Exponential growth and decay (Part 9): Amortization tables

This post is inspired by one of the questions that I pose to our future high school math teachers during our Friday question-and-answer sessions. In these sessions, I play the role of a middle- or high-school student who’s asking a tricky question to his math teacher. Here’s the question:

A student asks, “My father bought a house for $200,000 at 12% interest. He told me that by the time he finishes paying for the house, it will have cost him more than $500,000. How is that possible? 12% of $200,000 is only $24,000.”

Without fail, these future teachers don’t have a good response to this question. Indeed, my experience is that most young adults (including college students) have never used an amortization table, which is the subject of today’s post.

In the past few posts, we have considered the solution of the following recurrence relation, which is often used to model the payment of a mortgage or of credit-card debt:

$A_{n+1} = r A_n - k$

With this difference equation, the rate at which the principal is reduced can be simply computed using Microsoft Excel. This tool is called an amortization schedule or an amortization table; see E-How for the instructions of how to build one. Here’s a sample Excel spreadsheet that I’ll be illustrating below: Amortization schedule. My personal experience is that many math majors have never seen such a spreadsheet, even though they are familiar with compound interest problems and certainly have the mathematical tools to understand this spreadsheet.

Here’s a screen capture from the spreadsheet:

The terms of the loan are typed into Cells B1 (length of loan, in years), B2 (annual percentage rate), and B3 (initial principal). Cell B4 is computed from this information using the Microsoft Excel command $\hbox{PMT}$ :

$=\hbox{PMT}(\hbox{B2}/12,\hbox{B1}*12,-\hbox{B3})$

This is the amount that must be paid every month in order to pay off the loan in the prescribed number of years. Of course, there is a formula for this:

$M = \displaystyle \frac{Pr}{12 \displaystyle \left[1 - \left( 1 + \frac{r}{12} \right)^{-12t} \right]}$

I won’t go into the derivation of this formula here, as it’s a bit complicated. Notice that this formula does not include escrow, points, closing costs, etc. This is strictly the amount of money that’s needed to pay down the principal.

The table, beginning in Row 8 of the above picture, shows how quickly the principal will be paid off. In row 8, the interest that’s paid for that month is computed by

$=\hbox{B8} * \$ \hbox{B}\$\hbox{2}/12$

Therefore, the amount of the monthly payment that actually goes toward paying down the principal will be

$= \$\hbox{B}\$\hbox{4} - \hbox{C8}$

Column E provides an opportunity to pay something extra each month; more on this later. So, after taking into account the payments in columns D and E, the amount remaining on the loan is recorded in Cell F8:

$= \hbox{B8} - \hbox{D8} - \hbox{E8}$

This amount is then copied into Cell B9, and then the pattern can be filled down.

The yellow graph shows how quickly the balance of the loan is paid off over the length of the loan. A picture is worth a thousand words: in the initial years of the loan, most of the payments are gobbled up by the interest, and so the principal is paid off slowly. Only in the latter years of the loan is the principal paid off quickly.

So, it stands to reason that any extra payments in the initial months and years of the loan can do wonders for paying off the loan quickly. For example, here’s a screenshot of what happens if an extra $200/month is paid only in the first 12 months of the loan:

A definite bend in the graph is evident in the initial 12 months until the normal payment is resumed in month 13. As a result of those extra payments, the curve now intersects the horizontal axis around 340. In other words, 20 fewer months are required to pay off the loan. Stated another way, the extra payments in the first year cost an extra $\$200 \times 12 = \$2400$ . However, in the long run, those payments saved about $\$536.82 \times 20 \approx \$10,700$ !

Exponential growth and decay (Part 8): Paying off credit-card debt via recurrence relations

The following problem in differential equations has a very practical application for anyone who has either (1) taken out a loan to buy a house or a car or (2) is trying to pay off credit card debt. To my surprise, most math majors haven’t thought through the obvious applications of exponential functions as a means of engaging their future students, even though it is directly pertinent to their lives (both the students’ and the teachers’).

