# Part II: Transition from Anti-Derivatives to Integrals

In my last post, I outlined the road trip scenario I used to prime kids to think about areas under curves. It had nothing to do with anti-derivatives. And that’s important to keep in mind. This post is going to try to outline how we made the intellectual leap from areas under curves to anti-derivatives. [Update: I wrote this during 40 minutes of free time I had in school today where I didn’t want to do other things I was supposed to be doing. So I didn’t get to writing the part of the lesson where we make The Leap. This post ends where we’re literally all primed to make the leap. But indeed! I will make the leap with you! But in Part III. Which will be written. Soon. I hope.]

The road trip introduced this idea that kids can approximate how far someone traveled using a left-right-midpoint Riemann Sum approximation (we did not give it that name…). It arose naturally from the roadtrip scenario.

We also made the conclusion that if we had more data, we could get a better approximation for how far someone traveled. To remind you, we started with this data:

and then we got more data:

That’s going to be our transition. We are now going to give infinite velocity data! [1]

Wonderfully, kids had no problem with doing this. The reason I highlighted question 1c is because I was very intentional about including that question. When students graphed 1e, they often would draw:

I didn’t correct anyone while they were working. And it was nice to hear a few groups have the requisite conversation about why we needed to connect the points. Afterwards, when we debriefed as a whole, this was something I highlighted. We knew the position at every moment in time, including at t=2.31 (as asked in 1c).

Kids continued on with Questions 2, 3, and 4. They flew through these, actually.

These were golden. Let me say that again: these were golden. [2] It was amazing to watch kids:

• Parse the connection between velocity graphs and position graphs
• Understand the idea of negative velocity
• Think about the fact that we have to specify an initial position in order to create a position graph
• Draw a connection between motion on a number line and a graph of position v. time
• Understand what distance and displacement are, and see the difference between the two

Seriously, just watching kids work through these problems was… well, I’ll just say this feels like something I’m super proud of creating. It didn’t take much time to do but gave us so much fodder.

We didn’t need a lot of time to debrief these four questions. I had students highlight a few things, and I made sure we brought up the fact that we were drawing line segments for the position graphs, and not something curvy. Because constant velocity means position is changing at a constant rate, it’s linear. So for example, the position graph for 3c would look like what I have below. It isn’t a parabola.

But there was one huge thing we had to go over. With the roadtrip, we drew a connection between area and how far someone went. Most kids, as they were doing these problems, didn’t think about area. I wanted kids to think about area. So in our debrief, I explicitly asked them what our huge insight from the roadtrip was (area as distance travelled!), and if we could apply it to one of the problems.

So first I went back to question 2c. And I asked students how they calculated their answer:

Kids said she went backwards a total of 24 units. So they did $100-24$. And then I explicitly had them draw the connection to what we did with the roadtrip. This is when we talked about it being area, but “signed” to represent the backwardsness.

To be clear, some students had already been thinking in this way (about area/signed area) when working on these problems. But most hadn’t been, so we had to bring that idea to the forefront.

Then I had a student talk through Question 3 with areas in mind:

And finally, I asked groups to discuss how we could understand distance and displacement from the velocity graph.

There is one more part to this packet I had my kids work on that I will outline in my next blogpost in the series! But here’s an editable .docx of the file I made [2018-05-02 Velocity Graphs]. And here’s the document to view here:

Stay tuned for Part III.

[1] Both @calcdave and I stumbled upon the same approach for this!

[2] The only note is that a few students didn’t realize the time interval was 1/2 hour for Question 4. And it involves a fair amount of calculation.

# Part I: Transition from Anti-Derivatives to Integrals

An Important Prelude: On one of the early days in my calculus class, I have groups imagine a roadtrip. It’s one of the very best things I do, because it problematizes the idea of how something can be going at a particular speed at a moment in time. Like a speed involves a rate of change of position in a time interval, but we don’t have a time interval at a moment in time. So saying something like “we were going at 58mph at 2:03pm” suddenly goes from a statement kids accept to a problematic statement. And at that moment, we’re ready for calculus.

