### Modeling Daily Temperature with a Sinusoidal FunctionUnderstanding temperature variations over a day can be simplified using mathematical models. A sinusoidal function is particularly useful for this purpose because it captures the periodic nature of temperature changes. This section will guide you through deriving a sinusoidal equation to represent daily temperature variations.**Problem Statement :**- The high temperature for the day is 79 degrees.- The low temperature of 41 degrees occurs at 3 AM.- \( t \) represents the number of hours since midnight.Given these conditions, we need to find an equation for the temperature, \( D \), as a function of \( t \).The formula is generally in the form:\[ D(t) = A \sin(B(t - C)) + D \]Where:- \( A \) represents the amplitude of the function.- \( B \) adjusts the period of the sine wave.- \( C \) represents the phase shift.- \( D \) represents the vertical shift.**1. Amplitude (A):**The amplitude is half the difference between the maximum and minimum temperatures.\[ A = \frac{79 - 41}{2} = 19 \]**2. Vertical Shift (D):**The vertical shift is the average of the maximum and minimum temperatures.\[ D = \frac{79 + 41}{2} = 60 \]**3. Period:**A full cycle of temperature variation occurs over 24 hours.\[ \text{Period} = 24 \]The general sine function has a period of \( 2\pi \), therefore:\[ B = \frac{2\pi}{24} = \frac{\pi}{12} \]**4. Phase Shift (C):**Given that the low temperature of 41 degrees occurs at 3 AM, we use this to calculate the phase shift. Since the sine function normally reaches its trough at \( \frac{T}{4} \), where \( T \) is the period, we adjust for the shift:\[ t - C = \text{time for low temperature} \]\[ \frac{T}{4} = 6 \]\[ t - 3 = 6 \]\[ C = 9 \]Therefore, the equation becomes:\[ D(t) = 19 \sin \left(\frac{\pi}{12}(t - 9)\right) + 60 \]Thus

Write the equation of the temperature in terms of time.

Outside temperature over a day can be modelled as a sinusoidal function. Suppose you know the high temperature for the day is 79 degrees and the low temperature of 41 degrees occurs at 3 AM. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t. D(t) =

Outside temperature over a day can be modelled as a sinusoidal function. Suppose you know the high temperature for the day is 79 degrees and the low temperature of 41 degrees occurs at 3 AM. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t. D(t) =

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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### Modeling Daily Temperature with a Sinusoidal Function

Understanding temperature variations over a day can be simplified using mathematical models. A sinusoidal function is particularly useful for this purpose because it captures the periodic nature of temperature changes. This section will guide you through deriving a sinusoidal equation to represent daily temperature variations.

**Problem Statement :**
- The high temperature for the day is 79 degrees.
- The low temperature of 41 degrees occurs at 3 AM.
- \( t \) represents the number of hours since midnight.

Given these conditions, we need to find an equation for the temperature, \( D \), as a function of \( t \).

The formula is generally in the form:
\[ D(t) = A \sin(B(t - C)) + D \]

Where:
- \( A \) represents the amplitude of the function.
- \( B \) adjusts the period of the sine wave.
- \( C \) represents the phase shift.
- \( D \) represents the vertical shift.

**1. Amplitude (A):**
The amplitude is half the difference between the maximum and minimum temperatures.
\[ A = \frac{79 - 41}{2} = 19 \]

**2. Vertical Shift (D):**
The vertical shift is the average of the maximum and minimum temperatures.
\[ D = \frac{79 + 41}{2} = 60 \]

**3. Period:**
A full cycle of temperature variation occurs over 24 hours.
\[ \text{Period} = 24 \]

The general sine function has a period of \( 2\pi \), therefore:
\[ B = \frac{2\pi}{24} = \frac{\pi}{12} \]

**4. Phase Shift (C):**
Given that the low temperature of 41 degrees occurs at 3 AM, we use this to calculate the phase shift. Since the sine function normally reaches its trough at \( \frac{T}{4} \), where \( T \) is the period, we adjust for the shift:
\[ t - C = \text{time for low temperature} \]
\[ \frac{T}{4} = 6 \]
\[ t - 3 = 6 \]
\[ C = 9 \]

Therefore, the equation becomes:
\[ D(t) = 19 \sin \left(\frac{\pi}{12}(t - 9)\right) + 60 \]

Thus

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