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

In previous posts, I approached this problem using differential equations. There’s another way to approach this problem that avoids using calculus that, hypothetically, is within the grasp of talented Precalculus students. Instead of treating this problem as a differential equation, we instead treat it as a first-order difference equation (also called a recurrence relation):

$A_{n+1} = r A_n - k$

The idea is that the amount owed is multiplied by a factor $r$ (which is greater than 1), and from this product the amount paid is deducted. With this approach — and unlike the approach using calculus — the payment period would be each month and not per year. Therefore, we can write

$A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50$

Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months. The solution of this difference equation is

$A_n = r^n P - k \left( \displaystyle \frac{1 - r^n}{1-r} \right)$

green line

A great advantage of using a difference equation to solve this problem is that the solution can be easily checked with a simple spreadsheet. (Indeed, pedagogically, I would recommend showing a spreadsheet like this before doing any of the calculations of the previous few posts, so that students can begin to wrap their heads around the notion of a difference equation before the solution is presented.)

To start the spreadsheet, I wrote “Step” in Cell A1 and “Amount” in Cell B1. Then I entered the initial conditions: $0$ in Cell A2 and $2000$ in Cell B2. (In the screenshot below, I changed the format of column B to show dollars and cents.) Next, I entered $=\hbox{A2}+1$ in Cell A3 and

$=\hbox{B2}*(1+0.25/12)-50$

in Cell B3. Finally, I copied the pattern in Cells A3 and B3 downward. Here’s the result:

After the formula the algebraic solution of the difference equation has been found, this can be added to the spreadsheet in a different column. For example, I added the header “Predicted Amount” in Cell D1. In Cell D2, I typed the formula

$=2000*\hbox{POWER}(1+0.25/12,\hbox{A2})-50*(1-\hbox{POWER}(1+0.25/12,\hbox{A2}))/(1-(1+0.25/12))$

Finally, I copied this pattern down the Column D. Here’s the result:

Invariably, when I perform a demonstration like this in class, I elicit a reaction of “Whoa…. it actually works!” Even for a class of math majors. Naturally, I tease them about this… they didn’t believe me when I used algebra, but now it has to be true because the computer says so.

Here’s the spreadsheet that I used to make the above pictures: CreditCardDebt.

Exponential growth and decay (Part 7): Paying off credit-card debt via recurrence relations

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

$A_{n+1} = r A_n - k$

$A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50$

Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months.

green line

A full treatment of the solution of difference equations belongs to a proper course in discrete mathematics. In the previous posts, I demonstrated how this difference equation could be solved by directly finding $A_1, A_2, A_3, \dots$ and looking for a pattern.

In this post, I’d like to present an alternative method for deriving the solution. I’ll let the reader decide for him/herself as to whether this technique is pedagogically superior to the previous method. We will attempt to find a solution of the form

$A_n = a r^n + b$ ,

where $a$ and $b$ are unknown constants.Why do we guess the solution to have this form? I won’t dive into the details, but this is entirely analogous to constructing the characteristic equation of a linear differential equation with constant coefficients as well as using the method of undetermined coefficients to find a particular solution to a inhomogeneous linear differential equation with constant coefficients.

Substituting $n+1$ instead of $n$ , we find that

$A_{n+1} = a r^{n+1} + b$ .

So we plug both of these into the difference equation:

$A_{n+1} = r A_n - k$

$a r^{n+1} + b = r \left( a r^n + b \right) - k$

$a r^{n+1} + b = a r^{n+1} + r b - k$

$b = r b - k$

$k = (r-1) b$

$\displaystyle \frac{k}{r-1} = b$

We also use the fact that $A_0 = P$ :

$A_0 = a r^0 + b$

$P = a + b$

$P - b = a$

$\displaystyle P - \frac{k}{r-1} = a$

Combining these, we obtain the solution of the difference equation:

$A_n = \displaystyle \left( P - \frac{k}{r-1} \right) r^n +\frac{k}{r-1}$

Unsurprisingly, this matches the solution that was obtained in the previous two posts (though the terms have been rearranged).

