Main content
Current time:0:00Total duration:10:02

Video transcript

- [Voiceover] Let's think about another scenario that we can model with the differential equations. This is a scenario where we take an object that is hotter or cooler than the ambient room temperature, and we want to model how fast it cools or heats up. And the way that we'll think about it is the way that Newton thought about it. And it is described as Newton's Law of Cooling. And in a lot of ways, it's common sense. It states that the rate of change of temperature should be proportional to the difference between the temperature of the object and the ambient temperature. So let me write that in mathematical terms. So Newton's Law of Cooling tells us, that the rate of change of temperature, I'll use that with a capital T, with respect to time, lower case t, should be proportional to the difference between the temperature of the object and the ambient temperature. So that is a mathematical description of it. And once again, it's common sense. If something is much, much hotter than the ambient temperature, the rate of change should be pretty steep, it should be declining in temperature quickly. If something is much, much cooler, it should be increasing in temperature quickly. And if something is close, if these two things are pretty close, well maybe this rate of change shouldn't be so big. Now I know one thing that you're thinking. You're like, okay, if the temperature is hotter than the ambient temperature, then I should be cooling. My temperature should be decreasing. And a decreasing temperature would imply a negative instantaneous change. So how will this be a negative value in the case where our temperature of our object is greater than our ambient temperature? And the way that that would happen is, you would have to have a negative k. If you don't like thinking in terms of a negative k, you can just put a negative right over here and now you would have a positive k. Now it makes sense. If our thing is hotter, if it has a higher temperature than the ambient temperature, so this is a positive, then our rate of change will be negative, will be getting cooler. If it was the other way around, if our temperature of our object is cooler than our ambient temperature, then this thing is going to be a negative, and then the negative of that is going to be a positive, we're assuming a positive k, and our temperature will be increasing. So hopefully, this makes some intuitive sense. And our constant k could depend on the specific heat of the object, how much surface area is exposed to it, or whatever else. But now I'm given this, let's see if we can solve this differential equation for a general solution. And I encourage you to pause this video and do that, and I will give you a clue. This is a separable differential equation. So I assume you've had a go at it, so let's now work through it together. So, we just have to algebraically manipulate this so all my Ts and dTs are on one side. I should say, so all my capital Ts and dTs are on one side, this is going to be a little bit more confusing because I have a capital T and a lower case t. Capital T for temperature, lower case t for time. But hopefully we'll be able to work through it. And then I'm going to have all my time differentials and time variables on the other side. So one thing I could is I could divide both sides by T minus ambient temperature, minus T sub a. Remember this is just going to be a constant based on what our ambient temperature is. We're going to assume our ambient temperature doesn't change as a function of time, it's just such a big room that our cup of tea is not going to actually warm up the room. So that's just one of these assumptions that we're going to make. So if we do that, if we divide both sides by this, we are going to have... So I'm going to divide both sides, I'm going to do this in a new color. If I divide both sides by that, I get one over T minus T sub a, and let me multiply both sides times the time differential. So I'm going to have, that dT, our temperature differential. Times our temperature differential, is going to be equal to negative k times our time differential. So once again, to separate the variables, all I did was divide both sides by this, and multiply both sides by that. Now I can integrate both sides, we've seen this show before. So I can integrate both sides. And the integral of this is going to be the natural log of the absolute value of what we have in the denominator. And you can do u substitution if you want. If we said u is equal to T minus T sub a, then du is just going to be one dT, and so this is essentially, you could say the integral of one over u du, and so it would be the natural log of the absolute value of u, and this right over here is u. So this is the natural log of the absolute value of T minus T sub a, is equal to, and once again I could put a constant here, but I'm going to end up with a constant on the right hand side too so I'm just going to merge them into the constant on the right hand side. So that is going to be equal to, now here, this is going to be negative kt, and once again we have plus C. And now we can raise e to both of these powers, or another way of interpreting this is if e to this thing is going to be the same as that. So we can write this as, the absolute value, let me do that in that same blue color. We can write this as the absolute value of T minus T sub a is equal to e, something about e I always think of the color green. e to the negative kt plus C. This of course is the same thing as, this is equal to e to the negative kt, we've done this multiple times before. Negative kt times e to the C power. And we could just call this another arbitrary constant. If we called this C1, then we could just call this whole thing C. So this we could say is Ce to the negative kt. So at least it's starting to resemble what we did when we were modelling population. We'll see it's a little bit different. Instead of just temperature on this left hand side, we have temperature minus our ambient temperature. And so, we can do a couple of things. If, in a world, say we were dealing with a hot cup of tea, something that's hotter than the ambient temperature. So we could imagine a world where T is greater than or equal to our ambient temperature. So that means this is hot, or it's hotter, I guess we could say. So if we're dealing with something hotter than the ambient temperature, then this absolute value is going to be positive or the thing inside the absolute value is going to be positive. So we don't need the absolute value. Or the absolute value of it is going to be the same thing as it. And then we can just add T sub a to both sides, and then we would have our temperature, and I can even write this as a function of time, is going to be equal to this business, is going to be equal to Ce, let me do that in that same color. Ce to the negative kt plus T sub a. All I did is I'm assuming that this inside the absolute value is going to be positive, so the absolute value is not going to change the value. And I added T sub a to both sides to get this. So this right over here is going to be our general solution, in the case where we start with something that is hotter than the ambient room temperature. And if we want to look at the case where something is cooler than the ambient room temperature, so that's the situation, let's say T is less than our ambient room temperature. Then the absolute value of T, then this thing over here is going to be negative, and so the absolute value of it's going to be the negative of that. So then this up here results in T sub a minus T, that's going to be the same thing as the absolute value, it's going to be the negative of the negative. So then that is going to be equal to e to the negative k plus, actually let me just do it... T sub a minus T is going to be equal to Ce to the negative kt, so this is equal to that. I'm just assuming that T is less than T sub a. And so then, to solve for T, you could add T to both sides and subtract this from both sides. You would have T as a function of t is going to be equal to, let's see, if this went onto that side and this goes over here, you would have T sub a minus Ce to the negative kt. Did I do that right? So yep, that looks right. So this is the situation where you have something that is cooler than the ambient temperature. So this right over here, based on the logic of Newton's Law of Cooling, these are the general solutions to that differential equation. In the next video we can actually apply it to model how quickly something might cool or heat up.