# Engaging students: Introducing the number e

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 Loc Nguyen. His topic, from Precalculus: introducing the number $e$.

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

To be able to understand where the number e is produced in the first place, students need to understand how compound interest is calculated.  Before introducing the number e, I will definitely create an activity for the students to work on so that they can eventually find the formula for compounding interest based on the patterns they produce throughout the process.  The compound interest formula is F=P(1+r/n)nt.  From this formula, I will again provide students a worksheet to work on.  In this worksheet, I will let P=1, r=100%, t=1, then the compound interest formula will be F=(1+1/n)n. Now students will compute the final value from yearly to secondly.

When they do all the computation, they will see all the decimal places of the final value lining up as n gets big.  And finally, they will see that the final value gets to the fixed value as n goes to infinity.  That number is e=2.71828162….,

How has this topic appeared in the news?

To help the students realize how important number e is, I would engage them with the real life examples or applications. There were some news that incorporated exponential curves. First, I will show the students the news about how fast deadly disease Ebola will grow through this link http://www.npr.org/sections/goatsandsoda/2014/09/18/349341606/why-the-math-of-the-ebola-epidemic-is-so-scary.  The students will eventually see how exponential curve comes into play. After that I will provide them this link, http://cleantechnica.com/2014/07/22/exponential-growth-global-solar-pv-production-installation/, in this link, the article talked about the global population rate and it provided the scientific evidence that showed the data collected represent the exponential curve.  Up to this point, I will show the students that the population growth model is:

Those examples above was about the growth.  For the next example, I will ask the students that how the scientists figured out the age of the earth.  In this link, http://earthsky.org/earth/how-old-is-the-earth, the students will learn that the scientists used Modern radiometric dating methods to calculate the age of earth.  At this time, I will show them radioactive decay formula and explain to them that this formula is used to determine the lives of the substances such as rocks:

How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic?

To introduce to the students what the number e is, I will engage them with two videos. In the first video, https://www.youtube.com/watch?v=UFgod5tmLYY, the math song “e a magic number” will engage the students why it is a magic number.  While watching this clip, the students will be able to learn the history of e.  Also the students will see many mathematical formulas and expressions that contain e.  This will give them a heads up that they will see these in future when they take higher level math.  It is also pretty humorous of how Dr. Chris Tisdell sang the song.

In the second video, https://www.youtube.com/watch?v=b-MZumdfbt8, it explained why e is everywhere.  The video used probability and exponential function to illustrate the usefulness of e, and showed how e is involving in everything.  It gave many examples of e such as population, finance…  Also the video illustrates the characteristics of the number e and the function that has e in it.  Watching these videos will enhance students’ perception and understanding on the number e, and help them to see how important this number is.

Reference

http://www.math.unt.edu/~baf0018/courses/handouts/exponentialnotes.pdf

http://cleantechnica.com/2014/07/22/exponential-growth-global-solar-pv-production-installation/

http://www.npr.org/sections/goatsandsoda/2014/09/18/349341606/why-the-math-of-the-ebola-epidemic-is-so-scary

http://earthsky.org/earth/how-old-is-the-earth

# Engaging students: Compound interest

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 Andy Nabors. His topic, from Precalculus: compound interest.

What interesting (i.e., uncontrived) word problems using this topic can your students do now?

I would give the students a problem to find out where they should invest their money. They would be given several options with pros and cons and need to choose the best option for them and explain their reasoning. The problem would go something like this:

You are looking to put your graduation money, a total of $2,498, into a savings account. You have gone to several banks and found the interest rates and start-up fees for making an account there. Which bank is offering you the best deal? Which would you choose and why?  Bank Interest Rate Compounded Start-Up Fee Bank of America 2.5% daily$65 CitiBank 5% monthly $100 Comerica 3% weekly$50 JP Morgan 1.7% continuously $50 Wells Fargo 3.3% bi-annually$0

This material is Pre-Cal, so I assume the students are either juniors or seniors, so they may be looking at having to open a bank account of their own in their near future. Then this would be a relevant question for them to look into and figure out what exactly gives them the best option.

