I am thinking about re-organizing the first half of my physics classes to push much of kinematics until later — maybe much later! Here is what I may go with:

Start with vectors…displacements, forces, velocities (and for now, velocity is just a fancy word for speed)

This would include vector addition, vector components, an intro to the essential right-triangle trig, and applications including “river crossing” and equilibrium problems (hanging signs and such).

Next, a minimal introduction to kinematics…

This would include a definition of average velocity, and an examination of x vs t and v vs t graphs for constant positive velocity. Then, a look at constant acceleration: its x vs t graph, its v vs t graph and the definition delta V over delta t. BUT NOTHING YET ABOUT: area under v vs t, kinematic formulas, word problems, positive vs negative velocity and acceleration…

QUESTION #1 FOR MY COLLEAGUES: Is that enough kinematics to take on Newton’s laws? Am I forgetting something?

On to Newton’s Laws…

It seems crazy to me how long it can take me to get to these if I do a full treatment of kinematics (and even projectiles!) before Newton’s laws. If I adopt this new arrangement, I think I can start Newton’s laws in September. This does mean that I can’t spiral F=ma problems which require kinematics to get the acceleration. But eventually I can spiral the other way: put F=ma into kinematics problems.

On to…well, I am not sure…

At first, I was thinking that next I would go back and finish up kinematics and projectiles. Then, on to work and energy, followed by momentum. (I know others prefer to reverse those two…I am not religious about that.)

But then I got to wondering: could kinematics and projectiles even wait until after all of this? I might have to skip the derivation of W = delta KE but that feels minor. Could kinematics and projectiles come right before circular motion? From projectiles right into circular motion feels kind of natural.

So that leads to QUESTION #2 FOR MY COLLEAGUES: If you delayed kinematics in the way I am considering, when would you go back to it?

I don’t want to open comments on this blog (too much weird spam) so please find me on twitter @pckeller if you would like to chime in. Thank you!

And anyone else who wants an overview of what we do when we “do calculus”

AP Physics C is a calculus-based course. At my school, calculus is a co-requisite. It would be good if we could use the language of calculus right from the start. But most of the students have only seen a little calculus so far (or none at all). So over the last few years, I have written these blog posts to try to get them off to a running start.

Also, while it was not my intention, I think these posts would help anyone heading into their first calculus class to get an advance overview of what lies ahead.

Note to my students: For some of you, most of this will be a review – but definitely not all of it! But for others, this will all be new. Just take your time, read slowly, and feel free to ask lots of questions — you can save them for September or email me as you think of them! Or post them in our Google Classroom (which is already open).

One other note: we will be talking a lot about rotational physics this year. It would be helpful to have a clear understanding of radian measure. If you feel at all shaky, you should review these two posts: Angle Measurement for Pizza Crust Lovers and That Radian Feeling.

I hope you have a relaxing summer and that you are looking forward to another year of interesting physics.

Finally, finally we have arrived at the point
all of these posts have been building toward:

In
physics, we only occasionally care about “area”. But we often care about adding up an infinite
number of tiny little things. And when
we do, if we think of them as representing areas, we can use our Magic Theorem
to add them.

I want to demonstrate this using as
little physics as possible. So this next
example may seem a bit contrived. You
will see more significant examples as our year goes on (perhaps more in E &
M than in Mechanics). But this is a good one to begin with:

A beam is 2 meters long. It has a density that varies along its
length. As a function of distance from the left end, the linear density (in
kg/m) is given by the function:

Find the mass of the beam.

Before we jump in, let me say a little
about what we are hoping to do.

We have a quantity that varies
continuously: in this case, it’s the density.
(Also, please do not be intimidated by the fact that we are using the
Greek letter, lambda, to represent the density function. I think we do this because it is a LINEAR
density (mass/length) and so we want a letter that begins with L.)

If we chop our beam into short pieces, we can pretend the density is constant for a given piece. Really, density does vary but if the piece is small, density doesn’t change much. And we are eventually going to make the pieces reeeeally small. This is pretty much the same thing we did last time, when we pretended velocity was constant over a time interval.

We then multiply the length of each piece
by its density to find the mass of that piece and then we add them up.

So here’s the beam, nicely chopped:

We focus on one individual piece, say that blue one, and find its mass:

And we add up all those little masses:

As we have before, we can improve our
estimate by chopping into more pieces – or even better, an infinite number of
pieces:

Now let’s really examine that
expression. It sure looks a lot like a
Riemann sum! It looks like you are
getting ready to find the area under the curve:
f(x)=1+x-x^{2}/2by breaking it into
rectangles. And for that we can use the
Magic Theorem.