Exponential growth and decay (Part 6): Paying off credit-card debt via recurrence relations

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

$A_{n+1} = r A_n - k$

$A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50$

Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months.

green line

In yesterday’s post, I demonstrated that the solution of this recurrence relation is

$A_n = r^n P - k \displaystyle \left( \frac{1 - r^n}{1-r} \right)$ .

Let’s now study when the credit card debt will actually reach $0. To do this, we see $A_n = 0$ and solve for $n$ :

$0 = r^n P - k \displaystyle \left( \frac{1 - r^n}{1-r} \right)$

$0 = r^n \left(P + \displaystyle \frac{k}{1-r} \right) - \displaystyle \frac{k}{1-r}$

$0 = r^n \left( P[1-r] + k \right) - k$

$k = r^n \left( P[1-r] + k \right)$

$\displaystyle \frac{k}{P[1-r] + k} = r^n$

$\displaystyle \ln \left( \frac{k}{P[1-r]+k} \right) = n \ln r$

$\displaystyle \frac{ \displaystyle \ln \left( \frac{k}{P[1-r]+k} \right) }{ \ln r} = n$

That’s certainly a mouthful. However, this calculation should be accessible to a talented student in Precalculus.

Let’s try it out for $k = 50$ , $P = 2000$ , and $r = 1 + \displaystyle \frac{0.25}{12}$ :

Remembering that each compounding period is one month long, this corresponds to $86.897/12 \approx 7.24$ years, which is nearly equal to the value of $4\ln 6 \approx 7.17$ years when we solved this problem using differential equations under the assumption of continuous compound interest (as opposed to interest that’s compoounded monthly).

Exponential growth and decay (Part 5): Paying off credit-card debt via recurrence relations

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

$A_{n+1} = r A_n - k$

$A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50$

Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months.

green line A full treatment of the solution of difference equations belongs to a proper course in discrete mathematics. However, this particular difference equation can be solved in a straightforward fashion that should be accessible to talented Precalculus students. Let’s use the above recurrence relation to try to find a pattern. For $n = 1$ , we find

$A_1 = r A_0 - k = r P - k$ .

For $n = 2$ , we find

$A_2 = r A_1 - k$

$A_2 = r (rP - k) - k$

$A_2 = r^2P - rk - k$

$A_2 = r^2 P - k (1 + r)$

For $n = 3$ , we find

$A_3 = r A_2 - k$

$A_3 = r \left[ r^2 P - k(1+r) \right] - k$

$A_3 = r^3 - rk(1+r) - k$

$A_3 = r^3 P - rk - r^2k - k$

$A_3 = r^2 P - k \left( 1 + r+r^2 \right)$

At this point, we can probably guess a pattern:

$A_n = r^n P - k \left( 1 + r + r^2 + \dots + r^{n-1} \right)$

Using the formula for a finite geometric series, this simplifies as

$A_n = r^n P - k \left( \displaystyle \frac{1 - r^n}{1-r} \right)$ .

Indeed, though I won’t do it here, this can be formally proven using mathematical induction.

Exponential growth and decay (Part 4): Paying off credit-card debt

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a relative rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

In the previous two posts, I presented the general formula

$A = \displaystyle \frac{k}{r} - \left( \frac{k}{r} - P \right) e^{rt}$

which can be obtained by solving a certain differential equation. So, if $r = 0.25$ , $k = 600$ , and $P = 2000$ , the amount left on the credit card after $t$ years is

$A(t) = 2400 - 400 e^{0.25t}$ .

On the other hand, if the debtor pays $1200 per year, the equation becomes

$A(t) = 4800 - 2800 e^{0.25t}$

Today, I’ll give some pedagogical thoughts about how this problem, and other similar problems inspired by financial considerations, could fit into a Precalculus course… and hopefully improve the financial literacy of high school students.

green line Under the theory that a picture is worth a thousand words, let’s take a look at the graphs of both of these functions:

Students should have no trouble distinguishing which curve is which. Clearly, by paying $1200 per year instead of $600 per year, the credit card debt is paid off considerably quicker.

