Clairaut's Theorem (equality of "mixed" partial derivatives)
Suppose f(x,y) is a function of two variables, and the mixed partial
derivatives f_{xy} and f_{yx} both exist and are both
continuous. Then these mixed partial derivatives must be the same.
Certainly in Math 251, the hypotheses of the theorem will be
satisfied. There are examples (similar in nature to the bizarre
functions previously given) where things aren't the same. But in this
course, the mixed partials will be continuous and therefore will
agree. The verification of this result is in the textbook and uses the
Mean Value Theorem of 1 variable calculus. As I said, this result will
apply to almost all functions we will meet in 251. The result implies,
for example, that if we look at the "crowd" of all the possible third
partial derivatives of a function of two variables:
f_{xxx} f_{xxy} f_{xyx} f_{yxx} f_{xyy} f_{yxy} f_{yyx} f_{yyy}
it may seem that there are eight possibilities. But due to Clairaut,
there are only these four:
f_{xxx} f_{xxy} f_{xyy} f_{yyy}
The effect of the "concentration" gets even stronger as the number of
derivatives increases.
Return to the game
How does Clairaut influence my original question? Again, I was perhaps
not the most helpful person in leading the discussion, but eventually
the most relevant fact appeared. The function f(x,y)=(sin(y^{4})x7)^{3} is a cubic (degree 3) polynomial in
x. That is, it can be written as
(Stuff_{0})x^{0}+(Stuff_{1})x^{1}(+Stuff_{2})x^{2}+(Stuff_{3})x^{3}
where each of the "Stuff" terms is some function only involving y. An
x derivative, ∂/∂x, lowers the degree in x. And four x
derivatives will leave us with 0. If you toss a coin a large number of
times, it is overwhelmingly likely that there will be at least 4
heads, and therefore, in the differentiation choices, at least 4 x
derivatives. So since we can reorder these mixed partials in any way
we want, we could put those four derivatives first. And the result
will be 0. So, almost surely, if we toss a coin many times, and follow
the directed sequence of derivatives, the result will be 0.
What does differentiable mean in 1 variable?
What does f´(x)=Q mean? The definition we all tried to memorize
(for a while, anyway) went something like this:
lim_{w→0}(f(x+w)f(x))/w=Q 
f(x+w)=f(x)+Qw+Error 
A really silly (?) twovariable example
I gave a very artificial example to show that things can be quite
strange in more than 1 variable. I defined a function, f(x,y) to be 2x
if y=0, 3y if x=0, and 0 otherwise. To the right is an effort to show
the graph, z=f(x,y), of this strange function. The green line is
supposed to be tilted above the xaxis (the tilt or slope is 2), the
red line is tilted 3 over the yaxis, and the remainder of the graph
is flat at z=0, the xyplane.
Certainly if you slice the graph by y=0 (this is called a vertical trace in the textbook) then you get z=2x, and the partial derivative of f with respect to x exists and is 2. That is, ∂f/∂x(0,0)=2. Similarly, if you slice the graph by x=0 then you get z=3y, and the partial derivative of f with respect to y exists and is 3. That is, ∂f/∂y(0,0)=3.
This is a seriously silly function, but let me analyze it from the
point of view of differentiability at the point (0,0). Suppose that h
is a small perturbation in the first coordinate and k is a small
perturbation in the second coordinate. I would like to understand
f(0+h,0+y) in terms of the unperturbed value of f (that is, f(0,0)=0)
and also the firstorder part (this should be ∂f/∂x(0,0)h=2h
and also ∂f/∂y(0,0)k=3k). So I ideally would like
f(0+h,0+y)=f(0,0)+2h+3k+H.O.T.
But look: what if both h and k are not both 0? Then we are on
the grey part of the graph, and the equation above becomes
0=0+2h+3k+H.O.T. whenever h and k are not both 0. We also want
H.O.T. to be truly "higher order" so it might have stuff like
h^{3/2} or k^{2} or even a combined higher order term
like hk. None of these will cancel out the 2h+3k. This extremely silly
(but rather simple) function does not have the approximation property
which we would like in differentiable functions. And it is this
approximation property which is needed in many applications  this is
not "theory".
Repeating, looking at slices is not good enough to consider limits and continuity. Slices, even collections of slices in two perpendicular directions, just do not contain enough information about the function. In the case of f(x,y) and its partial derivatives, the key idea, both abstractly and computationally, turns out to be the two dimensional analog of (approximate) local linearity. If we "kick" the input to f in both x and y, we need to understand how the function "responds". The nicest response, similar to one dimension, would be the unperturbed response, f(x,y), then something proportional to h plus something proportional to k, and, finally, a higherorder error term.
By the way, if you don't like piecewise defined functions, you could consider the function defined by this formula: f(x,y)=xy/sqrt(x^{2}+y^{2}). It has properties which are similar to the piecewise function, but it is a bit harder to analyze. You could look at it using Maple, and then you might get some insight.
Differentiable in two variables
In mathematics, when something is wrong, one way to help is by making
a definition. The functions we want to consider are called
differentiable and have exactly the property that they can be
approximated nicely.
f(x,y) is differentiable at (x,y) if there are numbers
Constant_{1} and Constant_{2} so that for h and k
small,
f(x+h,y+k)=f(x,y)+Constant_{1}h+Constant_{2}+Error,
where the Error term→0 faster than h+k (so, faster than first
order).
Important results
Before hysteria strikes, here are two results which are verified in
the text. They are not difficult to check (again, the essential key is
the 1variable Mean Value Theorem), but we just don't have time in
class.
