Inverse Speed

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 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.)