Re: Problem

["John S. Denker" <jsd@MONMOUTH.COM>]

At 05:47 PM 9/18/01 -0400, Robert Cohen wrote:
> Suppose we made measurements of speed at some time increment and then
> figured out d(speed)/dt numerically at the point where speed=0.  What would
> we get?  Would it depend on the order we used (i.e., second-order vs.
> n-order)?

That's an excellent question.

The answer is no, there is no Nth-order approximation would do the job.

The sqrt() function just doesn't have a Taylor series expanded about the
origin.  The zeroth-order term is OK, but the first-order term doesn't
exist, the second-order term doesn't exist either, and ....  Do the math if
you don't believe me.  It's easy.

(Remember, this is relevant because speed in particular and vector norms in
general involve a sqrt in their definition.)

So here's how to look at the problem that provoked this thread:  In the
trajectory where there is no horizontal velocity, the velocity is a nice,
smoothly varying, differentiable function of time.  However, at the point
where speed=0, the _speed_ is not a nicely behaved function of time.  This
has absolutely no importance, because the laws of physics don't care about
the speed.  They care about the velocity and the acceleration.  In D=1,
(d/dt)speed is the same as acceleration EXCEPT at this peculiar point.  At
this peculiar point you cannot infer the acceleration from (d/dt)speed.

Don't be a worry-wart:
        The New York Yankees haven't scored any touchdowns!  Oh my!
     I think there are triacylglycerides in my butter dish!  Oh my!
        (d/dt)speed is not well behaved!
Yeah?  So?  Who cares?