So in this problem, we have a warehouse. Um, and there's no heating or cooling, um, or active heating or cooling in that in the warehouse, and it depending on the amount of installation, the time constant for the building may range from 1 to 5 hours to illustrate the effect installation having the temperature inside the warehouse soon the outside temperature various as a sine wave with a minimum of 16 degrees C and a maximum of 32 degrees C. So minimum occurs at 2 a.m. The maximum occurs at 2 p.m. Assuming the exponential term has died off. What is the lowest temperature inside the building of the time? Constant is one hour. And if it is five hours, um and what is the highest temperature? So we have our outside temperature. I'm gonna call it t not out is an average temperature of outside. Um, plus some change. We can figure out the average is 24 degrees C. On the change up or down is eight degrees C. The period is, um, not Let's see, this is, um, 1/12 of this is ours, not seconds. Um, so pi over 12 hours, Um so basically half a day. Um, so let's see here. Now we have, um and basically, this is so when we go for when we go, 12 hours we go, we go 24 hours we go from, you know, uh, you know, go through one sine wave eso Let's see here our equation described the temperature inside the building is D T d t times k times t plus k times t equals k times t not. So here. We're looking. This is the temperature inside the building. Um, we have a homogeneous solution that looks like an exponentially decaying function. Um, and we're gonna assume that that is a die off, by the time way, reach kind of peaks temperatures. And so then we need a particular solution, and they give us the particular solution in the book. Um, one particular solution is simply, um, a constant this constant here. So obviously, if we plug a constant into here, we'll get the constant term here and then the, uh, sign or the coastline term. Here we have a particular solution that looks like like this here with an amplitude of delta t. Um, again, this is just giving derived for you in the book, so I'm not gonna derive it here. Um, what we could do, though, is to make this to get peaks. It's hard to figure out where the maximum here is because of you know, the sign, plus the coastline term. But what we can do is we can write this an amplitude phase form so we can combine these into one sine function with the same, um, time dependence, but then a phase and then the amplitude would be the square root of this down here. So again, we, um it's just basically some trig identities to turn this into an amplitude phase form on. Why we do that is because, Well, we know that whatever this phases, because we're not asked for win win the temperature hit the peak. We just asked what the peak is. Well, we know the peak of a sine function is one, and the minimum is minus one. So we know that. Then the minimum occurs when, um, sign is one. Because we have a minus sign here and we can plug in our values for omega for delta T for tea. Not. And if we have K equals one. We get 16.26 degrees Celsius. And if we use, um, K equals 1/5. So the time constant of five hours, we get 19 point for 14 degrees Celsius. So if we have ah longer time constants are minimum temperature is higher. Now for the maximum Again with the maximum is when workers when this term here is a minimum. So we have one sign when sign the sine function is minus one. So then when sine function is minus one, we get t not plus delta t all over this room and again, we can plug in values for that. And if K equals one, we get 31.74 to B C and K equals 1/5. We get 28.86 degrees c. So again we can see that the max is higher. Um, and the if the time constants lower. So when we have a local over time constant, we have a greater swing from high to low. I remember from high went, you know, we almost almost reached the outside temperatures here. Here. If we have a longer time constant, which means more insulation, we don't don't quite. You know, we get we don't reach those peaks on the outside a zoo closely, so we get a much lower swing and temperatures here, then we do from the outside. So again, the key here to do this is because they didn't ask for the time. I mean, we could solve for the time if we wanted to, but because they didn't ask for the time. We just write this in the amplitude phase for him. And then we just know that this has a peak, as a maximum sign is, the man has a maximum value of one and a minimum value of minus one. So, um, that's whatever the time is. We'll figure out what that maximum value is. That either that it's either one or it's minus one. Yeah.