 Welcome back. We are in the middle of a problem a mixing problem or a change in dilution of a liquid problem or as we have been calling them tank problems. The problem that we set up right at the end of class and we'll pick up from that point and then I'd like for us to do one more problem before we advance in the text is problem 36 from this sheet. It is a problem in the book. The air in a room with volume 180 cubic meters contains point one five percent carbon dioxide initially so there's our time zero initial condition. Fresh air with point oh 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. Find the percentage of carbon dioxide in the room as a function of time what happens in the long run. So here's where we were at the end of class yesterday. We're letting C equal the amount of CO2 in the room. The rate of change of C is the rate of change of carbon dioxide in the room DC over DT is the rate at which carbon dioxide is coming in minus the rate at which it's going out. So the data from the problem. Fresh air with only point oh five percent carbon dioxide flows into the room at a rate of two cubic meters per minute so there's the concentration getting rid of the percent so it's point oh five percent so we've compensated for that. Two cubic meters per minute so this ought to sound like a rate at which carbon dioxide is coming into the room. So many cubic meters per minute sounds like what we want. Exiting at the same rate what is the concentration of carbon dioxide in the room at any point in time T. There are C cubic meters that's what C represented the amount of CO2 in the room at any time T. C cubic meters are coming dioxide equally dispersed in the whole volume of the room which is 180 cubic meters so there's the concentration issues or questions because we kind of got that and then we were over time yesterday and we left any issues with this differential equation. So let's do the numbers point zero zero zero five times two is what three zeroes two zeroes and one two zeroes and one minus C over ninety. It is a differential equation first order differential equation it's got the derivative of C in it and it's got C itself in it so it's got a first derivative first order differential equation. What did we do at this stage of our second example yesterday. The first example this piece was zero forget what it was salt something there was zero salt coming in because it was fresh water. Okay factor out the coefficient of C which I think was strange numbers one thirty over three something like that or three over one thirty anyway so now it's one over ninety right doesn't have to happen but I think you'll appreciate the fact that we do this now when we get to the integration of both sides so if we factor out negative one over ninety we're gonna have C and what's the other term when we divide point zero zero one by negative one over ninety what is it point oh nine that right so we'd be if we divide it by one over ninety we'd multiply it by ninety is that correct so point oh nine seems right but that doesn't mean it that is right so if you were to redistribute this negative one over ninety times that you'd be back here negative one over ninety times this hopefully you're back here so the next step is to separate so multiply both sides by dt kind of sends things in the direction they need to go so we're we've got DC's on the left side we need to move our C's over there we've got DT's on the right side if we have any T's we need to move them over there so it looks like we need to divide by so the C's and the DC's are separate from the T's which we don't have any in the DT's so we should be able to integrate both sides what's the integral of the left side C minus point oh nine so you would say let you equal C minus point oh nine if that's true D you equals DC so we've got one over you D you here so we're okay to do that we've got a constant we'll roll that to the right side what's the integral of this right side negative one over 90 T put the two constants together that's not a good constant is it let's make that a K because we've got C's in the problem so that's a K strange looking K what's next exponentiate both sides e to the natural log of something is that something over here we've got e to the and this is going to be times e to the K we've done that enough e is a number k is a number there's no variables in there so that's just another number right and we need to solve for B what are the values given to us in the problem that allows to solve for B time zero let me read that statement the air in a room with volume 180 cubic meters contains point one five percent carbon dioxide initially so C is the amount of CO2 so how do we handle that at time zero is it point zero zero one five or is it point zero zero one five times 180 right that second thing because it's an amount we want the amount so how many cubic meters not the percentage so it's point zero zero one five because it's point one five percent of the entire room which is 180 cubic meters so how many cubic meters of pure carbon dioxide do we have somebody tell me what that is point point two seven so we'll plug those into this equation so the amount is point two seven cubic meters that's initially that's at time zero so e to the zero is one so B is what point one eight so let's see if we've done what we're supposed to do find the percentage of carbon dioxide so we have solve for C in the room as a function of time I think we've done that the only