gistfile1.txt

### **Question 1-1: Period of the Function \( f(x) = 3 \cdot \sin\left(\frac{x}{2}\right) \)**

**Answer:**  
The period of the function is \( 4\pi \).  

**Explanation:**  
- The general form of a sine function is \( f(x) = A \cdot \sin(Bx + C) + D \), where:  
  - \( A \) = amplitude  
  - \( B \) determines the period  
  - \( C \) = phase shift  
  - \( D \) = vertical shift  
- The **period** of a sine function is calculated as:  
  \[
  \text{Period} = \frac{2\pi}{|B|}
  \]  
- In the given function \( f(x) = 3 \cdot \sin\left(\frac{x}{2}\right) \), \( B = \frac{1}{2} \).  
- Plugging in:  
  \[
  \text{Period} = \frac{2\pi}{\frac{1}{2}} = 4\pi
  \]  
- **Conclusion:** The function completes one full cycle every \( 4\pi \) units.  

---

### **Question 1-2: Height of the Water at a Beach**

**Correct Statements:**  
✅ **The difference between the highest and lowest water values is 6 feet.**  
❌ The difference between the highest and lowest water values is 12 feet.  
✅ **The water level is at its highest value 6 hours after midnight.**  
❌ The water level is at its highest value at midnight.  
❌ The water level is at its highest value 3 hours after midnight.  

**Explanation:**  
The function given is:  
\[
h(t) = 3 \cdot \sin\left(\frac{2\pi t}{12} - \frac{\pi}{2}\right)
\]  
- **Amplitude (A):** 3  
  - The difference between the highest and lowest water levels is \( 2A = 6 \) feet.  
- **Period:**  
  \[
  \text{Period} = \frac{2\pi}{\frac{2\pi}{12}} = 12 \text{ hours}
  \]  
- **Phase Shift:**  
  The function is shifted right by:  
  \[
  \text{Shift} = \frac{\frac{\pi}{2}}{\frac{2\pi}{12}} = 3 \text{ hours}
  \]  
  However, the negative sign inside the sine function means the **maximum** occurs at:  
  \[
  \frac{\text{Period}}{4} + \text{Shift} = 3 + 3 = 6 \text{ hours after midnight.}
  \]  
- **Conclusion:**  
  - The water varies between \(-3\) and \(+3\) feet (6 feet difference).  
  - The highest water level occurs at **6 hours after midnight**.  

---

### **Question 1-4: Trigonometric Function Graph**

**Answers:**  
1. **Midline:** \( y = 2 \)  
2. **Amplitude:** 3  
3. **Period:** \( 4\pi \)  
4. **Equation:** \( y = 3 \cdot \sin\left(\frac{x}{2}\right) + 2 \)  

**Explanation:**  
- **Midline:** The horizontal center of the graph (average of max and min).  
- **Amplitude:** Half the distance between max and min (height from midline to peak).  
- **Period:** The distance between two peaks (how long one cycle takes).  
- **Equation:**  
  - The graph resembles a sine function shifted up by 2.  
  - The amplitude is 3, and the period is \( 4\pi \), so \( B = \frac{1}{2} \).  
  - Thus, the equation is:  
    \[
    y = 3 \cdot \sin\left(\frac{x}{2}\right) + 2
    \]  

---

### **Question 1-5: Han’s Vertical Position on the Carousel**

**Answer:**  
\[
v(t) = 11 \cdot \sin\left(\frac{\pi t}{8}\right)
\]  

**Explanation:**  
- The carousel completes **1 revolution every 16 seconds**, so the period is 16.  
- The angular frequency (\( B \)) is:  
  \[
  B = \frac{2\pi}{\text{Period}} = \frac{2\pi}{16} = \frac{\pi}{8}
  \]  
- Han is **11 feet from the center**, so the amplitude is 11.  
- Since the carousel starts moving at \( t = 0 \), and Han starts at the midline (assuming standard position), we use a sine function:  
  \[
  v(t) = 11 \cdot \sin\left(\frac{\pi t}{8}\right)
  \]  

---

### **Question 1-6: Graph of \( f(\theta) = 3 \cdot \cos{(\theta + \pi)} \)**

**Graph Description:**  
- **Amplitude:** 3  
- **Phase Shift:** Left by \( \pi \)  
- **Midline:** \( y = 0 \)  
- **Key Points:**  
  - Starts at \(-3\) (since \( \cos(\pi) = -1 \)).  
  - Crosses midline at \( \theta = -\frac{\pi}{2} \).  
  - Reaches maximum at \( \theta = 0 \).  

**Explanation:**  
- The cosine function is shifted left by \( \pi \).  
- The amplitude is 3, so it oscillates between \(-3\) and \(+3\).  
- The graph looks like an upside-down cosine wave starting at \(-3\).  

---

### **Question 1-7: Daylight Hours in Glasgow**

**Answers:**  
1. **Amplitude:** 5.3  
   - **Meaning:** The number of daylight hours varies by **±5.3 hours** from the average.  
2. **Period:** 365 days  
   - **Meaning:** The daylight pattern repeats **every year (365 days)**.  
3. **Midline:** \( y = 12.28 \)  
   - **Meaning:** The **average daylight** over the year is **12.28 hours**.  

**Explanation:**  
The function is:  
\[
f(d) = 5.3 \cdot \cos\left(\frac{2\pi d}{365}\right) + 12.28
\]  
- **Amplitude (5.3):** The maximum deviation from the average.  
- **Period (365):** The time for one full cycle (matches Earth’s orbit).  
- **Midline (12.28):** The average daylight hours.  

---

### **Final Summary**
| Question | Key Answer |
|----------|------------|
| **1-1** | Period = \( 4\pi \) |
| **1-2** | Correct: 6 ft difference, highest at 6 hours |
| **1-4** | Midline: 2, Amplitude: 3, Period: \( 4\pi \), Equation: \( y = 3\sin\left(\frac{x}{2}\right) + 2 \) |
| **1-5** | \( v(t) = 11\sin\left(\frac{\pi t}{8}\right) \) |
| **1-6** | Shifted left cosine, amplitude 3, starts at \(-3\) |
| **1-7** | Amplitude: 5.3, Period: 365, Midline: 12.28 |

Let me know if you need any further clarification!