Preparation for Industrial Careers in the Mathematical Sciences: Building a Better Filter

The Mathematical Association of America recently published a number of promotional videos showing various mathematics can be used in “the real world.” Here’s the third pair of videos describing how mathematics is used for certain problems in materials science. From the YouTube descriptions:

Dr. Sumanth Swaminathan of W. L. Gore & Associates talks about his career path and the research questions about filtration that he considers. He works to understand the different waste capture mechanisms of filtration devices and to mathematically optimize the microstructure to create better filters.

Prof. Louis Rossi of the Department of Mathematical Sciences of the University of Delaware presents two introductory mathematical models that one can use to understand and characterize filters and the filtration processes.

Preparation for Industrial Careers in the Mathematical Sciences: Improving Market Strategies

The Mathematical Association of America recently published a number of promotional videos showing various mathematics can be used in “the real world.” Here’s the fourth pair of videos describing how mathematics is used in the world of finance. From the YouTube descriptions:

Dr. Jonathan Adler (winner of King of the Nerds Season 3) talks about his career path and about a specific research problem that he has worked on. Using text analytics he was able to help an online company distinguish between its business customers and its private consumers from gift card messages.

Prof. Talithia Williams of Harvey Mudd College explains the statistical techniques that can be used to classify customers of a company using the messages on their gift cards.

Preparation for Industrial Careers in the Mathematical Sciences: Creating More Realistic Animation for Movies

The Mathematical Association of America recently published a number of promotional videos showing various mathematics can be used in “the real world.” Here’s the first pair of videos describing how mathematics is used for computer animation. From the YouTube descriptions:

Dr. Alex McAdams, Senior Software Engineer at Walt Disney Animation Studios, talks about how mathematics is used to make realistic, yet art directable, animations.

Prof. Joseph Teran of the Department of Mathematics at UCLA gives an overview of the numerical linear algebra and iterative method techniques that are used to simulate physical phenomena such as water, fire, smoke, and elastic deformations in the movie and gaming industries.

Symphonic Equations: Waves and Tubes

This excellent and engaging video describes how sine and cosine functions can be applied to music.

Exponential growth and decay (Part 16): Logistic growth model

Today, I discuss the logistic growth model, which describes how an infection (like a disease, a rumor, or advertise) spreads in a population. In yesterday’s post, I described an in-class demonstration that engages students while also making the following formula believable:

$A(t) = \displaystyle \frac{Ly_0}{y_0+ (L-y_0)e^{-rt}}$ .

I’d like to discuss some observations about this somewhat complicated function that will make producing its graph easier. The first two observations are within reach of Precalculus students.

1. Let’s figure out the $y-$ intercept:

$A(0) = \displaystyle \frac{Ly_0}{y_0+ (L-y_0)e^{-r \cdot 0}} = \displaystyle \frac{Ly_0}{y_0+ L-y_0} = y_0$ .

In other words, the number $y_0$ represents the initial number of people who have the infection.

2. Let’s figure out the limiting value as $t$ gets large:

$\displaystyle \lim_{t \to \infty} A(t) = \displaystyle \frac{Ly_0}{y_0+ (L-y_0) \cdot 0} = \displaystyle \frac{Ly_0}{y_0} = L$ .

As expected, all $L$ people will get the infection eventually. (Of course, Precalculus students won’t be familiar with the $\displaystyle \lim$ notation, but they should understand that $e^{-rt}$ decays to zero as $t$ gets large.

3. Let’s now figure out the point of inflection. Ordinarily, points of inflection are found by setting the second derivative equal to zero. Though this can be done for the function $A(t)$ above, it would be a somewhat daunting exercise!

The good news is that the points of inflection can be found quite simply using the governing differential equation, which is

$A' = r A [ L - A] = r L A - r A^2$

Let’s take the derivative of both sides, remembering that $r$ and $L$ are constants:

$A'' = r L A' - 2 r A A'$

$A'' = A' (r L - 2 r A)$

So the second derivative is equal to zero when either $A' = 0$ or else $r L - 2 r A = 0$ . The first case corresponds to the trivial cases $A(t) \equiv 0$ and $A(t) \equiv L$ ; these constants are called the equilibrium solutions. The second case is the more interesting one:

$r L - 2 r A = 0$

$r L = 2 r A$

$\displaystyle \frac{L}{2} = A$

This suggests that, as the infection spreads throughout a population, the curve changes concavity at the time that half of the population becomes infected. In other words, the infection spreads fastest throughout the population at the time when half of the population has been infected.

The time at which the point of inflection occurs can be found by setting $A(t) = \displaystyle \frac{L}{2}$ and solving for $t$ :

$\displaystyle \frac{L}{2} = \displaystyle \frac{Ly_0}{y_0+ (L-y_0)e^{-rt}}$ .