variable on the right side is T what happens in the long run so I guess that means as T goes to infinity what happens to this thing right here so this would be e to the negative some huge number right which would throw it to the denominator so it's one over e to some huge number which approaches zero so this exponent is that correct approaches zero or e to this exponent approaches zero e to the exponent approaches zero which means that term approaches zero so in the long run it looks like C approaches point zero nine and there could be another part to this problem which we've already looked at in another example at T equals 60 minutes or 38 minutes or something what is the concentration of carbon dioxide in the room questions on that yes what do you say C approaches what well if this is approaching zero then point one eight times that disappear so C approaches point oh nine in the long run as T approaches sorry I didn't clarify that based on this problem I don't know that we need to go all the way through this next example but here's a type of tank problem that is not a room filled with air and we want to control the carbon dioxide that's our tank or actual physical tank this particular tank is a lake so our tank in here in this problem is a lake and we know approximately the volume of the lake and we've had a spill in the lake 500 gallons of pesticide is accidentally spilled into a lake with volume eight times ten to the seventh gallons and uniformly mixes with the water these are some assumptions that are made in this problem so this may not the results that we get may not be kind of real world but we're assuming that it kind of disperses fairly uniformly a river flows into the lake bringing 10,000 gallons of fresh water per minute we're assuming that the river doesn't have any pesticides of its own okay again another assumption but so we're assuming that no pesticides are coming in and the uniform mixture flows out of the lake flows over the dam to the lake at the same rate how long will it take to reduce the pesticide in the lake to a safe level of one part per billion so safe to like water skiing or swim in or safe to drink I don't know what safe is talking about but under certain conditions you would want to know what is either or safe to actually go in and swim safe to actually drink the water say for you to drink it I'm not going to be drinking it because I just don't have an affinity for pesticides they don't they don't work real well with me I don't have a palette for pesticides this one I can tell you a real-world situation that has become very frustrating to me and maybe one of the Wake County commissioners will be watching this cable telecast and actually do something about it my family and I like boating we have a boat we live three miles from Lake Wheeler which is not a huge lake but it's City of Raleigh reservoir three years ago I haven't been actually told what the circumstance is but some kind of runoff comes into the lake and so I would assume that we if we'd have some you know kind of gully washing storms that as long as they fix the problem that was coming into the lake that you know a couple months go by you have a couple of big storms it washes it out it cleans it out three years later the lake is still not open for water sports which is very annoying so something is going on with that situation either the runoff itself with the heavy rains are bringing more pollutants into the lake something is wrong and it's not being fixed so if you know anything about Lake Wheeler and the pollutants that are coming into Lake Wheeler in Raleigh which is three miles from my home I would appreciate kind of knowing about that or even better yet how about fixing it about fixing the problem okay so we've got pesticide do what yeah yes in bowling yesterday we learned that you always round down and so I told him that point nine repeat is the same as yeah I'm gonna get trouble now right yeah he didn't believe me then okay what's his name off the convincing coach kid kid yeah with two D's all right now if it's just point nine or point nine nine I can't convince him because that's not true but it is point nine repeating and he said it's still there is no such thing as rounding down by the way do you know what the proper term is since we're learning like proper terms exponentiate both sides proper we want to be proper mathematics what is it when you round down trunking so if you don't round up rounding is rounding up if you just ignore the decimals then that's called truncation we have to be proper so if P is the amount of pesticide then the rate of change of pesticide the rate the pesticide is coming in minus the rate the pesticide is going out all right so what do they tell us in the problem the rate that the pesticide is coming in well says a river flows into the lake bringing 10,000 gallons of fresh water per minute how much pesticide is that zero right so the concentration of pesticide is zero it is kind of irrelevant that it's being brought in at a rate of 10,000 gallons per minute but it's zero it's flowing out at the same rate what is the concentration of pesticide in the lake at any time T since it is changing P is the amount of pesticide right equally dispersed in the entire lake which is eight times ten to the seventh gallons so eight with seven zeros that right so we've got actually one of the simpler types because the rate at which it's coming in is zero so we've got just negative this term