We can find the anti-derivative ourselves or use Wolfram/TI/whatever to get:

And then see how much it changes:

We are done! And yes, that is also the area under that graph, but that’s not really the point. We were not interested in finding an area. We just wanted to know the mass. But as we have just seen, the same method that lets us find areas also helps us to add up the little pieces!

Well,
that’s all I wanted to say about the Fundamental Theorem and how it is used in
physics. If you have stayed with me
through this entire 5-part journey, thank you so much for your persistence. I find this to be pretty challenging
stuff. I will be happy to help you
through it again when we encounter this during our year in AP Physics. I hope the time you spent working through
these posts makes it easier for you when you see it again.

Here is an easy warm-up from your first year of physics class:

A car moves at a constant velocity of 6 m/s for 5 seconds. How far does it travel during that time?

When velocity is constant, you can multiply the velocity by the time and get the displacement:

But it is interesting to
look at a velocity vs. time graph that models this situation:

The area between the graph and the t-axis forms a rectangle. So when you are multiplying the velocity (6 m/s) by the time (5 s) you are also finding the area of that rectangle!

But is this really an area? Or is it more of an “area”? Look at the lengths of the sides and note the units. We expect the lengths of the sides of a rectangle to be measured in units of distance. But this one has one side measured in velocity units (m/s) and the other in units of time (s). And when you multiply them, you get a unit of distance, not a typical unit of area. So the “area” on the graph represents some other physical quantity we care about.

Now here’s something your physics teacher might ask THIS year:

A car’s velocity increases with time according to the function v(t)=t^{2}. How far does the car travel in the time from t = 0 to t = 5 s ?

Since we know how to find displacement when the velocity is constant, let’s adapt the idea we used in math class to generate an approximate answer.

Here is our plan:

Divide the time into intervals, each 1 second long.

Use the formula v(t)=t^{2}to find the velocities at the beginning of these intervals.

Use those velocities to find the displacements for each one-second interval and then add them up to find our approximate answer.

So you are pretending that instead of gradually increasing, the velocity stayed constant for 1 second at a time, jumping to the next value at the beginning of each second (depending on whether we want to over-estimate or under estimate our answer).

Why don’t we graph what thAT looks like on a velocity vs time graph?

“It’s like Déjà vu all over again” – Yogi Berra

I hope this looks familiar! It’s just like what we saw in the last post. Treating the motion as if if were a series of constant-velocity intervals is just like approximating the area under a curved boundary by using a bunch of rectangles!

I used a spreadsheet and put the results of these calculations in the chart below. Take a look at the overestimate and the underestimate. You might recognize those numbers…but once you understand what we are doing here, you won’t have to do these calculations at all!

So now we can say that the cart travels somewhere between 30 and 55 meters. You can probably guess what your teacher will say:

“I like your method, but I
wish it were more precise. Can you
divide the interval into more, finer intervals?
Like maybe 10 of them? Or even 50?

[It’s amazing how your math teacher and physics teacher are so in tune. It’s almost as if they were one person…]

Here’s what 50 rectangles
look like (again). Seriously, the only difference is that I labeled the x and y
axes to be velocity and time.

And if we want the exact
answer, we evaluate the limit:

Except we still don’t have to do it this way! We
can still use the Magic Theorem! Really, all that has changed is that we are
calling the function v(t) instead of f(x):

It is true that we have just
found an area under a graph. But what we
also found, and what we actually wanted
to find, is the answer to a physics question: how far did the car
move? It is just a helpful connection
that we can think of that as an area.

IT’S NOT JUST FOR DISPLACEMENT

Here are some other things we
will calculate this year. In each case, we can think of it as an area and then
use the Magic Theorem.

1. Impulse exerted by a force
that varies with time:

This is also the area under
the Force vs time graph.

2. Charge delivered by a current
that varies with time:

This is also the area under
the current vs time graph.

3. Work done by a force that
varies with position:

This is also the area under
the force vs. position graph.

Once you are comfortable with this, you can find these things without even drawing the graphs! Integrating will come to mean: adding up infinitely many tiny contributions to an overall total. And as you will see in the next (and last!) post in this series, that’s another powerful way use integration in physics.

The previous two posts have wrestled with the relationship
between an area (the “definite integral”) and antiderivative (the “indefinite
integral”). In short, they are related
by the Fundamental Theorem, or as I have been calling it, the Magic Theorem:

You can find the area under a graph by finding how much its
antiderivative has changed.

In mathematical notation:

But that still leaves us with the question:

Why do we care about area in physics?

I think the answer to that question is that really, we very
rarely care about the literal “area” as in area of some planar region, measured
in square meters. But we frequently care
about other physical values that we can represent as areas. And the methods we
use to find those areas can be extended to other useful things.

