Applications with Separable Equations (Differential Equations 14)

hey everybody welcome to another video I'm glad you're still watching things for sticking with me so we're gonna talk about applications right now we've discovered one of our basic techniques of solving differential equations is solving them by separating your variable separable equations right now we're gonna discover that man that's got a lot of applications for natural growth and decay so that means exponential growth and decay we'll talk about that a little bit so whenever I talk about differential equations with students and so your teachers probably knew the same thing in less I'm your teacher and then I definitely do the same thing is talk a lot about about a lot about applications because applications are really important for differential equations to see how these things work in real life that's the goal of math is to get it to apply to things otherwise it's fun and studying pure math can be fun but it's we really only confer usefulness in differential equations so with that in mind we're gonna talk a lot about applications of difference of equations as this course goes forward with that being said these things take a while and so I know these videos are a little bit longer than than normal like if I were teaching this in a regular classroom setting I probably wouldn't have enough time to do man we're gonna do ignite examples today on this video so if that's if you don't need all of that you can feel free to skip through some of them if you're just perfect on your applications that's fine but what I'm trying to do here is is do a lot of these applications because we do have the time that way when you get to them if you're a teacher and they probably are it is making you do a lot of applications or showing you some of them that you're you're a little more confident and a lot more competent you see when we're dealing with these application problems well not challenging mathematically to do and we're gonna see that they're not super challenging if you are comfortable with separable equations the problem is is that the setup just the wording of them can be a little funky you know what are they asking here that's the hardest part so I'll show you how to set them up and then we'll go through them I wouldn't say they're particularly the challenging after you set them up like most word problems once you get the equation if you know the technique it's gonna be fine so we'll say that a lot today if you feel comfortable feel free to move on and we'll talk about the another technique of solving differential equations in just a little while in a very next video I'll introduce the concept to you but for right now let's focus on being able to apply these differential equations and specifically the technique of separable equations to some applications so we're going to find out that in real life a lot of natural growth and decay can be modeled by that right there followed by the following so when we talk about natural growth the decay typically we have an exponential growth and an exponential decay so or those things have an exponent where or an exponential where a base is typically e and we're gonna see why that is right now so we'll go ahead and solve that differential equation but what this is saying to you is that hey stuff that grows exponentially now for natural growth and decay it's typically exponential it's modeled on this idea that the rate of growth or decay that's the rate of change the rate of how your population your amount whatever you talk about changing with respect to time is proportional to the amount that there is now that might sound a little funky at first but if you really think about it populations are gonna grow faster if there's more people to reproduce or bacteria or bunnies or whatever so that that's the idea that we're coping with right now is that hey the rate of change of your population or the amount that you have in nature typically the rate of change if it's unimpeded by like lack of food like like that that's gonna grow without bound exponentially so the rate of change of our population is is typically proportional and very simple models to how much there is or case this constant proportionality constant of proportionality or the constant of variation if you dealt with like that in intermediate algebra we just talk about stuff like that so the rate of change of our growth or decay rate of the growth or decay is proportional to the amount that there is that that does kind of need to make sense that yeah the more I have the faster we're gonna grow the less I have the store we're going to grow and with decay kind of the same thing happens with like radioactive decay will see that in similar examples as well so we're just gonna go through some examples and specifically I'm gonna focus on set up how the thought process works how this makes something that's exponential we're gonna see when we saw our very first one that oh yeah that does make exponential growth or decay pretty much every time so we'll look at that right now let's talk about a town that had 25,000 people in 1970 and then in 1980 it had 30,000 people if the population continues to grow exponentially what's the population going to be in 2010 barring any disease barring any people like large groups moving in or out if that's the same population like some set of people to start with what's it gonna be in 2010 now we're gonna start with this why well this is exponential growth or decay and you're gonna see it in just a little bit so let's start with all right let's assume our population growth the rate of change in our population the rate of change of our population with respect to time is proportionate or proportional to the amount of people that there are oh my gosh that's a that's a simple equation I've got a differential equation where I can get my variable hey DP and get the P on one side or DT on the other let's try that so 1 over P DP equals K DT you remember from last couple that we always want to keep our constants on the right hand side on the typically it's the exercise but on the independent variable size we're talking about so this looks a little strange where sometimes our X this X doesn't mean that it's the independent variable it means the amount that T the rate of change with populations over time time is the independent variable your time is the thing that is changing and our populations based on that the amount that we have is based on that so when we're doing our separable equations here we want the population on the left hand side we want time on the right hand side hey that's separate that's awesome now we can integrate so when we integrate both sides we get here it is this is the important reason this is why that II that natural number e shows up in all of these it's why it happens in nature is because or well we discovered this models the nature I suppose a better way say is because when you do this integral right here you're going to get natural log of f sub a theme that's base is e and when we solve that for people kind of get something on well exponent on e that's really kind of cool so natural log of the P we've done this a lot is equal to KT plus let's call it C 1 because we know what's going to happen when we do that e on both sides so take this as an exponent e that constant is going to change we're going to alter that a little bit so let's keep on going here since we have a natural log absolute value P equals some stuff with our independent variable T in there we know that absolute value of P itself is going to equal e to the KT plus C once remember what we do with this remember that if exponents are added we can create multiplication with a common base and switch sides here but this is the same thing common base is being added exponents added exponent means common base is being multiplied and then one more thing that we're gonna do we know that absolute value equals that plus and minus and now we can wrap this whole thing up into this larger if you will it could be smaller but larger arbitrary constant C remember that our population right now so P is our population is based on time so as time moves forward our population is changing that's why it's changing population with respect to time that ready to change the population inspect the time now we know that we can call this whole thing C equals e to the KT this should be very comfortable for you we just spent its ton of time talking about solving separable equations now it's our application time we're gonna go a little further though so now imagine if you would that our population at time zero is let's just call that initial population whatever it is so in this case would be twenty five thousand that's where we started our experiment from or observation from or whatever we're dealing with from we started from that that twenty five thousand let's say that it's just in general because what I'm trying to do for you is show you that these natural growth and decay where the changing population change in amount with respect to time is proportional are always going to be modeled on some sort of an exponential gear and we're seeing that we're seeing that there's a little recap that this is a separable equation we can pull our p on one side we can have the dt and the other when we take our integrals we are going to get an Ln solving that for ki you're going to get an exponential based on e that's really cool that's our base we can separate just like we normally did our exponents we can do our plus or minus we can wrap that up in a arbitrary constant C but now we're gonna get more specific so at time zero we are going to have an initial population whether it's 0 or 5 or 25,000 in this case let's plug that in so let's put this in here so at time 0 we are going to have an initial population that's what that says on population when you plug in 0 is P of 0 well do you see that e to the 0 is going to be 1 so whatever that barber trade constant is that whole thing that only stands for our initial population that's cool so if we rewrite this whole thing we said yeah the population with respect to time equals our arbitrary constant but that arbitrary constant sense we have an initial population here plug in 0 and you're going to get P of T equals your initial population well that means our C right here then we add so put that stuff in there this becomes 0 this is P of T our C since this is 1 is going to equal key of 0 so P of T gives us P of 0 nothing write this in 2 steps only thing easier so C stands for our initial population if we put this back in our original we now know that our population perspective time is no longer see we know that stands for initial population e to the KT no two