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Calculus: Radioactive Decay

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About this Lesson

  • Type: Video Tutorial
  • Length: 8:05
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 87 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus: Differential Equations (10 lessons, $13.86)
Calculus: Growth and Decay Problems (2 lessons, $3.96)

Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hopital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, "Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas" and of the textbook "The Heart of Mathematics: An Invitation to Effective Thinking". He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The "Journal of Number Theory" and "American Mathematical Monthly". His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

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Differential Equations
Growth and Decay Problems
Radioactive Decay Page [1 of 2]
Okay, so there's one thing about thinking about exponential growth where things are sort of growing and so forth, but there's also this notion of decay. And in fact, radioactive decay is a wonderful example of exponential decay. What's happening is you have some radioactive substance, and that radioactive substance is slowly decaying, but it decays at a rate that actually leads to exponential growth because radioactive substances have half-lives. So what's a half-life? It's just the amount of time required in order for you to have half as much material as you did before.
So, for example, if you consider Radium 266 - in fact, let me ask you to bring in some Radium 266 right now. Of course, it's very dangerous and you have to wear gloves and so forth - may I have Radium? Ah! Thank you. So here is our Radium. Now when you open this up, you see now what we have.
So you see, this is Radium 266 - don't actually ever touch this for real - and this actually has a half-life of 1590 years. That means if you wait 1590, you will have half as much substance. So it's a very slowly decaying substance. In fact, can you just take this back and just put it back there? So what that means is that if you wait 1590 years, if you look at what's left, you're going to see half as much. So for example, you can imagine if we were to fast-forward right now and we're now 1590 years later, what would we see? Let's see how much material is left. Can you please bring in the material? Here we go. So 1590 years later, and what do you see? Well, you see a pathetic graduate student, and you also half as much; and if we wait another 1590 years, you're going to see half as much again. So I think you get the idea.
So what's the point? The point is we are talking about radioactive decay. So let's see if we can model how much of the material is left at any particular point in time. We're getting less and less material. Now the population model actually remains the same. What we see is we p(t) - the population or the amount left at time (t) - is going to be the initial population times e^kt, where k is some constant. So the same model will actually work, but let's see what happens here.
So suppose, for example, the amount we started with - the 1590 years ago - was actually one hundred milligrams. So if I started with one hundred milligrams, what that would mean is that my initial population would be one hundred milligrams. That will allow me to figure out what my initial population is, it's one hundred. So I just insert that right there. So when I put in that piece of information, what I see is e^kt x 100. Now how can I figure out that constant k? Well, I know this has a half-life of 1590 years, which means if I wait 1590 years, what I'm going to see is half as much as I have right now. So that gives me another data point, and that's always required in order to figure out that constant. It's a second data point. So if I wait 1590 years, then the amount of Radium 266 I have is going to be half as much as I had to start with, so I have fifty. Therefore, I see 50(e^k), and then why did I have to wait here? I had to wait - wait, I'm doing this all backwards. Why did you let me do this? Aren't you watching this? This should be one hundred here, but this amount is five hundred, so I see what I should do. Okay, now I see what you're thinking. I apologize for that little outburst of personality. That I really meant to say was just this, right? That's what you were thinking I was going to say? Sure, that when you wait that long, that's how much is left. Right.
Now let's insert this into the formula. If I insert this into the formula, what I see here is the population will be fifty, and that will equal 100(e...to the...and now the time is that, so I have 1590 years - t is being measured, by the way, in years - times k. Now this is the equation, in fact. If I divide on this side by a hundred, I see just . So I see = e^1590K, and I can solve this by taking natural logs of both sides. The natural log of this is the natural log of , and the natural log of the exponential function kills off the exponential function, and I'm left with 1590k. So what's k? It just equals this quotient. So . So that's k. If I come back into the formula and insert that, what would I see? What I would see is p(t) = 100(e^k), which is the natural log of divided by 1590 - it's all in the exponent - times t. So that's the answer.
Now, in fact, you can actually write this if you wanted to without the e and the natural log thing. This is a perfectly fine answer, by the way, but let me just show you how you could actually finesse this a little bit. I have the 100 out in front - I'm not going to forget that. Now, this e thing, I'm going to put that in here, and let's write this as t divided by 1590, all times the natural log of . You see, if I do that, then I can actually see that this becomes a coefficient, which therefore can be brought up as an exponent because I'm looking at a natural log or a log in general. So this is now . So this actually equals e raised to a natural log power - those are inverse functions, so they sort of kill each other off - I'm just left with . So you could write it that way - that looks great. Or if you don't like the , you could write it this way: You could say that 100(2^-t) divided by 1590 because remember that is just two raised to the negative one power. So this is a perfectly fine answer, and you might like that because you can sort of see everything without the e's and the natural logs. Of course, if you actually wanted to calculate this on a calculator, it might be easier to use the e stuff because there's an e key and a natural log key. Otherwise, you've got to use a power key, which maybe you have and maybe you don't.
Anyway, this is the formula for the amount of material - the amount of Radium 266 that we have left over - at anytime t where t represents years. And you can see, with that negative power there, there is decay. You can see it here even easier: As t gets bigger and bigger and bigger, this gets smaller and smaller and smaller, and so we have less and less material because this is a radioactive substance, and so we're having decay, and so the population now is actually shrinking. So that is radioactive decay, which is again just an example of exponential...well, not growth, but exponential decay. I'll see you at the next lecture.

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