0:22

This is the subject of today's lesson,

Â dx/dt equals a times x where a is a constant.

Â This is a linear autonomous ordinary differential equation.

Â It is extremely important that you know this and its solution,

Â x equals the initial condition x nought times e to the at.

Â And what does this solution mean?

Â What does it look like?

Â Well, at time, t equals 0, it passes through x nought, the initial condition.

Â What it does from there depends on a.

Â If a is positive then you get what is called exponential growth.

Â You get an exponential function in t that is growing as t increases.

Â On the other hand, if a is negative then you have exponential decay.

Â Where the values of x are getting closer and closer to zero over time.

Â Much of the importance of this equation comes from its

Â ubiquity in modeling processes.

Â You may have seen this equation used to model radioactive decay.

Â If you look at, say, the radioactive isotope of carbon,

Â carbon-14, then the amount i of that isotope

Â changes to a degree that is proportional to

Â i where that constant of proportionality is negative.

Â 2:05

Likewise if you look at the amount of a drug or medicine that is in a body,

Â this changes over time according to the same differential equation.

Â Though perhaps with a different constant of decay.

Â Both of these have negative coefficients and

Â thus connote an exponential decay.

Â There are some situations where you have exponential growth, however.

Â If we look at the most simplistic model for population growth,

Â one in which the size of the population, P changes at a rate proportional to P,

Â that constant of proportionality is a birth rate.

Â And this doesn't work so well for modeling human populations, but for

Â certain settings, it's reasonable.

Â 3:02

On the other hand, continuously compounded interest

Â is a fairly realistic model, at least in the short term.

Â It says that the rate of change of the amount of money

Â you have invested is proportional to how much is there.

Â What is that constant of proportionality?

Â Well, that is an investment rate.

Â But what do we mean by continuous compounding?

Â Let's assume an annual interest rate of r and

Â an initial investment of money at time 0.

Â If we were to compound our interest yearly,

Â then after one year, the amount of money we would have would be our initial

Â investment plus that initial investment times the annual interest rate.

Â We could write that as the initial investment times quantity (1 + r).

Â If however, we compounded that interest monthly, then what would we get?

Â After one year we would have the initial investment times quantity

Â (1 + r/12), which is how much we would earn in January.

Â Then we would multiple all of that

Â by quantity (1 + r/12) at the end of February.

Â We would continue on until the end of the year, giving another factor of (1 + r/12).

Â That means at the end of the year we would have the initial

Â investment times quantity (1 + r/12) to the 12th power.

Â That's actually a little bit more than we would get from an annual compounding.

Â Now, of courseâ‰¤ monthly compounding is maybe not so

Â accurate because they have different days per month, so we can do it daily and

Â get the initial investment times quantity (1 + r/365) to the 365th power.

Â 5:02

I think you can see where this is going.

Â If we take the limit, then after one year we get the initial investment

Â times the limit as n goes to infinity of quantity (1 + r/n) to the nth power.

Â Now I know you've seen that limit before.

Â You may recall that it is, in fact, e to the r.

Â This is a hint that the standard linear differential

Â equation is really behind what's happening in continuous compounding.

Â Let's see how that falls out.

Â 6:03

Now let's change variables a little bit, and instead of looking at n going to

Â infinity, let us replace 1/n by h,

Â where h is a small amount going to zero.

Â Then the amount of money at time t + h is, from the above,

Â the amount of money at time t times quantity 1 + r times h.

Â And now, expanding out that multiplication,

Â we see a familiar sight from our definition of a derivative.

Â The coefficient in front of the h term

Â tells us the derivative of the amount of money, function of t.

Â And so we can see the differential equation,

Â d dt of the amount of money equals r times the amount of money.

Â That is our standard differential equation.

Â 7:08

Let's look at an unusual one coming from linguistics.

Â If we read English poetry from antiquity, we'll see some unfamiliar words.

Â And if you look at Chaucer, the Canterbury Tales,

Â many of the words are unfamiliar to us.

Â But if we move the clock ahead to Shakespeare's time,

Â then some of that poetry is familiar.

Â Some is unfamiliar.

Â Moving ahead a little bit further to Milton,

Â things are still a little obtuse but more,

Â the words make sense, and we could keep going with Blake and others.

Â What we might guess is the following linguistic model.

Â Namely, that the number of words in English remaining in common use

Â decreases at a rate proportional to the remaining words.

Â That means that words fall out of use.

