Re: Help w/calculus problem

The World Wide Wade wrote:
In article
<1169011085.937758.276920@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Russell" <russell@xxxxxxxx> wrote:

G Patel wrote:
Dean_Travers wrote:
Hello--

I'm currently studying arc length and have reached a problem at the end
of the chapter's exercises which has me stumped; I've tried a couple of
tutors and haven't had any luck. The problem is as follows:

Let y = f(x) be a smooth curve and suppose f'(x) is greater than or
equal to 0 on the closed interval [a, b]. Prove: there are numbers m
and M such that m is less than or equal to f'(x) is less than or equal
to M for all x in [a, b].

If anyone here would be kind enough to offer their insights, advice,
guidance--it would be greatly appreciated.



If y = f(x) is "smooth" that means it's continuously differentiable,
i.e., f '(x) is continuous.

And, it would appear that this condition is *required* for the
proposition to be true in general.

At least, I *think* I can come up with a counterexample, a
function that is differentiable and nondecreasing on [a,b]
yet whose derivative is unbounded there. My thought was
to take an infinite sum of smoothed almost-step-functions,
i.e. smoothing each step with a differentiable sigmoid over
some interval. The sigmoids would decrease both in width
and height as you approach b, so that their maximum slope
would blow up near b without f(b) itself blowing up. If care is
taken in the definition, f' could still be zero at b, was my
thought. Of course the derivative would not be continuous
at b.

This is probably spelled out in detail in some book that
I haven't read, of which there are many.

One way to do this: Consider triangles T_n of height n over the
points 1-1/n, making sure the bases of the T_n's are very small
and disjoint. We can do this so that the area of T_n < 1/2^n for
each n. Call the resultant sawtooth graph g. Note g is continuous
and nonnegative on [0,1). Set f(x) = int_[0,x] g(t) dt for x in
[0,1). Clearly f has a limit from the left at 1, namely the sum
of areas of the T_n's; let f(1) denote this value. Then f'(x) =
g(x) for all x in [0,1) by FTC, and a little work shows f'(1) =
0. So f'(x) exists and is nonnegative for all x in [0,1], and
f'(1-1/n) = g(1-1/n) = n for all n, so f' is unbounded.

Thanks! That's exactly why I posted, hoping that a
real mathematician would step up and say in a less
handwavy way what (I think) I just said. Damn, never
occurred to me that integration is my friend here.

.

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