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Next: Glossary: A Quick Guide Up: How to Ace Calculus Previous: Exponential Growth and Decay:

   
Exponential Growth and Decay: The Rise and Fall of Slime

These problems tend to be disgusting, involving either the growth of bacteria cultures or nuclear holocaust. But even the ugly side of life can be described by mathematics.

Basically, any situation where the rate of change of a function is proportional to the value of the function falls into this section. That's a mouthful. But for example, population growth fits this pattern. The rate of growth of a population of rabbits is greater when you have more rabbits. Radioactive decay behaves in a similar manner, though the amount of material that you have is shrinking over time instead of growing. The less radioactive material that you have, the slower it decays each minute. Letting N be the number of rabbits or the amount of plutonium that we have at any given time, the fact that the rate of change of the function (what we normally call the derivative of the function with respect to time ) is proportional to the value of the function can be stated mathematically as:

\begin{displaymath}\frac{dN}{dt} = kN.
\end{displaymath}

Here, k is just the proportionality constant, with actual value depending on the problem. In the case of a growth problem, such as the rabbit population, k will be positive. This is because $\displaystyle \frac{dN}{dt}$ must be positive since the function giving the number of rabbits in the population N is an increasing function. In the case of a decay problem, such as radioactive decay, k will be negative. This is because $\displaystyle \frac{dN}{dt}$ must be negative, so that the function giving the amount of radioactive material at any given time N is a decreasing function.

This equation $\displaystyle \frac{dN}{dt} = kN$ is a so-called differential equation. It's called a differential equation because it is an equation that involves derivatives. It is just about the simplest differential equation  you can come up with. (All right, all right, there are one or two simpler ones. Take $\displaystyle \frac{dN}{dt}= 0$ for instance.)

Unlike most of the differential equations out there, this one is solvable. That is to say, we can figure out exactly the general form of a function N that satisfies this equation.

First we ``separate the variables", putting everything that involves N on the left side of the equation and everything that involves t on the right side of the equation. We even treat the derivative $\displaystyle \frac{dN}{dt}$ as if it were a fraction, splitting the dN from the dt:

\begin{displaymath}\frac{dN}{N} = k \ dt
\end{displaymath}

Now we integrate both sides:

\begin{displaymath}\int \frac{dN}{N} = \int \ kdt
\end{displaymath}


\begin{displaymath}\ln N = kt +C
\end{displaymath}

(Notice we didn't bother with the absolute values on the N, since we know Nis positive here.)


Applying the exponential function to both sides, we obtain:


N = ekt+C = ekt eC = Aekt,

where we replace eC by A, since eC is just some constant anyway. Why use a fancy name like eC for it, when A works just as well?

So N = Aekt. Notice that at time t=0, N(0) = Aek0 = A. Since N(0) is the initial amount that we started with, A is just this initial amount, usually written as N0.

That means that our general formula for the solution to the differential equation $\displaystyle \frac{dN}{dt} = kN$ is given by N(t) = N0 ekt. This is called our exponential growth or exponential decay equation. Let's apply it in some representative examples.

Problem   Suppose that a colony of bacteria is growing in the corner of your shower stall. On June 1, there are 1 million bacteria there. By July 1, there are 7.5 million. Your shower stall can contain a total of 1 billion bacteria. When will you have to start taking showers down at the gym?

Okay, we know what you're thinking. You're thinking, ``Well actually, even before the entire shower stall is filled with bacteria, I'm not going to take showers in there. I mean, when it's half full, I'm not getting in there. And since I don't know exactly when I will be unwilling to get in there anymore, this problem is poorly stated and therefore I refuse to try to solve it."

But you have to be tough and get into the shower until the very end. THIS IS MATH. It's not for the lily-livered.


Solution. We know that the number of bacteria in this colony is given by N(t) = N0 ekt. Let's take June 1 as the time t= 0. Then, the initial number of bacteria is N0 = 1, 000, 000. So we have that

N(t) = 1,000,000 ekt.

We now need to determine k. But we know that on July 1, which corresponds to t = 30, the population N(30) = 7,500,000. Plugging this into our function, and solving for k, we have:


7,500,000 = N(30) = 1,000,000 ek(30)


7.5 = ek(30)


\begin{displaymath}\ln (7.5) = 30 k\end{displaymath}


\begin{displaymath}k = \frac{\ln (7.5)}{30} \approx 0.0672.\end{displaymath}

Plugging that value for k into our original function we now have:

N(t) = 1,000,000 e0.0672t.