To explain this, in this post we will step back and look at
how you can find area in math class before
you learn the Magic Theorem.

[Just for extra emphasis, I will say again: what follows is not a method you will need in AP Physics. In fact, except for a few days, you won’t even use this method in math class either. But if you want to understand what the Magic Theorem does for you, you have to understand what it replaces.]

In math class one day, your teacher might ask you to find the area between the x-axis and the function f(x)=x^{2}:

Now maybe rectangle areas are the only ones you know:

area = length times width

You are kind of stuck. But your teacher asks you to keep trying and says they are willing to accept an approximate answer. Oooh. Now there is a way forward here.

You could slice the region
into a handful of rectangles. Let’s start with five of them:

You could then calculate the
area of those rectangles and add them up to get a rough idea of the area of the
entire region. [Notice, I say we COULD
do this. It’s a tedious task to do by
hand, and in the end we won’t have to do it. But we could if we wanted to. As
you can see, I let Desmos do the calculations for us.]

In the diagram above, the
rectangles are INSIDE the region, so we have underestimated the area. We could have
chosen to put them OUTSIDE, thus overestimating:

So we can see that the area
we are looking for is more than 30 and less than 55. That’s an approximation, but it’s better than
no knowledge at all. It might even
satisfy your math teacher for a while. But then they come back to you with a
request:

“I like your method, but I
wish it were more precise. Can you
divide your graph into more, finer intervals?
Like maybe 10 of them? Or even 50?

Well, you can do this too.
Here are 10 “outside” rectangles:

And here, side by side, we
have 50 “inside” rectangles and 50 “outside”:

We can now make two
observations about what happens when you use lots of rectangles:

1. The difference between “inside”
and “outside” estimates becomes smaller and smaller.

2. The amount that you are
overestimating or underestimating by also gets smaller and smaller.

If we could some how let the
number of rectangles approach infinity, we would know the true area of this
region. In fact, that limit is the definition of the area of this region.

This way of finding area is
called the Method of Reimann Sums. The
area is often expressed using summation notation:

Setting up these expressions and then calculating the limit as the number of rectangles approaches infinity can be quite daunting. Mastering this method is a rite of passage in calculus class and I admit that it is possible you will not enjoy the process. It is good to understand what Riemann Sums are, but it is also really nice to not have to use them. That’s what is so magical about the Fundamental Theorem — it is so much easier that it feels like magic. Just to remind you of what we do:

1. We find a function which
has x^{2} as its derivative. For
example:

2. According to our magic
theorem, the area we want is the change in that antiderivative:

That’s not so bad! But also, now we have all the pieces in place to answer the question that started us down this road: why do we care about area in physics?

This is the lesson I share with my students in Google Classroom to set the stage for studying Faraday’s Law and induced voltages.

Today I am going to present you with a series of mysteries about motors. Take your time with these and make sure you understand the question before worrying about the answer.

1. Why do electric motors run at constant speed?

We know what makes a motor go: the wire carries current through a magnetic field. The field exerts a force on the wire.. The force accelerates the motor so the motor speeds up. But really think about the electric motors you have used. I’m talking about things like fans, cd players (or record players if you are old school), or even an electric can-opener. All of these devices begin to move when you close a switch, providing them with a current. So they definitely all accelerate — but only at first! It seems that they have an initial period when they are speeding up but that their speed then levels off at some constant value. They do NOT continue to accelerate. And I am asking you: why not?

This is a genuine question and not a trick. And the answer is not “friction”. Even a zero-friction motor would behave this way: initial speed-up followed by velocity leveling off to a constant. Why is that?

If you have a theory to explain this, please go ahead and post it as a comment. And it’s ok to be wrong! I’d be interested in seeing your ideas. But try to use physics-reasoning. The answer is not “magic” though it will certainly seem magical. Or you can just read on, spoilers ahead.

To investigate this mystery further, we are going to focus on a simple kind of motor. I call it “the straight line motor”. You are looking at a U-shaped circuit in a magnetic field. The bar on the right is a conductor that rests on the rails, completing the circuit. That bar is able to slide frictionlessly.