that's better similar to our first example the solution of this is going to be similar to what we've already done but let's take a look at the how we're going to answer the question how long will it take to reduce the pesticide in the lake to a safe level of one part per billion how do we handle that because how long will it take we're searching for T and then don't we have to put in capital P which is not like one part per billion capital P is an amount isn't it an amount so how do we do that okay that'll work the amount of pesticide over the entire volume of the lake right is one part per billion billion having how many zeros nine so that's what we want to actually plug in so that would be an amount of pesticide that we would want to know when it has been reduced to that amount which is a ratio of one part per billion so this value goes in here other than that it's similar to examples we've already done question on that one but we've looked at I guess kind of a typical tank problem that was a tank we had Brian coming into the tank we looked at a room full of air that's our tank and here's a lake that is representing a tank so it could be a variety of things here's an oak barrel previously held red wine okay and the barrel probably still has some red wine soaked into the wood and you want to run fresh water into it and how much of it remains after a period of time so that is the tank the tank is an oak barrel I like our county commissioners I really like them and I like Lake Wheeler and I like skiing and boating with my family probably more than I like my county commissioners okay let's forge into chapter seven section four my guess is that you already know a fair amount of situations dealing with exponential growth and decay which is what this is about I can hear the phone ringing from the you know it's not our jurisdiction it's City of Raleigh that's assuming that anybody actually watches this on cable TV right that's a huge assumption since we're talking about assumptions so let's say that we have a statement about a growth rate let's say the rate of change of population so we'll assume the population is growing dp over dt is directly proportional to the population at any point in time okay so anything that is kind of growing or actually decaying exponentially would follow follow this particular model we've already done this in terms of a differential equation but most of the time with these things you can take the statement we're going to do Newton's law of cooling in a moment you can take a statement of Newton's law of cooling take that English sentence translated into a mathematical sentence which is an equation that's all this is and then do what needs to be done in terms of separable differential equations so the rate of change of population is directly proportional to the population at any point in time so we would multiply both sides by dt not that we would do this we have actually already done this sorry so we do the separating the next thing we do is integrate both sides the integral of 1 over p dp is natural log p p is positive its population integral of k integrated with respect to t is kt we do have a constant exponentiate both sides e to the natural log of p is p this would be e to the kt times e to the c e to the c is a number I don't know that we went any further than this but we could say at time zero the population is typically referred to as p zero the initial population and if we plug those things in for p we plug in p zero and for t we plug in zero then it turns out that capital B is what the initial population so this equation becomes population at any time t is the initial population times e to the kt so you've seen that mathematical model in a variety of versions there's one kind of tailored to a population growth problem you may have seen it written like this you may have seen it written like that that's something that's growing exponentially and in fact that something is what for this particular model where have you seen this model before compound interest continuously compound interest in fact will kind of derive that today time permitting from the kind of more generic compound interest so continuously compounded interest that would be the model and that's anything that's growing exponentially everything this is kind of the future amount or final amount this is the principal amount which is the initial amount of money this is the interest rate which is the same thing as the rate at which that thing is growing one thing that the author addresses here let's go back to this let me at least make mention of this back to the original differential equation sometimes we call k the growth rate actually if you solve this equation for k you would get this so k although sometimes it's called the growth rate it's really probably better referred to as the relative growth rate because dp over dt is the rate of change of population with respect to time so that's kind of the growth rate but if you want to say what k is k is kind of the relative growth rate so it is based on the population at that point in time so it's not just dp over dt it's 1 over p times dp over dt so your author makes an issue out of that so probably it should be mentioned k is not really technically the growth rate it's the relative growth rate and we're not talking about the number of cousins and aunts and uncles that you have not those kind of relatives relative to the population at that particular point in time alright