Math Problem Statement

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature is 70 degrees at midnight and the high and low temperature during the day are 78 and 62 degrees, respectively. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t.

Solution

We are tasked with modeling the temperature D(t)D(t) as a sinusoidal function, given that:

  • The temperature is 70°F at midnight (i.e., t=0t = 0).
  • The maximum temperature is 78°F.
  • The minimum temperature is 62°F.
  • tt represents the number of hours since midnight.

Step 1: Determine the amplitude

The amplitude of the sinusoidal function is half the difference between the maximum and minimum temperatures. The amplitude AA is:

A=Max tempMin temp2=78622=8A = \frac{\text{Max temp} - \text{Min temp}}{2} = \frac{78 - 62}{2} = 8

Step 2: Determine the vertical shift

The vertical shift (midline of the sinusoidal function) is the average of the maximum and minimum temperatures:

Vertical shift=Max temp+Min temp2=78+622=70\text{Vertical shift} = \frac{\text{Max temp} + \text{Min temp}}{2} = \frac{78 + 62}{2} = 70

Step 3: Determine the period

Since the temperature varies over a 24-hour period, the period PP of the sinusoidal function is 24 hours. For a sinusoidal function, the period is related to the angular frequency ω\omega by the equation:

P=2πωP = \frac{2\pi}{\omega}

Substituting P=24P = 24, we get:

24=2πω    ω=2π24=π1224 = \frac{2\pi}{\omega} \implies \omega = \frac{2\pi}{24} = \frac{\pi}{12}

Step 4: Determine the phase shift

At midnight (t=0t = 0), the temperature is 70°F, which is the midline of the sinusoidal function. Since a standard cosine function starts at its maximum value, we need a phase shift to move the cosine curve so that it starts at the midline. The phase shift should be a quarter of the period (6 hours) because the cosine function reaches the midline at P4\frac{P}{4}. Thus, the phase shift is:

Phase shift=6 hours\text{Phase shift} = 6 \text{ hours}

Since the temperature is increasing after midnight, we use a negative phase shift to shift the cosine curve to the right.

Step 5: Write the equation

The general form of the sinusoidal function is:

D(t)=Acos(ωtϕ)+Vertical shiftD(t) = A \cos(\omega t - \phi) + \text{Vertical shift}

Substituting the values we found:

  • A=8A = 8
  • ω=π12\omega = \frac{\pi}{12}
  • ϕ=6\phi = 6 (phase shift)

Thus, the equation for the temperature is:

D(t)=8cos(π12(t6))+70D(t) = 8 \cos\left(\frac{\pi}{12}(t - 6)\right) + 70

This is the sinusoidal model of the temperature throughout the day.


Would you like more details or have any questions about this model? Here are some related questions for deeper exploration:

  1. How does changing the amplitude affect the temperature model?
  2. What if the maximum temperature occurred at a different time of day—how would the phase shift change?
  3. How could you modify this model to account for seasonal variations in daily temperatures?
  4. What is the significance of the angular frequency ω\omega in terms of the daily temperature variation?
  5. How would this model change if the high and low temperatures varied slightly throughout the week?

Tip: In sinusoidal functions, the phase shift helps to adjust when the function reaches its maximum or minimum points, effectively shifting the curve horizontally.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sinusoidal Functions
Amplitude
Period
Phase Shift

Formulas

Amplitude = (Max temp - Min temp) / 2
Vertical shift = (Max temp + Min temp) / 2
Period P = 24 hours, related to angular frequency ω by P = 2π / ω
Sinusoidal equation: D(t) = A * cos(ωt - φ) + Vertical shift

Theorems

Sinusoidal Function Properties
Cosine Function Phase Shift

Suitable Grade Level

Grades 10-12