things here that I want you to just make a note somewhere on your paper or somewhere in your mind that if K is positive we're gonna get exponential growth and if K is negative we're going to exponential decay but all of our problems let's start this way where the way that we're changing with respect to time is proportional there that way I hope you can see that that it's all structured just like that whether we're population or X or half-life or anything we're all going to be based on something like this where when you solve it down we've made this very general none of these numbers are in here anywhere it's just the idea that when you say yeah you know what at time 0 you're gonna have an initial amount well that initial amount that is going to be your C plug in 0 your get C equals your initial amount take your initial amount plug it back in for C and you have something that's very familiar a lot of times in calc 2 you see stuff like this even calc 1 if you did early transcendentals of course my videos aren't that mine are based on late transcendentalist which means that we deal with natural logarithms in calc 2 for us but first before here you see a lot of this in there and some of those word problems that came out of nowhere Newton's law of cooling where's that even come from this idea that the way that something cools is proportional on its temp to the difference between ambient and the temperature of that that body whatever you're checking well if it's proportional if the rate of change is proportional we can do an integral several places to do an integral and we're going to get that exponential very cool now we can do two things so we're going to deal with this pretty much all day today in this video so when we see this we know that it's based on some sort of natural growth and decay where the rate of change is proportional to the amount that you have whatever that means now we typically will do two things step number one is use one piece of information to solve for K and then once you've found K once you found K you can use the other piece of information to answer the question that's being asked so this is pretty much a structure for a lot of these examples let us go through that right now so the first sentence tells us something the first sentence says we had 25,000 people in 1970 well that's going to give us an initial population so in 1970 where a prop was starting our initial population is 25,000 okay that's very and 30,000 in 1980 so since this is based on a population change according to time we need to find out how much time has passed which is really easy from 1970 1980 you know that we had 10 years the past and then how much the population was at that time so that gives us another point of reference here that says that in 10 years or 10 years from when we started so a piece of zero here says at the start of your problem that's 1970 so I know it's not here zero because this time problem wasn't around at Year Zero but in 1970 that's the initial population so let's call that time zero let's call nineteen seven seventy this T equals zero so when our problems started that's why we get the initial population or twenty five thousand here so a population and started the problems twenty five thousand after ten years so nineteen eighty we had 30,000 that gives us enough information to solve for our case so when you're doing this on your own you need an initial population you also need another point of reference you need something after certain amount of time gave you some quantity over here let's plug all of that in so after a certain amount of time so after ten years we have 30,000 people I'm gonna fill the right-hand side and then I'll put that 30,000 we started with 25,000 in 1970 after ten years we had 30,000 starting here we had ten years thirty thousand people we're trying to solve for our case so this is going to use a lot of logarithms we're undoing an exponential because we have that K up there so the first thing we're gonna want to do is divide so 30,000 divided by 25,000 looking at one point two equals e to the 10 K and we're gonna see this pattern a lot where you do a natural logarithm so Ln because we get that base of e and then we're gonna divide by whatever that time is so we're gonna get K equals natural log of this ratio the ending amount divided by our starting amount and so natural log of and divided my starting amount divided my time that's almost always an equal your K and that's about equal to looks like 0.01 eighty two three and it goes forever try not to round that a lot sometimes especially when we're dealing with Exponential's those little little teeny differences in the hundreds and the thousands and sometimes even the ten thousands can make a huge impact on population or if you're checking out like how long ago fourteen the carbon-14 is decaying well that can make a huge difference in your time so try not to round that a lot we're gonna round occasionally and so some of my answers might be different depend on how we're rounding especially if on your calculator you store that number that's probably the best practice in applications trying to round to the very very last step so let's do a little recap just to make sure everything is clicking because for the later examples we're not going to spend all this time doing it so here's the recap of what's going on we have dealt with separable equations now we're talking about some applications of these separable differential equations we know that a lot of natural growth and decay is proportional to the amount that you have well when we start doing are solving differential equations by separable separating our variables here we are going to get exponential every single time and we did that with population as well come up with this we realized that man in solving this that arbitrary constant is actually representing our initial population or initial amount of whatever you're talking about that's pretty cool so when we deal with our application problems and we get down to this far we need to basically know two pieces of information what our starting amount is and how much we have after a certain amount of time the hat will let you solve for K once you solve for K you're basically just augmenting this original formula a lot notice how this right here this required no information for us to do at all we just sent outfits you get down - here we go okay now we can augment this we can augment it with our initial population which we're gonna do right now it's gonna be 25,000 and then whatever K you solved for after you do that there's only one very now you can calculate what our population is at any given time or how long it's going to take you to achieve a certain population and that's really how a lot of these word problems are structured so do your calculus or don't teach calculus I mean it's they cover the same every time plug in your initial population plug in your amount after a certain amount of time find your K take your cave take your initial population and augment that formula then you're required to answer one or two questions what the population is after certain amount of time or what the time is going to need to be to get a certain population that happens almost exclusively so we're going to finish our problem we're almost done erase all of our calculus because we've talked about it at length and then we're going to go through and solve it for whatever question we need to answer here so we know that our population according to time we had a initial population that's 25,000 et to the KT but we now know our K notice how our K is positive our K being positive means that our population is growing oh yeah Hey look here was 25,000 or 30,000 it should be more in 2010 if it was decreed negative we would be decreasing the population so this this fraction if you want mana this is gonna go way back in time for you but you remember how natural log looks remember how natural log is negative before you get to ln 1 @ ln 1 at 0 and after that is positive so that one more time natural logs look like that before your argument is 1 your negative after your argument is 1 your positive so this ratio tells you whether you are increasing or decreasing if this number is bigger than this number obviously you're gonna get more than one so your case can be positive it's showing you that you're growing that's kind of an obvious statement but that's how the math is working if this number is less than this number that means that your initial population has decreased over time well L n is something that's less than one is negative which means that you're going to be having this natural became when we solve for exponential of course we get our inverses are inverses would look like that for growth or that for decay that's already listed that for decay so anyhow has a little bit of explanation about why our K is positive or negative based on whether we are increasing or decreasing over time I hope that makes sense to you now now that we have our problems well our formulas structure for this specific problem when we can answer pretty much anything we want based on these two things whether we're solving for a population given a time or a time given a population in this question we're asking if the population grows exponentially here it is here's the constant that it's growing at find the population in 2010 please don't plug in 2010 because what that's gonna do is take two thousand ten years from where this problem started so if our T equals zero our initial population is based on 1970 our x gotta be based on that also so our T is not 2010 cluttered T is 40 because 2010 is 40 years past when we started the problem in 1970 where ever took our an initial population so 2010 translates to T equals 40 years let's just plug in 40 so the population after 40 years is what this represents we started with 25,000 we know we're going exponentially according to some constant of variation and we've given this 40 years to grow that's exactly what that formula means in plain English once we do that looks like we'll have around 50 1840 living in this talent ask yourself that's making sense you understand that do you understand that even though this says proportional when I take my integrals I'm going to get an exponential based on the fact that if that's the first power I'm always going to get an LM and solving that if you're always going to get it e raises some power do you understand that do you understand that it's going to take two pieces information to solve for your cave in initial population and that that is represented by your see that the same concepts here and you'll need an amount after a time once you get that we solve for K plug those two things in augment this formula