There’s another immediate take-away from these graphs — especially the graph for $k = 600$ , when the debt is being paid off over 7 years. Notice that the debt is being paid off very slowly in the initial years. Only in the latter years does the pace of paying off the loan pick up. So the moral of the story is: if you can afford to pay extra in the early years of a debt (credit card, mortgage, etc.), it’s much more important to pay off an extra amount in the early years than in the later years.

I believe this to be an important lesson for students to learn before they bury themselves deeply in debt as young adults… and Precalculus provides a natural vehicle for teaching this lesson.

Exponential growth and decay (Part 3): Paying off credit-card debt

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a relative rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

In yesterday’s post, I showed that the answer to this question was about 7.2 years. To obtain this answer, I started with the differential equation

$\displaystyle \frac{dA}{dt} = 0.25 A - 600$

which, given the initial condition $A(0) = 2000$ , has solution

$A(t) = 2400 - 400 e^{0.25t}$ .

I’ve read many Precalculus books; not many of them include applying exponential functions to the paying off of credit-card debt (or a mortgage on a house or car). Of course, yesterday’s derivation was well above the comprehension level of students in Precalculus. However, there’s no reason why Precalculus students couldn’t be given the general formula

$A = \displaystyle \frac{k}{r} - \left( \frac{k}{r} - P \right) e^{rt}$ ,

where $P$ is the initial amount, $r$ is the relative rate of growth, and $k$ is the amount paid per year. In other words, students could be given the formula without the full explanation of where it comes from. After all, many Precalculus textbooks give the formula for Newton’s Law of Cooling (the subject of a future post) with neither derivation nor explanation (though its derivation is nearly identical to the work of yesterday’s post), So I don’t see why also giving students the above formula for paying off credit-card debt isn’t more common.

Plugging in $k = 600$ , $r = 0.25$ , and $P = 2000$ into this equation again yields the function

$A(t) = 2400 - 400 e^{0.25t}$ ,

from which we find that it will take $t = 4\ln 6 \approx 7.2$ years to pay off the debt.

A natural follow-up question is “How much money actually was spent to pay off this debt?” By this point, the answer is quite easy: the lender paid $\$600$ per year for $4\ln 6$ years, and so the amount spent is

$\$600 \times 4 \ln 6 = \$2400 \ln 6 \approx \$4300$ .

When I teach this topic in differential equations, I let that answer sink in for a while. The original debt was only \$2000, but ultimately \$4300 needs to be paid over 7.2 years in order to pay off the debt.

The natural question is, “Why did it take so long?” Of course, the answer is that the debtor only paid the minimal amount — $50 per month, or $600 per year. It stands to reason that if extra money was paid each month, then the debt will be paid off faster at lesser expense.

To give one example, let’s repeat the calculation if the debtor paid twice as much ($100 per month, or $1200 per year). Then the amount owed as a function of time would be

$A(t) = \displaystyle \frac{1200}{0.25} - \left( \frac{1200}{0.25} - 2000 \right) e^{0.25t} = 4800 - 2800 e^{0.25t}$

To find when the credit card will be paid off, we set $A(t) = 0$ :

$0 = 4800 - 2800 e^{0.25t}$

$2800 e^{0.25t} = 4800)$

$e^{0.25t} = \displaystyle \frac{12}{7}$

$0.25t = \displaystyle \ln \left( \frac{12}{7} \right)$

$t = \displaystyle 4 \ln \left( \frac{12}{7} \right)$

$t \approx 2.16$

That’s certainly a lot faster! Also, the amount that’s spent over that time is also considerably less:

$\displaystyle 1000 \times 4 \ln \left( \frac{12}{7} \right) = 4000 \ln \left( \frac{12}{7} \right) \approx \$2156$ .

So, along with being a good way to practice proficiency with exponential and logarithmic functions, this problem lends itself for students discovering some basic principles of financial literacy.

Exponential growth and decay (Part 2): Paying off credit-card debt

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a relative rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or $600 per year), how long will it take for the balance to be paid?

