Inverse Speed

post by komponisto · 2011-03-27T05:57:36.035Z · score: 14 (17 votes) · LW · GW · Legacy · 57 comments

One must always invert.

- Carl Gustav Jacobi

I'm grateful to orthonormal for mentioning the following math problem, because it allowed me to have a significant confusion-dissolving insight (actually going on two, but I'll only discuss one in this post), as well as providing an example of how bad I am at math:

"[I]f you want to average 40 mph on a trip, and you averaged 20 mph for the first half of the route, how fast do you have to go on the second half of the route?" 

When I read this, my first thought was "Huh? If you spend an hour going 20 mph and then spend another hour going 60 mph, you've just gone 80 miles in 2 hours -- for an average speed of 40 mph, just as desired. So what do you people mean it's impossible?"

As you can see, my confusion resulted from interpreting "half of the route" to refer to the total time of the journey, rather than the total distance.

This misinterpretation reveals something fundamental about how I (I know better by now than to say "we") think about speed.

In my mind, speed is a mapping from times to distances. The way to compare different speeds is by holding time constant and looking at the different distances traversed in that fixed time. (I know I'm not a total mutant in this regard, because even other people tend to visually represent speeds as little arrows of varying length, with greater lengths corresponding to higher speeds.)

In particular, I don't think of it as a mapping from distances to times. I don't find it natural to compare speeds by imagining a fixed distance corresponding to different travel times. Which explains why I find this problem so difficult, and other people's explanations so unilluminating: they tend to begin with something along the lines of "let d be the total distance traveled", upon which my brain experiences an error message that is perhaps best verbalized as something like "wait, what? Who said anything about a fixed distance? If speeds are varying, distances have to be varying, too!" 

If speed is a mapping from times to distances, then the way that you add speeds together and multiply them by numbers (the operations involved in averaging) is by performing the same operations on corresponding distances. (This is an instance of the general definition in mathematics of addition of functions: (f+g)(x) = f(x)+g(x), and similarly for multiplication by numbers: (af)(x) = a*f(x).) In concrete terms, what this means is that in order to add 30 mph and 20 mph together, all you have to do is add 30 and 20 and then stick "mph" on the result. Likewise with averages: provided the times involved are the same, if your speeds are 20 mph and 60 mph, your average speed is 40 mph. 

You cannot do these operations nearly so easily, however, if distance is being held fixed and time varying. Why not? Because if our mapping is from times to distances, then finding the time that corresponds to a given distance requires us to invert that mapping, and there's no easy way to invert the sum of two mappings (we can't for example just add the inverses of the mappings themselves). As a result, I find it difficult to understand the notion of "speed" while thinking of time as a dependent variable. 

And that, at least for me, is why this problem is confusing: the statement doesn't contain a prominent warning saying "Attention! Whereas you normally think of speed as the being the (longness-of-)distance traveled in a given time, here you need to think of it as the (shortness-of-)time required to travel a given distance. In other words, the question is actually about inverse speed, even though it talks about 'speed'."

Only when I have "inverse speed" in my vocabulary, can I then solve the problem -- which, properly formulated, would read: "If you want your inverse speed for the whole trip to be 1/40 hpm, and your inverse speed for the first half is 1/20 hpm, how 'slow' (i.e. inversely-fast) do you have to go on the second half?" 

Solution: Now it makes sense to begin with "let d be the total distance"! For inverse speed, unlike speed, accepts distances as inputs (and produces times as outputs). So, instead of distance = speed*time -- or, as I would rather have it, distance = speed(time) -- we have the formula time = speed-1(distance). Just as the original formula converts questions about speed to questions about distance, this new formula conveniently converts our question about inverse speeds to a question about times: we'll find the time required for the whole journey, the time required for the first half, subtract to find the time required for the second half, then finally convert this back to an inverse speed.

So if d is the total distance, the total time required for the journey is (1/40)*d = d/40. The time required for the first half of the journey is (1/20)*(d/2) = d/40. So the time required for the second half is d/40 - d/40 = 0. Hence the inverse speed must be 0.

So we're being asked to travel a nonzero distance in zero time -- which happens to be an impossibility.

Problem solved. 

