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 section | Names |
---|---|
12.4 | Thomas Ehrlich Evan Fitzgerald |
12.5 | Jacelyn Gerges Andrew Harrison |
13.1, 13.2 | Pawan Harvu David Hsiung |
13.3 | Gibson Kim Kevin Kobilinski |
13.4 | Cosmo Kwok |
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!
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
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
Geometry, chirality, and prejudice
Cross product
A multiplication table
A formula for the cross product
Deception? Lie? Well, only inferentially ....
Me, dog, lamp post
More generally
Now with angles (a robot arm?)
Orthogonality
Resolving into parallel and perpendicular components
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 section | Names |
---|---|
12.1 | Valentyn Boginskey Joshua Bryan |
12.2 | Albert Chen Wei Chen |
12.3 | Alexander Crowell Christophe Doe |
I thank them in advance.
Vector space
{Inner|scalr|dot} product
{Length|norm|magnitude}
A function
A quadratic function
The triangle inequality
Examples (at last!)
From quantum mechanics.
R3
Vectors are directed line segments. Vectors are forces.
Why?
Components of vectors
Dot product
Law of cosines
(messed up in class!)
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
dist(a,b)=|a-b|
This is a great definition. Notice that:
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:
sqrt(A2+B2)+sqrt(C2+D2)≥sqrt((A+C)2+(B+D)2).
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:
A2+B2+2sqrt(A2+B2)sqrt(C2+D2)+C2+D2≥A2+2AC+C2+B2+2BD+D2
"Cancel" (subtract) everything that's equal on both sides. The result
is:
2sqrt(A2+B2)sqrt(C2+D2)≥2AC+2BD
Divide by 2:
sqrt(A2+B2)sqrt(C2+D2)≥AC+BD
And let's square again (amazing!):
(A2+B2)(C2+D2)≥(AC+BD)2
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:
A2C2+A2D2+B2C2+B2D2≥(AC)2+2(AC)(BD)+(BD)2
Let's cancel stuff again. The result is:
A2D2+B2D2≥2(AC)(BD)
Well, uhhhh ... this is:
A2D2-2(AC)(BD)+B2D2≥0
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
sqrt(A2+B2)sqrt(C2+D2)≥AC+BD
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 greenfie@math.rutgers.edu and last modified 9/7/2006.