In this post, I present the actual solution of this problem. In tomorrow’s post, I’ll give some pedagogical thoughts about how this problem, and other similar problems inspired by financial considerations, could fit into a Precalculus course.

Let’s treat this problem as a differential equation (though it could also be considered as a first-order difference equation… more on that later). Let $A(t)$ be the amount of money on the credit card after $t$ years. Then there are two competing forces on the amount of money that will be owed in the future:

The effect of compound interest, which will increase the amount owed by $0.25 A(t)$ per year.
The amount that’s paid off each year, which will decrease the amount owed by $\$600$ per year.

Combining, we obtain the differential equation

$\displaystyle \frac{dA}{dt} = 0.25 A - 600$

There are a variety of techniques by which this differential equation can be solved. One technique is separation of variables, thus pretending that $dA/dt$ is actually a fraction. (In the derivation below, I will be a little sloppy with the arbitrary constant of integration for the sake of simplicity.)

$\displaystyle \frac{dA}{0.25 A - 600} = dt$

$\displaystyle \int \frac{dA}{0.25 A - 600} = \displaystyle \int dt$

$\displaystyle 4 \int \frac{0.25 dA}{0.25 A - 600} = \displaystyle \int dt$

$4 \ln |0.25A - 600| = t + C$

$\ln |0.25A - 600| = 0.25 t + C$

$|0.25A - 600| = e^{0.25 t + C}$

$|0.25 A - 600| = C e^{0.25t}$

$0.25A - 600 = C e^{0.25t}$

$0.25 A = 600 + C e^{0.25t}$

$A = 2400 + C e^{0.25t}$

To solve for the missing constant $C$ , we use the initial condition $A(0) = 2000$ :

$A(0) = 2400 + C e^0$

$2000 = 2400 + C$

$-400 = C$

We thus conclude that the amount of money owed after $t$ years is

$A(t) = 2400 - 400 e^{0.25t}$

To determine when the amount of the credit card will be reduced to $0, we see $A(t) = 0$ and solve for $t$ :

$0 = 2400 - 400 e^{0.25 t}$

$400 e^{0.25 t} = 2400$

$e^{0.25t} = 6$

$0.25t = \ln 6$

$t = 4 \ln 6$

$t \approx 7.2 \hbox{~years}$

In tomorrow’s post, I’ll give some pedagogical thoughts about this problem and similar problems.

Exponential growth and decay (Part 1): Phrasing of homework questions

I just completed a series of posts concerning the different definitions of the number $e$ . As part of this series, we considered the formula for continuous compound interest

$A = Pe^{rt}$

Indeed, this formula can be applied to other phenomena besides the accumulation of money. Unfortunately, as they appear in Precalculus textbooks, the wording of questions involving exponential growth or decay can be either really awkward or mathematically imprecise (or both). Here’s a sampling of problems that I’ve collected from various sources:

One thousand bacteria on a petri dish are placed in an incubator, encouraging a relative rate of growth of 10% per hour. How many bacteria will there be in two days?

This is mathematically precise, as it relates to the differential equation $A'(t) = r A(t)$ with solution $A = P e^{rt}$ . The meaning of the value of $r$ is clear from dimensional analysis: the units of $A'(t)$ are $\hbox{bacteria}/ \hbox{hour}$ , while the units of $A(t)$ are $\hbox{bacteria}$ . Therefore, the units of $r$ must be $\hbox{hour}^{-1}$ . So saying that there’s a “relative rate of growth of 10% per hour” makes total sense.

Of course, when Precalculus students are solving this problem, they have no idea about what a differential equation is, making the word relative seem superfluous to the problem.

A sum of $5000 is invested at an interest rate of 9% per year. Find the time required for the money to double if the interest is compounded continuously.

What the problem is trying to say is “Let $r = 0.09$ .” But this is a horrible way to write this in ordinary English! After all, if we plug $r = 0.09$ and $t = 1$ into the formula, we obtain

$A = P e^{0.09 \times 1} \approx 1.09417P$

So it would appear that the interest rate after one year is about 9.417%, and not 9%.

