Your definition of “good at math” is totally a learnable skill. I used to want people to explain things to me intuitively too. After spending several years as a math tutor, I now just brute-force all new word problems with algebra and figure out why my intuition was wrong afterwards.
Seriously algebra is like a great big hammer and you just smash things with it.
I can’t quite tell if your comment is analogous to “move your arms and legs!” or if instead you’re just agreeing with what I said here (point 2), and mistaking the nature of my difficulty.
In any case, “wanting people to explain things to me intuitively” is not at all how I would put it. In my experience, asking for an “intuitive” explanation will often produce the very opposite kind of explanation from the kind I want: concrete and ad-hoc instead of abstract and general.
If I can’t figure something out, it’s because there’s a concept that I’m
missing, and what I want is to be told about that concept. The concept doesn’t have to be “intuitive”, it just has to be generalizable and (preferably) concisely expressible.
However, most people’s approach to explanation is antithetical to this. They try their best to avoid introducing new concepts, tortuously beating around the bush as a result. It’s not that I can’t follow their explanations, but rather that I can’t properly generalize from them, because they’re not telling me the “real story”.
In terms of your metaphor: my difficulties arise from lacking a hammer, not from being unwilling to smash.
My understanding has been that if you know how to set up a problem as a formula then you know how to make it out of toothpicks and rubber bands so you understand it.
This requires a certain ability to manipulate formulas:
2xy/(x+y)=2/(1/x+1/y)=1/(((1/x)+(1/y))/2)
as I’m sure you know.
(but I am 1-in-a-million atypical on this)
So like, the formula for concentrations goes like this:
If we add 1volume of xconcentration to 2volume of yconcentration then:
we are adding 1volume of stuff to 2volume of stuff and getting 1volume+2volume of stuff.
We are adding xconcentration x 1volume to yconcentration x 2volume of special stuff and getting xconcentration x 1volume+yconcentration x 2volume of stuff
so our final concentration is the amount of special stuff divided by the amount of stuff, or
(xconcetration x 1volume+ yconcentration x 2volume)/(1volume+2volume)=zconcentration
So, I mean, that’s all there is. That’s the formula, that’s how the problem works, that’s all that’s at issue.
It’s equivalent to the problem:
You spend 2volume time at yconcentration speed. How much time must you spend at xconcentration speed to attain an average speed of zconcentration?
Sorry, “move your arms and legs” was certainly not my intent. I wasn’t trying to make any awfully astute or insightful point at all, just disagreeing with your assessment that you cannot do this problem without an elaborate theoretical underpinning of what inverse speed means because you are (implied intrinsically) “bad at math”.
I can think of a couple of reasons that I find more plausible: you didn’t realize that “average speed” means “total distance/total time”, or you don’t have enough experience with word problems involving rates, or in general, to know where to start setting this up algebraically. Could be other reasons too. You’re clearly pretty bright and capable of thinking in the abstract, though, so “bad at math” doesn’t ring true, and anyway it isn’t specific enough to do anything about it.
I actually do identify with your desire for highly abstract and general explanations, and I used to think that way to a fault, but for whatever reason I’ve gotten more concrete and practical over time. It probably isn’t only the tutoring, but that’s certainly contributed: if, when faced with a problem you do not immediately know how to do, you stare off into space and go “hmmm,” the student quickly loses patience with you, but if you start writing and say “ok call this x and this y and here are some relationships we know must be true,” you look very competent, and half the time you realize you can just solve by substitution or something.
So, for the sake of argument, consider this explanation:
Speed is distance/time. Average speed is total distance/total time. We have numbers for speeds, but no numbers for distances or times. So call the distance traveled in the “first half” d, and the time it took to travel that distance t, and set this up:
20 = d/t
from which we get
20t = d
And then we write another equation, with the average speed of 40, we’ll want to set 40 equal to total distance/total time. We know that the total distance is twice the first half, so that’s 2d, but we don’t know the total time, except it must be some number added to the time for the first half. So we’ll call however long the second half took x, and the total time t + x.
40 = 2d/(t+x)
sub in 20t for d
40 = 2(20t)/(t+x)
Multiply both sides by (t+x) and distribute
40t+40x = 40t
x = 0
No time left for the second half of the trip. Speed = d/0 = undefined.
Now, maybe when you see it done this way the answer still feels wrong, or something, but there’s no reason why you couldn’t just set the thing up and solve it like I did. Is there? Am I still not getting where you’re coming from?
I wasn’t trying to make any awfully astute or insightful point at all, just disagreeing with your assessment that you cannot do this problem without an elaborate theoretical underpinning of what inverse speed means because you are (implied intrinsically) “bad at math”. …You’re clearly pretty bright and capable of thinking in the abstract, though, so “bad at math” doesn’t ring true
Well, it shouldn’t—this is a case of deliberate irony, since the linked comments (as well as the post itself) imply that I have considerable background in mathematics.
This post has a “secret agenda” that goes far beyond figuring out how to solve this particular problem. You in fact put your finger on a significant part of it with this:
if, when faced with a problem you do not immediately know how to do, you stare off into space and go “hmmm,” the student quickly loses patience with you, but if you start writing and say “ok call this x and this y and here are some relationships we know must be true,” you look very competent
This is a failure mode. In fact, this is exactly what Yudkowsky is warning about in posts such as Grasping Slippery Things Yes, in the short term, it may get a student through a word problem, or at a higher level, get you a publication. But, in the long term, it’s a bad habit, to the extent it prevents you from noticing your own confusion.
Now, maybe when you see it done this way the answer still feels wrong, or something, but there’s no reason why you couldn’t just set the thing up and solve it like I did. Is there? Am I still not getting where you’re coming from?
By way of clarifying where I’m coming from, I should emphasize that my interest is not in guessing teachers’ passwords. Yes,of course I am capable of performing the algebra you demonstrated, but doing so would not leave me with the feeling that the answer is obvious. Contrast with the following problem:
If your speed for the first hour of a trip was 20 mph and you want your average speed for the whole two-hour trip to be 40 mph, what must your speed be for the second hour?
Now that problem I understand. It’s not just that I can do the algebra involved; I can do it mentally. I probably wouldn’t have been tricked by it if it came up in conversation. This is what it means, to me, to “be able to do” a math problem. This is what knowledge feels like, as opposed to confusion.
You’re arguing that I can do without the “elaborate theoretical underpinning of what inverse speed means”, but I don’t want to do without that elaborate theoretical underpinning, because once I have that theoretical underpinning, the two problems become entirely symmetrical, and I can answer one just as easily as the other.
I don’t have a lot of time here because LeechBlock is about to cut me off, and I don’t want to make a well-written reply anyway because karma is kind of a whore. But you did say you didn’t have a hammer. And now you seem to be switching to “This hammer is bad! It breaks things!”
I agree btw that always brute-forcing problems could be a bad habit if you stopped there. Not infrequently, I pause afterwards and say, “Ok, that’s the answer, but I think I just did that the stupid way. Gimme a sec and I’ll see if I can tell you the smart way.” But it’s an adaptive habit in my line of work.
But you did say you didn’t have a hammer. And now you seem to be switching to “This hammer is bad! It breaks things!”
No, that’s not right. The problem is that it’s not a hammer at all, it’s a fake hammer. It lets you pretend you’re driving in nails when you’re really not. The hammer isn’t too powerful (“breaking things”), it’s not powerful enough. That might be okay if your only goal is to be seen “hammering”, but if you actually want to hammer the nails in, you need a real hammer.
When you don’t have the proper concepts, working out things with mere algebra lets you develop the concepts by focusing on constraints and other properties the concepts must have. Sure, it doesn’t force you to develop the concepts, but if you’re planning on doing so, it is extremely valuable for getting a grasp on this concept.
This is different than most slippery philosophical problems—the math actually fights back in a revealing way.
This is different than most slippery philosophical problems—the math actually fights back in a revealing way.
I don’t see any particular asymmetry, actually. (Which is no surprise when you realize that I consider mathematics to be rigorous philosophy.) Sometimes the way it fights back is (sufficiently) revealing, and sometimes it isn’t.