and then insult anything you want that you're going to be asked about it let's move on to the next one so a lot of times with these wheels this happens especially in half life you have a not necessarily a specific population that you after sir my time but it says that you've multiplied or you've decreased by a fraction or you have a certain percentage so I'm going to show you how to structure that problem because these are basically the two things that you have so imagine that we've got some bunnies and you know what bunnies do and after six months or after ten months they have increased six-fold that means there's six times as many bunnies as you started with after ten months we're not gonna go through this process of doing the integrals every single time again done that this is all about structuring the problem and understanding what it's asking you so let's go ahead and try to figure out how we can represent this problem in that formula because we know that this is based on population growth we're going to assume that it's based on this natural growth and decay so it's based on this formula here so that we did our whole integration same thing could happen here I just don't want to do it again well we need a couple pieces of information we need where we started we need how much we have after a certain amount of time when you're not given a specific a specific initial population or initial amount that you have you're probably being given a factor sort of problem like hey we increased by a factor of six or we decreased by a factor of 1/2 or we decreased by a factor of 5% what that means is that you kind of have an easier sort of a set up everything's gonna happen at once so if our bunny population increased six-fold that's a 600% increase or multiplied by six give yourself a very easy starting population like one now of course bunnies can't replicate with we're not going to go into the birds and the bees with you but so maybe you don't start with one day you start with a hundred or a thousand or a million or it doesn't matter because the the way that this is set up if we start with 100 bunnies in ten months so T is representing months here in ten months you will have six-fold six times that well what's 6 times 100 600 now you could be asked this question a different way we're doing it you have a six hundred percent growth after ten months or you know six times the amount of bunnies in ten months or if you're decaying you have half the amount of radioactive material in 5,700 four years or you have you want the level of radiation is 100 times which you what is okay so what do we want 1 over 100 for our level of radiation though that's a very common n so that's a sort of we're increasing or decreasing by a factor so a lot of these fall into two camps one you're given specific numbers specific initial populations and one that you're not you're given an increase in six-fold or two times or half the amount that goes into this idea so when you're given this this sort of an idea you need how much factor wise after a certain amount of time now I'm gonna get into the why I used to a number one rather than 600 the very first thing you would do to solve for K notice we can there's only one variable if there's gonna be really easy the first thing you would do is you divide by a hundred so if we divided both sides by 100 you are going to get 6 equals 1 e to the 10 K it doesn't matter how much you start with so think something easy make something like 1 then if you have to cut it in half well there'd be 1/2 if you have to multiply by 6 well there's six it makes things really really nice so I wouldn't even I wouldn't even write six hundred and 100 I would have definitely one write some crazy numbers like I don't know it's seven point two and fourteen point four that that's that's that's crazy what do I do that we know that if we're going to increased six-fold whatever we start with one is going to be multiplied by six that's how the second class of problems is structured hopefully that helps you because otherwise you go I don't I don't know how to where do I start how much do I start with doesn't matter doesn't matter because you're multiplying by six to figure out the amount that you're going to have in ten months that's kind of cool so let's solve it we know that all these are the same thing if they're based on natural growth and decay it's going to be exponential we know that we have an initial amount I know we have a certain amount after a certain amount of time with sulfur K so when we solve for K we'll have Ln of six equals by the way do you know that our case gonna be positive because our pending amounts larger that are started about it's going to be greater than one so natural log of 6 is going to equal 10 K so k equals natural log of 6 over 10 and that's about that's about 0.179 let's admit the problem here so if we have population with respect to time equals in initial population now here's the only only drawback is that if you don't have an actual initial population you really can't answer questions like how much are you gonna have in this amount of time because you don't know where you're starting so so those problems are sure off the table but they're not going to ask you that it's actually things like okay how long did it take to double or how long is it going to take to double or how long is it going to take to reach ten times the amount with those sort of questions we can't answer so we well we don't have an initial population we now can answer certain other questions so we have a our population is based on how much to start with but our constant of variation right now our constant of growth is 0.179 times how many months we have now they could tell you this suppose you started with ten bunnies how much will you have in whatever months you could do that but a lot of times if they're not giving you an initial population because you could have just use that earlier if they're not given an initial population they'll ask you things like how longs it take to double they don't have to but that's very common so we have looked at this we've said hey our populations increasing six-fold in ten months that lets us solve for K that lets us plug that back in now something we can use if I gave you an initial population and a time you can find out how many bunnies they're gonna be in that amount of months in this case for us we'll say how long did it take for the population to double should it be more than ten months or less than ten months well obviously I mean it increased six-fold in ten months to increase twofold or double should be significantly less than ten months and that's what it's asking so once we know that how long did it take to for a population level once we know that it doesn't matter where we start doubling is just going to be twice that amount whether you have one and two or a million and two million or five and ten it doesn't matter you're going to get this ratio of to one and then we can solve for T very nicely we know that we take a natural log so Ln two equals zero one one seven ninety if we divide by that one point one seven nine because at three point seven three point eight seven so these bunnies this population of bunnies based on this information would double in three point eight seven months so in under four months so you know that when you started with whatever this population was that in three point eight seven months you had double the bunnies is that making sense to you guys get in it if you do fantastic you're really grasping if not you might want to go back to what separable equations are see how they're solved every time when you have just a power one right right here see the natural log see that you're going to get an exponential every time because if it's the question is where's that coming from it's not okay right now okay that's coming from just an integral of one over your variable and you're getting Ln of that variable and then you're doing that e on both sides okay so that that's important for that to be okay in your head after that ask yourself do you understand of the two types of two main types of problems we're going to get then we need either an initial population and we need a population after certain amount of time or we need a factor sort of problem increased by six hundred percent or multiplied by six or cut in half after certain amount of time we always solve for our K first and then we use that with typically initial population or some other sort of factor to solve for a time or a time to solve for how much we have we have a lot more examples and I'm going to get back in just a minute let's do our next two examples this next one says and this is very common with this this radiocarbon dating an old femur so they found this old them has one sixth of the 14c or carbon-14 as a modern femur of the same species has what's the age and then they tell you a half-life of carbon-14 is 5,700 years what that means is this every 5,700 years the carbon-14 in this item changes into I think it's carbon-12 and not I can't remember but it depletes itself so it's a radioactive sort of carbon so it's it loses the amount so if they had like a gram of it then after 5,700 years in the half a gram after another 5,700 years there would be a quarter of a gram that is exponential decay that's what's happening here so theoretically you never never lose all of it it just cuts in half every 5,700 years so let's let's figure out how old this is and I want you to ask the question right now just see if you can classify this is this one of those problems where they're giving you a specific starting amount or is this one of the problems were they giving you a factor after a certain amount of time well if you think about it this is the factor after certain amount of time I mean it's literally dealing with half after time that that right there that's a factor sort of a problem now a lot of students running to the mistake here of going well I'm gonna use the one-sixth right off the bat let's talk about why we can't all right so are we going to be using exponential growth and decay of course should our can be positive or negative it should be negative because we're decaying here and specifically because you're starting with one and after a certain amount of time you have half that portion of what you had compared to what you started with is less than one Ln of less than one and zero you're gonna have a negative K here after certain number time so let's talk about why this is not OK so are we using exponential growth and decay of course the amount present after number of years is based on the initial amount and we're gonna have that same formula that we've been dealing with this is not going to give us for