o.k. I've taken to "amusing" myself lately by attempting to solve little geometrical problems I've set for myself. On the whole this has been successful, as measured both by the satisfactory completion of said problems (i.e. solving them), and by the "amusement" level associated with the process (e.g., not thinking about my foot, debts, work...).

HOWEVER, I've run into a stumbling block. I've come up with a solution to a problem which just doesn't seem like it could possibly be right. $20 to the first person who can show me where I went wrong. I'll lay out the problem, and then some basic things we'll need to agree on:

I thought it would be fun to see if, given some x, I could find some other number A, such that the line going between (A, 0) and (x, x^2) would be perpendicular to the function f(x)=x^2. A thousand words:

My thinking went something like this:

Axioms:

1. The slope of f(x) at x is equal to the derivative of f(x) evaluated at x.

2. if f(x)=x^n, f'(x)=nx^(n-1) [the "Power Rule"] 3. The slope of a line between two points is the quotient of the difference between y values and x values: Dy/Dx= (y2-y1)/(x2-x1)

4. If two lines are perpendicular, then their slopes are negative reciprocals of each other.

All of this should be, if not obvious, at least readily verifiable. Still with me? Then here's one argument, with justification for each step in brackets[]:

I. f(x)= x^2, therefore f'(x)=2x [stipulated, power rule]

II. The slope of our little green line, then, should be the negative reciprocal of the slope at x, or -1/2x [stipulated, 4]

III. The slope of our little green line should also equal (0-x^2)/(A-x) [stipulated, 3]

IV. Therefore, -1/2x= (0-x^2)/(A-x) [II., III.]

It's not terribly tricky to solve for A, but, for the sake of being explicit, here's how I did it:

1. -(A-x)/2x = (0-x^2) [mult. (A-x)]

2. -(A-x) = 2x(0-x^2) [mult. 2x]

3. -(A-x) = -2x^3 [dist.]

4. (A-x) = 2x^3 [mult. -1}

5. A = (2x^3) - x [sub. x]

Cool. So now we have a function that, when you put in some x, gives you back an A! Can you imagine the plot of this function? I did, and that was when I noticed that something fishy was going on. A thousand more words:

Problem: If (A, 0) is supposed to be the point where the line perpendicular to f(x) intersects the x axis, why would it ever be negative in the first quadrant? I imagine those lines getting closer and closer to vertical as one approaches the origin (f'(0)=0, and the negative reciprocal of this, while undefined, approaches -infinity, something like -1/0), but they would have to have a positive slope to get from anywhere on f(x) in quadrant one to the negative x axis.

So, there you have it. I can't see how to refute the math, but I can't see how it could be right either. For now there is just a hole where reason should be.

Better luck to you,

Paulman