Now, here's the interesting thing: I'll bet there are people reading this who (despite my best efforts) found the above explanation difficult to follow -- and yet had no trouble solving the problem themselves. And I'll bet there are probably also people who consider my explanation to be an example of belaboring the obvious.

I have a term for people in these categories: I call them "good at math". What unites them is the ability to produce correct solutions to problems like this without having to expend significant effort figuring out the sort of stuff I explained above.

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.

No indeed, I was not sorted into the tribe of "good at math".

I should find some sort of prize to award to anyone who can explain how to solve "mixing" problems in a manner I find comprehensible. (You know the type: how much of x% concentration do you add to your y% concentration to get z% concentration? et similia.)

57 comments

Comments sorted by top scores.

comment by [deleted] · 2011-03-27T13:32:54.726Z · score: 13 (13 votes) · LW(p) · GW(p)

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.

comment by komponisto · 2011-03-27T17:35:33.609Z · score: 1 (1 votes) · LW(p) · GW(p)

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.

comment by Will_Sawin · 2011-04-18T02:11:13.521Z · score: 2 (2 votes) · LW(p) · GW(p)

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?

comment by [deleted] · 2011-03-29T00:29:07.192Z · score: 1 (1 votes) · LW(p) · GW(p)

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?

comment by komponisto · 2011-03-30T15:44:47.423Z · score: 2 (2 votes) · LW(p) · GW(p)

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.

comment by [deleted] · 2011-03-30T16:19:41.479Z · score: 0 (0 votes) · LW(p) · GW(p)

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.

comment by komponisto · 2011-03-30T16:47:25.345Z · score: 0 (0 votes) · LW(p) · GW(p)

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.

comment by wnoise · 2011-03-30T17:27:28.132Z · score: 2 (2 votes) · LW(p) · GW(p)

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.

comment by komponisto · 2011-03-31T20:46:58.196Z · score: 0 (0 votes) · LW(p) · GW(p)

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.)

comment by [deleted] · 2011-03-30T19:13:28.205Z · score: 0 (0 votes) · LW(p) · GW(p)

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.

comment by komponisto · 2011-03-31T17:54:52.660Z · score: 1 (1 votes) · LW(p) · GW(p)

So wait, you’re saying algebra doesn’t work?

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?").

comment by [deleted] · 2011-04-01T12:29:15.239Z · score: 0 (0 votes) · LW(p) · GW(p)

"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.

comment by komponisto · 2011-04-01T17:55:44.796Z · score: 0 (2 votes) · LW(p) · GW(p)

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.

comment by [deleted] · 2011-04-01T18:53:06.008Z · score: 0 (0 votes) · LW(p) · GW(p)

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.

comment by komponisto · 2011-04-03T20:28:53.345Z · score: 1 (1 votes) · LW(p) · GW(p)

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!)

comment by [deleted] · 2011-04-04T00:23:22.274Z · score: 2 (2 votes) · LW(p) · GW(p)

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.

comment by komponisto · 2011-04-05T04:40:53.299Z · score: 0 (0 votes) · LW(p) · GW(p)

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.

comment by [deleted] · 2011-04-05T15:55:01.765Z · score: 0 (0 votes) · LW(p) · GW(p)

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."

comment by komponisto · 2011-04-05T22:13:48.247Z · score: 1 (1 votes) · LW(p) · GW(p)

(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 1/10 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:

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.

comment by Sniffnoy · 2011-04-06T01:05:55.889Z · score: 2 (2 votes) · LW(p) · GW(p)

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"...

comment by komponisto · 2011-04-06T02:20:44.476Z · score: 1 (1 votes) · LW(p) · GW(p)

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).

comment by Sniffnoy · 2011-04-06T02:54:00.790Z · score: 1 (1 votes) · LW(p) · GW(p)

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?

comment by komponisto · 2011-04-06T03:06:49.551Z · score: 1 (1 votes) · LW(p) · GW(p)

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 ![](http://www.codecogs.com/png.latex?V%20\\otimes%20W%5E\%20) .

comment by Sniffnoy · 2011-04-06T03:57:46.439Z · score: 1 (1 votes) · LW(p) · GW(p)

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...

