00:01
Hi there, so for this problem, we are given, we are told that the population develops according to the logistic equation, and the logistic equation is this one that is given right here.
00:13
So what we can do in order to, so for, to answer part a of this problem, what we need to do is to write that equation in the following form.
00:25
The rate of change of the population with respect to time is equal to.
00:30
To a concept of proportionality that we are going to call k, this times the population at any given time, this times one minus the population divided by the l, which is the current capacity.
00:42
Now, for part a, we need to determine that current capacity.
00:46
So we need to pass the equation that we are given into this form.
00:50
So, and what we can do is to take out as a common turn 0 .5 and the population at any given time.
00:59
This will give us 1 minus the population.
01:06
And of course, if we divide 0 .0025, divided by 0 .5, we obtain a value of 0 .005 times the population.
01:27
Now, this value right here, the inverse of that will give us the current capacity.
01:32
So the carrying capacity is equal to 1 divided by 0 .005.
01:38
So using our calculator, we obtain a value of 200.
01:45
So that is the carrying capacity, 200.
01:50
Now, for part b of this problem, we are asked about, for what interval is the population increasing? so we know that a function, is increasing if its derivative, for example, in this case, the derivative of the population with respect to time is greater than zero than the population is increasing.
02:29
So with that said, since we know that the population is a value greater than zero, a positive value.
02:37
The only way that this change to be to be positive is is this right here is greater than zero, which is this term right here, the remaining term right there.
03:05
So if we solve in there, so we want the interval.
03:10
And for what interval is the population increasing.
03:19
So that will be when one, rather than the population, so solving in here, that will be when the population is less than the current capacity.
03:33
When the population is less than the current capacity, then we will have that the population is.
03:44
Increasing.
03:48
Now for par c of this problem, we have something similar, but in this case, we want when this expression is less than zero, because that means that the population is decreasing.
04:03
So that will be when 1 minus p divided by the current capacity is less than zero, so that will give us when the population is greater than the current capacity.
04:16
Okay.
04:18
So, now, now, for part d of this problem, we are asked about the general solution for this.
04:26
Okay? so we start with the expression that we are given, this one right here.
04:37
Okay, so we have the derivative of the population with respect to time is equal to, okay, the concept of proportionality, 0 .5.
04:50
This times the population and this one minus p and then we can take the inverse of this to put explicitly the current capacity so that will be minus the population divided by 200 okay so now to solve this what we are going to do first is to just simply separate the variables so we can pass this to the left side.
05:25
So we will have the differential in the population.
05:28
This divided by the population times one minus the population divided by 200.
05:35
And then this, we take the integral of this.
05:39
Then this is equal to 0 .5 times the differential in the time.
05:43
We take the integral of that.
05:45
So with that said, we know that in order to evaluate the left hand side of this, this is integral.
05:54
What we are going to do is to use, well, we can write this in the following way.
06:02
We can write this as just 200 divided by population this times 200 minus the population.
06:16
So what we have done is to multiply and divide this term by 200.
06:22
And so we're left with this.
06:24
And now we can separate this in the following way.
06:27
1 divided by the population plus 1 divided by 200 minus the population.
06:35
So as you can see in here, we are now reduced to two integrals...