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Topic: Riemann Sums and Dimensional Analysis
Replies: 0

 Richard Sisley Posts: 4,189 Registered: 12/6/04
Riemann Sums and Dimensional Analysis
Posted: Apr 19, 1998 12:19 PM

In a recent post someone whose name I could not determine suggested an
interesting type of problem about applications of the integral concept:

I give a series of graphs (sketches), with descriptions of
quantities on each axis, and numbers on each axis. On each graph I
draw a point on the graph, and I shade in an area beneath a
portion of the graph. The directions are:

For each graph,
(a) say in words what the indicated point tells you (use
specific numbers)
(b) say in words what the indicated area represents, or
say "the area doesn't mean anything."

(Part a is less important than part b; it's there mainly to get them to
think about the meaning of the graph.)

The graphs are sketches of:

(1) x: time (months); y: price of gasoline (\$)
(2) x: time (days); y: daily wombat sales (ferrets/day)
[graph has a bump up during the Christmas shopping season]
(3) x: time (years); y: population of your town (thousands)
(4) x: time (years); y: birth rate (births/year) (thousands)
(5) x: time (days since impregnation); y: cow's daily milk
production (gal/day)
(6) x: time (days); y: the height of a plant (cm)
(7) x: time (days); y: daily sales at a store (\$)
(8) x: time (hours); y: rate of change of temperature (degrees F per
min)

I understand this suggestion to mean that the graphs whose sketches are
provided are potential integrands, and the question is whether an
integral with this integrand would have any meaning.

This type of question could be analyzed using the fact that integrals
can be approximated by Riemann sums. Here is an example of this type of
analysis applied to the first example:

If values of the integrand have unit designation "dollars (per gallon)",
then each of the products making up a Riemann sum estimate of the
integral would be the product of a number with designation "dollars per
gallon" and a number with designation "months." Thus, the products
making up this sum would have unit designation "dollars per
gallon-month." The Riemann sum would have the same designation, as
would the integral which it approximates. An integral which equals a
number of "dollars per gallon-month." would seem to have no meaning.

On the other hand, in the second example, values of the integrand have
unit designation "ferrets (sold) per day" and the products making up the
Riemann sum have designation "ferrets (sold) per day times days" =
"ferrets sold." An integral with this designation might be of interest
to the manager of a pet store.

The same kind of dimensional analysis can be applied to the examples
involving densities cited in the same post:

(2) [based on Harvard Calculus, p. 423 (I changed r from 1 to 1.5)]:

If the density of air at a height h is given by
P = 1.28e^(-0.000124h), where P is the density in kg/m^3
and h is the height above ground level in m,
find the mass of a cylindrical column of air 25 km high
with a diameter of 3 m.

Suppose a function models data points described as (height in meters
measured from the ground, the area of a cross section of this
cylindrical column of air at this height). Then an integrand function
whose values have unit designation "kg/m" can be formed by multiplying
this function (with values m^2) by the function which models the pairs
(heights above the ground, the density of the air at that height in
kg/m^3). The products making up Riemann Sums approximating an integral
would have unit designation "m^2 times kg/m^3 times m" or, "kg." The
integral would then measure the weight of the column of air.

Thanks to the unidentified donor for suggesting this analysis.

Sincerely,

Richard Sisley