$\displaystyle \frac{1}{2} = \displaystyle \frac{y_0}{y_0+ (L-y_0)e^{-rt}}$ .

$y_0 + (L-y_0) e^{-rt} = 2y_0$

$(L-y_0) e^{-rt} = y_0$

$e^{-rt} = \displaystyle \frac{y_0}{L-y_0}$

$-rt = \displaystyle \ln \left( \frac{y_0}{L-y_0} \right)$

$t = \displaystyle - \frac{1}{r} \ln \left( \frac{y_0}{L-y_0} \right)$

This technique for finding the points of inflection directly from the differential equation is possible whenever the differential equation is autonomous, which loosely means that the independent variable does not appear on the right-hand side.

Exponential growth and decay (Part 15): Logistic growth model

Today, I introduce the logistic growth model, which describes how an infection (like a disease, a rumor, or advertise) spreads in a population. For example:

Sources: http://www.xkcd.com/1220/ and http://www.xkcd.com/1235/

In yesterday’s post, I described an in-class demonstration that engages students while also making the following formula believable:

$A(t) = \displaystyle \frac{L}{1 + (L-1)e^{-rt}}$ .

Where does this formula come from? Suppose that a disease is spreading in a population of size $L$ . It stands to reason that the rate at which the disease spreads is proportional to the number of possible contacts between those who have the disease and those who don’t. If $A(t)$ is the number of people who have the disease, then $L-A(t)$ is the number of people who don’t have the disease. Therefore, the product $A(t) [ L - A(t) ]$ is the number of possible contacts between those who have the disease and those who don’t. This leads to the governing differential equation

$A'(t) = c A(t) [ L - A(t) ]$ ,

where $c$ is the constant of proportionality. This is often rewritten by letting $c = \displaystyle \frac{r}{L}$ , or $r = cL$ :

$A'(t) = \displaystyle \frac{r}{L} A(t) [ L - A(t) ]$

$A'(t) = r A(t) \displaystyle \left[1 - \frac{A(t)}{L} \right]$

The good news is that this differential equation can be solved using separation of variables, just like the governing differential equations for continuous compound interest, paying off credit card debt, radioactive decay, and Newton’s Law of Cooling. The bad news is that it’s a lot harder to calculate the required integrals! After all, the right-hand side, after distributing, has a term containing $A^2$ , which makes this differential equation non-linear.

Solving this differential equation is a bit tedious, and I don’t feel particularly obligated to re-invent the wheel since it can be found several places on the Internet. Suffice it to say that integration by partial fractions and some very tricky algebra is necessary to solve for $A(t)$ and obtain the solution above. Among several different sources (which likely use different letters than the ones I’m using here):

Wikipedia: http://en.wikipedia.org/wiki/Logistic_function#In_ecology:_modeling_population_growth

MathWorld: http://mathworld.wolfram.com/LogisticEquation.html

Exponential growth and decay (Part 14): Logistic growth model

Today, I introduce the logistic growth model, which describes how an infection (like a disease, a rumor, or advertise) spreads in a population. Before I actually present the formula to my students, I usually perform a 8- to 10-minute demonstration to convince students that the formula actually works. This demonstration works well with between 15 and 45 students; I have personally not attempted this demo with a class larger than 45.

I wish I could take credit for the idea behind this demonstration. I’m afraid I can’t remember who told me the idea behind this demo about 15 years ago, but I’m thankful to him or her for this idea, as I’ve used it with great success over the years when teaching Precalculus and even when teaching Differential Equations.

Here’s the demonstration:

1. The class period before the demo, I ask my students to bring their calculators to class.

2. On the day of the demo, I prepare slips of paper with the numbers 1, 2, 3, etc. I hand these to my students as they take their seats before class starts (and, as needed, to students who arrive late).

3. I tell the class that we’re going to model how a rumor gets spread. On the chalkboard, I write down the numbers 0, 1, 2, …, up to $N$ , the number of students in the class that day. Invariably, I get asked, “What’s the rumor?” In response, I’ll playfully point to someone in the front row and say, “The rumor is about him.”

4. I point out that, at time 0, only one person has heard the rumor…. me. I’m person number 0 (confirming the popular belief of my students). So I’ll cross out the 0 on the board and mark on a table that only one person has heard the rumor so far. (Here’s the spreadsheet that I’ve used to keep track of this information while simultaneously making a graph of the data: logisitic).

5. I begin to spread the rumor. To spread the rumor, I use my calculator to get a random number between $0$ and $N$ . This can be done by just using the built-in random number generator found on many calculators and then multiplying by $N+1$ . (After all, there are $L= N+1$ people in the room: $N$ students plus one instructor.) The part after the decimal point is not important; the number before the decimal point represents the next person to hear the rumor.

