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Exponential models & differential equations (Part 2)

AP.CALC:
FUN‑7 (EU)
,
FUN‑7.F (LO)
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FUN‑7.F.1 (EK)
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FUN‑7.F.2 (EK)
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FUN‑7.G (LO)
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FUN‑7.G.1 (EK)
Given the general solution P=Ceᵏᵗ and the conditions P(0)=100 and P(50)=200, we find the solution to an exponential modeling problem.

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  • piceratops seed style avatar for user AshHeels
    In the final solution, couldn't you yoink the ln(a) term out of the exponent and simplify to P=a*C*exp(t*K/a)?
    It would look a lot cleaner.
    (11 votes)
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  • leaf blue style avatar for user Trent Kierstead
    Is the constant (C) we solved for not specific to that initial condition at t0? How can we be sure that the constant doesn't change as t --> inf?
    (4 votes)
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    • purple pi purple style avatar for user doctorfoxphd
      We cannot be sure. In fact in population biology, it is rare for a population to follow its own growth curve for very long due to the changes in nutrients, resources, predators, ambient temperature, available mates, suitable substrate/living space, and on and on. So you have to keep in mind that we are MODELING a population and that the model is only as good as the assumptions. So we build a model and compare with reality, build a better model and compare with reality, and so on. Usually the model becomes complex to address population growth with different light regimes or temperatures or nutrient gradients. After a while, the set of models will be a good predictor. Then someone upsets the petri dish or has a toxic spill or maybe there is a tsunami...
      (14 votes)
  • blobby green style avatar for user cicadatorlan
    We only know two points : (0;100) and (50;200).
    Wouldn't there possibly be more than this one exponential function ( P(t)=100e^(ln(2)·t/50) ) passing through them (and thus being valid modelisations too) ?
    (5 votes)
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    • female robot grace style avatar for user tyersome
      I don't think there are any other purely exponential functions that could be use to model this. You could always add some sort of additional term e.g. P(t)=M + Ce^k·t ... (M could even be a function of time ...)

      However, I would not do this unless you have some data that requires making an extra assumption (in this example, adding another term to the equation). This conservative approach, often referred to as "Occam's Razor"(http://math.ucr.edu/home/baez/physics/General/occam.html), is in essence the idea that you should never make things more complicated than they need to be.

      If it is not clear why, think about what value you would want to use for M ... if we have no information, why not use 0? More importantly, exponential functions actually do a good job of modeling many processes, are well understood, and are relatively easy to manipulate -- if we add extra terms we lose those benefits. Finally, even if we have more data and can make our model fit better by adding extra term(s) we may in fact loose predictive power by "overfitting" (https://en.wikipedia.org/wiki/Overfitting).
      (4 votes)
  • blobby green style avatar for user Cameron Chartrand
    what if we're give n two points but not the initial value?
    (3 votes)
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  • leafers tree style avatar for user Abby
    What would happen/how would we model if people were also immigrating and emigrating at a constant rate?

    Would we end up with dp/dt = kP + (rate immigrating) - (rate emigrating)?
    (3 votes)
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    • blobby green style avatar for user Ross Henderson
      Not quite. That might be true if somehow the immigrants and emigrants were a part of the population that didn't factor into the growth rate (maybe, but even that seems iffy).

      We still have dP/dt = kP, but P is dependent upon dPi/dt - dPe/dt, the immigration rate and the emigration rate. This could get complicated if these rates are also dependent upon the population P, which seems likely. But let's assume they're basically linear - say 5% in and 2% out per year. So the original P becomes P(1+3t/100). Note the unit of this t is years, not days. And now you can try the same logic Sal applied, with that pesky second factor to work through. You'd still need an initial population and another data point to get a specific solution.
      (3 votes)
  • blobby green style avatar for user mail.nguyen15
    How do you determine that K is equal to growth rate?
    (2 votes)
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  • blobby green style avatar for user christopher.j.k141
    How does Sal know that at t = 0, P = 100, and that at t = 50, P = 200? These values seem to be arbitrarily obtained?
    (2 votes)
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  • blobby green style avatar for user claire.sullivan.sullivan
    How would you model an equation where population does not go up at even intervals. For example: (0,85), (1,700), (2,850) and so on.
    (1 vote)
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  • leaf red style avatar for user Auper Bhadra
    In the previous video you let e^k as constant c. So I think we don't need two pair of value. One is ok as we need to find only one unknown that is k.
    (1 vote)
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  • blobby green style avatar for user alicebir28
    I think there is a contradiction there; "e^kt should not be the e^k * e^t and this be equal to (e^k * C = 100)?"
    (1 vote)
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Video transcript

- [Voiceover] In the last video, we established that if we say the rate of change of a population with respect to time is going to be proportional to the population, we were able to solve that differential equation, find a general solution, which involves an exponential. That the population is going to be equal to some constant times e to some other constant, times time. And in the last video we assumed time was days. So let's just apply this, just to feel good that we can truly model population this way. So let's use it with some concrete numbers. Once again, you've probably done that before. You've probably started with the assumption that you can model with an exponential function, and then you used some information, some conditions, to figure out what the constants are. You probably did this earlier in pre-calculus or algebra class. Let's just do it again, just so that we can feel that this thing right over here is useful. So let's give you some information. Let's say, that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of. And let's say that at time equals 50, the population, so after 50 days, the population is at 200. So notice it doubled after 50 days. So given this information, can we solve for C and k? I encourage you to pause the video and try to work through it on your own. So this first initial condition's pretty straighforward to you, because when t is equal to zero, P is 100. So we could say, based on this first piece of information, we could say that 100 must be equal to C times e to the k times zero. Well that's just going to be e to the zero. Well e to the zero is just one. So this is just the same thing as C times one, and just like that we have figured out what C is. We can now write that the population is going to be equal to 100e to the kt. So you can see, expressed this way, our C is always going to be our initial population. So, e to the kt. And now, we can use this second piece of information. So our population is 200, let's write that down. So our population is 200, when time is equal to 50. After 50 days. So 200 is equal to 100e to the k times 50, right? t is now 50. Let me just write that. k times 50. Now we can divide both sides by 100, and we will get, two is equal to e to the 50k. Then we can take the natural log of both sides. Natural log on the left hand side, we get the natural log of two. And on the right hand side, the natural log of e to the 50k, well that's just going to be, that's the power that you need to raise e to, to get e to the 50k. Well that's just going to be 50k. All you did was took the natural log of both sides. Notice, that this equation that I've just written expresses the same thing. Natural log of two is equal to 50k, that means e to the 50k is equal to two, which is exactly what we had written there. And now we can solve for k. Divide both sides by 50, and we are left with, k is equal to the natural log of 2, over 50. And we're done. We can now write the particular solution that meets these conditions. So we can now write that our population, and I can even write our population as a function of time, is going to be equal to 100 times e to the, now k is natural log of two over 50, so I'll write that. So, natural log of two over 50, and then that times t. If you assume rate of change of population is going to be proportional to population, you assume this information right over here, this function is what is going to describe your growth of population.