all of these examples pretty much we altered a little bit for cooling now which one do we use well in order to figure out your K with a factor sort of a problem like you're increasing by the fact or decreasing by factor in order to figure out your K you need the factor but you also need the time so even if we knew that we were starting with let's say 1 and decreasing tu-160 amount we don't know how long that took us so you can put one here in one sixth year but you don't have that T that's literally what the question is asking how long did it take to go from the full amount to one six the amount based at that so this is down a rabbit trail we can't get out of use something when you're talking about factoring factor sort of problems are you increasing to a factor or decreasing by to a fraction that's a factor use something that gives you both that factor and the time so almost all the time and half-life problems you start with the half-life I'm saying almost because there's there's always an exception I mean they're word problems so there's lots of different cases here but try this first start with your half-life first that will find your K because it's giving you how much you start with a hundred percent how much you have after a certain amount of time a factor of that fifty percent so we can just do after five thousand seven hundred years if we started with 100 percent or one gram or you do a million here we'd have half of that so 50 percent or one half or half a million if I said a million this right here puts our half-life in perspective according to this formula it says after certain amount of time this is your vector that's how to solve for your K once we've done that then we can go back and answer the other questions so let's do it natural log of 1/2 all look at that you got to know this this right here that you have you decreased is going to be your K is negative Ln of a number that's less than one is going to be negative equals 5700 k if you divide by 5700 K is about its nasty negative zero point 1 2 3 zeros 1 2 1 6 0 5 let's say one more time because we saw in the last example of it if if this ratio where you start with 2 where you end with if this number increases your kid is positive because that ratio will be more than 1 ellen:oh more than 1 is positive if the ratio is less than 1 so this amount of decreases well then your Ln is less than 1 and you can have a negative number Ln of native the numbers between 0 & 1 are negative your case gonna be negative since we're dealing with such a long time here this K is really important to write a lot of a lot of numbers are being around it I always say just don't round it store that number in your calculator and use it later so let's othman to our formula so the amount of radioactive material that we're going to have this C 14 is based on the starting amount eetu the now we have a k times however many years no i mentioned also that on these problems the next thing is going to ask you for one of two situations how much you're going to have after certain amount of time or how much time you need to reach a certain amount this is the latter of those two so this is a factor sort of where you had a certain amount left after a certain amount of time or you have a certain amount that you're gaining after certain mount of time and then ask you the next question there's no amount that we have here I don't I don't know what that how much milligrams the head of that but they're asking you well if if the femur has won six of the original amount then what's the time now I can fill that out well that's let's try the trainer here on your own you can that's and that's based on this factor idea so going from a mount to a factor of it so or a fraction of it in this case so increasing by a factors what I mean so six fold to fold twice the amount five times ml or half the amount one-third the amount whatever here we say you know what this bone were it had a hundred percent or one one hundred percent that's the same thing of its c14 when that this organism died it stopped in taking that that that carbon element and then it started losing that so how much it has now is one-sixth the amount and if we solve for T that's saying how long does it take based on this constant of decay exponential decay for a hundred percent of your carbon-14 to be reduced to one six bit amount so that's it's asking that well it's only it's only a fraction of it so how we how we get in that let's do the natural logarithm well we're gonna get negative 0.0001216 o5t and if we divide by that nasty K T is about hey that's a negative number natural logarithm of no vs. and one our negative between 0 & 1 or negative and that's a negative number so when we divide the negative by negative we're gonna get an actual year I'm out here I got around fourteen thousand seven hundred thirty four years so according to the math here if this is all accurate and if the the amount of c14 hasn't altered drastically in the atmosphere for the last forty seven ten years or when if it has altered drastically if it was continuous then this would have happened so we know that the exponential decay would for you guys went decay like this we know that the king would be negative and it sure is because we've had half Ln of one half as negative and then we can use it saying how do we go from this to it before to just a fraction of that or a multiplication factor here so how do we go from 1 to a factor of 1 times 1/6 divided by your K they're done we get about fourteen thousand seven hundred thirty four years next one so we need to give this a drug to a kid to maybe help them with some sort of disorder or help them with something or unnecessary or whatever it may be so the certain drug has to be at this this proportion of forty five milligrams per kilogram so that means the heavy you are the more that your body needs for this to be effective and the lighter you are well the less that you need so obviously we don't really want to over those people but we want it to be effective so we're gonna have to structure this to be effective and we want this to be effective for one hour so as soon as it drops to below forty five milligrams per kilogram of body weight it's gonna stop being as effective as it needs to be does that make sense so we're were at a certain point the amount in drug in your and the body here is not gonna work over that point it works we don't want to go too much because then it's going to work too long so imagine if you had to had necessarily you're necessary in density the person who that's people hundred and you just go I'm just gonna give them a ton of drugs they're gonna be under for a long time maybe forever because there's a half-life to drugs there's just amount of time that passes and the drugs start where they wear off so if the half-life is five hours how long how much drugs do we need to make this effective for this particular weight child so that they wear out for one hour you don't want us randomly give them drugs because either they're not going to be as effective or they're gonna be too effective and have the effect way too long so let's start looking at this as what type of problem it is let's up let's see em firstly if this has a half-life it probably structured on this natural decay let's start with our same sort of effect so how about we have the amount of the drug after time is based on the initial amount that you're giving and it's exponential growth or decay now this is a half-life think about whether that K should be positive or negative it's got a half-life it's decaying so it's dropping that case should be negative and we're gonna see that just a little bit secondly thinking about whether this is an amount idea at first or if it's a sort of a multiplying by a factor idea well let's look this looks like an amount so we need 45 milligrams per kilogram the effective okay how much to begin for a 50 kilogram child so the fix we're after wear off after one hour I see an hour that's a time but I don't have an initial amount I also don't have how much is there after that hour so this is not an initial value problem it's in this initial amount of problem this is a unit have light so the path length of this drug is five hours using a half-life will solve your ok for you so this starts off with a you're multiplying by a factor idea so we start with one you're having one half after five hours do you see how that's working do you see how if you have the full amount Hunter said and you have half of it 50% for one half after five hours the only thing that's left is K those half length problems work like that almost exclusively so start with the half-life idea solve for your K and then go back and try to answer your question let's let's do that now if you haven't done it I would suggest you guys do it of course I'll do it right now so if our half-life is 5 hours I know that from the full amount 100% which is 1 1 and 100 cents same thing we wouldn't have half after 5 hours easy to solve notice how your case gonna be negative element of 1/2 equals 5k man we've got a 10k and a 5k I like running so we have K equals element 1/2 over 5 that's about oh let's see negative it shoots negative element of numbers less than 1 or between 0 and 1 our negative point 1 3 8 6 let's now augment our formula you'll notice that I didn't put an initial note because this questions it's not really asking it's not really asking how much is going to be there after certain amount of time this question is asking this how much do you need to start with in order for this to be gone or two for the this drug to wear off after one hour so this is the third case it's rare that this only using word River it doesn't happen as much as the other ones where you're looking for a starting amount to satisfy how much you're gonna have after certain amount of time so there's really this three cases two of them happened often where you you're asked how long does it take to reach a fraction or factor of a certain amount how much time it takes or given a certain amount of time how much are we gonna have this is the last one doesn't happens off with another two but it says okay you have to have a certain amount after a certain amount of time what do you need to start with and that's what this is asking you well we got to be a little bit smart about figuring that out so we're gonna be solving that we know that we want this drug to be ineffective after one hour so we know that's going to be one that's what this says so we need this to wear out after one hour so after one hour we're gonna have to have a certain amount how much sure to start with you need to figure out what that certain amount is so what this drug is doing I don't know what the drugs actually doing