Topic 2.1: Defining Average and Instantaneous Rates of Change at a Point

The camera that keeps up for one instant

This is the very first Crew Dragon launch, at two forty-nine in the morning in March 2019, with nobody on board except a test dummy named Ripley. There it goes. By sixty seconds it’ll be about eight kilometers up and still speeding up, and I’d like to show it to you from a very strange camera.

Let’s ride along with an imaginary camera that climbs right beside the rocket at one steady speed the whole time, never faster and never slower. Keep your eye on where the rocket sits in the frame.

Look at that. Right at sixty seconds the rocket stops dead in the frame, just for an instant, and then it pulls away again. It’s like a car on the highway that matches your speed and seems to hang beside your window. At that one instant, our camera is climbing exactly as fast as the rocket.

What I’d like to know is how fast our camera is climbing. By the end of this lesson you’ll use our model of this flight to work out that speed right down to the last digit, and then you’ll get to fly the camera yourself. That speed has a name you’ll meet on almost every page of this course, the derivative, so let’s go and find it.

The speed that matches

Let’s put the flight on a graph, with time going across and altitude going up. I’ll call the altitude capital H, because little h has another job coming. A camera that climbs at one steady speed draws a straight line on this graph. This one climbs about a hundred thirty-five meters every second, the average for the whole first minute. But look at sixty seconds, where the curve is climbing much more steeply than this line.

Now I’d like you to aim a straight line through the rocket’s point at sixty seconds, so that it just touches the curve right there, the way our camera’s path would. At three hundred meters per second the camera is level with the rocket at sixty seconds, but it keeps going and slips right past, so that’s too fast. The little graph underneath tracks the gap, how far the rocket is above the camera, and it crosses zero right at sixty seconds.

At two hundred sixty the camera is ahead of the rocket just before sixty seconds, and the rocket overtakes it right at sixty, so that’s too slow. Somewhere in between there’s exactly one speed where the camera climbs up to the rocket, touches it, and falls behind again without ever passing it, and that’s our camera. You know the word for a line like that from circles, a tangent line, and the camera’s speed is exactly its slope.

To find a slope you usually need two points, so let’s borrow a second one, h seconds later, at sixty plus h. The slope between them is the change in altitude divided by h. As h shrinks, the second point slides in toward the first, and the slope goes from two eighty-two point eight to two seventy-nine point eight, then two seventy-nine and a half.

The step h can be negative too, which puts the second point before sixty seconds. What sign do you think the slope will have then?

Let’s put the second point one second before sixty. The change in altitude comes out negative as well, so the slope is still positive. From this side it climbs toward the same two seventy-nine and a half. From both sides the slopes close in on one number, the same move you made with limits in Unit 1.

Let’s write down what we just did. The second altitude minus the first, over h, is a difference quotient, the slope of the line through both points, which is the average rate of change over that little window.

Then we let h go to zero, and we call the number those slopes approach the derivative of H at sixty. We write it H prime of sixty. The derivative at a point is the instantaneous rate of change, our camera’s speed, and the definition comes with one condition, that the slopes really do settle on a single number.

For any function f and any input a it looks like this, and it’s one of the two ways the AP exam writes it. Right now we only know H prime of sixty is about two seventy-nine and a half, because watching digits settle is a strong hint, but it isn’t proof. Let’s see how algebra turns a hint like that into an exact answer.

Exactly six

Let’s try it on a curve with friendly numbers, f of x equals x squared, at the point where x is three. With h equal to one the second point is at four, where f is sixteen, so the slope is sixteen minus nine over one, which is seven.

Shrink h to a tenth and the slope is six point one, and a hundredth gives six point zero one. From the left, with h negative, we get five point nine and then five point nine nine. It sure looks like six, but looking isn’t proving, so let’s do the algebra and keep h as a letter the whole way.

The difference quotient is three plus h, squared, minus nine, all over h. Squaring three plus h gives nine plus six h plus h squared, and the nines cancel, which leaves six h plus h squared on top.

Every term on top has an h in it, so we can divide by h, and that’s allowed because h never actually equals zero, it only gets close. Now let h go to zero, and six plus h goes to exactly six, so the derivative of x squared at three is six, with no rounding anywhere.

If we call this quotient Q of h and graph it, we get the line six plus h with a hole right at h equals zero, because there Q is zero over zero. Yet from both sides, it heads straight for six. Let’s make sure that picture sticks. Read these four statements about the quotient and the derivative at three, and pick the one that’s true.

The other way to write it

There’s a second way to write the same limit, because the moving point has two names. If it sits at x, then x is three plus h, so the gap x minus three is just h, and its height, three plus h all squared, is just x squared. When h goes to zero, x goes to three.

Written with x, the slope from three out to x is x squared minus nine, over x minus three, another average rate. The derivative is its limit as x goes to three. The top factors into x minus three times x plus three, the x minus threes cancel, and x plus three heads to six, the same number we got with h.

These are two equivalent ways to write the derivative at a point. One measures the step h and the other names the point x. The exam uses both, so you want to recognize either one on sight.

Let’s see if you can spot it. Find the limit on this card that’s exactly f prime of five, for f of x equals x squared.

We do, you do

Let’s do one together with a little more going on, p of x equals x squared minus four x plus one, at three. First we need p of three, which is nine minus twelve plus one, so negative two.

Next we need the height at the nearby input, so we replace every x with three plus h. Squaring gives nine plus six h plus h squared, and distributing negative four gives negative twelve minus four h. When we collect the constants and the h terms, we get negative two plus two h plus h squared.

