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((2\pi)/(365)[x - 80])`

(c)
`Hours= 1.794`

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\ + (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((2\pi)/(T)(x -n))`

Where

`\mu = Average\ Value`

`\mu = 12\ hr 8\ min`

Convert to hours

`\mu = 12 + (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((2\pi)/(T)(x -n))`

`f(x) = 12.133 + 2.883sin((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((2\pi)/(365)[x - 80])`

`f(41) = 12.133 + 2.883sin((2\pi)/(365)[41 - 80])`

`f(41) = 12.133 + 2.883sin((2\pi)/(365)[-39])`

`2\pi = 360^\circ`

So:

`f(41) = 12.133 + 2.883sin((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`