# In 2001, Windsor, Ontario received its maximum amount of sunlight, 15.28 hrs, on June 21, and its least amount of sunlight, 9.08 hrs, on December 21. Due to the earth's revolution about the sun,...

In 2001, Windsor, Ontario received its maximum amount of sunlight,
15.28 hrs, on June 21, and its least amount of sunlight, 9.08 hrs, on
December 21.

1. Due to the earth's revolution about the sun, the hours of daylight function is periodic. Determine an equation that can model the hours of daylight function for Windsor, Ontario.
2. On what day(s) can Windsor expect 13.5 hours of sunlight?

embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

Since the phenomenon is cyclical we can model with a sinusoid. The model is `y=Asin(B(t-h))+k` where A is the amplitude; B is derived from the period, t is the time in days, h is the horizontal translation and k the vertical translation. (y=k is the midline.) y represents the amount of sunlight in hours.

A: The amplitude is `A=("max"-"min")/2` so `A=(15.28-9.08)/2=3.1`

B: `B=(2pi)/p` where p is the period. The amount of sunlight should have a period of 1 year or roughly 365 days. Then `B=(2pi)/365`

h: h is the phase shift or horizontal translation. We take t=0 to be Jan. 1. The maximum of the sine function occurs one quarter of the period away from the start. This would translate to day 91; here the maximum occurs at day 171 (June 21 is day 172 but we are taking Jan 1 to be t=0, so t=171) thus there is a phase shift of 80 days to the right.

h=80

k: k is the midline or the arithmetic mean of the maximum and minimum. So `k=(15.28+9.08)/2=12.18`

Our model is `y=3.1sin((2pi)/365(t-80))+12.18`

The graph:

(b) If y=13.5:

`13.5=3.1sin((2pi)/365(t-80))+12.18`

` ` `3.1sin((2pi)/365(t-80))=1.32`

`(2pi)/365(t-80)=sin^(-1)(.4258)`

`t-80~~(.4399)(365/(2pi))`

`t~~105.6`

But we also have to use the other possible angle for sine:

`t-80~~(2.702)(365/(2pi))`

`t~~236.9`

Thus Windsor will get 13.5 hours of sunlight on day 106 and day 237.