Wavefronts, Box Diagrams, and the Product Rule: A Discovery Approach
John W. Dawson, Jr.
The Two-Year College Mathematics Journal, Vol. 11, No. 2. (Mar., 1980), pp. 102-106.
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Wed Aug 22 11:12:14 2007
Wavefronts, Box Diagrams, and the Product
Rule: A Discovery Approach
John W. Dawson, Jr.
John W. Dawson, Jr., received his S.B. in mathematics from
M.I.T., and his Ph.D. from the University of Michigan. He joined
the Penn State faculty in 1972, transferring to York as an
assistant professor in 1975. He is a specialist in mathematical
logic.
Of the many differentiation rules encountered in calculus, the product rule is
perhaps the first to shake students' naive faith in the "simplicity" of mathematics.
It is easy to refute their expectation that (fg)' = f'g', but counterexamples often
seem more to bewilder than enlighten. Moreover, the correct formula is not easy to
guess, except for the special case when one factor is constant; and although many
students accept the formula (cg)' = cg' as agreeing with their intuition, few realize
that this special case itself contradicts the equation (fg)' = f'g'. Later, having
learned the correct general formula, some students laboriously apply it to all
products, ignoring the erstwhile special case, or failing to recognize it as such!
Such observations suggest that few students develop any real understanding of
the product rule. Rather, having no idea why the rule takes the form it does, most
students simply memorize it, adding it to their growing stockpile of unrelated facts.
Of course, at some point they have probably seen the product rule derived through
algebraic manipulation of the defining difference quotient. The usual proof is
concise and rigorous, and students will probably assent to the correctness of each
step. Nevertheless it is a hollow demonstration, one that verifies rather than
explains, for it conveys no idea of how the formula, or the proof itself, might ever
have been discovered. In particular, the rationale for transforming f ( x + Ax)g(x
Ax) - f ( x ) g ( x ) to [ f ( x + Ax) - f ( x ) ] g ( x Ax) f ( x ) [ g ( x Ax) - g ( x ) ] via the
simultaneous addition and subtraction of the term f ( x ) g ( x Ax) is mystifying to
students; too often the step is justified merely as a clever trick that works.
In short, the product rule provides a classic example of how a logically rigorous
"proof" may fail to be psychologically convincing or enlightening. There is, however, a more heuristic approach to the product rule that leads to discovery of both
the formula and its proof. This approach invokes the geometric representation of
products as areas, an idea later to prove so important in integral calculus, while
stressing certain physical considerations important in their own right. At the same
time the mystery surrounding the algebraic manipulations is dispelled. Indeed, the
only mystery left unexplained is the conspicuous absence of such an approach from
most calculus texts.
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The very word "calculus" provides a fitting point of departure: Latin for
"pebble," it alludes to the counting stones used in early abacus-like devices, but it
also calls to mind the ripple created by a pebble cast into water, and so suggests the
familiar problem of computing the rate of increase of the area enclosed by a
radially expanding circular wavefront. Although this problem appears in nearly
every calculus text, it is usually considered only after introduction of the chain rule,
and all too often it is presented as a routine exercise devoid of any particular
significance. However, merely by factoring a difference of two squares, it is easy to
show directly that
and far from being surprising, the result is readily conjectured, since the area inside
the ripple increases through the engulfing of space by the expanding wavefront,
which at any given instant has length 27rr and is moving outward at speed d r / d t
(Figure 0). Similarly, a radially expanding spherical membrane engulfs 3-dimensional space at a rate equal to the product of its instantaneous surface area and its
instantaneous velocity. The physical intuitions involved are quite natural and
general*, and students might well pass on to consider the rate at which area is
engulfed by an "arbitrary" radially-expanding simple closed curve. It is again
Figure 0.
dA = C dr
dt
dt
*Similar considerations underlie the basic integral theorems of vector analysis, a classical subject
notoriously obscured by modem rigorous treatments. The outstanding text by Schey cited below
presents a lucid informal account of those topics from a physicist's point of view, while vividly
describing the bewilderment and hostility that such material often evokes from students previously
exposed to abstract presentations.
reasonable to conjecture that the rate should be proportional to the length of the
curve, but exploration of the assumptions involved could quickly lead to such
deeper concepts as rectifiability, orientability, and homology. (As in any discovery
approach, the instructor must be well-grounded in his discipline to anticipate and
lead discussion of such matters!) In any case, we are here concerned with the less
ambitious aim of conjecturing the product rule.
