 Welcome back today's lecture 23 math 241. We are going to wrap up chapter 7 section 3 These types of problems. Obviously, they're separable differential equations. That's why they're in this section They take on a variety of different names. This textbook calls them mixing problems or Other textbooks might refer to them as a change in dilution of a liquid I think it's pretty appropriate to just call them tank problems. Okay because we're talking about some closed room actual physical tank some specific domain where we've got some Substance brine Fresh water something flowing into that tank Something flowing out of the tank the homogeneous mixture flowing out of the tank and we're going to determine when it's safe to Drink the stuff that's in the tank when it's safe to go back into the lake that had a pesticide spilled into it When the room is was that was formerly a real smoky room when it's Breathable air that's not going to pollute your lungs those kind of things now We don't start off with the real interesting problems We'll start off with kind of the real basic problems and hopefully we'll get to some Examples by the end of class today that are actually kind of solving some some real world type problems So they are separable differential equations Thus the location in the book This book calls them mixing problems change in dilution problems or I just like that name for some reason tank. It's just kind of got a concise Ring to it so we'll call them tank now not all problems that you discuss in a math class that refer to a tank are In fact what I would call tank problems, but where you've got Something going in something coming out at the same rate Then we're going to classify those as tank problems So We're going to have Something that we're talking about in the problem salt if it's a brine mixture coming in a Pesticide that we have spilled in a lake Carbon dioxide in a room we're going to have some substance that we're going to be talking about so we're going to Let some letter equal that substance actually Let's be a little more clear than that just We're going to let a equal the amount of substance salt carbon dioxide pesticide in the tank at some time t Well if a is the amount of this stuff Whatever it is that we're talking about In the tank at time t. What is d a over dt? What's that sound like to you? Okay, it's a derivative of this stuff, but let's put right. It's the rate of change of a with respect to t So it's an instantaneous rate of change So if you think about how this stuff in the tank might be changing it's going to be changed by And as far as memorization of a formula, this is about as easy as it gets for this class It's going to be the positive portion of this is going to be the rate at which a whatever this substance is is coming in Minus the rate at which a is going out pretty simple rate in Positive minus rate out because it's leaving So two components of each rate depending on the kind of how the problem is stated, but the rate at which this substance is coming in will be based on the concentration of that substance times its actual flow and we want the same Same things over here the concentration that is leaving the tank and the flow That it's leaving the tank now in this type of problem that we're going to deal with today and for this particular section The flow of liquid coming into the tank and the flow of liquid going out of the tank will be the same So we'll have the same amount in the tank now. There are other tank problems where this will Differ so you have possibly more coming into the tank Then you have going out of the tank and then you have to compensate for that in the equation, but for us The flow in and the flow out will be the same So there's our equation that will kind of get us started in these tank problems And we'll look at a variety of different Tank problems or mixing problems as they're referred to in the book Again, you've got a good example on page 518 in the book And Welcome you to look through that that I think this is a good textbook By the way, we are probably going to Look at other textbooks that are out there on the market We've got a committee looking at it, but you don't have to worry about it because once you've started the calculus sequence You will finish the sequence in the book that you started it, but if someone Next semester you're in calc 3 and a friend of yours your younger brother or sisters in calc 1 They potentially will have a different book, but then you'll everybody that started in this book will be allowed to finish in that Hopefully you appreciate the fact that that seems fair All right first example. These are problems that are in your book Let's look at this one first and then there's another version similar to this actually problem 38 sorry, you can see the kind of the Binding of the book I try to smash it on the on the Screen to get it clear, but it doesn't quite want to smash the way. I'm afraid I'm going to push too hard and break the glass First example a tank. I like that because it actually calls it a tank a tank contains a thousand liters of brine With 15 kilograms of dissolved salt. So that's what starts the problem We know how much salt is in the tank at time zero pure water Enters the tank at the rate of 10 liters per minute The solution is kept thoroughly mixed That's a little bit of a of a stretch for some of these problems. How do you keep the mixture in the tank? You know homogeneous, you know, why is it not less salty at the wherever the water is coming in the fresh water and more salty elsewhere That's where mathematical models although they attempt to model the situation may not actually model it completely So we'll assume. There's some mechanism