This tells us the number of bacteria at any time. When will this equal 1,000,000,000? Set it equal to 1,000,000,000 and solve for t:

1,000,000,000 = 1,000,000 e0.0672t,


1,000 = e0.0672t,


\begin{displaymath}\ln (1,000) = 0.0672t,\end{displaymath}


\begin{displaymath}t = \frac {\ln {(1,000)} }{0.0672} \approx 102.8 \mbox{ days}.\end{displaymath}

So, on September 10, that shower is one giant black slimy mass. Get ready to move.


Problem   A nuclear bomb set off by terrorists has made New York City uninhabitable. Many people argue that this is not a change. But suppose the residual cobalt levels are 100 times safe levels. If cobalt has a half-life  of 5.37 years, how soon before people can live in New York again?

Solution: This is called a decay  problem. Not because our urban centers are in decay, but because radioactive material decays over time, so that the remaining amount decreases. So, what do we do here? Let N(t) be the amount of cobalt in the city at time t, t years after the initial explosion. Then N(t) = N0 ekt where N0 is the initial amount of cobalt released into the city and k is our decay constant, which must be negative. In most decay type problems, N0 is given to you. But here, we don't even know that. However, we have been given the half-life  of cobalt. We can use that to determine the decay constant k.

Since the half-life of cobalt is 5.37 years, we know that if we start with an initial amount of cobalt N0, we will have N0 / 2 of it left at the end of 5.37 years. Therefore N0/2 = N(5.37) = N0 ek 5.37.

N0/2 = N0 ek 5.37. Solving for k:

1/2 = ek 5.37


\begin{displaymath}\ln(1/2) = k 5.37
\end{displaymath}


\begin{displaymath}k=\frac {\ln(1/2)}{5.37} \approx -0.129
\end{displaymath}

Therefore, N(t) = N0 e-0.129t gives the amount of remaining cobalt at any given time.

Now we want to know when it will be safe for people to return to New York. Initially, cobalt levels were 100 times safe levels, so we would like to know when the amount of cobalt will be $\displaystyle \frac{N_0}{100}$. At that time, the amount of cobalt will have dropped down to a safe level. How do we determine when that time is? We just set our function giving the level at any time equal to that amount and solve for t:


\begin{displaymath}N(t) = N_0 e^{-0.129t} = \frac{N_0}{100}\end{displaymath}


\begin{displaymath}e^{-0.129t} = \frac{1}{100}\end{displaymath}


\begin{displaymath}-0.129t = \ln (\frac{1}{100}) \end{displaymath}


\begin{displaymath}t = \frac{1}{-0.129}\ln (\frac{1}{100}) \approx 35.7 \mbox{ years}.\end{displaymath}

Not bad, although personally, we're waiting at least 36 years, just to be on the safe side.


Problem   The Case of the Big Cheese .

``Well, look who the cat dragged in," said Sergeant Woffle, as Detective Horns walked into the room. `` If it isn't Sheer Luck Horns."

``Still talking in cliches, I see," said Horns. ``And the sarcasm needs work. What's the m.o.?" He looked down at the crumpled body lying in a heap on the floor. Woffle shrugged.

``He's Hiram Fentley, heir to the Fentley feta cheese fortune. Found dead at 2:30 A. M. Somebody decided they had had enough Fentley."

``What else?"

``Body temperature  at 3 A.M. was 85o and at 4 A.M. was 78o. Don't ask me how I got it. You don't want to know."

``All right, Sarge, so when did he die?"

``Don't ask me, Horns, you're the detective."

``Well then, let me give you a free lesson, and maybe you won't spend the rest of your career taking temperatures. If a body is cooling in a room with air temperature R then the rate of change of the temperature of the body, call it `T', is proportional to the difference between the body temperature and the room temperature, namely T - R."

``So the differential equation  governing temperature T says:


\begin{displaymath}\frac{dT}{dt} = k(T-R)\end{displaymath}

where R is the room temperature. Of course, this isn't the standard differential equation that governs exponential growth or decay."

``Of course."

``But as is done with the standard equation, we can separate the variables:

\begin{displaymath}\frac{dT}{T-R} = k dt\end{displaymath}

Integrating both sides, we get:


\begin{displaymath}\ln (T-R) = kt +C\end{displaymath}


T-R = ekt+C= ekt eC = ekt A,

where we replaced eC with Ajust because it is prettier."

"You always had an eye for the capital letters,"interjected Woffle. Horns ignored him.

``So,

T(t) = Aekt + R,

where R is the room temperature. In our case, we have $\displaystyle T(t) = Ae^{kt} + 70. $"

``I follow you."

