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Message |
| Dave L. Renfro |
 Posted: Wed Jul 12, 2006 11:32 am |
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Guest
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Karl M. Bunday wrote (in part):
[quote:ee4729b6d3]I'm a math team coach and homeschooling parent looking
for some extra depth for the math learners in my care.
Most of the young people on my math team get their
main math instruction from the U. of MN Talented Youth
Mathematics Program, using the Stewart Early Transcendentals
textbook with an excellent supplementary materials. Even
though they've got some good instruction already, I'm looking
for suggestions of calculus problems that really make a
learner THINK about calculus concepts and go beyond the
usual school homework exercise.
[/quote:ee4729b6d3]
Browse and search the AP-Calculus forum
http://mathforum.org/kb/forum.jspa?forumID=63
Caltech Calculus Lecture Notes
http://www.math.caltech.edu/classes/ma1a/index.html#lect
http://www.math.caltech.edu/classes/ma_9/index1.html#lect
Indiana University-Purdue University Fort Wayne
Honors Calculus web page (maybe some ideas here)
http://www.ipfw.edu/math/h-calculus/
MIT's Fall 2002 Honors Calculus Lecture Notes
http://tinyurl.com/ranuv
Here's an example that I posted in the AP-Calculus
forum on 15 March 2006 at
http://mathforum.org/kb/message.jspa?messageID=4542260
Let [[x]] denote the greatest integer less than
or equal to x, and define f(x) by
f(x) = x * [[-(x^2)]].
Computing the derivative of f'(0) leads us to
compute the limit as h --> 0 of the expression
[ f(0 + h) - f(0) / h ],
which, for h not equal to 0, can be rewitten
as [[-(h^2)]].
Note that this is defined for h=0, as we were able
to cancel the h in the denominator. However, plugging
in h=0 gives an incorrect result! Even though we
can plug in h=0, _we're_supposed_ to take the limit
as h --> 0, and this gives -1. Thus, f'(0) = -1.
The neat thing about this example is that if you
simplify sufficiently so that you can plug in h=0,
and then plug in h=0, you don't get the correct
result.
This example is given in:
David Sanders, "A cautionary counter-example",
Mathematical Gazette 59 #407 (March 1975), 44-45.
Dave L. Renfro |
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| Dave L. Renfro |
| Posted: Wed Jul 12, 2006 11:36 am |
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Karl M. Bunday wrote (in part):
[quote:2b9e25ef5c]I'm a math team coach and homeschooling parent looking
for some extra depth for the math learners in my care.
[/quote:2b9e25ef5c]
Dave L. Renfro wrote (in part):
[quote:2b9e25ef5c]Browse and search the AP-Calculus forum
http://mathforum.org/kb/forum.jspa?forumID=63
[/quote:2b9e25ef5c]
Oh, I see you already know about this one!
(Your post is repeated there.)
Dave L. Renfro |
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| Ken Pledger |
| Posted: Wed Jul 12, 2006 9:05 pm |
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In article <5v_sg.7484$cd3.1995@newsread3.news.pas.earthlink.net>,
"Karl M. Bunday" <kmbunday@yahoo.de.com> wrote:
[quote:4de04094a4].... I'm looking
for suggestions of calculus problems that really make a learner THINK about
calculus concepts and go beyond the usual school homework exercise....
[/quote:4de04094a4]
This classic little fallacious argument may be too easy for them.
Since x^2 is positive, you would expect
the integral from -1 to +1 of 1/(x^2) dx
to be positive. However, the usual calculation using an antiderivative
gives
[-1/x] from -1 to +1
= - 1 - (- (- 1))
= - 2 which is negative.
Why?
Ken Pledger. |
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| Proginoskes |
| Posted: Thu Jul 13, 2006 2:38 am |
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Ken Pledger wrote:
[quote:3397119f66]In article <5v_sg.7484$cd3.1995@newsread3.news.pas.earthlink.net>,
"Karl M. Bunday" <kmbunday@yahoo.de.com> wrote:
.... I'm looking
for suggestions of calculus problems that really make a learner THINK about
calculus concepts and go beyond the usual school homework exercise....
This classic little fallacious argument may be too easy for them.
