Question

The city of Windsor, Ontario, receives its maximum amount of sunlight of 15.28 hrs on June 21, and its least amount of sunlight of 9.08 hrs on December 21. (8 marks)

a. 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.

b. On what day(s) can Windsor expect 13.5 hours of sunlight?

https://gyazo.com/a9a16160463bbc11c41be51673da500c   

Answer 1

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.

Answer 2

Max value= 15.28 Min value= 9.08
Range = 15.28 - 9.08 = 6.2.... therefore, the function goes from -3.1 to 3.1 with zero in the middle.

K= 3.1

However, the function has been shifted vertically...
15.28 - 3.1= 12.18
D = +12.18

365 days in radians....

27T 365 B=-= 0.0172

june 21 is the day t....=172

sin(-) = sin(-*172-C)[equation] s11 365

ANOTHER FORM

assume that the orbit is circular and the daylight hours can be modeled as a sine wave. Since we are given times of maximum and minimum, we will use a cosine function. Let "d" represent hours of daylight.

15.28+9.08 +(15.28-9.08) *2*cos( is the number of days after June 21 365 d 12.18+3.10 cos 365 Solhving for t that will give 13.5 hours of davlight 13.5- 12.18+3.10cos 265 27t COS_ 1.32 3.10 27t 365 ,1.32 3.10 365 1.32 27 CoS 3.10 365 365*cos105.1 days] k-1,1.32 27T 65 days either way from 21 June gives .. August 25.. April 17