Your Concern That It Is Awful

 to be me is misplaced.

Nine days ago, in the evening, and after throwing off the bear shoulders, I grew a celesta. My lungs were glass bells. Washboard, guitar string hips.  Accordion bellow arms.  I waltzed the orange light as she slipped upward through  my hands, opening to gray-blue. Alone with the cool-waisted dusk, I ran wood wrists against my grooved, silver legs.  I sang.

But you were not worried about the dreams.  It is the machine-work you do not like. It is plain on your face.  "Strange."  Say it again.

"I would hate to think { } caused you to think this way."
It has not.  My mind is what I have made.  There are two persons in your frown, one broken and one fixed, and I am neither.   I know it seems to you I am unwell. And if you were wrong?  How would you check?  I do not say to you, "my wood wrists."  I said nothing about the Snow Queen.  I said, that I reason. That it is not difficult, it is not intelligence.  The times I am misunderstood are oftenest when I use words in an ordinary way.  I do not expect people to think "on the same level."  I expect them to mean what they say, and when they choose to declare truth, when the words are "therefore" and "because", I expect those words to hold.  I am not the one who has chosen them.  This is strange to you when I say it, that I choose to reason.  This conversation is no longer about me.

You write it down,
"... the idea that what I understand to be thought; what it means to me to be alive and possess a mind, is somehow fundamentally wrong or broken..."  You like this sentence.

"Are you sure you've always thought this way?"

"Yes.  In fact, I think this way so completely, that I did not even realize--"

"I mean, even when you were younger?  When you were 10?"

Oh.  I see. The window has shut.
"Yes."

This question is irrelevant; I am not broken.  And I am not a child.  I have had a lifetime to consider; to choose what I will think and how.  Those things about my mind, body and heart which are beautiful, of which I am the most proud, did not exist when I was 10.  I built them.  It is the one thing  I have done which is the most beautiful.

Vector surfaces

I haven't forgotten about the schoolgirl problem or Snell's Law.  I (finally) have a  correct general solution for Kirikman's schoolgirls: a puzzle consisting of simple blocks with which any of the particular solutions can be built.

I intend to finish Applied Vector Analysis in a month.  8 Chapters remain; most of the math I need is here.  

• DifferentiationSpace curves
  Surfaces
  Tangent, curvature, torsion

• Divergence, Gradient, Curl

• Integration
  Line, surface and volume integrals
  Gauss' (Divergence) Theorem
  Green's Theorem
  Volume-to-surface; surface-to-line transformations
  Irrotational Fields
• Orthogonal Curvilinear Coordinates
  Cripple Special coordinate systems

• Fluid Mechanics:
  Equations of continuity, motion, and energy
  Fluid States
  Steady flow and streamlines
  Vortex flow and circulation

• Electromagnetic Theory:
  Electromagnetic field
  Maxwell's equations
  Potential
  Energy and the Poynting vector
  Static fields

Rad.
I'll take notes here.

PARAMETRIC SURFACES:  NORMAL VECTOR

 The normal N to any surface represented by a vector function r(x,y) is given by

     
where
     


I. Explicit function of x, y
Let a surface S be given by the equations
     
where x, y, z are the coordinate axes.  Then the position vector of any point on the surface is given by the vector function
     
Compute the partial derivatives,
     

 Compute the cross product:
     

And the magnitude, for normalization:
     

The normal vector to any point on the surface is then
     


II. Implicit function of x, y.
I'm still not comfortable with partial derivatives of implicit functions.  I lose the intuition quickly, and return to the definitions often.
Fix it, Ryan.
Okay.  The following result makes me happy.

"

Problem 4-11  Find a unit normal vector n to a surface S that is represented by                                                

"

Go one step at a time.  This is a surface.  Any surface can be represented by a (series of) function(s) of two variables.  So, the given function has two independent variables.  Phi implicitly defines a function of two variables.  By convention, I assume the independent variables are listed first, and are therefore x and y.

Phi defines z implicitly as a function of x and y.


Differentiate with respect to x:
     

Solve for the partial of z with respect to x:
     

Differentiate with respect to y, and solve for the partial of z with respect to y:
     

Now, we have implicitly defined z as a function having the same form as the explicit equation in section I. Substitute the conspicuously Jacobian values (1) and (2) into the equations from I, and we now have the desired partial derivatives:
     

Again, the cross product:
     
 The relationship is clearer, and normalization simplified, if this is rewritten as:
     

Normalize!
     

Goodbye, extra z-partial.

     

Notes:
1. It is not clear to me why the implicit function is differentiated without reference to the unit vectors i, j and k. "Because it's implicit".  Yeah.  I get it.  But rendering z = z(x,y) as a vector function is independent of the function's status as implicit or explicit.  The answer is about measure, and I need to find it.

