Exponential growth and decay (Part 12): 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. 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

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

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.

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.

Exponential growth and decay (Part 10): Half-life

While these formulas are easy to state, not many high school teachers are aware of the physical principles from which they arise. The basic idea is that if amount $A$ of a radioactive substance (carbon-14, uranium-235, brain cells) is present, the rate at which the substance decays is proportional to the amount of the substance currently present. This can be rewritten as a differential equation, since the rate at which the substance decays is $dA/dt$ . So we find that

$\displaystyle \frac{dA}{dt} = - k A$

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):

$\displaystyle \frac{dA}{A} = -k$

$\displaystyle \int \frac{dA}{A} = - \displaystyle \int k \, dt$

$ln |A| = -k t + C$

$|A| = e^{-kt+C}$

$|A| = e^{-kt} e^C$

$A = \pm e^C e^{-kt}$

$A = C e^{-kt}$

To solve for the constant, we usually use the initial condition $A(0) = A_0$ , a number that must be given in the problem:

$A(0) = C e^{-k \cdot 0}$

$A_0 = C \cdot 1$

$A_0 = C$

Plugging back in, we obtain the final answer

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

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.

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

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

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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.

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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.

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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.