We close the switch. The current flows. And a quick application of the right-hand rule shows that we get a magnetic force, pushing the bar to the right. The bar speeds up, but again, only initially. The speed levels off to a constant value. We still don’t know why…

Before we resolve this mystery, I want to mention that this motor I have described is a real thing! You can see how to make one here: https://www.youtube.com/watch?v=Yvfz51moMaQ

And there is a related item called a “linear motor”. It is slightly more complicated but it illustrates the same principles. You can see a few here: https://www.youtube.com/watch?v=ifBPgVDvvK0

The Solution to MM#1

Let’s look at that circuit again, this time doing a few calculations:

We see we have a 12-volt battery. The resistance of the circuit is a constant, let’s say 3 ohms. We close the switch, expecting a 4-ampere current, which we could use to find the magnetic force (F=ILB) and then the resulting acceleration (a = F/m). But if you could watch the current meter, you would see something we were not expecting, something important:

The current is only 4 amperes for an instant, right after we close the switch. After that, the current decreases as the motor speeds up.

For real motors, there is a start-up current that is the one you would predict based on the battery voltage and circuit resistance. And then there is the operating current which is much lower. In fact, if this little motor of ours had no friction at all, the current would drop all the way down to zero!

You might not believe what you are seeing. So as the motor speeds along, you reach out with your hand and block it. Strangely, the current shoots back up again, back to the start-up value. You let go, and the motor once again goes through that speeding-up period. And as it does, the current once again drops down to the operating level. We now have evidence that lets us unravel that first mystery:

As a motor moves faster, the current reduces until there is only enough magnetic force to balance friction (or whatever mechanical force there is opposing the motion). If there are no opposing forces, the current drops to zero.

[Note: This is actually a very fun experiment to do. You hold the motor in your hand, watching the current meter. If you block the motor from spinning, the current shoots up. When you let go, the current goes back down. It is strangely compelling and more exciting than I’ve drawn it.]

You should take some time to make sure you understand both the original mystery and our solution. But even if you understand it, it is quite possible that you have another question. In fact, you SHOULD have another. When you are ready, read on.

2. What makes the current drop when the motor moves faster?

I hope you believe me when I tell you that the current does drop. (And I hope to show you this in class one day.) But what makes it drop? We don’t change the battery. We use the same wires with the same resistance. And yet the current drops as the motor spins faster, almost as if the motor and battery were in communication. You can imagine the motor talking to the battery:

“OK, I’m just sitting here, so when the switch closes, send me all the current you can. Ok, now I am starting to spin….you can ease off the current. I’m going even faster! Less current please…Ok, I’m at my top speed now, feeling good — OH NO, SOMETHING IS BLOCKING ME! Send more current! Give me all you got! Oh, wait, I’m moving again….you can ease off the current…ok, I’m good.”

Of course, that’s not really what happens. But when you do the experiment, it sure seems like that. So what is really happening?

Here is another explanation, one that does not require talking motors. And this one is really ALMOST true…

The Nearly-Correct Solution to MM#2

Inside every motor is a kind of invisible “Anti-battery”. It is a voltage source, separate from the actual external battery, and it has some interesting properties:

1. It only provides voltage when the motor is moving.

2. The faster the motor moves, higher the “anti-battery” voltage.

3. The “anti-battery” always opposes the voltage of the external battery.

It’s like a battery placed the wrong way in a flashlight so that it opposes the other batteries, causing a reduction in the net voltage. That is why I am calling it an “anti-battery”. (It is also called a “back EMF” but I like “anti-battery” better.)

But how does this explain the mystery of the dropping current?

Well, when the motor is starting from rest, there’s no anti-battery, no opposition. So the net voltage would be at its highest and you would get the most current. Then, as the motor moves faster, the anti-battery voltage gets higher, so the net voltage gets lower…and the current drops, just as we have seen. But if you block the motor, the anti-voltage goes away and once again your current shoots back up to the maximum value. Remember, the anti-battery only has voltage when the motor is actually moving.

I believe the explanation I am offering you would be approved by the physics authorities if I agreed to make two minor changes. I think they would let me use the name “anti-battery”. But they would insist on the following:

1. Delete the word “invisible”. The anti-battery is absolutely visible if you know where to look.

2. Delete the word “Inside”. The anti-battery is not inside the motor. It’s also not outside the motor. Hmm…

This brings us to the third motor mystery.

3. Where is this “anti-battery”?

I’m not ready to give this one away just yet. Let me see if I can jog your memory.

You did a lab this year where you used a generator to store charge on a capacitor. The plan was to use the stored charge to light a lightbulb. It worked…but something else also happened during that lab if you were not quick enough on the switch. Do you remember what that was?

After the lab was finished, I took a generator and connected the wires directly to a battery. Do you remember what happened?

Then, one of you held a generator that I had connected to a second generator. We each took turns cranking our generators. Do you remember what happened?

Think back to those events and you will not be surprised to hear the solution to this third mystery.