There remain deeply mysterious unsolved problems in mathematics, for which merely fiddling with existing tools has not produced answers. The point of view I take (which is implicitly advocated by this post) is that whenever you have a problem that you can’t solve, it’s because your existing tools are inadequate, and you need to develop better tools. How does one develop better tools? Well, you can hope to discover them by accident in the course of analyzing the unsolved problem, or you can try to develop them systematically by figuring out how to better solve problems you are already able to solve. The latter is my preferred approach.
(I would also recommend this comment for context.)
So wait, you’re saying algebra doesn’t work? Because it definitely does. It’s nice that it keeps my hands busy, but I also sincerely believe I’m helping my students by teaching them to approach problems this way. Algebra works equally well for the other problem you suggested:
If your speed for the first hour of a trip was 20 mph and you want your average speed for the whole two-hour trip to be 40 mph, what must your speed be for the second hour?
You just set it up with your unknown in the numerator instead of the denominator:
20 = d/1
40 = (d+x)/2
Or say another problem, one you might not recognize as similar if you’re stuck on speed and inverse speed: My dad is borrowing my mom’s hybrid car to drive to Miami. She’s very hung up on what the display says her mileage is, and she’ll get grumpy if it’s less than 50 mpg. My dad drives like a hyperactive child and gets 25 mpg on his way to Miami. What mileage will he need to get on his way back to make 50 mpg overall?
25 = d/g
50 = 2d/(g+x)
Solve the system for x, turns out he can’t use any gas on the way back. His only hope is to reset the mileage calculator and drive like a sane person on the way home. (Or drive around the neighborhood really efficiently several hundred times after he gets home and hope she doesn’t notice the extra miles.)
(Or convert the Insight into a plug-in in my cousin’s garage.)
What I’m saying is that algebra is a fully general tool for solving word problems, and you should embrace it. Thanks for reminding me to train my intuition too, though. And I’m sorry if I came off as patronizing.
No—I’m afraid you misunderstand. This is by no means a question of “algebra” versus “non-algebra”: in order to solve such problems, algebra necessarily has to be performed in some manner. The point is that the calculations you present are too bare to serve the goal of dissolving confusion; they do not invoke the sort of higher-level concepts that are necessary for me to mentally store (and reproduce) them efficiently.
Let me explain. It is in fact ironic that you write
Or say another problem, one you might not recognize as similar if you’re stuck on speed and inverse speed
because “recognizing different things as similar” is exactly what higher-level concepts are for! As it turns out, I have no trouble recognizing the other problem as similar, because the difficulty is exactly the same: it asks about the inverse of the quantity that you need to think in terms of in order to solve it. That is, it asks about “mileage” when it should be asking about “gallonage”.
The problem is not that “algebra”—solving for an unknown variable—is required. The problem is that the unknown variable is in the denominator, and that’s confusing because it means that the quantity associated with the whole journey is not the sum (or average) of the quantities associated with the parts of the journey. That is, x miles per gallon on the first half does not combine with y miles per gallon on the second half to yield x+y (or even, say, (x+y)/2) miles per gallon for the whole trip. As a result, the correct equation to solve is not 25+x = 50, nor (25/2)+(x/2) = 50. Instead, it’s something else entirely, namely (1/25)(1/2) + (1/x)(1/2) = 1⁄50, or equivalently 50x/(x+25) = 50.
Now, there are actually two ways of making this comprehensible to me. One would be the way I’ve been talking about, which is to switch from talking about miles per gallon to talking about its inverse, gallons per mile. That way, the quantities do combine properly: x gpm for the first half and y gpm for the second half yields (1/2)x+(1/2)y gpm for the whole trip. (This is how I was able to write down the equation above!) The other way would be to explicitly teach the rule for combining mileages on parts of a journey to obtain the mileage for the whole journey, which is that x mpg on the first half combines with y mpg on the second half to yield 2xy/(x+y) mpg for the total.
Now I think the first way is preferable, but the second way would also be tolerable, and it illustrates the distinction between an “intuitive” explanation and what I’m seeking. There’s nothing particularly “intuitive” about the formula 2xy/(x+y); it’s just something I would have to learn for the purposes of doing these problems. But it makes the algebra make sense. What I would do is regard this as a new arithmetic operation, and give it a name, say #, and I would learn x#y = 2xy/(x+y). Then, when you asked me
My dad drives like a hyperactive child and gets 25 mpg on his way to Miami. What mileage will he need to get on his way back to make 50 mpg overall?
I would set up the equation
25#x = 50
which automatically converts to
50x/(25+x) = 50
which I could then easily solve.
So do you see what I mean? The issue isn’t algebra, it’s having the right higher-level concepts to organize the algebra mentally. In this context, that means either thinking about the inverse of the quantity asked for (inverse speed instead of speed, or “gallonage” instead of mileage), or else learning a new operation of arithmetic (that is, asking “well, what’s the general rule for combining speeds on segments of a journey?” or “what’s the general rule for combining mileages on segments of a journey?”).
“recognizing different things as similar” is exactly what higher-level concepts are for!
I am fully on board with this. Higher-level concepts ftw. Of course, just setting up the algebra can help you recognize two problems as similar too. By “setting up the algebra” I mean “accepting the statements made in the word problem and translating them into math, introducing symbols to represent unknown quantities as necessary”.
As it turns out, I have no trouble recognizing the other problem as similar, because the difficulty is exactly the same: it asks about the inverse of the quantity that you need to think in terms of in order to solve it. That is, it asks about “mileage” when it should be asking about “gallonage”.
Well yes, I expect you did recognize it, in context; I did everything I could to make the similarity explicit. And naturally, having recognized it as the same problem, you can then solve it by the same method you used for the first problem. But does having “inverse speed” in your arsenal really make it natural to approach this problem in the same way? I guess I’ll have to take your word for it.
Now, there are actually two ways of making this comprehensible to me. One would be the way I’ve been talking about, which is to switch from talking about miles per gallon to talking about its inverse, gallons per mile. That way, the quantities do combine properly: x gpm for the first half and y gpm for the second half yields (1/2)x+(1/2)y gpm for the whole trip. (This is how I was able to write down the equation above!)
Yeah, I get that you can do it that way. I agree that it’s kinda neat, even. It’s sort of like how you calculate the total resistance of resistors wired in parallel. The only thing I take issue with is that you cannot, or should not, do this problem without it.
The other way would be to explicitly teach the rule for combining mileages on parts of a journey to obtain the mileage for the whole journey, which is that x mpg on the first half combines with y mpg on the second half to yield 2xy/(x+y) mpg for the total…There’s nothing particularly “intuitive” about the formula 2xy/(x+y); it’s just something I would have to learn for the purposes of doing these problems. But it makes the algebra make sense. What I would do is regard this as a new arithmetic operation, and give it a name, say #, and I would learn x#y = 2xy/(x+y).
Ok, using formulas by rote totally is a failure mode, so I would be against that. (Btw on the subject of “intuitive explanations”, so we can avoid arguing about terms, I understand that to mean not “an explanation that appeals to the intuition you already have,” but “an explanation that fixes your intuition.” Explanations like that are fun and cool, but I don’t insist on having one before I do every new problem.)
So do you see what I mean? The issue isn’t algebra, it’s having the right higher-level concepts to organize the algebra mentally.
I guess I don’t get why you can’t just use “average rate of change = (total change in one variable)/(total change in other variable)” as your organizing concept, introduce symbols to represent the various quantities, and grind it out. That approach is certainly useful for much more than just finding average speeds or average mileages in this particular special case. You don’t need a new formula; you don’t even need a new concept. You can just use the ones you’ve already got.
But does having “inverse speed” in your arsenal really make it natural to approach this problem in the same way?
The idea in my arsenal is not just inverse speed, of course; it’s inverses in general, and the fact that finding a rate may first require finding the inverse of that rate.
I recognized the two problems as similar not merely because the context implied that they were constructed to be similar (in point of fact the mileage one contained a large amount of distracting detail), but because they involve the same difficulty—I “got stuck” at the same point, for the same reason: not knowing the relationship between the “partial” rates (those associated with the segments of the journey) and the “total” rate (that associated with the whole journey).
