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Exponential Growth and Decay
Calculus 2 · Chapter 5: Differential EquationsSection 5.4 · lesson 25 of 5412 min · free
← Separable Differential EquationsNext lesson: The Logistic Model →
In this lesson
One model, four applications.
- Solve \frac{dy}{dt}=ky and interpret the sign of k
- Convert doubling time and half-life into a rate constant
- Apply the law to populations, decay, interest and cooling
- Extract k from data before predicting anything
Transcript
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MAT-265 - Calculus 2 · Section 5.4
Exponential Growth and Decay
Four situations. A bacterial culture multiplying. A radioactive sample decaying. A bank balance compounding. A cup of coffee cooling. They seem to have nothing in common. But every one of them is governed by the same differential equation. d y d t equals k y. The rate of change is proportional to the amount present. And we already know how to solve that. It's separable — section five point three handled it. So this section is not really about new mathematics. It's about recognizing one model in four different disguises.
The Law
One Formula Behind All of It
Here's the result, and it's worth committing to memory. The unique solution of the initial-value problem d y d t equals k y, with y at zero equal to y naught, is y of t equals y naught, e to the k t. Here y naught is the initial amount, and k is the proportionality constant. The rate constant. A positive k gives growth. A negative k gives decay. Every single application in this section is a special case of that one formula. Look at the picture. Both curves start at the same y naught. The green one has positive k, and climbs without bound. The red one has negative k, and decays toward zero — approaching the axis but never reaching it. One model. Two completely different behaviors. And the only thing that differs is the sign of k.
Application 1
Population Growth and Doubling Time
First application. Population growth. When resources are unlimited, a population reproduces at a rate proportional to its size. So k is positive. The model predicts the characteristic explosive increase of unchecked growth. Now, that positive rate constant is often re-expressed in a more intuitive form. The doubling time. That's the time required for the population to double. Set y equal to two y naught. You get two equals e to the k T sub d. So the doubling time is log two, over k. And here's why that matters conceptually. A population with a fixed percentage growth rate doubles on a fixed schedule — regardless of its current size. That's what makes exponential growth so deceptively rapid. The doubling doesn't slow down as the numbers get large. Let's do one. A bacterial culture starts with five hundred cells, and grows to two thousand after three hours. Find the population after six hours. Here y naught is five hundred. First find k, using the data point at t equals three. Two thousand equals five hundred, e to the three k. So e to the three k is four. Three k is log four. And k is log four over three. About zero point four six two. Now evaluate at t equals six. y at six is five hundred, e to the six k. Which is five hundred times four squared. Eight thousand cells. And notice the shortcut hiding in there. The population quadrupled in three hours. So it quadruples again in the next three hours. Two thousand, to eight thousand. No calculation needed.
Application 2
Radioactive Decay and Half-Life
Second application. Radioactive decay. Radioactive atoms disintegrate at a rate proportional to the number present. So here k is negative. The most natural way to describe the rate is the half-life, T. The time for half the material to decay. And half-life is the standard language in physics and chemistry for a precise reason. It's independent of the sample size. A gram and a kilogram of the same isotope each lose half their mass in exactly the same time. That's what makes radioactive decay the basis of dating methods across archaeology and geology — measuring the surviving fraction of an isotope reveals how much time has elapsed. To link T and k, set y at T equal to half of y naught. You get e to the k T equals one half. So k equals log one half, over T. Which is minus log two, over T. Let's date something. Carbon-fourteen has a half-life of about five thousand seven hundred and thirty years. An artifact retains twenty-five percent of its original carbon-fourteen. Estimate its age. First find k from the half-life. Minus log two over fifty-seven thirty. About minus one point two one times ten to the minus four, per year. The condition is that y of t equals zero point two five y naught. So e to the k t equals one quarter. Take the natural log. k t equals log one quarter, which is minus two log two. Solve for t, and the log twos cancel beautifully. You get two times fifty-seven thirty. Eleven thousand four hundred and sixty years. And that makes sense — exactly two half-lives. Because twenty-five percent is half of a half.
