Topic 1.3: Rates of Change in Linear and Quadratic Functions

One shot

Picture yourself on the gun deck of a warship about two hundred and fifty years ago, with the sun going down and the whole sea turning gold around you. Somewhere out there in the haze, more than half a mile away, there’s an enemy ship, and from where you’re standing it’s barely a smudge on the horizon.

Your crew loads the gun, aims it as well as anyone can, and fires. Let’s freeze it right there, two seconds into its flight, because right now nobody on this deck knows if it’s going to make it.

So what’s your call? Is this shot going to hit her? Let’s see where it lands. And it lands in the water a long way short, which means the other ship now knows exactly where you are.

So how do you fix the next shot? Real gunners spent years building a feel for it, but the first two seconds of that flight already held everything you need. By the end of this lesson, you’ll be able to tell where a cannonball is going to land. All you’ll need is its first two seconds in the air. And you’ll use that to hit a ship you can barely see.

The pattern hiding in that flight also decides how far a car needs to stop, which will matter to you a lot more than cannons ever will. So let’s replay that shot slowly and watch it one second at a time.

One second at a time

Here’s the same shot from the side. The ball leaves the barrel sixteen feet above the water, and I’ll drop a marker wherever it is at the end of every second. After one second it’s a hundred sixty feet up, and after two seconds it’s at two hundred seventy-two. So during the first second it climbed a hundred forty-four feet, and during the next second only a hundred twelve, which means it’s still rising, but it’s already climbing less.

Those two climbs are all that the first two seconds give us, so before we let it fly any further, I want you to guess how far the ball climbs during the third second. Let’s see what it actually does. In the third second it climbs eighty feet, then forty-eight, and then just sixteen, so from one second to the next, the climb shrinks by exactly thirty-two feet.

And it keeps going after the top, where the ball drops sixteen feet, then forty-eight, then eighty, and each change is still thirty-two less than the one before. Notice how little its height changes near the top. It spends half its flight, five of its ten seconds, in the top quarter of its climb, which is exactly why a basketball player going up for a dunk seems to hang in the air.

So the height changes by a different amount each second, but watch what happens to the change itself. It drops by thirty-two, then by thirty-two again, and again, every single time. That thirty-two is gravity. Gravity takes the same thirty-two feet per second off the ball’s upward speed every second, so the ball leaves the gun rising at a hundred sixty feet per second, and one second later it’s rising at only a hundred twenty-eight.

So at every moment of the next second, it’s rising thirty-two feet per second slower than at the same moment one second before, and over a whole second that adds up to exactly thirty-two fewer feet of climb. Now look at the water right under each marker. Sideways, it covers exactly two hundred forty feet each second, so that rate never changes at all.

And look at what the first two seconds were telling us. Climbs of a hundred forty-four and a hundred twelve mean the step is thirty-two, so if you keep taking away thirty-two, you can write down the whole flight before the ball gets there. It comes back to deck height after ten seconds, two thousand four hundred feet out. That’s well short of the ship. So sideways the ball moves the same distance every second, and up and down the change itself changes at a steady rate, so let’s take this shot to the whiteboard.

The slope between two moments

Here’s the same shot on the whiteboard, starting with the heights we measured, sixteen feet, a hundred sixty, two seventy-two, three fifty-two, four hundred, and finally four sixteen at the top. The climb in each second is the new height minus the old one, so one sixty minus sixteen is a hundred forty-four, and two seventy-two minus one sixty is a hundred twelve. Then it climbs eighty, forty-eight and sixteen.

Now take each climb minus the one before it, so a hundred twelve minus a hundred forty-four is negative thirty-two, eighty minus a hundred twelve is negative thirty-two again, and it stays negative thirty-two all the way down. On a graph those heights are these dots, and they follow a parabola, h of t equals negative sixteen t squared plus one sixty t plus sixteen.

To measure how fast the height changes between two moments, we use the average rate of change, the change in height divided by the change in time. Over the first second, that’s a hundred forty-four feet in one second. Every climb we found was already an average rate of change. On the graph, it’s the slope of the line through two dots. That line is a secant.

That’s how AP questions describe it too, so let’s write it down. The average rate of change over an interval is the slope of the secant line through its endpoints. Now try one that’s longer than a second. At t equals one the ball is one hundred sixty feet up, and at t equals four it’s at four hundred. What’s the average rate of change of its height between those two times, in feet per second?

