7 8A Exponential Models w Differential Equations

Añadido: 2026-05-11

[00:00:13]all right sorry about that here we go uh topic seven eight more differential equations but this time we're going to be talking about exponential models um some of you may have heard of newton's law of cooling you do not need to know newton's law of cooling for the ap exam they won't refer to it they may use it but they're not going to refer to it as newton's law of cooling but you will have some story problems uh growth and decay type problems that type of thing so let's take a look at what it might look like first of all if they talk about the rate of change of y the rate of change of y is proportional to y so if you remember you're from back in oh probably ninth or tenth grade direct proportion inverse proportion these are directly proportional to each other so if you see proportional it'll be a k so proportional to y is k times y okay so you have exponential growth in dk model theorem this theorem right here c times e to the kt this is the equation that you want to just memorize to make all of these easier okay so instead of having to take a derivative and solve just know that the i'm sorry the anti-derivative the integral um that the anti-derivative of the integral will give you this okay if you memorize that and not have to work it out it'll save you a lot of time okay so c is your the initial value okay k is the proportionality constant um and exponential growth is when your k is positive and decay is when your k is negative so just memorize this and these problems are a piece of cake and there's not ton of them so let's just take a look it says the rate of change of y is proportional to y so that's what you want to look for the rate of change of y is proportional to y so anytime you see that you want to use the anti-derivative okay this right here is anti-derivative is equivalent to this so we can just memorize that i'll prove it to you later if you really want to okay all right so we're going to use this so the first thing you do is you plug in the first point that's given okay so the first point that's given to us is 0 2. so we're going to go ahead and plug that in so 2 equals c times e k times our t value of 0.

[00:03:18]so e to the 0 is 1 so that means that c is equal to 2. okay so that's the first thing that you're going to do the next thing you're going to do is you're going to plug in your second point which is what 0 4 right so that gives us four is equal to two times e to the k oops not zero sorry sorry t is equal to two so it gives us this and that's basically telling us that e to the 2k is 4 divided by 2.

[00:04:09]okay we're trying to find k we're trying to find k oh and if you didn't realize i'm sorry i skipped right over this this 2 right here is this 2 right here okay so the first thing we did step one is plug in the first point and we found c the second step is to plug in c with the second point that's given so that's what we've done so far okay so that c goes here the second point that goes with it t and y we're trying to find k how do you solve this well this is an exponential equation exponential equations you can change to log form or you can take the natural log of both sides so i believe that when we learned it we took an exp to solve an exponent natural equation we changed it to log form okay so we're just gonna go with that um the natural log a log is equal to the exponent right log is equal to the exponent e is your base well isn't natural isn't that redundant right there so we really don't need to write that so isn't that exponential form to logarithmic form let's check e to the 2k equals 2. remember circle method e to the 2k equals 2. exponential form logarithmic form okay but again we don't need that because that's redundant so basically it's telling us that k is equal to one half the natural log of two so if we go back up to our original equation our solution would be y equals our c value which you calculated to be a 2.