Now we subtract p of three, and this is the step where most people slip, because we’re subtracting negative two, which means adding two. The twos cancel and we’re left with two h plus h squared over h, which is two plus h, so as h goes to zero the derivative is two. A quick check on the graph agrees, because at three the parabola is climbing, and not very steeply.

Now I want you to try one on your own. A tank drains so that its depth, D of t, is twelve minus t squared meters, t hours after the valve opens. Use the definition to find D prime of two, in meters per hour.

The units come straight from the quotient, meters on top and hours on the bottom, so the derivative is in meters per hour. Two hours after the valve opens, the depth is falling at four meters per hour, and the minus sign is the word falling.

The line that touches

Back to x squared at three, because three different things live at that one point. f of three is nine, the height of the point three, nine, and f prime of three is six, a slope. Put them together in point-slope form and you get a whole line, the tangent line, y minus nine equals six times x minus three.

Now let’s zoom in on the point. At ten times the curve is already hugging the line, and at a hundred times you can’t tell them apart. That’s why we get to call the tangent’s slope the slope of the curve itself.

Near the point, the line is a good stand-in for the curve. At three point one it predicts nine plus six times point one, which is nine point six, and the real value of three point one squared is nine point six one. Let’s test that idea on y equals x cubed at the origin, where the line y equals zero runs right through the curve.

The curve y equals x cubed crosses the line y equals zero here, from below to above. Is that line still the tangent at the origin?

A tangent line is the line that matches the curve’s slope at a point, the way our camera matched the rocket’s speed. The rocket’s tangent happened not to cross, but a tangent is allowed to.

The fastest instant of the 9.58

We’re in Berlin for the 2009 World Championships, and Bolt is settling into the blocks in lane four for the hundred-meter final. Here he goes. Nine point five eight seconds, the world record he set that night, which works out to an average of about ten point four meters per second for the whole race. He started from zero in the blocks, so somewhere in the middle he must have been going a lot faster than that. Researchers timed him at every ten-meter mark to find out how much.

Here are those times, measured from the gun, so the first ten meters include his reaction to it. Between sixty and seventy meters he covered ten meters in eighty-one hundredths of a second. That’s about twelve point three meters per second, the fastest ten meters in the table. That’s a difference quotient, with a at six point two nine seconds and h equal to point eight one seconds, an average over a window just like the rocket’s slopes.

Let’s try one from the table. What was Bolt’s average speed from twenty meters to thirty meters, to the nearest tenth? To get his speed at one instant we’d shrink h toward zero, but a table can’t do that, because there’s nothing in it between the ten-meter marks.

Picture a camera on a rail beside the track, set to eleven meters per second. Bolt catches it, then simply runs away from it. Now you drive the camera. Set its steady speed so that Bolt stays right in the middle of the frame as he passes sixty-eight meters, then run it.

The research team also tracked him with a laser and modeled his run as one smooth curve. That model’s fastest instant is twelve point three four meters per second, at about sixty-eight meters. Every speed we worked out from the table is an average over a window, and his speed at one instant is the slope of the tangent.

When there’s no answer

Every derivative so far came with that condition, that the slopes settle on one number. Let’s look at a graph where they don’t, the absolute value of x, at zero. For positive h, the absolute value of h is just h, so every slope from the right is exactly one. For negative h it flips sign, so every slope from the left is exactly negative one. The two sides never agree.

There’s no single number for the limit to be, so f prime of zero doesn’t exist, and no one line through that corner matches the graph on both sides. On the exam you’d justify it like this. The difference quotient approaches one from the right and negative one from the left. The one-sided limits are different, so the limit doesn’t exist, and neither does f prime of zero.

How the exam asks it

When the AP exam hands you a limit like the one I’m writing here, the first move is to recognize it as a derivative in disguise. Match it to the pattern. We’re cubing two plus h, so the function is x cubed and our starting input is two. The eight is f of two, which makes this whole limit f prime of two.

Two plus h, all cubed, expands to eight plus twelve h plus six h squared plus h cubed. The eights cancel, the quotient becomes twelve plus six h plus h squared, and f prime of two is twelve.

When a question asks what a derivative means, it wants a full sentence, like this one. At sixty seconds after liftoff, the rocket’s altitude is increasing at about two hundred seventy-nine and a half meters per second.

It says when, what’s changing, which way, how fast, and in what units. It never says the rocket climbs that many meters in the next second, because we saw it really climbs about two eighty-two point eight.

This next one is exactly how the exam asks it. This limit is f prime of a for some function f and some number a, and I want you to find the pair.

Fly the camera

Let’s go back to the rocket, because I promised you’d fly that camera. The yellow line is your first guess from the very start, flown as a camera, and the little gap graph shows how it does. Now steer your camera’s speed until it climbs up to the rocket at sixty seconds and just touches it, without crossing.

For our smooth model of this flight, the same algebra you did with x squared, only longer, turns the quotient into one number plus terms that all carry an h. As the step shrinks to zero they vanish, so the camera’s speed is exactly two hundred seventy-nine point five one five two eight meters per second.

The camera’s path is the tangent line, and your first guess sits right beside it in yellow. With the numbers rounded, it starts at about eight thousand ninety-one meters at sixty seconds and climbs two seventy-nine and a half meters every second.

Now pick a moment of your own, forty-five seconds or seventy-five, and fly a camera that just touches the rocket there. Our camera finally matches the rocket because we found the limit of the average rates. That limit is the derivative, and it gives the slope of the tangent line. That limit exists only when both sides agree on one number.

At the start I asked how fast our camera was climbing. Now you know how to get that speed exactly, with the same algebra, just longer, and you’ve flown the camera yourself. Next time we’ll slide that point along the whole flight and watch the derivative become a function of its own.