0:19

Okay, let's continue to Chapter 3, okay?

Â In Chapter 2, we have seen several classes of first

Â order differential equations which allow the analytic

Â methods of finding the general solutions, okay?

Â Here in Chapter 3, we'll introduce several simple physical or

Â biological phenomenon which lead to first order differential equations.

Â 1:16

In formulating real world problems in mathematical terms, we,

Â in most cases, need some idealization and

Â simplification in order for the resulting model is mathematically tractable, okay?

Â 1:34

When a problem involves a rate of change, that is derivative of something,

Â its modeling naturally lead us to a differential equation, okay?

Â Modeling sometimes involves conceptual replacing

Â of a discrete process via continuous one as we can

Â see in the study of population dynamic below, okay?

Â As the first such example, okay,

Â I will consider the so-called radioactive decay.

Â 2:23

In other words, if I denote by Q(t), capital Q(t),

Â okay, the total amount of certain radioactive material at time t, okay?

Â Then it means, okay, its rate of change, okay, that is Q',

Â okay, is proportional to the present amount Q(t).

Â 2:59

So that the radioactive material, okay, the total amount,

Â satisfies this very simple first order linear

Â differential equation, Q' = -r times Q(t), right?

Â 3:17

This is also separable, okay?

Â You can solve it very easily to obtain its general solution,

Â will be Q(t) = Q nought times exponential -r times t.

Â And the Q of nought, this is Q of 0, okay?

Â Say, the initial amount of that radioactive material, okay?

Â 3:44

In particular, in this situation, we call the time period during

Â which the mass of a certain radioactive material is reduced to one-half of

Â its original mass is called the half-time of this material, okay?

Â 4:47

So let's denote Q(t), okay?

Â Then the total amount of Einsteinium 237 after t days, okay?

Â The units of the time here is days, okay?

Â So it means Q(12), after 12 days, okay?

Â And lose one-third of its total mass.

Â That means, how much is it remaining?

Â Two-thirds is remaining, two-thirds of the original amount.

Â So Q(12) must be equal to two-thirds times Q(0).

Â Q0 is the initial amount.

Â 5:26

On the other hand, the general solution of Q(t) that

Â is Q0 times exponential -r of t, right?

Â So Q(12) will be Q0 exponential -12 times r.

Â That should be equal to two-third times Q0.

Â So you can divide through by Q0 which is non 0.

Â That means that you have exponential -12r is equal two-third, okay?

Â Let me write it here, okay?

Â So you get exponential, right?

Â 6:05

Negative 12 times r, that is equal to two-third.

Â Take logarithm on both sides then from the left.

Â Log of exponential something is something.

Â So you get simply negative 12 times r, that is equal to log two-third,

Â is the same as the negative log of reciprocal of two-third.

Â That is three-halves.

Â So okay in fact you have there for

Â r is equal to 1 over 12 times the log of three-halves right.

Â Okay that is equal to r okay.

Â Okay right here, okay r is equal to log of three-half,

Â log of three-half over 12 okay.

Â 6:53

What we're interested in is we are interested in the half-life of of

Â Einsteinium 237.

Â Okay so let tau be the half-life of Einsteinium 237.

Â What's the half life means?

Â It is the time necessary for the Einsteinium 237

Â loses one-half of its total mass, right?

Â 7:19

So, Q of tau must be one half of cube nought, that's the initial amount.

Â That should be equal to Q nought times e to the negative r of tau,

Â when t is equal tau, okay?

Â We already found r here, right?

Â So, divide this equation through Q nought then you will get,

Â you will get divided through by the Q nought, and

Â you have one-half is equal to e to the minus r times tau.

Â Take log on both sides, you'll a have log of one-half is equal to- r of tau,

Â okay, and that is equal to, log of one-half is the same as -log of 2, right?

Â Okay, so that, okay you can read it out,

Â the half-life tau should be log 2 over r, right?

Â So tau is equal to, as here, right?

Â Tau is equal to log 2 over r, okay?

Â Which is the same as, because r = log of three-halves over 12, right?

Â We already computed it, okay?

Â So that we have tau = 12 times log of 2 over log of three-halves,

Â through some scientific calculator,

Â you can approximate the value of tau to be approximately 20.5 days.

Â That is the half-life of Einsteinium 237, okay.

Â As the second example okay,

Â let's consider the so called the carbon dating.

Â Okay it was invented by the chemist W Libby who got

Â the Nobel Prize in chemistry.

Â 9:16

The half-life of radioactive

Â isotope C-14 is 5600 years.

Â Then, determine the age of a wooden fossil if it

Â contains only 5% of the original amount of C-14.

Â 9:52

So, now we know that the half-life of C-14 is 5600 years okay.

Â What does that mean, that means half of Q nought is equal to,

Â right, Q nought times exponential,

Â negative 5600 times r, okay, right here, this equation.

Â 10:35

Now the problem we are confronting is,

Â determine the age of the wooden fossil if it contains

Â only 5% of the original amount of C-14.

Â What is 5% of it?

Â This is 1 over 20 the original amount, right?

Â So you simply get, okay 1 over 20 times Q0 is equal to Q0 times e to

Â the -r times unknown time t, okay.

Â 11:11

Remember that we already computed the decay constant r as log 2 over 5600.

Â So now from this equation 1 over 20 times

Â Q nought is equal Q nought times e to the -r of t.

Â Take logarithm on both sides, try to find the t then,

Â t is given by log of 20 over r.

Â So using this quantity r down there, you have the time

Â needed is 5600 times log of 20 over log of 2, okay?

Â