# Lesson Video: Exponential Growth and Decay Models Mathematics • 9th Grade

In this video, we will learn how to model exponential growth and decay arising from the differential equation 𝑦′ = ±𝑘𝑦.

13:35

### Video Transcript

In many natural phenomena, quantities grow or decay at a rate proportional to their size. For example, investing money in a bank account which pays compound interest is a good example of exponential growth. Though perhaps not quite at the rate most savers would like. At the other end, atmospheric pressure decreases exponentially with increasing height above sea level, an example of exponential decay. In this video, we’ll learn how to model exponential growth and decay arising from differential equations and use these models to make predictions about future behavior.

In general, we say that if 𝑦 of 𝑡 is the value of a quantity 𝑦 at time 𝑡, and if the rate of change of 𝑦 with respect to 𝑡 is proportional to its size, 𝑦 of 𝑡, then d𝑦 by d𝑡 is equal to 𝑘 times 𝑦 for real constants 𝑘. This is sometimes called the law of natural growth, for values of 𝑘 greater than zero, and the law of natural decay, for values of 𝑘 less than zero. And we can solve this differential equation generally. Remember, whilst d𝑦 by d𝑡 isn’t really a fraction, under certain circumstances we do treat it a little like one, and we rewrite our differential equation as one over 𝑦 d𝑦 equals 𝑘 d𝑡.

We now integrate both sides of this equation. Using the standard result that the integral of one over 𝑦 is the natural log of the absolute value of 𝑦, we see that we have the natural log of the absolute value of 𝑦 plus 𝐴 equals 𝑘 times 𝑡 plus 𝐵, where 𝐴 and 𝐵 are constants of integration. We combine these constants by subtracting 𝐴 from both sides. And let’s call this new constant 𝐶. So, the natural logarithm of the absolute value of 𝑦 is equal to 𝑘 times 𝑡 plus 𝐶. We now raise both sides of this equation as a power of 𝑒, such that 𝑒 to the power of the natural log of the absolute value of 𝑦 equals 𝑒 to the power of 𝑘𝑡 plus 𝐶.

Well, 𝑒 to the natural log of 𝑦 is 𝑦. And we don’t need the absolute values here because we’re talking about a population. It must be greater than zero by definition. Then, using the laws of exponents, we see that we can rewrite 𝑒 to the power of 𝑘𝑡 plus 𝐶 as 𝑒 to the power of 𝑘𝑡 times 𝑒 to the power of 𝐶. But of course, 𝑒 to the power of 𝐶, given that 𝐶 is already a constant, is itself a constant. So, let’s call that 𝑃. And we find that the general solution to the differential equation d𝑦 by d𝑡 equals 𝑘 times 𝑦 is 𝑦 equals 𝑃 times 𝑒 to the power of 𝑘𝑡.

Now, in fact, this 𝑃 has a special significance. We can identify what this significance is by imagining the scenario when 𝑡 is equal to zero. Clearly, since 𝑡 equals zero is the initial time, we know that 𝑦 equals 𝑃 times 𝑒 to the power of zero to be the initial value of 𝑦. However, 𝑒 to the power of zero is one. So, we find that when 𝑡 is equal to zero, 𝑦 is equal to 𝑃. This means that 𝑃 is the initial value of the function. Now, we redefine a little and we say that, in general, the solution to the initial value problem d𝑃 by d𝑡 equals 𝑘 times 𝑃 for 𝑃 of zero equals 𝑃 nought is 𝑃 of 𝑡 equals 𝑃 nought times 𝑒 to the 𝑘𝑡th power. In order to see how to work with these equations, let’s have a look at a specific example.

The population of rabbits on a farm grows exponentially. If there are currently 245 rabbits and the relative growth rate is 23 percent, find a function 𝑛 of 𝑡 to describe the number of rabbits after 𝑡 years.

In this question, we have our variables defined. We’re told that 𝑡 is the number of years, and 𝑛 of 𝑡 is the function that describes the number of rabbits. We know that since the population grows exponentially, we can use the equation d𝑃 by d𝑡 equals 𝑘 times 𝑃 to describe it. In fact though, we’ll rewrite this using our variables such that d𝑛 by d𝑡 equals 𝑘 times 𝑛. We could have written this using function notation such that 𝑛 prime of 𝑡 equals 𝑘 times 𝑛. But personally, I prefer to work with Leibniz notation.