Indeed, if we read the problem at face value that the interest rate is 9% per year, then it stands to reason that, after one year, we have

$P(1.09) = P e^{r \cdot 1}$

$1.09 = e^r$

$\ln 1.09 = r$

In a nutshell, saying that there is “an interest rate of 9% per year” can easily be interpreted to mean that the annual percentage rate is 9% year, and this can be a conceptual barrier for literally-minded students.

I don’t have a good solution for this impasse between ordinary English and giving clear directions to students about what numbers should be used in the formula. But I do think that it’s important for teachers to be aware of this possible misunderstanding as students read their homework questions.

Engaging students: Solving exponential equations

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Isis Flores. Her topic, from Precalculus: solving exponential equations.

How could you as a teacher create and activity or project that involves your topic?

For students who have not seen exponential functions before the overall concept might be difficult to comprehend if there is not concrete example that they get to experience. In a mathematics exploration course, TNTX 3100, there was a unique and concrete experience which aided in grasping the concept of exponential functions. I believe that pre-calculus students would benefit just as much if not more from doing a similar activity. The activity itself is to model radioactive decay with m&m’s. Students would be given a set number of m&m’s in a cup. Students would then shake the cup and turn the contents out onto a plate. Those m&m’s with the “m” side up get to be eaten and the number record along with how many times the cup has been turned over (this represents years). Students will continue this pattern until they are out of m&m’s. Students will then take their recorded data and plot it in order to further analyze what is happening and try to come up with a mathematical model for the data. This activity is great in the sense that it involves something concrete, and edible, but also because students get to experiment and a bit of science is included in the process. To shake things up students should be in groups and each group should get to run the experiment at least twice. At the end of the unit it would be a neat idea to ask students to come up with their own representation of exponential functions and maybe try a few of them out as a class.

How can this topic be used in your students’ future courses in mathematics or science?

After students go through pre-calculus they might take courses which will require them to have some base knowledge of exponential functions. In calculus students will need said base knowledge in order to comprehend what occurs when taking derivatives of exponential functions. Students will also be exposed to “e” and having an understanding of exponential functions will aid them in comprehending what the mathematical definition of “e” is and to recognize its form. Students will also use exponential functions when analyzing interest rates and investments, which is something they may need when they at a later stage in their life (i.e. planning for retirement or calculating college loans). In science students will explore radioactive decay, half-life, and even capacitor discharge all of which will require them to have a good grasp on exponential functions. If students truly understand exponential functions not only will they be able to solve problems presented to them in their science courses, but it will give them an advantage towards actually comprehending what is happening and being able to visualize it, as in the case with capacitor discharge. Such comprehension which goes beyond computations ensures that students are truly learning, and not just mindlessly memorizing steps or formulas.

How has this topic appeared on the news?

A topic which has been on the news radar for a period of time is population growth, which behaves exponentially. It would be quite interesting to perhaps introduce the topic of exponential functions with a news article which speaks about the increase population growth, (see nytimes.com link under references) and have students attempt to model said growth. A more exciting news link, at least from a student perspective, was the Red Bull Stratos Jump. The jump was performed form 128,000 feet and was to be sort of an advertisement for the energy drink, Red Bull (which has the slogan “Red Bull gives you wings). Students can explore the exponential decay of atmospheric pressure vs. altitude and have a short clip of the jump be the engage for the lesson. This news topic will definitely interest students since it is not something that occurs a lot, and a few of them might have actually watched the live jump.

References:

TNTX 3100 course

http://gauss.vaniercollege.qc.ca/pwiki/index.php/The_Exponential_Function_and_Its_Applications_in_Science

http://www.nytimes.com/2014/03/27/nyregion/population-growth-in-new-york-city-is-reversing-decades-old-trend-estimates-show.html?_r=0

http://www.redbull.com/cs/Satellite/en_INT/Video/Exclusive-What-Felix-Saw-Red-Bull-Stratos-Live-Jump-POV-021243270932859