Since x^2 is positive, you would expect
the integral from -1 to +1 of 1/(x^2) dx
to be positive. However, the usual calculation using an antiderivative
gives
[-1/x] from -1 to +1
= - 1 - (- (- 1))
= - 2 which is negative.
Why?
[/quote:3397119f66]
It's an IMP_____ INTEGRAL.
--- Christopher Heckman |
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| Karl M. Bunday |
| Posted: Sun Jul 16, 2006 3:31 pm |
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Fermat wrote, responding to my request:
[quote:d5891d2567]I do not see Kleppner Kolenkow in there. If you want some good
problems, involving calculus, try this. And you can try Hassani for
more physical applications.
[/quote:d5891d2567]
Thanks. I looked up those authors and will seek out their books.
--
Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345
Learn in Freedom (TM) http://learninfreedom.org/
remove ".de" to email |
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| Karl M. Bunday |
| Posted: Sun Jul 16, 2006 3:32 pm |
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Ken wrote, providing information I sought:
[quote:b447abb7bb]This classic little fallacious argument may be too easy for them.
Since x^2 is positive, you would expect
the integral from -1 to +1 of 1/(x^2) dx
to be positive. However, the usual calculation using an antiderivative
gives
[-1/x] from -1 to +1
= - 1 - (- (- 1))
= - 2 which is negative.
Why?
[/quote:b447abb7bb]
I appreciate the problem. Thanks for your help.
--
Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345
Learn in Freedom (TM) http://learninfreedom.org/
remove ".de" to email |
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| Karl M. Bunday |
| Posted: Sun Jul 16, 2006 3:35 pm |
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Dave wrote, in reply to my questions:
[quote:5cadb6cd62]I'm a math team coach and homeschooling parent looking
for some extra depth for the math learners in my care.
Most of the young people on my math team get their
main math instruction from the U. of MN Talented Youth
Mathematics Program, using the Stewart Early Transcendentals
textbook with an excellent supplementary materials. Even
though they've got some good instruction already, I'm looking
for suggestions of calculus problems that really make a
learner THINK about calculus concepts and go beyond the
usual school homework exercise.
Browse and search the AP-Calculus forum
http://mathforum.org/kb/forum.jspa?forumID=63
[/quote:5cadb6cd62]
I see your P.S. noting that you saw my posting there. (I posted there first, but
you saw it later because of asynchronous posting to a moderated email list. I
thought it would be a good idea to post that question there, Dave, because I had
seen your name in the archives [smile].)
[quote:5cadb6cd62]Caltech Calculus Lecture Notes
http://www.math.caltech.edu/classes/ma1a/index.html#lect
http://www.math.caltech.edu/classes/ma_9/index1.html#lect
[/quote:5cadb6cd62]
Very good. Thank you.
[quote:5cadb6cd62]Indiana University-Purdue University Fort Wayne
Honors Calculus web page (maybe some ideas here)
http://www.ipfw.edu/math/h-calculus/
MIT's Fall 2002 Honors Calculus Lecture Notes
http://tinyurl.com/ranuv
Here's an example that I posted in the AP-Calculus
forum on 15 March 2006 at
http://mathforum.org/kb/message.jspa?messageID=4542260
Let [[x]] denote the greatest integer less than
or equal to x, and define f(x) by
f(x) = x * [[-(x^2)]].
Computing the derivative of f'(0) leads us to
compute the limit as h --> 0 of the expression
[ f(0 + h) - f(0) / h ],
which, for h not equal to 0, can be rewritten
as [[-(h^2)]].
Note that this is defined for h=0, as we were able
to cancel the h in the denominator. However, plugging
in h=0 gives an incorrect result! Even though we
can plug in h=0, _we're_supposed_ to take the limit
as h --> 0, and this gives -1. Thus, f'(0) = -1.
The neat thing about this example is that if you
simplify sufficiently so that you can plug in h=0,
and then plug in h=0, you don't get the correct
result.
This example is given in:
David Sanders, "A cautionary counter-example",
Mathematical Gazette 59 #407 (March 1975), 44-45.
[/quote:5cadb6cd62]
That's an interesting problem. Thanks for the reference.
--
Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345
Learn in Freedom (TM) http://learninfreedom.org/
remove ".de" to email |
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