but what this drug is decaying and it says and ineffective that it says that if you are at forty five milligrams per kilogram that is the threshold for effectiveness more than that your effective less than that your ineffective so the bare minimum amount would need to be 45 kilograms for half of 45 milligrams sorry for how many kilograms that body is in this case it's 50 so we multiply 45 times 50 that's the amount of the drug that needs to be in the body at a barest minimum and then right under that so like a second later it's going to start becoming ineffective that is 2250 2250 milligrams think through that again it needs this much per kilogram hey kid weighs that much you're gonna need this much or more to be effective right after that it starts being ineffective so what this question is asking you is I need that amount to be present after one hour so it's a slightly different take on this I need this much I mean this wants to be there after one hour the question is how much do you need to start with and that's a valid that's a great question for the snare because you if you know how much you need after an hour you're gonna want to know how much to give them an hour beforehand and that's a good thing what's the initial amount of that drug that you need to give not particularly hard to solve none of these are really all that hard to solve it's just that the setup can be difficult so in this case it's yeah half left great selfie okay but where do you plug your numbers in well you need 2,250 milligrams of the stroke for it to be effective that's the threshold here and you want that to be in the body after one hours so that it starts wearing off after one hour you said I want this to be my end after one hour of course we're gonna administer more than that because the drug is decreasing over time it's got a half-life to it so if we take 2250 and we divide by this whole side that's gonna give us an amount of the drug I got a brown 2585 so we would administer for this child's weight we would administer 2585 milligrams of whatever this is it's going to start decreasing immediately it's got a half-life of five hours after one hour it will reach the point of 2250 which is exactly the proportion of 45 milligrams two kilograms for this particular body so this this drug is going to start being less effective at that point that's the way that these there's three cases two more very I smoked that before there's really two main ones this is the third one kind of hidden in there but it can really trip you up if you're not paying attention to what's being asked here I hope that would make sense to you and we'll come back with a couple problems in a minute all right back at it let's talk about another one and I want you to try if you can to do it on your own so we have managed to have two of them up here see if you can read through these see if you can figure out if it's a you're multiplying by a factor over a certain amount of time or if you're giving an initial value and you're given a time where you reach another value anyhow see if you can structure these appropriately start with your exponential growth or decay models see if you can find the K first augment your formula and see if you can go through and try it at least on that one this one is Newton's law of cooling and so it's a little different let's show you the derivation of where that comes from but if you've seen it before maybe even started if not it's okay we're gonna start right now on this so something happened and we have this level of cobalt that's dangerous so Cobalts this radioactive element and had a half-life five point two seven years right now the amount of cobalt there is 100 times what it should be and that's not good because you don't want that so when will it be okay well the world we do let's classify it is this a problem where we start with an initial amount or it's just a problem where we have this there's a you're taking it amount times some sort of factor so some sort of multiplication over certain my time most half-lives are the star the ladder so let's start with this exponential growth of decay we have half-life we're going to be decaying we still have an exponential model so with our half-life half-life automatically gives you how much you had to start with how much you have as a spent number times a factor so this multiplication concept over amount of time so we had 100% to start with and this is whatever couple we had and we have a half of it after five point two seven years that lets us sulfur Kate really nice notice that when we have this number smaller than this number your case can be negative so element of one-half equals five point two seven K and so K is element of one-half over 527 and that's about negative zero point one three one let's start changing our formula so now that we have our K we can start answering other questions notice how we can't answer the how much is actually there so we can't go in or would you probably want to go back and measure how much actual cobalt we we have so we know it's radioactive decay we know we're decaying this exponential decay of radioactive element we'd had that half-life we now know our K and now we don't have an initial value but we do have something else right now we know the level is 100 times what it should be now this is weird but we have how much we have now this initial value is 100 times what it should be if our initial value is 100 times what it should be then if it's 100 what we want is one hey yeah what we have now is 100 times what we wanted to be we want it to be a hundred times less or you can think about it this way you hoping you can see this leaving this up here hopefully you can see this if you divide both sides by 100 you say what we have now is one 100% that's not good we want it to be 100 times less than that so 100th of what it should be now do you see these are identical no matter how you think about it we have an amount we want it to be a fact it's that time some sort of number this factor less so this is a factor sort of problem locations half-life and then we don't have an initial value they have to be some sort of a how much you have times a factor gives you how much you need in a certain amount of time we're gonna find that time right now so we had this restriction okay we had our K we know that I need a hundred times less so this is a hundred times what I want it to be let's figure out how long it takes to get there if we do Ln of 1 over 100 of course we're going to get a negative number here which is fine because we're gonna be dividing by a negative number divided by that negative that's about 35 years so whenever this event happened 100 times the level of what we want for COBOL the 35 years later that will be an acceptable level level it's all based on the tip amount that you have time sort of some sort of factor and how long it took to get there okay next one so this has to do with mittens on cooling I spend a little bit more time on this problem the previous ones because this is the first one we had with Newton's law of cooling so I won't show you where that actually comes from in like a kouchi class or even some Calculon classes you might have seen Newton's law of cooling but chances are unless it I don't even know if I did it showed you the derivation of where that comes from you kind of just in that level of the class you kind of just get to this and you go yeah and I wake my hands and then magically we get Newton's law of cooling in there and you go what's the subtraction you go well I mean it's that's the sort of proportional to the difference between ambient and the temperature of the body but we're gonna get a little more specific here because that's the class so let's talk about this so your so the cam right now is 25 degrees Celsius you know I drink that maybe you want to be ice cold mmm good drink soda probably shouldn't it's better cold warm soda nasty so to save energy you're gonna be outside because it's cold outside right it might be cold outside where you are right now it let's say it's zero degrees Celsius so you're gonna take that and put it out there and you're gonna wait and so because you're the type of person that you are like like me it would be very specific you want to drink it when it's five degrees Celsius at that perfect temperature so not where you open and get the ice on it and just like man it's like a soda slushy I hate that but right before that boom so good well you measure the temperature because you're precise as I am and after 20 minutes it's 15 degrees Celsius so you want it perfect so you're going to time how long it's gonna take you to reach five degrees Celsius Newton's law of cooling is pretty cool and what Newton's law of cooling says is that the temperature the rate of change the temperature the rate of change of the temperature based on time is proportional to the difference between the ambient temperature and the temperature of the item you're trying to cool or heat that makes sense I suppose I mean if you had if you had some sort of if you had the soda cans somewhere even colder it's going to get colder faster but if you had it warmer it's gonna be colder slower so the way that something cools or warms is of course dependent on the the surrounding temperature we all know that if you drink coffee if you go out on a warmer day with a cup of coffee it stays nice and hot for a long time if you go out when it's icy cold you kind of chug it down because you're not signing to cold burn your mouth that's really awful but we know that the Weidman we know that the way that things cool or heat depends on the temperature around it and specifically the difference between what the temperature is outside and the temperature of the thing you're putting into it we'll say so how we're going to start this we're going to start with that just this basic idea that from from Newton's law of cooling the rate of change of your temperature with respect to time is proportional that means you get a constant to the ambient temperature minus the temperature of the item so in plain English the rate of change of the temperature of your item the object with respect to time is proportional that's always K times two the difference between the air or the medium that your your items in minus the temperature of your up to your actual object let's go through the calculus and see why we have a subtraction in here let's see why we have the formula we do so is this still a separable equation absolutely a is not variable a your ambient temperature is a constant so if your room is 75 degrees we're going to assume it's 75 degrees for the remaining remainder of this or if it's zero degrees outside we're gonna assume that zero degrees for the life of