Growth and decay problems

There are many ways in science in which exponential increase or decrease, exponential growth, occurs. And it’s very useful, therefore, to be familiar with some of the terminology and reasoning associated with that phenomenon. Here’s a first example. It’s of a word problem type. As you can see, not just an equation you have to solve, but a situation described in words. I love these kind of problems, personally, but some students find them a little tricky. So we have a sheet of material. And the thickness of this sheet is 1/10 of a millimeter. You’re going to fold it a certain number of times over on itself. How many times must you fold it so that the thickness is at least one meter?

Quite an increase. Well, to analyze the problem, you see that when you fold it once, you will have doubled the thickness. So the thickness will be 2 times 1 over 10 in millimeters. When you fold it over again, you will double that thickness. So you will get 2 squared over 10 millimeters, and so on. You’ll see– could be proved by induction– that after n folds, the thickness of your material will be 2 to the power n over 10 millimeters. Now you’re aiming for a thickness of at least one meter. That’s 1000 millimeters. So you want the number of folds n to be such that 2 to the power n over 10 will be greater than 1000, which is 10 cubed.

In other words, 2 to the n has to be greater than 10 to the fourth. You could solve this inequality explicitly. That means that n has to be greater than 4 log to base 2 of 10, which you could look up on a calculator. However, this problem is sufficiently simple that you can find the n by hand. It’s not very hard to calculate 2 to the power 13. You already have 2 to the power 6 is 64. And you find that 2 to the 13th is over 8000. And 2 to the 14th is over 16000. And your answer, therefore, is n equals 14 folds. Notice how exponential growth is awfully fast. Just 14 folds got you up to a meter.

Now more generally, problems of exponential growth and decay occur frequently in science, in physics and biology and other places. And there’s a definition that goes with this, a useful definition that you should retain. If you have a quantity y that depends on time, so y of t, it is said to have an exponential law, either of growth or decay, if the y of t is described as a function of the following type– y naught, or y0, times the exponential of k t. Here, y0 is taken always to be a positive constant. Now we say that it’s growth if k is positive and decay if k is negative. Why?

Because if k is positive, the function y is an increasing function of time. And it’s decreasing if k is negative. Incidentally, we note that the constant y0 is simply the value of y at time 0. It’s the amount of y you have at the beginning of the discussion. Here’s an example of how this kind of concept occurs in, again, a word problem. You have a certain bacterial population that experiences exponential growth. And you observe– you can’t count these bacteria, but by looking at the mass or the weight or whatever, you see that its initial size has doubled in 3 hours. And the question is, how many times will the initial population have been multiplied by after 9 hours?

Well, here’s how to solve this problem. You have to carefully use all the given information. First of all, it’s said to exhibit exponential growth, this population. That means, by definition, that y of t is of the form y0 e to the k t. t here is in hours. Now you’re also given that y of 3, that’s the population after 3 hours, is twice the initial population. So that tells you that 2 y0 is equal to y0 e to the 3 k. You’ll notice that the y0 will cancel on both sides of this equation, and you’ll be able to solve for k, which we do. We find k is ln 2 over 3, a positive number.

Once we know the value of k, then we know that the population after 9 hours is equal to y0 times e to the 9 k. We put in the k that we found, we do a little simplification using the laws of logarithm that we’re familiar with, and we find 8 y0. In other words, after 9 hours, the population is multiplied by 8. And that’s the answer to the question that’s been posed. Let’s look now at an example in which things decrease rather than increase– decay. We have a radioactive substance. And these substances, in fact, behave as exponential decay. We observe for this particular substance that the 200 grams we had initially reduces to 50 grams after one hour.

The question is, how much of the substance will be left after 3 hours? We have to carefully use all the information in these short sentences. Here’s how it works. First of all, we know that y of t is of the form y0 e to the k t. We’re also told that 200 grams reduces to 50 grams in one hour. That means that the 200 grams we had initially, the value of y0, when multiplied by e to the k times 1, will give you 50. We find e to the k equals a quarter.

We easily solve that simple equation, as we solved earlier ones, to find that k is minus ln 4– a negative number, because here, it’s a matter of decay rather than growth. Once we know the value of k, we can plug it in to find y of 3. After a little simplification, we find 50 over 16 grams. Now let me ask you a question. For the same radioactive substance, at which point will there be none left at all? When will the substance be completely gone? The answer is never, because a function like exponential of k t, although it goes down to 0, it decays towards 0 asymptotically, it never attains 0. So a radioactive substance is never completely gone. It’s eternal.

It doesn’t have a finite life. And yet oddly enough, it has a half life. Now that sounds like a paradox. Let me explain. A well-known definition in physics is the half life of an exponentially decreasing quantity– that means the time required for the quantity to become half of what it was initially. That turns out not to depend on the actual quantity. It’s an intrinsic constant linked to the substance itself. Let’s look at the definition more carefully. You have a y of t, quantity y depending on t, that is decaying exponentially. That means, if you look at the graph of y, it’s going to be the graph of an exponential function with a negative exponent.

And we know that it’s going to go down towards 0 asymptotically. And its initial value is y0 at time t. Now if you look at y0 over 2, that’s a strictly positive quantity. And there will be, at some point, some time at which you hit that value. I’m going to call that time tau. You know, tau is another beautiful Greek letter that we like to use, especially for time– tau. Why is there such a tau? Well, there is such a tau, because the function y goes down to 0, and eventually, the graph must cross the value y0 over 2. And furthermore, there’s only one such point, because the function is strictly decreasing. And what defines the point tau?

Well, what defines it is that the value of y at tau– which, on the one hand, is equal to y0 e to the tau k– is 1/2 the initial value, 1/2 y0. So the y0 cancels from both sides, and you find a simple equation for tau, which you proceed to solve. And you find that tau is equal to minus k over ln 2. That’s a positive number, by the way. And that’s called the half life of the radioactive substance. It’s the usual measure by which you measure the persistence of a radioactive element.