let's do a growth model I actually had some great data when I was in DC traveled with the park scholars on one of their learning labs to DC and this was given out at the basically the the Fed the Federal Reserve building it's got federal debt at the end of years 1940 and then it's got predictions up through 2013 and then the summary of receipts surpluses and deficits from 1789 all the way up to what they're projected to be in 2013 I think some of this data is probably blown out of the water recently with some bailouts and deficit spending so to predict where we would be in 2010 or 12 or 13 probably not gonna be a real good problem but I'll hang on to this maybe it'll be valid at some point in time I like real data so here's some real data so I decided to switch to this real data so this is real data from the college board this is the cost of attending a public four-year college or university and I am have a son that's here right now and so I'm right in the middle of realizing the cost of attending a public four-year college or university we don't did you think that faculty like their kids got a cut they could come here for a reduced rate no so legislators if you're watching sorry but that would be a nice little perk you know to let our children come here at a reduced rate that's what somebody was telling me earlier I didn't really personally think that but that would be nice where do you where you from Pennsylvania Pennsylvania I like Pennsylvania so well we have private colleges in North Carolina that have reciprocating agreements like that that's Davidson your third child third child now he had to talk but that'd be neat just to you know free books you know or free tuition or something like that would be a nice perk for university faculty alright so let's use some data let's take this data from 95 96 let's call this 95 data and let's take another data point so this will be our time zero our initial data because that's what we're starting our clock in this particular point in time in this chart and let's go to four years later so we're going 95 to 99 so if we're calling this 95 data and this 99 data this would be time for right four years after our initial data point so we're going to use this model it's not population so I won't use capital P so our two data points time zero in time four I'll call it y sub four we won't actually solve put it in like that but so can't we just plug in y zero into the mathematical model right the exponential growth model so we're assuming that it's growing exponentially that doesn't mean that it's like really rapidly growing but it could be you know fairly gently growing but it still is exponential in nature so we'll plug that in for y zero so once that's plugged in for y zero then we go to our second data point and if we plug in our second data point what's that going to allow us to solve for okay so the new y value so we'll plug in t equals four and the new y value the new cost of attending a public four-year college or university and now we want to solve for k so we divide both sides by d exponentiating we're actually going to de-exponentiate at some time we call that something else because if you de exponentiate what are you actually doing taking the natural log of both sides is there a verb for that I don't know if there's a verb for that we gotta come up with login login I don't know think be creative and come up with a new name but we do when we have a variable in the exponent position we've got to get it out of there somehow so that would be taking the natural log of both sides and because natural log and and raising e to the x are inverse functions of one another that's why this works so nicely so the natural log of e to some power is in fact that power right and that what you're looking for when you're looking for the natural log you're looking for the power you would raise e to get that there's a little understood e there so what power would you raise e to get e to the 4k that's a self-answering question you would raise e to the 4k to get e to the 4k divide both sides by four and we'll fire up our little machines take that quotient take the natural log of it divide that by four this should be a growth problem so we should expect k to be positive right k is positive which the amount of money in this table is growing as we work our way down the table so we would expect it to be positive what is k now it's not k for all of this data but it's k for the two data points that we looked at 0.045 truncated or rounded truncated and I don't know it kind of depends on the problem we'll do a decay problem that I think we'll have to get to the seventh or eighth decimal place to actually get some meaningful data so it kind of depends on what you're talking about so we would take this now back up to our model so y equals so this table went up to 2005 it was we're using the left number so up to 2005 so let's predict the cost of attending a public four-year college university again there are some problems or assumptions with this we use two data points they may not be in line with the rest of the data point so using the two that we have predict the cost in 2009 so 2009 would be t equals what for us with time zero was 1995 so 14 so this is basically a button pushing problem from this point so the cost if we use this model and assume that it's exponential growth what would that be 16 16,000 I don't know I'm guessing 12,700 roughly how do you think we did let me show you the chart