Setting the Stage Now: In calculus, I had gotten my kids to take tons of derivatives, and then taught them about antiderivatives and how to take them. But of course the whole time they were doing antiderivatives, they were asking “but why are we doing this?” I got this question a few times, but I just said “you’ll see very soon… probably next week… I just want to get some of this algebraic thinking out of the way so we can focus on concepts.” [1]

The Problem: I looked back at what I’ve done in the past about how to draw the connection between the antiderivative and signed areas together, and had a bunch of pretty terrible methods. I didn’t like any of them. So I put out a call on twitter while brainstorming myself. [2] I got a ton of responses, three of which stood out to me. The first one is something I kinda did at the start of the year already (before we did any calculus), so I’m excited for when my kids see the connection…

The second came was from the appropriately named @calcdave:

What’s incredible is that when I tweeted out asking for help, I had already started brainstorming … and it was so amazing that we had such similar ideas that I had to take a photo of my scratch work to send David:

I also got a DM from Brett Parker (@parkermathed) with a screenshot of how he digs into the idea. And what was his idea?

Yup. Driving in a car. Which of course got me thinking of the start of the year when I basically transitioned to derivatives using a road trip.

What I did: Part I

So I told kids we were going to put a pin in anti-derivatives for a short while to go on a short detour. And I used Brett’s setup, but changed it to be a followup from the roadtrip we had talked about earlier this year. We read the setup together, and then I gave each group a few minutes to talk about part (a).

Surprisingly, this was not an easy question for kids. Many didn’t instantly think distance=(rate)(time). Additionally they didn’t know what assumption to make about the speed for the minute that passed between 7:10 and 7:11. I emphasized the approximate part of the question, and really told kids they would need to make an assumption.

When we debriefed, most kids suggested Alex was probably driving 69mph for the minute, so they did $69\cdot\frac{1}{60}$.

Some suggested he was maybe going 70mph for most of the minute, so they suggested $70\cdot\frac{1}{60}$. And with that, I suggested that maybe he was going 68mph for most of the minute… We did the calculations for all three and saw they were pretty similar.

(We also had discussed that in that minute, perhaps Alex started at 70mph, went to 100mph, and then slowed down to 68mph… We just didn’t know. So we were making and using an assumption, but one that is pretty reasonable.)

Then kids in their groups went to the next three parts. And each group was assigned an assumption: average speed, left hand speed, or right hand speed.

The only errors I saw kids make in part (b) was not taking the different time intervals into account. Since we did one example which had a time interval of $\frac{1}{60}$ hour, some groups were using that time interval for everything. But pretty rapidly most kids got there.

In our debrief, I wrote out the calculation kids did for the left hand assumption, and then asked students what I would have to change to do the calculation for the right hand assumption. (That was written in yellow.) This question was a key question to ensure kids understood the difference.

Then we talked why this was an approximation? (We had to make assumptions about the speed of the car at all the times inbetween the times we were given.) And then kids said that to get a more accurate distance for Alex, we’d need more data.

And that’s precisely what they got when they flipped the sheet over.

I assigned some groups the left hand assumption, and some groups the right hand assumption. I thought with so many data points, kids would be like “argh! I have to do all these calculations!” But no, there were no audible groans (I don’t think…) And they plotted the points and then did the calculations. I told kids to write out all their calculations instead of just saying how far Alex drove in the first minute, the second minute, the third minute, etc. No everyone listened. Shame. They lost out. Because when we debriefed, we saw:

Yup. The fact we could factor out the 1/60th is key. It made this all go so much faster. Only a few kids noticed that when doing the calculations. And then we compared that with the right hand assumption:

The same answer! Why? Kids saw that the sum inside the brakets (after factoring out the 1/60) would be the same because the starting speed and ending speed were both the same.

So we’re done, right? NO. We had two more moves to make.

First, we looked at the graph.

And I asked: “if we were doing the left hand assumption, what would our velocity graph look like?” And we concluded that for each 1 minute interval, Alex was driving at a constant rate. So it would look like this:

Second, I asked: Between 7:05 and 7:06, how much are we assuming Alex went? (Kids answered $67\cdot\frac{1}{60}$) Here’s where I did some talking. I could probably have asked kids to think geometrically and had them come up with this, but I was running out of time.

