Math Problem Statement

In a certain​ city, the number of hours of sunlight on the summer solstice is 15.788 and the number of hours of sunlight on the winter solstice is 8.276. Answer parts​ (a) through​ (c) below. Assume that summer solstice occurs on the 172Superscript nd day of the​ year, and that there are 365 days until the next summer solstice. Question content area bottom left Part 1 ​(a) Find a sinusoidal function of the form y equals Upper A sine left parenthesis omega x minus variant phi right parenthesis plus Upper B that fits the data. yequalsnothing sine left parenthesis nothing x minus nothing right parenthesis plus nothing             ​(Simplify your answer. Type an exact answer in terms of pi​.)

Solution

We need to find a sinusoidal function of the form:

y=Asin(ωxϕ)+By = A \sin(\omega x - \phi) + B

where:

  • The maximum number of daylight hours is 15.788 (on the summer solstice, x=172x = 172).
  • The minimum number of daylight hours is 8.276 (on the winter solstice, x355x \approx 355).
  • The period is 365 days (one full cycle from summer solstice to the next summer solstice).

Step 1: Find the Amplitude AA

The amplitude of a sinusoidal function is given by:

A=maxmin2A = \frac{\text{max} - \text{min}}{2}

A=15.7888.2762=7.5122=3.756A = \frac{15.788 - 8.276}{2} = \frac{7.512}{2} = 3.756

Step 2: Find the Midline BB

The midline (vertical shift) is the average of the maximum and minimum values:

B=max+min2B = \frac{\text{max} + \text{min}}{2}

B=15.788+8.2762=24.0642=12.032B = \frac{15.788 + 8.276}{2} = \frac{24.064}{2} = 12.032

Step 3: Find the Angular Frequency ω\omega

The period of the function is 365 days, so we use the standard formula:

ω=2πperiod\omega = \frac{2\pi}{\text{period}}

ω=2π365\omega = \frac{2\pi}{365}

Step 4: Find the Phase Shift ϕ\phi

Since the function should reach its maximum (15.788 hours) at x=172x = 172, we set up the equation:

sin(ω172ϕ)=1\sin(\omega \cdot 172 - \phi) = 1

Since sin\sin reaches its maximum at π2\frac{\pi}{2}, we set:

ω172ϕ=π2\omega \cdot 172 - \phi = \frac{\pi}{2}

Substituting ω=2π365\omega = \frac{2\pi}{365}:

2π365172ϕ=π2\frac{2\pi}{365} \cdot 172 - \phi = \frac{\pi}{2}

344π365ϕ=π2\frac{344\pi}{365} - \phi = \frac{\pi}{2}

Solving for ϕ\phi:

ϕ=344π365π2\phi = \frac{344\pi}{365} - \frac{\pi}{2}

ϕ=344π365182.5π365\phi = \frac{344\pi}{365} - \frac{182.5\pi}{365}

ϕ=161.5π365\phi = \frac{161.5\pi}{365}

Final Equation

y=3.756sin(2π365x161.5π365)+12.032y = 3.756 \sin\left(\frac{2\pi}{365} x - \frac{161.5\pi}{365}\right) + 12.032

Would you like a breakdown of any specific step?

Related Questions

  1. How do you determine the period of a sinusoidal function?
  2. Why is the sine function used instead of cosine in this problem?
  3. How would the function change if the number of daylight hours varied in a different pattern?
  4. Can we use a cosine function instead of a sine function for this problem?
  5. How would the function change if the period were 366 days (leap year)?

Tip

A phase shift helps align a sinusoidal function to real-world data, ensuring that the peaks and troughs occur at the correct times.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sinusoidal Functions
Amplitude
Angular Frequency
Phase Shift

Formulas

A = (max - min) / 2
B = (max + min) / 2
ω = 2π / period
sin(ωx - φ) = 1 at x = 172

Theorems

Sinusoidal Model for Real-World Data
Trigonometric Identity for Maximum Value

Suitable Grade Level

Grades 11-12