Question

In New York City at the spring equinox there are 12 hours 8 minutes of daylight. The longest and the shortes 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 trigonamic function to model the hours of daylight hours on certain days of the year in New York City

Answer 1

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)

$A = 2.883$  --- Amplitude

$T = 365$ ---- Period

(b) Trigonometry function

$f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])$

(c) $Hours= 1.794$

Step-by-step explanation:

Given

$Average\ Sunlight = 12hr\ 8 min$

$Variance = 2hr\ 53min$

Solving (a): Amplitude (P) and Period (T)

The amplitude is the amount of time the longest and the shortest day vary.

So

$A = 2\ hr\ 53\ min$

Convert to hours

$A = 2\ + \frac{53}{60}$

$A = 2+0.883$

$A = 2.883$

The period (T) is the duration i.e 1 year

$T = 1\ year$

Assume no leap year

$T = 365$

Solving (b): Trigonometry function

The function follows a sinusoidal pattern and the general form is:

$f(x) = \mu+ Asin(\frac{2\pi}{T}(x -n))$

Where

$\mu = Average\ Value$

$\mu = 12\ hr 8\ min$

Convert to hours

$\mu = 12 + \frac{8}{60}$

$\mu = 12 + 0.133$

$\mu = 12.133$

$A = 2.883$  --- Amplitude

$T = 365$ ---- Period

$n = Equinox$

$n = March\ 21$

$March\ 21st = 80th\ day$

So:

$n= 80$

The function becomes:

$f(x) = \mu+ Asin(\frac{2\pi}{T}(x -n))$

$f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])$

Solving (c): Fewer daylight hours will Feb. 10 have.

$Feb\ 10 = 41st\ day$

So:

$f(x) = 12.133 + 2.883sin(\frac{2\pi}{365}[x - 80])$

$f(41) = 12.133 + 2.883sin(\frac{2\pi}{365}[41 - 80])$

$f(41) = 12.133 + 2.883sin(\frac{2\pi}{365}[-39])$

$2\pi = 360^\circ$

So:

$f(41) = 12.133 + 2.883sin(\frac{360}{365}[-39])$

$f(41) = 12.133 + 2.883sin(-38.466)$

$f(41) = 12.133 - 2.883*0.6221$

$f(41) = 10.339$

The fewer daylight hours is the calculated as:

$Hours= Average - f(41)$

$Hours= \mu - f(41)$

$Hours= 12.133 - 10.339$

$Hours= 1.794$