Re: what "REALLY" is derivative?

"Luke Wu" <LookSkywalker@xxxxxxxxx> wrote in message
news:1134507920.370743.43740@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Hi,
>
> Back in grade 11 when learning about parabolas, a teacher told me that
> there are lines that can touch the parabola only at one point, for
> every point on the curve (tangents), and that we needed to wait for
> more advanced courses to find the equation of these lines.
>
> I thought this was nonsense, because if two curves (parabola and line)
> intersect, then it's easy to solve for the slope of the line (because
> we knew how to solve the intersection of two lines- "this can't be much
> different" i thougth).. this is what I did:

Hey - good for you! If only more students were as keen as this :)

Below I'll try and clarify exactly what you're doing and why it's working.
Maybe this will help something click...

>
> equation of parabola: y = ax^2 + bx + c
>
> point slope equation of line: y - y1 = m(x - x1)
>
> what I did was substitute the point (q , aq^2 + bq + c) for (x1,y1) in
> the slope equation
>
>
> y - aq^2 - bq - c = m(x - q)

OK - this is the equation of a straight line passing through your chosen
point on the parabola. As m isn't constrained yet, this straight line could
represent a tangent line or a chord, depending on where (how many separate
points) it intersects the parabola...

>
> then substitute the y = aq^2 + bq + c

If you really did this you would just get straight back to x=q, which is no
surprise!

What you actually do is you substitute y = ax^2 + bx + c. You are doing
this to find points (x,y) which are on both the parabola, and on your
straight line. Hopefully for a special value of m we will see there's only
ONE intersection, because the line is a tangent line. In general though we
would expect the straight line to be a chord and intersect the parabola in
two places, right?

>
> ax^2 + bx + c - aq^2 - bq - c = m(x - q)
>
> a(x^2 - q^2) + b(x - q)= m(x - q)
>
> (x-q) [ a(x+q) + b ] = m(x - q)

OK, this equation gives the x-values for the intersection of your straight
line and the parabola, right? This is a quadratic equation in x, as we
would expect since in general there are two such points of intersection...

Of course, we know that x=q is always one such value, due to the way you set
up the equations for the line and parabola. Since all these equation are
*polynomials*, as you note below, we would expect (x-q) to be a factor of
the polynomial, i.e. it "divides out" of your equation.

>
> THEN I DID THE UNTHINKABLE and divided both sides by (x-q)

OK - if we want to find the SECOND point of intersection of your line and
parabola, then we look for the other solution to the quadratic above. You
have spotted that (x-q) is a factor - the remaining factor will give the
remaining solutions to the equation (that's how polynomials work). So the
other solution for x will satisfy...

>
> m = a(x+q) + b and that is the slope of a line touching parabola at x
> = q

.... and here you've jumped a bit! What you've actually got here is the
x-value for the *second* point of intersection of your original line and
parabola. Now you say that in fact you only want there to be one point
(corresponding to x=q), and this only happens when m = 2aq + b.

Another way of viewing this is that the line and parabola are intersecting
in two coincident points - i.e. the quadratic equation you are solving has
(x-c)(x-c) as a factor and so has a double root.

You could also think of your method like this: m = a(x+q) + b, is giving
you the slope of the chord of the line passing through the parabola at
points (q, ???) and (x, ???). As x gets closer and closer to q, the chord
gets closer and closer to the tangent line, and m gets closer and closer to
2aq+b. I.e. the gradient appears from this as a natural limit.

It so happens that function you are taking the limit of is continuous at
x=q, and so you can easily see the limit just by setting x=q. !! None the
less, the process you've gone through is essentially that of looking at
CHORDS connecting points on the parabola, and then identifying the special
case where the chord becomes a tangent point (either through looking at
limits, or by looking for double root solutions to your equations.

I hope this helps you see the relationship between what you've done
yourself, and the limiting process now being used to define derivatives in
your course.

>
> I had no clue what a limit was, or a derivative, but I used the ideas
> learned about the intersection of two lines, and used it for the
> intersection of a parbola and a line (and just ASSUMED that x-q != 0)

Well hopefully I've expressed above more clearly what you're doing? (In
"assuming" x != q, you are looking for the SECOND point of intersection of
your line and parabola, and then you backtrack a bit to get the condition
that in fact the second point of intersection is coincident with the first
(known) point. This is the condition for a tangent line which just
intesects the parabola once.)

>
> I then tried this for other curves successfully, and when I started
> learning calculus (2 years later), I couldn't get past the fact that
> the limit seemed to me (for reasons shown above) as an excuse to allow
> us to divide f(x) - f(a) by x - a.

The reason this works so well for the functions you try is that you are
focusing on polynomials, and as you realise, f(x) - f(a) will always have
(x - a) as an explicit factor, as x=a is a solution of f(x) - f(a) = 0.

>
> How is there always a x-a (for us to cancel out) within f(x) - f(a) if
> f is differentiable at 'a'. Does anyone know how this really works?

Well, for polynomials it works for the reason explained above.

More generally, if a function is differentiable at a point, what this
amounts to is that the function can be thought of as "approximately linear"
at that point. What I mean by this is that if you looked at the graph of
the function under a strong enough microscope it would look like a straight
line. Obviously it would not really be a straight line, but the idea is
that it for x close to a it should look like:

y = f(a) + m(x-a) + "other bits of smaller order than (x-a)"

where m is the gradient we are interested in at the point (a, f(a)). You
see that this is saying the function is like the "tangent line" plus a
residual deviation bit.

Whenever the function f is like this, we can rearrange the above to get:

(y - f(a)) / (x-a) = m + "other bits..."/(x-a)

and the second term on the right becomes arbitrary small as x gets close to
a, which is why the limit definition of the derivative works...

So the division you like doing will in general give you a function of x
which is continuous at x=a, and as x-->a it will tend to m, which is the
derivative of f at x=a. (This is exactly the limit definition of the
derivitive.) For certain functions (e.g. polynomials) the form of this
quotient will be an expression you can get hold of and easily see it's value
at x=a, but really the limit is the way to approach this for more general
functions.

I don't really know if all this will help you or just make things more
confusing! Anyway I tried :)

Regards,
Mike.

> Have any of you wondered how solving limits (be in with e/d or using
> limit laws/continuity without e/d) always works out perfect (we always
> get a factor x-a or h that cancels out or we always get the magical
> absolute value term with e-d limit proofs)? I know if f(x) - f(a) is a
> polynomial, it is zero when x = a, so there must be a factor x-a (by
> factor theorem) ... but really.. how "really" does this work?
>
> My post sounds unintelligible because I can't shape my
> confusions/thoughts properly to explain it clearly (and English is not
> my native tongue). My "funky" grade 11 self discovered method is
> clashing with the limit of difference quotient definition of a
> derivative. My extreme inquisitiveness (yes extreme, which is why I
> always do bad in school, because I can't get over the "WHYYYYY?" even
> if I"m told the WHYYY?? is coming later on).
>
> Can anyone shed some light on my confusion? What "really" is a
> derivative?
>
> Any insights or guidance would be greatly appreciated.
>


.

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