again and I just I just picked out some values I didn't pick them out ahead of time so according to this in 05 it's already 12,1 so we use data from several years ago maybe the growth rate has increased kind of looks like it did from here to here right a little more rapidly than what we found over this four-year stretch so it's kind of a little bit of a guessing game which data do you use to help you predict so it might be a little underestimate based on what we see as real data from 2005 but it's truthfully it's not that far off from what I'm paying this year I think it's probably about 14 so it's not ridiculously that far off okay I have a decay problem I think probably have time for that and we've really got the rest of the week to do this section and all the examples that are in it so we've got here's what we have yet to do we have exponential decay we'll do one of those at least as much of it as we can then we'll go to Newton's law of cooling which has its own statement which has its own differential equation and then we'll go to continuously compounded interest actually let's switch the order up let's not let's do this one this is a decay problem because it has half-life in it so something is decaying over time as opposed to the money being paid out to the colleges was growing as the table went down so this is a carbon dating problem so if you're like me and a geeky math major in college this was the only kind of dating that I could actually get was carbon dating so it has it's especially meaningful to me that which kind of reminds me if a couple of you met my wife and you know I'm thinking I look at her and she's just so pretty I'm thinking what were you thinking you know me and you mean gosh was it just like a bad day for you bad week maybe a bad you know a couple of months just having trouble seeing so carbon dating brings back some good memories so radioactive carbon 14 C 14 has a half-life of five thousand seven hundred and fifty years that's why this particular element is used to date things so what's half-life me okay how long it takes this thing that we're talking about this element this isotope to decay into half of what it was originally so if you started with a hundred grams of this nasty stuff five thousand seven hundred and fifty years later it has decomposed to 50 grams so this is a very slow decay rate because it's so slow that's why it's useful in carbon dating now carbon dating has its flaws as well if you read an article in a science magazine or a journal and it says using carbon dating we have determined that the age of this whatever it is is 10 billion years old probably not meant for that purpose because we don't know what the atmosphere was like you know several hundred thousand years ago if in fact the planet was even in existence then so this if you get something to be several thousand years old that's probably pretty valid with this technique but hundreds of millions or billions of years old this is not a good technique for that so C 14 half-life of 5750 so we're going to have a problem to deal with here but the percentage of carbon 14 present in the remains of plants and animals can be used to determine age archaeologists found okay but so our real problem starts right there so let's see if we can find out what K is for this stuff C 14 so we don't know how much we're going to start with let's call it why zero we do know that 5750 years later this original sample of why zero has decayed or decomposed to one half of why zero so you don't have to know what you start with you can make it up if you want to you can say we started with a hundred and it decomposed to 50 and it still works but if you divide both sides by why zero that's gone so we'll take the natural log of both sides natural log of e to a power is that power doesn't look like a negative number but the natural log of one half is a negative number so we'll divide that by 5750 and I can't remember the number of zeros but you get a certain amount of zeros five zeros and then one two oh five or negative four okay so three zeros one two oh five and actually that may not be enough decimal places to keep for K depending on the level of accuracy that you want but now we know in our particular model for this stuff C 14 there is the decay rate so it's decaying very very slowly all right so back to the problem so that works for any problem that deals with carbon 14 archaeologists found that the linen wrapping from one of the Dead Sea Scrolls which were found about late 40s I think 1947 somewhere in there right around 1950 along the shore in a cave up along the shores of the Dead Sea so the linen wrapping from one of the Dead Sea Scrolls had lost 22.3% of its carbon 14 how old was the linen wrapping so we will assume that this linen wrapping was used to wrap the Dead Sea Scrolls right after the plant that made from which they made the linen wrapping was actually harvested so originally it had 100% of its carbon 14 because very soon after it ceased to be a living plant it was made into the linen wrapping so originally it had a hundred percent we're looking for how old this is it had lost 22.3% I don't want to put 22.3 here so I want to know how much it still had so if it had lost 22.3 it still had 77.7 so it had gone from a hundred percent of its C 14 to 77% so a hundred percent is one so I'll just leave that out and 77.7% would be that we're out of time today we got it set up we'll pick up from this point on this problem tomorrow