

College Algebra: Half-Life



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  • Type: Video Tutorial
  • Length: 11:08
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 120 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
College Algebra: Exponential Growth & Decay (4 lessons, $5.94)






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Now when you think about decay, it sort of sounds gross like bodies sort of decaying and stuff, but really decay does happen a lot in nature, especially with radioactive substances. Usually with decay, what you have is halve life. That means that you have some radioactive substance that after some time has half as much and then at that exact same time has half as much and half as much and so on. So, let me try to illustrate this with a sort of simple visual example. Of course, if you're going to use a radioactive substance you have to wear gloves. So radioactive substance might look something like this and then in a certain period you have half as much. Well, now you have that. If you wait even longer, you would have half as much as that. So, really, maybe a better example would be this.
Let's say the half-life of this was 5 years. That means that in 5 years you would have half as much. In another 5 years you would have half as much, in another 5 years you have half of that and so on. You would always something, a radioactive decay, so it keeps going on. Anyway, here's an actual, real world type radio decay problem that we can look at together. Now, it turns out that a nuclear, radioactive substance by the name of Sodium 24 is a radioactive isotope that doctors use to put into the body and monitor certain functions. Anyway, assuming that we have a patient where we've inserted 4 micrograms of this Sodium 24 into the body--this is good, we're trying to help the patient this is not gross. Then the amount of micrograms in the body remaining after T hours is given by actually a decaying function, this is less and less in the body, it slowly decays.
Of course, that person will always have some in the body for as long as the person lives, unfortunately, but so what. Let's say this is the amount remaining. In fact, in honor of the fact that this is radioactive I will wear the gloves through the entire lecture. So, the amount remaining in the body after T hours is given by 4e^-.046t. So if you give me the time, I'll be able to figure out how much is left in the body. We're starting with 4. I remind you we started with 4. Okay, so the first question we could ask, for example, is to sketch a graph of this. What does the graph of this function look like? Now, it should have this decay type feature. If we look at this really fast, we can actually see what the graph of this looks like. Let's see, so the first thing we should do is just plot e^x or e^t in this case. Let's forget about the minus sign and all this other stuff.
Now, there's a 4 out in front, so let me first of all just put in the e^x function, to remind you what the exponential function looks like. So, there's the exponential function. Now, the 4, what does that do? Well, it takes this point, which is at a height of 1 and moves it to 4 and it sort of takes everything else and multiplies it by 4. So we get a much more dramatic looking thing. So, it raises it up a lot. Okay, so that's 4 times e^x. But that negative sign there and that coefficient--what the negative sign does, remember, is it flips over the y-axis. So, in fact, it flips and this factor up there is going to elongate a little bit. So the final picture might look something like this.
Roughly speaking, that's the graph. You notice the decay. See, as time goes on, we have less and less of the substance, but there's always something of the substance because it's radioactive. So, it never actually touches the zero line. So, there's the graph of that. You get a sense of that thing decaying, so that's pretty good. Now, what I'd like to do is say, "Okay, what is the amount of Sodium 24 that remains in our patient's body after 5 hours?" So, how would you find that? Well, I'm asking for what is the amount after 5 hours? So, that's just asking for A(5). That's pretty easy. It's just 4 times e^-.046 times 5. You can plug that into a calculator, -.046 multiply it by 5 and take e to that power and then multiply it by 4 and what you get is, numerically, .7945 micrograms. Okay, so that means, remember we started with 4 micrograms in the beginning and now we're down after 5 hours to .7945 micrograms. So, you see it's leaving the body reasonably quickly.
Now another question we could ask is, what is the half-life of this radioactive substance? Now remember half-life is the amount of time required so that you have the amount cut in half. So, how could you figure that out? Well, we know that we started with 4. So, if we started with 4, then what would the amount be where we would have half? Well, plainly, 2. So, the question really is, what is the time, how long do I have to wait for the amount that remains to be equal to 2, since I started with 4? So, I can set up an equation that looks like this. 2 = 4e^-.046t.
