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About this Lesson
 Type: Video Tutorial
 Length: 10:03
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 109 MB
 Posted: 06/27/2009
This lesson is part of the following series:
College Algebra: Full Course (258 lessons, $198.00)
Algebra: Exponential and Logarithmic Functions (36 lessons, $49.50)
College Algebra: Exponential Growth & Decay (4 lessons, $5.94)
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found athttp://www.thinkwell.com/student/product/collegealgebra. The full course covers equations and inequalities, relations and functions, polynomial and rational functions, exponential and logarithmic functions, systems of equations, conic sections and a variety of other AP algebra, advanced algebra and Algebra 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 UTAustin 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, padic 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.
About this Author
 Thinkwell
 2174 lessons
 Joined:
11/14/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
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It turns out that actually exponential functions even go on in the refrigerator. Here's a question I want us to think about right now. Suppose you have an icy cold refrigerator and suppose that the temperature was 34° in there. Now, you put a soda can into it and there it is. That soda can of course was outside for awhile, so it's actually pretty warm. In 5 minutes the temperature of the soda can drops from 75° all the way down to 65°. Huge drop in just 5 minutes. So that's some serious cooling going on. Now, we can ask ourselves, "Well, can you figure out a variety of things." For example, can you figure out what the temperature of the soda would be in 30 minutes? Could you figure out when the soda will be a particular temperature? Suppose you want to drink it exactly when its 36°. How would you go about that?
So actually figuring these things out can be done using some basic physics principles. In fact, you would use what's known as Newton's Law of Cooling. Newton was everywhere, by the way, he just made a fortune on cooling and calculus you wouldn't believe it. Anywayand his laws, he has all these laws too. Anyway, so it turns out that his law of cooling is the following. The temperature after t minutes of an object that's placed in an environment where the environment has constant temperature, like for example your refrigerator in an ideal sense, could actually be found by just knowing a few things. First of all, you have to know the temperature of the constant environment. So, we'd call that A.
So, A just represents the temperature, for example, of the refrigerator. It's constant. Then you add T[0], which is the initial temperature of the object that you're putting into this environment of constant temperature and subtract off that constant temperature of the environment. Then you have e^kt, where t is the time, again, in minutes and k is some constant. Now k is some constant that depends upon the object you're putting in. In this case it's a soda. So, that constant depends upon the can, is it aluminum, is it plastic, whatever and so on. Actually that constant is determined heavily by the object itself. So we don't know what that is. That's sort of a mystery thing right now. But if you want to find the temperature after t minutes you plug into this formula, where this is the temperature of the ambient space, this is the initial temperature of the object and k is this constant.
So, if you actually want to use this thing, you have to know all this information. Now we do know, for example, that the temperature of the refrigerator, I told you, was 34°. So this is 34. We also know that the soda can, initially when it was put in, I reminded you was 75. So we can actually fill in a lot of these things for our soda can example. I would see what? Well, this number here I said was 34°. Actually, I'm not going to put the little degree symbol there, but of course these are degrees. Plus, and then I have 75 and I subtract off 34, so that's going to be 75  34 is 41. So I have 41 and then I have e to the minus, but wait a minute. I don't know what the constant is. Shot. Well, this seems now like we're in trouble. Because if I don't know what that number is, I can't actually answer these question.
But there was one piece of information that we were given that we haven't used yet. We new sort of one data point. Because were told that 5 minutes later, after we put the thing in this thing cooled down to a nice frosty temperature of 65°. So, that means that this must satisfy the fact that when t = 5 minutes, the temperature must by 65°. Well, that gives us an equation and allows us to solve for the constant. So, just knowing that one little data point, I can now figure out what k is and then answer the other questions. So, let's actually input that. We know that when t is 5 minutes, 5 minutes later this temperature is 65. So what that tells me is 65 = 34 + 41e^5k, because the time is 5. Well notice that's an equation just in terms of k. I can now find that constant. It eluded me for so long, now I've got it.