Theorem If f(x,y) is differentiable, then the partial derivatives of f(x,y) exist, and Constant_{1}=∂f/∂x(x,y) and Constant_{2}=∂f/∂y(x,y).
Theorem If ∂f/∂x and ∂f/∂y are both continuous then f(x,y) is differentiable (in the approximation sense defined above).
Please realize that essentially all functions we will meet in Math 251 will satisfy the hypotheses of the preceding theorem, so these functions will be differentiable and will have suitable approximation properties. In fact, here is a sort of easy example, although there is some computation involved.
Linear approximation: an example
Here we looked at something like
f(x,y)=sqrt(x^{4}y^{2}+2xy3). Notice that
f(2,3)=sqrt(2^{4}3^{2}+2·2·33)=
sqrt(169+123)=4. This is an example in a calculus class, and it was
chosen so that f(2,3) was nice.
Then
∂f/∂x=(1/2)sqrt(x^{4}y^{2}+2xy2)^{1}(4x^{3}+2y)
and
∂f/∂y=(1/2)sqrt(x^{4}y^{2}+2xy2)^{1}(2y+2x).
We can evaluate these derivatives at (2,3):
∂f/∂x(2,3)=(1/2)(1/4)(4·2^{3}+2·3)=(38)/8
and ∂f/∂y(2,3)=(1/2)(1/4)(2·3+2·2)=2/8.
If we want a linear approximation to f(2.03,2.98), then we may
use the following formula:
f(2.03,2.98) is approximately f(2,3)+∂f/∂x(2,3)(.03)+∂f/∂y(2,3)(.02).
Here the change in x from 2 to 2.03 means that h is .03 and the change
in y from 3 to 2.98 means that k is .02. The linearized approximation
gives us 4+(38/8)(.03)+(2/8)(.02) which is 4.1475. The "true value"
(well, up to 10 decimal places) of f(2,3) is 4.147314409.
Tangent planes
We can get a bit more out of the slicing picture. The vector i+∂f/∂xk is tangent to the curve in
R^{3} gotten by fixing y on the surface z=f(x,y), and the
vector j+∂f/∂xk is tangent to the curve in
R^{3} gotten by fixing x on the surface z=f(x,y). If the
surface is nice and smooth (that is, if the function f(x,y) is
differentiable) people agree that the two vectors determine a plane
which is tangent to
y=f(x,y). To write the equation of a plane, we need a point and a
normal vector.
Suppose we're at the point (x_{0},y_{0},f(x_{0},y_{0})). The normal vector will be perpendicular to both i+∂f/∂x(x_{0},y_{0})k and j+∂f/∂x(x_{0},y_{0})k. So we need to compute the cross product: [i+∂f/∂xk]x[j+∂f/∂xk]. So:
( i j k ) det( 1 0 ∂f/∂x)=[∂f/∂x]i [∂f/∂y]j +k=a normal vector ( 0 1 ∂f/∂y)
Then if you remember the data needed to write a plane (a point and a
normal vector) we can write the equation of a tangent plane.
If f(x,y) is differentiable, then the equation of a plane tangent to z=f(x,y) when x=x_{0} and y=y_{0} is (zz_{0})=∂f/∂f(x_{0},y_{0})(xx_{0})+∂f/∂y(x_{0},y_{0})(yy_{0}).I did an example, but as you will soon see, there's a better way of thinking about this stuff and maybe even computing the things we need.
I did begin the nebula discussion in this class, but I decided to put it all in "one place" so what I did on this day is glued together with the stuff for Wednesday's class (October 1).
Two half planes
I defined a piecewise function to exercise "our" intuition.
( y if x>0 f(x,y)= ( (2x if x<=0First, this function does depend on y. So we computed:
The behavior on the two sides of the yzplane (where x=0) is different. More globally, the "rear" halfplane (where x<0) is a half of a plane whose equation is z=y. This equation gives a plane tilted at 45^{o} to the yaxis. The front half, where x>0, was a plane which was tilted up as y increased. The graph that is shown is Maple's version. There are some real problems with how the program shows this graph. The jumps, which we investigated carefully, are connected because what Maple does is just connect dots. So it connects the jumps even when these are not part of the graph of the function (in 1 variable graphs, this "connecting" idea can be turned off, but I don't know how to turn it off in several variables). There are no vertical line segments in this graph! I strongly recommend that you try to graph this function yourself, and rotate the Maple plot systematically to understand what's going on.
How to define this function
Use the following, please:
>f:=(x,y)>piecewise(x>y,y,2*x);
This command tells Maple that if x>0, then the value of f(x,y) is
y. If the statement is false, the value is 2x. Then I loaded the
package plots and just used the commands
plot3d and contourplot.
The contour "lines"
The contour lines are also not sketched too well. Most particularly,
the contour "line" f(x,y)=0 is very peculiar. It actually consists of
the yaxis together with the positive xaxis. Maple doesn't
want to draw anything like that, so it actually omits a line segment
in this Tshaped contour line. I tried various options with
contourplot but I could not get the T contour (C=0) drawn
correctly. The other contours are again for integer level sets. The
level sets for C>0 are horizontal half lines in the first
quadrant. The level sets fo C<0 have two pieces. One part is a
horizontal half line in the fourth quadrant, and one part is a whole
vertical line in the left halfplane. This may be hard to visualize. I
urged people to try to educate their intuition. The left lines are
closer together than the horizontal halflines. This is all very very
peculiar, but the worst is to come.