Ok, using formulas by rote totally is a failure mode
Not necessarily, but that isn’t even the point here. I wouldn’t actually need to memorize 2xy/(x+y), because I could easily derive it by taking inverses. (Indeed I wouldn’t be satisfied until I understood the derivation.) The point is that the existence of a rule for combining mileages (or speeds) -- and indeed, the abstract concept of a binary operation other than the usual operations of arithmetic—be acknowledged.
The only thing I take issue with is that you cannot, or should not, do this problem without it...I don’t get why you can’t just use “average rate of change = (total change in one variable)/(total change in other variable)” as your organizing concept...You don’t need a new formula; you don’t even need a new concept. You can just use the ones you’ve already got.
What you’re missing is that the avoidance of new concepts is not a desideratum. If I can’t figure out how to solve a problem, an explanation that uses only concepts I already had available will never be satisfactory, because there was some reason I couldn’t figure it out, and such an explanation necessarily fails to address that reason.
In this case, a = b/c wasn’t enough for me. That’s just a fact. The speed problem crashed my brain until I came up with the concept of inverse speed. Now, in retrospect, now that I have a satisfactory understanding of these problems, I can go back and look at the solutions that just use a = b/c, translate them into my way of understanding, and come up with a way that I could have written down those same solutions that just use a = b/c while privately having a more sophisticated interpretation of what I was writing, so that I could appear to be doing the same thing as everyone else. But I wouldn’t be doing the same thing as everyone else.
Now, when I say I couldn’t have done it without this higher-level conceptual understanding, do I mean that literally, in the sense that no amount of mere fiddling with variables would have eventually allowed me to stumble upon the correct answer? Of course not. However, that wouldn’t be satisfactory. For one thing, it would be difficult for me to do that quickly: if this had been some sort of two-minute test (or worse, a conversation among mathematically knowledgeable people, with status at stake, where you have only a few seconds), I would probably have been out of luck. But more importantly, I wouldn’t “believe” the solution, or to put it differently, I wouldn’t feel that I myself had solved it, but rather that “someone else” had and that I was taking their word for it. There would be a feeling of discomfort, of inadequacy. I would acutely sense that I was missing some insight. And, as it turns out, I would be entirely right! The insight I would be missing would be that described in this post and these comments, which is a legitimate and powerful insight that makes things clearer. I wouldn’t want to do without it, even if I could manage to do so.
Oh, I am perfectly happy for you to like your way better. It’s a good way of doing the problem. But it seems like asking for trouble to say this:
If for any reason anyone is ever tempted to describe me as “good at math”, I will invite them to reflect on the fact that an explicit understanding of the concept of “inverse speed” as described above (i.e. as a function that sends distances to times) was a necessary prerequisite for my being able to solve this problem, and then to consider that problems of this sort are customarily taught in middle- or high school, by middle- and high school teachers.
if you actually mean this:
But more importantly, I wouldn’t “believe” the solution, or to put it differently, I wouldn’t feel that I myself had solved it, but rather that “someone else” had and that I was taking their word for it. There would be a feeling of discomfort, of inadequacy. I would acutely sense that I was missing some insight. And, as it turns out, I would be entirely right! The insight I would be missing would be that described in this post and these comments, which is a legitimate and powerful insight that makes things clearer. I wouldn’t want to do without it, even if I could manage to do so.
You know, they seem to be saying different things, to me.
And I don’t agree that fiddling with variables is somehow cheating, or that it’s a “fake hammer.” It only gets easier to understand a problem once you know how to find the right answer.
If I can’t figure out how to solve a problem, an explanation that uses only concepts I already had available will never be satisfactory, because there was some reason I couldn’t figure it out, and such an explanation necessarily fails to address that reason.
Huh, is that actually true? Plenty of times I have failed to understand a problem just because I misinterpreted it—you know, like holding the wrong variable constant? If I introduced a new concept every time, my picture of the world would get pretty baroque. It might be better to go in and knock down old, wrong concepts instead, or just clarify the ones I have, or remind myself that yes, they really do apply here.
Anyway. The idea of solving for a different variable (like you’re doing here with your inverse rates, taking time as a function of distance instead of distance as a function of time) often seems to confuse my students. I wonder if it comes from introducing the concepts of dependent and independent variables too early, and teaching them how to do all kind of operations with functions, but only spending a couple of days talking about how to find the inverse of a function.
it seems like asking for trouble to say this:...if you actually mean this:
I really don’t think there’s much of a difference. Yes, there was some rhetorical exaggeration in the first quote; to make it literally accurate I should have written about a necessary prerequisite for comfortably or reliably solving the problem. But since I take it for granted that a solid understanding is the goal, the difference between barely being able to solve it and not being able to solve it at all isn’t particularly significant from my point of view.
There’s a real psychological phenomenon having to do with difficulty that I’m pointing to here, something that goes beyond mere aesthetic preference (as important as that might also be, to me). I have a history of having trouble with problems of this type, at multiple points during my life. Believe me, I’ve been exposed to the standard explanations—the three-row tables and so on. I’ve tried to learn them. I’ve even had sporadic success. But it just doesn’t work reliably. There is some piece of knowledge or cognitive habit that these explanations presuppose that is always left implicit, but that I don’t actually possess. To the extent that I can understand them, it is always by figuring out the solution in my own way and then “translating”—a process which greatly increases the effort required. Failing that, I’m stumbling in the dark and guessing—which sometimes works, and sometimes doesn’t.
If I introduced a new concept every time, my picture of the world would get pretty baroque. It might be better to go in and knock down old, wrong concepts instead, or just clarify the ones I have, or remind myself that yes, they really do apply here
I think most people’s picture of the world is probably not “baroque” enough. However, yes, it’s always possible to overdo anything; the way you avoid having too many concepts is by constantly upgrading the ones you have—making sure they’re as general and powerful as possible.
This is my conception of mathematical research: “concept R&D.” Whereas for most people, it’s about solving problems (by any means, and then you’re done once they’re solved), for me, it’s about developing concepts that make the solution easy.
The idea of solving for a different variable (like you’re doing here with your inverse rates, taking time as a function of distance instead of distance as a function of time) often seems to confuse my students. I wonder if it comes from introducing the concepts of dependent and independent variables too early, and teaching them how to do all kind of operations with functions, but only spending a couple of days talking about how to find the inverse of a function.
My personal saying is bite the abstraction bullet early. My experience is that students have trouble with the very idea of a function, and will try to avoid actually learning it (and similarly abstract ideas) for as long as they can—instead seeking lower-level “shortcuts” that allow them to get the right answer (sometimes). They need to be broken out of this as early as possible, lest they show up in a calculus class incapable of understanding the meaning of “f(x+h)”, as all too often happens.
That’s what I suspect is going on here: what’s confusing them is that they’re not used to thinking about independent and dependent variables at all: they think of d = rt as a rule for manipulating symbols, not as a representation of a relationship between (abstract) quantities. It’s like confusing numbers with numerals. (You can ultimately think of mathematics as rules for maniplulating symbols, but in that case the rules are much more complicated than “d = rt”! That’s their problem: they want the rules to be both concrete and simple, but they can’t have it both ways!)
There is some piece of knowledge or cognitive habit that these explanations presuppose that is always left implicit, but that I don’t actually possess. To the extent that I can understand them, it is always by figuring out the solution in my own way and then “translating”—a process which greatly increases the effort required. Failing that, I’m stumbling in the dark and guessing—which sometimes works, and sometimes doesn’t.
Hm, hell if I know. I can tell you that your talk about speed as a “mapping of times to distances” seemed downright weird to me—I intuitively think of the relationship as being two-way, and to say that you’re going faster means both that you’ll go further in the same length of time and that you’ll travel the same distance in a shorter period. If the price of chocolate bars goes down, I can buy the same number of chocolate bars for less, or I can buy more chocolate bars for the same amount of money.
Is it possible that you love functions too much? Not every situation has a natural choice of independent and dependent variables, after all. It’s not any more meaningful to say that pressure depends on volume than that volume depends on pressure; pV just equals nRT.
