Math 291 diary, fall 2006
In reverse order: the most recent material is first.

More volunteers
These students have volunteered (well, they are in 291, darn it!) to answer questions next time about the 251 assigned homework problems in these sections:

Textbook sectionNames
12.4Thomas Ehrlich
Evan Fitzgerald
12.5Jacelyn Gerges
Andrew Harrison
13.1, 13.2Pawan Harvu
David Hsiung
13.3Gibson Kim
Kevin Kobilinski
13.4Cosmo Kwok

Monday, September 25

Brief repetition of PLANE curves

Motion in a straight line
The bug doesn't feel the sides.

Motion in a circle
The moving bug always feels a transverse force.

Differentiation and three kinds of multiplication for vector-valued functions
Scalar multiplication
Dot product
Cross product

An ongoing example: the right circular helix

Space curves: the beginning
Tangent vector, unit tangent vector, speed

Trick #1: defining the principal normal

The definition of curvature

The binormal

The Frenet frame

How the binormal changes

The definition of torsion
Bending out of a plane

The derivative of the normal

The Frenet-Serret formulas

Check everything with the helix!

Wednesday, September 20

The dot product of a vector with a cross product

The volume of a parellelopiped

The vector triple product equals a determinant equals a volume

Curves in space and in the plane

A differentiable curve has differentiable coordinates

The derivative is velocity

Length is speed

Direction is tangent to the curve

Tangent vector and tangent line

Distance=rate·time as an approximation

Definition of distance traveled

And the length of a curve

And the intractibility of computation

How a plane curve bends

What's more; what's less

The bewildering effect of speed on "perceived" bending

Test cases
circles, line, parabola

The rate of change of angle

The rate of change of angle with respect to arc length

The formula defining curvature

Monday, September 18

Recursive definition of determinant

Too big! There must be other ways ...

More vector algebra

Is a point on a plane determined by three given points?

Description of a plane with a given point and a given normal vector

Description of a line through a given point in a given direction

Distance from a point to a plane

Other distances

Thursday, September 14

Word of the day
remiss careless of duty; lax, negligent Homework problems done, quite well, by terrific students!

Geometry, chirality, and prejudice


Cross product

A multiplication table

A formula for the cross product

Deception? Lie? Well, only inferentially ....

Wednesday, September 13

Word of the day
ineluctable Against which it is useless to struggle. "Applications"
I stated the Cauchy-Schwarz inequality, the triangle inequality, and a sort of reverse triangle inequality.

Me, dog, lamp post

More generally

Now with angles (a robot arm?)


Resolving into parallel and perpendicular components

These students have volunteered (well, they are in 291, darn it!) to answer questions next time about the 251 assigned homework problems in these sections:

Textbook sectionNames
12.1Valentyn Boginskey
Joshua Bryan
12.2Albert Chen
Wei Chen
12.3Alexander Crowell
Christophe Doe

I thank them in advance.

Monday, September 11

Abstract math

Vector space

{Inner|scalr|dot} product


A function

A quadratic function

The Cauchy-Schwarz inequality

The triangle inequality

Examples (at last!)
From quantum mechanics.

Vectors are directed line segments. Vectors are forces. Why?

Components of vectors

Dot product

Law of cosines
(messed up in class!)

Wednesday, September 6

The instructor tried to waste time. He made bravura (arrogant?) capsule summaries of calc 1 and calc 2, and then tried to indicate how things might be different in Rn for n>1.

Word of the day
bravura Showy; ostentatious.

Calc 1

Calc 2
Lots and lots of detailed schemes for antidifferentiation. Wow!
More intricacies about sequences and series. More wow!!
This stuff is all worthwhile, but everyone would agree that the highlights of the course are:

Several variable calculus
Several here means "more than one".

Distance in R1
We looked at distance in the real line. The distance between a and b is defined by
This is a great definition. Notice that:

  1. dist(a,b) is always non-negative, and is equal to 0 exactly when a=b.
  2. dist(a,b)=dist(b,a): this is called symmetry.
Then we had a discussion of how dist(a,b) and dist(b,c) and dist(b,c) should be related. We considered some examples. If we want an equality relating these distances, then the result sort of depends on whether one number is between the other two. But if we want a result that is always true, then we need an inequality. Inequalities are good. Most data in the real world is an approximation, accurate measurements are very difficult ... inequalities better describe what really is.
  1. For any points a and b and c in R1, dist(a,b)+dist(b,c)≥dist(a,c).
Those three "rules" for distance seem nice.