The suicidal bug
Now comes some of the harder stuff. I asked people to imagine that
some bugs were "walking" on the graph of z=f(x,y). The green bug, whose path is shown to the
right, strolls along in a path which is roughly circular around the
origin. This bug runs into trouble at any point on the positive
yaxis, where there's a drop. It also has problems along the negative
yaxis, where again there is a big difference in heights. This is a
very small bug. I have tried to indicate this by a sort of light
reddish color surrounding these halflines. The
blue bug walks from the right halfplane to the left
halfplane. It is careful to cross only at the origin. The blue
bug is totally safe, and never comes across any severe height
differences. So I would like to discuss (and name [define], since it
is a math course!) the differences the bugs encounter more precisely.
Limits in one dimension
In one variable, limits are relatively simple. To define
lim_{x→a}f(x) we look at how x gets close to a from both
sides. There are some standard pictures and standard examples of
bad situations. Below are a few, to remind you.
Bad limiting behavior in dimension 1  

A jump
(y=x+7 for x<3, and 2x otherwise.) 
Many wiggles
(y=sin(1/x) for x positive, y=0 otherwise.) 
Several variables
In several variables limiting behavior can be quite complex, much more
than with one variable. I tried to give a few examples.
Many straight line limits exist
I asked students to consider the function
f(x,y)=xy/(x^{2}+y^{2})
This is an algebraic
formula which behaves is a strange fashion for (x,y) near (0,0). We
could try some values, but we can also take advantage of the
appearance of x^{2}+y^{2}. Almost always that's a
signal to at least attempt to understand things in polar coordinates
 that is, to take advantage of circular symmetry.
Since x=r cos(θ) and y=r sin(θ), we know that
x^{2}+y^{2}=r^{2} and
xy=r^{2}cos(θ)sin(θ). Therefore
f(x,y)=xy/(x^{2}+y^{2})=cos(θ)sin(θ)
The
value of f(x,y) only depends on the angular part of the polar
coordinate representation of (x,y) and not at all on the radial
component. The graph is made up of a bunch of half lines all parallel
to the xyplane, radiating out from the zaxis. These halflines, since
cos(θ)sin(θ)=(1/2)sin(2θ), all have height between 1/2
and +1/2.
A Maple graph of the surface over the first quadrant (x>0
and y>0) is shown to the right. I also attempted, with the help of
a stalwart student accomplice, to "draw" the surface kinetically. The
student "volunteer" held one end of a bungee cord under some tension
(both in the student and the cord!) while the calculus instructor held
the other end and walked around the student. The calculus instructor
raised and lowered the cord twice and the student was asked to keep
the end of the cord at the same level as the instructor's
end. Therefore along every angle a limit existed, but as the angle
changed, the limits changed. There were infinitely many
different limits possible along straight line approaches to
(0,0).
Always 0 on a straight line approach
The final example of misbehavior was the following function:
( 1 if y=x^{2} and x>0 f(x,y)= ( ( 0 otherwiseThis function has only two values, 0 and 1. Certainly if you "walk" towards 0 on a straight line approach in the second, third, and fourth quadrants, the function values are all 0 and therefore the limit is 0. What's not so obvious perhaps is the behavior of the function on straight line approaches in the first quadrant.
QotD
I introduced ∂ by just using
it.
Example 1 If F(a,b,c)=a^{2}b3bc^{3} then
∂F/∂a=2ab and ∂F/∂b=a^{2}3c^{3} and
∂F/∂c=9bc^{2}.
Example 2 If F(a,b,c)=a^{3}sin(7b5c^{2}) then ∂F/∂a=3a^{2}sin(7b5c^{2}) and ∂F/∂b=a^{3}cos(7b5c^{2}) and ∂F/∂c=a^{3}cos(7b5c^{2})(10c).
Limits, 1 dimension Here we've got a function of one variable, and we want to define and understand lim_{x→a}f(x)=L. The actual definition, frequently stated but rarely stressed in calc 1 classes, is the following: (and, yes, the Greek letters ε and δ are almost always used)
Limits, 2 dimensions
Derivative, 1 dimension

Slicing and partial derivatives
Now we can define partial derivatives. Please realize that everything
we are doing can be done in any number of variables (want a
picture of 703 dimensions?) but I'll stick with 2 dimensions here
because I can draw pictures and I like pictures.
So look at a graph of z=f(x,y). We can slice this in various ways. For
example, we could slice this by a plane
perpendicular to the y axis with y fixed. This will give sort
of an zx curve. We could "lift" that curve up and just consider it as
a function of one variable, x, and then look at the derivative. That's
∂f/∂x. This notation, with a sort of curly d, is not
terrific, and it can be (as we will see) quite confusing. But it is
what almost everyone uses. Similarly, we could slice by a plane perpendicular to the x axis with x fixed
and consider the derivative of the resulting curve or function. That
will be ∂f/∂y. Here are the formal definitions if you would
like them:
lim_{h→0}(f(x+h,y)f(x,y))/h=∂f/∂x lim_{k→0}(f(x,y+k)f(x,y))/k=∂f/∂y 
I did some silly examples and then actually computed some partial derivatives of 1/sqrt(2πt)e^{x2/t}. This function occurs in many applications analyzing diffusion, including heat and fluid mixing.
This is lecture #7, and I've "covered" up to section 14.3, so I am already (!) a section behind the syllabus.
Here r(t)=a*cos(t)i+a*cos(t)j+btk, the position vector. The velocity vector is r´(t)=a*sin(t)i+a*sin(t)j+bk. The length of this velocity vector is sqrt([a*sin(t)]^{2}+[a*sin(t)]^{2}+b^{2}). This simplifies because we know that sin^{2}+cos^{2}=1. There are very few other curvature computations which are so simple.