It always fascinates me to find how different people are in the types of explanations that work for them. For example, all that “distracting detail” in my mileage story would be motivating detail for me, and for many of my students I’ve found that the more concrete I can make a problem, the more sense it makes to them.
I intuitively think of the relationship as being two-way, and to say that you’re going faster means both that you’ll go further in the same length of time and that you’ll travel the same distance in a shorter period.
I might have thought the same, before the experience of being confused by this problem revealed otherwise.
Do you find the following to be equally easy to answer, intuitively?
(1) You spend one hour going 10 mph and then one hour going 20 mph. What’s your average speed?
(2) You go one mile at 10 mph and then one mile at 20 mph. What’s your average speed?
Perhaps you do; but (at least prior to this discussion) I wouldn’t have.
Not every situation has a natural choice of independent and dependent variables, after all. It’s not any more meaningful to say that pressure depends on volume than that volume depends on pressure; pV just equals nRT.
However, you cannot talk about rates—that is, derivatives—without making a choice: dp/dV is as different from dV/dp as speed is from inverse speed.
Which brings me to the following:
I can tell you that your talk about speed as a “mapping of times to distances” seemed downright weird to me
Well, as it turns out, it’s inherent in the very definition!
The derivative of a function at a point is defined to be the linear map that best approximates the function near that point. So if we have a function x = f(t) that maps times t to distances x, the derivative f’(t) -- the “speed”—at time t is by definition also a mapping from times dt to distances dx (given by the formula dx = f’(t)dt).
Hence, there’s nothing idiosyncratic about my way of thinking. It might be “sophisticated”, but it’s hardly “weird”. Of course, it has been my repeated experience that perspectives labeled “sophisticated”, “advanced”, or “abstract” are those that I tend to find most natural.
However, I think the exoticity here is actually pretty minimal. Consider how people visually represent speed: they usually draw arrows whose length represents the distance traveled in a fixed time interval. To represent a speed that is twice as fast, they will make the arrow twice as long, not half as long.
Do you find the following to be equally easy to answer, intuitively?
I am equally confident that I can give a right answer to them both, but one of them makes the calculations easier to do in my head. Here’s what I might say if you sprung each of these on me:
(1) You spend one hour going 10 mph and then one hour going 20 mph. What’s your average speed?
“I go ten miles and then twenty miles. 30 miles/2h = 15 mph”
On the SAT, problems are arranged from easiest to hardest, not by the difficulty of the concepts involved, but according to how many students get them wrong. If two questions use the same concepts and procedures, but one gives an answer that “looks right” (is a whole number, for example), there will be a difficulty difference between them. This one would be right at the beginning of the SAT, because it’s the same answer you get by doing the problem in a naive way: you see two numbers and the word “average”, so you just average them.
(2) You go one mile at 10 mph and then one mile at 20 mph. What’s your average speed?
“I take 1⁄10 of an hour and then 1⁄20 of an hour, so that’s two miles over, um, 3⁄20 hours… so 40⁄3 mph? Yeah, I guess that’s between 10 and 20.”
The math is a little trickier, and the answer isn’t a whole number, so I’m sure it would take a few more seconds to come up with, but I did the problem in basically the same way, by dividing distance by time. (Of course, I’m assuming that given the distances involved, you know how to get the times, but any high school chemistry student knows you can flip your conversion factors if you need to.) This one would definitely go at the end of the SAT, not only because of the weirdness of the answer(+), but because it requires you to recognize exactly what question is being asked.
So intuitively I find neither problem harder to understand. I know that going an hour at 20 mph is totally different from going a mile at 20 mph. Just about everybody knows that, if they think about it. The difference is that you can get a right answer on the first problem without understanding it.
However, you cannot talk about rates—that is, derivatives—without making a choice: dp/dV is as different from dV/dp as speed is from inverse speed.
Well, yes, you would have to differentiate with respect to one or the other variable, but you can do either just as well; the relationship doesn’t force you. And having found your dp/dV, you could flip it over to get dV/dp. This seems like it might be a pitfall of function notation, actually; if I tell you that V(p) = nRT/p, you can tell me that V’(p) = -nRT/p^2, but you’re forced to differentiate with respect to p, and it’s probably not so easy to make the jump to seeing that dV/dp = -V/p and dp/dV = -p/V. Maybe it’s no coincidence that my Calc I students sometimes learn how to perform the chain rule, but don’t figure out what it actually means until they learn to do implicit differentiation? I dunno, just thinking aloud here. (thinking a-type?)
Well, as it turns out, it’s inherent in the very definition!
Is that the only way to define a derivative? I know it’s one way, and it works, but is that the only way?
However, I think the exoticity here is actually pretty minimal. Consider how people visually represent speed: they usually draw arrows whose length represents the distance traveled in a fixed time interval. To represent a speed that is twice as fast, they will make the arrow twice as long, not half as long.
Not sure this is a good example. It’s a lot more natural to have lengths of arrows correspond to distances than to times… since, you know, they actually are distances. But if you consider that people often say “coming quick” to mean “coming soon”(++), it seems like there’s an instinctive association between higher speeds and shorter times as well.
(+)You have no idea how much kids hate fractions. When they see fractions they just don’t even try.
(++)Is this a Southern thing? When I was a kid people would say “Christmas is coming quickly!” and I would think “It’s not coming any more quickly than it was before. It’s coming at a rate of one day per day.”
(2) You go one mile at 10 mph and then one mile at 20 mph. What’s your average speed?
“I take 1⁄10 of an hour and then 1⁄20 of an hour, so that’s two miles over, um, 3⁄20 hours...
Very interesting. When I read this, it struck me as a “good-at-math” person’s thought process, and after reflecting on it, I think I know why:
You went directly from “one mile at 10 mph” to “1/10 of an hour”—skipping right over what is for me the most important step in the whole solution: the conversion from 10 mph to 1⁄10hpm. I’m guessing you didn’t even realize there was a step missing here, did you?
It’s a fairly abstract step, of course: it involves explicitly performing an operation on rates, which as discussed previously, are mappings (functions). But the point is, if you talk to me about “one mile at 10 mph”, my natural, intuitive reaction is “ERROR: SYNTAX”. The operation “10 mph” does not accept “one mile” as an input (nor vice-versa: “one mile” doesn’t accept “10 mph” either). A quantity with “mph” needs a number of “hours”; a quantity with “miles” needs something with “miles” in the denominator.
(Strictly speaking, thanks to a mathematical construction known as the tensor product, anything can operate on anything else—but the result will in general be a new kind of thing. For example, if mph acts on miles, the result will be labeled miles^2/hour.)
Now, you write:
Of course, I’m assuming that given the distances involved, you know how to get the times, but any high school chemistry student knows you can flip your conversion factors if you need to.
but “knowing that you can” do something (or even “knowing how”) is different from being able to do it without explicitly thinking about it as a separate step!
It’s interesting that this parenthetically-mentioned assumption of yours is, for me, the entire sticking point, and the subject of this post.
Now that you mention high-school chemistry, let me tell you another interesting thing: I used to be on the other side of this discussion, once upon a time—or so it would have appeared. That is, I used to ridicule high-school chemistry for this “dimensional analysis” business, satirizing it by elaborately solving problems such as “if there are 5 apples in each barrel, and you have 6 barrels, how many apples do you have?” via “conversion factors” and cancellation of “barrels”. It seemed to me that this was just a technique for mechanizing these problems for the benefit of slow students who couldn’t just see that obviously if you have 6 barrels of 5 apples, you must have 30 apples in total. (Perhaps exactly analgously to the way that you, unlike me, can just “see” that if you go one mile at 10 mph, you took 1⁄10 of an hour.)
I now realize, however, that that wasn’t my true rejection. What I actually objected to about “dimensional analysis” was that it was an ad-hoc, discipline-specific kind of mathematics that chemistry people were using which lacked a theoretical justification in math class. The latter, you see, had never provided any conceptual foundation for treating “5″, “5 apples”, and “5 barrels” as different kinds of mathematical objects. Sure, there were expressions with “unlike terms” (such as x, y, and xy) that you couldn’t just “add together”, but those unlike terms always stood for different amounts of the same kind of thing: abstract numbers, or numbers reprenting one particular kind of quantity. So where did these chemistry people get the idea that they were allowed to perform symbolic algebra on units, which after all aren’t numbers at all?