So what would I do first? Well, first I'll bring this 34 to this side and that gives me a 31 and that equals 41e^5k. I'll divide both sides by the 41 here and I see 31 divided by 41 equals e^5k. How do I solve an exponential of this sort where the variables upstairs in the attic? I want to get it on ground floor, so I'll take the natural log. If I take the natural log, I see ln(^31/[41]) = ln(e^5k). Now of course I use the property of exponents inside of logarithms to realize that can be pulled out in front as a coefficient. That's the whole point of taking logs, to bring that unknown down to the first floor. When I do that, what do I see? I see the ln(^31/[41]) = 5kln(e). Well, what' the ln(e)? Well the ln(e) is just 1. So, in fact, that term goes away. So, I see this just equals 5k.
If I want to solve this all I've got to do now is divide by 5. So, I see that k, that constant, equals ln(^31/[41]) all divided by 5. We can plug that into a calculator and see what the value is. It turns out the value is .0559 stuff. So that's the constant. I haven't answered any questions yet, but I now have a complete formula that I can use. So all this work was just to find this mysterious constant that we knew existed, but we didn't know what it was. Now let's see if I can do some fancy erasing here. I'm going to try to do this. This is going to be sort of a little experiment on the fly here. You want to see the experiments, don't you? There we go. Except I just threw away my constant, but I'll remind you what it is. Here we go.
Now I can actually insert this constant right into this value. Let me just do that with a different color. That'll be sort of fun. In fact, right here this exponent has now become .0559t. See that's actually that constant. I put that in. See the minus sign is the minus sign from here. Okay, great. Well there's the formula. So now we can start answering questions. For example, the first question I want to know is what will be the temperature after 30 minutes? Well that says, "What's the temperature 30 minutes later?" So that's just T(30). So, I plug in 30 for t and what do I see? 34 +41e to what power? Well, e^.0559. That's the constant we just found, multiplied by 30 minutes. So multiplied by 30. You can use a calculator to actually compute that with the exponential function there and you'll see 41.6°. Does that make sense?
Well, just think about it. It's very, very, very qualitative. It started off at 75, 5 minutes later it was 65. So 30 minutes later, it should be even colder than that and certainly it is. Notice it can be too cold. What if I got an answer of 25°? Could that make sense? Absolutely not. Because remember the refrigerator itself is 34. How could a refrigerator cool something even colder than the thing is itself? Not possible. So, in fact, this answer seems reasonable. It seems within the realm of reason. Okay, now I could ask another question though. Suppose I want to drink it at the very moment it's 36°? I want a real icy cold one. What would I do?
Now I want to find the time when the temperature is 36. So I set this equal to 36 and I solve for t. So I say, "36 = 34 + 41e^.0559t." I want to solve that for t. So the first thing I'll do is bring this over to this side, subtract it. So, if I do that, I would just see a 2 and if I divide by the 41, I would see ^2/[41] = e^.0559t. So, again, what did I do? I just subtracted 34 here, that gave me a 2, but then I also divided by the 41 to get me down here. If I take the natural log of both sides, what would I see? Well the natural log of both sides would give me the following. Natural log of (^2/[41]) and what would I have on the right? On the right I would take the natural logI'm going to all this in one step. The natural log of all this, but that now comes down as a coefficient and the natural log of e is 1. So, repeating the steps that I did previously, I just see .0559t. So what's t? t equals ln(^2/[41]) all divided by .0559. You can compute that on a calculator and you'd get 54.03 minutes. So, almost an hour, almost an hour I'd have to wait.
Does that seem reasonable? Well, it seems at least within the realm of reason, because we knew that 5 minutes in I was at 65°, 30 minutes in I was 41°. So this has to be longer than 30 minutes and certainly that is. So, what would happen is 54 minutes later, you would open up your refrigerator, have an icy cold one and all set.
Exponential and Logarithmic Functions
Word Problems Involving Exponential Growth and Decay
Newton's Law of Cooling Page [2 of 2]
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