he level of water over time.

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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What’s the level of water over time.
### Understanding Sinusoidal Functions: Tsunami Example

A tsunami is approaching the San Francisco Bay! Here's a detailed breakdown of how the water level changes:

- The water first dips below its normal level.
- Then, it rises to a height equal to that below the normal level on the other side.
- Finally, it returns to its normal level after 20 seconds.

The key information we have:
- **Normal water depth:** 15 meters
- **Maximum water depth:** 26 meters

Let's write a sinusoidal function to represent the water level over time.

#### Step-by-Step Solution:

1. **Identify the amplitude:**
   The amplitude (\(A\)) is the difference between the maximum and the midpoint (normal level) of the wave.
   \[
   A = 26 \, \text{meters} - 15 \, \text{meters} = 11 \, \text{meters}
   \]

2. **Determine the period:**
   The period (\(T\)) is the time it takes for the wave to complete one full cycle. Given that the water returns to its normal level after 20 seconds:
   \[
   T = 20 \, \text{seconds}
   \]

3. **Calculate the frequency:**
   The frequency (\(f\)) is the reciprocal of the period.
   \[
   f = \frac{1}{T} = \frac{1}{20 \, \text{seconds}}
   \]

4. **Find the angular frequency:**
   The angular frequency (\(\omega\)) is calculated using:
   \[
   \omega = \frac{2\pi}{T} = \frac{2\pi}{20 \, \text{seconds}} = \frac{\pi}{10} \, \text{radians per second}
   \]

5. **Determine the vertical shift:**
   The vertical shift (\(D\)) is the normal water level.
   \[
   D = 15 \, \text{meters}
   \]

6. **Capture the phase shift:**
   We can use either sine or cosine as our base function. If we use cosine, which starts at the maximum value, our function does not need a phase shift.

#### Final Sinusoidal Function:
Using cosine (since it starts at the maximum):
\[
y(t) = A \cdot \cos(\omega t)
Transcribed Image Text:### Understanding Sinusoidal Functions: Tsunami Example A tsunami is approaching the San Francisco Bay! Here's a detailed breakdown of how the water level changes: - The water first dips below its normal level. - Then, it rises to a height equal to that below the normal level on the other side. - Finally, it returns to its normal level after 20 seconds. The key information we have: - **Normal water depth:** 15 meters - **Maximum water depth:** 26 meters Let's write a sinusoidal function to represent the water level over time. #### Step-by-Step Solution: 1. **Identify the amplitude:** The amplitude (\(A\)) is the difference between the maximum and the midpoint (normal level) of the wave. \[ A = 26 \, \text{meters} - 15 \, \text{meters} = 11 \, \text{meters} \] 2. **Determine the period:** The period (\(T\)) is the time it takes for the wave to complete one full cycle. Given that the water returns to its normal level after 20 seconds: \[ T = 20 \, \text{seconds} \] 3. **Calculate the frequency:** The frequency (\(f\)) is the reciprocal of the period. \[ f = \frac{1}{T} = \frac{1}{20 \, \text{seconds}} \] 4. **Find the angular frequency:** The angular frequency (\(\omega\)) is calculated using: \[ \omega = \frac{2\pi}{T} = \frac{2\pi}{20 \, \text{seconds}} = \frac{\pi}{10} \, \text{radians per second} \] 5. **Determine the vertical shift:** The vertical shift (\(D\)) is the normal water level. \[ D = 15 \, \text{meters} \] 6. **Capture the phase shift:** We can use either sine or cosine as our base function. If we use cosine, which starts at the maximum value, our function does not need a phase shift. #### Final Sinusoidal Function: Using cosine (since it starts at the maximum): \[ y(t) = A \cdot \cos(\omega t)
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