Oscillating growth rates Some species have growth rates that...

Oscillating growth rates Some species have growth rates that oscillate with an (approximately) constant period $P$. Consider the growth rate function $$N^{\prime}(t)=A \sin \left(\frac{2 \pi t}{P}\right)+r$$. where $A$ and $r$ are constants with units of individuals/year. A species becomes extinct if its population ever reaches 0 after $t=0.$ a. Suppose $P=10, A=20,$ and $r=0 .$ If the initial population is $N(0)=10,$ does the population ever become extinct? Explain. b. Suppose $P=10, A=20,$ and $r=0 .$ If the initial population is $N(0)=100$, does the population ever become extinct? Explain. c. Suppose $P=10, A=50,$ and $r=5 .$ If the initial population is $N(0)=10,$ does the population ever become extinct? Explain. d. Suppose $P=10, A=50,$ and $r=-5 .$ Find the initial population $N(0)$ needed to ensure that the population never becomes extinct.

Oscillating growth rates Some species have growth rates that oscillate with an (approximately) constant period $P$. Consider the growth rate function $$N^{\prime}(t)=A \sin \left(\frac{2 \pi t}{P}\right)+r$$. where $A$ and $r$ are constants with units of individuals/year. A species becomes extinct if its population ever reaches 0 after $t=0.$
a. Suppose $P=10, A=20,$ and $r=0 .$ If the initial population is $N(0)=10,$ does the population ever become extinct? Explain.
b. Suppose $P=10, A=20,$ and $r=0 .$ If the initial population is $N(0)=100$, does the population ever become extinct? Explain.
c. Suppose $P=10, A=50,$ and $r=5 .$ If the initial population is $N(0)=10,$ does the population ever become extinct? Explain.
d. Suppose $P=10, A=50,$ and $r=-5 .$ Find the initial population $N(0)$ needed to ensure that the population never becomes extinct.

Key Concepts

Initial Conditions and Phase Effects

The initial conditions of a system, such as the starting population size, as well as the phase of any oscillatory components at the start, play a significant role in the subsequent behavior of the model. These factors can determine whether early declines lead to extinction or if the system can sustain periods of low growth without reaching a critical threshold. The interplay between initial values and the timing of the oscillations is therefore key to predicting the long-term dynamics.

Population Extinction Threshold

This key concept revolves around determining when a population falls to zero, leading to extinction. In mathematical models, reaching an extinction threshold means that once the population number becomes zero, the species cannot recover. Analyzing when and how the population might hit this threshold is essential for understanding the sustainability and survival conditions within a given model.

Stability Analysis in Dynamical Systems

Stability analysis examines whether the solutions to a differential equation remain bounded and avoid undesirable states like extinction. In the context of oscillatory growth rates, it involves assessing the relative influence of the oscillatory amplitude, constant growth terms, and initial conditions to determine if the population will persist over time or eventually collapse. This analysis is fundamental in evaluating the resilience of systems under periodic fluctuations.

Differential Equations in Population Dynamics

This concept involves using differential equations to model how populations change over time. In such models, the rate of change of a population is expressed as a function of the current population and environmental or intrinsic factors. Differential equations provide a framework to understand and predict long-term behaviors such as growth, stabilization, or collapse (extinction) of populations.

Oscillatory Forcing

Oscillatory forcing refers to incorporating periodic functions, such as sine or cosine, into a differential equation to represent cyclical external influences, like seasonal variations. These oscillations can cause the growth rate to vary over time, sometimes promoting growth and at other times causing declines, and are crucial for analyzing scenarios where the environmental conditions fluctuate regularly.

Transcript

00:01 Okay, here we have a base function for the rate of a specific oscillating growth rate for a species, and we're going to figure out if that species will go extinct base up this formula, or voice up this function.

00:12 So the first thing we're going to do is we're going to figure out n of 0.

00:16 So n of 0, oh, sorry, n of t.

00:25 So to figure that out, we're actually going to integrate this.

00:28 If we do that, we get a times p over 2 pi, cosine of 2 pi t over p and over here we have a negative because we have this is going to be negative then we have plus r t plus c we're going to figure out what plus c is based on each one so if we get that we get n of t our first scenario we're going to have p equals 100 or sorry p equals 10 a equals 20 r equals zero and n of zero equals 10 if we start with that, a times b is 200, but because we're being divided by 2 pi, we're getting 100 over pi.

01:37 Then 2 pi t over p, 2 divided by 10 is 1 5th.

01:42 So we have cosine of pi t over 5.

01:47 This is going to be negative.

01:50 R is equal to 0.

01:52 So we get plus c.

01:54 To figure out plus c, we're going to figure out what's inside.

01:57 So we're going to do 10 equals 100 cosine of 0 divided by pi plus c.

02:07 Because this is equal to 0, this is a negative.

02:10 We're going to get 10 plus 100 pi equals c.

02:19 So this c is 10 plus 100 over pi.

02:27 Based on that information, we can figure out if this is ever going to be extinct.

02:34 So to see that, all we have to do is look at this.

02:37 This value, because we're trying to get it to go extinct, we're trying to make it equal to zero.

02:42 This value has to be negative.

02:44 The most negative this can be is when cosine is equal to positive 1.

02:48 So we get negative 100 over pi plus 10 plus 100 over pi.

02:55 If we try that, these two will just cancel and we're left with 10.

02:59 So this will not ever be extinct.

03:01 So ne for not extinct.

03:05 For the next scenario, we have p equals 10, a equals 20, r equals 0, and n of 0 is equal to 100 this time.

03:15 So this problem is exactly the same.

03:18 So we can make the exact same formula function, but replace all of our tens with a hundredths.

03:25 So we solve negative 100 over pi, cosine pi t over 5, plus 100 plus 100 pi.

03:34 But the same thing happens.

03:36 Because the highest value that cosine can have is 1, the highest negative value that we can have for this first one, is negative 100 over pi.

03:44 That would make these two cancel and it equal to 100.

03:48 So this will never drop below 0 as well.

03:51 So this is also never going to be extinct.

03:57 Our next scenario has p equals 10, a equals 50, r equals 5, and n of 0 is equal to 10.

04:09 If we have this instead, we're going to have n of t equals negative 500, which simplifies to 250, because remember we're dividing it by 2 .5 times cosine of pi t over 5, same as last time, plus 5t plus c...

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