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## AP®︎/College Calculus AB

### Unit 7: Lesson 7

Exponential models with differential equations- Exponential models & differential equations (Part 1)
- Exponential models & differential equations (Part 2)
- Worked example: exponential solution to differential equation
- Differential equations: exponential model equations
- Differential equations: exponential model word problems

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# Exponential models & differential equations (Part 2)

AP.CALC:

FUN‑7 (EU)

, FUN‑7.F (LO)

, FUN‑7.F.1 (EK)

, FUN‑7.F.2 (EK)

, FUN‑7.G (LO)

, 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.

## 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.(11 votes)- 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)(55 votes)

- 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)
- 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)

- 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)- 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)

- what if we're give n two points but not the initial value?(3 votes)
- 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.(7 votes)

- 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)- 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)

- How do you determine that K is equal to growth rate?(2 votes)
- 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)(5 votes)

- 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)
- He just chose some conditions for the example (Note at4:42"And you assume this information right over here"). If you had a specific population to study those values would change.(4 votes)

- 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.(2 votes)

- 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)
- Nope. We need a second pair of values in order to find k. Otherwise we will just find the value of k in terms of C, and that wouldn't be specific enough.(2 votes)

- 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)
- Oh no I've noticed that it is k*t not k+t(1 vote)

## 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.