If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Trig word problem: modeling daily temperature

Sal solves a word problem about the daily change in temperature by modeling it with a sinusoidal function. Created by Sal Khan.

Want to join the conversation?

  • male robot donald style avatar for user Anish Upadhayay
    My equation was T(t) = 7.5sin(pi/12(t + 10) + 10.5
    So... basically the only difference between his and mine is the plus or minus 10. If t was AM in F(t) and we want it to be AM (midnight), we have to go back 10 hours. Going back 10 hours is going to the left and going to the left 10 hours is adding 10 to t. That is my reasoning, but with his, it would seem that it is hours after PM because he subtracts 10. Subtracting 10 means 10 hours to the right. AM plus 10 hours is PM, not 12 AM

    Am I wrong with my reasoning or is he wrong?
    (37 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Yamanqui García Rosales
      Think about what happens to the graph. The graph starts right now at AM, but we want the graph to start at AM (10 hours earlier, as you said). So what he have graphed so far needs to be 10 hours in the future, so the graph needs to move to the right. That is why you have to replace t with t - 10.

      In your equation, you are moving the graph 10 hours to the past, so your equation says, "What is the temperature t hours after PM".
      (35 votes)
  • aqualine ultimate style avatar for user nobleman5055
    Ok this video is very confusing can somebody please simplify it? Sometimes Sal breaks things down so much that it actually at first is pretty confusing.
    (27 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Adam Skonieczny
      He started at 10am because 10am is the at the mid line of the function- like a sine function. Then he graphed a sine function from that (and the other information that the problem gives) and since all trig functions are periodic- he simply needed to shift the function 10 hours to get the real formula that the question was asking. So he just attacked a problem by making it simpler :) Hope this helps
      (23 votes)
  • leafers sapling style avatar for user giokim
    how do i know if it might be a sine function or a cosine?
    (14 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user Ain Ul Hayat
    At , When Sal substitutes -10 isn't the time then going to be 8 p.m ?
    The question asks for temp after midnight which is after 12 a.m, shouldn't it be -14 then? if he is going to shift it to the right at 12 a.m.
    I don't understand can someone please explain?
    (13 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user carlschneiderlogging
    so, maybe i misconstrued something on the addition of 10 also (as others have stated), but if the function f(x) starts at 10 am, wouldn't the shift be 14 hours ahead (7.5sin(x-14)) + midline instead of "........(x-10)+midline)? please correct me if I'm mistaken, it happens often haha
    (11 votes)
    Default Khan Academy avatar avatar for user
    • duskpin ultimate style avatar for user Lydia
      It helps (for this example) if you switch to millitary time instead of two 12-hour cycles.
      10.00 minus 10 hours would be 0.00 (midnight) and/or 10.00 plus 14 hours equals 0.00 (digital clocks usually switch to 0.00 after 23.59 instead of showing 24.00)
      Therefore the equation "......(x-10))+midline" is correct, and the equation "......(x+14))+midline" is a correct solution.
      (4 votes)
  • blobby green style avatar for user Chuck Finley
    Did any one else come up with -7.5 sin(Pi/12x + Pi/6) + 10.5? I did not use the initial simplification that Sal employed. The temperature is below the average prior to 10 a.m. To model the initial decline and the subsequent rise towards the average the sine reflected in the line of the equilibrium temperature seemed appropriate, hence the -sin. I then simply phase shifted with x+2 in order for the avarage to be reached at t-2 or 10 p.m. of the previous day. Obviously that includes hours not under consideration, but it is just a way of making the graph line up.
    (9 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Akira
      Both graphs are the same so yours should be correct. You shifted the graph to the left by 2 whereas Sal shifted the graph to the right by 10.
      You might want to compare the two graphs using graphing calculator such as Desmos.com.
      (3 votes)
  • female robot ada style avatar for user Harini
    I dont get it...💀
    (8 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Festavarian
    In the graph of first function F(t) you can clearly see what happens to the temp. after midnight. What is the point of the T(t) function other than to make midnight =0 hours, or is that the only point of T(t)?
    (8 votes)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user nobleman5055
    What does he mean when he say "When the argument of the function=0" ?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • cacteye green style avatar for user Martin
    At Sal says that you could model it with either but what would the cosine version look like?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user InnocentRealist
      Here are the 4 equations I got (all from the original graph with t=0 at 10am (That's all you need.)

      7.5sin((pi/12)(t+14))+10.5; (sin(t*pi/12 + 7pi/6)).(t=0 (midline) for sine is 10am.)
      7.5sin((pi/12)(t-10))+10.5; (sin(t*pi/12 - 5pi/6)).
      7.5cos((pi/12)(u+8))+10.5; (cos(u*pi/12 + 2pi/3)). (u=0 (max) for cosine is 4pm.)
      7.5cos((pi/12)(u-16))+10.5; (cos(u*pi/12 - 4pi/3)).