I said: “where do we see 67 represented on this graph?” (Kids said the height of the velocity graph from 7:05 to 7:06.) I then said: “where do we see the $\frac{1}{60}$ on this graph?” (Kids said the 1 minute length.) So I drew a rectangle. Slowly. And then shaded it in. Slowly. And turned around. Slowly.

Yes. Audible gasps from many. I then said: “What about 7:04 to 7:05? What was our calculation for the distance Alex traveled?” And after a few more, we saw:

Kids saw. Heck yeah. They got that the approximate distance that Alex traveled was algebraically calculated in one way. And then they saw that number had a graphical representation. It was awesome.

I left by showing the original velocity points graphed. Reminded them of our left hand assumption, and that it was just an assumption, though a pretty good one. And then I drew the curve below. And dropped the microphonic device and left.

All in all, this was a pretty detailed blow-by-blow. And since I did it in two different classes, and things unfolded slightly differently in each, this is an amalgam of what happened. But it’s a pretty solid recap of the story I wanted to tell and how it was told. (And it’s a testament to the help of my (twitter) frands.) Most of the time, students were working. But a lot of great conversations happened as a whole class. It is a long post, but the question we worked on was only one page [editable version to download: 2018-04-30 A Road Trip Reprise] And it probably only took 20 minutes total from start to finish. A really exciting 20 minutes for me.

Stay tuned for how I used the idea that @calcdave and I both stumbled upon to make the connection between area and anti-derivatives. Right now kids have seen that there is a connection between area under a velocity curve and the distance someone has traveled. There is still no connection to anti-derivatives. That’s coming up.

[1] To be honest, I made the decision years ago to do tons of antiderivatives before introducing the integral. I wanted all the algebraic work done and solid before we introduced the concept of the integral. I didn’t want kids messing with the idea of signed area, new notation, and tough antiderivatives all at the same time. I still kind of think this is the right decision, but some doubts have crept in. I just really hate when I have to say “you’ll see why… promise…”

[2] Apparently I asked this on my blog ages ago in 2009.

# For a sphere, why is the derivative of the volume the equation for the surface area?

Yeah, so the title of the post says it all. I am teaching a standard calculus course, and I wanted my kids to see why this beautiful thing holds true.

It’s not a coincidence. And in fact, a circle also has a nice property: the derivative of the area of a circle ($A(r)=\pi r^2$) gives you the circumference of the circle ($C(r)=2\pi r$). So yesterday I decided I wanted to come up with a short investigation that at least exposes my kids to this idea.

After working for around 90 minutes this morning, I ended up with a packet, and these things on my desk which I’m going to use for illustrations (blocks, dumdums, and tape):

UPDATE: I was shopping yesterday and found these gems. YAAAS!

I’ll post the packet I whipped up below. It goes through the standard argument, so in that way it’s nothing special. But in the past when I taught the course, I used to just kinda stand up at the board and give a 5 minute explanation. But I wasn’t sure who was really grokking it, and I was doing too much handwaving.

The big picture trajectory:

*At the start of class, but way before doing this activity, I’m going to have kids recall what a derivative is graphically (the slope of a tangent line), and then how we approximated it before we used limits (the slope between two points close to each other). And from that, I’m going to remind kids of the formal definition of the derivative:

* I may also start class with this problem, suggested to me on twitter by Joey Kelly (@joeykelly89). It’s a classic problem that was featured on xkcd, but oh so unintuitive and surprising!:

*Way later in class, I will transition to this activity. The first idea is to get kids to see the connection between the volume of a sphere and the surface area of a sphere. And then again for the area of a circle and the circumference of a circle.

*Then I try to get kids to understand what’s going on with the sphere first… followed by the circle.

*Then I show kids the “better explained” explanation. Why? Because at this point, kids are spending a lot of time thinking about the algebra, and I’m afraid they might have lost the bigger picture. The algebra focuses on one “shell” of the sphere, or one “ring” of the circle. But how does it all fit together? [@calcdave sent me this video, which I’d seen before but forgot about, which has the same argument… this is where the licorice wheels above come into play.]

*Finally, I problematize what they’ve learned. I have them mistakenly make a conjecture that the derivative of the volume of a cube is going to be the surface area of the cube, and the derivative of the area of a square is going to be the perimeter of the square. But quickly kids will see that isn’t quite true. So they have to tease out what’s happening.