Now, in fact, we have a value for 𝑘. We know that the relative growth rate is 23 percent. That’s 0.23 as a decimal. And we can therefore say that d𝑛 by d𝑡 equals 0.23 times 𝑛. Now remember, once d𝑛 by d𝑡 is absolutely not a fraction, there are certain circumstances where we treat it a little like one. Here, we can rewrite our differential equation as one over 𝑛 d𝑛 equals 0.23 d𝑡. We’re then going to integrate both sides of this equation. The integral of the left-hand side is the natural log of 𝑛 plus some constant of integration 𝑎. And the integral of the right-hand side is 0.23𝑡 plus a different constant of integration 𝑏.

Now, note that usually the integral of one over 𝑛 would be the natural log of the absolute value of 𝑛. But we know the number of rabbits can’t be negative, so we don’t actually need to include that here. Combining our constants of integration by subtracting 𝑎 from both sides, and we see the natural log of 𝑛 is equal to 0.23𝑡 plus 𝑐. We then raise both sides to the power of 𝑒 to find that 𝑒 to the power of the natural log of 𝑛 is equal to 𝑒 to the power of 0.23𝑡 plus 𝑐. Well, 𝑒 to the natural log of 𝑛 is simply 𝑛.

And then, we use the laws of exponents to rewrite the right-hand side as 𝑒 to the power of 0.23𝑡 times 𝑒 to the power of 𝑐. But of course, 𝑒 to the power of 𝑐 is itself a constant. Let’s call it 𝑃. So, 𝑛 is equal to 𝑃 times 𝑒 to the power of 0.23𝑡. Now, we’re almost done. There were initially 245 rabbits on the farm. In other words, when 𝑡 is equal to zero, 𝑛 is equal to 245. So, 245 equals 𝑃 times 𝑒 to the power of zero because 𝑒 to the power of zero is one. So, we find 𝑃 to be equal to 245. And we found our function for 𝑛. We can say that 𝑛 of 𝑡 is equal to 245𝑒 the power of 0.23𝑡.

Now, it’s probably quite clear that we didn’t need to set up and solve a differential equation here. We could have simply stated the general form for an equation that describes exponential growth and decay. It’s the general solution to d𝑃 by d𝑡 equals 𝑘 times 𝑃 for the initial value of 𝑃 is equal to 𝑃 nought. It’s 𝑃 of 𝑡 equals 𝑃 nought times 𝑒 to the power of 𝑘𝑡. Essentially, 𝑃 nought is the initial value of the function, and 𝑘 is the growth rate. Of course, it’s always useful to know how to derive this formula, when necessary.

We’ll now look at an example of a question where we can use the general solution to the differential equation without fully deriving it.

A doctor injected a patient with 13 milligrams of radioactive dye that decays exponentially. A doctor injected a patient with 13 milligrams of radioactive dye that decays exponentially. After 12 minutes, there were 4.75 milligrams of dye remaining in the patient’s system. Which of the following is an appropriate model for the situation? Is it A) 𝑓 of 𝑡 equals 13 times 0.0805 to the power of 𝑡? B) 𝑓 of 𝑡 equals 13 times 𝑒 to the power of 0.9195𝑡. Is it C) 𝑓 of 𝑡 equals 4.75 over one plus 13 times 𝑒 to the power of negative 0.83925𝑡? Or D) 𝑓 of 𝑡 equals 13 times 𝑒 to the power of negative 0.0839𝑡.

Remember, we can model exponential growth and decay by using the formula 𝑃 of 𝑡 equals 𝑃 of nought times 𝑒 to the power of 𝑘𝑡, where 𝑃 nought is the initial value of 𝑃, and 𝑘 is the rate of growth or decay. So, let’s list what we actually have. Since the answers in this question are listed in terms of 𝑓, we’re going to change this equation to 𝑓 of 𝑡 equals 𝑓 nought times 𝑒 to the power of 𝑘𝑡, where 𝑓 is the amount of dye that’s actually in the patient’s system after 𝑡 minutes. Now, we’re told that the doctor injected the patient with 13 milligrams of radioactive dye. In other words, that’s the initial amount of dye in the patient’s system. So, 𝑓 nought is 13. And we can therefore say that 𝑓 of 𝑡 is equal to 13 times 𝑒 to the power of 𝑘𝑡.