comment by komponisto · 2011-04-06T05:32:06.373Z · score: 1 (1 votes) · LW(p) · GW(p)

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.)

comment by Vladimir_M · 2011-03-27T09:23:51.043Z · score: 13 (13 votes) · LW(p) · GW(p)

If you're confused by this kind of thing, you could probably give your intuitions a good training by studying the basics of electrical circuits (i.e. Ohm's law, Kirchoff's laws, the relations between current, voltage, resistance, and power, the Thevenin-Norton transformations, the series and parallel combinations of resistors, etc.). This will stretch your brain very nicely with a whole bunch of problems where you must be careful about what's being held constant or otherwise your intuition leads you into awful contradictions. (For example, power is proportional to resistance if you hold the current constant, but inversely proportional to it if you hold the voltage constant.) The concepts are very simple mathematically and don't require any background beyond very basic physics.

comment by AandNot-A · 2012-09-04T15:41:20.733Z · score: 4 (4 votes) · LW(p) · GW(p)

Where would one find that kind of exercises, online?

comment by jsalvatier · 2011-03-27T06:36:24.193Z · score: 8 (8 votes) · LW(p) · GW(p)

Similar issues arise with gasoline usage. Gallons per mile is probably a more natural unit for milage than miles per gallon since the quantity of miles traveled probably stays more constant than the number of gallons used. If you got a car that was twice as efficient, you probably would not drive twice as far.

comment by David_Gerard · 2011-03-27T08:45:36.039Z · score: 5 (5 votes) · LW(p) · GW(p)

Litres per 100 kilometres is actually the unit of choice in much of the world. (Kilometres per litre is a unit of choice in many other parts.)

comment by Oscar_Cunningham · 2011-03-27T09:44:22.057Z · score: 21 (23 votes) · LW(p) · GW(p)

For some reason no one does the obvious cancellation to end up in m^2. This even has an intuitive meaning, it's the cross-section that a line of fuel would need so that as you travelled along it you'd be "picking it up" at the same rate you were burning it.

comment by JGWeissman · 2011-03-28T18:08:09.663Z · score: 6 (6 votes) · LW(p) · GW(p)

"Miles per gallon"/"kilometers per litre" and their inverses are more convenient when people will be integrating that information with "dollars per gallon" / "Euros per litre" to get the information they really care about: how much they will be spending on fuel. "Gallons" / "Litres", though they are volumes, are being used abstractly to talk about an amount of gas, in the same way we could have abstractly used mass to refer to the same thing.

comment by JoshuaZ · 2011-03-27T16:39:18.288Z · score: 6 (6 votes) · LW(p) · GW(p)

That's not nearly as helpful for most people. They don't think in terms like that as having an intuitive meaning simply because cars don't work that way. When people think of problems of this sort they frequently use other data about the reference classes to help them. If you alter the behavior of the reference class they will generate confusion. For example, that's why so many people have trouble with "If some doctors are men, and some men are tall, does it follow that some doctors are tall?" but don't have trouble with "If some US Presidents were Republican and some Republicans were women, does it follow that some US Presidents were women?" Don't underestimate how much people use the actual behavior of a class to guide them rather than just the section of it that you have abstracted.

comment by David_Gerard · 2011-03-27T11:29:33.852Z · score: 2 (2 votes) · LW(p) · GW(p)

Distance-per-volume seems intuitively comprehensible to people and they can usually do the simple arithmetic to answer the simple questions "how much range do I get from this fill?" or "how much petrol do I need to go this far?" If you told people their fuel consumption in "square metres" they'd look at you like you were some sort of nerd. (I like it otherwise, so I'm likely some sort of nerd also.)

comment by Oscar_Cunningham · 2011-03-27T14:16:39.425Z · score: 4 (6 votes) · LW(p) · GW(p)

If you told people their fuel consumption in "square metres" they'd look at you like you were some sort of nerd.

:-) I am aware of this.