So ds/dt=(sqrt(a^{2}+b^{2})) (the length of the velocity vector, which is the speed) and we get the unit tangent vector by dividing the components of r´(t) by ds/dt. SO T(t)=(1/sqrt(a^{2}+b^{2}))(a sin(t)i+a cos(t)j +bk). According to what we did last time, if we differentiate this we should get a vector perpendicular to T(t). So d/dt(T(t))= =(1/sqrt(a^{2}+b^{2}))(a cos(t)ia sin(t)j+0k). If you take the dot product of this with T(t) you will get 0 (the strangelooking signs make that true). But this is d/dt(T(t)) and I want for curvature d/ds(T(t)). The Chain Rule suggests that I divide d/dt(T(t)) by ds/dt to get d/ds(T(t)). If I do this, the result is (1/(a^{2}+b^{2})(a cos(t)ia sin(t)j+0k). Wow. We are not done yet. This should be κN: that is, it should be the curvature, a positive number, multiplying a unit vector, and this is the unit normal vector. If you stare at what we have you should eventually see the following:
dT/ds=(1/(a^{2}+b^{2})(a cos(t)ia sin(t)j+0k)=[a/(a^{2}+b^{2})](cos(t)isin(t)j)
Here what is in blue is a scalar, multiplying
what is in green which is a unit
vector. Therefore what is in blue is κ,
the curvature, and what is in green is the
unit normal vector. By the way, this is why everyone (and me too, darn
it, me too!) uses formulas such as those in the book to compute these
things, because going directly from the definition is too darn
complicated.
If κ=[a/(a^{2}+b^{2})] for the helix, how does this match up with our forecasts? Here is the scorecard.
Here are some pictures of various helices produced by Maple (the plural of "helix" is
"helices"). The pictures below were produced using the command spacecurve([a*cos(t),a*sin(t),b*t],t=0..6*Pi, axes=normal,color=black,thickness=2,scaling=constrained,numpoints=180); The procedure spacecurve is loaded as part of plots using the command with(plots);. I used the option scaling=constrained in order to "force" Maple to display the three curves with similar spacing on the axes. Otherwise the x and y variables would be much altered in each image. I hope that these pictures give some idea of what the curvature and torsion represent.

The curvature of the twisted cubic
O.k., I tried to compute the curvature of
r(t)=ti+t^{2}j+t^{3}k. I used a formula from section 13.4:
κ=r´xr´´(t)/(r´(t))^{3}.
Even though this formula looks weird, it is much better to use than
trying the definitions (I have tried using the definitions with this
example, and the computations are terrible).
So r´(t)=1i+2tj+3t^{2}k and r´´(t)=0i+2j+6tk. Now for the cross product computation:
 i j k  det 1 2t 3t^{2} = det2t 3t^{2} i– det1 3t^{2} j + det1 2tk  0 2 6t   2 6t  0 6t  0 2And then I evaluated the 2by2 determinants, so we saw that the vector on the top of the formula for κ is (12t^{2}6t^{2})i6tj+2k=(6t^{2})i6tj+2k. But we need the magnitude or length of this for the top, and this is sqrt(36t^{4}+36t^{2}+4). Wow. The bottom is "just" the threehalves (!) power of the length of the velocity vector, and this is (1+3t^{2}+9t^{4})^{3/2}.
Therefore (as they write in textbooks) the curvature of the twisted
cubic is
sqrt(36t^{4}+36t^{2}+4)

(1+3t^{2}+9t^{4})^{3/2}
Now, having done this ludicrous computation, I will admit that I get almost nothing out of it. Here is this ridiculous formula, and what does it tell me? Well, maybe a little bit. If t is very large positive or very large negative, it seems to say that the curvature gets small (look at the "net" power of t on top, maybe a t^{2}, and on the bottom "net" a sort of t^{6}). I guess this means that the curve gets flatter as t→∞. Maybe this is interesting. (It sort of resembles y=x^{2} that way.)
How to think of all this?
What would I like "you" (especially the engineer and physics "yous")
to take away? Another formula from the textbook (section 13.5) resembles exactly what we already saw in two dimensions:
r´´(t)= (d^{2}s/dt^{2})T +
κ(ds/dt)^{2}N
This is a decomposition of the acceleration vector into
tangential and normal components, and actually has some interesting
information about physical quantities. Remember that F=ma, and mass is
a scalar, so that the Force an object "feels" is exactly a scalar
multiple of its acceleration. So let me look at two special cases.
Motion in a straight line
Imagine a ball bearing (?) moving in a straight tube. I claim that the
tube never "feels" any force from the walls of the tube. Why? Well,
since for a straight line κ=0, the coefficient of the normal
component of the acceleration, κ(ds/dt)^{2}, is always
0, for any motion, and therefore, since force is a scalar
multiple of acceleration, the force also must have zero normal
component. So the ball bearing can be pushed along or back in the
tube, but it is never pushed against the wall if you
insist that the ball bearing move in a straight path.
Motion in a circular arc
Now a more complicated thought experiment. Imagine the ball bearing
constrained to move in a piece of a circular tube. All we know is that
the object is moving, and its motion is part of a circular arc. I
claim that then/ no matter what, the ball bearing is "feeling"
a transversal (normal) force from the "walls" of the tube. Why? Well,
again the normal component of the acceleration (a scalar multiple of
the force acting on the object) is κ(ds/dt)^{2}. Well,
κ is not 0 since it is 1/(radius of the circle). And, since the
object is moving, we also know that ds/dt is not 0. Therefore the
product isn't 0, and the normal component of the acceleration isn't 0.
This is, of course, related to the wonderful primitive experiment of quickly spinning a bucket of water on a rope  and the water is "pushed" into the bucket by the "centrifugal force". Hey, that force is a scalar multiple of the normal component of the acceleration of the bucket's motion. It really works.
