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### Course: Differential equations>Unit 1

Lesson 5: Exponential models

# Exponential models & differential equations (Part 2)

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.

## Want to join the conversation?

• 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.
• The solution is `P(t) = 100·e^[ln(2)/50 t]`. The natural logarithm inside the exponential is not alone, and therefore it cannot be "yoinked" out.

What you can do, is change the base of the exponent, like so:
``P(t) = 100·e^[ln(2)/50 t]P(t) = 100·e^[ln(2)·t/50]P(t) = 100·( e^(ln(2)) )^(t/50)P(t) = 100·2^(t/50)``

Which is lot cleaner, and it reflects the fact that the population doubles every 50 days.

(http://s3.postimg.org/92nu88jpv/2015_05_28_23_35.png)
• 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?
• 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...
• 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) ?
• 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).
• what if we're give n two points but not the initial value?
• Degrees of freedom check.
Two equations: P1=Ce^(kt1) and P2=Ce^(kt2)
Two unknowns: C, k
0 Degrees of freedom - you're good to go.
• 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)?
• 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.
• How do you determine that K is equal to growth rate?
• k is not in fact the growth rate!

The growth rate can be determined by calculating:
dP/dt=ln(2)/50•100e^((ln(2)/50)•t)=2ln(2)•e^((ln(2)/50)•t)
• How does Sal know that at t = 0, P = 100, and that at t = 50, P = 200? These values seem to be arbitrarily obtained?
• He just chose some conditions for the example (Note at "And you assume this information right over here"). If you had a specific population to study those values would change.
• 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)
• If you are using exponential population growth as a potential model for the growth given by your numbers, then you will only be given an approximation to this population change.