2

I am enjoying this mathematics book for the general public called Measurement by Paul Lockhart. For the most part, I am happy with his metaphors and intuitive explanations of the different concepts, but I occasionally come along something that I wish was discussed more rigorously. There are also times when I can't tell whether something is part of general mathematical practice or is just his own idiosyncratic invention.

In particular, I want your help in interpreting what Lockhart calls the "Leibniz d operator". I had better quote an example passage from the book to show you how he treats this concept:

How can we take the information $$a^{2}=b^{2}+3$$ $$c=2a+b$$ and determine the ratios $da:db:dc$ ? The direct approach would be to give the mixing board a kick, check out the variation of the sliders, and figure out where their proportions are heading as the kick gets smaller. But here's the point: we don't actually need to go through this laborious process. Instead, we can simply apply the d- operator to both sides of the equations: $$d(a^{2})=d(b^2+3)$$ $$dc=d(2a+b)$$ After all, if two variables are always equal their rates must also be equal. Expanding these accordingly, we obtain the differential equations $$2a\:da=2b\:db$$ $$dc=2da+db$$ So for instance, at the moment when a = 2, b = 1, and c = 5 (which does in fact satisfy our original equations and so qualifies as an actual moment), we have $$4\:da=2\:db$$ $$dc=2\:da+db$$ Thus at that precise instant, b is moving twice as fast as a, and c is moving four times as fast. In other words, the ratio $da:db:dc$ is $1:2:4$. We now have a simple and direct method for solving any problem concerning relative rates of change$-$just $d$ everything!

"Kicking the mixing board" is his metaphor for using the difference quotient. I recognize and understand this concept of the derivative from my first exposure to calculus. I am familiar with the $\frac{dy}{dx}$ notation, but this is not with respect to any variable; it is just this $d$-ing something arbitrarily that is unfamiliar to me.

I suppose my question is a reference request related to this idea of $d$-ing both sides of an equation. I think I understand what he is doing, but I want to read a more rigorous discussion of how $d$ can act like some sort of function or operator. Sorry if this is a dumb question, but where can I find more information about this interpretation of taking a derivative?

Eric Haney
  • 769

2 Answers2

4

The previous answer does arrive at the same conclusion as Lockhart, but actually I think Lockhart has a different interpretation in mind.

Instead of studying $x'(t) = dx/dt$, Lockhart wants to consider $dx$ and $dt$ separately. He does not view them as infinitesimal quantities (as Leibniz did), but as the "abstract, nonnumerical velocity" of $x$ and $t$, respectively. Instead of always considering rate of change in $x$ with respect to $t$, we drop this "reference" and consider the rate of change (velocity) of $dx$ alone. Lockhart is careful to emphasize that this $dx$ is not a number; only when we form a ratio like $dx/dt$ do we get numbers, by comparing two quantities. But the point is that the differentiation laws go through in the same way regardless of what reference we choose, and in some ways this approach is simpler.

One example considered by Lockhart is $p = t^2$, which in his $d$-operator convention has the rate $dp = d(t^2) = 2t\ dt$. Here $dt$ is the "reference" speed of time; the numerical value of $dp/dt$ will depend on what units we measure $p$ and $t$ by, but the functional form is not affected. And this is true for any quantities of course; time is not special in any way.

A motivation for this view is that when comparing the rates $dx$ and $dy$ of two quantities $x$ and $y$, we really don't need a third "reference" quantity (a unit) like time $t$. The rates $dx$ and $dy$ are directly measurable against each other, as the ratio $dx/dy$, and this ratio is the same regardless of what unit we happend to measure time in. Therefore $dx$ and $dy$ are meaningful on their own.

Personally, I rather like this idea, but it does take some getting used to. I'm still working on it :)

Roland
  • 276
1

What's going on is $d$ is supposed to mean "an infinitesimal change in".

To make it clearer what's happening, imagine we had the following situation. The quantities $a$, $b$, and $c$ all depend on time. Let's write $a(t)$ to mean the value of $a$ at time $t$. We also write $b(t)$ and $c(t)$. Furthermore, you know that they solve those two equations: $$a(t)^2 = b(t)^2 + 3$$ $$c(t) = 2a(t) + b(t)$$ Now, take the derivative of both equations. $$2a(t)a'(t) = 2b(t)b'(t)$$ $$c'(t) = 2a'(t) + b'(t)$$ Now let's say when $t=7$ we have $a(7) = 2$ and $b(7)=1$ and $c(7) =5$. Then, $$2a'(7) = b'(7)$$ $$c'(7) = 2a'(7) + b'(7)$$ which is practically the equation found in the passage cited.

Now in the above calculation, replace $a(t)$ with just $a$, and $a'(t)$ with $da$. (similarly for $b$ and $c$.) We do the same calculation.

Any calculation you see similar to that one in the book, you can understand by pretending that the quantities you're thinking about depend on time, and then taking time derivatives. But then you realize that the calculation has nothing to do with "time", really. $t$ can just be any parameter. In the words of the book you're reading: suppose $t$ is the value of some slider. Then the above calculation works just as well.

Tim Carson
  • 381