It was for the same reason that I resisted vectors, when they were introduced in physics class before I had been properly exposed to the mathematical subject of linear algebra: you’re not allowed to invent new mathematics outside of mathematics class (which in my mind serves as the Department of Anti-Compartmentalization).
Now if you say “What? How crippling that would be to physics and chemistry!”, you’re missing the point. The problem wasn’t with physics and chemistry, the problem was with math class. (Indeed, often physics and chemistry were too accomodating to the lack of mathematical prerequisites, such as in avoiding calculus, which is utterly silly.) The logical foundation for “dimensional analysis” is multilinear algebra, and so I should have learned multilinear algebra in math class before being asked to do “dimensional analysis” in chemistry class.
So, you can see that my apparently having been on the other side, once upon a time, was in fact nothing other than an instance of the same thing: a need for the proper theoretical foundations to be in place before I can “understand” something.
Is that the only way to define a derivative? I know it’s one way, and it works, but is that the only way?
It’s the most general way(+), hence the best. All other ways are either equivalent to this (and just as abstract) or don’t make sense outside of a restricted setting.
(+) Perhaps not technically true, but close enough to the truth for our purposes here.
So to summarize, basically komponisto needs to learn to always think of bijections as always accompanied by their inverses, in particular when that bijection is given by multiplication by a nonzero real number[0], as will always be the case when the mapping in question is a nonzero derivative and you’re only working in one dimension, and more generally to not always think of relations as one-way functions?
OK, but it’s still important to understand how this plays out in the 1-dimensional case. These aren’t incompatible, one’s just a special case. Though I’m not seeing the relevance of that particular isomorphism here, as I don’t see just what it is here that would naturally be interpreted as an element of that first space in the first place?
OK, but it’s still important to understand how this plays out in the 1-dimensional case
Well, yes! That’s what I seek to do, as opposed to regarding the 1-dimensional case as a separate magisterium, compartmentalized away from the general case.
I don’t see just what it is here that would naturally be interpreted as an element of that first space in the first place?
Here V is distances, and W is times. If something has the label “distance”, it’s an element of V; if it has the label “time”, it’s an element of W; and if it has the label “time^-1”, it’s an element of W. Something with the label “distance/time” is then an element of
 .
Here V is distances, and W is times. If something has the label “distance”, it’s an element of V; if it has the label “time”, it’s an element of W; and if it has the label “time^-1”, it’s an element of W*.
Oh, OK. For some reason I was thinking the scaling was wrong for that to work. Of course, if you travel 3 miles in 2 hours, that’s 3 mi \otimes 1⁄2 h^-1, not 3 mi \otimes 2 h^-1...
That’s right: (1/2)h^-1 is the map that takes a time and gives its coordinate with respect the basis {2h}, which is the one being used here to define the speed.
(General rule: a/b means you input b to get a. So, since our coordinate-computing map should input 2h and output 1, it is written 1/(2h), or (1/2)h^-1.)
Your definition of “good at math” is totally a learnable skill. I used to want people to explain things to me intuitively too. After spending several years as a math tutor, I now just brute-force all new word problems with algebra and figure out why my intuition was wrong afterwards.
Seriously algebra is like a great big hammer and you just smash things with it.
I can’t quite tell if your comment is analogous to “move your arms and legs!” or if instead you’re just agreeing with what I said here (point 2), and mistaking the nature of my difficulty.
In any case, “wanting people to explain things to me intuitively” is not at all how I would put it. In my experience, asking for an “intuitive” explanation will often produce the very opposite kind of explanation from the kind I want: concrete and ad-hoc instead of abstract and general.
If I can’t figure something out, it’s because there’s a concept that I’m missing, and what I want is to be told about that concept. The concept doesn’t have to be “intuitive”, it just has to be generalizable and (preferably) concisely expressible.
However, most people’s approach to explanation is antithetical to this. They try their best to avoid introducing new concepts, tortuously beating around the bush as a result. It’s not that I can’t follow their explanations, but rather that I can’t properly generalize from them, because they’re not telling me the “real story”.
In terms of your metaphor: my difficulties arise from lacking a hammer, not from being unwilling to smash.
My understanding has been that if you know how to set up a problem as a formula then you know how to make it out of toothpicks and rubber bands so you understand it.
This requires a certain ability to manipulate formulas:
2xy/(x+y)=2/(1/x+1/y)=1/(((1/x)+(1/y))/2)
as I’m sure you know.
(but I am 1-in-a-million atypical on this)
So like, the formula for concentrations goes like this:
If we add 1volume of xconcentration to 2volume of yconcentration then:
we are adding 1volume of stuff to 2volume of stuff and getting 1volume+2volume of stuff.
We are adding xconcentration x 1volume to yconcentration x 2volume of special stuff and getting xconcentration x 1volume+yconcentration x 2volume of stuff
so our final concentration is the amount of special stuff divided by the amount of stuff, or
(xconcetration x 1volume+ yconcentration x 2volume)/(1volume+2volume)=zconcentration
So, I mean, that’s all there is. That’s the formula, that’s how the problem works, that’s all that’s at issue.
It’s equivalent to the problem:
You spend 2volume time at yconcentration speed. How much time must you spend at xconcentration speed to attain an average speed of zconcentration?
Sorry, “move your arms and legs” was certainly not my intent. I wasn’t trying to make any awfully astute or insightful point at all, just disagreeing with your assessment that you cannot do this problem without an elaborate theoretical underpinning of what inverse speed means because you are (implied intrinsically) “bad at math”.
I can think of a couple of reasons that I find more plausible: you didn’t realize that “average speed” means “total distance/total time”, or you don’t have enough experience with word problems involving rates, or in general, to know where to start setting this up algebraically. Could be other reasons too. You’re clearly pretty bright and capable of thinking in the abstract, though, so “bad at math” doesn’t ring true, and anyway it isn’t specific enough to do anything about it.
I actually do identify with your desire for highly abstract and general explanations, and I used to think that way to a fault, but for whatever reason I’ve gotten more concrete and practical over time. It probably isn’t only the tutoring, but that’s certainly contributed: if, when faced with a problem you do not immediately know how to do, you stare off into space and go “hmmm,” the student quickly loses patience with you, but if you start writing and say “ok call this x and this y and here are some relationships we know must be true,” you look very competent, and half the time you realize you can just solve by substitution or something.
So, for the sake of argument, consider this explanation:
Speed is distance/time. Average speed is total distance/total time. We have numbers for speeds, but no numbers for distances or times. So call the distance traveled in the “first half” d, and the time it took to travel that distance t, and set this up:
20 = d/t
from which we get
20t = d
And then we write another equation, with the average speed of 40, we’ll want to set 40 equal to total distance/total time. We know that the total distance is twice the first half, so that’s 2d, but we don’t know the total time, except it must be some number added to the time for the first half. So we’ll call however long the second half took x, and the total time t + x.
40 = 2d/(t+x)
sub in 20t for d
40 = 2(20t)/(t+x)
Multiply both sides by (t+x) and distribute
40t+40x = 40t
x = 0
No time left for the second half of the trip. Speed = d/0 = undefined.
Now, maybe when you see it done this way the answer still feels wrong, or something, but there’s no reason why you couldn’t just set the thing up and solve it like I did. Is there? Am I still not getting where you’re coming from?
Well, it shouldn’t—this is a case of deliberate irony, since the linked comments (as well as the post itself) imply that I have considerable background in mathematics.
This post has a “secret agenda” that goes far beyond figuring out how to solve this particular problem. You in fact put your finger on a significant part of it with this:
This is a failure mode. In fact, this is exactly what Yudkowsky is warning about in posts such as Grasping Slippery Things Yes, in the short term, it may get a student through a word problem, or at a higher level, get you a publication. But, in the long term, it’s a bad habit, to the extent it prevents you from noticing your own confusion.
