Topic 7.8: Exponential Models with Differential Equations

Proportional to itself

Think of a population, where twice as many people means twice as many births each year. In general, when the rate of change of a quantity is proportional to the size of the quantity, we get the most important differential equation in this unit. In symbols, that’s dy dt equals k y, so the more there is, the faster it changes.

Let’s run the four steps. We divide by y and multiply by dt, so each side has just one variable. Since we divided by y, notice that y equals zero is a solution too, because if there’s none of it, it never changes. Integrating, one over y gives the natural log of the absolute value of y, and the right side gives k t plus C.

To undo the log, raise e to both sides. Then e to the k t plus C splits into e to the C times e to the k t, a product. Writing it as a sum instead is a classic slip, so I’ll cross that one out. Dropping the absolute value puts a plus or minus in front, and plus or minus e to the C is just a constant, so let’s call it A.

The plus C is already in, so we can use the point now. At t equals zero, e to the zero is one, so A is the starting amount, y naught. That gives the formula, y equals y naught e to the k t. Positive k means growth and negative k means decay, from the same one-line rule about the rate.

Let’s put it to work. Bacteria grow at a rate proportional to how many there are, with two hundred at the start and six hundred after three hours. How many are there at six hours?

Let’s go to Hadley, on the Moon, where Apollo 15 landed in 1971. Two differential equations play out right here, and one of them is still running right now. This is where Dave Scott dropped his hammer, and its velocity changes by negative one point six two meters per second, every second, a rate that depends on nothing but time.

Every possible toss is one of these parallel lines. Scott let go from rest, so the hammer’s velocity is zero plus the integral from zero to t of negative one point six two, which is negative one point six two t. More than a hundred meters away, the astronauts left something that’s been changing in a very different way ever since. This finned cylinder is a nuclear generator called SNAP-27, with a core of plutonium-238.

Every plutonium atom has the same tiny chance of decaying each year, so the more there is, the faster it decays, and its rate of decay is proportional to the amount. That’s our equation with a minus sign, dN dt equals negative k N, where N is the amount of plutonium left. Here k is positive, and the minus sign does the decaying, so the amount left is N naught e to the negative k t.

Its half-life is eighty-seven point seven years, so after that long, e to the negative k t is one half. Take the natural log of both sides. The log of a half is negative the log of two, so the minus signs cancel, and k is the log of two over eighty-seven point seven, about point zero zero seven nine per year.

How much of it is left? By 2026 it had been about fifty-five years, and e to the negative k times fifty-five is about point six five. That means roughly sixty-five percent of that plutonium was still on the Moon in 2026, still decaying and still warm, even though NASA switched the experiments off in 1977.

Let’s find the year when this generator is down to a tenth of its plutonium. Starting from July 1971, with a half-life of eighty-seven point seven years, work it out, then set the calendar to that year and see if you’re right. Notice how slowly the tail goes, because the less plutonium is left, the slower it decays.

Newton’s equation, the exam way

Back to Newton’s iron, where the gap above the air shrinks at a rate proportional to the gap itself, just like the plutonium. Now we can solve his equation. Like the circles, it has a whole family of solutions, and the reading of a hundred ninety-six at time zero picks out ours.

Let’s write out the separation and the starting value carefully, because those steps earn the credit, and a cooling-pie question on the 2026 exam used this same method. First, separate, by dividing by H minus four and multiplying by dt. That line alone is worth a point, and without it you can’t earn any of the others.

Integrating gives the natural log of the absolute value of H minus four equals negative k t plus C, and each of those two antiderivatives earns its own point. Use the starting point right away. At t equals zero the heat is a hundred ninety-six, so C is the log of a hundred ninety-two. In that 2026 question, leaving out the plus C cost the last two points, so write it as soon as you integrate, before you use the starting temperature.

Undo the log, and the right side becomes a hundred ninety-two times e to the negative k times t. Because the iron starts at a hundred ninety-six, above the air, H minus four is positive. Drop the bars and add the four back, and that gives the particular solution, H equals four plus a hundred ninety-two e to the negative k t.

One more reading pins down k. At five minutes the iron reads a hundred, so take away the air’s four, and the gap is ninety-six, half of a hundred ninety-two. We have e to the negative five k equal to one half. Taking the natural log gives negative five k equals negative the natural log of two, so k is the natural log of two over five, about point one three nine per minute.

Put that k back in and look at one five-minute stretch: there the exponential becomes e to the negative natural log of two, which is one half, so every five minutes the gap gets multiplied by another half. That’s why the readings every five minutes go a hundred ninety-six, a hundred, fifty-two, twenty-eight, the gap halving every time. If you met this forge in precalculus, you found that pattern in the data, and now you know why it had to be there.

If a question asks you to justify that the temperature is decreasing, give the sign of the derivative and explain why that sign stays negative. Because k is positive and H minus four, a hundred ninety-two e to the negative k t, is positive for every t, dH dt is negative for every t, so H is decreasing.

It also tells you the iron never quite reaches four, because e to the negative k t gets tiny but never hits zero, so the curve hugs the air temperature forever. The bismuth bead freezes at eighty-one on Newton’s scale. Work out when that happens, set the alarm for that minute, and let the iron cool to see if the bell rings just as the bead freezes.

Run the clock backward

Now it’s your turn to be Newton. On a different night, we’ve walked into the yard too late, and someone has already taken the iron out of the fire. It isn’t glowing anymore, and our thermometer reads forty, at nine forty-seven by the clock. It’s the same iron in the same wind, so it follows the same formula, with k still about point one three nine per minute. This fire always sends the iron out at a hundred ninety-six.

How many minutes ago did this iron come out of the fire? Work it out, drag the clock back to that moment, and watch the iron rewind.

There it goes, back to nine thirty-five and back into the fire, exactly when your formula said it left. Forensic scientists use a more careful version of this same idea to estimate the time of death from a body’s temperature. On the AP exam, cooling and warming problems like this one keep showing up.

To solve a separable differential equation, separate the variables and integrate both sides with one plus C. The starting point fixes the constant, and once you solve for y, check where your formula and the original equation make sense.

The plus C gives you a whole family of solutions, and one point picks out the single particular solution, sometimes on a restricted domain. Whenever a rate is proportional to the amount, dy dt equals k y, and the answer is always y naught e to the k t, growing when k is positive and decaying when it’s negative. You started this lesson watching Newton run a cooling iron backward to read the heat of a fire, and you just did it yourself, from one rule and one reading.