Date: Nov 15, 2012 4:11 PM
Author: Dave L. Renfro
Subject: Re: Safety Glasses in Algebra?

Robert Hansen wrote:

http://mathforum.org/kb/message.jspa?messageID=7923725

> http://www2.wnct.com/news/2012/nov/14/3/lenoir-county-school-uses-interactive-lab-learn-al-ar-2780878/

> "One glance at her safety glasses, and one might think
> 7th grader Anne-Wesley Taylor is busy in science class.
> But this lesson on water quality is actually a new,
> interactive way to understand a sometimes confusing
> subject: Algebra."


I haven't listened to the video and don't really have
a comment about the teaching method (getting students
engaged is good, assuming it's in a way that leads to
appropriate learning), but I did want to complain about
something I see way too often (and have complained about
before), which is the tendency of advocates of "the latest
new thing" to misrepresent the past. The first sentence of
the article follows:

** When many of us were in school, math class was about
** word problems and memorization.

Since when was math class about memorization? Math has always
involved the least amount of memorization of any subject I
can think of, with the possible exception of P.E. classes.
And, now that I think about it, I took a number of multiple
choice tests on volleyball rules and other sports rules
in my high school P.E. class, tests whose preparation for
involved nothing but memorization.

As someone who always had great difficulty with memorization
(I had to transfer to another undergraduate university due to
Foreign language requirements, I got a 60 (under 70 was an F)
on my 3rd quarter 9th grade English report card because I was
making 30s to 50s on the spelling tests our class began taking
that quarter, I almost failed a supposedly easy classics elective
because I couldn't remember the various the painting and
sculpture and architecture styles we needed to distinguish
on tests, etc.), I'm EXTREMELY AWARE of the amount of
memorization in various subjects. Sure, I often forgot
things in math too (e.g. is the derivative of u/v equal
to (u'v - uv')/v^2 or (uv' - u'v)/v^2), but almost always
you can "see the complete picture" by filling in the missing
parts by using some alternate method. For example, in the case
of the quotient rule, see which of the two possibilities work
for the case of u/v = 1/x (whose result you know by using the
power rule).

Dave L. Renfro