that keeps this Saltiness in the tank pretty much homogeneous from top to bottom By the way pure water how much salt if we do let a equal the amount of salt in the tank And we talk about the amount of salt that's coming in If it's only pure water coming in how much salt is coming in the tank? Zero okay, so we do have to pay attention to what we let our Letter our variable represent if a is the amount of salt Then the rate at which salt is coming in and if it's pure water would be zero Pure water enters the tank at a rate of 10 liters per minute The solution is kept thoroughly mixed and drains from the tank at the same rate So that'll be true in all the problems we examine rate in or the flow in and the flow out will be the same How much salt is in the tank? after T minutes and then once we get that equation we can throw in 20 minutes to see after that 20 minute time frame how much salt is there so we start with 15 Kilograms of salt in the tank and then hopefully if we bring in fresh water, and we look at this 20 minutes later There'll be less salt in the tank So we're going to let a equal the amount of salt So we want to pay attention to the salt that's coming in and the salt that's going out So the rate at which salt is coming in Well, it's pure water entering the tank So pure water the concentration of salt would be zero And then we can put in the actual liquid flow coming in it doesn't matter because it's times zero But there's no salt coming in, but we've got 10 liters per minute of this fresh water The rate at which salt is going out Well, we know the stuff leaving the tank The flow is actually the same as what's coming in it's 10 liters per minute What we need is the amount of salt In the tank at any time t it is written on the sheet of paper What is the amount of salt at any time t? It's a a is the amount of salt. So we have a kilograms of salt out of What's the tank size thousand liters? So let's see if it sounds like I'll go ahead and put the units in there a kilograms is What we have at any point in time obviously that changes as the clock ticks in this problem If we reduce liters over liters we're left with kilograms per minute. Does that sound like the rate at which? Salt could be leaving the tank sounds like a rate. So this is also Kilograms per minute, but since it's zero that part kind of disappears Now the other thing we're not using thus far is that initially the tank had 15 kilograms of dissolved salt. Where do you think that will come into play? Initial value so at times it once we get our equation at time zero a will be 15 In fact, we're going to write that down. We'll use that later at time zero a the amount of salt is 15 we'll plug that in later So zero times this is just zero. So we have zero minus this Which is just minus a? Over what about a hundred reduce the 10 with the thousand and that units kilograms per minute That's the rate at which salt is leaving the tank. We don't have any salt entering the tank So we should have at this point in time And this is an easier first example because of the fact that the first piece is missing. It's zero We've got a separable differential equation. We have the first derivative of a and we have a itself So typically when we have a separable differential equation. What's the next step? Separate the a's and the da's from the T's and the DT's right So we multiply both sides by DT and divide both sides by a Or multiply by 1 over a So the a and the 1 over a knockout we're still left with a negative I Didn't give you a chance at the before class to see if anybody had a wet had a web assigned question Do we have any of those pending? I'll try to gear how many of these we do All right, so we've got negative one one hundredth DT does that look right? next step integrate both sides So what's the integral of 1 over a integrated with respect to a? natural log of a Normally we'd put absolute value in there A is an amount right? We're not going to have a negative here. So we don't have to worry about that We have a plus C or a K here. We'll roll that to the other side What's the integral of some number integrated with respect to T? T right we'll put a plus C there You can see why it's in this particular location in the book even though it's a different kind of problem Eventually we get back to the point where it's a separable differential equation. What's next? So for a so for a how do we get kind of out from under the natural log? Good you're catching on to that word. That's good. So if two things are equal e to those Quantities in the equation are also equal e to the natural log of a is a And if we have a sum in the exponent position that kind of goes backward Algebraically to the product of two things that have that same base base e in this case So e is a number C is a number. So there's nothing variable about that. That's just something Some number So now that we've kind of set it up rate which salts coming in rate at which salts going out Get rid of things that aren't really there like the zero and Separate integrate solve for a now we throw in at time zero The amount of salt in the tank is 15 so basically that says that B is what? B is 15 Because this is e to the zero which is one Sorry equation is a which varies B is now 15 So the questions that are asked at the end of this problem how much salt is in the tank after t minutes That's where we are now, right? We have an equation that gives us the amount of salt in terms of t and Then the specific problem after 20 minutes. I usually kind of check my equation to make sure that I Did the problem in terms of minutes and now I'm asked to give an answer in terms of minutes So I want my units to agree So we had leaders per minute, and we're asking for minutes so that agrees So at time 20 our units are in agreement and we have pure water coming into the tank So we would expect there to be less than 15 I mean you can tell from the equation that it's it's decreasing, right? Either the negative something times t is a decay type curve So we better get something less than 15. I don't remember what the answer is to this 14.9 point nine nine Nine nine three. I thought you said nine nine repeating. That's a different issue which we probably ought to take up with that 14.93 right nine nine three so we haven't had a whole lot of Watering down the brine yet, right with only 20 minutes or anybody else getting the same number Okay, I don't remember that number Okay, I'm remembering less. Okay, so let's we want a consensus here 12 now do we have a consensus? 