Teaching Techniques of Integration, Part 2

Nov 10

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Well,  from the comments here and elsewhere, my previous post certainly seems to be stirring up a discussion worth having!  That’s good.  All points of view are welcome.

On to trigonometric substitution tomorrow.  A colleague of mine had the following fantastically fun idea for their class.  Click N’ Clack, the Tappet brothers of Car Talk on NPR, were posed a very practical call-in question by Rich, the driver of an 18-wheeler. The Great Fuel Tank Problem. You can download the 2 min clip here.

In short, his fuel gauge was inoperative and he was using a dipstick to measure fuel levels in his cylindrical tanks.  Inserted from the top of the tank, the full fuel level was 20 inches, so he marked the stick at 10 inches to measure half a tank.  His question was “Where should I mark the stick for a quarter tank?” because he was pretty sure that 5 inches was not right.

One can solve this problem with integral calculus.  What’s great is that one can also solve this problem in many other very creative ways too, capitalizing on the special geometry of the tank (my favorite one involves not pen and paper, but a soda/beer can, toothpick, and 4 drinking glasses).  But the most direct calculus solution (in the style of the traditional curriculum) is to set up an indefinite integral computing the volume up to a fuel level h, compute an expression for it in terms of elementary functions of h using techniques of integration (like trigonometric substitution) and set up an equation to solve for h.  What’s unfortunate is that the equation turns out to be transcendental.  It is most likely not possible to express the value of h as a formula in special values of the elementary functions.  By why would you need to?  He doesn’t need to know where to mark the stick to arbitrary precision.  At best he needs to know it within a 1/16th of a inch.  Using numerical methods at this step is necessary and more than sufficient.

On the other hand, if you were solid on the concepts of calculus, you would recognize that the indefinite integral you cook up \int_0^h f(y)\,dy (where f(y)=\sqrt{100-(y-10)^2} for example) is a continuous function of h, being the indefinite integral of an integrable function.  If an upper bound for the integral \int_0^{h_1} f(y)\,dy, computed by a Riemann sum for example on a computer, is lower than the volume of a 1/4 tank of fuel for some h_1 (for example h_1=5.5) and a lower bound for the integral \int_0^{h_2} f(y)\,dy is higher than that for another value h_2 (say 6.5) then the desired height must lie between h_1 and h_2 by the Intermediate Value Theorem.  Bracketing in, on could find an estimate for desired value of h in short order.  Applying techniques of integration here only adds more steps to derive an equation that still must be analyzed by numerical methods.  I just wonder if it would be better to teach students how to use this kind of reasoning and the tools at their disposal to solve similar problems, at the expense of toning down the emphasis on rote computations and algebra.  This still teaches problem solving, but of a different sort, using different tools.

The comments have raised a number of issues/concerns that I would like to address in further posts.  But for now… signing off.  Good night.

Posted on November 10, 2011, in Uncategorized. Bookmark the permalink. 2 Comments.

  1. Virgil Pierce | November 10, 2011 at 7:38 am

    I was asked a very similar question by one of my student’s grandparents (parents day on campus and they came by my office specifically to ask this question). Grandpa needed to take a medication three times a day, but the total dose for one day was one basically cylindrical pill (of small height compared to radius) — I believe there was a sidebar story here about how he got the pills from a Mexican pharmacy where the standard dosage was different than an American one:

    He could of course cut the pill into 3 pie shaped pieces — however if the pill is small enough (and this one was), this becomes an engineering nightmare and is basically not possible to do accurately (one would have to locate the center of the pill with higher and higher accuracy as the radius decreases and you can show that the induced percent error in the volumes of the pieces would be significant).

    So Grandpa was designing a device in his machine shop that would hold two razor blades in place and could be driven down on the pill to divide it into 3 equal volume parts. The question was: how far apart to hold the razor blades.

    The further question was: suppose we made this device, would the percent error induced on the pieces be any better than cutting pie shaped pieces? The last question is not so important, as in any case the device would be easier to use than the pie shaped analogue. (you can find in pharmacies devices like this which cut a pill into halves by driving a blade).

    The solution ends up needing a Newton’s method or similar to find an approximate solution to a trig=poly equation and has now become a staple of my engineering mathematics course for numerical methods.

    A nice further question deserving of some work is how the problem needs to be modified by a elliptically-cylindrical pill.

    Reply
    • Virgil Pierce | November 10, 2011 at 8:06 am

      oh. I forgot. A secondary solution, which eliminates much of the error of cutting 3 pie shaped pieces, is to cute 6 pie shaped pieces. Your first cut divides the pill into half and can be done with very little error. Each half (A and B) is then divided into 3rds, then the smallest from A is paired with the largest from B, the largest from A with the smallest from B, and the two middle sized pieces are paired.

      This does two things which help to control the error: First it is easier to locate the middle of a line segment which is what you are doing in cutting the parts A and B into thirds, then you are reducing the variation in the solution by combining the over estimate with the under estimate from each piece.

      The downside is that you are more likely to grind your pill into powder.

      Reply

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