Topic 2.5: Exponential Function Context and Data Modeling

The pop

It’s six in the morning in Key West, at the very end of the Florida Keys, and someone is walking out of that little shed with a balloon that’s taller than they are. They let it go, and a small white box on a long string rides up underneath it, and the same thing happens at hundreds of weather stations around the world, all at the same two times every day.

Look at the balloon now. Up here the sun has already come up, while the island below is still waiting for sunrise. It keeps climbing for almost two hours, up past the clouds and far above the airliners, and the whole way up, it keeps getting bigger.

It left the ground less than two meters wide. Now it’s about eight, stretched as thin as a soap bubble. Look how far below us the ground is now. So how high do you think we are?

And at thirty-three point four kilometers, this is what happens. Every one of these balloons pops eventually, and up here the air is so thin that you’d barely hear it. By the end of this lesson, you’ll be able to take the readings this one balloon sent down on its way up. And from those, you’ll call where tomorrow morning’s balloon is going to pop. You’ll know it before that balloon even leaves the ground.

The same fraction every kilometer

That little box is measuring the air pressure, the weight of all the air above it pressing down. The higher the balloon climbs, the lower the pressure gets, and the helium inside spreads out until the rubber can’t stretch any more. So let’s rewind and watch the readings come in. The box radioed down the pressure every second, more than four thousand readings, and I’ve kept one for each kilometer.

At the ground it read a hundred and one point six kilopascals. That’s the normal pressure of air at sea level. So how is it dropping? Subtract each reading from the next, and the air loses eleven kilopascals, then ten, then nine, eight and seven, so it isn’t losing the same amount each kilometer.

But watch what happens when we divide instead. Ninety point six over a hundred and one point six is zero point eight nine, and every ratio after that lands within a hair of it. Every kilometer up, the air keeps the same fraction of its pressure, about eighty-nine percent, and when equal steps in the input multiply the output by the same factor, that’s exactly what we mean by exponential.

Here are those readings as a table, the height in kilometers on the left and the pressure in kilopascals on the right, one row for every kilometer from the ground up to five. Ninety point six minus a hundred and one point six is negative eleven, and the rest keep shrinking, negative ten point one, nine point one, eight point two, seven point four.

Dividing instead, ninety point six over a hundred and one point six is zero point eight nine two, and the rest are zero point eight eight nine, eight eight seven, eight eight five and eight eight three. The differences change a lot while the ratios barely move, so what stays the same from one kilometer to the next is the factor. This is exactly how the exam wants you to argue it, by showing the ratio for more than one pair of readings, because one ratio on its own proves nothing.

Since every kilometer multiplies by the same factor, three kilometers up should be a hundred and one point six times zero point eight nine, three times over, about seventy-one point six, and the real reading was seventy-one point four. So the whole climb fits in one line, P of h equals a hundred and one point six times zero point eight nine to the h, where h is the height in kilometers. The starting value goes in front and the ratio goes in the base, and that’s the first way to build an exponential model, from an initial value and a ratio.

Look at that zero point eight nine again. It’s the part of the pressure the air keeps each kilometer, so the air loses the other eleven percent. Lots of people see an eleven percent drop and put zero point one one in the base, but that sky would lose eighty-nine percent of its air every kilometer, with almost nothing left three kilometers up.

So the base is always what’s left after one step. That’s why we call b the growth factor, even when it makes things shrink. The growth factor belongs to one step of the input, so over five kilometers the factor is zero point eight nine to the fifth, about zero point five six, and the air loses forty-four percent, not five times eleven.

Let’s try a different sky. A weather station high in the Andes reads seventy-five kilopascals, and the pressure drops twelve percent for every kilometer you climb above it. Which model fits?

Two readings

Often all you get is two readings, so say we only know that at two kilometers the pressure was eighty point five, and at seven kilometers it was forty-three point one. That’s two equations with the same two unknowns, and the move is to divide the second one by the first, because the a on top cancels the a on the bottom, and b to the seventh over b squared leaves b to the fifth.

Take the fifth root on the calculator, and b comes out to about zero point eight eight two, so the air keeps about eighty-eight percent each kilometer, close to what we found before. Then put b back into the first equation. Eighty point five divided by b squared gives a, about a hundred and three.

And look at what just happened. The model worked out the pressure at the ground, about a hundred and three, from two readings nowhere near the ground. The real reading there was a hundred and one point six. That’s the second way to build a model, from two input-output pairs by solving the system, and it’s exactly how the exam asks it.

