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

Video transcript

Sal:The longest day of the year in Juneau, Alaska is June 21st.
It's 1,096.5 minutes long. Half a year later when
the days are at their shortest the days are
about 382.5 minutes long. If it's not a leap year
the year is 365 days long and June 21st is the
172nd day of the year. Write a trigonometric function that models the length L of the T-th day of the year. It's going to be L as a function of T, assuming it's not a leap year. I encourage you to pause
this video and try to do this on your own before
I try to work through it. Let me give a go at it. 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 use as an intermediary variable that will help
set it up in a simpler way. Where U is days after June 21st. Let's just think about this a little bit. June 21st. If we're thinking about in
terms of U, U is going to be equal to zero because it's
zero days after June 21st. But if we're thinking
about it in terms of T, June 21st is the 172nd day of the year. 172. What's the relationship between U and T? It's shifted by 172 days. U is going to be equal to T minus 172. Notice when T is 172, U is equal to zero. Let's figure out L of
U first and then later we can just substitute U with T minus 172. First of all what's happening
when U is equal to zero? Let me write all of this down. What is happening when U is equal to zero. U equals zero is June 21st
and that's the maximum point. What trig function hits its maximum point when the input into the
trig function is zero? Well sine of zero is zero
while cosine of zero is one. Cosine hits its maximum point. It seems a little bit easier
to model this with a cosine. It's going to be some amplitude times cosine of, let's say, I'll write some coefficient
C right over here. Actually just let me use a
B since I already used an A. Some coefficient over
here times our U plus some constant that will shift the
entire function up or down. This is the form that our
function of U is going to take. Now we just have to figure out
what each of these parts are. First let's think about what the amplitude and what the mid line is going to be. The mid line is essentially how much we're shifting the function up. Let's get our calculator out. The mid line is going to be half way between these two numbers. We could say 1,096.5 plus 382.5 divided by two gets us to 739.5. That's what C is equal
to. C is equal to 739.5. Now the amplitude is how much
do we vary from that mid line. We could take 1,096 minus this or we could take this minus 382.5. Let's do that. Let's take 1,096.5 minus what we just got, 739.5 and we get 357. This is how
much we vary from that mid line. A is equal to 357. This right over here is equal to 357. What's B going to be equal to? For that I always think,
"What's the behavior of the "function, what's the period
of the function going to be?" Let me make a little table here. Let's put some different inputs from U. When U is zero we're zero
days after June 21st, we're at our maximum point. And we already said that what
we want the cosine function to evaluate to at that
point is essentially we want it to evaluate as 357 times
cosine of zero plus 739.5. Now what's a full period?
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. You go all the way to 365, when U is 365 we should
have completed a period. We should be back to that maximum point. This should essentially be
357 times cosine of two pi. If we were just thinking in terms of a traditional trig function, if
we just had a theta in here, you complete a period every two pi. This should be equivalent to
what I'm writing out over here. Plus 739.5. One way to think about it is B times 365 should be equal to two pi. Notice this is going to be B times 365, so let's write that down. B times 365. That's the input into the cosine function needs to be equal to two pi. Or B is equal to two pi over 365. B is equal to two pi over 365. We are almost done. We
figured out what A, B and C are and now we just have to
substitute U with T minus 172 to get our function of T. Let's just do that. We get -- we deserve a little
bit of a drum roll now. L of T is equal to A, which is 357 times cosine
of B two pi over 365. Two pi over 365 times not U, but now we're going
to write it 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 mid line. Plus 739.5. We are done. It seems like a very
complicated expression but if you just break it
down and think about it, make the point that we're
talking about the extreme point, either the minimum or the maximum. Make that when the input into
our function is zero or two pi. Zero is actually the easiest one. Then later you can worry about the shift. Hopefully you found that helpful.