By way of clarifying where I’m coming from, I should emphasize that my interest is not in guessing teachers’ passwords. Yes,of course I am capable of performing the algebra you demonstrated, but doing so would not leave me with the feeling that the answer is obvious. Contrast with the following problem:
If your speed for the first hour of a trip was 20 mph and you want your average speed for the whole two-hour trip to be 40 mph, what must your speed be for the second hour?
Now that problem I understand. It’s not just that I can do the algebra involved; I can do it mentally. I probably wouldn’t have been tricked by it if it came up in conversation. This is what it means, to me, to “be able to do” a math problem. This is what knowledge feels like, as opposed to confusion.
You’re arguing that I can do without the “elaborate theoretical underpinning of what inverse speed means”, but I don’t want to do without that elaborate theoretical underpinning, because once I have that theoretical underpinning, the two problems become entirely symmetrical, and I can answer one just as easily as the other.
I don’t have a lot of time here because LeechBlock is about to cut me off, and I don’t want to make a well-written reply anyway because karma is kind of a whore. But you did say you didn’t have a hammer. And now you seem to be switching to “This hammer is bad! It breaks things!”
I agree btw that always brute-forcing problems could be a bad habit if you stopped there. Not infrequently, I pause afterwards and say, “Ok, that’s the answer, but I think I just did that the stupid way. Gimme a sec and I’ll see if I can tell you the smart way.” But it’s an adaptive habit in my line of work.
No, that’s not right. The problem is that it’s not a hammer at all, it’s a fake hammer. It lets you pretend you’re driving in nails when you’re really not. The hammer isn’t too powerful (“breaking things”), it’s not powerful enough. That might be okay if your only goal is to be seen “hammering”, but if you actually want to hammer the nails in, you need a real hammer.
When you don’t have the proper concepts, working out things with mere algebra lets you develop the concepts by focusing on constraints and other properties the concepts must have. Sure, it doesn’t force you to develop the concepts, but if you’re planning on doing so, it is extremely valuable for getting a grasp on this concept.
This is different than most slippery philosophical problems—the math actually fights back in a revealing way.
I don’t see any particular asymmetry, actually. (Which is no surprise when you realize that I consider mathematics to be rigorous philosophy.) Sometimes the way it fights back is (sufficiently) revealing, and sometimes it isn’t.
There remain deeply mysterious unsolved problems in mathematics, for which merely fiddling with existing tools has not produced answers. The point of view I take (which is implicitly advocated by this post) is that whenever you have a problem that you can’t solve, it’s because your existing tools are inadequate, and you need to develop better tools. How does one develop better tools? Well, you can hope to discover them by accident in the course of analyzing the unsolved problem, or you can try to develop them systematically by figuring out how to better solve problems you are already able to solve. The latter is my preferred approach.
(I would also recommend this comment for context.)
So wait, you’re saying algebra doesn’t work? Because it definitely does. It’s nice that it keeps my hands busy, but I also sincerely believe I’m helping my students by teaching them to approach problems this way. Algebra works equally well for the other problem you suggested:
You just set it up with your unknown in the numerator instead of the denominator:
20 = d/1
40 = (d+x)/2
Or say another problem, one you might not recognize as similar if you’re stuck on speed and inverse speed: My dad is borrowing my mom’s hybrid car to drive to Miami. She’s very hung up on what the display says her mileage is, and she’ll get grumpy if it’s less than 50 mpg. My dad drives like a hyperactive child and gets 25 mpg on his way to Miami. What mileage will he need to get on his way back to make 50 mpg overall?
25 = d/g
50 = 2d/(g+x)
Solve the system for x, turns out he can’t use any gas on the way back. His only hope is to reset the mileage calculator and drive like a sane person on the way home. (Or drive around the neighborhood really efficiently several hundred times after he gets home and hope she doesn’t notice the extra miles.)
(Or convert the Insight into a plug-in in my cousin’s garage.)
What I’m saying is that algebra is a fully general tool for solving word problems, and you should embrace it. Thanks for reminding me to train my intuition too, though. And I’m sorry if I came off as patronizing.
No—I’m afraid you misunderstand. This is by no means a question of “algebra” versus “non-algebra”: in order to solve such problems, algebra necessarily has to be performed in some manner. The point is that the calculations you present are too bare to serve the goal of dissolving confusion; they do not invoke the sort of higher-level concepts that are necessary for me to mentally store (and reproduce) them efficiently.
Let me explain. It is in fact ironic that you write
because “recognizing different things as similar” is exactly what higher-level concepts are for! As it turns out, I have no trouble recognizing the other problem as similar, because the difficulty is exactly the same: it asks about the inverse of the quantity that you need to think in terms of in order to solve it. That is, it asks about “mileage” when it should be asking about “gallonage”.
The problem is not that “algebra”—solving for an unknown variable—is required. The problem is that the unknown variable is in the denominator, and that’s confusing because it means that the quantity associated with the whole journey is not the sum (or average) of the quantities associated with the parts of the journey. That is, x miles per gallon on the first half does not combine with y miles per gallon on the second half to yield x+y (or even, say, (x+y)/2) miles per gallon for the whole trip. As a result, the correct equation to solve is not 25+x = 50, nor (25/2)+(x/2) = 50. Instead, it’s something else entirely, namely (1/25)(1/2) + (1/x)(1/2) = 1⁄50, or equivalently 50x/(x+25) = 50.
Now, there are actually two ways of making this comprehensible to me. One would be the way I’ve been talking about, which is to switch from talking about miles per gallon to talking about its inverse, gallons per mile. That way, the quantities do combine properly: x gpm for the first half and y gpm for the second half yields (1/2)x+(1/2)y gpm for the whole trip. (This is how I was able to write down the equation above!) The other way would be to explicitly teach the rule for combining mileages on parts of a journey to obtain the mileage for the whole journey, which is that x mpg on the first half combines with y mpg on the second half to yield 2xy/(x+y) mpg for the total.
Now I think the first way is preferable, but the second way would also be tolerable, and it illustrates the distinction between an “intuitive” explanation and what I’m seeking. There’s nothing particularly “intuitive” about the formula 2xy/(x+y); it’s just something I would have to learn for the purposes of doing these problems. But it makes the algebra make sense. What I would do is regard this as a new arithmetic operation, and give it a name, say #, and I would learn x#y = 2xy/(x+y). Then, when you asked me
I would set up the equation
25#x = 50
which automatically converts to
50x/(25+x) = 50
which I could then easily solve.
So do you see what I mean? The issue isn’t algebra, it’s having the right higher-level concepts to organize the algebra mentally. In this context, that means either thinking about the inverse of the quantity asked for (inverse speed instead of speed, or “gallonage” instead of mileage), or else learning a new operation of arithmetic (that is, asking “well, what’s the general rule for combining speeds on segments of a journey?” or “what’s the general rule for combining mileages on segments of a journey?”).
I am fully on board with this. Higher-level concepts ftw. Of course, just setting up the algebra can help you recognize two problems as similar too. By “setting up the algebra” I mean “accepting the statements made in the word problem and translating them into math, introducing symbols to represent unknown quantities as necessary”.
Well yes, I expect you did recognize it, in context; I did everything I could to make the similarity explicit. And naturally, having recognized it as the same problem, you can then solve it by the same method you used for the first problem. But does having “inverse speed” in your arsenal really make it natural to approach this problem in the same way? I guess I’ll have to take your word for it.
Yeah, I get that you can do it that way. I agree that it’s kinda neat, even. It’s sort of like how you calculate the total resistance of resistors wired in parallel. The only thing I take issue with is that you cannot, or should not, do this problem without it.
Ok, using formulas by rote totally is a failure mode, so I would be against that. (Btw on the subject of “intuitive explanations”, so we can avoid arguing about terms, I understand that to mean not “an explanation that appeals to the intuition you already have,” but “an explanation that fixes your intuition.” Explanations like that are fun and cool, but I don’t insist on having one before I do every new problem.)
I guess I don’t get why you can’t just use “average rate of change = (total change in one variable)/(total change in other variable)” as your organizing concept, introduce symbols to represent the various quantities, and grind it out. That approach is certainly useful for much more than just finding average speeds or average mileages in this particular special case. You don’t need a new formula; you don’t even need a new concept. You can just use the ones you’ve already got.