12.281 So I knew that 20 minutes wasn't a long time, but it didn't seem like it could It would knock that down a little bit further than that. So anybody else run it through 12.281 Check that out Nicole and see what happened Questions on that problem with that setup find it okay parenthesis situation We're going to do some things other than just salt in a tank But let's take a look at the setup of problem 38 which is similar to this By the way, let's let's address that point nine nine repeating Which is what I thought Nicole said if we ever talked about in here that point nine nine repeating is Actually 100% equal to one Tell a couple of you didn't like that at all. Maybe we haven't talked about that Let's talk for a moment about that that those two things are equal Do you think most of you are thinking this is smaller, right? This is smaller than one. So if it's smaller than one, what would you add to it to get one? So something Can you tell me what you would add to point nine nine repeating decimal to make it equal to one? Which you can't by the way because they are equal Well, then why does my interpersonal teacher if you get like an eighty nine point nine repeat? She doesn't round it up to a nine. Well, she's mathematically challenged She doesn't do that because they are the same and it's not a trick. There's nothing there's not a mathematical trick about it Since we have calculators out because we just finished a problem. What is One ninth punch that into your calculator Point one repeating right? What's two divided by nine? Repeating okay, we know three ninths is one third, which is point three repeating well with this particular pattern When we get down here to eight ninths, we would expect that to be and you can check this with your calculator That's point eight repeating. Oh, well for goodness sakes What's nine ninths? That would be point nine repeating by this particular pattern, right? But nine ninths is really equal to one But that's not good enough like some of you aren't buying that So let's say we don't really know what it is. It's something it's in What if I multiplied both sides by ten? That's legal. I've got two things that are equal. I'm not saying what in is yet But what is this side? This side times ten is ten in what is this side? How do you multiply a decimal number by ten? Don't you just move the decimal place over so if ten in is that and In is that those are two equations now subtract Ten in minus in would be nine in What's nine point nine repeating decimals not a trick not a trick? nine point nine repeating minus the point nine repeating that Would be nine right divide both sides by nine and In is one. What did we start with in was that what did we end up with and is that so it is true Point nine repeating decimal is in fact 100% equal to one There's nothing you can add to it that would make it one it already is one So what is? Let's take that just a step further. What common fraction is that? That's point two five, which is really one-fourth So the point nine repeating kind of is a little bit Mysterious because it's repeating and we all have very finite mines mine is a whole lot more finite than yours Because of my age But because we have finite mines we are we have a hard time Wrapping our minds around the fact that point nine if you let it repeat. It's no different at all from one Sorry felt like it was time to address that All right, this example kind of comes off of problem 35 nicely It's also in your book. These are problems that are on page 520 from your book a Tank contains a thousand liters of pure water this time a little bit different brine that contains point oh five kilograms of salt per liter So that's its concentration Enters the tank at a rate of five liters per minute brine we have another Tube coming in here to the tank brine that contains point oh four kilograms of salt per liter of Water that's its concentration enters the tank at ten liters per minute. So we have actually two things Coming in to the tank The solution is kept thoroughly mixed and drains from the tank at a rate of 15 So again, you'll see we've got 10 liters per minute coming in another Port that gives five liters per minute. So that's a total of 15 liters per minute coming in We've also got 15 liters per minute going out How much salt is in the tank after two minutes and after one hour? Well, I do see a little bit of a problem with hours and minutes, but we'll just instead of using hours We'll just plug in 60 minutes So let a equal the amount of salt in the tank at time t rate of change of a Is the rate of change of in this case salt? Read at which salt is coming in we do have some brine emptying into the tank That couple different ports coming in Let's go ahead and write down the initial conditions at time zero, which we won't use right away But we'll come back and use later how much salt is in the tank at time zero Here's