Solving a Sinusoidal Equation for Hours of Daylight

• June 14th 2011, 11:36 AM
starshine84
Solving a Sinusoidal Equation for Hours of Daylight
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.

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?

This is what I've done so far....

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.

a= 3.1

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

365 days in degrees = 365/360= 1.01
b = 1.01

Max value of sine normally occurs at 90 degrees and minimum normally occurs at 270 degrees.
- June 21st is day 170... 170/1.01 = 168 degrees (Max occurs at 168 degrees)
- December 21st is day 350... 350/1.01 = 347 degrees (Min occurs at 347 degrees)
- 168 - 90 = 78 degrees
c = -78 degrees because the max and min points have been shifted 78 degrees to the right.

Therefore the equation would be y = 3.1 sin (1.01(x - 78)) + 12.18
OR
3.1 sin (1.01x - 78.78)???

Then, assuming that is right, to solve for 13.5 hours for question b I would make y 13.5 and solve for x.

y = 3.1 sin (1.01x - 78.78) + 12.18
13.5 = 3.1 sin (1.01x - 78.78) + 12.18
1.32 = 3.1 sin (1.01x - 78.78)
0.425806451 = sin (1.01x - 78.78)
sin^-1 0.425806451 = 1.01x - 78.78
25.201719199 = 1.01x - 78.78
103.9817199 = 1.01x
102.9521979 = x

Therefore, they would have 13.5 hours of sunlight on day 103?
• June 14th 2011, 02:58 PM
skeeter
Re: Solving a Sinusoidal Equation for Hours of Daylight
$B = \frac{2\pi}{365}$
June 21 is day $t = 172$
$\sin\left(\frac{\pi}{2}\right) = \sin\left(\frac{2\pi}{365} \cdot 172 - C\right)$
solve for $C$ ... note that using cosine instead of sine will reduce the phase shift.