Just a little bit more ...
I hate to leave this subject since there is much more to be told, and
everything turns out to be amazingly useful in a wide variety of
applications. I hate to leave the subject, but the course is very
dense. So: just a little bit more. Think of an airplane flying along a
curve in the sky (the dashed red line). Then the unit tangent and
normal vectors are as shown. There's another vector, B, called the
binormal vector, which is TxN. Since T and N both have
length 1 and they are perpendicular, then B also has length 1 (sine of
a right angle is 1). And B is perpendicular to both T and N. The path
is determined by two functions. One of these functions is curvature
which says how much the plane is being pulled from a straight line. It
measures how much the T is changing in the N direction. Thegreen surface on the tail or the rudder of the
plane is the major object determining κ. But the motion is
threedimensional. The way the binormal, B, changes, says how much the
plane is twisted out of the two dimensional plane determined by T and
B. The rate of change of B is called the torsion, and is
usually denoted by τ, the Greek letter tau. It is determined
mostly by the angles that the blue control
surfaces have with the wings. When a plane takes off or is landing,
these surfaces have relatively high angle to the wings.
This is all a big simplification, but it is mostly
correct.
There is more about all this in the textbook if you are interested.
We move on to one of the major topics of the course. The word "several" is almost technical in mathematics, and means "more than 1". We will start with an almost ludicrously simple function.
x^{2}+y^{2}
Here f(x,y)=x^{2}+y^{2}. This is a function defined by
a formula (essentially all of the functions we'll consider in this
course will be defined by formulas). The notation means that the input
to the function is an ordered pair of numbers, (x,y), and the output
is one number. Here the output for the ordered pair (2,3) is 13.
Formalities: domain and range
The domain will be the collection (the "set") of all possible
inputs. Just as in calc 1, if the function is defined by a formula,
then the domain will be all inputs for which the function makes
sense. The usual restrictions that will concern us are:
f(x,y)=x^{2}+y^{2}  

Domain I think all pairs (x,y) of real numbers, all of R^{2}. 
Range Since squares are nonnegative, certainly the values of this function are nonnegative. And f(0,0)=0, and f(sqrt(A),0)=A for A positive. I am just verifying precisely that the range is all nonnegative real numbers. 
f(x,y)=1/(yx^{2})  
Domain So this example is chosen to illustrate the restriction about not dividing by 0. The domain is all pairs (x,y) of real numbers for which y is not equal to x^{2}. Geometrically, this means all points of R^{2} which are not on the parabola y=x^{2}. 
Range Well, 0 isn't in the range (it isn't the reciprocal of any number). But everything else is: check this by just looking at what happens to (0,A), which gives 1/A for all nonzero A's. 
f(x,y)=sqrt(yx^{2})  
Domain So this example is chosen to illustrate the restriction about square roots. The parabola y=x^{2} divides R^{2} into two pieces. One piece contains, say, the point (3,4) ("below" the parabola). This point has yx^{2}=43^{2}=5<0, so (3,4) is not in the domain of this function. The domain is the "other" piece of R^{2} and also those points which are on the curve y=x^{2}. 
Range The range is all nonnegative numbers. Again, to check this you could look at what happens to (0,A) for A>=0. 
f(x,y)=ln(yx^{2})  
Domain I still must "throw out" the part of R^{2} which is below the parabola. But here inputs to ln must be positive, so the domain does not include the curve y=x^{2}. The domain is all of the points in R^{2} which are above the parabola. 
Range The range is the range of ln, which is all real numbers. 
Kinds of graphs
Let me return to the
simplest of the functions I just considered:
f(x,y)=x^{2}+y^{2}. There are various graphs which are
commonly used. Maybe the simplest is to consider the points (x,y,z) in
R^{3} which satisfy the equation
z=x^{2}+y^{2}: this is usually called the graph of
the function. A Maple representation
of this graph is shown to the right, and the procedure which produced
it is plot3d, part of the plots package. This is rather a simple
function, and I hope you can see the shape of this surface. It is a
cup, axially symmetric around the zaxis. It is called a
paraboloid.
I looked at this graph and studied curves for fixed values of x and y (called traces in the textbook. These are just (?) parabolas opening up. Piecing them together to get this parabolic cup is not totally obvious.  
Another kind of plot, or, anyway, some geometric clue to the nature of
the function, can be gotten by looking the contours of f(x,y). There
are topographic maps (say, used by hikers) which give a
twodimensional representation of the information in the surface
picture above. Pick a constant, C, and look at the (implicitly
defined) "curve" f(x,y)=C. I put quotes around the word "curve"
because maybe it doesn't have to be a neat nice curve. (An example was
discussed in class, and is below.) To the right is a collection of
contours for f(x,y)=x^{2}+y^{2}. These contours
correspond to the positive integers 1, 2, 3, 4, 5, and 6.
Please notice how these contours, which are at evenly spaced "heights", get closer together as the threedimensional graph gets steeper. Of course, if the contours are not labeled with the values of the constants, I can't tell if the function is increasing or decreasing! This picture was made with contourplot, another part of the Maple package, plots. 
The origin in this paraboloid graph is a local and absolute minimum. I briefly sketched a graph of f(x,y)=x^{2}y^{2}, which is an upsidedown version of what was just drawn. So it has a local and absolute minimum. That's not too new. But let's consider something weird and wonderful.