The Solution to MM#3

The “anti-battery” is not some invisible or even visible thing INSIDE the motor. The anti-battery IS the motor itself, now acting as a generator. What you saw in our old lab was that motors and generators are the same physical devices: essentially, just coils of wire, mounted in a magnetic field. When you provide voltage, the device acts like a motor — it spins. When you make it spin, it acts like a generator — it produces a voltage.

All we are saying now is that even when the device is being used as a motor, it still acts like a generator. As it spins, it generates a voltage. And that voltage fights the battery voltage, always. So the spinning motor has lower current.

You may be thinking: wouldn’t it be nice if we could get that generated voltage to assist the battery rather than fight it? Well, it would be, but the laws of physics prevent that. The induced “anti-voltage” always opposes the battery voltage. You will be hearing more about this later.

You may also be thinking: if a spinning motor simultaneously acts like a generator, does that mean that a spinning generator also acts like a motor? And the answer is yes! It does! But this time, it’s a magnetically induced force opposing an external force. That is why those generators got dramatically harder to spin when you closed the switch! [That is part of the reason we do labs: to give you physical experience and memories of observed phenomena.]

This interplay between motors and generators is fascinating but also challenging to understand. We will revisit this again later. But first we have one more question to ask:

How does the spinning wire act like a generator in the first place?

That is the topic we will be exploring for the rest of this unit. Stay tuned…

The College Board would like you to believe that they are on your side. To help convince you of this, they have provided many resources to help you prepare for the SAT. One essential resource: actual practice SATS. You can get them here. Taking real practice tests is certainly the key element in an effective prep plan. So it is nice of the College Board to provide you with them. And they even give you answers and solutions. How kind of them!

But one way to tell that they are not sincerely trying to helping you is by examining those math solutions. Yes, the solutions are scrupulously correct. A bit dry maybe. Sometimes stark and intimidating, but nevertheless, quite correct.

But there is an omission so glaring, so disingenuous, that the only explanation is that they want to claim the appearance of being helpful while still maintaining their own agenda. (World domination? Eventually. But for now, they would like to be the official end-of-high-school test in every state.)

If they were honest, the solutions would open with something like this:

Every SAT problem can be solved by a formal application of the mathematical concepts and procedures you have been learning in high school. We believe that this knowledge is important for success in many fields of study in college. So we are presenting solutions that exercise those skills in that formal way you have seen. If you learn to solve these problems in this particular way, your score on the SAT will certainly improve.

I believe the College Board would not object to that last paragraph. But here is what I wish they would say next:

But we also recognize that many of the problems on the SAT can be solved in alternative ways, using less formal (but more creative) mathematical approaches. We present those solutions here as well. Learning to solve problems in this manner will also help your score to improve. While these methods occasionally seem to circumvent some of the required math knowledge, they also promote a flexibility and creativity that will serve you well in your future studies.

I am emphatically not “anti-math” in any way. I like algebra! But if we want to help people to improve their SAT scores, we should not be holding back this information. And if you happen to be a little shaky in the formal methods of algebra, that is hard to fix in a shorter (say a few months) time frame. Learning another way to play can really open some doors for you. There is no way that the College Board does not know about these methods. But you see no evidence of that in their solutions. I guess they want to help you improve but only if you are willing to do it their way.

To illustrate what I am talking about, I am going to provide a few examples from the most recent practice test, #10. (I have a longer set of these for tests 1 – 8 here and I will try to update the collection soon.)

I hope that is a large enough sample to convince you that these approaches are potentially useful. I have seen them work a substantial number of times on every SAT that has ever been released. Also, it is good to have more than one way to approach a question. That gives you a way to stay in the game even when you don’t feel you know the “right” way to do a problem.

From Test #10, Section 3:

Again, I am not saying that these are the fastest solutions. Please don’t feel that you have to email or tweet at me with your algebra. These are ALTERNATIVE solutions! (And not every question has one…that I have found yet!)

1.You could try each answer, plugging in until you find one that works.

2. Ignore the algebra and just play around until you find how many weeks it will take: You after the initial $60 payment, you still owe $240. At $30 per week, it will take 8 weeks. Aha! Now plug in w=8 into each answer choice…

3. From the table, we can get a sample value. For example, when x = 10, the answer is supposed to come out to $21.89 . So try x = 10 in each answer choice. If you don’t get $21.89, it’s wrong.

5. Make up a number for x…like x=4. You get the square root of 144 which is 12. Now put x=4 into each answer choice….

6. Seriously, carefully trying each answer is quicker and easier than doing the algebra…

9. Go to the graph and find a point that is on the line. I used (-2,-2). That means that we can put x=-2 and y=-2 into the equation and it should work. Only one does! (If more than one worked, I would try another point.)