As it turns out, a more useful measurement would be square millimetres since this takes values of a reasonable size. Also, if you multiply by the amount of kilometres you travel you'll get the fuel consumption in litres, i.e. mm^2=L/km.

comment by Theist · 2011-04-21T22:00:54.202Z · score: 1 (1 votes) · LW(p) · GW(p)

Thank you for that humorous insight. I am entertained by the knowledge that my car has a fuel efficiency of 0.0784 mm^2.

comment by [deleted] · 2011-03-27T07:15:33.172Z · score: 6 (6 votes) · LW(p) · GW(p)

how much of x% concentration do you add to your y% concentration to get z% concentration?

Dimensional analysis is an important tool.

Here's an example. Suppose the problem is: "How much 85% alcohol do you have to add to 1 kilogram of 40% alcohol to get 55% alcohol?" Ethanol has a cute property where "Mixing equal volumes of ethanol and water results in only 1.92 volumes of mixture", which would be a distraction here, so let's specify that we're working with mass fractions, aka alcohol by weight here.

We're starting with 1 kg of 40% alcohol, which contains .4 kg alcohol in 1 kg of total fluid. We're adding N kg of 85% alcohol, which contains (.85 N) kg alcohol in N kg of total fluid. So we'll end up with (.4 + .85 N) kg of alcohol in (1 + N) kg of total fluid. We want 55% alcohol, so this gives us an equation:

(.4 + .85 * N) kg alcohol / (1 + N) kg total fluid = 55% alcohol

We specified that "55% alcohol" means dividing kilograms of alcohol by kilograms of total fluid, so the division is correct here. (Dimensional analysis means more than just making sure that you don't try to add kilograms to liters. Dividing kg total fluid by kg alcohol would give us a dimensionless number, but not one that could be referred to as "percent alcohol". Same for dividing kg alcohol by kg water.)

Now we can nuke the units and grind out the math:

(.4 + .85 * N) / (1 + N) = .55
(.4 + .85 * N) = .55 * (1 + N)
.4 + .85 * N = .55 + .55 * N
.4 + .85 * N - .55 * N = .55
.85 * N - .55 * N = .55 - .4
.3 * N = .15
N = .15 / .3
N = .5

Verify:

1 kg of 40% alcohol contains 1 * .4 = .4 kg alcohol
.5 kg of 85% alcohol contains .5 * .85 = .425 kg alcohol

.4 kg alcohol + .425 kg alcohol = .825 kg alcohol
1 kg total fluid + .5 kg total fluid = 1.5 kg total fluid

.825 kg alcohol / 1.5 kg total fluid = .55 YAY!
comment by janos · 2011-03-27T13:22:11.035Z · score: 10 (10 votes) · LW(p) · GW(p)

The way I'd try to do this problem mentally would be:

Relative to the desired concentration of 55%, each unit of 40% is missing .15 units of alcohol, and each unit of 85% has .3 extra units of alcohol. .15:.3=1:2, so to balance these out we need (amount of 40%):(amount of 85%)=2:1, i.e. we need twice as much 40% as 85%. Since we're using 1kg of 40%, this means 0.5kg of 85%.

comment by RobinZ · 2011-03-27T16:10:29.834Z · score: 2 (2 votes) · LW(p) · GW(p)

That's clever! Changing your frame of reference is a useful tool - there are a lot of problems which become simpler if you use measurements from a 'zero' that you pick.

comment by Benquo · 2011-03-27T14:38:57.019Z · score: 3 (3 votes) · LW(p) · GW(p)

To put it more generally, for the specified class of problem, you have 2 mixtures of the same 2 substances. Classify one substance (e.g. grams of salt) as the numerator, and either the other substance (e.g. grams of water) or their sum (e.g. grams of saltwater) as the denominator. But be sure to pick one representation and stick with it!

Add the numerators and denominators separately to express the problem algebraically, then solve.

comment by Alicorn · 2011-03-27T15:04:18.192Z · score: 4 (6 votes) · LW(p) · GW(p)

Like the concept (and upvoted). Hate the "inverse" terminology - when I'm already confused by a math problem, I don't want to add something that makes me flip a concept around every time I want to use it. Especially since not everybody will understand the same thing you do when they read "speed" - I usually think of speed as a thing you use to get to a specific destination, so fiddling with the distance doesn't make sense unless you're fiddling with your route, which is beyond the scope of the problem.

comment by komponisto · 2011-03-27T17:46:42.511Z · score: 1 (1 votes) · LW(p) · GW(p)

Like the concept (and upvoted). Hate the "inverse" terminology - when I'm already confused by a math problem, I don't want to add something that makes me flip a concept around every time I want to use it.

