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Main content
Current time:0:00Total duration:9:58
AP.CALC:
FUN‑7 (EU)
,
FUN‑7.H (LO)

Video transcript

so we left off in part one getting pretty close to finding our n of T that satisfies the logistic differential equation where its initial condition is between zero and K and now we just have to really just do some algebra to finish things up so we left with this that for our n of T this must be true now we could use a little bit of logarithm properties to rewrite this left-hand side as the logarithm of have the logarithm of something minus logarithm of something else that's going to be the first something the logarithm of the first something divided by the second something one minus n over K and this is of course going to be equal to all this business that we have it's going to be equal to actually let me just write it it's going to be equal to equal to R times T plus C plus C and now what we could do this is the same thing as saying that that e to the RT plus C is going to be equal to this thing right over here this this the natural log of this is equal to the exponent that I have to raise e to to get to this right over here so I can just write that we could just write or another way of thinking about as we're taking we could take e to if this is equal to that we could take e to this power on the left hand side and e to this power on this right hand side and they should be equal so e to this power is just going to be what's inside the parenthesis it's just going to be n it's just going to be n over 1 minus 1 minus n over K I'll do that in green color so you can just keep track of where things came from is equal to e to this business is equal to e to the R T are out of the T and white are T plus C plus C now this I could rewrite if I want if I want to if I have e to something plus something else I could rewrite this as to the RT times e times e to the C and just to simplify things this is just going to be another constant here and I could you know if this is C I could call it C 1 but I'm just going to call that a constant so I could just say that that's going to be equal to some constant times e to the RT and now we just need to solve for n and once again if at any point where while we're working on this you get inspired feel free to solve for n so let's see one way to solve for n let's see we could take the reciprocal of both sides of this we're going to get 1 let me draw a line here just so you know that we're so let me draw a line so if we take the reciprocal of both sides of this we're going to get 1 over N 1 minus n over K over light-green n is equal to is equal to and so let's you know we could say it's equal to it's equal to 1 over C times e to the negative RT but 1 over C that's just going to be another constant so I could write 1 over C here when I take the reciprocal but once again this is going to be another constant so I'm going to be a little bit of hint a little bit hand wavy says well ok we're going to get another constant here I could have called you know this C 1 this C 2 I could call this C 3 if I want to make it clear that these are not going to be the same number this is e to this power this is the reciprocal of that actually maybe I'll do that just to make it a little bit a little bit clearer that would have been C 1 right over there and so this is going to be sorry the reciprocal of this is C 3 and e to the negative the reciprocal of U to the Archies e to the negative RT e to the negative e to the negative R e to the negative RT and let's see if we divide the numerator and the denominator by n or if we divide i wondering when we divide this term by n we're going to get so this term by n is going to be 1 over n and then this term by n is just going to be minus 1 over K so this is just going to be minus 1 over K is equal to this is equal to this business so copy and paste it's going to be equal to that this is good algebra practice here now let's see we could take this one over K add it to both sides so let's do that so let me just cut so let me cut and paste it I'm going to add it to both sides so that should be a plus 1 over K so plus 1 over K and now to solve for n I just take the reciprocal of both sides so I'm going to get I'm going to get N and I'll write it in kind of the function notation and of T make sure we make make my tea in white since I've been taking the trouble all this time of rewriting this in white is equal to 1 over is going to be equal to 1 over all of this business is going to be equal to 1 over all of this business so copy and paste is going to be equal to that and that by itself that's by itself is already interesting that by itself is already interesting so I could write it like this and if I want if I don't like let's see well yeah I could just if I don't like having this K you know kind of a fraction in a fraction I could rewrite it as actually maybe I'll do it over here and of n of T is equal to I'll just multiply the numerator and the denominator by I'll just multiply it actually let me just let me just leave it like that for now and what I'm going to do what I'm not what I'm now going to do is I'm going to say well look you know we're assuming that n of 0 is n sub-nought so we're assuming that n of 0 and of 0 is equal to n sub-nought so let's write this thing let's solve for the constant let's figure out what what this could be if we know what our initial condition is so n of 0 n of 0 is going to be equal to it's going to be equal to 1 1 over when T is 0 this is just going to be equal one so it's going to be our constant C three plus one over k plus one over the maximum population that our population that our environment can handle and that's going to be equal to and not and now we can solve for our constant so we get yeah I'm probably gonna need a lot of real estate for this I can take the reciprocal of both sides again so this is something that we're doing a lot of C 3 plus 1 over K is equal to 1 over N naught this is equal to 1 over n not just took the reciprocal of both sides and so we get our constant C 3 is equal to 1 over N naught minus 1 over K minus 1 over K and so we can rewrite our solution which we'll call the logistic function we get we get this is fun now n of T n of T is equal to 1 over our constant is this so it's going to be let me copy and paste this so copy and paste it's going to be that that's our constant so it's going to be that times e to the negative RT to the negative R times T plus 1 over K plus 1 over K and if we don't like having all of these denominators in the done all of these fractions in the denominator why don't we multiply everything times the numerator and the denominator by n naught K so I'm going to multiply the numerator times n not K and I'm going to multiply the denominator by n naught K and not K and so what do we get this is all going to be equal to in the numerator I have n naught times K and naught times K and in the denominator I'm going to have let's see if I multiply this term right over here times n kay I'm going to have K if I multiply this term times n not K I'm going to have and not so it's going to be minus an naught minus n naught times e to the negative RT times e to the negative RT negative RT and then if I multiply this times and not K I'm going to get n naught so plus and not and there you have it we have found a solution for the logistic differential equation we will call this logistic function and in future videos we will explore it more and we will see that it actually does if you were to plot this I encourage you to do so either on the internet you could try Wolfram Alpha if you're on your graphing calculator you will see that it has the exact properties that we want it to have it starts at n naught it starts to increase at increasing rates but then it starts to slow down as we reach our maximum population and so that is actually a very neat function you know if you model a population with this you can kind of start to make predictions about what might the population be a time you know whatever and so hopefully you found that satisfying
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