My solutions:

I haven’t taught this yet. So it could be a complete disaster. I don’t have a sense of timing. I don’t know how much of this is me and how much will be them. I am just hoping tomorrow isn’t a disaster! Fingers crossed!

# There might be light at the end of the Chain Rule tunnel… maybe.

This is going to be a half-formed post. I wanted to get a conceptual way for kids to grok why the chain rule works in calculus. But without doing too much handwaving. And I wanted something visual.

The hard part is: if we have a function $g(f(x))$, we can approximate the derivative at a particular point by doing the following.

Find  two points close to each other, like $(x,g(f(x))$ and $(x+0.001,g(f(x+0.001))$.

Find the slope between those two points: $\frac{g(f(x+0.001)-g(f(x))}{(x+0.001)-x}$.

There we go. An approximation for the derivative! (We can use limits to write the exact expression for the derivative if we want.)

But that doesn’t help us understand that $\frac{d}{dx}[g(f(x)]=g'(f(x))f'(x)$ on any level. They seem disconnected!

But I’m on my way there. I’m following things in this way: $x \rightarrow f \rightarrow g$

Check out this thing I whipped up after school today. The diagram on top does $x \rightarrow f$ and the diagram on the bottom does $f \rightarrow g$. The diagram on the right does both. It shows how two initial inputs (in this case, 3 and 3.001) change as they go through the functions f and g.

At the very bottom, you see the heart of this. It has $\frac{\Delta g}{\Delta f}\cdot\frac{\Delta f}{\Delta x}=\frac{\Delta g}{\Delta x}$

And then I thought: okay, this is getting me somewhere, but it’s to abstract. So I went more concrete. So I started thinking of something physical. So I went to how maybe someone is heating something up, and in three seconds, the temperature rises dramatically. The temperature measurements are made in Farenheit, but you are a true scientist at heart and want to see how the temperature changed in Celcius.

And then of course when I got home, I wanted to see this process visualized, so I hopped on Geogebra and had fun creating this applet (click here or on the image below to go to the applet). These sorts of input-output diagrams going from numberline to numberline are called dynagraphs. You can change the two functions, and you can drag the two initial points on the left around. (The scale of the middle and right bar change automatically with new functions you type! Fancy!)

And of course after doing all this, I remembered watching a video that Jim Fowler made on the chain rule for his online calculus course, and yes, all my thinking is pretty much recapturing his progression.

This, to be clear, is about the fourth idea I’ve had as I’ve been thinking about how to conceptually get at the chain rule for my kids. The other ideas weren’t bad! I just didn’t have time to blog about them, but I also abandoned them because they still felt too tough for my kids. But I think this approach has some promise. It’s definitely not there yet, and I don’t know if I’ll have time to get there this year (so I might have to work on it for next year). But I know to get there I’ll have to focus on making the abstract very tangible, and not have too many logical leaps (so the chain of logic gets lost).

If I’m going to create something I’m proud of, kids are going to have to come out saying “oh, yeah… OBVIOUSLY the chain rule makes sense.” Not “Oh, I guess we did a lot of stuff and it all worked out, so it must be true.”

A blogpost of unformed thoughts, and an applet. Sorry, not sorry. This is my process!

# POP! Popcorn Optimization Problem

I’m in the middle of optimization in my calculus class now. I had a “long block” (every seven school days, I see a class once for 90 minutes) and for the second half of that long block, I like to do something slightly different. Since I knew my kids hadn’t seen or done the traditional “box optimization problem” in precalculus (since I taught them last year also!), I decided to do that.

This might jog your memory if you don’t know what I’m talking about. You take a piece of paper. You cut out four squares (the same size squares) from the corners. You then fold up the four flaps and tape the box shut. There you go!

You can probably tell that the box’s volume is going to be based on the original paper you start with, and the size of the square you decide to cut out. The question is: what’s the largest volume you can get for this box.

If you cut out a teeeny tiiiiny square, you’re going to have a very large base for the box, but almost no height. And if you cut out a giant square, you’re going to have a large height but a teeeeeeeny tiiny base. And somewhere between a teeeeeny tiny square and a giant square is going to be the perfect square to cut out which will give you the largest volume.

