Problem 189 - Precalculus

Question

For the following exercises, use this scenarion of a city has fisen over a 20 -year interval. Its population may be modeled by the following function: \(y=12,000+8,000\) sin \((0.628 x),\) where the domain is the years since 1980 and the range is the population of the city. Over this domain, when does the population reach \(18,000 ? 13,000 ?\)

Step-by-Step Solution

Answer

1981.35, 1983.17 for 18,000; 1980.2, 1984.86 for 13,000.

Step 1: Understanding the Sinusoidal Function

The function given is a sinusoidal function: \[ y = 12,000 + 8,000 \sin(0.628x) \] where \( x \) represents the years since 1980. The function oscillates around the midline (12,000) with an amplitude of 8,000. This means that the maximum population is 20,000 (12,000 + 8,000) and the minimum is 4,000 (12,000 - 8,000).

Step 2: Finding When Population Reaches 18,000

To find when the population reaches 18,000, set the equation equal to 18,000:\[ 18,000 = 12,000 + 8,000 \sin(0.628x) \]Subtract 12,000 from both sides:\[ 6,000 = 8,000 \sin(0.628x) \]Divide both sides by 8,000:\[ \sin(0.628x) = 0.75 \]Find the angle whose sine is 0.75. The principal value is \( \sin^{-1}(0.75) \approx 0.8481 \; \text{radians} \).

Step 3: Solving for x (Years Since 1980) When sin(x)=0.75

For \( \sin(0.628x) = 0.75 \), solve for \( x \):\[ 0.628x = 0.8481 + 2\pi k \quad \text{or} \quad 0.628x = \pi - 0.8481 + 2\pi k \]where \( k \) is an integer for the number of cycles. Solve the first equation for the first solution:\[ x = \frac{0.8481}{0.628} \approx 1.351 \]And for the second equation:\[ x = \frac{\pi - 0.8481}{0.628} \approx 3.170 \]

Step 4: Finding When Population Reaches 13,000

To find when the population is 13,000, set the equation equal to 13,000:\[ 13,000 = 12,000 + 8,000 \sin(0.628x) \]Subtract 12,000 from both sides:\[ 1,000 = 8,000 \sin(0.628x) \]Divide both sides by 8,000:\[ \sin(0.628x) = 0.125 \]Find the angle whose sine is 0.125. The principal value is \( \sin^{-1}(0.125) \approx 0.1258 \; \text{radians} \).

Step 5: Solving for x (Years Since 1980) When sin(x)=0.125

For \( \sin(0.628x) = 0.125 \), solve for \( x \):\[ 0.628x = 0.1258 + 2\pi k \quad \text{or} \quad 0.628x = \pi - 0.1258 + 2\pi k \]Solve the first equation for the first solution:\[ x = \frac{0.1258}{0.628} \approx 0.2 \]And for the second equation:\[ x = \frac{\pi - 0.1258}{0.628} \approx 4.86 \]

Step 6: Convert to Years Since 1980

The solutions \( x \approx 1.351, 3.170, 0.2, \text{ and } 4.86 \) represent the number of years since 1980. So, the population reaches 18,000 at approximately 1981.35 and 1983.17, and 13,000 at approximately 1980.2 and 1984.86.

Key Concepts

Population Modeling

Population modeling is a crucial tool used by demographers and city planners to understand and manage the dynamics of urban growth. It involves using mathematical equations to represent population changes over time. In this particular problem, a sinusoidal function is used to model the population of a city over a 20-year period starting from 1980.

The function given is: \( y = 12,000 + 8,000 \sin(0.628x) \)

Where:

• \( y \) is the population at any given year \( x \), since 1980.
• 12,000 represents the average population around which oscillations occur.
• 8,000 is the amplitude, showing the extent of population fluctuation above and below the average.
• 0.628 is a constant that affects the wavelength and frequency of the sinusoidal curve.

Such a model highlights how populations can fluctuate seasonally or cyclically due to various factors like birth rates, migration, or even economic changes.

Amplitude

Amplitude in a sinusoidal function, like the one used for population modeling, is a crucial element that represents the height of the peaks and the depths of the troughs from the midline of the function. It tells us how much the population swings up and down from the average.

In our equation: \( y = 12,000 + 8,000 \sin(0.628x) \)

The amplitude is given as 8,000. This means:

• The population can increase by up to 8,000 from the average (reaching 20,000).
• Similarly, it can decrease by the same amount, dipping to a minimum of 4,000.

The amplitude is significant as it helps predict the maximum and minimum population sizes. Understanding amplitude assists planners in foreseeing possible growth surges or declines, helping in efficient resource allocation and infrastructure planning.

Trigonometric Equations

Trigonometric equations involve finding angles or values that satisfy relationships involving trigonometric functions like sine, cosine, and tangent. In our scenario, we use the sine function as part of the equation to model population changes.

The goal is often to solve for the variable \( x \) when given specific conditions, such as when the population reaches certain levels like 18,000 or 13,000.

To find when the population equals a particular value:

• Set the sinusoidal equation equal to the population value.
• Rearrange to isolate the sine term.
• Use the inverse sine function to find the angle corresponding to the solved sine value.
• Finally, solve for \( x \), noting that since sine has a periodic nature, there can be multiple solutions within one cycle, typically \( 0.628x = \theta + 2\pi k \) and \( 0.628x = \pi - \theta + 2\pi k \).

These steps allow you to determine the specific years in which the population reaches the desired thresholds. Understanding trigonometric equations is key to solving numerous real-world problems, including wave patterns and population cycles.

Angle Measurement

Angle measurement is fundamental when working with trigonometric functions, as it refers to the size of an angle in various units, typically degrees or radians. In trigonometric problems like our population model, angles are usually measured in radians.

Radians provide a natural way of describing angles for calculus and trigonometric functions because they directly relate arc length to radius. Here's a brief conversion:

• 1 full circle = \( 2\pi \) radians = 360 degrees.
• 1 radian ≈ 57.296 degrees.

When solving for \( x \) in trigonometric equations, the solutions initially found are in radians, such as \( \theta = \sin^{-1}(0.75) \approx 0.8481 \; ext{radians} \) for one part of our example.

Understanding radians and how to convert between radians and degrees is critical for both interpreting and solving trigonometric equations effectively. This knowledge facilitates accurate modeling and predictions based on sinusoidal functions.