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Current time:0:00Total duration:10:55

Trig word problem: modeling daily temperature

Video transcript

Voiceover:In Johannesburg in June, the daily low temperature is usually around 3 degrees Celsius and the daily high temperature is around 18 degrees Celsius. The temperature is typically halfway between the daily high and the daily low at both 10 a.m. and 10 p.m. and the highest temperatures are in the afternoon. Write a trigonometric function that models the temperature capital T in Johannesburg lowercase t hours after midnight. So let's see if we can start to think about what a graph might look like of all of this. So this, let's say this is our temperature axis in Celsius degrees. So that is temperature, temperature, and I'm actually going to first, I'm going to do two different functions. So that's my temperature axis. And this right over here is my time in hours. So, that's lower case t, time in hours and lets think about the range of temperatures. So the daily low temperature is around 3 degrees Celsius. So lets actually, and the high is 18, so lets make this right over here 18, this right over here is 3 and we can also think about the midpoint between 18 and 3 that we hit at both 10 a.m. and 10 p.m. 18 plus 3 is 21 divided by 2 is 10.5. So the midpoint, or we can say the mid line of our trigonometric function is going to be 10.5 degrees Celsius. Let me draw the mid line. So we're going to essentially oscillate around this right over here, we're going to oscillate around this. The daily high is around 18 degrees Celsius. The daily high is around 18 degrees Celsius and the daily low is around 3 degrees, 3 degrees Celsius, just like that. So we're going to oscillate around this mid line. We're going to hit the lows and the highs. Now to simplify things because we hit this 10.5 degrees at 10 a.m. and 10 p.m., to simplify this I'm not going to tackle their question that they want immediately, the hour in terms of t hours after midnight. I'm going to define a new function. F of t, F of lowercase t which is equal to the temperature, temperature in Johannesburg, we're assuming everything is in Johannesburg. Temperature t hours after, I'm going to say t hours after 10 a.m. The reason why I'm picking 10 a.m. is because we know that the temperature is right at the mid line at 10 a.m. t hours after 10 a.m. Because if I want a graph F of t at t equals zero that means we're at 10 a.m., that means that we're halfway between, they tell us, we're halfway between the daily low and the daily high. Now what is the period of this trigonometric function going to be? Well after 24 hours we're back to 10 am, so our period is going to be 24 hours. So we put 24 hours there and then this is, halfway is 12 hours. So what happens after 12 hours? After 12 hours we're back at 10 p.m. where we're back at midway between our lows and our highs. And then after 24 hours we're back at 10 a.m. again so those are going to be points on F of t. And now lets think about what will happen as we go beyond, as we start at 10 a.m. and go forward. So as we start at 10 a.m. and go forward they tell us that the hottest part, the highest temperatures are in the afternoon. The afternoon is going to be around here. So we should be going up in temperature and the highest point is actually going to be halfway between these two so it's going to be 6 hours after 10 a.m. which is 4 p.m. so that's going to be the high at 4 p.m. So let me draw a curve, draw our curve like this. So it'll look like this. And then our low, so now we're at 10 p.m. and then you go 6 hours after 10 p.m. you're now at 4 a.m. which is going to be the low. This is 18 hours after 10 a.m. you're going to be at your low temperature, roughly right over there. And your curve will look something like this. So what would be, before we even try to model T of t, what would be an expression, and obviously we keep going like that we can even go hours before 10 a.m. This is obviously, just keeps on cycling on and on forever. Now, what would be an expression for F of t and I encourage you once again to pause the video and try to think about that. Well, one thing that you say, well this could be a sine or a co-sine function, actually you could model it with either of them but it's always easiest to do the simplest one. Which function is essentially at it's midline when the argument to the function is zero. Well the sine of zero is zero and if we didn't shift this function up or down the mid line of just a sine function, without it being shifted, is zero. So sine of zero, zero and then sine begins to increase and oscillate like this. So it feels like sine is a good candidate to model it with. Once again you could model it with either but I have a feeling that this is going to be a little bit simpler. Now lets think about the