Question:
In New York City at the spring equinox there are 12 hours 8 minutes of daylight. The longest and the shortest days of the year vary by 2 hours 53 minutes from the equinox. In this year, the equinox falls on March 21. In this task, you'll use a trigonometric function to model the hours of daylight hours on certain days of the year in New York City.
- Find amplitude and the period of the function
- Create a trigonometric function that describes the hours of sunlight for each day of the year
- Then use the function you built to find how fewer daylight hours February 10 will have then March 21
Answer:
(a)
[tex]A = 2.883[/tex] --- Amplitude
[tex]T = 365[/tex] ---- Period
(b) Trigonometry function
[tex]f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])[/tex]
(c) [tex]Hours= 1.794[/tex]
Step-by-step explanation:
Given
[tex]Average\ Sunlight = 12hr\ 8 min[/tex]
[tex]Variance = 2hr\ 53min[/tex]
Solving (a): Amplitude (P) and Period (T)
The amplitude is the amount of time the longest and the shortest day vary.
So
[tex]A = 2\ hr\ 53\ min[/tex]
Convert to hours
[tex]A = 2\ + \frac{53}{60}[/tex]
[tex]A = 2+0.883[/tex]
[tex]A = 2.883[/tex]
The period (T) is the duration i.e 1 year
[tex]T = 1\ year[/tex]
Assume no leap year
[tex]T = 365[/tex]
Solving (b): Trigonometry function
The function follows a sinusoidal pattern and the general form is:
[tex]f(x) = \mu+ Asin(\frac{2\pi}{T}(x -n))[/tex]
Where
[tex]\mu = Average\ Value[/tex]
[tex]\mu = 12\ hr 8\ min[/tex]
Convert to hours
[tex]\mu = 12 + \frac{8}{60}[/tex]
[tex]\mu = 12 + 0.133[/tex]
[tex]\mu = 12.133[/tex]
[tex]A = 2.883[/tex] --- Amplitude
[tex]T = 365[/tex] ---- Period
[tex]n = Equinox[/tex]
[tex]n = March\ 21[/tex]
[tex]March\ 21st = 80th\ day[/tex]
So:
[tex]n= 80[/tex]
The function becomes:
[tex]f(x) = \mu+ Asin(\frac{2\pi}{T}(x -n))[/tex]
[tex]f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])[/tex]
Solving (c): Fewer daylight hours will Feb. 10 have.
[tex]Feb\ 10 = 41st\ day[/tex]
So:
[tex]f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])[/tex]
[tex]f(41) = 12.133 + 2.883sin(\frac{2\pi}{365}[41 - 80])[/tex]
[tex]f(41) = 12.133 + 2.883sin(\frac{2\pi}{365}[-39])[/tex]
[tex]2\pi = 360^\circ[/tex]
So:
[tex]f(41) = 12.133 + 2.883sin(\frac{360}{365}[-39])[/tex]
[tex]f(41) = 12.133 + 2.883sin(-38.466)[/tex]
[tex]f(41) = 12.133 - 2.883*0.6221[/tex]
[tex]f(41) = 10.339[/tex]
The fewer daylight hours is the calculated as:
[tex]Hours= Average - f(41)[/tex]
[tex]Hours= \mu - f(41)[/tex]
[tex]Hours= 12.133 - 10.339[/tex]
[tex]Hours= 1.794[/tex]