The idea in my arsenal is not just inverse speed, of course; it’s inverses in general, and the fact that finding a rate may first require finding the inverse of that rate.
I recognized the two problems as similar not merely because the context implied that they were constructed to be similar (in point of fact the mileage one contained a large amount of distracting detail), but because they involve the same difficulty—I “got stuck” at the same point, for the same reason: not knowing the relationship between the “partial” rates (those associated with the segments of the journey) and the “total” rate (that associated with the whole journey).
Not necessarily, but that isn’t even the point here. I wouldn’t actually need to memorize 2xy/(x+y), because I could easily derive it by taking inverses. (Indeed I wouldn’t be satisfied until I understood the derivation.) The point is that the existence of a rule for combining mileages (or speeds) -- and indeed, the abstract concept of a binary operation other than the usual operations of arithmetic—be acknowledged.
What you’re missing is that the avoidance of new concepts is not a desideratum. If I can’t figure out how to solve a problem, an explanation that uses only concepts I already had available will never be satisfactory, because there was some reason I couldn’t figure it out, and such an explanation necessarily fails to address that reason.
In this case, a = b/c wasn’t enough for me. That’s just a fact. The speed problem crashed my brain until I came up with the concept of inverse speed. Now, in retrospect, now that I have a satisfactory understanding of these problems, I can go back and look at the solutions that just use a = b/c, translate them into my way of understanding, and come up with a way that I could have written down those same solutions that just use a = b/c while privately having a more sophisticated interpretation of what I was writing, so that I could appear to be doing the same thing as everyone else. But I wouldn’t be doing the same thing as everyone else.
Now, when I say I couldn’t have done it without this higher-level conceptual understanding, do I mean that literally, in the sense that no amount of mere fiddling with variables would have eventually allowed me to stumble upon the correct answer? Of course not. However, that wouldn’t be satisfactory. For one thing, it would be difficult for me to do that quickly: if this had been some sort of two-minute test (or worse, a conversation among mathematically knowledgeable people, with status at stake, where you have only a few seconds), I would probably have been out of luck. But more importantly, I wouldn’t “believe” the solution, or to put it differently, I wouldn’t feel that I myself had solved it, but rather that “someone else” had and that I was taking their word for it. There would be a feeling of discomfort, of inadequacy. I would acutely sense that I was missing some insight. And, as it turns out, I would be entirely right! The insight I would be missing would be that described in this post and these comments, which is a legitimate and powerful insight that makes things clearer. I wouldn’t want to do without it, even if I could manage to do so.
Oh, I am perfectly happy for you to like your way better. It’s a good way of doing the problem. But it seems like asking for trouble to say this:
if you actually mean this:
You know, they seem to be saying different things, to me.
And I don’t agree that fiddling with variables is somehow cheating, or that it’s a “fake hammer.” It only gets easier to understand a problem once you know how to find the right answer.
Huh, is that actually true? Plenty of times I have failed to understand a problem just because I misinterpreted it—you know, like holding the wrong variable constant? If I introduced a new concept every time, my picture of the world would get pretty baroque. It might be better to go in and knock down old, wrong concepts instead, or just clarify the ones I have, or remind myself that yes, they really do apply here.
Anyway. The idea of solving for a different variable (like you’re doing here with your inverse rates, taking time as a function of distance instead of distance as a function of time) often seems to confuse my students. I wonder if it comes from introducing the concepts of dependent and independent variables too early, and teaching them how to do all kind of operations with functions, but only spending a couple of days talking about how to find the inverse of a function.
I really don’t think there’s much of a difference. Yes, there was some rhetorical exaggeration in the first quote; to make it literally accurate I should have written about a necessary prerequisite for comfortably or reliably solving the problem. But since I take it for granted that a solid understanding is the goal, the difference between barely being able to solve it and not being able to solve it at all isn’t particularly significant from my point of view.
There’s a real psychological phenomenon having to do with difficulty that I’m pointing to here, something that goes beyond mere aesthetic preference (as important as that might also be, to me). I have a history of having trouble with problems of this type, at multiple points during my life. Believe me, I’ve been exposed to the standard explanations—the three-row tables and so on. I’ve tried to learn them. I’ve even had sporadic success. But it just doesn’t work reliably. There is some piece of knowledge or cognitive habit that these explanations presuppose that is always left implicit, but that I don’t actually possess. To the extent that I can understand them, it is always by figuring out the solution in my own way and then “translating”—a process which greatly increases the effort required. Failing that, I’m stumbling in the dark and guessing—which sometimes works, and sometimes doesn’t.
I think most people’s picture of the world is probably not “baroque” enough. However, yes, it’s always possible to overdo anything; the way you avoid having too many concepts is by constantly upgrading the ones you have—making sure they’re as general and powerful as possible.
This is my conception of mathematical research: “concept R&D.” Whereas for most people, it’s about solving problems (by any means, and then you’re done once they’re solved), for me, it’s about developing concepts that make the solution easy.
My personal saying is bite the abstraction bullet early. My experience is that students have trouble with the very idea of a function, and will try to avoid actually learning it (and similarly abstract ideas) for as long as they can—instead seeking lower-level “shortcuts” that allow them to get the right answer (sometimes). They need to be broken out of this as early as possible, lest they show up in a calculus class incapable of understanding the meaning of “f(x+h)”, as all too often happens.
That’s what I suspect is going on here: what’s confusing them is that they’re not used to thinking about independent and dependent variables at all: they think of d = rt as a rule for manipulating symbols, not as a representation of a relationship between (abstract) quantities. It’s like confusing numbers with numerals. (You can ultimately think of mathematics as rules for maniplulating symbols, but in that case the rules are much more complicated than “d = rt”! That’s their problem: they want the rules to be both concrete and simple, but they can’t have it both ways!)
Hm, hell if I know. I can tell you that your talk about speed as a “mapping of times to distances” seemed downright weird to me—I intuitively think of the relationship as being two-way, and to say that you’re going faster means both that you’ll go further in the same length of time and that you’ll travel the same distance in a shorter period. If the price of chocolate bars goes down, I can buy the same number of chocolate bars for less, or I can buy more chocolate bars for the same amount of money.
Is it possible that you love functions too much? Not every situation has a natural choice of independent and dependent variables, after all. It’s not any more meaningful to say that pressure depends on volume than that volume depends on pressure; pV just equals nRT.
It always fascinates me to find how different people are in the types of explanations that work for them. For example, all that “distracting detail” in my mileage story would be motivating detail for me, and for many of my students I’ve found that the more concrete I can make a problem, the more sense it makes to them.
I might have thought the same, before the experience of being confused by this problem revealed otherwise.
Do you find the following to be equally easy to answer, intuitively?
(1) You spend one hour going 10 mph and then one hour going 20 mph. What’s your average speed?
(2) You go one mile at 10 mph and then one mile at 20 mph. What’s your average speed?
Perhaps you do; but (at least prior to this discussion) I wouldn’t have.
However, you cannot talk about rates—that is, derivatives—without making a choice: dp/dV is as different from dV/dp as speed is from inverse speed.
Which brings me to the following:
Well, as it turns out, it’s inherent in the very definition!
The derivative of a function at a point is defined to be the linear map that best approximates the function near that point. So if we have a function x = f(t) that maps times t to distances x, the derivative f’(t) -- the “speed”—at time t is by definition also a mapping from times dt to distances dx (given by the formula dx = f’(t)dt).
Hence, there’s nothing idiosyncratic about my way of thinking. It might be “sophisticated”, but it’s hardly “weird”. Of course, it has been my repeated experience that perspectives labeled “sophisticated”, “advanced”, or “abstract” are those that I tend to find most natural.
However, I think the exoticity here is actually pretty minimal. Consider how people visually represent speed: they usually draw arrows whose length represents the distance traveled in a fixed time interval. To represent a speed that is twice as fast, they will make the arrow twice as long, not half as long.