So the question is: given a specific piece of paper, what size square do you need to cut out to get the maximum volume.

This question has been done to death in middle school classes, in Algebra II classes, in Precalculus classes, and in Calculus classes. And I recognize that this post is just another rehashing of the same old problem. But I remember reading about a teacher who did a variation of this by including popcorn. And I wanted to do the same. No surprise, when I looked it up, it was dear Fawn. But I had such a lovely time in class today watching this unfold that I wanted to share the specific sheet I made up for kids to do this.

Teacher Moves / Outline

This activity requires students knowing and using the quadratic formula. My kids (standard level calculus) are pretty weak with algebra, so I started the class with a “do now” that had kids use the QF. So I recommend that.

Show kids the popcorn. (I had two different flavors.) Show your excitement about the activity. (I was genuinely excited!) Get this psyched. Hand out the worksheet but nothing else.

Put a three minute timer on the board. Explain the problem. Show kids a piece of cardstock with 4 squares drawn on it. Show kids a second piece of cardstock with those same four squares cut out and the flaps folded up so it looks like a box (but untaped, so you can unfold it too). Tell them the volume they create is the amount of popcorn they are going to get. And that you aren’t going to overfill their boxes — just to the brim. Tell them they have 3 minutes to work with their partner to come up with the best size square they want to cut out. And they are not allowed to do any calculations. Just visual estimation.

At that point, give cardstock, ruler, scissors, and tape to kids. Do not let kids start until you press “GO” on the timer. Then… GO!

After three minutes, my kids were done. They measured the side length of the square they cut out and recorded it on the worksheet. They then cut and taped. They weren’t allowed to get their popcorn until they did one more thing… some math…

It was super important to me that kids didn’t measure anything, except for the side length of the square, to do these problems. Why? Because this is where I want kids to recognize the side length of the square is the height (that was obvious to all my kids), but also that when calculating the length and width, they were going to be doing 216-2x and 279-2x (where x was their side length). Only a few kids didn’t get the 2bit (they only subtracted x), but I sent them back to their seats to rethink their length and width and they immediately got it. It was actually awesome to hear their OOOOOOHHHHH moment. But yeah, no measuring. They have to use their brainzzz to come up with the length and width with what they are given!

Only after checking their volume with me, and I said it was correct, could they fill their boxes with popcorn.

As an aside, when writing this activity, I had to decide what level of scaffolding I wanted to give for this. I decided not much. So I didn’t include any diagrams. (Well, I did put two on the very last page of the worksheet in case a kid needed some additional help. Turns out no one did.) I also initially wrote the worksheet to be in inches, but then changed to centimeters, and then after thinking a bit more, I changed to millimeters. Why? So kids don’t have to deal with fractions (inches) or decimals (centimeters), and we could keep our eye on the prize. It also made the volume huge — and so kids would have to do a little work to get the correct window when graphing.

At this point, I sent them back to their seats with popcorn in their box to then solve the general case. Close to the end of class, I posted the different volumes students got by estimation (it was a tiny class today… kids were absent or at sports).

Overall, I spent about 35 minutes on this in class. One pair finished completely. All the others are at the place where they are in the middle of the calculus work (close to being done).

# Play! Create! Adult!: My Second TMC17 Recap Post

Here are some more TMC17 notes!

I love the idea of having kids engaging in recreational math. I don’t have much time to encourage that in my curriculum — or at least the only way I’ve found for that to happen is with my explore math project [posts 1, 2, 3; website]. Some kids get some extra math problems to work on at math club (usually problems from math competitions or brilliant.org), and kids do math problems on our math team. But that isn’t the spirit of what I want to bring to my school. I want to get kids just fooling around with math for fun! Tinkering! Thinkering! Building! Collaborating! So that’s why I fell in love with Joey Kelly (@joeykelly89)’s my favorite presentation. Where he shared with us Play With Your Math.

