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Suppose we alter the differential equation in Exe…

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Problem 23 Easy Difficulty

In a seasonal-growth model, a period function of time is introduced to account for seasonal variations in the rate of growth. Such variations could, for examples, be caused by seasonal changes in the availability of food.
(a) Find the solution of the seasonal-growth model
$ \frac {dP}{dt} = kP \cos(rt - \phi) P(0) = P_0 $
where $ k, r, $ and $ \phi $ are positive constants.
(b) By graphing the solution for several values of $ k, r, $ and $ \phi, $ explain how to the values of $ k, r, $ and $ \phi $ affect the solution. What can you say about lim $ _{t \to \infty} P(t)? $

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Chapter 9

Differential Equations

Section 4

Models for Population Growth

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okay in this question we're looking at a seasonal growth model part A says dp D. T equals K. P. Co sign R t minus I think that's fi Yeah and P not as the initial condition. So in order to solve this we can do separation of variables. So bring the peas to one side and bring the tease to the other side. K co sign R T. S fi D. T. And then integrate. So the integral of one over P is going to be log P. The integral of the K. Cosan. You can use U substitution and you should get K sign of our tea minus phi divided by our and then you have to add a constant of integration. So I'll just call that C. Then you wanna exponentially both sides to solve for P. P equals E. To the K. Sign R T minus phi divided by our plus C. And you can rewrite this pretty commonly when you're working with Constance and exponential you'll want to separate the E. To the see constant By doing the product of two exponential and this here is just another constant. So because I never specified what the constant is. I'm just going to keep writing it as C. But it's not the same value as it is in the previous step That's just there for convenience. And we're going to solve for C by plugging into equal zero And sending that equal to p. Not so p of zero equals C. E. To the K. Sign Aarti will go to zero. So we're just left a negative phi here that equals peanut. So we're solving for C. See becomes do not eat a negative K. Sign negative thigh. You could cancel out the negative sign With the in front of the K. and the negative sign in front of the five because sign of negative phi equals negative sign of five because sign is an odd function but I'm going to leave it like that so that we can right pft equals peanut E. To the negative K. Sign of negative phi E. To the K sign of our t minus phi. Or all together. Now P not E. K. sign are T -5. Oh I am missing an over our at some point. Yes, there was supposed to be an over are here. So I'm going to rewrite this as instead of just K. I'm just gonna take the K. Over are here. Okay. Over our and then we have sign of our t minus phi I minus sign of negative phi. And that is the solution to the equation. The next part asked to graph several examples and see what happens. So let's do that here. I've used mathematica to plot a bunch of different graphs. Um In this case we're going to have that the we make this a little bit smaller so that's easier to see. Mhm. Yeah. Okay. So in these examples, the red graph represents when the constant is 0.5 blue Is when the constant is equal to one and orange. I know that screen but there's no orange option here, that's going to be one of the constants equal to two. So the first plot Is when we're changing K between .51 and two. So you can see that as K gets larger. These graphs get more stretched out vertically and that's because K is more about scaling, it's a scaling factor are has the effect of kind of stretching out diagonally. So Um when are is .5 you get a big stretch and then when r is to you get a shrinking how? But that kind of makes sense because that's part of the sine term. And so that will describe how quickly the period is. But also it's in front of the the sign term and here Mhm. Yeah. Okay, thank you. The next part is I call this s because running down five would be difficult in Mathematica but that's going to represent another um period shift and there's a little bit of a stretch too because you have the constant term negative sign to the negative S and the exponential. So as you change S they'll have a effect on the total. And the last thing I plotted was the initial point but question doesn't ask specifically about that. Okay, what can you say about the limit and as T approaches infinity of pFT. Well in these examples you can see that there's not one final limit which makes sense because it's seasonal growth, you expect things to go up and then down, but it will settle down along a point which is going to be it's going to be um sign you soil around peanut E two K over our. So it's going to be constantly going above and below this value. Which makes sense because as Sign goes from negative 1 to 1, it's never going to converge to a single point. All right, that's it.

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Problem

Suppose we alter the differential equation in Exe…

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Calculus: Early Transcendentals

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Caleb Elmore

Kristen Karbon

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Caleb Elmore

Kristen Karbon

James StewartCalculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Caleb Elmore

Kristen Karbon

James Stewart

Calculus: Early Transcendentals