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# 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.

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• 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? •  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".
• 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. • 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
• how do i know if it might be a sine function or a cosine? • 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? • 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 • 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.
• 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)? • 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. • At Sal says that you could model it with either but what would the cosine version look like? • 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).  