Applications 3 and 4
Interest, and Newton's Law of Cooling
Third application. Continuously compounded interest. An investment earning interest at an annual rate r, compounded continuously, grows so that its rate of increase is proportional to the current balance. So here k equals r, and it's positive. The balance follows A of t equals A naught, e to the r t. Two thousand dollars invested at five percent for ten years gives two thousand, e to the zero point five. About three thousand two hundred ninety-seven dollars and forty-four cents. Continuous compounding is an idealized limit — in reality interest is credited at discrete intervals, monthly or daily. But as those intervals shrink toward zero, the discrete compound-interest formula converges to the exponential model. Which makes A naught e to the r t both a clean approximation for frequent compounding, and the theoretical ceiling on how fast a given rate can grow a balance. Fourth application. Newton's Law of Cooling. An object's temperature changes at a rate proportional to the difference between its temperature and the surrounding ambient temperature. The same law governs warming — a cold drink in a warm room heats by the identical mechanism. Now, this is not quite d y d t equals k y. But a simple substitution converts it. The law says d T d t equals k, times T minus T sub s. Let y be T minus T sub s — the temperature difference. Since T sub s is constant, d y d t equals d T d t. And the equation becomes the familiar d y d t equals k y. Translating back, T of t equals T sub s, plus the quantity T naught minus T sub s, times e to the k t. That substitution trick — shifting by a constant to recover the standard model — is widely useful. Let's use it. Coffee at ninety degrees Celsius is placed in a twenty degree room. After five minutes it has cooled to sixty. Find its temperature after ten minutes. T sub s is twenty. T naught is ninety. So the initial difference is seventy. The model is T of t equals twenty, plus seventy e to the k t. Use the five-minute reading. Sixty equals twenty plus seventy e to the five k. So e to the five k is four sevenths. Now evaluate at ten minutes. e to the ten k is e to the five k, squared. So four sevenths, squared. Sixteen forty-ninths. T at ten is twenty plus seventy times sixteen forty-ninths. Which is twenty plus one sixty over seven. About forty-two point nine degrees. And notice the long-run behavior. As t goes to infinity, the exponential term vanishes, and T approaches twenty degrees. Room temperature. Exactly as physical intuition demands.
Reference
One Model, Four Applications
Let's put all four in one table, because the pattern is the point. Unlimited population growth. The quantity is the number of individuals. k is positive. Formula: y naught e to the k t. Radioactive decay. The quantity is mass or atom count. k is negative, and equals minus log two over the half-life. Compound interest. The quantity is the account balance. k equals r, positive. A naught e to the r t. Newton cooling. The quantity is the temperature difference, T minus T sub s. k is negative. The mathematics never changes. Only the interpretation does. And that's the genuine power of the model. Rather than learning four separate formulas, you learn one and re-label its symbols. There's also a recurring two-step pattern worth naming. First, use a known data point to solve for the rate constant k. Then use that k to predict the quantity at another time. Almost every applied problem in this section is some version of those two steps. Now, something to think about. A striking feature of exponential decay is that the half-life is constant. It always takes the same amount of time for the quantity to halve, no matter how much is present. A sample takes fifty-seven thirty years to drop from one hundred percent to fifty. Another fifty-seven thirty to drop from fifty to twenty-five. Another to reach twelve point five. Why does that constant halving time follow directly from the form y equals y naught e to the k t? And why would it not hold for a quantity that decreased by a fixed amount each year, rather than a fixed fraction?
Key Takeaways
The Errors Are Never in the Formula
The errors in this section are rarely in the formula itself. They're almost always in extracting k from the given data. A frequent mistake is treating a stated percentage rate, or a half-life, as if it were k directly. The half-life is not k. It must be converted, by k equals minus log two over T. Similarly, a quantity that triples every four years does not have k equal to three. The correct k comes from solving three equals e to the four k. Giving k equals one quarter log three. A second common slip is a sign error. Forgetting that decay requires k negative — which then produces a growing answer for a shrinking quantity. The safeguard is simple. Always solve for k explicitly from a known data point before predicting anything. And sanity-check the sign against whether the quantity should be rising or falling. One final caution. The unlimited-growth model is an idealization. Real populations eventually hit resource limits. So — five takeaways. The law is y equals y naught e to the k t, the unique solution of d y d t equals k y. And the sign of k decides growth or decay. Doubling time is log two over k. Half-life gives k equals minus log two over T. Neither one is k itself. Always solve for k explicitly from a known data point before predicting anything. Newton's cooling becomes the standard model after substituting y equals T minus T sub s. Shifting by a constant is a widely useful trick. And the unlimited-growth model is an idealization — real populations hit resource limits. Which is exactly what section five point five addresses. Next up: The Logistic Model.
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About Exponential Growth and Decay
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