A saddle Consider the function f(x,y)=x^{2}–y^{2} (I'm trying to exaggerate the minus sign typographically in this since that's the most interesting part). The traces are again "just" parabolas. But here, when, say, y=2 (a plane transversal or perpendicular to the yaxis), we get f(x,2)=x^{2}4. So the curve z=x^{2}4 is a parabola opening up. As we change the y's in this sort of slice, the bottom of the parabola moves down for big y. What about the other traces, with x=a constant? Well, if x=3, then f(3,y)=9y^{2}. In the plane perpendicular to the xaxis given by the equaltion x=3, the curve is z=9y^{2}. a parabola opening down. And by thinking we can see the top of these parabolas moves up when x is large. It is difficult for a novice to see how to put these curves together. To the right is a Maple graph of this function for 3≤x≤3 and 3≤y≤3. When you look at this on the computer (please try!) you can rotate it and magnify it, and things might become more clear.  
The contour curves of this function are shown to the right.
Consider x^{2}y^{2}=3, for example. One point on this curve is x=2 and y=1, so look at the curve which goes through the point (2,1). This curve is a hyperbola. The hyperbolas (each has two pieces) corresponding to positive values of the function open left and right. On the other hand, there is a family of hyperbolas opening up and down. These correspond to negative values of the function. And there's a special number. If x^{2}y^{2}=0, then, since x^{2}y^{2}=(x+y)(xy), the points on this "curve" correspond to the two straight lines x=y and x=y. (The strange little box near (0,0) in the contour plot is because Maple thinks that very small +/ numbers which it samples should also be 0. I am sorry about that. It may not be obvious looking at the curvy surface above that there are two straight lines on the surface. But there indeed are.  
The point at the origin has a new kind of behavior, not found in one
variable calculus. In certain directions it is a maximum. In other
directions it is a minimum. This sort of point will be called a
saddle point (not too strange a name) or a minimax.
You could imagine that there will be a whole range of behaviors in three hundred variables!  

Arc length
If speed is the magnitude of the velocity vector, then since
distance=rate·time, well, with variable speed, we need to chop
up the time interval and compute and add up pieces of distance. This
is the basic idea behind the definite integral, so it makes sense to
compute the distance along a curve from time t_{1} to time
t_{2} by this:
∫_{t1}^{t2}v(t) dt.
O.k., this is good (all math is good!) but maybe some
examples ... will show how silly it is.
Textbook example
Here is an example taken from a textbook. Suppose
r(t)=<12t,8t^{3/2},3t^{2}>. What is the length
of the curve from t=0 to t=1?
This is a typical artificial problem in a textbook. We know that
r´(t)=<12,8(3/2)t^{1/2},6t> so that the
speed is
r´(t)=sqrt{12^{2}+12^{2}t+36t^{2}}. Now
we "remember" that distance=rate·time, and that the magnitude
of the velocity vector is ds/dt, the speed. So the total distance
traveled is
∫_{0}^{1}sqrt{144+144t+36t^{2}} dt.
Now look at the coincidences. We can pull out the 36 from under the
square root, and the result is
6∫_{0}^{1}sqrt{4+4t+t^{2}} dt, and,
what a coincidence, 4+4t+t^{2} is (2+t)^{2} and the
integral becomes 6∫_{0}^{1}(2+t) dt=6(2t+(1/2)t^{2}]_{0}^{1}
and this is 6[2+(1/2)].
More realistically ...
Almost every combination of two or three functions, even rather
"simple" functions that you try to use as a position vector, will
yield something that can't be antidifferentiated in terms of familiar
functions. For example, look at the following mess:
> a:=t>t^3; 3 a := t > t > b:=t>cos(t); b := t > cos(t) > c:=t>exp(t); c := t > exp(t) > int(sqrt(a(t)^2+b(t)^2+c(t)^2),t=0..1); bytes used=4000036, alloc=3407248, time=0.14 bytes used=8007244, alloc=5241920, time=0.30 1 /  6 2 2 1/2  (t + cos(t) + exp(t) ) dt  / 0And that means Maple tried but can't figure out anything useful to reply except to echo the integral back to the questioner. My next step is
> evalf(%); 1.971185471which asks for an approximate (numerical) evaluation of the integral. That's almost always what's going to be necessary. Oh well: the situation in textbook problems is much too nice.
Curvature is a number which measures how a curve bends
I tried to analyze the idea of how a curve bends. Curvature will be a
measure of this bending. I began an analysis that was apparently first
done by Euler in about 1750. I'll study plane (2dimensional)
curves. A curve is a parametric curve, where a point's position at
"time" t is given by a pair of functions, (x(t),y(t)). Equivalently,
we study a position vector, r(t)=x(t)i+y(t)j. Here x(t) and y(t) will
be functions which I will feel free to differentiate as much as I
want.
There are some special test cases which I will want to keep in mind.
#1: a straight line
A straight line does NOT bend, so it should have curvature 0.
#2: a circle
A circle should have constant curvature, since
each little piece of a circle of radius R>0 is congruent to each
other little piece, and, in fact, the curvature should get large when
R gets small (R is positive), and should get small when R gets large
(and looks more like a line locally).
#3: "the" parabola
Even y=x^{2} might be a good test to keep in mind, since there
the curvature should be an even (symmetric with respect to the yaxis)
function of x, and should be bellshaped, with max at 0 and limits 0
as x goes to +/∞. The curve actually bends most near the
origin, and far away, when x is large positive or large negative, even
though the curve is very steep, it gets really flat.
The problem of understanding or defining curvature is to somehow extract the geometric information from the parameterized curve. That is, if a particle moves faster, say, along a curve, it might seem like the same curve bends more. So what can we do? Somehow the geometry of the curve should be separated from the dynamics (kinetics?) of motion along the curve.