He and a friend created it. Right now it has 15 sheets of paper that can be printed out, each with a challenge. The name, inspired. Design wise, fantastic. But the problems are captivating, easy to dive into, and many have this open-endedness that can lead to obsession. When I was at the Desmos Fellowship a couple weeks ago, they had these for us to work on as a way to get to know each other. Each table had a different one and we were encouraged to play, and meet others who were playing, and then move to a different table and meet and play when we felt like it. The one I spent all my time on, trying to come up with a strategy? One that I know will get my kids in competitive mode? Poster 5:

I liked getting to know people and I liked these problems! At TMC we were given poster 14 and I became obsessed. And eventually, I solved it (and a second more complicated one). But it took A LONG TIME and I DIDN’T CARE. I refused to go play boardgames at gamenite until I had climbed this mountain!

I need to brainstorm if and how I am going to use these in my school. Some initial ideas:

1. Leave copies of these in the library for kids to use. Or put many copies of all of them on a bulletin board for kids to take, so when they’re board and standing there, they just grab one and start thinking.

2. Use these when I need to fill a long block (we have double periods one out of every five times we meet our kids) and I don’t have a good idea.

3. Plan an Upper School math night, where we gather at a space in the school, do math, order pizza. Like PCMI’s “pizza and math” (was that what it was called? we can do better!). These can be the amuse bouche or the main event!

Math Art!

Speaking of recreational math, at TMC17 there was so much math art. I just wanted to share some of it!

Captivating! I hope at some point to learn how to make crochet coral. It feels like once I get in the rhythm, it could be so soothing. Actually, I wonder if it would be fun to have a MAKER MATH club where we make math stuff together. And create our own math art gallery. Things like the things shown here, but also like these, and origami (demaine and lang), and a menger sponge made of business cards, and design and 3d print these optical illusions, and carefully color in pictures from Patterns of the Universe, and create our own mathart coloring pages. If you are reading this and have ideas of things that we could make, let me know in the comments! You probably can tell this is something I’m actually totally *feeling* (FYI, for me, the definitive math art page is @mathhombre’s page here.)

So @rawrdimus gave a my favorite on how to adult. He was teaching calculus and wanted to keep his seniors engaged. So he came up with this project that had kids pick a few houses and figure out what they’d need to buy it. He was the banker (a hilarious banker) and gave them two different mortgage options (a 15 year and a 30 year, with different interest rates) and they had to figure out their monthly payments.

I know come the spring, the kids in my calculus class will have their attention wane. So I think something like this could work (this investigation on wealth inequality worked a few years ago)! But right now it’s a little bit like trying to put a square peg into a round hole. I need it to have some more calculus before I do something like this though. Maybe we’ll spend some time talking about e or we’ll do something with summing (in)finite geometric series, and maybe seeing that as a riemann sum? I think it’s totally doable — I just need to think a bit more! But if you want to get a sense of why I’m trying to make this happen, just watch Jonathan’s presentation and you’ll totally get it. (Here’s his blogpost.)

# The Formal Definition of the Derivative, or Why Holes Matter

Lucky you! Two calculus posts in one day. Mainly because I don’t want some of these ideas to disappear in my hiatus from teaching it. This one deals with our favorite topic: the formal definition of the derivative.

$\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{(x+h)-(x)}$

I see that expression and my mind goes to the following places:

• Doing a bunch of tedious algebraic calculations for a particular function in order to find the derivative.
• I “see” in the expression the slope of two points close together.
• I envision the following image, showing a secant line turning into a tangent line

And I think for many teachers and most calculus students, they think something similar.

However I asked my (non-AP) calculus kids what the $h$ stood for. Out of two sections of kids, I think only one or two kids got it with minimal prompting. (Eventually I worked on getting the rest to understand, and I think I did a decent job.) I dare you to ask your kids and see what you get as a response.

What I suspect is that kids get told the meaning of $\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{(x+h)-(x)}$ and it gets drilled into their heads that they might not fully understand what algebraically is going on with it.

It was only a few years ago that I came to the conclusion that even I myself didn’t understand it. And when I finally thought it all through, I came to the conclusion that all of differential calculus is based on the question: how do you find the height of a hole? I started seeing holes as the lynchpin to a conceptual understanding of derivatives. I never got to fully exploit this idea in my classes, but I did start doing it. It felt good to dig deep.

The big thing I realized is that I rarely looked at the formal definition of the derivative as an equation. I almost always looked at it as an expression. But if it’s an equation…

$f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{(x+h)-(x)}$

… what is it an equation of? An equation with a limit as part of it?! Let’s ignore the limit for now.

