Answer to Question #103430 in Trigonometry for tamara

Question #103430
Can hours of daylight data be modeled as a sinusoidal function for every location on earth? Explain.
1
Expert's answer
2020-02-20T12:54:55-0500

As we know that earth is completing one round around the sun in "365\\dfrac{1}{4}" days on average.

The longest day is June 21, on this day sun rise at 4:47AM and sunset at 7:38PM, so the daylight hour is 14hour 51 min. So daylight hour is 14.85hour.

And on the shortest day is December 21, on this day sun rise at 7:24AM and sunset at 4:54PM, so the daylight hour is 9.48 daylight.

So the amplitude of the daylight hour = "\\dfrac{14.85-9.48}{2}=2.685" hours

mid line of the ="\\dfrac{14.85+9.48}{2}=12.165hours"

The average time period of the cosine function ="365"

So, the cosine function without the horizontal shift "f(x)=2.685\\sin(\\dfrac{2\\pi}{365}t)+12.175"

So, if we are taking any random day for example Jan 14,

It is the 207th day after the longest day.

So, "f(207)=2.685\\sin(\\dfrac{2\\pi}{365}207)+12.175=9.734hours"

Which is reasonable as compared to 21Dec, from the shortest day, which is just 3 weeks near to the shortest day.


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Comments

James Way
14.08.21, 10:23

Thank you so much! I was given a data set of the daylight hours of a couple of cities (Tokyo,Yakutsk and Adelaide). Thanks to this article, I did not need to plot the graphs on Desmos to calculate the amplitude. I also have a question. How does the amplitude vary with the geographical location of the cities? What is the relation? Thank you once again!

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