You’d think it wouldn’t matter much which two readings you pick. But it matters a lot, so let’s go back up and see why. Here’s the whole flight, every kilometer from the ground to the pop, and here’s the model through the two lowest readings, the ground and five kilometers.

It fits the bottom beautifully, but above ten kilometers it drifts away from the readings, and at thirty kilometers it says two point eight, when the balloon measured one point one. That model would call the pop at almost forty-two kilometers, about eight point four kilometers too high. Now take two readings much farther apart, the ground and thirty kilometers. This one hugs the top of the flight and calls the pop at thirty-three point four, right where it happened.

It sags below the readings in the middle. That’s because the air doesn’t thin at exactly the same rate all the way up. So spread your two readings out, and make sure they cover the part you care about.

Here are just the two readings. The ground reads a hundred and one point six, and thirty kilometers reads one point one three. Using the ground reading and the reading at thirty kilometers, what’s the growth factor b?

All of them at once

Two readings throw away the other thirty-two, so calculators have a tool that uses every reading at once, called exponential regression. It finds the one curve of the form a times b to the h that stays closest to all the readings together, and for this balloon it gives a hundred and twenty-two times zero point eight five six to the h.

Wait, a hundred and twenty-two? The ground reading was a hundred and one point six, so did the calculator get it wrong? It didn’t, because near the ground the air keeps about eighty-nine percent each kilometer, and up high, where it’s coldest, only about eighty-four, so no single exponential passes through every reading.

The regression trades a miss at the bottom for a better fit over the rest of the flight, so its a is just that best curve’s value at h equals zero, not the first reading. And on the exam, saying you ran a regression earns nothing as a reason that data are exponential. The reason is the ratios, and the regression is just a tool for building the model.

Back in the sky, here’s the regression’s curve against every reading, and it reaches this balloon’s popping pressure at thirty-three point four five kilometers, about eighty meters from where it really popped. The top of Everest would be up here, with only about a third of the air above it. And when the balloon popped, more than ninety-nine percent of the air was below it. Around five point seven kilometers the pressure is exactly half what it was at the ground, which means half of the whole atmosphere is below this line, lower than the top of Kilimanjaro.

Near the ground, our model is a hundred and one point six times zero point eight eight seven to the h. That’s the eighty-nine percent from our table, with one more digit. The ring in the sky marked where half the air is gone. Where does our model put that height?

Try a few powers on the calculator. Zero point eight eight seven to the power five point eight comes out to almost exactly one half. The model halves the pressure every five point eight kilometers, close to the ring at five point seven. We want that one half inside the model. Raise zero point eight eight seven to the five point eight first, then to the h over five point eight.

A power of a power multiplies the exponents. Five point eight times h over five point eight is just h. This is still our model. Inside the parentheses is one half. The same model is a hundred and one point six times one half to the h over five point eight.

The exponent counts the halvings. At five point eight kilometers it’s one, so half the air is left. At eleven point six it’s two, so a quarter is left. Both lines are the same function. The top one says what happens every kilometer. The bottom one says how far until half the air is gone.

This idea is hiding inside one of the longest-running online games. In Old School RuneScape your total experience doubles every seven levels, and level eighty-five takes about three point two six million points. Which level takes twice as much?

Scientists often write the air’s pressure with a special number that comes from a question Jacob Bernoulli asked in 1683, about money earning a hundred percent interest for a year, paid in n equal pieces. Paid all at once, a dollar becomes two, paid monthly, about two sixty-one, daily, two seventy-one, and even paid every minute it only reaches two point seven one eight, a number it never passes.

That number is e, about two point seven one eight, and it turns up all over science, our sky included. Our model can use e too. Try e to the minus zero point one two on your calculator. It comes out to about zero point eight eight seven, our growth factor.

Our model was a hundred and one point six times zero point eight eight seven to the h. Put e to the minus zero point one two in for the zero point eight eight seven. A power of a power multiplies the exponents. The same model is a hundred and one point six times e to the minus zero point one two h.

Your far-pair model was one hundred one point six times zero point eight six one to the h. E to the minus zero point one five is about zero point eight six one. Which is the same model?

Newton’s wind

The air’s pressure was heading toward zero the whole way up. But a lot of real things level off at some other number, and one of the first people to write down how to handle that was Isaac Newton.