Without the limit, we have an average rate of change function, between $(x,f(x))$ and $(x+h,f(x+h))$. And since we have removed the limit, we really have a function of two variables.

$AvgRateOfChange(x,h)=\frac{f(x+h)-f(x)}{(x+h)-(x)}$

We feed an $x$ and $h$ into the function, and we get an output of a slope! It’s the slope between $(x,f(x))$ and $(x+h,f(x+h))$!

Let’s get concrete. Check out this applet (click the image to have it open up):

On the left is the original function. We are going to calculate the “average rate of change function” with an x-input of 1.64 (the x-value the applet opens up with).We are now going to vary h and see what our average rate of change function looks like: $f(1.64,h)=\frac{f(1.64+h)-f(1.64)}{h}$. That’s what the yellow point is.

Before varying h, notice in the image when h is a little above 2, the yellow “Average Rate of Change” dot is negative. That’s because the slope of the secant line between the original point $(1.64, f(1.64))$ and a second point on the function that is a little over 2 units to the right is negative. (Look at the secant line on the graph on the left!)

Now let’s change h. Drag the point on the right graph that says “h value.” As you drag it, you’ll see the second point on the function move, and also the yellow point will change with the corresponding new slope. As you drag h, you’re populating points on the right hand graph. What’s being drawn on the right hand graph is the average rate of change graph for all these various distances h!

Here’s an image of what it looks like after you drag h for a bit.

Notice now when our h-value is almost -3 (so the second point is 3 horizontal units left of the original point of interest), we have a positive slope for the secant line… a positive average rate of change.

The left graph is an $x-f(x)$ graph (those are the axes). The right graph is a $h-AvgRateOfChange$ graph (those are the axes).

Okay okay, this is all well and dandy. But who cares?

I CARE!

We may have generated an average rate of change function, but we wanted a derivative function. That is when h approaches 0. We want to examine our average rate of change graph near where h is 0. Recall the horizontal axis is the h-axis on the right graph. So when h is close to 0, we’re looking at the the vertical axis… Let’s look…

Oh dear missing points! Why? Let’s drag the h value to exactly h=0.

The yellow average rate of change point disappeared. And it says the average rate of change is undefined! 0/0. We have a hole! Why?

(When h=0 exactly, our average rate of change function is: $\frac{f(x+0)-f(x)}{(x+0)-x}$ which is 0/0. YIKES!

But the height of the hole is precisely the value of the derivative. Because remember the derivative is what happens as h gets super duper infinitely close to 0.

We can drag h to be close to 0. Here h is 0.02.

But that is not infinitely close. So this is a good approximation. But it isn’t perfect.

And this is why I have concluded that all of differential calculus actually reduces to the problem of finding the height of a hole.

Here are three different average rate of change applets that you might find fun to play with:

one (this is the one above)     two     three

In short (now that you’ve made it this far):

• Look at the formal definition of the derivative as an equation, not an expression. It yields a function.
• What kind of function does it indicate? An average rate of change function. And in fact, thinking deeply, it actually forces you to create a function with two inputs: an x-value and an h-value.
• Now to make it a derivative, and not an average rate of change, you need to bring h close to 0.
• As you do this, you will see you create a new function, but with a hole at h=0.
• It is the height of this hole that is the derivative.

PS. A random thought… This could be useful in a multivariable calculus course. Let’s look at the average rate of change function for $f(x)=x^2$:

$AverageRateofChange(x,h)=\frac{(x+h)^2-x^2}{h}$

Let’s convert this to a more traditional form:

$z=\frac{(x+y)^2-x^2}{y}$

Now we have a function of two variables. We want to find what happens as h (I mean y) gets closer and closer to 0 for a given x-value. So to do this, we can just visually look at what happens to the function near y=0. Even though the function will be undefined at all points where y=0, visually the intersection of the plane y=0 and the average rate of function should carve out the derivative function.

If this doesn’t make sense, I did some quick graphs on WinPlot…

This is for $f(x)=x^2$. And I graphed the plane where y=0. We should get the intersection to look like the line $f'(x)=2x$.

Yup. Cool.

I did it for $f(x)=\sin(x)$ also… The intersection should look like $f'(x)=\cos(x)$.