n 2001, Windsor, Ontario will receive its maximum amount of sunlight, 15.28 hrs, on June 21, and its least amount of sunlight, 9.08 hrs, on December 21.
a. 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?
Can someone please help me find the second-day using degrees, please?
I figured out a which is y-3.1(360/365(x-79.75))+ 12.18
I also figured out one of the days that Windsor can expect sunlight by setting the equation = to 13.5
13.5 =3.1sin(360/365(x-79.75))+12.18
1.32= 3.1sin(360/365(x-79.75))
sin(30/365(x-79.75)=1.32/3.1
360/365(x-79.75)= sin-1(1.32/3.1)
360/365(x-79.75) = 25.20171985
x-79.75= 360/365(25.20171985)
x-79.75= 0.9863013699(25.20171985)
x-79.75= 24.861876017
x= 104.6118702
so the first day would be 105
Can someone please help me find the second day using degrees please?
if sin(z) = 1.32/3.1 = 0.4258
then sin(180-z) is also 0.4259=8
so, solve 360/365(x-79.75) = 180-25.20171985
I think you would want daylight hours, not necessarily sunlight hours, and that the number of daylight hours would follow a sinusoidal pattern. Let's go with a sine function
your range = 15.28-9.08 = 6.2
so the amplitude is 3.1
looks like you are going from June 21 (start of summer) to Dec 21 (start of winter) from a max to a min in 1/2 year
So the period is 1 year , or 365 days
period = 2π/k
365 = 2π/k ----> k = .017214
sofar we have daylight hours = 3.1sin .017214 t , where t is the number of days
but we want our min to be 9.08 so we need
daylight hours = 3.1sin .017214 t + 12.18
BUT, June 21 is 172 days into the year, so when t = 172 we need 15.28 as our answer
We need a phase shift!
daylight hours = 3.1sin .017214(t + c) + 12.18
15.28 = 3.1sin .017214(172 + c) + 12.18
1 = sin .017214(172+c)
I know sin (π/2) = 1 , so
.017214(172+c) = π/2
172+c = 91.25
c = -80.75
daylight hours = 3.1sin .017214(t - 80.75) + 12.18
testing for Dec 21 or t = 355
daylight hours = 3.1sin .017214(355 - 80.75) + 12.18
= 9.08 , nice!
https://www.wolframalpha.com/input/?i=y+%3D+3.1sin%28.017214%28t+-+80.75%29%29+%2B+12.18+from+0+to+365
I'm sorry but 9.08 is not the answer. The answer should be 236.70. I'm having a hard time working that out as well
2nd part, when is hours = 13.5
3.1sin .017214(t - 80.75) + 12.18 = 13.5
sin .017214(t - 80.75) = .425806...
.017214(t - 80.75) = .43985....
(t - 80.75) = 25.55
t = 106.3 <----- day 106 or April 16th
The 9.08 answer part was just my test to check that the equation is correct.
I got an answer of 106 days, you claim the answer is 236.7
As you can see from the graph, there are two times in the year when there are 13.5 number of daylight hours, I simply found the first one.
The graph shows a 2nd answer of 13.5 hrs at the approximate time of 236 days
To find the second day using degrees, you can use the same process as you did to find the first day. Let's go through the steps:
1. Start with the equation: 13.5 = 3.1*sin(360/365(x-79.75)) + 12.18
2. Solve for sin(360/365(x-79.75)). Subtract 12.18 from both sides:
1.32 = 3.1*sin(360/365(x-79.75))
3. Divide both sides by 3.1:
sin(360/365(x-79.75)) = 1.32/3.1
4. Use the arcsin function to find the angle:
360/365(x-79.75) = arcsin(1.32/3.1)
5. Convert the angle from radians to degrees:
360/365(x-79.75) = (arcsin(1.32/3.1)) * (180/pi)
6. Solve for x by multiplying both sides by 365/360:
x - 79.75 = (arcsin(1.32/3.1)) * (180/pi) * (365/360)
7. Simplify the equation:
x - 79.75 = (arcsin(1.32/3.1)) * (1.013889)
8. Add 79.75 to both sides:
x = (arcsin(1.32/3.1)) * (1.013889) + 79.75
9. Calculate the value using a calculator:
x ≈ 90.4378218
Therefore, the second day, when Windsor can expect 13.5 hours of sunlight, corresponds to a value of approximately 90.44 degrees.