 Welcome back to our lecture series, Math 1050, College Algebra for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misteldine. In this video, well, the first video for lecture 44, I wanna talk some more about exponential models. In the previous lecture, number 43, we talked a lot about exponential growth and decay. I wanna look at some variations of this. In this video, I wanna talk about what's known as Newton's Law of Cooling. The idea is the following. If you were to take a very hot substance, like a hot cup of chocolate or something, and put it in a very cold reservoir, the temperature of the environment would cool down the temperature of our hot object. But it wouldn't, and so it's kind of like a decay model of exponential, exponential is kind of like what the radioactive decay we talked about previously, but it wouldn't decay towards zero, it only would decay towards the temperature of the underlying room. And so, if you think about this, you know, your X and Y axis, your temperature starts off like really hot, but then it cools down, but it doesn't, again, it doesn't cool down towards zero, you're gonna cool down towards some asymptotic value that potentially could be non-zero. That is, it would be the temperature of the surrounding area, we'll call that T sub S for a moment. And this gives us Newton's Law of Cooling, which the temperature of our object, it'll equal some coefficient A, which we'll talk about what that means in just a second. We're gonna get E times negative KT plus TS, which as we try to unravel this right here, TS is gonna represent the surrounding temperature. So if we place our hot cup of chocolate, say in the refrigerator, or we take our hot apple pie and put it on the window seal there, it's gonna cool down based upon the surrounding temperature there. When it comes to this exponential function, could you do see E to the negative T there, right? It's gonna be exponential decay because of this negative number that's in place there. As we've seen previously, the plus K, or in this case, plus TS, a part of this exponential function gives us the location of the horizontal asymptote because this plus TS is gonna give us some type of vertical shift up or down on the graph and that'll move the asymptote from the X axis to this location right here. That's why we wanna use this. So Newton's Law of Cooling is basically exponential decay with a horizontal shift, excuse me, a vertical shift to move the horizontal asymptote. So it doesn't cool more than the surrounding environment. It also works with exponential heating that you could be rapidly heating towards this environment. So the farther away you are from the surrounding, you're gonna grow rapidly, but then it'll cool down as you get close. I should say the rate on which it's heating will change based upon how close you are to that asymptote, all right? So if TS is our surrounding environment, what this number K here is, K is gonna measure the rate in which this object cools down or heats up. We'll focus just on cooling problems. Different materials will cool at a different rate like a hot piece of iron, if we shove it in water, will cool down very differently than if you take a hot ceramic and put it in the cool water or something like that or a hot apple pie. So K has a lot to do with the substance that is cooling down in general. And then what about this number A right here? This is the coefficient. And this coefficient is gonna be the difference between T sub naught, which will be our initial temperature with the surrounding temperature TS. And we've talked about this in previous videos as well. As we tried to take the y-intercept, which is gonna be T sub naught here, we take the difference between T naught and TS right here. That gives us this stretch value of A, the vertical stretch. Again, we've done this as we try to graph exponential functions previously. So that's how we're gonna figure these things out. So A is gonna be the difference. It's kind of like the elbow room we have between the current, the initial temperature and the surrounding temperature. We have the surrounding temperature there. And then this K has to do with the rate of cooling of the substance. So imagine we have a cheesecake and it's taken out of the oven with an ideal internal temperature of 165 degrees Fahrenheit. And it's placed in a refrigerator with just 35 degrees Fahrenheit. After 10 minutes, the cheesecake has cooled to 150 degrees Fahrenheit. If we must wait until the temperature is cooled to 70 degrees before we can eat it, how long do we have to wait for that to happen? So we can model the cooling of this cheesecake using Sir Isaac Newton's law of cooling using the formula right here. So what data do we already have? So we have that the initial temperature T naught is equal to 165 degrees Fahrenheit. We have that the surrounding temperature that is the refrigerator is kept at a chilled 35 degrees Fahrenheit. We also have the observation that after 10 minutes, the temperature of the cheesecake lowers to be 150 degrees Fahrenheit. So using this information, we can build our model. So some things we can do. So A is supposed to be the difference of T naught and TS. So we're gonna take 135 and subtract from it 35 degrees. So we, I'm sorry, we have, got a little ahead of myself there. We got 165 take away 31, no, 35 degrees, there we go. That's gonna give us the 130 is as our A value right there. So our model will look something like the temperature T is gonna equal A, which is 130 times E to negative KT plus 35. So we don't know what K is yet. That's what we have to solve for in this situation. That's where the observation of what happened at 10 minutes is significant for us. We can solve for it by like, okay, the temperature was 150 