For example, in the figure below, I would tell that person #35 of my class of 37 students was the next to hear the rumor. (If my random number is 0, I’ll privately cheat and get until I get a random number other than 0. I only permit the possibility of cheating on the first step so that the data fits the predicted curve as accurately as possible.)

6. At this point, I’ll X out the number of the next person to hear the rumor (in this case, 35), and I then ask how many people have heard the rumor. Obviously, two people have heard the rumor. So I’ll note on the table that two people have heard the rumor after one step.

7. Now we repeat the process. I get a new random number, and I ask the first student to pull out his/her calculator to get a random number too. But there’s an important rule: if you get a number that’s already been called, that’s OK. This models what really happens when a rumor (or disease) spreads in a population — it’s perfectly possible to hear the rumor twice.

8. We repeat the process — X’ing out numbers that have been previously called and students calling out the next person to hear the rumor — until the entire class hears the rumor. At some point, it becomes easiest to ask students to only call out if they get a number that hasn’t been called yet. Invariably, there’s always one person at the end who hasn’t heard the rumor yet, and this student is often the subject of some good-natured ribbing. Eventually, a chart like the following is produced.

9. Students immediately see that this is a different type of function than pure exponential growth. It actually does start off looking like exponential growth, but at some point the curve levels off. This makes sense because there’s a limiting value of $L=N+1$ (in this case, 38), which can’t happen for a model like $A = A_0 e^{rt}$

10. The punchline is that the spreadsheet secretly computes the actual curve predicted by the logistic growth model. The numbers are actually located in column C, which is conveniently hidden beneath the graph. The function is

$A = \displaystyle \frac{L}{1 + (L-1)e^{-rt}}$ ,

where $r = \ln 2$ . (Had each person the rumor to two different people at each step, then $r$ would have been equal to $r = \ln 3$ .) Here’s the graph, superimposed upon the data collected from class. I can do this pretty quickly in class because the curve is actually already drawn in the figure above… but it’s drawn in gray, the same color as the background. By changing the color to black, the graph becomes clear:

I never expect the curve to exactly fit the data, but it should come pretty close. After this fairly dramatic revelation, my students are completely sold that the mathematics that I’m about to show them actually works.

Exponential growth and decay (Part 13): Newton’s Law of Cooling

In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. In the previous posts in this series, I considered financial applications. In today’s post, I’ll discuss Newton’s Law of Cooling, which describes how quickly a hot object cools in a room at constant temperature. (This law is not to be confused with Newton’s Three Laws of Motion.)

While Newton’s Law of Cooling is easy to state, not many high school teachers are aware of the physical principles from which they arise. The basic idea is that the rate at which a hot object cools is proportional to the difference between its current temperature and the surrounding temperature. After solving the appropriate differential equation, the temperature $T(t)$ of the object is found to be

$T = S + (T_0 - S) e^{-kt}$ ,

where $t$ is the time, $S$ is the constant surrounding temperature, and $k$ is a constant that depends on the object.

Of course, students in Algebra II or Precalculus (or high school physics) are usually not ready to understand this derivation using calculus. Instead, they are typically given the final formula and are expected to use this formula to solve problems. Still, I think it’s important for the teacher of Algebra II or Precalculus to be aware of how the origins of this formula, as it only requires mathematics that’s only a year or two away in these students’ mathematical development.

This is the third application of exponential functions considered in this series; the previous two were continuous compound interest and radioactive decay. Unlike these previous two applications, Newton’s Law of Cooling can actually be demonstrated in class to engage students. All that’s required is the appropriate classroom technology and a standard-issue temperature probe.This stands in sharp contrast to the previous applications of exponential functions considered in this series. While students can easily envision making money via compound interest, no one will actually give them the money during class. And certainly I don’t encourage performing a real demonstration of radioactive decay with, say uranium-235, during class time! (There are ways of simulating radioactive decay using M&Ms or other manipulatives, however.)

A simple Google search yields thousands of webpages describes multiple classroom activities for Newton’s Law of Cooling. Some activities merely require collecting data and performing a regression fit to an exponential curve; such an activity would be appropriate for middle-school students. Other activities are more explicit about using Newton’s Law of Cooling. Here’s a sampling:

These links are aimed at a students at a variety of levels. Indeed, it’s possible to use a graphing calculator to plot the numerical derivative $T'(t)$ as a function of time, use linear regression to solve for the constant $k$ , and then produce the exponential curve using this value of $k$ . Several years ago, I saw an effective demonstration of this idea at the Joint Mathematics Meetings in which the presenters covered these aspects of Newton’s Law of Cooling in less than 10 minutes. (Naturally, additional time is needed when students perform these activities for themselves.)