Â But this model states that words fall out of use according to a linear

Â differential equation where W is the amount of words remaining in

Â usage and alpha is a decay constant.

Â Assuming this model, let's answer the following question.

Â If we find 20% of the words in Milton's poem to be unusual,

Â then what fraction of Chaucer's poetry did Shakespeare's audience recognize?

Â Whew, that sounds difficult.

Â What can we do with that?

Â Well, let's say that t corresponds to the year.

Â We're going to be interested in two functions of t.

Â The first, W sub M, is the number of

Â words in Milton's usage currently used at time t.

Â That is, according to the solution to this differential equation,

Â W sub M at 1667, the initial condition, times e to the minus

Â alpha times the number of years that have elapsed since 1667.

Â Likewise, we'll be interested in W c of t,

Â the number of words from Chaucer's time still in common usage.

Â This has the same solution with 1667 being replaced by 1400.

Â Most importantly, the alphas are assumed same because this is all English.

Â Now what is it that we are given?

Â We are told that the number of words from Milton's time in common use today,

Â let's say in 2012, is 80% of those originally available.

Â Plugging in the solution to the differential equation for W sub M when t

Â equals 2012, and then dividing by the initial condition

Â gives us a single equation that has only alpha as an unknown.

Â We can therefore solve for alpha and

Â we get a quantity that is about 6.5 times 10 to the -4.

Â Now, if we use that alpha in our solution for

Â W sub C of t, then what can we get?

Â Well, our goal is to find the fraction of words from

Â Chaucer's time that people in Shakespeare's time would have understood.

Â That is, W c of 1600 divided by W c of 1400.

Â Plugging in our solution to the differential equation for W sub C

Â 11:18

Well, that's pretty cool.

Â But is that actually true?

Â Well, I don't think so.

Â This simple model that we've written down ignores so many features of linguistics,

Â such as the creation of new words.

Â Such as the evolution of words.

Â Words are constantly changing their meaning.

Â So that even though a word does not fall out of common usage,

Â it's really a different word with an altogether different meaning.

Â Lastly, there is whole scale evolution of language.

Â I'm speaking American English which is in some respects very different then

Â British English.

Â If you look at, say, computer languages, you can see, even over short time scales,

Â very rapid evolution in the language itself.

Â This brings us to an important conclusion,

Â that, even though our mathematics was perfectly accurate,

Â a perfectly accurate solution is only as good as the model on which it's based.

Â If you have a bad model,

Â it doesn't matter how good your math is, you can't conclude truth.

Â So with that in mind, let's talk about zombies, and

Â we'll model the zombie apocalypse as follows.

Â The rate of change of the infected population

Â is proportional to the uninfected population.

Â If we denote by u(t) the uninfected population,

Â and z(t), the infected population, then p,

Â the net population size, is a constant.

Â That is u plus z equals p.

Â 13:21

Now, what is happening to dU/dt?

Â Let's think about that.

Â U is P- Z.

Â So differentiating gives us

Â dU/dt = dP/dt- dZ/dt.

Â Now, dP/dt, we know is 0, since P is constant.

Â And we know dZ/dt = rU.

Â And so we see that our zombie model is really just

Â the linear differential equation in disguise for the uninfected population.

Â We know the solution U(t) is the initial condition, U nought, times e to the -rt.

Â 14:32

Now, before you dismiss this differential equation as being ridiculous for

Â modeling zombies, consider.

Â It may also be used to model the spread of disease, or

Â the spread of propaganda, or the adoption of new technology in a population.

Â This simplistic equation is not

Â rich enough to really model these complex phenomena.

Â But it's a good first approximation.

Â It's even better when applied to heat.

Â Newton's law of heat transfer is essentially the same equation.

Â It says the rate of change of temperature with respect to time is proportional

Â to the difference in temperature with the ambient environment.

Â So in our zombie equation, this constant of proportionality r was an infection

Â rate, and P was a constant population size.

Â In Newton's law of heat transfer, the constant is a thermal conductivity.

Â How easy is it for heat to spread?

Â And the constant A is an ambient temperature.

Â Same equation, very different interpretations.

Â That's one of the beauties of differential equations.

Â And so we see one simple differential equation gives rise to so

Â many applications.

Â But what happens if we change that differential equation?

Â How will we determine the solution?

Â What is it going to be good for?

Â In our next lesson, we'll consider some other classes of ordinary

Â differential equations, their solutions, and their applications.

Â