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

This activity would be similar to one we did in 4050. At the time of this activity, they would not know the formula for compound interest. I would put the students in pairs and pose the question “Suppose you have $1000 that earns 8% interest. How much would you have at the end of 2 years if the interest was compounded: a.)annually b.) biannually c.)quarterly d.)daily”. Then the students would work in pairs to figure out the answers and I would instruct the students to find a pattern as they worked to make it easier. The students would eventually discover the formula for compound interest compounded for any number. They would then be asked how many times the money would have to be compounded to put out the highest total. The students would discover that the higher number, the more total, but as the compounded numbers increased, the difference between the outputs would decrease. So we could then say that there is a limit to how much the output could be, and that limit would be compounded infinitely. Then we could take the limit and find out what the formula is for finding compound interest compounded continuously. How can technology be used to effectively engage students with this topic? Students need to know how to do their own research in their future for things like buying a house or car, choosing whether or not to rent or buy, or other things where they are having to find the best deal and fit for them. This activity would have students researching different banks. They would be asked to find out the details on certain banks’ interest rates. They would need to find out about fees and how many times the interest is compounded. They would need information about at least three banks, and then would need to research on independent sites which bank would be the best to start an account with from the banks they chose. Then they would choose a bank for them based on their own findings and calculations, and would choose a bank based on what an online article said. This would let students form their own opinions based on data they found, and weigh that data against the opinions of others. Their findings and opinions may not match up, and that’s why this activity would benefit them. It’s important that students learn to not take the opinions of others as fact, but do their own research to find the best deal. # Engaging students: Compound interest 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 Daniel Littleton. His topic, from Precalculus: compound interest. How has this topic appeared in the news? In a publication entitled Business Insider, Sam Ro published an article entitled “Every 25 Year Old In America Should See This Chart” on March 21, 2014. In this article Ro stated that in past times companies would offer pension plans to long term employees in order to support them in retirement. He goes on to state that in modern times employees need to contribute to retirement plans such as a 401K or an IRA in order to save for retirement. These plans function by the mathematical principle of compound interest. While the mechanics of compound interest are not presented in the article an illustration is shown how individuals who save their money through this formula accumulate a greater amount of money over time. He even presents a situation in which one individual can save money for a less amount of time than another and still accrue a greater total amount of savings because of compound interest. This illustration, presented below, can be a useful tool in engaging students in the possibilities that compound interest could have in their own futures. This information was collected from the following web page on Friday, April 04, 2014; http://www.businessinsider.com/compound-interest-retirement-funds-2014-3. How can this topic be used in your students’ future courses in mathematics or science? Compound interest is introduced at the Pre-Calculus level of secondary education. At the Post-Secondary Education level compound interest is a concept that is included in several areas of study. For example, students that wish to study business will need to have a mastery of compound interest. Additionally, those studying finance or economics will constantly use the principle of compound interest in their computations. Not only does this formula come into play in the mathematics of monetary systems, but also in the workings of political science as well. Those that wish to pursue political aspirations will need a firm understanding of economics and the means by which funds can be grown over time. As is evident, compound interest is a mathematical formula, but like many realms of mathematics it affects multiple realms of interest and practice in a real world environment. What interesting word problems using this topic can your students do now? There are an innumerable amount of problems that can be presented to students involving compound interest. One could deal with the monetary worth of valuable or precious items. For instance, “A necklace is appraised at$7200. If the value of the necklace has increased at an annual rate of 7.2%, how much was it worth 15 years ago?” This question is asking the student to solve for the original principle of the necklace, rather than the accrued value which is given. Another problem could be “A sum of $7000 is invested at an interest rate of 7% per year. Find the time required for the money to double if the interest is compounded quarterly.” This problem requires the student to determine the amount of time necessary for the investment to yield the desired amount. These are only two problems that I have presented that will allow the students to practice the concept of compound interest. There are undoubtedly multiple others that could be written with the same effect. # Compound Interest and Mary Poppins What would have happened if Michael Banks had invested his tuppence in a savings account? FiveThirtyEight has the answer. # 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 fi nishes 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 $A_n = r^n P - k \left( \displaystyle \frac{1 - r^n}{1-r} \right)$ 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 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.

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

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. 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

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. 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 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 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. 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.