amplitude. Well how much do we vary, what's our maximum variance from our mid line? So here we are 7.5 above our mid line. Here we are 7.5 below our mid line. So our amplitude is 7.5 and actually let me just do that in a different color just so you see where things are coming from. So this is 7.5, this is 7.5, so our amplitude looks like it's 7.5. And now, what is our period? Well we've already talked about it, our period is 24, 24 hours. This distance right over here is 24 hours which makes completes sense, after 24 hours you're at the same point in the day. So we would divide 2 pie by the period, divided by 24 times t. And if you forget, hey you know divide 2 pie by the period here, you could just remind yourself that what t value will make us go from, so when t is equal to zero the whole argument to the sine function is going to be zero, that's when we're over here. And then when t is equal to 24 the whole argument is going to be 2 pie. So we would have made one rotation around the unit circle if you think about the input into the sine function. Now we're almost done. If I were to just graph this, this would be, have a mid line around zero but we see that we've shifted everything up by 10.5. So we have to shift everything up by 10.5. Now, this is, we've just successfully modeled it and we could simplify a little bit we could write this as pie over 12 instead of 2 pie over 24. But this right over here models the temperature in Johannesburg, t hours after 10 a.m., after 10 a.m. That's not what they wanted, they want us to model the temperature t hours after midnight. So what would T of t be? We're going to have to shift this a little bit. So lets just think about it a little bit, let me just write it out. So T of t, this is now, we're modeling t hours after, after midnight. So we're going to have this same amplitude. We're still going to have the same variance from the mid line. So it's going to be 7.5 times sine of, I should be doing it in the same color so you see what I'm changing and not changing. Times the sine of, instead of 2 pie over 24 I'll just write pie over 12, instead of writing t I'm going to shift t, either to the right or left. Actually you can shift in either direction because this is a periodic function we're going to have to think about how much we're shifting it. So t is going to be plus or minus something right over here. I'm going to shift it, plus 10.5. Plus 10.5, now this is always a little bit, at least in my brain I have to think about this in a lot of different ways so that I can make sure that I'm shifting it in the right direction. So here at 10 a.m. we were at this point. When t is equal to zero, this is zero hours after 10 a.m. But in this function when is 10 a.m. Well in this function 10 a.m., let me write it this way. 10 a.m. is ten hours after midnight. So T, capital T of 10, this is 10 hours after midnight should be equal to F of zero because here the argument is hours after 10 a.m. so this is 10 a.m. this right over here represents temperature, temperature at 10 a.m. and this over here, cause capital T function, this is hours after midnight. This is also temperature at 10 a.m. So we want T of 10 to be the same thing as F of zero. Or another way of thinking about it, when F of zero, this whole argument is zero. So we want this whole argument to be zero when T is equal to 10. So how would we do that? Well this is t minus 10, notice T of 10, you put a 10 here this whole thing becomes a zero. This whole thing becomes zero and you're left with 10.5 and over here F of zero well the same thing, this whole thing becomes a zero and all you're left with over here is 10.5. So T of 10 should be F of zero. So if we wanted to graph it, we've already answered their question. If we put a 10 here, the argument to the sine becomes zero, these two things are going to be equivalent. But lets actually graph this. So T of 10, so if we're graphing capital T, T of 10 so this is 6, 12 lets see so this is maybe, so 10 is going to be someplace around here. So T of 10 is going to be the same thing as F of zero. So it's going to be like that and then it's just going to, and then we've essentially just shifted everything to the right, everything to the right by 10. That makes sense because zero after whatever hours you are after 10 a.m. is going to be 10 more hours to get to that same point after midnight. So your curve is going to look, so this is going to be shifted by 10, this is going to be shifted by 10 and your curve is going to look something like, let me see my, this is going to be shifted by 10 so you're going to get, this is going to be at 16 hours, so it's going to look something like that. And of course it will keep oscillating like that. So essentially we just have to shift it to the right by 10, the argument we have to replace t with t minus 10 to do it and this was the logic y.