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Current time:0:00Total duration:9:58

AP.CALC:

FUN‑7 (EU)

, FUN‑7.H (LO)

-(Voice over) So we left off in part one getting pretty close to finding our N of T that satisfies the logistic
differential equation where it's 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. 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. The logarithm of something minus the 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, is 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 E to the R T plus C is going to be equal to this thing right over here. 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 could just write that. We could just write, or another way of thinking about it, we could take E, if this is equal to that, we could take E to this
power on this 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 one minus, one minus, N over K. I'll do that in a green color, so you can keep track of
where things came from. Is equal to E to this business. Is equal to E to the R T. R, I'll do the T in white. R T plus C, plus C. Now this I could rewrite if I want to. If I have E to something
plus something else, I could rewrite this as to the R T times E times E to the C, and
just to simplify things, this is just going to be
another constant here, and I could, if this is C, I could call it C one, 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 R T, and now we just need to solve for N. And once again, at any point 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. If we could take the reciprocal of both sides of this, we're going to get. Let me draw a line here
just so you know we're. Let me draw a line. So, if we take the reciprocal
of both sides of this, we're going to get one over N, one minus N over K over
light green N is equal to, is equal to, and so let's, we could say it's equal to, it's equal to one over C
time E to the negative R T. But one over C, that's just
going to be another constant. So I could write one over C here, and I take the reciprocal. And once again, this is just going to be another constant, so I'm going to be a
little bit of hand waving, and say well O.K we're going to get another constant here. I could have called
this C one, this C two. I could call this C three if I want to make it clear that these are not going to be the same number. This is E to the this power. This is the reciprocal of that. Actually, maybe I'll do that just to make it a little bit clearer. That would have been C
one right over there, and so this is going to be, sorry, the reciprocal of this C three and E to the negative, the reciprocal of E to the R
T is E to the negative R T. E to the negative, E to the negative R, E to the negative R T. And let's see, if we divide the numerator and the denominator by N, or if we divide E. I want to think, if we divide this term by in we're going to get. So this term by N is
going to be one over N, and then this term by N is just going to be minus one over K. So this is just going to be minus one over K is equal to this. 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 one over K. So, plus one 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. N of T, actually let me make my T in white since I've been taking the trouble all this time of rewriting this in white, is equal to one over. Is going to be equal to one
over all of this business. Is going to be equal to one
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 kind of a fraction in a fraction, I could rewrite it as. Actually, maybe I'll do it over here. N 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 leave
it like that for now. And what I'm going do, what
I'm now going to do, is say, "Well look. "We're assuming that N
of zero is N sub knot." So we're assuming that N of zero, N of zero, is equal to N sub knot. So let's write this thing. Let's solve for the constant. Let's figure out what this could be if we know what our initial condition is. So N of zero, N of zero, is going to be equal to, is going to be equal to one, one over. When T is zero, this is just going to be equal to one, so it's just 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 N knot, and now we can solve for our constant. So we get, let's see I'm probably going to need a lot of real estate for this, so I can take the reciprocal
of both sides again, so this is something that
we're doing a lot of. C three plus one over K is
equal to one over N knot. Is equal to one over N knot. Just took the reciprocal of both sides, and so we get our
constant C three is equal to one over N knot minus one over K. Minus one over K. 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 one
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 R T, to the negative R times T, plus one over K, plus one over K. And if we don't like having
all of these denominators in, all of these fractions in the denominator, why don't we multiply everything times the numerator and the denominator by N knot K. So I'm going to multiply the numerator times N knot K, and I'm going to multiply
the denominator by N knot K. N knot K. And so what do we get? This is all going to be equal to. In the numerator I have N
knot times K, N knot times K. And in the denominator I am going to have, let's see if I multiply this term right over here times N knot
K I'm going to have K. If I multiply this term times N knot K, I'm going to have N knot, so it's going to be minus N knot, minus N not, times E to the negative R T, Times E to the negative R T, negative R T, and then if I multiply this times N not K, I'm going to get N knot. So plus N knot. 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 what it actually does. If you were to plot this, and I encourage you to do so, either on the internet, you could try Wolfram Alpha, or 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 knot. It starts to increase
at an increasing rate, but then it starts to slow down as we reach our maximum population. And so that is actually
a very neat function. If you model a population with this, you can kind of start to make predictions about what might the
population be at time whatever. And so, hopefully, you
found that satisfying.

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