2.  In the implicit function, it doesn't matter what x, y and z are. One of the three partial derivatives of φ is nonzero, or it does not meet the general definition of a surface.  This nonzero partial corresponds to the dependent variable.  I assume the partial derivative of φ with respect to z is nonzero.  If it turns out to be x or y, the result can be computed in the same fashion.
The convention is for simplicity: list the dependent variables last.  It saves the following step: suppose I intend to solve problem 4-11 and I require the dependent variable in the third position.  Then I will solve the system

instead, where w is the dependent variable.  Map the variables x, y, z in problem 4-11 to u, v, w accordingly as the surface function is known.  I do not have to change a letter of the solution.

Double integration equivalence

Time to clarify my assumptions about integration.

The double integral,

 is shorthand for:
             

We begin at the innermost integral.   Integrate with respect to x, holding  y constant.  Then integrate the result with respect to y (holding x constant).  Each step is an ordinary, single-variable integral.  We may also change the order of integration:
 

Just remember to keep x-bounds with x, and  y-bounds with y.

These are convenient results, and after hammering through partial differentiation, integrating in any order makes intuitive sense.  This is the next chapter of Advanced Calculus, but I think the result is self-evident, and I want to clarify that assumption so I can find any mistakes.

  I have heard equality (2)  referred to as Fubini's Theorem, but the statement before me is, the two integrals are equivalent.  The only Theorem I have any business with must be the Fundamental Theorem of the Calculus.  If I break up (2) into infinite sums, the sum on the right side of the equal sign must be the same as that on the left.  Right?

Variable bounds, paramaterization, partial summation with discrete and continuous quantities are a few pages away.  I  need a working, base example: a continuous function with constant bounds.  Let me build that now with the simplest case: a dobule integral of the form (1), with constant bounds.

A FUNCTION OF A SINGLE VARIABLE

Let F(x) be a function whose derivative, F'(x) = f(x)  is continuous over the closed interval [a, b], i.e.
              
Then the Fundamental Theorem of Calculus states


PROOF:  This approach is abstract, but concise.  By continuity, we can write the integral as the limit of a sum


Where δ is the largest subinterval and ξ_k lies in the subinterval [x_k - x_k-1].
Now, we choose for our  ξ_k the points in each subinterval where the line tangent to F(x)  has the same slope as the segment joining the endpoints, F(x_k ) and  F(x_k-1):

             (Rise/Run) (Run)             =   (Rise)

Then the right hand side of (4) becomes,


Rearranging the terms,  we have

    _____________________________________________________


And the theorem is proved.  Let me try to extend this method myself.

A FUNCTION OF TWO INDEPENDENT VARIABLES

Beginning with equation (1):
     
First, consider the inner integral:
     
What relationship does f(x,y)dx have to F(x)?  It is the partial derivative F₂₁(x):
     

The result of the first integration will then be  F₂(x).  This is my principal assumption. Specifically,
     
I'm not sure how I feel about that.  The assumption is sufficient to make the sums work out.  And to use the Fundamental Theorem, we must already know the relationship between f and F.  I don't see a way around equality (8).   But I already know that F₂₁(x, y,) and F₁₂(x, y) are identical for 'nice' functions, and the FTC is just going to give that information back to me.

That's why  I need to work out the concrete terms, determine if it holds, and if so, how.  Otherwise I'm thinking in a circle.  I will use the notation of partial derivatives, but I did not begin with this assumption.  I tried several ways of indexing the sums, and in the end found that "indexing" was the same thing as labeling partial derivatives.  The FTC makes this process clear.

I.  Integrate with respect to x
     

Choose the ξ_k so that
     
This is a directional derivative.  There is no slope in the y direction because I'm not varying y. Hence the equality.  This is the crux.  Not-varying one of two independent variables is exactly the same thing as taking a partial derivative. The x intervals sum and cancel to a discrete value, independently of the y intervals.  Multiplication and addition are commutative and associative, so the sums and products of these values can be arranged in any order.  Taking the integrals separately is also exactly the same thing as saying the integration can be done in any order. x and y never mix in the infinite sums.  That's not a theorem; that's what we're doing.  This is what I mean about Fubini's theorem, whatever the hell it is.  Integrating the independent variables separately is how we insure there is only one answer.  Onward!

Now we have


Rearranging the terms,

   __________________________________________________________________



II.  Integrate with respect to  y
I will integrate the general form:
     
For the purposes of integration, this is the same as
     
and can be applied to (7)'.   The process is now the same as above.
     
Choose the ζ_k  so that
     

That's still a derivative on the left! (5)'' can alternately read,
                 
Thus we have


And the last line can again be rearranged so that all but two terms cancel:

   ______________________________________________________________



III. Equivalence
Now compute the result:


... with the happy result that giving one order of integration entails the equivalence of the other.

HOW'S MY DRIVING?

   I have the feeling double integration is applicable in a variety of circumstances, leaving this approach insufficient for many cases.  Or maybe I made mistakes, and it's all wrong.  Somebody is letting a 12-year old drive a semi, and both of those guys are me.  Tell me how my driving is.  Leave me a comment.  Correct a mistake.  Prattle on about Space-Wogs.

Fuckin' space wogs.  Always framping the scrumpets.

{Thanks again to CodeCogs for free LaTeX equation editing and hosting the resulting, tiny bitches images.}