(Thanks for the upvote, first of all.) I'm not exactly sure what you mean here. We could easily give inverse speed its own name, like "slowness". But would that help? Your wording makes it sound almost like you might be objecting to the very idea of adding "the inverse of speed" to your concept inventory, in which case I'm wondering what aspect of the post you actually liked!

comment by Alicorn · 2011-03-27T18:29:37.993Z · score: 0 (0 votes) · LW(p) · GW(p)

To you, "speed" means thing X. To solve the problem, you need thing Y, which is the inverse of thing X. To me, speed means thing Y, so if you went around saying "inverse speed", I would not only have to perform an "inverting" operation on "speed" every time the term came up, I would wind up with thing X after I did it, which is not useful to solving the problem.

comment by komponisto · 2011-03-27T20:10:08.884Z · score: 1 (1 votes) · LW(p) · GW(p)

Wait -- really? Are you saying that you interpret "speed" to mean exactly what I mean by "inverse speed"? Hours per mile instead of miles per hour? So that when someone says they're going 5 miles per hour, you don't think "they'll have gone 5 miles after one hour" but rather "1/5 of an hour will have passed after they've gone one mile"? And if I wanted you to think "they'll have gone 5 miles after one hour", I would have to say "they're going 1/5 hours per mile"?

That's not what I got from your original comment at all. When you said

I usually think of speed as a thing you use to get to a specific destination, so fiddling with the distance doesn't make sense unless you're fiddling with your route

-- well, I didn't really understand what you meant, but I thought you were talking about thinking in terms of a two-dimensional map with different routes to the same place having different distances.

I suppose in retrospect I can make sense of it by interpreting the phrase "specific destination" as referring to keeping the distance fixed (and letting time vary).

But I still find it hard to believe that's what you mean -- for one thing, it would imply that you shouldn't have been confused by the original problem at all, and instead should be more confused by something like this:

"If you want to average 40 mph on a two-hour trip, and went 20 mph for the first hour, what should your speed be for the second hour?"

(If you find this easier than the other problem, then that means your idea of speed is like mine rather than being inverse to it.)

comment by Alicorn · 2011-03-27T21:59:17.550Z · score: 1 (1 votes) · LW(p) · GW(p)

So that when someone says they're going 5 miles per hour, you don't think "they'll have gone 5 miles after one hour" but rather "1/5 of an hour will have passed after they've gone one mile"?

I don't think any of those things. I wonder about how far they're going at that speed, and if the answer is "up the block" I think "oh, they'll be there soon" and if the answer is "to the moon" I think "that's going to take forever". I do not naturally think in numbers.

And if I wanted you to think "they'll have gone 5 miles after one hour", I would have to say "they're going 1/5 hours per mile"?

No, if you want me to think that they will have gone five miles after an hour, you tell me they're going someplace five miles away and it'll take them an hour to get there.

I thought you were talking about thinking in terms of a two-dimensional map with different routes to the same place having different distances.

Well, I was, but this was incidental to my point.

I suppose in retrospect I can make sense of it by interpreting the phrase "specific destination" as referring to keeping the distance fixed (and letting time vary).

Yes.

you shouldn't have been confused by the original problem at all

I was not confused in the way you were. I was confused in a different way, which has nothing to do with how I read the English word "speed" and everything to do with how my brain generates error messages when presented with math problems.

"If you want to average 40 mph on a two-hour trip, and went 20 mph for the first hour, what should your speed be for the second hour?"

Well, this is also confusing (my brain generates "sixty" automatically, but I don't actually know if that's an answer to this problem or just the result of seeing "40", "20", and "average" in that order, and I would have to do work to find out). It is not differently confusing than the first problem. I don't get any farther or stop any earlier before I want to seek assistance. (I don't even know if this is the same problem or not.)

comment by komponisto · 2011-03-30T16:23:47.796Z · score: 2 (2 votes) · LW(p) · GW(p)

I wonder about how far they're going at that speed, and if the answer is "up the block" I think "oh, they'll be there soon" and if the answer is "to the moon" I think "that's going to take forever". I do not naturally think in numbers.

