# Inverse Speed

One must always in­vert.

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 140 hpm, and your in­verse speed for the first half is 120 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.)

• 27 Mar 2011 13:32 UTC
13 points

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.

• 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) = 150, 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!)

• 4 Apr 2011 0:23 UTC
2 points
Parent

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 110 of an hour and then 120 of an hour, so that’s two miles over, um, 320 hours… so 403 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 110 of an hour and then 120 of an hour, so that’s two miles over, um, 320 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 110 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 110 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?

[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

$V \\otimes W^\* \\to L\(W,V\$)

(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?

• 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) .

• 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 12 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.)

• 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.

• Where would one find that kind of ex­er­cises, on­line?

• 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.

• 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.)

• 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.

• “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.

• 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.

• 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.)

• 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.

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

• 27 Mar 2011 7:15 UTC
6 points

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!

• 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%.

• 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.

• 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.

• 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.

• 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!

• 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.

• 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 15 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.)

• 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 15 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.)

• 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.)

• 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.

• 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).

• 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.

• 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”)

• 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.

• 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.

• 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 > 12. 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 > 13; and S1 /​ St > 14 if the dis­tance is a fourth; and so on.

• 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.

• 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.

• 9 Jan 2014 12:04 UTC
0 points

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.

• 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).

• 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.