Re: Calculus doesn't apply to the unchanging curve

On Jun 22, 3:37 pm, Virgil <Vir...@xxxxxxxxx> wrote:
In article <idpeSFY=Q35UrgrSyrEP4IuPF...@xxxxxxx>,
 "[Mr.] Lynn Kurtz" <ku...@xxxxxxxxxxxxxxx> wrote:





On Sat, 21 Jun 2008 22:00:28 -0600, Virgil <Vir...@xxxxxxxxx> wrote:

In article <br9dSLC+YR0DzY56iipxnxmgR...@xxxxxxx>,
"[Mr.] Lynn Kurtz" <ku...@xxxxxxxxxxxxxxx> wrote:

On Sat, 21 Jun 2008 17:04:08 -0700 (PDT), BURT <macromi...@xxxxxxxxx>
wrote:

On Jun 21, 3:58 pm, Virgil <Vir...@xxxxxxxxx> wrote:

There are changing slopes when changing tangent lines to a circle, and
while such lines have slopes, circles don't.- Hide quoted text -

- Show quoted text -

Thankyou Virgil. Well put.

Mitch Raemsch

Not! The slope of a curve at a point a is *defined* as f'(a),
presuming it exists. Circles and other curves have slopes except at
points where the tangent line is vertical.

--Lynn

In my day, curves could have derivatives at some points, but they were  
slopes of the appropriate tangent lines , not slopes of the curves
themselves.

When did that change?

I don't think it was ever that way. You can't define the notion of
tangent line without using the notion of the limit of slopes of secant
lines i.e, the derivative. In other words, the tangent line through
(a,f(a)) is the line through that point with slope f'(a). How else
would you define the notion of "tangent". The derivative of the curve
*defines* the slope of the tangent line, not the other way round.

As an aside, the fact that Mitch seems to agree with you should give
you pause...

--Lynn

You are missing at least a part of my point, which is that it is lines,
not curves, for which slopes were originally defined.- Hide quoted text -

- Show quoted text -

A zero dimensional point cannot have a changing rise and run. It does
not have a "change" in height to differentiate. That is proof that
there can be no derivative.

Mitch Raemsch
.

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