this this example so that's a constant that is a variable oh my god just that variable let's separate this let's leave our to this constant on the right and let's move this thing to the left let's do an integral you know how to do integrals based on these separable equations that's a natural log because that's our only variable but notice how we have that negative so that's important here so when we do the natural log do it you sub in your head or just think that I'm gonna get natural log of the absolute value of a minus T but I'd be dividing by that negative the derivative that's negative 1 we're gonna get a negative on the right hand side we get the familiar KT plus c1 let's let's start moving around some of our our signs so let's make this negative go away from left hand sides divide by a negative or multiply we have an element you should know what to do now since we have a natural logarithm trying to solve for T here let's take that and put a so absolute value of a minus T equals e to the negative KT minus c1 we also have these exponents are being added you're adding a negative so let's separate that let's get rid of an absolute value what else can we do well that lead to that plus or minus e to the negative see what looks pretty nasty let's just call that an arbitrary constant then let's solve for T because a is not even a variable is your is your ambient temperature the surrounding temperature that you have let's solve for T the temperature of the item that we want now now that we've got that let's see what we can do with our problem so that that's a workable equation we just did our our integrals based on that separating our variables we got down to solving for our temperature and this looks pretty familiar I hope that this looks familiar so when we when you had like a calc 2 level you just saw something like this where you had ambient temperature minus some sort of changing temperature based on the surrounding air and then that's that's pretty neat we're now able to solve for stuff like see here let's see what we have so this is not one of the factor type of problems where you're multiplying something times a factor and you have you have some sort of I had one I now have one half after certain time this is an initial initial value sort of a problem where we start at a temperature source so the can is 25 degrees that's a temperature so the the item that we're talking about to so we can the temperature the soda can is 25 degrees okay to say manager we put outside at 0 degrees our ambient temperature 0 degrees so we can fill out 25 degrees we can fill out ambient temperature and that's how we start at time 0 so do you remember what we need to have for things that are not those factor time problems we need to have an initial value and then we also need to have a value after time so our initial value at time 0 at time 0 we know that the temperature is 25 degrees we know that the ambient doesn't change it's never gonna change is constant at 0 degrees and so we can fill this out and we can solve this for C so if our temperature and time 0 right when you put it out there the temperature the soda can is five degrees so at T of zero so time is zero we have 25 our ambient is zero we have negative Siva we have e to the negative K times zero because that's our first little secondly we put it out there does that make sense I hope that makes sense to you that our temperature McCann's this our ambient temperatures this we don't have our C but we can solve it and with initial values like this your first job is to solve for your C then we use the next piece of information to solve for y or K that's how we did it in the first one what was it one or two examples so if we solve this for C we go okay well 25 equals well this is all one needed C so C equals negative 25 which means that if we plug this back in if we start augmenting this this problem the temperature based on time is equal to our and mint - our arbitrary constant so basically that the negative of our our temperature for initial temperature e to the negative KT let's keep on going well we're certainly negative we got addition we can also put in the ambient right now since that's not changing let's just make that zero it's a really convenient problem because if it wasn't zero of course you'd have a number there so if you put this inside a refrigerator where it wasn't zero degrees Celsius or something you'd have a different situation any harder not really it's just going to be a number right there so not a not a huge issue now now we've used just the initial value of this we knew that it was 25 degrees at time zero we knew the ambient temperature we were able to solve for C now on this on these type of problems is class of problems you need another piece of information you know what the temperature is after certain amount of time why if I know the temperature after certain amount of time I can solve for K let's do that we know that we know that after 20 minutes we're gonna have a temperature of the soda can of 15 degrees Celsius so we'd have 15 degrees Celsius after 20 minutes and we see that we can solve for K here so we divide both sides by 25 so 15 divided by 25 as three-fifths equals e to the negative 20k we're solving for K so do your natural log Ln of three-fifths is going to be negative negative twenty is also negative so you go wait a minute I've thought I thought that we were gonna be decreasing we are because in the formula itself we have that negative K so when you find this K it's positive because when you plug it in it's going to change to negative so that negative isn't showing up here because we already know that this from our formula where you have that negative K in there if your temperature was increasing actually you get a negative K here and then that would become positive so you put the soda can and a microwave I'm not a microwave another to put an oven and say the same temperature why you do that I don't know what would you warm so that whatever so if you did that well then you have this ratio of temperature would be increasing this would be more than one this would be a positive this would be a positive divided by negative the K would be negative but when you plugged it in it would be a positive case showing you growth that's what would happen there for us we have our K is about zero point zero two five five so let's go ahead let's change our let's change our formula that we're working with and then we can finally answer the last question so our temperature with respect to time is based on an initial or sorry an ambient temperature that's zero our initial temperature based on a time of zero but the opposite of that and now we have negative whatever our K was and this asks when will it be at five degrees Celsius so this is the other opportunity for us to answer the question based on initial value initial Italy started him out that we had we did the other one with the bunnies where we said oh well how no I guess that was a little different first example is this one where we said how much will there be in a certain amount of time like bunnies were sort of like that how much will there be in a certain amount of time here and we can find him out this is asking how long it will take to achieve this amount so we put this all together we say okay now after 20 minutes 15 degrees Celsius or soft fur ok cool when will it be 5 degrees Celsius when is a time so how long is it going to take for us to be at 5 let some plug-in that five for the temperature that we wanted to be if we solve for T of course we're going to divide by 25 first Ln on both sides and if we divide by negative zero point zero two five five we get this Ln or 1/5 sure divided by negative sure that makes sense because negative divided by negative is gonna give us a positive time and that's about looks like 63 minutes after 20 minutes is 15 degrees after 63 minutes it's gonna be a 5 degrees that can't wait that long I can't wait that I'd probably put it over ice you know so we can save myself the whole math problem it's really I mean you get the soda like I don't want to do math to show you my soda probably should get nice anyway I hope it's making sense to you I hope you understand the concepts that we're trying to say this there's two main classes of these problems where you have a x factor solver k and then you can answer a variety push it up that already have an initial value use that solve for k and then you have another one sorry use that assault for your scene and then you have another set where you have an amount after time to solve your K the the only other one I give you that kind of special case where we had the how much do you need start with to achieve an amount after a certain time those are not as common they do happen let's move on to a couple more let's get start on two more problems barometric pressure is the the way that the pressure changes as we consider the weight of gravity oh sorry they grew the weight that gravity puts the atmosphere on to us so the lower you are towards sea level the more pressures there because gravity sort of stacks the atmosphere on top of us or around us and pushes it down but the higher we get well the less effect that the atmosphere you're gonna have there's less gravity the further we go away from our earth and the atmosphere is less and less dense so there's less and less pressure so that's measured in in inches of mercury and so what we know is that barometric pressure is given by this differential equation the rate of change of barometric pressure with respect to how many miles you're above the sea level is given by negative 0.2 whatever the pressure is well what that means for us is that as we go higher our pressures decreasing let's go ahead and let's obey the initial pressure so barometric pressure at 0 miles above so at sea level is 29.92 inches of mercury what's it gonna be 15 inches so at 15 inches we either need special training or breathing apparatus to be able to manage that so some sort of special stuff and all about but I think it's kinda cool so let's figure out where that is let's go through all the calculus let's separate our variables that see what what actually happens here when we start doing the the math behind and see what happens to our numbers how this is going to be exponential decay as we go higher and higher in our atmosphere according to our pressure so let's start with what they've given us already is it differential equation can you separate the variables let's do it so we're going to move our B we're gonna leave our constant we're going to move our DX and we're going to integrate them we're gonna do that inverse of a natural logarithm which is e to the negative zero point two x times e to the c1 we know