Distance in R2
Well, I went through the usual diagram to try to motivate the algebraic definition of distance in R2. Look, please, at the diagram to the right. In the plane, points correspond to ordered pairs of numbers. So a point p might correspond to an ordered pair, (x1,y1), and q might correspond to (x2,y2). Then the point (x1,y2) is the vertex of a right triangle whose hypotenuse is a line segment connecting p and q. One leg of the right triangle is on a line where all the first coordinates are x1, and the length of that leg is given by the one dimensional formula, |y1-y2|. The other leg is on the line where all the second coordinates are y2, and the length of that leg is |x1-x2|. Then by Pythagoras, the hypotenuse has length sqrt(|x1-x2|2+|y1-y2|2). And usually the absolute values are discarded since we are squaring the quantities. Therefore we officially define:
dist(p,q)=sqrt((x1-x2)2+(y1-y2)2) if p has coordinates (x1,y1) and q has coordinates (x2,y2).

Are rules 1 and 2 and 3 above still valid? Well, 1 is true because, first, the square root is always non-negative. And if the square root defining dist is equal to zero, then since the terms inside it are also non-negative, the only way the square root can be zero is if both x1-x2=0 and y1-y2=0. The coordinates must be the same, so p and q are the same points.

Rule 2, symmetry, is also true and equally simple to verify, since squares of W and -W are the same. But rule 3 is a different sort of thing.

Rule 3 for distance in R2
Well, if p has coordinates (x1,y1) and q has coordinates (x2,y2) and r has coordinates (x3,y3), then I want to know if dist(p,q)+dist(q,r) is always ≥ dist(p,r).

The geometric point of view
This is what I personally prefer, and this is what a student in the class saw almost immediately. The length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides. And that is exactly what Rule 3 states. So rule 3 is clearly true!
The algebraic point of view
Maybe other people don't necessarily accept the picture, and don't "see" the result. Let us try to verify algebraically that
sqrt((x1-x2)2+(y1-y2)2)+sqrt((x2-x3)2+(y2-y3)2) if p≥sqrt((x1-x3)2+(y1-y3)2)>
I don't think this is totally, immediately "clear". In fact, this inequality, if you want to worry, is a statement about six variables. Six! I asked if there was any way to transform it into another algebraic statement whose correctness we would accept immediately. We began working. We squared, we rearranged ... it was a mess. Then I suggested that maybe we should try to rewrite things a bit:
Change x1-x2 to A     Change y1-y2 to B     Change x2-x3 to C     Change x2-x3 to A    
Then x1-x3 will become A+C and y1-y3 will become B+D.

The inequality we would like to verify now is:
At least we have only four variables now, rather than six. Now square: A2+B2+2sqrt(A2+B2)sqrt(C2+D2)+C2+D2≥(A+C)2+(B+D)2
This step is reversible because we know that the inequality we began with has only non-negative terms, and square roots and squaring will preserve such inequalities. Now "expand":
(A+C)2+(B+D)2 becomes A2+2AC+C2+B2+2BD+D2.

We want to verify that:
"Cancel" (subtract) everything that's equal on both sides. The result is:

Divide by 2:

And let's square again (amazing!):
But (AC+BD)2=(AC)2+2(AC)(BD)+(BD)2
and (A2+B2)(C2+D2)=A2C2+A2D2+B2C2+B2D2
Therefore the inequality we want is:
Let's cancel stuff again. The result is:

Well, uhhhh ... this is:
and this is the same as:
(AC-BD)2≥0 and this is true because squares are non-negative. Sigh.

What is going on?
Any student who followed this is either crazy or brilliant or both. We have verified that the triangle inequality (rule 3 for distance) is true for the distance formula in R2. These inequalities are all quite famous, actually. I just took a rather "low tech" way to verify them. The inequality
which we also verified in passing has its own name. It is called the Cauchy-Schwarz inequality. It is important.

The approach I just used is what could have been done maybe two or three hundred years ago. I will try to show you what the heck is going on in a more modern fashion Monday.

Read the text!
Please read the first few sections of chapter 12, as on the Math 251 syllabus.

Maintained by and last modified 9/7/2006.