the statement that's going to tell us that a tank contains a thousand liters of pure water Zero right So at time zero we have zero salt in the tank it's pure water Now some of you might be thinking that you know pure water does have trace amounts of salt in it. We're not We're not visiting that particular proposal. Okay, it's pure pure means without salt So the rate at which salt is coming in we've got two Concentrations in two flow rates coming into the tank. Here's the first statement brine that contains point oh five kilograms of Salt per liter that's its concentration enters the tank at a rate of five liters Per minute I usually like to stop and see if that sounds like The rate at which salt could be coming in so if you again strike out liters over liters You're left with kilograms per minute sounds like a rate at which salt could be coming in the tank We're still dealing with rate in because we have some more coming in brine that contains point oh four Kilograms of salt per liter enters through the tank enters the tank at a rate of ten liters per minute So again Cross out liters over liters you've got kilograms per minute So now we've talked about everything that's coming into the tank We want to subtract the rate at which salt is going out Well, we do know it's going out at 15 liters per minute What is the concentration of salt in the tank at any point in time? How much salt is in the tank at any time is a So we have a kilograms of salt Uniformly distributed amongst the what 1,000 liters So there's concentration of salt in the tank at any time t and that's the flow at which it's going out So we've got point oh five times five Nothing variable in this at all and then point oh four times ten and then we're going to add those together What's the result is that point? two five and Point four point six five is that right and I was going to stop at this point But this is different enough from our first example that I think we need to see this one through to the end Can we reduce that at all? Three five goes in there three times In five goes in there 200 so we do have a separable differential equation Now you don't have to do this And we didn't have to do this in the first problem because this was zero right so we just had negative Some number and so we went right into the stage where we separated and integrated Did all that but we can't really do that yet because we have a sum on the right side What we would really like is to have some nice clean Function of a that we could then divide both sides by and Then have a natural log on the left and whatever on the right so what I would recommend is Factoring out you can do the problem without doing this, but I think it's helpful at this stage Whatever is the coefficient of a? factor it out so we have a By itself and it's going to be easier to move when a doesn't have any type of strange coefficient So if we factor it out of this term, that's clean that gives us a Now we've got to see what this other term would be if we factor out three over two hundred Well, if you factor something out, you're really dividing by it, right? So what's point six five divided by? three over two hundred yeah It's going to be negative I've already kind of accounted for that because I want this negative that's out in front times this negative to take us back To the positive where we were here 43 and a third repeating 130 Thirds okay, you see how that comes into play later in the problem. I think we'll revisit that particular number Before we're done with this problem All right now. I think we're in the position where we can separate Thanks, and then once we get them separated They're going to be easier to integrate than they would have been had we not factored out the code lead coefficient of a So we would want to multiply both sides by dt Divide both sides by the term that has a in it Sometimes I say things and it reminds me of something that somebody else has said Have I mentioned that Brian Regan is my favorite comedian in here to any of you know who Brian Regan is He's I mean he's really really funny, but he's also clean Very rarely will you hear anything? Dirty coming out of his mouth His science experiment in high school was he forgot it He woke up the morning of the science experiment He's had six months to work on it and he forgot it and he realized it's due today So he got a styrofoam cup and he went and put some dirt in it and so he took it in and his teacher said Brian, what's your project? He goes it's it's a cup of dirt and These he says it really funny and then she goes well explain it. He goes well. It's a cup with dirt in it You know I'm sorry anyway, so he goes just give me an F and move on to the next student But when I said something that Brian if you get a chance to Listen to Brian Regan. He's hilarious and he's clean And he's going to be in Durham September 25th next year and I've already bought my tickets It's a cup With dirt in it All right, so we've got things separated So we would then want to integrate both sides And I don't know if you can appreciate this at this point in the problem But if we did not factor out negative three over 200 This would be a let you equal such-and-such kind of problem So I mean it's doable, but I Just think it's easier to factor out that ugly coefficient Get it done kind of algebraically rather than have to deal with the calculus that results So what's the integral of 1 over a minus this number? Again, we're not going to have to contend with the situation where we're going to take the natural log of a negative number So we're kind of dropping that there is a possible C We're going to throw that to the other side the integral of a number times t is this Number integrated