the rate, showing how it fluctuates throughout the year. By observing the coefficients and argument of the cosine function, we can determine the amplitude and the period of the wave. The amplitude, or the fluctuation size, is determined by the coefficient in front of the cosine, while the period is influenced by the frequency inside the cosine, which is driven by \(\frac{\pi t}{6}\), indicating a full cycle every 12 months. Utilizing this model allows us to make accurate predictions about resource consumption over time.

Average Value of Functions

The average value of a function over an interval provides a singular value that represents the mean of the function's values. To calculate this for the function \( \left(11.21 - \cos \left(\frac{\pi t}{6}\right)\right)\), we integrate the function over its period and then divide the result by the length of the interval.
The formula for the average value \( f_{avg} \) of a continuous function \( f(t) \) over the interval \([a, b]\) is:
\[ f_{avg} = \frac{1}{b-a} \int_a^b f(t) \, dt \]
For our gasoline consumption model, integrating from 0 to 12 (representing an entire year) followed by dividing by 12, provides the average monthly consumption, which was found to be 11.21 billion gallons per month.
This average can be useful for creating policies or making predictions, as it gives a baseline against which fluctuations can be gauged.

Scalar Multiplication

Scalar multiplication is a basic concept in vector mathematics, where each component of a vector is multiplied by a scalar (a single real number). Although scalar multiplication more traditionally applies directly to vectors, the concept can be illustrated similarly with scalar functions in calculus.
In the context of our function, multiplication by 10 billion (\(10^9\)) acts as a scalar multiplication. This helps scale the sinusoidal function for practical application, adjusting the model's units to match real-world measurements (e.g., billions of gallons).
Thus, the base function \( 11.21 - \cos \left(\frac{\pi t}{6}\right) \) becomes \( \left(11.21 - \cos \left(\frac{\pi t}{6}\right)\right) \times 10^9 \), making the output suitable for understanding total gasoline consumption at any month \( t \). Understanding how these multiplications adjust the scale of a function is crucial for translating mathematical models into tangible predictions.

Definite Integrals

Definite integrals are fundamental in calculus for calculating the exact area under a curve over a specific interval. This makes them crucial tools for determining accumulated quantities, like total gasoline consumption, from a rate function.
The definite integral of \( \left(11.21 - \cos \left(\frac{\pi t}{6}\right)\right) \) from \( t = 0 \) to \( t = 12 \) provides the total gasoline consumption for the year. This is calculated as:
\[ \int_0^{12} \left(11.21 - \cos \left(\frac{\pi t}{6}\right)\right) \, dt \]
This integral, previously computed as 134.52 billion gallons, represents the entire year's consumption.\
Additionally, integrating from \( t = 3 \) to \( t = 9 \) yields the consumption from April to September. The use of definite integrals allows us to capture the precise quantity consumed within specified time frames and determine averages via additional division, such as finding the average monthly consumption during selected periods.