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

let's think about another scenario that we can modeled with differential equations and 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 Newton's described as Newton's law of cooling 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 lowercase 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 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 much cooler it should be declawed it should be increasing in temperature quickly and if something is close well maybe the rate if the 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 is like okay if if the 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 shouldn't this 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 now if you don't like thinking in terms of a negative K you could just put a negative right over here 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 within the negative of that is going to be a positive or 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 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 algebra algebraically manipulate this so all my T's and DTS are on one side or I should say so all my capital T's and D capital T's on one side and this is going to be a little bit more confusing because I have a capital T and a lowercase T capital T for a temperature lowercase 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 do is I can divide both sides by capital T minus ambient temperature minus T sub a remember this is just going to be constant based on what our ambient temperature is we're going to sue our ambient temperature doesn't change as a function of time that there's just it's just 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 let me do some new color if I divide both sides by that I get 1 over capital t minus T sub a and let me multiply both sides times the time differential so I'm going to have so that DT DT so our temperature differential times our different temperature differential is going to be equal to negative K negative K times our time times our time differential so once again to separate the variables all I did is 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 you could do use substitution if you want if we said U is equal to capital t minus T sub a then D U is just going to be 1 DT and so this is essentially the you could say say the integral of 1 over u D U and 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 capital t minus T sub a is equal to n once again I could put a constant here but I'm going to end up with a constant on the right hand side 2 so I'm 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 K T and once again we have plus C and now we can take we can raise e to both of these powers or another way of interpreting this is that 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 EW I always think of the color green e to the negative KT plus C plus C this of course is the same thing as as this is equal to e to the negative KT we've done this multiple times before negative KT times e to the C times e let me do that in that same so times e to the C power and we could just call this what we could just call this another arbitrary constant let's see if we call this C 1 C 1 then we could just call this whole thing C so this we could say is C e C e e to the negative KT negative KT so at least it's starting to resemble what we did with your modeling population but we 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 in a world say we're dealing with a hot cup of tea something that's hotter than the ambient temperature so we could imagine a world where tea is let's say 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 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's going to be the same thing as it and then we could just add T a to both sides t sub maeda 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 C e let me do that in the same color see e to the negative KT negative KT plus T sub a plus T sub a all I did is I'm assuming that this inside the absolute value is going to be positive so you 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 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 montt 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 appear 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 that is going to be equal to e e to the negative K a let me do that actually I'll do it in the same color I think you get the point negative KT plus all right so let me just do it T sub a minus T is going to be equal to C e to the negative KT so this is equal to that I'm just assuming the T is less than T sub a and so then to solve for T to solve for T you could add T to both sides and subtract this from both sides you would have capital T as a function of T is going to be equal to let's see if this went on to that side and this goes over here you would just have the you would have T sub a minus minus C e to the negative KT did I do that right so yep that looks right so this is the situation where you're you have something that is cooler cooler than the ambient temperature so this right over here based on the logic of Newton's law of cooling are the two are two or I guess this is these are the general solutions to that differential equation now in the next video we can actually apply it to model how quickly something might cool or heat up