Well, neither do I (I naturally think in terms of operations and transformations), so that's not the relevant distinction. The relevant distinction is between "wow, they're already far away" (speed) vs. "wow, they got there quickly" (inverse speed).

Let me see if I can generate, in your mind, something analogous to the confusion that existed in my mind. I probably won't succeed, but the idea of attempting is too interesting to resist.

Here are two questions that are easy to answer:

(1) If I travel for an hour and spend a lot of time on the first part of my journey, how much time will I spend on the second part? (Answer: not much.)

(2) If I travel a mile and go a large distance during the first part of my journey, how far will I have to go during the second part? (Answer: not very far)

And now here are two questions that are confusing:

(3) If I travel for an hour and go a large distance during the first part of the journey, how much time will I have to spend on the second part? (Answer: Huh? That depends on how much time you spent going that large distance during the first part.)

(4) If I travel a mile and spend a long time on the first part of the journey, how much distance will I have to cover on the second part? (Answer: Huh? That depends on how much distance you covered during that long time you spent on the first part.)

comment by jwhendy · 2011-04-01T01:58:42.445Z · score: 3 (3 votes) · LW(p) · GW(p)

Interesting post. Just had me pondering it all day and I thought I'd propose another way to go about this: visually. Open THIS in another tab. What you have is a plot of distance vs. time. The black line is shown with a constant slope of 20mph.

I have plotted two points on that line, (t1,d1) and (t2,d2). Imagine these as snapshots of your journey. You travel along the black line and at some moment check your clock and odometer. These "checkpoints" would be like t1/t2 and d1/d2. You compare your travel time to your distance traveled and determine that d/t = 20mph.

At each of these snapshots, you decide to maximize your speed over the same distance traveled so far. If you were traveling along on the black line, this means that you take a sharp left and travel infinitely fast along the dotted line until your distance is twice what you'd traveled so far. Since you traveled infinitely fast, you spent no time and your new coordinates are either (t1, 2d1) or (t2, 2d2).

Note that both of these points lie on a new line, shown in green. This line's slope indicates the final rate achieved. We can now see that:

r = (2d1)/t1 = (2d2)/t2

But d1/t1 and d2/t2 are equal to d/t = 20mph because they lie on the original constant sloped line. So we know that the green line's slope is 2 d/t = 2 20mph = 40mph.

Just thought it might be interesting to see this portrayed another way. I've shown what happens if you travel infinitely fast and use a vertical line to head on up to the final distance. If you traveled less than infinitely fast, your line would lie between the original rate (20mph) and our shown theoretical maximum. This area is the solution set and is highlighted in light green.

comment by komponisto · 2011-04-01T06:55:15.725Z · score: 1 (1 votes) · LW(p) · GW(p)

Very nice! This illustrates the idea that at any time during the journey, the average speed up to that point constrains the possible average speeds for the whole journey.

I thought I'd point out that merely the first point (t1, d1) suffices to construct the second (green) line, since the lines both start at the same point (the origin).

comment by jwhendy · 2011-04-01T15:25:21.331Z · score: 0 (0 votes) · LW(p) · GW(p)

Thanks! And yes, I could have left it at one point on each line since the origin counts, but thought two points might help drive the point (no pun intended!) home, lest one point appear to be a "lone solution" -- two helps show that the green line is actually a maximum to a whole solution set rather than just a line created from one data point.

comment by handoflixue · 2011-04-26T23:45:19.324Z · score: 2 (2 votes) · LW(p) · GW(p)

Thank you so much for finally giving words for my intuitive understanding of the problem, which matches yours. I always assumed it just meant I was "doing it wrong" o.o

(I knew how to solve it, it's just not intuitive to me to conclude "impossible" instead of "60 mph")

comment by CronoDAS · 2011-03-29T04:25:33.665Z · score: 2 (2 votes) · LW(p) · GW(p)

This particular problem is indeed a big pain in the ass, for pretty much the reason you specify: you have to realize that "half the trip" means "half the distance" and then set up the algebra in different way than you usually do.