that we can separate those exponents by multiplication common bases if we know that we can do the plus and minus to remove our absolute value we have our e to the C sub one we can wrap this all up in a nice arbitrary constant so the barometric pressure based on how many miles we're above sea level is some arbitrary constant e to the negative zero point to X let's talk about the other arbitrary constant so our arbitrary constant should end up being our initial pressure or the pressure at zero and we can see it right now so when we plug in zero miles above sea level and we said yes there are zero miles below sea level this whole thing is 1 and we notice that C is our initial pressure if you wanna think about that way or the pressure at zero miles it's 29.92 inches of mercury so this is that amount it's right there for so very much like initial population so let's automate our formula we know that barometric pressure with respect to miles above sea level is our base amount 29.92 inches of mercury e to the negative zero point two x and now it's really easy to answer the next question so this was problems are kind of getting easier here we just do some basic calculus we don't run the same exact model that we've had before but now it's asking us hey when's our going to be fifteen inches when how many miles above the sea level do we need to be in order to reach fifteen inches of mercury so are this is one of those ones or says how long or what what do we need to be at to achieve a certain amount we already know where we start so if we put in our 15 for how much we want our pressure to be in this case we didn't even have to solve for K look they gave it to us they gave us the cave that was pretty nice we'd have to do anything for that so they're gonna give you a difference in equation and they not even giving you a K you don't even need that value that's it's already given to you as well look what you do need though I mean they didn't give me the K they'll also give you the initial value so they've basically done that work for you all we need to do is some basic integral and then we have to understand the idea that we're solving for the amount of miles to reach that 15 inches of mercury here let's divide notice how you have a ratio here that's less than one your natural logarithm is going to be negative but when you divide by negative you'll get a positive and we'll have that many miles above the sea level so natural log of 15 over 29.92 equals negative zero point two X if we divide by negative zero point two we're gonna get about these three points something three point four five miles if you want to change that to feet just multiply x thousand two hundred eighty and get around 18,000 feet so right at about 18,000 feet we start getting really pretty so low pressure really hard to breathe be well not hard to breathe but you're not you feel like you need to breathe more and more because there's less density up there's less pressure that means there's less density of the the molecules we need to survive as far as our ears concerns so we probably need some sort of breathing apparatus or if you get altitude sickness that can happen and as we start getting higher and higher so interesting problem we get we now know that above that that amount of three point four five miles above sea level so eighteen thousand roughly feet we started getting some pretty low barometric pressure all right big mouth person discovers a secret or a treasure if you want and after one week 10,000 people in your town of a hundred thousand to know about the discovery of the treasure earth now of course the more people that know about your treasure the more likely that someone's probably gonna rob you because not everyone is as integrity as we all try to be so let's see about how long it's gonna take you don't half the people know know this the only reason why we're doing this problem it's not that it's specifically hard but the setup is a little weird so we need to kind of start from scratch here we're not gonna really be able to jump into our full exponential I didn't tell me think about this the rate of change of those who know our proportion of those who don't now when we look at this sort of a problem this is not a multiplying by a factor problem this is a you start with a certain amount of people who know that's zero at time zero and then all of a sudden someone's discovers it and then after one week we have 10,000 people so this is one of those problems where we have an initial value and then after certain amount of time we have something else we have another amount that we know so that's structured this is not a you multiply by a factor idea and then we're going to say how long and how long so fine at a time how long does it take for this amount of no but let's start by trying to find out a differential equation so the rate of change of those people who know is proportional those people who don't know start with the rate of change the rate of change of people who know with respect to time that's the rate of show how many people the rate of change of people knowing with respect to time is performed not use K let's use T for treasure know about the treasure if you will so the rate of change of the people who know about the treasure with respect to time is proportional which is pi cut use K we're gonna have a constant of variation here this constant proportionality it's proportional to well what's it proportional rate change of those you know are forcible to those who don't know well we don't know how many people earn the talent at the beginning that's how many people don't know but every time someone knows it it gets subtracted from that amount so one person knows you'd subtract one from the amount of 100,000 and you find people who don't know if 10,000 people know then 90,000 people don't know so the rate of change of those who know about the treasure is proportional to those who don't know those who don't know are the people who are left from those who do know that's the hardest part of this problem is understanding that the rate of change of those who know about the treasure with a strength of time is proportional to there's that many people in the town there's that many people who know so subtract them proportional those who do not know that's the amount of people who don't know about the treasure after that it's a separable equation you can separate the variables I would encourage you to do that right now separate the variables to do your do your integral and then the last little bit here so let's move our tea with our DT BT for treasure little T for time if we integrate both sides on the right-hand side we get ke T plus C sub one on the left-hand side we have the same sort of an idea with Ln idea we've had before we'd have an Ln absolute value 100,000 minus T but we have been negative due to the use of that we got in there and the derivative of the inside being negative let's start moving some stuff around so natural log of absolute value 100,000 minus the treasure knowers equals negative KT minus C sub 1 same thing we've done before let's make our our natural logarithm go away with that e raise both sides let's separate this when we do it we've done many times before let's do a plus and minus here so 100,000 minus T equals plus or minus this thing that's going to become our main arbitrary constant here's my hope for you right now my hope is that all this make it absolutely perfect sense to you because you've done it so much but more than that that you understand the concept this is the hardest part understanding how to represent the rate of change is based on some sort of a proportion or the comparison of another amount that's the hardest part when inventing your own differential equation it's been saying how the structure works now let's keep on going so after we have this well we can really I mean you can solve that for T if you want to but you you don't even really need to because we're only really concerned about how long how long until half no notice that if we solve for T right now to solve for big T those you know the treasure we solve for those who know the treasure we're gonna be asking how long and just a little bit longer so we'd have to end up altering the formula anyway you can lean it just like this if you really want to no need to solve for that team I'm gonna show you that right now so just another idea that we can start plugging in the stuff to solve for R K and then solve our final answer so whenever we're software C than arcaded or balancer so whenever we're doing these you're given an initial value and then hey you know how much up to certain amount of time your first fool is a soul for the C with your initial value and then solve for your K with the piece of information that's given you an amount after certain amount of time so right now we've got to solve for that C so let's look at how that works a person with the big mouth discover the treasure in a town of a hundred thousand after one week ten thousand people knew about the treasure so wait a minute where's the initial value how many people know at times zero zero so right when the sit right as The Situation's happening someone goes treasurer will write at that moment no one knew that person discovered it right after time zero so at time zero nobody knew let's plug that in so at time zero nobody nearly so one hundred thousand minus zero equals C times e to the negative K times zero we always use the initial value to solve for R seeing families of problems like this RC here is the size of the town then the amount of people who don't know about the treasure initially or specifically one hundred thousand let's alter our our formula now so we know that 100,000 minus the amount of people know about the treasure equals 100,000 e to the negative KT team now we're ready to use a second piece of information to solve for the K and then we can find out how long it takes up half the people know so let's do that so the second piece of information is after one week so we know a time 10,000 people knew so we know a team don't think we won't know was that cake-like ass off for that 10,000 people new out of the town of 100,000 based on the fact that 100,000 people didn't know when we started and it took one week so T is based on weeks right here took one week for that to happen in other words 90,000 people didn't know based on that 100,000 people we started with after one week that gives you a fraction there a ratio a proportion of people who didn't know or people who didn't know two people who didn't know at the start after a certain amount of time do you see your logarithms gonna be negative you see them do you see that right now our K is negative and so when we when we