Topic 5.1: Using the Mean Value Theorem

Two photos

It’s a wet night in the Scottish Highlands, and this driver has just passed a yellow sign. It means average speed cameras are watching the next stretch of road. The limit here is seventy, so the driver eases off to sixty-eight for the first camera, and it snaps a picture of the car and the exact time. Then come four miles of dark road with no cameras at all, and nobody sees what this car does out there.

At the second camera the driver is back down to sixty-eight, perfectly legal, exactly three minutes after the first picture. The driver is under the limit at both cameras and still gets a ticket, because four miles in three minutes is an average of eighty miles an hour. Nobody saw this car do anything wrong. Yet math can prove that somewhere in that dark stretch, at one exact instant, its speedometer read eighty.

By the end of this lesson, you’ll be able to prove that yourself, and write it so clearly that an AP grader would give you every point. You’ll also get to try to beat this camera, and I’ll warn you now, the math wins. Let’s start with the one thing the cameras really do know, which is how long the car took.

Eighty, exactly

The cameras know that the road between them is four miles long, and that this car took three minutes to cover it. Three minutes is a twentieth of an hour, and twenty of those make a whole hour. At this pace that’s twenty times four miles, eighty miles an hour. That’s what the ticket is for, and the camera didn’t need any calculus to work it out. It just divided.

My promise was bigger, though. I said the speedometer itself showed exactly eighty at some instant, and that’s a claim about a moment nobody saw. Before I show you, what’s your honest guess? Somewhere between those two cameras, did this car’s speedometer ever show exactly eighty?

Let’s give this car a ghost, a second car that drives the whole section at exactly eighty, never faster and never slower. It leaves the first camera level with our driver. Eighty is the average, so the ghost reaches the second camera at the same moment too. Our driver starts at only sixty-eight, so the ghost pulls ahead, and the gap between them grows.

Then the driver speeds up. Watch the gap grow more and more slowly, until for one instant it stops growing at all, and right then, the speedometer reads eighty. If you’ve ever driven next to another car, that makes sense, because the gap only holds still while you’re both going the same speed. Our driver blows past the ghost at about ninety-two, then brakes hard to look innocent at the second camera. On the way, the gap holds still once more, and the speedometer reads eighty again.

Let’s graph the gap between the cars, because its shape shows why their speeds had to match. The gap starts at zero at the first camera and ends at zero at the second, because the cars are level at both. Below the line, the car is the one ahead. Anything that leaves zero and comes back has to turn around somewhere, and right where it turns, it isn’t changing for an instant.

When the gap isn’t changing, the two cars have the same speed, and the ghost’s speed is the average, eighty. Our driver hit eighty twice, but the math only guarantees at least one matching instant, and another trip might have just one, or several. We never had to see the dark stretch. Any trip that starts and ends level with its ghost has an instant like this. Only an impossible car gets away, and you’ll meet two soon.

The ghost comes back

This idea isn’t really about cars, so let’s see it on a graph, with a Falcon 9 rocket and its ghost. If you took precalculus with us, you’ve met this ghost before. In two minutes the rocket climbs about forty-one kilometers, so a ghost climbing a steady three hundred forty-five meters per second ties it at both ends.

On the graph, the ghost’s path is the secant line, and its slope is the rocket’s average climb rate. The rocket’s climb rate at one instant is the slope of the tangent line, and at thirty seconds the tangent is still much flatter than the ghost’s line. Now I’d like you to find that instant yourself, by sliding the tangent along the curve until it runs exactly parallel to the ghost’s line.

Now watch the ghost’s lead as they race. It grows for over a minute, then at sixty-nine seconds it stops growing, exactly at the instant you found. Look how far apart they are, with the ghost almost thirteen kilometers higher, yet for this instant both rockets climb at exactly the same rate. There’s an even simpler picture. Slide the ghost’s line down without tilting it, and the point where it just touches the curve is our instant.

Now try to beat it yourself. Draw any flight you like from the pad up to forty-one kilometers at two minutes, and try to draw one with no parallel tangent. Every smooth curve does this, and that drawing, a tangent parallel to the secant, is the whole theorem.

Let’s carry the rocket’s altitude onto the board. It gains about forty-one kilometers in two minutes, so its average climb rate is about three hundred forty-five meters per second. At the matching time, call it c, about sixty-nine point three seconds, the tangent has that same slope, so the rate at that instant equals the average rate.

The Mean Value Theorem

We call this the Mean Value Theorem, because mean means average, just like the mean of your test scores. It says that somewhere inside the interval, the instantaneous rate of change equals the average rate of change.

Before we claim that a matching tangent exists, we check two conditions, that f is continuous on the closed interval from a to b and differentiable everywhere inside it. Then there’s at least one c between a and b where f prime of c equals f of b minus f of a, over b minus a. The right side is the secant’s slope, the average rate, and the left side is the tangent’s slope at c, the rate at one instant.

Our car gave us two matching instants, so at least one leaves room for both, and each one lies strictly between a and b, out on the road, never at a camera. The promise only holds when f is continuous all the way to the ends and differentiable in between. In a few minutes you’ll try to cheat it, and you’ll see why.

