Exponential growth and decay (Part 2): 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 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.

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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:

  1. The effect of compound interest, which will increase the amount owed by 0.25 A(t) per year.
  2. 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}

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

Leave a comment

2 Comments

  1. Stephen

     /  May 18, 2016

    My only issues is that loan interest is not compounded continuously, isn’t it compounded daily? At least in all the credit cards i’ve used the Terms state a daily accrual rate or the daily periodic rate…calculated…oh wait this is a math forum we know how to do that part lol.

    Reply
  1. Exponential growth and decay: Index | Mean Green Math

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