Exponential growth and decay (Part 12): Newton’s Law of Cooling

While Newton’s Law of Cooling is easy to state, not many high school teachers are aware of the physical principles from which they arise. The basic idea is that the rate at which a hot object cools is proportional to the difference between its current temperature and the surrounding temperature. In other words, in a room at a temperature of $70^\circ$F, a very hot object at $270^\circ$F will cool twice as fast than when its temperature drops to $170^\circ$F.

The above paragraph can be rewritten as a differential equation. Let $T(t)$ be the temperature of the object at time $t$ , and let $S$ be the (constant) surrounding temperature that surrounds the object. Since the rate at which the substance decays is $dT/dt$ , we find that

$\displaystyle \frac{dT}{dt} = - k (T - S)$ ,

where $k$ is a constant that depends on the object. The negative sign on the right-hand side isn’t strictly necessary, but it’s a reminder that amount present decreases as time increases.

This differential equation can be solved in several ways, including separation of variables (below, I’ll be sloppy with the constant of integration for the sake of simplicity):

$\displaystyle \frac{dT}{T-S} = -k$

$\displaystyle \int \frac{dT}{T-S} = - \displaystyle \int k \, dt$

$ln |T-S| = -k t + C$

$|T-S| = e^{-kt+C}$

$|T-S| = e^{-kt} e^C$

$T-S = \pm e^C e^{-kt}$

$T-S = C e^{-kt}$

$T = S + C e^{-kt}$

To solve for the constant, we usually use the initial condition $T(0) = T_0$ , a number that must be given in the problem:

$T(0) = S + C e^{-k \cdot 0}$

$T_0 =S + C \cdot 1$

$T_0 - S = C$

Plugging back in, we obtain the final answer

$T = S + (T_0 - S) e^{-kt}$

The careful reader will notice that this derivation essentially parallels the previous derivations for continuous compound interest and for radioactive decay. In other words, these phenomena from apparently different realms of life have similar solutions because the governing differential equations are very similar.

Exponential growth and decay (Part 11): Half-life

One way of writing the formula for how a radioactive substance (carbon-14, uranium-235, brain cells) decays is

$A(t) = A_0 e^{-kt}$

In yesterday’s post, I showed how this formula is a natural consequence of a certain differential equation. Of course, students in Algebra II or Precalculus (or high school chemistry) are usually not ready to understand this derivation using calculus. Instead, they are typically given the final formula and are expected to use this formula to solve problems. Still, I think it’s important for the teacher of Algebra II or Precalculus to be aware of how the origins of this formula, as it only requires mathematics that’s only a year or two away in these students’ mathematical development.

There is another correct way to write this formula in terms of half-life. Sadly, my experience is that many students are familiar with these two different formulas but are not aware of how the two formulas are connected. As we’ll see, while yesterday’s post using differential equations is inaccessible to Algebra II and Precalculus students, the derivation below is entirely elementary and can be understood by good students in these courses.

Let $h$ be the half-life of the substance. By definition, this means that

$A(h) = \displaystyle \frac{1}{2} A_0$

Substituting into the above formula, we find

$\displaystyle \frac{1}{2} A_0 = A_0 e^{-kh}$

Let’s now solve for the constant $k$ in terms of $h$ :

$\displaystyle \frac{1}{2} = e^{-kh}$

$\displaystyle \ln \left( \frac{1}{2} \right) = - k h$

$\displaystyle -\frac{1}{h} \ln \left( \frac{1}{2} \right) = k$

Let’s now substitute this back into the original formula:

$A = A_0 e^{-kt}$

$A = A_0 e^{ -\left[ -\frac{1}{h} \ln \left( \frac{1}{2} \right) \right] t}$

$A = A_0 e^{\ln \left( \frac{1}{2} \right) \cdot \frac{t}{h}}$

$A = A_0 \left[e^{\ln \left( \frac{1}{2} \right)} \right]^{t/h}$

$A = A_0 \displaystyle \left( \frac{1}{2} \right)^{t/h}$

This is the usual half-life formula, where the previous base of $e$ has been replaced by a new base of $\displaystyle \frac{1}{2}$ . For most applications, a base of $e$ is preferred. However, for historical reasons, the base of $\displaystyle \frac{1}{2}$ is preferred for problems involving radioactive decay. For example,

$A(2h) = A_0 \displaystyle \left( \frac{1}{2} \right)^{2h/h}$

$A(2h) = A_0 \displaystyle \left( \frac{1}{2} \right)^{2}$

$A(2h) = \displaystyle \frac{1}{4} A_0$

In other words, after two half-life periods, only one-fourth (half of a half) of the substance remains.