In 1701 he published a short paper, without even signing it, about measuring how hot a fire is, when no thermometer he had could survive that kind of heat. So he set a glowing block of iron out in a steady wind, with small bits of metal on top, and timed how long each one took to freeze. As the clock ticks the glow fades, the lead freezes first, and then the bismuth.

He measured heat on his own scale, where melting snow is zero and your body is twelve, and in our version of his experiment, at equal ticks of the clock, the iron reads a hundred ninety-six, a hundred, fifty-two, twenty-eight, sixteen and ten. The ratios drift, zero point five one, five two, five four, five seven, six three, so these readings aren’t exponential, at least not yet.

Newton realized the iron was cooling toward the temperature of the air, so what shrinks by the same fraction every tick is the gap between the iron and the air. He wrote that the excess of the iron’s heat above the air’s falls in a geometric progression while the times go up in an arithmetic one, so the gap shrinks by the same factor in every equal stretch of time.

Now you find the floor. Drag it up from zero until all the ratios come out the same. With the floor at four, the gaps are a hundred ninety-two, ninety-six, forty-eight, twenty-four, twelve and six, halving on every tick of the clock. Let’s write that down.

Here’s Newton’s rule as a model. It’s four plus a hundred ninety-two times one half to the t, where t counts ticks of the clock. That plus four lifts the whole exponential up by four, so it levels off at the air’s temperature instead of zero, a transformation that comes straight from the context. Subtracting four from every reading is adding negative four, and adding a constant like that is how you reveal a pattern hiding in the data.

Here’s a real cup of coffee, measured every minute for ten minutes in a room at twenty-four degrees Celsius. Divide each reading by the one before, and the ratios all sit near zero point nine seven. So it looks exponential. If you hand these readings straight to the calculator, it happily fits a curve.

Now ask that model about two hours from now. It says the coffee will be at about two point seven degrees, in a room that’s at twenty-four degrees. The calculator just put your coffee in the fridge. It did that because nobody told it the room was there.

Subtract the room’s twenty-four degrees, fit the gaps, and add the twenty-four back, and you get twenty-four plus fifty-eight point two times zero point nine five eight to the t, which settles onto the room temperature like a real cup. The context tells you the floor, and the regression can’t, so before you fit anything, ask what the data are leveling off toward.

Exactly how the exam asks it

Before we keep our promise, let’s try a question from the 2026 AP Precalculus exam, about a car that loses value every year, and it uses almost everything we did today. At the end of 2020 the car was worth twenty-seven thousand two hundred dollars, and at the end of 2025, fourteen thousand eight hundred, with time counted in years since the end of 2019.

With V of t equal to a times b to the t, in thousands of dollars, that’s twenty-seven point two equals a times b, and fourteen point eight equals a times b to the sixth, the same two-reading move as the balloon. Dividing the second by the first cancels a and leaves b to the fifth equals zero point five four four, so b is about zero point eight eight five, and a is about thirty point seven.

So the car keeps about eighty-eight and a half percent of its value each year, and loses about eleven and a half. And a, about thirty thousand seven hundred dollars, is the model’s value when the car was bought, at the end of 2019. The last part is about the domain, because the owner plans to donate the car to a mechanic school once it’s worth two thousand dollars, and at that moment it loses all its value.

So in which year does the owner donate the car? Try whole numbers of years on the calculator until the value drops below two. So the model only makes sense from t equals zero to about t equals twenty-two point four, and stating that domain with its reason from the context is exactly what earns the point on that question.

Call the pop

Now let’s keep the promise. It’s the next morning, the nineteenth of February, at the same little shed, and a new balloon is ready to go up. In the sky is the model you built from yesterday’s balloon, the regression through all thirty-four readings, and nothing at all from today’s flight. You get to decide how much helium goes in. The fuller the balloon, the less room the rubber has left to stretch, so the sooner it pops.

Set the fill, read the pressure where that balloon pops, then drag your call to the height where your model reaches that pressure, and launch it. Here’s the balloon they really launched that morning, and the model built from the day before says it should pop at thirty-two point eight kilometers. And it popped at thirty-two point eight. That’s about twenty meters from the call, with a model built entirely from yesterday’s readings.

The model has limits, though. It’s only good from the ground up to about thirty-three kilometers, because that’s as high as the readings went. Above that, there’s no data to back it up. You checked that equal steps give the same ratio. You built the model from two readings, or from all of them.

When the data level off, you find the floor first. You took one balloon’s readings and called where the next one would pop. You were off by about two school buses. That’s what an exponential model is for.