4.3.3

Professor: So we're going to consider the solution of the logistic differential equation a bit differently than the textbook does because in the textbook they actually prove this by showing a general strategy for solving logistic differential equations, and they show that this equation is valid. However, they don't show that to you until after this question, so I'm actually going to appeal to that solution closed form in and through this question. 

So let's consider the population of whitetail deer in the state of Kentucky. The Kentucky Department of Fish and Wildlife Resources sets guidelines for hunting and fishing in the state. Before the hunting season in 2004, it estimated a population of 900,000 deer. And this person, George Johnson, notes that, "A deer population that has plenty to eat and is not hunted by humans or other predators will double every three years." Now we will evaluate this claim-- we'll come back to that. 

This observation corresponds to a rate of increase of r equals a natural log of 2 over 3, which is 0.2311, so the approximate growth rate is 23.11% per year. This assumes the population grows exponentially, which is reasonable, at least in the short-term, with plenty of food and no predators. Now they also report deer densities for 32 counties in Kentucky, average of which is 27. 

So you've got 27 deer per square mile and the state has 39,032 square miles. So we determine their carrying capacity is 1,072,764 deer. All right, so for this approximation of this application, we are going to take our initial amount of 900,000, a carrying capacity of 1,7072,764, and a growth rate of 0.2311. We want to substitute these into the logistic equation and make an initial-value problem. So we know that the form of these is dp dt equals rp 1 minus p over k so that our initial-value problem is dp dt equals 0.2311 times p times 1 minus p over our carrying capacity of 1,072,764 with a p of 0 equal to 900,000. 

OK. Now solve this. This is where we diverge from the text because we're going to appeal to that solution that is in a closed form that is p of t equals p0 times-- sorry, our initial amount times our carrying capacity times e to rt over k minus p0, how far we are away from our carrying capacity with our initial amount, plus p0 e to the rt. So that our closed-form solution will be p of t equals p0, that's 900,000 times k times e to the rt divided by k minus p0 e to the r-- oh, actually scratch that-- plus p0, which is our 900,000, e to the rt, and we know r 0.2311 t. 

Now if we simplify a few things and actually factor out a 900,000 from denominator, we end up with 1,072,764 e to the 0.2311 t. Now I'm going to subtract that 1 million figure minus 900,000 and then divide by 900,000. That's how I'm getting 0.19196 plus e to the 0.2311 t. So factoring out the 900 from this leaves us with just that exponential term, subtracting those then dividing by 900 gives us that figure there. And that is our solution. 

So next we want to use this model and determine what the population in three years is. Well, that would be taking that as our function but we need to find p of 3. And I'll leave that to you, but I've already evaluated this function at three years and that is 978,830 deer. Now this man George Johnson predicted that in three years the population would double. That's what we need to compare here, and it doesn't appear that the population has doubled because doubling time assumes that we have unlimited growth. A logistical model assumes that we approach the carrying capacity. Very big difference, very big difference. 

So that is the population-- this more realistic model would predict in three years. Now suppose the population managed to reach 1.2 million deer. What does the logistical equation predict will happen to the population in this scenario? So let's go ahead with-- go back to this 0.2311 p, and we have 1 minus p divided by 1,072,764. Well, if we evaluate for $1.2 million, we would see that 1 minus that figure, 1.2 million divided by 1,072,764, that is less than 0. Which means the population decreases at that point.