[00:06:48]times e to the kt and our k is one half the natural log of 2 all of that times t now this has an added step this is the solution and again you could put that one half up here and it could be the ln of the square root of two however you want whatever log properties you want again if it's multiple choice they might put that one half up here and call that the square root of two that's your exponent properties for logarithms um but in this particular one it says what is the value of y when t equals three so when you see what is the value of y when t equals three you have to know and this is going back to the last couple sections as well when they ask you a question like this um you know that goes back to the you have to know that you're finding the particular solution in the last couple of topics they said find the particular solution this one says find the y value which means you need the particular solution so this is the particular solution so then our last step is going to be to actually find the value of y when t is equal to 3. so y is equal to 2 times e to the one-half times the ln of 2 right and all of that is going to be multiplied by 3. again these are usually non-calculator problems if this is a free response question you can leave it anything that can be typed into a scientific calculator you can leave this is an acceptable answer on a free response however on a multiple choice it's not going to look like that we have to do some simplifying here so we could simplify it to 2 times e this 3 gets multiplied by the half okay um that can get turned into e to the ln of two to the three halves and let's see if you remember what all of this is equal to all of that is equal to just all of this here these bases are the same this is base e this is base e it's just equal to two to the three halves so that's two to the two halves that's two to the first times two to the three halves which equals two to the what to the 5 halves now different ways you can take the square root of 2 we don't know the square root of 2 we could do 2 to the fifth so it's the square root i don't need that two i guess of 32 which reduces to four square roots of two and four so four squares of two could be a multiple choice answer okay 4 times the square root of 2 is approximately 5.657 so graphically this is what our example looks like that we're doing right now um let's go over to let's just kind of check our answer they're not going to give you the graph but i gave it to you just so you can see that our answer is correct when you go over to t equals three you go up here you're a little bit more than five a little bit less than six so 5.657 again this will not um it could be a calculator problem on the multiple choice but again on a free response stop here always stop if you can type it into a scientific calculator if you really want to go back and simplify later on if you have time then do that okay we have a couple other applications for growth and dk i'm sure you all have heard of half life um so let's take a look at this this particular example um you have you're starting with 15 grams uh how long will it take for 15 grams again how long will it take um it's asking you to solve something in this problem which means we need to find the particular solution okay we need to find the particular solution so let's go ahead and let's see what they give us you start with 15 grams right so that is 0 15 and we're talking about what is that calcium casein i'm sure you guys know more than i do that's right here here is its half life 30 years 30 years so 30 years later how much would you have seven and a half right half of what you started with okay so again we're going to use um our equation y equals c times e to the kt because again we're talking about exponential growth and exponential decay so just memorize this and will save you a lot of time all right so we're going to start with our first point um y is 15 when this is k times zero so basically any time that's zero that's your c that's your initial amount so you really don't even need to do that work zero means initial amount c is initial condition so our equation is y equals 15 times e to the kt okay so we're going to then use our second point which is uh seven and a half after 30 years so that's 30 okay divide both sides by 15 well it's half-life so you're going to get a half okay so um again i showed you how to um change it from exponential form to log form again another way is just taking the natural log of both sides that is also another method for solving an exponential equation you take your pick so the natural log of one half the natural log of e to the 30 k is 30k so k is one over 30 times the natural log of one half so our equation our solution our particular solution is y equals 15 e to the 1 over 30 times the natural log of one-half that's our k times t okay so then when we go how long will it take for the 15 grams that we started with to decay to one gram how long will it take they're asking us to solve for t so i'm going to take this equation i'm going to write it up here so they must be giving us y they want us to have 1 gram 15 times e um i'm going to go ahead and take take change that's the natural log of one half [Music] see this down here this is i'm gonna move this t here if you don't mind i can move that t there because it makes it a lot easier if i do that and now i can have one-half the natural log of one-half all of that that t to the 30th that's the exponent properties for logarithms it's in the front i can move it to the exponent okay actually the exponent out one half sorry guys long day there okay so this x this coefficient can go up to the exponent of the one half exponent properties of logarithms okay so how does that help me well again we have a formula and again i'll go ahead and write it down for those of you who don't remember that if you have v log of base b of x that's just equal to x so this is base e this is base e so the bases are the same so it's just equal to this so that's one of the log properties way back when so if we continue with this we get 1 equals 15 and all of this is equivalent to what one half times t divided by 30.

[00:17:24]now bad bad math multiplying these two together that's bad math so i have to divide by 15 and i have an exponential equation which i can turn to log form or simply just take the log of both sides that's just my preference again completely up to you and because you take the log of both sides you can now move that in the front so for those of you who hated your logarithms unit in 10th grade you probably won't like this this section so that gives me the natural log of 1 over 15 equals t over 30 times the natural log of one half again i'm trying to solve for t um what would you do first i know i guess completely up to you i'm going to go ahead and i'm going to take the natural log of 1 over 15 and i'm going to divide it by the natural log of 1 half and then i'll multiply both sides by 30. and again this is this is going to be a calculator problem so t is approximately equal to you can go ahead and check it on your calculator i would practice typing this in very carefully excuse your fractions you want to don't want to mess up your parentheses t is going to be approximately 117.207 years okay um so this is usually a multiple choice question um that's a lot of work for 1.2 points um you know something you might want to do last okay all right i'm going to stop this video right here