Problem 1 write the odes of the following models in each...

Problem 1. Write the ODEs of the following models: In each case, write the general solution, draw graphs, and compute the initial value, i.e. y(0). a) Malthus Model b) Radioactive Decay c) Newton's Law of Cooling d) Wolves & Rabbits Model e) Population Growth in Later Stages of Growth f) Verhulst Model Problem 2. The population of a country grows according to a logistic law. In year 1980, the population was 10 million and in year 2000 it was 11 million. If the maximal capacity is 20 million, what will be the population in the year 2050? Problem 3. The population of a country grows according to a logistic law. Show that the graph indicating the population bends exactly at half of the maximal capacity. Problem 4. The population of rabbits in an area is initially 1000 rabbits. Normally, it doubles every 3 years. They bring N wolves to the area. If each wolf eats 1 rabbit per week, for what values of N can we keep the rabbit population under control?

Transcript

00:01 So i'm going to say some general things about the most general form of periodic function that you might study, say, in electrical signals.

00:11 So the most general way to write this function is as an amplitude a times either a cosine or sign function.

00:21 And then also there is an offset.

00:25 Sometimes that parameter b is called bias.

00:27 The phase inside of the oscillating function means that your function does not necessarily start at the t -equal zero.

00:42 And the period is the t, the big t within that argument.

00:48 So we'll go ahead and just get what this looks like.

00:51 The b is sometimes called bias or baseline as well as offset, and it represents a shift of the zero line above y equals zero.

01:07 Let me try to draw this.

01:09 We'll not draw it quite like offset from the sign function.

01:20 So this looks like a sign function and this means that phi is equal to minus pi over two, i believe, plus or minus pi over two.

01:32 So it doesn't start like a cosine.

01:35 The period is the time in between peaks or troughs, depending how you want to define that.

01:45 You don't want to use the crossings of the midline because there are two of those per period.

01:52 And the amplitude a is how much of a swing you go beneath and above the center line.

02:04 So let's take a look at a practical example, maybe.

02:08 Let's say that there is a population of rabbits, which is oscillating.

02:20 Now, we know they can't go below zero.

02:22 What does it mean to have a negative number of rabbits? that doesn't make much sense.

02:28 So we're going to take a look, and let's say that at time 30 weeks, the rabbits have reached their maximum value.

02:42 3 ,500.

02:46 So we'll call that r.

02:48 And 90 weeks, it's reached as minimum value of 500.

03:06 So we'd like to see what kind of an oscillating function we could write for this.

03:12 And what we know is that the half period, because we're going from a peak to a trough, is 60 weeks.

03:20 So the time in between 30 and 90 is t over 2.

03:27 And so the period is 120 weeks, which means that the zero crossing happens a quarter of a period before the 3 ,500.

03:42 And 30 weeks later, it crosses again.

03:47 So what we basically have is a sign function with the period equals 120.

03:58 And we can find, the position in the middle by taking the max plus them in and dividing by two.

04:13 So that's the idea is that the sign function averages out to the midline.

04:20 So that's 2000.

04:26 Let's see, we colored that black up there, so we will also do that 2000.

04:33 And then the amplitude is the swing either up or down, amplitude, amplitude.

04:38 We can see now is 3 ,500 minus 2 ,000 or 1 ,500, and it also swings below that by the same amount.

04:49 So we can write a function for the rabbits as a function of time is equal to the amplitude ,500, and we have a perfect sign function of 2 pi over 120 plus 2 ,000.

05:21 So as another example, let's suppose we had a population of foxes.

05:32 So f as a function of time is equal to minus 80 cosine of pi, t minus 10 plus 60 ,120.

05:51 We can immediately pick out some important things.

05:56 We can first of all see that the phase is not equal to zero, not equal to zero, because there is something adding to the time inside of the cosine, and also there's a negative sign on front.

06:13 So this will not start off as a cosine function...

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