Now let’s graph the sideways distance, which grows by two hundred forty feet every second, so its graph is a straight line, x of t equals two forty t. Draw a secant over the first second, over five whole seconds, or over just half a second, and the slope is two hundred forty every time, because every secant lies right on the line. That’s what makes a function linear. Its average rate of change is the same over any interval you pick, anywhere on the graph.

The rate of the rate

Now back to the height, and this time I want the slope of every single second at once. I’ve split the flight into ten one-second intervals, and I’ll draw the secant line on every one of them.

Then I’ll lift the slope of each secant off the curve and stand it up as a dot, a hundred forty-four, a hundred twelve, eighty, all the way down to negative one forty-four. And look at that, the rates land on one perfectly straight line, which means the rates themselves form a linear function of time.

That happens for every quadratic. Its average rates of change over consecutive intervals of the same length can always be given by a linear function. Every step down that line is the same, negative thirty-two, so the average rates of change are changing at a constant rate, and it’s the same thirty-two you saw out at sea.

Here’s something for you to try. With one-second intervals every step is negative thirty-two, so take the tower and change the width of the intervals until every step is negative sixteen.

Now let’s go back to whole seconds and run the sideways motion through the same machine. Every secant has a slope of two hundred forty, so the dots sit in a flat row with no steps at all. For a linear function the rates never change, so they form a constant function, and the exam has a careful way of saying that, they’re changing at a rate of zero.

Why the step never changes

Every quadratic has a steady step like the cannonball’s, even the ones with no gravity in them. So let’s find out what it is about squaring a number that keeps the step the same. Let’s build squares out of tiles, one by one, then two by two, three by three, four by four and five by five, and count the tiles in each, one, four, nine, sixteen and twenty-five.

To get from one square to the next, you add an L-shaped border along two sides, and those borders hold three tiles, then five, then seven, then nine. Those are the odd numbers, and each border holds exactly two more tiles than the one before it. And you can see why. Every border is two arms with a tile in the corner, and each time the square gets one tile wider, each arm gets one tile longer, so the border always grows by two, no matter how big the square gets.

Galileo found these same odd numbers around 1604 in a ball rolling down a ramp, and in 1638 he published them as a rule for how things fall. Our cannonball does it too. After the top it falls sixteen feet, then forty-eight, eighty and a hundred twelve, sixteen times the odd numbers, so every tile is worth sixteen feet, and the two extra tiles in each border are the thirty-two.

Now let’s check any quadratic, a x squared plus b x plus c, on an interval that runs from x to x plus h. Its average rate of change is f of x plus h, minus f of x, all over h, and when I expand the square, the a x squared, the b x and the c all cancel.

Every term that’s left has an h in it, so dividing by h leaves two a x plus a h plus b. That’s a linear function of x, just like the tower showed. Each time you slide one interval to the right, x grows by h, so two a x grows by two a h, and that’s the step, the same every time.

The two is the two arms of the border, a is what each tile is worth, and h is the length of each interval, which is why halving the interval halved the step. For the cannonball, a is negative sixteen, so one-second intervals give negative thirty-two, and half-second intervals give negative sixteen, exactly what you found on the tower. So the step depends on how long your intervals are, and that’s exactly where AP questions like to set a trap.

Twice as fast

The same pattern decides something you’ll care about the first time you drive at night. On this test road, the moment those lights turn red is the moment a driver sees trouble ahead and has to stop. Here’s a car doing thirty miles an hour. When the light turns red, the driver needs a moment to react before the brakes even bite, and the car stops seventy-five feet past the line.

Now put the same driver on this road at sixty miles an hour, twice as fast. How far past the line do you think the car will stop?

Now watch the second car, because at the spot where the first car is already stopped, this one is still doing fifty-seven miles an hour, and it doesn’t stop until it’s two hundred forty feet past the line. While the driver reacts, the car keeps going at full speed, so the thinking distance simply doubles, from thirty feet to sixty, but the braking distance goes from forty-five feet to a hundred eighty, four times as far.

That’s because at twice the speed the brakes need twice as long to stop you, and the whole time you’re covering ground twice as fast, and two times two is four, just like a square twice as wide holds four times the tiles. Now let’s line up four cars at thirty, forty, fifty and sixty and stop them all at once. They stop at seventy-five feet, a hundred twenty, a hundred seventy-five and two hundred forty.

From above, the stopping points make a staircase, with steps of forty-five feet, then fifty-five, then sixty-five. Each step is ten feet longer than the one before, which is the cannonball’s fingerprint again, a change that changes by a constant amount, with nothing falling anywhere in sight, so stopping distance is a quadratic function of speed.