We looked at θ, the angle that the velocity vector r´(t) makes with respect to the xaxis. How does θ change? After some discussion it was suggested that we look at the rate of change with respect to arclength along the curve: that is the same as asking for the rate of change with respect to travel along the curve at unit speed, and therefore somehow the kinetic information will not intrude on the geometry. This is the same as asking a bug to travel along the curve at unit speed, and to describe the rate of change of θ (the turning of r´(t), the velocity vector) as the bug strolls along the curve.
The quantity s and the quantity t
Arc length, s, on a curve is computable with a definite integral:
sqrt(x´^{2}+y´^{2}) integrated from
t_{0} to t with dt is the arc length. As shown before, this is
rarely exactly computable with antidifferentiation using the usual
family of functions. But ds/dt is just
sqrt(x´^{2}+y´^{2}) by the Fundamental
Theorem of Calculus. And the Chain Rule suggests that d*/ds(ds/dt)=d*/dt if * is any quantity
of interest, such as θ. So to get d*/ds from d*/dt, multiply the
latter by 1/(ds/dt) which is
1/sqrt(x´^{2}+y´^{2}).
By drawing a triangle we see that θ is arctan of y´/x´: θ=arctan(y´/x´). We can find dθ/dt: it is (y´´x´x´´y´)/(x´^{2}+y^{2})^{2}. This uses the formula for the derivative of arctan, the Chain Rule, and the quotient rule. Then the previous results say that
Back to the test cases
The straight line We compute for some constants A and B and C and D: x(t)=At+B x´(t)=A x´´(t)=0 y(t)=Ct+D y´(t)=C y´´(t)=0so that the top of the curvature formula is 0. κ=0 for any straight line.  
A circle Suppose we have a circle of radius R centered at the origin. Then parametric equations are not too difficult to get: x(t)=Rcos(t) x´(t)=Rsin(t) x´´(t)=Rcos(t) y(t)=Rsin(t) y´(t)=Rcos(t) y´´(t)=Rsin(t)Let me look first at the bottom of the curvature formula. This is (x´^{2}+y´^{2})^{3/2}. Notice that x´^{2}+y´^{2}=R^{2}([sin(t)]^{2}+[cos(t)]^{2})=R^{2}, so the 3/2´s power is R^{3}. The top is y´´x´x´´y´ and this is Rcos(t)(Rsin(t))Rcos(t)Rcos(t)=R^{2}·1. The whole curvature formula is then R^{2}/R^{3} which is 1/R. When R is large, this is near 0. When R is close to 0 and positive, this is large positive. And the curvature is constant which it should be since the circle has the same local geometry at every point.  
A parabola Here I want to look at y=x^{2}. A simple parameterization is good. So let's try: x(t)=t x´(t)=1 x´´(t)=0 y(t)=t^{2} y´(t)=2t y´´(t)=2Now the bottom of κ, (x´^{2}+y´^{2})^{3/2}, becomes (1+4t^{2})^{3/2}, and the top, y´´x´x´´y´, is just 2. So κ=2/(1+4t^{2})^{3/2}. Here the "local geometry" definitely changes from pointtopoint. The most curvy (?) part of the graph is at the origin. As x→+/∞, although the graph gets steeper and steeper, the curve locally actually gets more and more flat: the curvature→0. 
The smaller figure is supposed to be the curvature of the parabla. 
I defined the unit
tangent vector, T, to be a unit vector in the direction of
r´(t). If θ is the angle that r´(t) makes with the positive
xaxis, then T must be cos(θ)i+sin(θ)j.
Also r´(t)=(ds/dt)T, where ds/dt is the length of r´(t),
the speed. Now differentiate the formula for r´(t) using
one of the product rules we had stated earlier. The result is
r´´(t)=(d^{2}s/dt^{2})T+ (ds/dt)d/dt(T). But T is
cos(θ)i+sin(θ)j, and differentiation with respect to t is the
same as differentiation with respect to s multiplied by ds/dt. But
differentiation with respect to s gives (sin(θ)i+cos(θ)j)
multiplied by the derivative of θ with respect to s, and this is
κ. All this put together is:
We have decomposed acceleration into the normal and tangential directions. This can be significant physically and can help to understand physical situations. Here are two "extreme" cases.
Straight line motion: no transverse forces
Moving in a circle: always normal force
Silly example
An answer 
Now let's look at space curves. The geometry of these curves, as seen from the point of view of calculus (called "differential geometry of space curves") is a subject which originated in the 1800's. The material presented here was stated in about 18501870. When I learned about this material in college, it mostly seemed rather abstract and useless  stuffy formulas no intelligent person would care about. Like a number of other judgments about merit that I've made in life, this is completely wrong. The geometry of curves has within the last few decades become very useful in a number of applications: robotics, material science (structure of fibers), and biochemistry (the geometry involving the structure of big molecules such as DNA): amazingly useful formulas and ideas!
The right circular helix
I'll carry along is a right circular helix as a basic example.
x(t)=a cos(t) y(t)=a sin(t) z(t)=b tThe quantities a and b are supposed to be positive real numbers. This helix has the zaxis as axis of symmetry. It lies "above" the circle with radius a and center (0,0) in the (x,y)plane. The distance between two loops of the helix is 2π b. The "b" changes the pitch or angle of the screw threads modeled by the helix.
How should curvature behave for the helix? We discussed this and decided κ should have these properties:
Now let's analyze space curves
If r(t)=x(t)i+y(t)j+z(t)k (the position vector), then
r´(t)=x´(t)i+y´(t)j+z´(t)k=(ds/dt)T(t) is called the velocity vector.
