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Re: [Phys-L] solve equation w/o calculus



On 12/19/2016 02:19 PM, Carl Mungan wrote:

here is a simpler [solution] that someone else came up with.

https://www.usna.edu/Users/physics/mungan/_files/documents/Scholarship/DescentRampTrack.pdf

The key passage in the .pdf document is:

3x^2 - 2yx + (4 - y^2) = 0 (4)

x = (y ± 2*sqrt(y^2 - 3)) / 3 (5)

But as Y decreases in value, these two values of X converge toward
one common value that must evidently give the minimum

I don't want to get a reputation as a tough grader, but I would not
give full credit for that analysis. Generally, I interpret words like
«evidently» or «obviously» as red flags, marking flaws in the argument.

Normally, a quadratic equation has /constant/ coefficients, in which
case the place where the determinant goes to zero must indeed be the
minimum (or maximum).

We could try to argue by analogy in equation (4), but the analogy does
not automatically hold. The problem is, y is a function of x ... and
the coefficients are functions of x, too!

With 20/20 hindsight we know it doesn't matter /in this case/. However,
it does matter in general, and it's not particularly «evident» even in
this case, unless you assume what you're trying to prove ... or unless
you carry out the missing steps.

If you want full credit, I need to see something like this:

Replace x --> √3 + dx y --> 1/√3 + dy (A)
so
3 dx = dy + 2 √(dy^2 + 2√3 dy) (B)

Check the work by verifying that (dx,dy) = (0,0) is a solution.

Consider what happens to the RHS when √dy is very small. Throw away
terms that are small squared, such as dy. Then to leading order, dx
is proportional to √dy, as advertised.

Note that if equation (4) had taken a slightly different form, such as

3x^2 - 2x√y + (4 - y^2) = 0 (4')

then the place where the discriminant goes to zero would not be the
extremum. So the extra work of looking at the lowest-order terms
is absolutely necessary. You don't know what terms can be thrown
away unless you do the work.

I would tell students that expanding things to lowest order is valuable
in a tremendous variety of situations. Often a complicated situation
can't be understood exactly. Being able to understand it to first order
is a big improvement over not understanding it at all. Furthermore,
sometimes the lowest order expansion tells you everything you need to
know (as in this case).

Lowest-order expansions are a step up from the more primitive (but still
valuable) notion of estimating things to order of magnitude.

Substituting x --> (something) + dx is the typical first step. It's like
pawn to king four. Pretty soon you can get that far blindfolded.

I would also tell students that you can do lots of first-order expansions
with just algebra, but this gives you a reason to sign up for the calculus
class, because calculus gives you systematic, convenient ways of finding
Nth-order approximations to lots of things that you wouldn't have a handle
on otherwise.

Newton invented calculus for a /reason/. He invented it because he wanted
to do physics with it.