``Sure you do. Now, we'll use the particular temperature readings that you took to determine both of the constants A and k. It's our choice as to when t=0, so let's choose t=0 when the first temperature reading was taken, at 3:00 A.M.

Then we know 85 = T(0) = Ae(k)0 + 70. So 85 = A+70 and A=15.

Therefore

T(t) = 15ekt + 70.

We also know 78 = T(1) = 15ek + 70. So 15ek = 8.


\begin{displaymath}e^{k} = \frac{8}{15}\end{displaymath}


\begin{displaymath}k= \ln (\frac{8}{15}) \approx -0.6286.\end{displaymath}

So our function governing the body temperature is:


T(t) =15e-0.6286t + 70."

``Assuming he didn't just step out of a sauna, Fentley's temperature was 98.6o at the instant he was murdered. So we set our temperature function equal to this and solve for t to determine exactly when he was murdered."


98.6 = 15e-0.6286t + 70


28.6 = 15e-0.6286t


\begin{displaymath}\frac{28.6}{15} = e^{-0.6286t}\end{displaymath}


\begin{displaymath}\ln {\frac{28.6}{15}} = -0.6286t\end{displaymath}

So,

\begin{displaymath}t = \frac{\ln {\frac{28.6}{15}}}{-0.6286} = -1.02 \text{ hours.}\end{displaymath}

``So what does all that mean?" said Woffle.

``It means," said Horns, ``that Hiram bit the big one around 2:00 A.M."

``Okay, Horns, that's all good, but how did he die?"

Horns leaned over, wedged his fingers in the dead man's mouth and yanked out a large piece of cheese. ``If I don't miss my guess," he said, ``Hiram suffocated to death on this."

``You mean he just choked on a piece of cheese he was eating?"

``No, Woffle, that's not what I mean. No person in their right mind, let alone a cheese baron, would put a hunk of cheese this big in his mouth."

``So, someone stuffed it in there? But it's too big. They could never get it in."

``Ah, but don't you see, Woffle, there is one way to get it in."

``What's that?"

``It was a simple twist of feta."


Well, after all this violence, slime, and nuclear holocaust, let's see if we can come up with a nice sweet growth or decay problem, without this dark side.

Problem (The Gardening Shop Problem.)   A lovely elderly woman named Adelaide has invested her life's savings in a gardening shop specializing in daisies. Unfortunately, due to a drought, the daisies do badly, and the shop is going to go under unless it has an infusion of cash. Adelaide has a choice of borrowing $5,000 from a loan shark at 23%  for one year, compounded  continuously, or $5,000 from her son at 24% for one year compounded quarterly. Assuming that her legs will be broken, in either case, if she doesn't pay up at the end of the year, which deal should she go with?

Look, we're sorry about the broken legs part, but we just want to get across this idea that growth and decay problems are about the slimy underbelly of life, the evil that lurks just beneath the surface of our bucolic everyday existence. Anyway, if we get her the better deal, maybe she'll be able to pay off the loan.


Solution: First, let's see what her son is charging her for one year. He wants 24% for the year, which is 6% per quarter. At the end of the first quarter, she will owe her son $5,000(1+.06). Since we are compounding quarterly, it is on this amount that we compute the interest for the next quarter. At the end of the second quarter, she will owe $5,000 (1+.06)2. The amount she owes at the end of the third quarter will be 1.06 times this amount,

5,000(1+.06)3.

And at the end of the fourth quarter, which is the end of the year, she will owe

\begin{displaymath}5,000(1+.06)^{4}= \$6,312.39.
\end{displaymath}

rounded up to the nearest penny. (Her son always rounds up.)


While we are at it, let's just mention that if you are investing Pdollars at a rate of interest of r (r expressed as a decimal) for a period of t years, compounded n times a year, then the amount of money that results is P(1 + r/n)nt. The argument is essentially the same as the one we used above.


Now, let's see what the loan shark is going to charge. Since the loan shark uses continuous compounding, the amount owed is

\begin{displaymath}5,000e^{.23(1)}= \$6293.00.
\end{displaymath}

So she gets the better deal from the loan shark. It's a good thing she knows calculus. (Just so you know, it all worked out in the end; after she created a catchy homepage on the Web, her flower shop became a major corporation, gobbling up the family-run flower businesses all over the country and eventually her son came to work for her where she humiliated him on a daily basis.)


General formula to remember: If P dollars is invested at an annual rate of interest  r for a period of t years, compounded  continuously, then the amount of money that results is

Pert.


next up previous contents
Next: Glossary: A Quick Guide Up: How to Ace Calculus Previous: Exponential Growth and Decay:
Joel Hass
1999-05-26