11. Draw a neat diagram of a circle centered at (5,7) with a radius of 2. Find any point on that circle. For example, 2 units to the right is (7,7). Then, see which equation works when you use x=7, y=7…

14. Again, trying their answers is a lot easier than doing the algebra. But this question is obnoxious. It targets the students who do not know that the square root symbol means the POSITIVE number that squares to give you what you want.

20. I cannot believe that they did not mention this solution: you could solve for u and t. That seems to require algebra, but you can also do it by playing with numbers. We need numbers whose sum is 5 and difference is 2. Let’s see: 4+1=5 but 4-1=3…too big. 2+2=4 but 2-2=0…too small. Uh oh. Fractions? 3.5 + 1.5 = 5. 3.5-1.5 = 2. Nice. You still have to do the arithmetic though…

OK that’s all for now. I will eventually post more of these for Section 4 and for Test #9. And I am pondering a way to turn this into a group project. I know a good number of fellow SAT nerds. So here is a heads-up to them: I may be reaching out soon to see if you want to help build a collection of these in some free, publicly available format. Stay tuned…

In light of the on-going controversy regarding admission to New York City’s selective high schools, there has been some discussion of bringing test prep to the middle school classroom. At first, test prep in the math classroom may sound like an awful idea. Should we take what is frequently, sad to say, a joyless, stultifying slog and add in a large helping of stress-inducing rote drill to prepare for the SHSAT, PSAT, SAT, what have you? That can’t be a good idea. And yet…

Maybe we should work on fixing the joyless, stultifying part. I have taught (oh, lord) a ton of SAT prep. None of it involves rote drills. It does involve a number of ideas and methods that absolutely help make math more joyful. And many, many students have come into my SAT class announcing that they don’t like math only to later report that they “kind of like” SAT math. So what is happening in SAT class that isn’t happening often enough in school? Here are some possible answers:

1. SAT math is subversive. And being subversive is fun. You don’t have to solve the problem the “right”
way! You can play, guess, draw pictures,
make lists…and if it works, you get full credit! School math is often presented like some
deranged cooking class where everyone learns to cook the same dish in the exact
same way. No invention, no experimentation, no surprises. (And it’s rarely
anything anyone wants to eat anyway.) We
should be encouraging mathematical improvisation and creativity – even when it
doesn’t “work”. After all, how often do
we really care where the vertex of the parabola is? Isn’t the more valuable experience the
search? A math problem that only has one
solution path isn’t really a “problem” — it’s more like a “chore”. No one likes doing chores.

2. SAT math rewards trial and error. School math often treats trial and error as a
tool for the weak. But trial and error
is a fabulous way to begin a problem when you don’t see an algebraic path. It is not merely an algebra-evasion technique
(though on the SAT, it can certainly work that way). Trial and error can serve
as an on-ramp to the algebraic solution.
I have used trial and error this way with students in SAT class, physics,
precalc, calculus, and multivariable calc.
To keep this rant brief (well…) I will expand on and demonstrate this
idea in a separate post. But to give you
the flavor:

1.
You take a guess (or pick an answer if it is a multiple choice test)

2.
You mess around with your guess, applying whatever line of reasoning you can
come up with to show that your guess is wrong (as it usually will be).

3. You replace your guess with a variable and then use your earlier reasoning as a template to come up with the algebraic equation or equations that will lead to an answer.

I have shown this method to students as young as 6^{th} grade and to colleagues who have been teaching for decades. I can tell you that most of them got a kick out of this. Again, it feels subversive. And who doesn’t like the feeling that you are getting away with something? But really, you’re not. You are just separating the problem into two phases, doing the reasoning before you attempt to apply the algebra. But it only works if you are willing to take a guess. “Could this be the answer?” is an amazingly helpful question, and not just on a standardized test.

3. SAT math rewards making up numbers for variables. This one is an SAT classic that should really
be a staple of school math as well. Many
students see an algebraic equation as a sequence of letters and symbols,
probably including an “=” somewhere in there. We want them to see it as a statement, a
sentence written in the language of algebra, one that might be true if the
variables have the “right” values.

This gives us an efficient way of finding
whether two expressions are equivalent.
Make up numbers (ideally, slightly “unusual” ones), put them into both
expressions, and see if they come out equal.
And yes, this method can generate false positives. But it never
generates false negatives – if you get two different numeric answers, then your
two algebraic expressions were definitely not equivalent.

This idea has many uses. First of all, we can teach students that
arithmetic is a testing bench for algebra.
Any time you are not certain whether something is allowed, try it with
numbers! We all have taught plenty of
students who believe that (x + y)^{2} = x^{2} + y^{2}. So let x =4 and y = 6 and see what happens!