Now I have to solve this for t. Okay, so that's an exponential equation. I can divide, first of all, both sides through by the 4 to get rid of that. So, I see a . So, I see = e^-.046t. Now, how do you solve and exponential equation? Well, if you have the variable up on top, you don't want it in the attic, you want it on the ground floor. So, I'll take the natural log of everything and use the property of exponents and logs and this can be pulled out as a coefficient. So, if I natural log both sides, I see ln = lne^-.046t. Now I use properties of logs and exponents. That tells me if I have an exponent in a log, I can pull that out in front and it becomes a coefficient, which is exactly the point of taking logs. When I do that, I still have the ln on this side, but now on this side I see -.046t and then it's multiplied by the natural log of e. But what is the natural log of e? The natural log e, natural log is just log[e] and I'm taking out of e. So what's the exponent I have to raise e to in order to get e? Well, it's 1. The natural log of e is 1. So, I'm not even going to write it.
Well, now solving for t is a piece of cake. I just divide everything through by -.046. So I see t = ln() divided by -.046. Now before I compute that, let's just think and say, "There could be a problem here." Because t is time. It's the waiting time to have half-life. I see a negative sign there. So, maybe this is going to be weird answer. How could the time be negative? How can I go backwards in time in order to have half as much? Sounds like there could be a problem, until you remember that if you're taking the natural log of a number that's less than 1, that natural log will be negative. So, in fact, the top here is actually negative. So, negative over negative, that's going to be okay, we should get a positive answer. You could plug this into a calculator and you would see that we'd have 15.068 hours. So, that's the half-life. In fact, if you wait 15.06 hours, the amount you have should be half of what you have.
Okay. Great. Except I have to admit I'm a little bit worried about one little teeny thing here. You know what I'm worried about? So, let's just think about this now. This is all live math. This is great and safe math because I'm wearing my gloves. If the half-life here is 15, then how come when I wait just 5 hours, I'm already done to a fraction? That doesn't sound right to me. So, let's recalculate that and see if we've made a mistake. What I'm going to do is I'm going to compute this number once again. Maybe I made a mistake when I computed that. So, let's see if I can--oh no. Okay, so let's compute this again. So, we take -.046 and we multiply it by 5. Now I'm going to take e to that thing. So, I take e to -.046 times 5. I did make a mistake, because that value that I have written here is just this e part, I never multiplied it by 4. I never multiplied this by 4. Look how I caught my own mistake--so if I multiply it by 4. Awe, I get a much happier answer. I get 3.17 stuff micrograms.
Now, that makes a lot more sense and look how I caught it. I caught it by not just reporting the answers, but by thinking about what's going on. If I have to wait 15 hours to have half as much and I start with 4, 15 hours later, I'm only going to have 2 left. How come 5 hours later I'd have .79 left? That doesn't make sense. I should have more than half left over and sure enough, I do have more than half. So notice I was able to catch an error I made just be thinking through my answers. It's a really important lesson actually.
Terrific. We found out the half-life. Now here's a question--I could ask a little follow-up question, a little bonus. The bonus is, how long would we have to wait in order for us to have 1 microgram left? Well, there's two ways of doing this problem. One is to set up this thing. If you have 1 microgram left, you just set A to equal 1 and you solve this for t, just like we did before with the half-life--taking logs and so forth. Or you could be a little bit tricky and realize that once you know the half-life, how many half-lives do you have to wait if you start with 4 to get down to 1? Well 1 half-life gets you down to 2, another half-life will get you down to 1. So, in fact, all you have to do is say, that's going to be 2 half lives. So you multiply this answer by 2. So it would just be 2 times the half-lives and of course that's 30.13 something hours.
Once you know the half-life you can actually figure out all sorts of things or you might want to practice just doing this method of solving this and seeing the exact same answer. In either case you can see that in the real world these radioactive substances actually do obey the exponential laws of decay. So now that our experiment is over, try these. If you don't have the gloves, be careful.
Exponential and Logarithmic Functions
Word Problems Involving Exponential Growth and Decay
Half-Life Page [2 of 2]

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