As a relatively young child (I think I was about 10 years old), I used a computer software program that taught me the General Method To Solve Any Word Problem You May Have To Do For Homework. I don't remember what it was called, and I don't know if I can actually explain the algorithm over the Internet, but I think I can credit my ability to successfully execute that algorithm with a lot of my later success in school.

comment by Gray · 2011-03-31T19:21:16.921Z · score: 1 (1 votes) · LW(p) · GW(p)

I don't know about anyone else, but whenever I encounter a problem like this, involving rates, I immediately do a dimensional analysis. I just took a stab at the problem and got a contradiction. I think this is equivalent to what you're saying.

And I think I'm dawning on the same realization you're having, though I don't exactly follow your analytical explanation in terms of "mappings", and that is that I shouldn't have had to start writing down equations to understand that the problem is impossible. So that's why I want an intuitive understanding for why the problem is impossible. I don't quite have that yet.

comment by Gray · 2011-03-31T22:19:23.100Z · score: 2 (2 votes) · LW(p) · GW(p)

Okay, here's my analysis:

The characters:

St, Tt, Dt: The average speed, time interval and distance for the total trip. S1, T1, D1: The average speed, time interval and distance for the first half of the trip. S2, T2, D2: The average speed, time interval and distance for the second half of the trip.

Now, running through the equations with the numbers plugged in gives me the strange equation: T1 + T2 = T1. This is a contradiction, because it is premised that T2 > 0 (which can be inferred from denying infinite speeds).

But I want an intuitive understanding, so I decided to start over using generic constants rather than specifying 40mph and 20mph in the problem. So I start with computing the total average speed for the trip (this comes from the definition of average speed):

St = (S1 T1 + S2 T2) / (T1 + T2)

We know that D1 = D2 (because they are both half the distance of the whole trip), and thus S1 T1 = S2 T2 (from the formulas for speed, time and distance), so therefore:

St = 2 S1 T1 / (T1 +T2)

T1 + T2 = (2 S1 / St) T1, if T1 + T2 != 0

All of these values need to be positive numbers, so S1 / St > 1/2. Otherwise, the problem yields a contradiction. The numeral 2 in the above equation comes from the fact that D1 is half the distance of Dt, so it seems likely that if the problem were stated differently such that D1 was a third of Dt, then it would need to be true that S1 / St > 1/3; and S1 / St > 1/4 if the distance is a fourth; and so on.

comment by sixes_and_sevens · 2011-03-27T09:46:29.468Z · score: 1 (1 votes) · LW(p) · GW(p)

Only tangentially related, but this reminded me of of when I noticed Hubble's Constant had the unit of inverse-seconds, and for the life of me couldn't figure out what that meant.

comment by Kutta · 2011-03-27T17:37:35.787Z · score: 0 (0 votes) · LW(p) · GW(p)

I also gave it a puzzled look but the realization that you're supposed to multiply the constant with a distance hit me immediately thereafter.

comment by [deleted] · 2014-01-09T12:04:04.507Z · score: 0 (0 votes) · LW(p) · GW(p)

If an ever you find yourself working with percentages always recalculate them as fractions. In fact with solutions you are working with probabilities of encountering a given molecule (loosely speaking).

So with mixing solutions: You have X amounts of P(x) solution, and you want to add any amount Y of P(y) solution so it will in the end be a P(z) solution. Now you are adding weighted probabilities.

I taught my then elementary-school-level sister how to convert percentages to fractions and fractions to percentages and she has never since used specialised algorithms for percentages.

comment by Prismattic · 2011-12-05T22:20:09.826Z · score: 0 (0 votes) · LW(p) · GW(p)

This post made me think of this. Perhaps it will be of interest.

(On reflection, it seems the Cheaptalk post appeared one week before your post, which makes me wonder if it wasn't the stimulus for it).

comment by cousin_it · 2011-03-27T10:52:38.285Z · score: 0 (0 votes) · LW(p) · GW(p)

Thinking that "first half of the route" refers to time is a reading comprehension problem. When I first encountered the problem, I noticed the possible ambiguity (huh, did they refer to time or to distance?), then resolved it (in case of time they'd have referred to "first half of day" instead), then wrote an equation in my mind and solved it.