Re: Sloppy integration method leads to confusion


"Chris" <a@xxxxx> wrote in message
news:dkkp1d$dbs$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Ok. sorry for the long quotes but they seem to be because i seem to be
> failing to express myself properly.
>
> I consider my understanding of integration to be sound. Im confident in
> integrating standard integrals and more tricky ones in sqrts in the
> denominator (although these all seem to be given to you now) and ones
> where you substitute things like x = a sin(theta) or sinh(theta) etc. I
> dont think I have got my understanding wrong and if I have made it sound
> like I have thats wrong. Its just know up to know Ive been thinking of
> integrating both sides wrt the same variable and getting the answer with
> indivisible dy/dxs. There hasnt been any need to bust up dy/dx. But now I
> encounter a problem where the sheet gives d[dy/dx] = const dx and they go
> to d[dy/dx]/dx = const which is d^2y/dx^2 = const.

This just mean the 2nd derivative wrt x, that's all. In the former
notation, they are using the d[f(x)]/dx notation where f(x) just happens to
be the first derivative, dy/dx, of some function. The second notation,
d^2y/dx^2 does not make as much intuitive sense (sort of like sin^2(x)
doesn't make much "sense" for the square of sin(x)) but it is a concise and
widely used notation, so better get used to it.

> The thing about learning why this works would be nice and hence my wanting
> to learn this, but I fear from other replies, one from Darell, is that it
> is beyond the scope of what you should expect to learn.

You're blowing this a little out of proportion, I'm afraid. I apologize if
I have contributed to your fear. If its any consolation, know that some of
the results encountered in a standard calculus I course have _proofs_ which
utilize means beyond the scope of the course. For that matter, some of the
results obtained in a beginning algebra course have proofs which are well
beyond the scope of the course.

Right now, if you're in a standard calc series and are studying indefinite
integration, in all probability it is simply a matter of _definition_ that:

dy/dx=f(x) -> dy=f(x)dx -> y=int[f(x)dx]+C

I doubt your text gives any _rigorous_ reasons why you can move dx around
like that.

> (I am not a mathematician so Im expected to just do integrations by
> learning the method rather than the understanding, whereas I try to the
> best of my ability to understand why things are ok but sometimes you just
> have to accept things work and how to use them.)

I believe this is indeed one of those times, though I certainly admire your
desire for further understanding.

> Since I havent been taught about moving bits of dxs, dys around on their
> own this lead to the question, but I fear this is something which requires
> me to see a proper tutor.
>
> (I do concede that what i said about the chain rule doesnt inspire
> confidence.<...>

The chain rule has two recognizable forms:

dy/dx = dy/du * du/dx

In this form it appears the du's simply "cancel." Not so. This goes back
to my previous post. This particular notation is algebraically convenient,
but _not_ because the du's cancel, but rather because this is equivalent to
the other form of the chain rule:

d/dx[f(g(x))] = f'(g(x))g'(x)

So, _in practice_ it appears the du's cancel, so don't feel too bad about
drawing slashed through them across the fraction bar. Sometimes fluency
must be achieved at the cost of something else.

--
Darrell



.

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