Problem 65 - Calculus Single Variable

Question

The lowest monthly normal temperature of Nome is \(4^{\circ} \mathrm{F}\) and occurs at the end of December \((t=12)\). The highest monthly normal temperature of Nome is \(51^{\circ} \mathrm{F}\) and occurs at the beginning of July \((t=6)\). Find a model of temperature \(T\) as a function of time \(t\) that has the form \(T(t)=\) \(b+A \sin (\omega t+\phi)\).

Step-by-Step Solution

Answer

The temperature model is \( T(t) = 27.5 + 23.5\sin\left(\frac{\pi}{6}t\right) \).

Step 1: Identify the Components

The temperature model has the form \( T(t) = b + A\sin(\omega t + \phi) \). We need to determine the values for \( b \), \( A \), \( \omega \), and \( \phi \) using the given information: The lowest temperature is at \( t=12 \) and the highest is at \( t=6 \).

Step 2: Calculate the Amplitude and Midline

The amplitude \( A \) can be calculated as half of the difference between the maximum and minimum temperatures: \( A = \frac{51 - 4}{2} = 23.5 \). The midline \( b \) is the average of the maximum and minimum temperatures: \( b = \frac{51 + 4}{2} = 27.5 \).

Step 3: Determine the Frequency

The frequency \( \omega \) depends on the periodicity of the seasonal temperatures. There are 12 months in a year, so \( \omega = \frac{2\pi}{12} = \frac{\pi}{6} \).

Step 4: Find the Phase Shift

Given the highest temperature occurs at \( t=6 \), where the sine function equals 1, we use \( \omega t + \phi = \frac{\pi}{2} \). Therefore, \( \frac{\pi}{6} \cdot 6 + \phi = \frac{\pi}{2} \). Solving for \( \phi \) gives \( \phi = \frac{\pi}{2} - \frac{\pi}{6} \cdot 6 = 0 \).

Step 5: Write the Final Temperature Model

By substituting the calculated values into the model, we find the temperature function: \[ T(t) = 27.5 + 23.5\sin\left(\frac{\pi}{6}t\right) \]. This is the required sinusoidal model for the temperature \( T(t) \).

Key Concepts

Amplitude

Amplitude is a key feature in trigonometric functions, representing how much variation there is from the midline of the wave. In our temperature model, the amplitude is calculated as the average distance from the highest to the lowest temperature.
This determines how extreme the temperatures get over the year. Mathematically, it is found by taking half of the difference between the maximum and minimum values.
For the model provided, the amplitude is calculated as:

• Maximum temperature = 51°F
• Minimum temperature = 4°F

Hence, the amplitude \( A \) is computed as:\[A = \frac{51 - 4}{2} = 23.5\]This value of 23.5°F indicates that the temperature can rise or fall by 23.5°F from the average temperature over the year. Understanding amplitude helps us grasp the extent of temperature fluctuations.

Frequency

Frequency in the context of trigonometric functions signifies how often a specific pattern or cycle repeats over a given period of time. In this temperature modeling problem, frequency is linked to how the temperatures vary throughout a typical year.
We know temperatures swing from high to low in a cycle that spans all 12 months. Thus, we need to set up the frequency to reflect this periodicity.
To find the frequency \( \omega \), we use the formula for annual cycles:

• The time for a full cycle (one year) is 12 months.

This gives the calculation:\[\omega = \frac{2\pi}{12} = \frac{\pi}{6}\]This part of the formula tells us the temperature pattern will repeat every year. The frequency, therefore, helps identify the regular interval of the temperature changes over time.

Phase Shift

Phase shift is another critical component in trigonometric modeling, accounting for the horizontal shifting of the wave. It reflects how the model aligns with actual data points, like the highest or lowest temperature months.
In this exercise, we ascertain the phase shift by looking at the point when the highest temperature occurs, tied to the cycle's peak point in July \( t=6 \).
The sine function reaches its maximum value at \( \frac{\pi}{2} \), giving:

• \( \omega t + \phi = \frac{\pi}{2} \).
• For July: \( \frac{\pi}{6} \times 6 + \phi = \frac{\pi}{2} \).

Solving for the phase shift \( \phi \):\[\phi = \frac{\pi}{2} - \frac{\pi}{6} \times 6 = 0\]This result shows the sine wave doesn’t need further shifting to match the data because reaching its peak at \( t=6 \) naturally aligns with the peak temperature.

Temperature Modeling

Temperature modeling using trigonometric functions enables capturing the natural oscillations inherent in seasonal temperature changes. We use the standard form: \( T(t) = b + A \sin(\omega t + \phi) \).
This equation modeled the annual temperature fluctuation in Nome by taking into account its seasonal highs and lows, calculated through amplitude, frequency, and phase shift.
The full expression for temperature \( T(t) \) based on our calculations becomes:\[T(t) = 27.5 + 23.5 \sin\left(\frac{\pi}{6}t\right)\]Here:

• \( b = 27.5 \) represents the average temperature (midline).
• \( A = 23.5 \) is the amplitude.
• \( \omega = \frac{\pi}{6} \) is the frequency for a yearly cycle.
• \( \phi = 0 \) indicates no phase shift is necessary.

This sinusoidal model mirrors the natural ebb and flow of temperature in Nome, Alaska, offering a precise prediction of temperature over the months. Understanding each component's role like amplitude and frequency helps in creating accurate models for various natural phenomena.