10 minutes later. You get 130 E to the negative K. That time was 10 minutes plus 35. So what we wanna do is we wanna solve for K in this situation here. So what can we do? We can subtract 35 degrees from both sides of the equation. On the left-hand side, you're gonna get 150 take away 35, which is 115 degrees. We're gonna get 130 equals, or it's 130 times E to the negative 10 K right here. So divide both sides by 130, whoops, 130. So those cancel out right there. That then gives us that E to the negative 10 K is equal to, well, 115 over 130, which we can simplify the fractions. This becomes 23 over 26. I'm hesitating to use decimals because I don't want to get a rounding error. So I wanna keep things fraction. You don't have to reduce the fraction if you don't want to. Now to get rid of the natural exponential on the left-hand side, we need to take the natural log of both sides so that these cancel. That would then give us negative 10 K is equal to the natural log of 23 over 26. I mean, you could expound that if you wanted to be the natural log of 23 minus the natural log of 26, but you can keep it as fraction as well. It doesn't make much of a difference. Now to continue, divide both sides by negative 10. And so we end up with our K value. K is equal to, well, again, we have the natural log of 23 minus the natural log of 26 over negative 10, for which we're gonna try to estimate this thing, right? We do wanna have a lot of decimals so that we get accuracy here. I mean, we could pay attention to significant digits if you wanted to, but I'm not gonna worry about that in this mathematics lecture. We get the decimal 0.0123. Lovely if the next digit was a four. But anyways, we'll just round to four decimal places. So our K value is 0.0123. So now coming back to our model from before, right? We see that T equals 130 times E to the negative 0.0123T plus 35. Notice that because we used our T value as 10 minutes earlier, our rate is actually looking, it's being measured per minute. So we wanna measure these things in the future by minutes. We can always convert to hours and days if we had to, but be aware of T value, the way that we found for K here, this is gonna force T to be in minutes right here. All right, so what was the question again? When will the temperature reach 70 degrees? So that's what we wanna figure out. So now what we're tasked to do is we need to solve. So we need to solve for T in the equation. Well, when the temperature is 70 degrees, and then using our model here, so we just plugged in 70 in for capital T. So we have E, or 130 times E to the negative 0.0123. And then we're gonna times that by T, which we don't know T as yet plus 35. So what we're gonna do is we're gonna start solving for T, very much how we did over here, right? Logarithms are gonna come into play. So what can we do? We can subtract 35 from both sides. Again, you're gonna see that the arithmetic is essentially the same here. Some of the numbers are different, but again, the process will be the exact same way. So we get 35, which is 70 take away 35, is equal to 130 times E to, I'm just gonna write negative KT for the moment if that's okay, so I don't have to write the number over and over and over again. Divide both sides by 130, like so. If you want to reduce the fraction, you can. It's not super critical that you do so, especially since we're gonna throw this into the calculator in just a second, but you know, 35 divided by 130, there is a common factor of five. You could simplify it to be seven over 26, if you so choose. Take the natural log of both sides. We're gonna get negative KT is equal to the natural log of seven over 26. Or if you prefer, you could write that as the natural log of seven minus the natural log of 26. In which case, now we have to divide both sides by K. And so we end up with T is gonna equal, again, the natural log of seven minus the natural log of 26. This will sit above K, which honestly, K was this value from before, right? So you're gonna get a negative 10 times that divided by the natural log of 23 minus the natural log of 26, which although there's two subtractions of 26 there, it's not gonna do you a whole lot of good, just leave it alone. You can't just cancel the 26. The natural log of 26 crossed both sides of the equations and they're not divisors there. I meant to say across the fraction bar there. But anyways, you throw this number into your function or you could also have done just the natural log of seven over 26 divided by your K value from before the negative K, that was appropriate as well, whatever. But when we put this in our calculator, we're gonna end up with, rounding to the nearest minutes, we're gonna get 107 minutes, which when you have that timeframe, you usually probably wanna insert hours into there, 16 minutes in an hour, right? So this would be one hour and 47 minutes. Now, one has to be careful about this moment here in time, right? So we're taking 107 minutes from when we first placed the temperature, we placed the cheesecake inside of the refrigerator. But remember, after 10 minutes we measured it, it's temperature like, oh, it was, what was it, 135 or something like that? Yeah, excuse me, 150. Has it cooled down to after 10 minutes? So really, we should probably slash 10 minutes off at this time right here and be like, oh, 97 minutes in an hour and 37 minutes from now. That's really how we should view this. So we have to make sure we understand what was the initial point of time so that as we're using this model, we know that from this moment forward, how much time does it take? So after you did the 10 minute measurement of temperature, we need to wait another 97 minutes. So go watch a very short movie and come back and get your cheesecake in a little bit.