with respect to t is a number Times t and we'll put both constants together and on the right side So a little different than our first example because we had zero salt coming into the tank We want to solve for a what's next Exponentiate both sides e to the natural log of this thing is this thing And on the right side e to this power Write it as a product e to the c is a number So I'm going to call this something else just call it b and then we've got this negative 130 over 3 we'll add it to both sides So I'm just going to rewrite that same equation up here All right, we need to find out what b is What was our initial condition? Time zero the amount of salt in the tank was zero right because it was fresh water. So that means b is negative 130 over 3 Got to get zero. So this is b times 1 right, which is just b So b is negative 130 over 3 So we could write I just write in the positive term first the negative term second We could factor out and I think it's probably Worth while to factor this out Not that you have to do that, but I think that'll Be the last I want that to be the last thing we do on this problem is to examine kind of our final equation The questions asked at the end of the problem. How much salt is in the tank after t minutes? That's what we now have After one hour so that would be we want minutes since we put minutes into this equation So we'll put 60 minutes in for t Do the arithmetic? The calculators do the dirty work. See if we get a consensus on this one 25.7 Got a couple of those Did you get that Chandler? 20 still got it and what was our unit kilograms? Yes, you'll assume that you we got that okay Set it aside for a minute All right, let's go back to this equation Can you visualize from this equation? What happens as you let the clock? roll indefinitely is There a limiting value for the amount of salt in the tank What'd you say 130 over 130 over three Is that true? What happens as t approaches infinity? What happens to that term? That's negative the exponent is negative right and doesn't three over 200 times some huge number It eventually gets huge in magnitude, but it's negative right so what's e to some negatively very large number? Practically zero so this term disappears as T goes to infinity so it's one minus zero So this is just one so it's 130 over three so as t approaches infinity The amount of salt in the tank According to what was given to us in this problem It's closer and closer and closer to 130 over three which is what what we say that was 43 and a third Kilograms of salt so it kind of has a limiting value whether you go a hundred and eighty hours 3,216 hours eventually that's kind of it's limiting Concentration in terms of the amount of salt in the tank all right. Let's set up one other problem I was hoping we'd get all the way through at least three problems, but it doesn't look like what that's going to happen What I think is the next best I want to do one that even though it's called a tank problem It isn't really a tank So let's look at this one right here at least to set it up The air in a room so the room is our tank in this problem The volume excuse me of the room is a hundred and eighty cubic meters Initially the room contains point one five percent carbon dioxide fresher air With only point zero five percent carbon dioxide Flows into the room at a rate of two cubic meters per minute and The mixed air flows out at the same rate We don't want the room to get too much air and explode okay So we flow into the room at two cubic meters per minute and it flows out at the same rate That'll be true in all the problems we do at this point in time Find the percentage of carbon dioxide in the room as a function of time What happens in the long run so we kind of did that long run thing on the previous problem all right? So let's let's see Be the amount carbon dioxide In the room the room is our tank So DC over dt is the rate of change of carbon dioxide in the room So we want to pay attention to the rate at which carbon dioxide is coming in and Then subtract out the rate at which carbon dioxide is going out of the room fresher air With only point oh five percent carbon dioxide flows into the room. So that's the rate or the flow rate Or now the flow rate is two Cubic meters per minute and we have a concentration in terms of percent point oh five percent So how do I take care of that chamber? Well point zero five percent of the air that's coming in now We're gonna end up multiplying these eventually anyway, but let's make sure we write point zero five percent Properly one zero zero three zeros and then a five right so that's point zero five percent So if it were point oh five that'd just be five percent, but it's point zero five percent So you can tell how much and and that doesn't really have a unit So that means we're stuck with cubic meters per minute Does that sound like the rate at which carbon dioxide can be coming into the room so many cubic meters of carbon dioxide? Per minute a very very small amount, but it does sound like a rate Now the rate at which carbon dioxide is going out well What is the amount of carbon dioxide in the room at any time t? see And that amount of carbon dioxide at any time T is equally distributed in the whole room So it's we want to get a percentage a kind of a concentration So the room is what 180? Cubic meters. This is the amount of cubic meters of carbon dioxide The room itself the volume of the room which is our tank in this problem So those are going to be gone so there's a concentration of carbon dioxide And then it's leaving at the same rate So there's our equation Okay, I said we'd set it up and we'd stop and we'll pick up at this point in time tomorrow on this problem