Toward that end, let h(t) and w(t) denote the height and width, varying as
functions of time t, of a rectangle having one pair of adjacent sides lying along the
axes of a Cartesian coordinate system. Any product may be represented as the
(signed) area of such a rectangle, and we may think of the sides not lying along the
axes as forming an L-shaped wavefront. By analogy with the ripple, at any time t
the side of length w(t) will be moving with speed ht(t), and the side of length h(t)
will be moving with speed w'(t) (Figure 1).
Figure 1. [ h ( t ) w ( t ) r = h l ( t ) w ( t ) + h ( t ) w f ( t ) .
Hence the total rate of engulfing (or
disgorging) of area should be w(t)h'(t) h(t)w'(t). To prove the correctness of this
result, we draw the rectangles corresponding to times t and t + At. In case all of
h(t), hf(t), w(t), and wf(t) are positive, we obtain Figure 2, in which the L-shaped
region representing the difference of the areas (numerator of the difference quotient) is left unshaded.
+
h(t
+ At)
w(t)
Figure 2. A ( t + A t )
-
A(t) = h(t
+ At)w(t + A t ) - h ( t ) w ( t ) .
There are two quite natural ways (Figures 3 and 4) to
decompose that region into a pair of rectangles, each yielding an alternative form
of the identity invoked in the algebraic proof. Of course, there are other cases to
consider, and it would be well for students to draw the figures corresponding to
other sign combinations; nonetheless the algebraic proof is perfectly general. The
illustration (Figure 5) for the special case is also enlightening, since it is obvious
that if one dimension is held fixed, the area of the rectangle can still change
through the independent motion of the other dimension.
w(t
+ At)
h(t
+ At) - h ( t )
+ At) - w ( t )
A ( t + At) - A ( t ) = [ h ( t + A t ) - h ( t ) ] w ( t + At) + h ( t ) [ w ( t + A t ) - w ( t ) ] .
w(t
Figure 3.
h(t
Figure 4. A ( t
+ At)
+ At) - A ( t ) = [ h ( t + At) - h ( t ) ] w ( t )+ h ( t + At)[w(t + A t ) - w ( t ) ] .
w(t) Figure 5. [cw(t)l' = cwf(t). Two further observations are in order. First, in seeking to discover the form of
the product rule, one may wish to consider examples of functions f # F and g # G
for which f' = F' and g' = G' but (fg)'# (FG)'; such examples show that no
formula for (fg)' involving only f' and g' can possibly be correct, thereby providing
a simple demonstration of a non-existence theorem. Second, box diagrams are
useful in many other contexts-in the author's opinion, their range of applicability
is at least as wide as the much-touted Venn diagrams. As an example, consider
Bayes' theorem. The formula
P(A ( B )=
P(A)P(BIA)
P(A)P(B I A) + P(S - A)P(B I S - A)
corresponding to events A, B in a sample space S is difficult for most students to
remember, and its analytic derivation means little, if anything, to them. Yet after
struggling mightily in vain attempts to train students to solve Bayesian problems,
the author was amazed to discover how readily the ideas were assimilated when
they were presented in the context of a box diagram rather than a formula. For
example, consider the twins problem:
Roughly 3/4 of all twins are fraternal, while only 1/4 are identical.
Among identical twins, half are girls and half boys, while among
fraternal twins, half are mixed sets and 1/4 each consist of just boys
or just girls. Given that a set of twins consists of two boys, what is the
chance those twins are fraternal?
Many students never learned to fit such a problem into Bayes' formula, but after a
single lecture most could quickly draw the diagram below and use it to obtain the
solution. Many even remarked how easy such problems were to solve!
3/4 Fraternal
1/4 Identical
3/16
Figure 6 . P(Fraternal1 both boys) = 3/16
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REFERENCE
H. M. Schey, Div, Grad, Curl, and All That, Norton, 1973.
3
= 5