[MUSIC] Okay, let's continue to Chapter 3, okay? In Chapter 2, we have seen several classes of first order differential equations which allow the analytic methods of finding the general solutions, okay? Here in Chapter 3, we'll introduce several simple physical or biological phenomenon which lead to first order differential equations. They provide us a good motivation for our interest in differential equation, okay? Mathematical modeling is a process of simulating reality by using mathematical language, okay? In formulating real world problems in mathematical terms, we, in most cases, need some idealization and simplification in order for the resulting model is mathematically tractable, okay? When a problem involves a rate of change, that is derivative of something, its modeling naturally lead us to a differential equation, okay? Modeling sometimes involves conceptual replacing of a discrete process via continuous one as we can see in the study of population dynamic below, okay? As the first such example, okay, I will consider the so-called radioactive decay. It is known that the radioactive material decays at the rate proportional to its present amount. In other words, if I denote by Q(t), capital Q(t), okay, the total amount of certain radioactive material at time t, okay? Then it means, okay, its rate of change, okay, that is Q', okay, is proportional to the present amount Q(t). So let's call the proportional constant to be -r, where r is a positive number denoting the decay rate, okay? So that the radioactive material, okay, the total amount, satisfies this very simple first order linear differential equation, Q' = -r times Q(t), right? This is also separable, okay? You can solve it very easily to obtain its general solution, will be Q(t) = Q nought times exponential -r times t. And the Q of nought, this is Q of 0, okay? Say, the initial amount of that radioactive material, okay? In particular, in this situation, we call the time period during which the mass of a certain radioactive material is reduced to one-half of its original mass is called the half-time of this material, okay? This is a very important concept related to the radioactive decay, okay? For example, I have to consider the following, okay? The Einsteinium 237 loses one-third of its mass in 12 days, okay? Knowing that, find its half-life, okay? This is a typical word problem. We say that Einsteinium 237 which is radioactive material, it loses one-third of its mass in 12 days. So let's denote Q(t), okay? Then the total amount of Einsteinium 237 after t days, okay? The units of the time here is days, okay? So it means Q(12), after 12 days, okay? And lose one-third of its total mass. That means, how much is it remaining? Two-thirds is remaining, two-thirds of the original amount. So Q(12) must be equal to two-thirds times Q(0). Q0 is the initial amount. On the other hand, the general solution of Q(t) that is Q0 times exponential -r of t, right? So Q(12) will be Q0 exponential -12 times r. That should be equal to two-third times Q0. So you can divide through by Q0 which is non 0. That means that you have exponential -12r is equal two-third, okay? Let me write it here, okay? So you get exponential, right? Negative 12 times r, that is equal to two-third. Take logarithm on both sides then from the left. Log of exponential something is something. So you get simply negative 12 times r, that is equal to log two-third, is the same as the negative log of reciprocal of two-third. That is three-halves. So okay in fact you have there for r is equal to 1 over 12 times the log of three-halves right. Okay that is equal to r okay. Okay right here, okay r is equal to log of three-half, log of three-half over 12 okay. What we're interested in is we are interested in the half-life of of Einsteinium 237. Okay so let tau be the half-life of Einsteinium 237. What's the half life means? It is the time necessary for the Einsteinium 237 loses one-half of its total mass, right? So, Q of tau must be one half of cube nought, that's the initial amount. That should be equal to Q nought times e to the negative r of tau, when t is equal tau, okay? We already found r here, right? So, divide this equation through Q nought then you will get, you will get divided through by the Q nought, and you have one-half is equal to e to the minus r times tau. Take log on both sides, you'll a have log of one-half is equal to- r of tau, okay, and that is equal to, log of one-half is the same as -log of 2, right? Okay, so that, okay you can read it out, the half-life tau should be log 2 over r, right? So tau is equal to, as here, right? Tau is equal to log 2 over r, okay? Which is the same as, because r = log of three-halves over 12, right? We already computed it, okay? So that we have tau = 12 times log of 2 over log of three-halves, through some scientific calculator, you can approximate the value of tau to be approximately 20.5 days. That is the half-life of Einsteinium 237, okay. As the second example okay, let's consider the so called the carbon dating. Okay it was invented by the chemist W Libby who got the Nobel Prize in chemistry. The half-life of radioactive isotope C-14 is 5600 years. Then, determine the age of a wooden fossil if it contains only 5% of the original amount of C-14. We know, again let's remind the rules here. Q(t), the total amount is Q0 initial amount times e to the negative r of t. It's given by that. So, now we know that the half-life of C-14 is 5600 years okay. What does that mean, that means half of Q nought is equal to, right, Q nought times exponential, negative 5600 times r, okay, right here, this equation. Take logarithms of both side. Then you will get this decay rate r is equal to log 2 over 5600, okay. Now the problem we are confronting is, determine the age of the wooden fossil if it contains only 5% of the original amount of C-14. What is 5% of it? This is 1 over 20 the original amount, right? So you simply get, okay 1 over 20 times Q0 is equal to Q0 times e to the -r times unknown time t, okay. Remember that we already computed the decay constant r as log 2 over 5600. So now from this equation 1 over 20 times Q nought is equal Q nought times e to the -r of t. Take logarithm on both sides, try to find the t then, t is given by log of 20 over r. So using this quantity r down there, you have the time needed is 5600 times log of 20 over log of 2, okay? Which is approximately 24,203 years, okay? That's the age of roughly the approximate age of this wooden fossil.