Secondly, it inoculates students from the classic standardized test question: “which of the following expressions is equivalent to…” , a type of question that deserves to be subverted.

And thirdly, once again it gives us a way
to separate the reasoning from the algebra.
You can make up numbers for the variables, work out an answer, and then
put your variables back into the answer choices. It’s probably the SAT method that seems most
devious. Even in physics class, often my beginning students are flummoxed by
questions like:

A
car has mass m and speed v. A second car
has half the mass and three times the speed. Find the ratio of the second car’s
kinetic energy to the first car’s kinetic energy.

The problem is not that these students
don’t know the formula for kinetic energy. It’s that they have not yet
developed the ability to think about the ratios and proportions involved. When they are fluent, they will be able to
look at the formula, K = ½ mv^{2} and think through the problem with
ease. But while they are still
developing, isn’t it nice to have an alternative scaffold available?

Make
up numbers for m
and v and calculate the kinetic
energy.

Cut
the m
in half and triple the v and recalculate the kinetic energy.

Find
the ratio of your two answers.

Beginners are often surprised that this
works, and even skeptical, thinking that there was something special about the
numbers they made up. So let them make up a new set of numbers. It
seems like magic. And again, it serves as an on-ramp to the algebraic discussion.

Two more points and I’ll call it a rant

1. None of what I am recommending will
take time away from your curriculum. These are attitudes and strategies that
can overlay whatever you are currently teaching. Still, take a hard look: it
isn’t all joyful now and it isn’t all as essential as you may think. But we can argue about that another time. I would teach these methods even if the SAT
and its ilk were to vanish. It’s a more
fun way to do math.

2. Test prep in middle school is not
going to magically bring diversity to the elite high schools. It might help,
but I would guess that it won’t be nearly enough. We are going to have to do more.

Even after all these years, I still find logarithms to be challenging, compelling, and surprising. This post is about an idea that snuck up on me this morning. And it snuck up even though:

I have written: Log_{2} 8 = 3 because 2^{3} = 8 on many whiteboards and many review sheets.

I have learned (and taught!) the classic collection of assorted log properties

I have read David Mermin’s lovely piece “Logarithms!” (and blogged about it here.)

And yet, there is a basic idea about logarithms that I never really thought about (until this morning). So here it is:

The integer part of the logarithm of an integer is one less than the number of digits the integer has when it is represented in the same base as the logarithm itself.

You could write:

Floor[Log_{b} N] + 1 = the number of N’s digits when N is represented in base b.

In other words, the logarithm counts the digits — and not just in base-10. You just have to round down and add 1. I would like to explain this, if for no other reason than to convince myself that it is true!

Starting with base 10:

Let’s look at some common (base 10) logarithms that we can
find without a calculator:

Log 1 = 0 …because
10^{0} = 1

Log 10 = 1 …because
10^{1} = 10

Log 100 = 2 …because
10^{2} = 100

Log 1000 = 3 …because
10^{3} = 1000

Log 10000 = 4 …because
10^{4} = 10000

I think you get the idea. If we construct a number, say N, by writing a 1 followed by k zeroes, then the common (base 10) logarithm, log N = k. And the number will have k + 1 digits: the k zeroes and the 1 in front of them. So for each of these, adding 1 to the common logarithm gives you the number of digits.

Now let’s look at one that you do need a calculator for.
I’ll report three decimal places:

And it makes sense: 375 is more than 100 so its common logarithm is greater than 2. But 375 is also less than 1000 so its common logarithm is less than 3. It is more than the smallest 3 digit number and less than the smallest 4 digit number. And if you take the common logarithm, which was 2.574, round it down and then add 1, you get the number of digits. Nice.

Let’s try another (again reporting 3 decimal places):

Log (8787) = 3.944 … because 10^{3.944} ≈ 8787 (though with a little more round-off error, but still…)

And this makes sense too: 8787 is more than 1000 so its
common log is greater than 3. But it is
less than 10000 so its common log is less than 4. In fact, it is closer to 10000 so its log is
close to 4, but still less. And if we
round down 3.944 and add 1, again we get the number of digits in the number we
started with.

OK, I admit that I kind-of knew that already. But for some reason, this next step caught me by surprise…let’s move away from base 10 and see what happens.

Other bases and other logarithms

On and off this year, I have been playing with James Tanton’s invention, “Exploding Dots”. If you have not yet seen this, you really should take a look. You can get started here. But really, Google it – those dots are not just exploding, they are proliferating!

