Comment on Inverse Speed

Thu, 31 Mar 2011 17:54:52 GMT

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?").

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