I am equally confident that I can give a right answer to them both, but one of them makes the calculations easier to do in my head. Here’s what I might say if you sprung each of these on me:
“I go ten miles and then twenty miles. 30 miles/2h = 15 mph”
On the SAT, problems are arranged from easiest to hardest, not by the difficulty of the concepts involved, but according to how many students get them wrong. If two questions use the same concepts and procedures, but one gives an answer that “looks right” (is a whole number, for example), there will be a difficulty difference between them. This one would be right at the beginning of the SAT, because it’s the same answer you get by doing the problem in a naive way: you see two numbers and the word “average”, so you just average them.
“I take 1⁄10 of an hour and then 1⁄20 of an hour, so that’s two miles over, um, 3⁄20 hours… so 40⁄3 mph? Yeah, I guess that’s between 10 and 20.”
The math is a little trickier, and the answer isn’t a whole number, so I’m sure it would take a few more seconds to come up with, but I did the problem in basically the same way, by dividing distance by time. (Of course, I’m assuming that given the distances involved, you know how to get the times, but any high school chemistry student knows you can flip your conversion factors if you need to.) This one would definitely go at the end of the SAT, not only because of the weirdness of the answer(+), but because it requires you to recognize exactly what question is being asked.
So intuitively I find neither problem harder to understand. I know that going an hour at 20 mph is totally different from going a mile at 20 mph. Just about everybody knows that, if they think about it. The difference is that you can get a right answer on the first problem without understanding it.
Well, yes, you would have to differentiate with respect to one or the other variable, but you can do either just as well; the relationship doesn’t force you. And having found your dp/dV, you could flip it over to get dV/dp. This seems like it might be a pitfall of function notation, actually; if I tell you that V(p) = nRT/p, you can tell me that V’(p) = -nRT/p^2, but you’re forced to differentiate with respect to p, and it’s probably not so easy to make the jump to seeing that dV/dp = -V/p and dp/dV = -p/V. Maybe it’s no coincidence that my Calc I students sometimes learn how to perform the chain rule, but don’t figure out what it actually means until they learn to do implicit differentiation? I dunno, just thinking aloud here. (thinking a-type?)
Is that the only way to define a derivative? I know it’s one way, and it works, but is that the only way?
Not sure this is a good example. It’s a lot more natural to have lengths of arrows correspond to distances than to times… since, you know, they actually are distances. But if you consider that people often say “coming quick” to mean “coming soon”(++), it seems like there’s an instinctive association between higher speeds and shorter times as well.
(+)You have no idea how much kids hate fractions. When they see fractions they just don’t even try.
(++)Is this a Southern thing? When I was a kid people would say “Christmas is coming quickly!” and I would think “It’s not coming any more quickly than it was before. It’s coming at a rate of one day per day.”
Very interesting. When I read this, it struck me as a “good-at-math” person’s thought process, and after reflecting on it, I think I know why:
You went directly from “one mile at 10 mph” to “1/10 of an hour”—skipping right over what is for me the most important step in the whole solution: the conversion from 10 mph to 1⁄10 hpm. I’m guessing you didn’t even realize there was a step missing here, did you?
It’s a fairly abstract step, of course: it involves explicitly performing an operation on rates, which as discussed previously, are mappings (functions). But the point is, if you talk to me about “one mile at 10 mph”, my natural, intuitive reaction is “ERROR: SYNTAX”. The operation “10 mph” does not accept “one mile” as an input (nor vice-versa: “one mile” doesn’t accept “10 mph” either). A quantity with “mph” needs a number of “hours”; a quantity with “miles” needs something with “miles” in the denominator.
(Strictly speaking, thanks to a mathematical construction known as the tensor product, anything can operate on anything else—but the result will in general be a new kind of thing. For example, if mph acts on miles, the result will be labeled miles^2/hour.)
Now, you write:
but “knowing that you can” do something (or even “knowing how”) is different from being able to do it without explicitly thinking about it as a separate step!
It’s interesting that this parenthetically-mentioned assumption of yours is, for me, the entire sticking point, and the subject of this post.
Now that you mention high-school chemistry, let me tell you another interesting thing: I used to be on the other side of this discussion, once upon a time—or so it would have appeared. That is, I used to ridicule high-school chemistry for this “dimensional analysis” business, satirizing it by elaborately solving problems such as “if there are 5 apples in each barrel, and you have 6 barrels, how many apples do you have?” via “conversion factors” and cancellation of “barrels”. It seemed to me that this was just a technique for mechanizing these problems for the benefit of slow students who couldn’t just see that obviously if you have 6 barrels of 5 apples, you must have 30 apples in total. (Perhaps exactly analgously to the way that you, unlike me, can just “see” that if you go one mile at 10 mph, you took 1⁄10 of an hour.)
I now realize, however, that that wasn’t my true rejection. What I actually objected to about “dimensional analysis” was that it was an ad-hoc, discipline-specific kind of mathematics that chemistry people were using which lacked a theoretical justification in math class. The latter, you see, had never provided any conceptual foundation for treating “5″, “5 apples”, and “5 barrels” as different kinds of mathematical objects. Sure, there were expressions with “unlike terms” (such as x, y, and xy) that you couldn’t just “add together”, but those unlike terms always stood for different amounts of the same kind of thing: abstract numbers, or numbers reprenting one particular kind of quantity. So where did these chemistry people get the idea that they were allowed to perform symbolic algebra on units, which after all aren’t numbers at all?
It was for the same reason that I resisted vectors, when they were introduced in physics class before I had been properly exposed to the mathematical subject of linear algebra: you’re not allowed to invent new mathematics outside of mathematics class (which in my mind serves as the Department of Anti-Compartmentalization).
Now if you say “What? How crippling that would be to physics and chemistry!”, you’re missing the point. The problem wasn’t with physics and chemistry, the problem was with math class. (Indeed, often physics and chemistry were too accomodating to the lack of mathematical prerequisites, such as in avoiding calculus, which is utterly silly.) The logical foundation for “dimensional analysis” is multilinear algebra, and so I should have learned multilinear algebra in math class before being asked to do “dimensional analysis” in chemistry class.
So, you can see that my apparently having been on the other side, once upon a time, was in fact nothing other than an instance of the same thing: a need for the proper theoretical foundations to be in place before I can “understand” something.
It’s the most general way(+), hence the best. All other ways are either equivalent to this (and just as abstract) or don’t make sense outside of a restricted setting.
(+) Perhaps not technically true, but close enough to the truth for our purposes here.
So to summarize, basically komponisto needs to learn to always think of bijections as always accompanied by their inverses, in particular when that bijection is given by multiplication by a nonzero real number[0], as will always be the case when the mapping in question is a nonzero derivative and you’re only working in one dimension, and more generally to not always think of relations as one-way functions?
[0]Or in other words, “division is available”...
Who said I think of relations as one-way functions? I think of them as what they are, namely subsets of the Cartesian product.
As for division, I’m very happy to trade it in for an intuitive understanding of the canonical monomorphism
(which, in concrete terms, means the ability to view something labeled “mph” as a linear map from the space of times to the space of distances).
OK, but it’s still important to understand how this plays out in the 1-dimensional case. These aren’t incompatible, one’s just a special case. Though I’m not seeing the relevance of that particular isomorphism here, as I don’t see just what it is here that would naturally be interpreted as an element of that first space in the first place?
Well, yes! That’s what I seek to do, as opposed to regarding the 1-dimensional case as a separate magisterium, compartmentalized away from the general case.
Here V is distances, and W is times. If something has the label “distance”, it’s an element of V; if it has the label “time”, it’s an element of W; and if it has the label “time^-1”, it’s an element of W. Something with the label “distance/time” is then an element of  .
Oh, OK. For some reason I was thinking the scaling was wrong for that to work. Of course, if you travel 3 miles in 2 hours, that’s 3 mi \otimes 1⁄2 h^-1, not 3 mi \otimes 2 h^-1...
That’s right: (1/2)h^-1 is the map that takes a time and gives its coordinate with respect the basis {2h}, which is the one being used here to define the speed.
(General rule: a/b means you input b to get a. So, since our coordinate-computing map should input 2h and output 1, it is written 1/(2h), or (1/2)h^-1.)