      (If you extend the graph out another 24 hrs and relabel the x axis with 4pm = 0, you get the cosine graph and its proper 0 point).
      (6 votes)

Video transcript

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 in the daily   low at both 10:00 a.m. and 10:00 p.m.  and the highest temperatures are in the   afternoon right 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 then this right  over here is my time in hours so that's   lowercase T time in hours and let's  think about the range of temperatures   so the daily low temperatures around  3 degrees Celsius so let's actually   in the high is 18 so let's 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 it 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 midline  of our trigonometric function is going   to be 10 point 5 degrees Celsius  so let's let me draw the midline   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 the   3 degrees Celsius just like that so  we're going to oscillate around this   midline we're going to hit the lows and  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 tea   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 where we're   assuming everything is in Johannesburg  temperature T hours after I'm going to   say two hours after 10:00 a.m. and  the reason why I'm picking 10 a.m.   is because we know that the temperature  is right at the midline at 10:00 a.m. T   hours after 10:00 a.m. because if I want  to graph f of T at T equals zero that   means we're at 10:00 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 we're back we're going to be back to  10 a.m. so our period is going to be 24   hours so let me put 24 hours there and  then this is half ways 12 hours so what   happens after 12 hours after 12 hours  we're back at 10 p.m. where we're back   at the 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 let's think   about what will happen as we go beyond  as we start at 10 a.m. and go forward   so to go start at 10 a.m. and go for  they tell us that the hottest part 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   six 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 look like   this and then our low so now we're at 10  p.m. and then you go six 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. after 10 a.m. you're going to be at  your low temperature roughly right over   there and your curve your curve will  look something something like this so   what would be before we even try to  Model T of T let's what would be an   expression and obviously we keep going  keep going like that we can even go you   know hours before 10:00 a.m. this is  obviously just keeps on cycling on and   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 you say this is could  be a sine or a cosine function actually   you could model it with either of  them but it's always easiest to do   the simplest one which which function  is essentially at its midline at its   midline when 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 midline of   just a sine function without it being  shifted is is zero so sine of 0 is 0   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 this  is going to be a little bit simpler   now let's think about the amplitude well  how much do we vary what's our maximum   variance from our midline so here we  are 7.5 above our midline here we are   7.5 below our midline 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 point 5 this is 7 point 5 so our  amplitude looks like it's 7.5 and now   what is our period well we are we've  already talked about it our period is   24 24 hours this distance right over  here is 24 hours which makes complete   sense after 24 hours you got the same  point in the day so we would divide 2 pi   by the period divided by 24 times T and  if you forget hey you know divide 2 pi   by the period here you could just remind  yourself that what are.what T value will   make us go from so when T is equal to 0  the whole argument to the sine function   is going to be 0 that's when we're over  here and then when T is equal to 24 the   whole argument is going to be 2 pi so we  would have made one rotation around the   unit circle if we think about the input  into the sine function now we're almost   done if I were to just graph this this  would be this would be have a midline   around 0 but we see that we've shifted  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   modelled it and we could simplify  a little bit we could write this   as pi over 12 instead of 2 pi 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  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 let's just think about  it let me just write it out so T of T   so T of T this is now we're modeling  T hours after after midnight so we're   going to have the same amplitude we're  just going to have the same variance   from the midline so it's going to be 7.5  times sine of which we do the same color   so you see what I'm changing and not  changing times the sine of instead of   2 pi over 24 I'll just write pi over 12  instead of writing T I'm going to shift   T either to the to the right or the left  and actually you could shift in either   direction because this is a periodic  function we 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  ten point five plus ten point five 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 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 T 10 a.m. is 10 hours after   midnight so T capital T of 10 this is 10  hours after midnight should be equal to   should be equal to f of 0 because here  the argument is hours after 10 a.m. so   this is 10 a.m. this right over here  represents temperature temperature at   10:00 a.m. and this over here if because  capital did this capital T function this   is hours after midnight this is also  temperature at 10:00 a.m. so we want   T of 10 to be the same thing as f of 0  or a st. another way of thinking about   it when F of 0 this whole argument is  0 so we want this whole argument to be   0 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 0 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 point five so T of 10 should be F of  zero so if we wanted to graph it we've   already answered their question we've  if we put a 10 here the argument to   the sign becomes zero these two things  are going to be equivalent but let's   actually graph this so T of 10 so if  we're graphing capital T T of 10 so this   is 6 12 let's 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 0 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 and that makes sense because 0 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 if I this  is going to be shifted by 10 so you can   get I'm just going to get it's going to  be at 16 hours so let's see it's going   to look something something like that  and of course it'll 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 why.