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Sunrise, Sunset, Daylight Hours, and Trigonometry

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Sunrise, Sunset, Daylight Hours, and Trigonometry Sunrise, sunset, daylight hours, and trigonometry Phil Petrocelli, mymathteacheristerrible.com March 25, 2019 Anything that is intrinsically periodic can be described by trigonometric functions. Here is an inter- esting one. Question. At our city’s latitude, we see 16 hours of daylight on the 172nd day of the year and 8.5 hours of daylight on the 350th day of the year. What is f(t), the number of hours of daylight expected on day number t of each year? Solution. As I stated in the intro, this is an interesting function that is intrinsically periodic1. (Sorry not sorry, flat-earthers, but) Since the number of daylight hours we see in any particular place varies with the orbit of Earth around the Sun, we certainly have a cyclical relationship. Fortunately, coming up with our function doesn’t require celestial mechanics or anything like that, rather, it’s just like making a table of values by recording how much daylight we see every day for a year. What’s even more interesting is, that, since we know one year is 365.25 days, and we were given how many hours of daylight we see on both the shortest and longest days of the year, we have quite enough to develop our function! Three pieces of data! You may recall that for most cyclical relationships, a sine function or a cosine function will work. The general forms of both are: f(t) = A sin B(t C) + D (1) − f(t) = A cos B(t C) + D (2) − Where: A = Amplitude 2π B = period C = phase shift, also called horiz. shift or offset D = vertical shift, also called center line 1Also called cyclical, also called sinusoidal 1 Here, we can start with D. The center line is always the midway point between the maximum and minimum values of the function. An easy way to calculate this is to average the value of f(t) at its maximum and minimum. In our case: max + min D = 2 16 + 8.5 = 2 24.5 = 2 = 12.25 hours Next, we can have a look at A. A is the difference between the maximum of f(t) and the center line, D. In our case: A = max D − = 16 12.25 − = 3.75 We can easily move on to B. B is calculated using the period of our function, which we know to be 365.25 days. So: 2π B = period 2π = 365.25 At this point, all the easy calculations are out of the way. Now we have to choose whether we are going to use a sine function or a cosine function for our f(t). Note that A, B, and D will all be the same for either sine or cosine. Let’s think about our daily experience of observing number of hours of daylight. The most relevant question is: Does either the day with the most hours of daylight or the least hours of daylight occur on the first day of the year? Right. The answer to that question is certainly no. Now, think about the graphs of sine and cosine at t = 0, for C = 0. We know that the sine graph starts at y = D and increases, while the cosine graph starts at y = D + A and decreases. Please verify that these relationships between starting values hold for our general definitions of f(t) [(1) and (2)] above. Here, we do not have f(1), we simply have f(172) = 16 and f(350) = 8.5. So, for simplicity, let’s choose a cosine function and let’s say that its starting point is t = 172, where the maximum of f(t) occurs. With this intuitive step, we can recognize that the cosine function will be shifted right by exactly 172, and that means our angle, B(t C) will necessarily be 2π (t 172). Hence, our function can be stated as a cosine function, with our value − 365.25 − for C being 172. So: f(t) = A cos B(t C) + D − 2π = 3.75 cos (t 172) + 12.25 · 365.25 − � � We can verify with a calculator (in radians mode!) that this function gives us the correct values stated in the given information. 2 Please also verify that this function can also be written as a sine function. There are two ways to do it. The easiest way to do so is to simply apply the cofunction identity for sine: π sin θ = cos θ 2 − � � The other way to do it is to solve for C in the sine function by using the given information. Let’s use that f(350) = 8.5: 2π f(t) = 3.75 sin (t C) + 12.25 · 365.25 − � � 2π f(350) = 3.75 sin (350 C) + 12.25 · 365.25 − � � 2π 8.5 = 3.75 sin (350 C) + 12.25 · 365.25 − � � 2π 3.75 = 3.75 sin (350 C) − · 365.25 − � � 2π 1 = sin (350 C) − 365.25 − � � 3π 2π = (350 C) 2 365.25 − 3 365.25 · = 350 C 4 − 4383 = 350 C 16 − 1217 = C − 16 − 1217 = C 16 C = 76.0625 So, the sine version of our function is: 2π f(t) = 3.75 sin (t 76.0625) + 12.25 · 365.25 − � � Done. 3 Reporting errors and giving feedback I am so pleased that you have downloaded this study guide and have considered the techniques herein. To that end, I am the only writer and the only editor of these things, so if you find an error in the text or calculations, please email me and tell me about it! I am committed to prompt changes when something is inaccurate. I also really appreciate it when someone takes a moment to tell me how I’m doing with these sorts of things, so please do so, if you feel inclined. My email address is: [email protected]. Please visit https://mymathteacheristerrible.com for other study guides. Please tell others about it. Please donate I write these study guides with interest in good outcomes for math students and to be a part of the solution. If you would consider donating a few dollars to me so that these can remain free to everyone who wants them, please visit my PayPal and pay what you feel this is worth to you. Every little bit helps. My PayPal URL is: https://paypal.me/philpetrocelli. Thank you so much. 4.
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