But we also know that after 12 minutes, there were 4.75 milligrams of dye in the patient’s system. In other words, when 𝑡 is equal to 12, 𝑓 is equal to 4.75. So, let’s substitute these values into our function and solve for 𝑘. That’s 4.75 equals 13𝑒 to the power of 12𝑘. We’ll solve for 𝑘 by first dividing both sides by 13. And we see that 4.75 over 13 must be equal to 𝑒 to the power of 12𝑘. Next, we find the natural logarithm of both sides of this equation. This is useful as it’s the inverse operation of raising something to the power of 𝑒. And so, these two operations will essentially cancel one another out. In other words, the natural log of 𝑒 to the power of 12𝑘 is just 12𝑘.

So, we find that the natural log of 4.75 over 13 equals 12𝑘. We’ll solve now by dividing both sides by 12. And so, 𝑘 is equal to a 12 of the natural log of 4.75 over 13. Typing this into our calculator and we obtain 𝑘 to be equal to negative 0.083900, and so on. Correct to three significant figures, that’s 𝑘 is approximately equal to negative 0.0839. And substituting this back into our equation gives us that 𝑓 of 𝑡 equals 13 times 𝑒 to the power of negative 0.0839𝑡. And we see that that corresponds to the answer D. 𝑓 of 𝑡 is 13 times 𝑒 to the power of negative 0.0839𝑡.

Once we have these models, we can use them to make predictions about the future behavior. Let’s have a look at an example of this form.

At the start of an experiment, a scientist has a sample which contains 250 milligrams of a radioactive isotope. The radioactive isotope decays exponentially at a rate of 1.3 percent per minute. a) Write the mass of the isotope in milligrams, 𝑚, as a function of the time in minutes, 𝑡, since the start of the experiment. And b) Find the half-life of the isotope, giving your answer to the nearest minute.

Remember, we can model exponential growth and decay using the formula 𝑃 of 𝑡 equals 𝑃 nought times 𝑒 to the power of 𝑘𝑡, where 𝑃 nought is the initial value of 𝑃, and 𝑘 is the rate of growth or decay. In our case, there are 250 milligrams of the sample at the start of the experiment. So, 𝑃 nought must be equal to 250. We said that 𝑘 is the rate of growth or decay. Since the isotope is decaying, this value is going to be negative. And 1.3 percent as a decimal is 0.013. So, 𝑘 is negative 0.013. We can therefore say that the function that describes the mass in terms of minutes is 𝑚 equals 250 times 𝑒 the power of negative 0.013𝑡.

We’re now going to look at part b). Here, it helps us to understand the definition of the phrase half-life. It’s the time taken for the radioactivity of the isotope to fall to half of its original value. In our case, that’s half of 250. That’s 125 milligrams. We need to work out for what value of 𝑡 𝑚 is equal to 125. We therefore set 𝑚 equal to 125 and solve for 𝑡.

We see that 125 equals 250 times 𝑒 to the power of negative 0.013𝑡. We then divide both sides of this equation by 250. 125 divided by 250 is 0.5. So, we see that 0.5 equals 𝑒 to the power of negative 0.013𝑡. We then take the natural logarithm of both sides of this equation to get the natural log of 0.5 equals the natural log of 𝑒 to the power of negative 0.013𝑡. But of course, the natural log of 𝑒 to the power of negative 0.013𝑡 is just negative 0.013𝑡. And so, our final step is to divide both sides of this equation by negative 0.013. That gives us 𝑡 equals 53.319 minutes which, correct to the nearest minute, is 53 minutes.

In this video, we’ve learned that we can solve differential equations of the form d𝑃 by d𝑡 equals 𝑘𝑃 by writing them in the form one over 𝑃 d𝑃 equals 𝑘 d𝑡 and integrating both sides. When we do, we see that the solution is of the form 𝑃 equals 𝑃 nought 𝑒 to the 𝑘𝑡, where 𝑃 of zero equals 𝑃 nought. We saw that whilst it’s useful to be able to derive the formula, it’s absolutely fine to quote it and apply it to problems involving exponential growth and decay.