get R K we should have a well we'll have a positive K but when we plug it in will be negative the amount of people who don't know is decreasing about people who do know it's increasing and so let's solve that let's solve for K so when we divide by 100,000 on both sides we'll get nine tenths if we do Ln on both sides that gives us negative KS and I said she was one one week so K is negative Ellen of nine tenths that's around so our K is positive point one zero five four this is negative negative times negative is a positive let's alter our formula to make sure that we have this stuff down so we have 100,000 minus T equals 100,000 some of the people who don't know e to the negative because our K was negative right here so our K is positive yes but when we plug it into our function that worked well our equation that we're trying to use we have a negative of whatever we found so negative zero point one zero five 14 now we can use it answer the last part so how long will it take us for half the people to know our town is 100,000 people so how much is half of that well it's 50,000 people so we're looking for the time that this is 50,000 or how could you do it differently yeah you can make a portion out of it and use it like a wonderful defined by factor problems this one I just didn't do that way so 100,000 minus 50,000 easy to that same exact stuff getting kind of lazy I suppose when I subtract this we get that he grant if we divide by a hundred thousand we're gonna get one-half oh my gosh it is a factor sort of problem it just had some numbers in there so we got that Helen on both sides that's gonna be element one-half divided by that negative point one zero five four we have a negative divided by a negative and let's see a line of 1/2 divided by that's six point five seven weeks or about 46 days if you multiply by seven and so either way you want to go and that's the way we structure that so I hope these are making sense I hope you understand the idea that all these are be some basic differential equations are all separable we can always get an ln of that one variable that means we're going to get an e that means exponential growth or decay and the sign of our constant of variation there it one last one we'll talk about Newton's law of cooling one more time there'll be kind of a shorter bunks about and explored that and then we'll be good to go let's go ahead and tackle our last problem so our last well be here let's say and I've always been interested in like the crime-scene investigator sort of stuff how much did the FBI I was like I do criminal justice for like a year enjoyed it but then I just didn't didn't want to go through them I'd like to teach in math more so went that route but I've always been fascinated by it and so I'd watch the like CSI and CIS and all that sort of stuff I think it's kind of cool but that's sort of gruesome and messy and things but it always interests me how interested me how they found the time of death and so when I did this problem with that that was pretty cool the first time years ago but this is sort of the idea that they would go through they would check temperatures and they would do news all cooling based on the ambient temperature so we're going to do that one last time so let's say you walk into the room as a as a homicide detective or ins as an investigator and you see that the temperature of the room is thermostat controlled and it's 70 degrees someone's looking comfortably and the body was found at 12 a.m. told the morning that was 80 degrees so you took a internal body temperature of 80 degrees and you kept on checking it and after one hour the temperature has dropped 75 degrees and the room has stayed the same temperature that is key you have to assume that the room is not increasing temperature or decreasing and staying the same that's important that's got to be held constant so considering that the normal human body temperature is 98.6 degrees can we estimate the time of death that's gonna be perfect no remember that it going back to the very very beginning none of this is really perfect because differential equations isn't concerned with perfect differential equations most of the time is concerned with approximating good enough so that it's useful you see if we wanted to be perfect we'd have equations that are so crazy then we couldn't solve them and the real world has lots of variables in it so is the room going to be perfect at 70 degrees Fahrenheit probably not in fact as the body loses heat the room gains heat so it's probably increasing temperature minuscule e depending the size of the room so when we find this out at the very end it's an estimate it's fairly accurate but it's an estimate so what we're going to do we're installing new is law of cooling I'm not going to reinvent the wheel we've already done that in this video just a little while to go what we're going to do is take this and it's a very quick example just kind of a out window it's a fun walking on you know on this but an interesting one of these so the way that the temperature of a body cools or any object is what I mean by body is a proportion it's proportional to the difference between the ambient temperature and the initial temperature of that that body we found that out a little early so now that we don't have any more real calculus to do this is one of those family of problems where what you think about it is it where I have a multiplication of my original amount times a factor and I'm trying to over a certain amount of time or have I given an initial value and then the amount after a certain amount of time what's the second one we are going to use the initial value to solve for C and then the second piece information that gives us temperature after time to solve for K and then we'll be able to figure out well when was this body at 98.6 degrees so let's do the first one so the first thing was say the room is 70 degrees know that my ambient temperature 70 degrees the bodies found at 12 a.m. at 80 degrees so 12 a.m. don't put in 12 for the time because that would say 12 hours earlier started our starting time is 12 a.m. at T of 0 which is 12 a.m. our body's temperatures 80 degrees so at T zero the body's temperature is 80 degrees and the ambient was 70 degrees let's click that it so T of zero that's 80 equals we know that the ambient 70 - I don't know it C is that we're trying to find right now e to the negative K times zero for the last time in our video when you're given initial value like this solve for C first well let's see 80 minus 70 is 10 this is all 1 so 10 equals negative C C was negative 10 which means simply change our formula a little bit we know the ambient 75 we know that I'm subtracting C since C is negative we'd have plus 10 e to the negative KT now that we've found our arbitrary constants no longer arbitrary it's based on our initial value now we can go ahead and use a second piece of information considering that our starting time when we found this this body it was a 12 MST T equals 0 after 1 hour so at t equals 1 the body 75 degrees that's the second piece of information so the first piece solve for C second piece solve for K ok so it's 75 degrees so T of Milan equals 75 degrees oh sorry wrong to you t1 equals 75 degrees ambient doesn't change and that happened after one hour and we can see that we easily solve for K here so we'll subtract 70 let's divide by 10 if we do the Ln on both sides we can element of one half or people must make it okay so k equals negative Ln 1/2 another 1/2 is negative so our case is any positive but look since we have a formula that negative K we're going to have a negative number up there so we get let's see what is that 1/2 looks like about zero point six nine let's open our form let's put that back in there so from here we already knew our seed we now know our cave we know that that's a negative case so for a case positive that we found we have negative of that number and now let's let's a salt for anything that we want as far as if I said hey what's the temperature going to be at 3 a.m. that's three hours past so we could just do three in there and figure out what the temperature body is going to be but what we want to know is when was the temperature 98.6 so that would give us a time of death well then our 90s point six is gonna go here we'll solve for our team I'd like you guys to do that on your own I don't want to do it for you I want to make sure that you at least you have done one we'd subtract our seventy of course we divide by 10 of course we do the element on both sides and then we would divide by negative 0.69 and you might run into an issue here wait a minute Arty's negative how long the woman's you get a negative negative one point five two hours since RTS based of hours tell the world you get a negative time well this body is cooling which means it died about an hour I have to go so if you wanted to find the time of death you probably would do minutes so when you have any sort of time like negative one point five two hours keep the negative 1 that's negative one hour and then just multiply 0.5 two times 60 because there's 60 minutes in an hour this would be one hour and thirty one minutes ago roughly about 1 hour 31 minutes so that's prior to 12 a.m. so that means that if the body was found at 12 a.m. all this stuff happened that an hour and 31 minutes prior so what would that be 1029 so 1029 then an hour 29 minutes body was found so that's how these problems work man I hope this has made sense I hope you understand that this is just an application of our differential equations where we have a very simple process we have separable equation so we separate variables we do an integral I know I've done a lot of examples it's a very long video but I hope at least some of them helped you if you can make it through all of them that's great I mean that's you really help you on some of these applications let me know if the video is helpful leave some comments if you would if these these application problems are actually helping because they take so long they take a lot to actually make them I don't want to waste my time of your time so if you're enjoying them and they are profitable to you and they're helping you understand let me know and if not let me know also either way so I hope you all are all doing well I hope this is making sense we should hit in the next video to get onto another technique of solving differential equations which is pretty cool I'll see you for that you

Channel: Professor Leonard

Views: 52,758

Keywords: Math, Differential Equations, Separable Equations, Separable Differential Equations, Exponential Growth, Exponential Decay

Id: VZChL3Pqioo

Length: 110min 7sec (6607 seconds)

Published: Wed Oct 17 2018