One thing that Exploding Dots teaches you almost immediately and organically is how to express numbers in other bases. And something interesting happens when you look at the base 2 logarithms as you look at the base 2 representations…

So if a number N
happens to be an integer power of 2, say 2^{k},
then we know two things:

1. Its base-2 logarithm, log_{2} N = k

2. Its base-2 representation will consist
of k zeroes and a leading 1. And that
makes k + 1 digits.

Now let’s get some help from a
calculator.

Log_{2} 5 = 2.322 …because 2^{2.322} ≈ 5

Does it make sense? Well, since 5 is more
than 4, we expect its base-2 logarithm to be more than 2. But 5 is less than 8, so we expect its base-2
logarithm to be less than 3. So this
looks right.

But we are not done! If we explode some dots, we can find the base 2 representation.

5 = 1 0 1 and that makes 3 digits – which
is also what we get when we round down the base-2 logarithm and then add
1!

OK, let’s check a big number:

500 expressed in base 2 is 1 1 1 1 0 1 0
0 (which you might want to check…)

So if you want to know the number of digits in the base-b
representation of a number, find the base-b logarithm, round it down to its
floor and then add one. The logarithm is
the digit counter!

So how are areas related to antiderivatives? In other words (and symbols):

This is the fundamental question. When you know the answer, you will know the “Fundamental Theorem of Integral Calculus”. But the main idea behind this all is something you have seen before in physics class (but not used quite this way).

In your first physics class, while studying kinematics, you learned that the area under a velocity vs time graph tells us the change in the object’s position. To help jog your memory, play the video below:

k1a

The upper graph shows the velocity as a function of time for an object that happened to have a constant velocity of 10 m/s.

The lower graph shows the position vs time for the same object. The velocity is constant so the position graph is linear with a constant slope.

On the velocity graph, you can see the area being swept out from t = 3 s to t = 7 s. And on the position graph, you can see how far the object has moved during the same time interval. Try stopping the animation at different times. Examine the graphs until you have convinced yourself of the following claim:

For a given time interval, the area under the Velocity vs. time graph will be equal to the change in position during that time interval.

YOU MAY BE SKEPTICAL…

After all, that was just with constant velocity. But what if the velocity is changing? Like so…

k2a

The velocity is increasing linearly and the position is increasing quadratically. [Remember, velocity is the derivative of position with respect to time. And the derivative of a 2nd-degree polynomial is a linear function.]

But even so, once again we see:

For a given time interval, the area under the Velocity vs. time graph will be equal to the change in position during that time interval.

STILL SKEPTICAL? STAY TUNED…

The velocity does not have to be changing linearly. In fact, it does not even have to be velocity and position that we are talking about. It can be any two quantities where one is the derivative of the other. Here’s one where the first graph is the rate at which water flows into a container and the second graph is the volume of water in that container:

k3a

Stop the animation at various points to check our claim:

If f(x) is the derivative of F(x), then the area under the graph of f(x) is equal to the change in F(x) over that same interval.

When we used this in 1st-year physics, we were using the area under the DERIVATIVE (velocity vs time) to find the change in the original function (position vs. time). But now that we know how to find anti-derivatives, we can do this the other way around, using the change in the ANTI-DERIVATIVE (position) to find the area under the graph of the original function (velocity vs time).

And that answers the question we started with. What does area have to do with antiderivatives? The area under the graph is equal to the change in the antiderivative. That is called the Fundamental Theorem. We write:

WHY IS THIS USEFUL?

Well, suppose you want to find the area under some graph. This theorem gives you a procedure you can follow:

Step 1: Find an antiderivative.

Step 2. See how much the antiderivative changed by over the interval.

For example, suppose we want to know the area under the graph of f(x)=x^{2} between x=1 and x = 2, as illustrated below.

Let’s see. Can we find an antiderivative? Oh, yes, this one we did already.

F(x)=(1/3)x^{3} is an antiderivative of f(x)=x^{2}.

So now we just have to see how much F(x) has changed over the interval from x = 1 to x = 2.

F(3) = (1/3)(2)^{3} = 8/3

F(1) = (1/3)(1)^{3} = 1/3

F(3) – F(1) = 7/3

But what if I can’t find the antiderivative?

Well, that could be a problem. That’s why you will spend so much time developing this skill in AP Calculus. But I will reassure you again: in AP Physics, the antiderivatives you need will be manageable. And there’s always Wolfram Alpha…

Wait! I don’t really understand how you can calculate the area of something with curvy boundaries. What does that even mean?

And I don’t understand: why does the area under the velocity graph equal the displacement? Do we just have to accept that?

And I still don’t see why this is so important? Are there lots of areas to be found in AP Physics?

These questions turn out to all be related! See next post, Part III of our discussion of areas and integrals.