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# Trig word problem: length of day (phase shift)

CCSS.Math:

## Video transcript

the longest day of the year in Juneau Alaska is June 21st it's one thousand ninety six point five minutes long half a year later when the days are at their shortest the days are about three hundred eighty two point five minutes long if it's not a leap year the year is 365 days long and June 21st is the hundred and seventy second day of the year right a trigonometric function that models the length L of the teeth day of the year so it's going to be L as a function of T assuming it's not a leap year so I encourage you to pause this video and try to do this on your own before I try to work through it so let me give a go at it so instead of first starting at L of T I'm going to actually start with L as a function of U where u is another variable I'll just use this kind of an intermediary an intermediate intermediary variable that'll help set it up in a little bit of a simpler way where u is days days after days after June June 21st so let's just think about this a little bit June 21st June 21st if we're thinking about in terms of U U is going to be equal to 0 because it's 0 days after June 21st but if we're thinking about in terms of T June 21st is 170 second day of the year so 172 so what's the relationship between U and T well it's shifted by 172 days U is going to be equal to T t minus 172 notice when T is 172 when T is 172 U is equal to 0 so let's figure out L of U first and then later we can just substitute U with T minus 172 so first of all what's happening when u is equal to 0 let me write all this down so what is happening when u is equal to 0 well u equal 0 is June 21st and that's the maximum point so what trig function hits its maximum point when the input into the trig function is zero well sine of 0 is 0 while cosine of 0 is 1 cosine hits its maximum point so I'd liked it seems a little bit easier to model this with a cosine so it's going to be some amplitude some amplitude times cosine times cosine of of and let's say I'll write some coefficient C right over here actually let me just use a B since I already used an a some coefficient over here times R u times R u plus some constant that'll shift the entire function up up or down so this is the form that our function of U is going to take and now we just have to figure out what each of these parts are so first let's think about the amplitude and what the midline is going to be the midline is essentially how much we're shifting the the function up so let's get our calculator out so the midline is going to be halfway between these two numbers so we could say 1096 0.5 plus three hundred eighty two point five divided by two divided by two gets us to seven hundred and thirty nine point five so that's what C is equal to C is equal to seven hundred and thirty nine point five now the amplitude is how much do we vary from that midline so we could take 1096 minus this or we could take this minus three hundred eighty two point five so let's do that so let's take one thousand ninety six 0.5 minus what we just type well we just got seven hundred thirty nine point seven hundred thirty nine one hundred thirty nine point five and we get 357 so this is how much we vary from that midline so a is equal to three hundred and fifty-seven so this right over here is equal to three hundred and fifty-seven so what's B going to be equal to and for that I always think about well what's the behavior of the function what's the period of the function going to be let me make a little table here so so when when you let's put some different inputs from you from you when you is 0 where 0 days after June 21st were at our maximum point and we already said that what we want what we want the cosine function to evaluate to at that point is essentially we want it to evaluate as 357 times cosine cosine of zero plus seven hundred thirty nine point five now what happens what's a full period well a full period is a year at a year we get to the same point in the year which is I guess a little bit of common sense so you go all the way to 365 you go all the way to when you is 365 we should have completed a period we should be back to that maximum point so this should essentially be 357 times cosine times cosine of 2 pi when do we when do we if we were just thinking in terms of a traditional trig function if we just had a you know just a de in here you complete a period every 2 pi so this should be equivalent to what I'm writing out right over here plus seven hundred thirty nine point five so one way to think about it is B times 365 should be equal to 2 pi notice this is going to be B times 365 so let's write that down B times 365 times 365 that's the input into the cosine function needs to be equal to 2 pi or B is equal to 2 PI over 365 B is equal to 2 PI over 365 and we are almost done we figured out what a B and C are now we just have to substitute U with T minus 172 to get our function of T so let's just do that so we get we deserve a little bit of a drumroll now L of T is equal to is equal to a which is 357 357 times cosine of B 2 PI over 365 2 PI over 365 times not you but now we're going to write in terms of T we want to think about day of the year not days after June 21st so times T minus 172 and then finally plus our midline plus seven hundred thirty nine point five and we are done so it seems like a very complicated expression but if you just break it down and kind of think about it make the point that we're talking about the extreme point either the minimum or the maximum make that something make that when the input into our function is zero or 2pi all right zero is actually the easiest one and then later you can worry about the shift hopefully that found you found that helpful