Now let’s slow the driver’s reaction down to a second and a half. That’s closer to how long real drivers take when something surprises them, and a phone in your hand makes it longer still. Every car needs a lot more road, but the second step is still exactly ten feet, because reaction time only changes the thinking part, and that part grows in equal steps.

Now it’s your car, and you react as fast as our first driver. The road is closed one hundred fifty feet past the light, so use the staircase to find the fastest speed you can drive and still stop before the barrier.

Which way it bends

Let’s bring the tower back for the stopping distances. On each ten-mile-an-hour stretch the secant gets steeper, four and a half, five and a half, then six and a half feet per mile an hour, so the average rates of change are increasing. When that happens the graph bends upward, like the inside of a bowl, and we call it concave up.

The cannonball is the opposite. Its rates keep decreasing, a hundred forty-four, a hundred twelve, eighty, even while the ball is still rising, and its graph bends downward, so we call it concave down. On the AP exam you’ll see it stated like this. If the average rate of change over equal-length intervals is increasing for all small intervals, the graph is concave up, and if it’s decreasing, the graph is concave down.

Those words, for all small intervals, are there for a reason. This function matches a parabola at every whole number, so a table that checks once per unit sees rates that grow by the same amount every time. But shrink the intervals and a wobble appears in between, so a table can only tell you about the intervals it actually measures.

Let’s see if you can use that rule. The average rates of change of a function g over equal-length intervals keep decreasing on the interval from two to seven, however small you make the intervals. What can you say about the graph of g there?

Reading a table like an AP reader

AP questions love to hand you a table instead of a formula, so let’s read one together, and before anything else, notice that the inputs go up by two each time. Ten minus two is eight, twenty-six minus ten is sixteen, and then the jumps are twenty-four and thirty-two, so the jumps themselves grow by eight each time.

So by how much does the average rate of change change from one interval to the next? Think about those steps of two. That’s also why you always check the input steps first. In this little table the outputs rise by two every row, but the inputs jump by one, then two, then one, so the rates go two, one, two, and it isn’t linear at all.

Here’s a table that fits neither pattern. Its outputs are two, six, eighteen, fifty-four and a hundred sixty-two, and its inputs are one apart, so its rates are just the jumps, four, twelve, thirty-six and a hundred eight.

Those rates change by eight, then twenty-four, then seventy-two, so it’s neither linear nor quadratic, although its rising rates do fit a graph that’s concave up. And when the inputs are just zero, one, two, three, we call that list of outputs a sequence, and the change from one term to the next is its average rate of change over an interval of length one.

Here’s the kind of table an AP question might hand you, except this one is real, a page of Galileo’s notes, which the historian Stillman Drake read as how far a ball had rolled down a ramp after each of eight equal stretches of time. His distances grow by thirty-three, then ninety-seven, then a hundred sixty-eight, and the changes of those changes all sit close to sixty-five, remarkably steady for measurements made more than four hundred years ago. So on the AP, you’d justify a quadratic model by writing that the changes in the average rates of change over consecutive equal-length intervals are approximately constant.

Hit the ship

Our first shot came back down to deck height after ten seconds, two thousand four hundred feet out, and the enemy ship is sitting at two thousand eight hundred eighty. The ball moves two hundred forty feet sideways every second, so to reach the ship it has to stay up for twelve seconds, and the pattern tells us how to make that happen.

Twelve seconds means six up and six down, and on the way up each second’s climb is thirty-two feet less than the one before, ending with a climb of sixteen feet just before the top. So how much does the ball need to climb in its very first second? Count back from that last climb of sixteen feet.

So we raise the barrel and add a little powder until the ball climbs a hundred seventy-six feet in its first second, and we fire. And that’s a direct hit from more than half a mile away, and all it took was a pattern in the numbers!

There’s a second ship farther out, at three thousand three hundred sixty feet, and this one is all yours. The ball still goes two hundred forty feet a second sideways. How much does it need to climb in its very first second to hit her?

What to remember

So let’s pull it together. The average rate of change over an interval is the slope of the secant line between its endpoints. For a linear function, like the sideways motion, that rate is the same on every interval, so the rates change at a rate of zero. For a quadratic, like the height or a car’s stopping distance, the rates over equal intervals form a linear function, so they change at a constant rate.

And that’s because there’s a square inside every quadratic, and a square grows by borders that are two tiles bigger every time. And when those rates increase over all small intervals the graph is concave up, and when they decrease it’s concave down. The next time you watch a ball fly or a car brake hard, check how much it changes each second and then how much that change changes, because now you know what those numbers are telling you.