What it never tells you is where c is, only that c exists, and that was all we needed for the ticket. If you do want to know where, you solve f prime of c equals the average rate. For the rocket, a calculator lands on that same sixty-nine point three seconds.

Let’s find a c we can compute, with f of x equals x squared on the interval from one to three. Since x squared is continuous and differentiable everywhere, the theorem applies, and step one is the secant’s slope, nine minus one over three minus one, which is four.

The derivative of x squared is two x, so two c equals four, and c equals two. Two is inside the interval, so it counts, and there’s its tangent, parallel to the secant. Two happens to be the midpoint, which always happens with a parabola, so don’t let it trick you into thinking c is always in the middle.

Let’s try one together. For f of x equals x cubed on the interval from zero to two, find the c the theorem promises, to three decimal places.

Try to beat the camera

Back to the glen, because now you get to try to beat this camera. You start at the first camera doing sixty-eight, and you can drive the four miles any way you like, speeding up and braking as often as you want. Your mission is to reach the second camera without your speedometer ever showing your own average speed. You won’t know your average until you finish, so the replay will find it for you.

Hold the accelerator to speed up, let go to ease off, and see if you can get through without ever hitting your own average. Let’s break each condition and see what the car would have to do. The first cheat changes speed in zero time, from sixty straight to a hundred. Sixty for half the time and a hundred for the other half still averages eighty. But your speed leaps right over eighty, and on the distance graph that leap is a sharp corner.

The second way is even sillier, because you drive at sixty the whole way and teleport a mile up the road halfway through. That’s three miles of driving plus the one-mile jump, so you still arrive in three minutes. Your speed never reads eighty, and on the graph the teleport is a break in the curve.

Those two cheats are exactly what the theorem’s conditions rule out. Continuous means no teleporting, and differentiable means the distance graph has a slope at every instant inside, which a corner doesn’t. Now you cheat. Drive the section again, use one of the two cheat buttons, and see what the replay says.

On the exam, a cheat looks like the absolute value of x from negative one to one, where the secant is flat. The tangents have slope negative one on the left and one on the right, and the corner has no tangent at all, so nothing is ever as flat as the secant. Careful, though, because when a function breaks the rules, the theorem just stops promising.

Look at this one, the absolute value of x, minus x squared. Its secant is flat again, and its corner at zero still breaks the differentiability condition, yet you can see two horizontal tangents, at minus a half and plus a half. So when a condition fails, the theorem stops promising, but a matching tangent might still exist, like these two.

Let’s see if you can spot the cheats. Which of these four functions is the theorem guaranteed to work for, on its interval?

Write it like a grader wants

Now you can write the proof I promised at the start, the way the AP exam wants it. Call the car’s distance past the first camera s of t, with t in minutes. A real car can’t teleport or change speed instantly, so s is differentiable, and therefore continuous. Now we calculate the average rate with the actual numbers. Four miles over three minutes gives four thirds of a mile per minute, or eighty miles per hour.

Then name the theorem and its conclusion. There’s a time c strictly between zero and three where s prime of c equals four thirds of a mile per minute, which is eighty miles per hour. Last, say it in plain English with units. At one instant between the cameras, this car was doing exactly eighty miles per hour, ten over the limit.

Now suppose a student writes only this: by the Mean Value Theorem, s prime of c equals eighty miles per hour. If you were grading it, what would you need the student to explain before you accepted it?

Students drop that first line all the time, and on the exam it costs the point even when everything else is right. That’s the promise kept. From two photos and a stopwatch, you just proved what that car did when nobody was watching.

On the exam

The exam loves to ask this with a table. This one gives a skydiver’s downward velocity at five moments, and the problem tells us that v is differentiable. The speed is forty at five seconds, fifty-two at both twelve and twenty seconds, and twelve at thirty, after the parachute opens. We don’t know the speeds in between. The first question asks whether, at some time between zero and five seconds, the acceleration must be exactly eight meters per second squared.

Acceleration is the derivative of velocity, so this is about v prime, and we use the Mean Value Theorem on v itself. The average rate of v from zero to five is forty over five, which is eight meters per second squared. Since v is differentiable, it’s continuous too, so the Mean Value Theorem guarantees a time between zero and five when the acceleration is exactly eight.

The exam also asks whether there must be a time when the acceleration is exactly zero, so which interval guarantees it? Then comes a sneaky follow-up that catches a lot of people. Must there be a time when the speed itself is exactly forty-five? Which theorem would you use for that one?

Keep this rule for questions like these. A value of the function between two values you know points to the Intermediate Value Theorem, and a derivative equal to an average rate points to the Mean Value Theorem.

Here’s the last one on your own. From twenty to thirty seconds the speed falls from fifty-two to twelve, so what acceleration does the Mean Value Theorem guarantee on that interval?

Somewhere, exactly eighty

When something changes smoothly, at least one instant inside an interval has a rate exactly equal to the average rate. On a graph, that’s a tangent parallel to the secant, strictly inside the interval, and you only get the promise when the function is continuous and differentiable.

It tells you c exists, not where, and to find it, you solve f prime of c equals the average. Somewhere on that road, at an instant no camera saw, that car was doing exactly eighty, and now you can prove it. Next time, we’ll use the same kind of thinking to find the highest and lowest points a function can reach.