Here T(t) is called the unit tangent vector and is a unit vector in
the direction of r´(t). ds/dt is the speed, and is
sqrt(x´(t)^{2}+y´(t)^{2}+z´(t)^{2}), the
length of r´(t). We use ds/dt also to convert derivatives with respect
to t to derivatives with respect to s, as in two dimensions using the
Chain Rule.
Since T(t)·T(t)=1 differentiation
gives
T´(t)·T(t)+T(t)·T´(t)=0. But the dot product is
commutative, so this is 2T´(t)·T(t)=0 or just
T´(t)·T(t)=0.
This means that T´(t) and T(t) are always
perpendicular. In fact, we are interested in dT/ds, which is the same
as (1/(ds/dt))T´(t) (it is usually easier to compute T´(t) directly,
however, and "compensate" by multiplying by the factor
1/(ds/dt)).
Think about this sentence, which states a fact I'll use again and
again in this course:
Any
nonzero(!) vector is equal to the product of its magnitude times a
unit vector in its direction.
For dT/ds, the magnitude is
defined to be the curvature, κ, and the unit vector is
defined to be the unit normal N(t). This essentially coincides
with what's done for plane curves, when curvature was defined to be
d(θ)/ds.
I will continue with this material next time. I confess that I am currently about 1/2 lecture behind the syllabus, but the formulas and examples and concepts involving curvature are important and complicated. Please read the textbook carefully and try the homework problems in sections 13.3, 13.4, and 13.5. This is all work, but please do this! I will compute the curvature for the helix next time, and discuss some of the consequences of all of this stuff before continuing with the syllabus.,
Parametric curves: a nuisance?
So I began with some irritating examples.
Distinguishing between geometric aspects of the path of a particle and the dynamic aspects of the particle will take up most of our next class. This won't be obvious.
Two problems in the textbook
I tried to discuss two problems from the textbook. These pictures are
copied from page 743 of the text.
Problem 5 of section 13.1
Match the space curves in Figure 8 with
their projections onto the xyplane in Figure 9.
Answer
How can we "solve" this problem? Well, I actually think most people
can solve the problem, but I think describing the process is somewhat
difficult. It is difficult because, for example, the pictures in
Figure 8 are twodimensional. And they are images of some idea
threedimensional situation. What can we see? Well, in Figure 8 (A) I
see a straight line. I believe that the straight line will not be made
"curvy" if we project into the xyplane. Therefore the only candidate
which matches is the noncurvy (!) picture (ii) in Figure 9. What
else? Well, maybe the picture in (B) seems to be helical (more precise
information about a helix later). That's sort of a motion around a
cylinder whose axis of symmetry is, in this case, the xaxis. If we
push this down and neglect the z coordinate, I think what we get is
picture (i). This leaves (C) matching up with (iii). If you think
about a particle moving around the path indicated in (C), maybe
you see the xy"squashing" of it moving back and forth along the
parabolic path in (iii). I was asked if there was somehow "proof" that
this was (iii), and, well, we're told that the triples of pictures
match up, and (C) and (iii) are what's left. It is certainly true that
something like (C) could have a more complicated squashing than what
is in (iii).
Problem 6 of section 13.1
Match the space curves in Figure 8 with the following vectorvalued
functions:
(a) r_{1}=<cos(2t),cos(t),sin(t)>
(b) r_{2}=<t,cos(2t),sin(2t)>
(c) r_{3}=<1,t,t>
Answer
Well, the linear formula is (c), and if we just
consider the first two coordinates, the 1 and the t, that would seem
to describe (ii), so I think that (c) is the algebraic version of
(A). As for (b), the first two coordinates are t and cos(2t) which
seem to describe (i). And if you consider the second and third
coordinates of (b), they are cos(2t) and sin(2t). The sum of the
squares of these is 1 (cos^{2}+sin^{2}=1) and the
first coordinate in (b) just moves the point along, so we get the
helix in (B). That leaves r_{1} in (a) to describe (C) and
(iii). Let's consider the first two coordinates of (a), which are
cos(2t) and cos(t). A trig identity you may/may not remember is
cos(2t)=2(cos(t))^{2}1. If we take x as cos(2t) and y as
cos(t), this becomes the equation x=2y^{2}1 which surely
looks like (iii). And the third coordinate, sin(t), of (a) makes the
image go up and down, up and down. So (a) is consistent with both (C)
and (iii). I think this third curve and formula are a bit interesting
and strange.
Pictures of a point in motion
Twisted cubic
The twisted cubic is the curve
r(t)=<t,t^{2},t^{3}>. It is one of the beginning
curves shown to students in this subject, with the warning that here
is the pictures can be deceptive. A Maple command created a
3dimensional viewing box of the curve. I rotated this box in various
ways and exported the images shown below.
spacecurve(<t,t^2,t^3>,t=5..5,color=black,axes=normal,thickness=2);
A perspective on real life Frequently, similar difficulties have occurred in dissections of bodies, when slides of different cells are prepared. Depending on the angles of the slices, very different pictures are seen. 
Robots and molecules
I also remarked that people who wanted to understand the conformation
(geometry) of biologically interesting macromolecules will need to
learn how to look at curves in R^{3}. Also, people who want to
describe motion of robot arms necessarily need the ideas we will describe.
The derivative
If r(t)=<a(t),b(t),c(t)> is a vector function of t, we can
consider (1/h){r(t+h)r(t)}. Inside the {} is a difference of vectors,
and that's therefore a vector. Then multiplying by 1/h: that's a
scalar multiple. So the result is a vector. It may happen that the
limit as h>0 of this quotient may or may not exist. If it does,
we'll say that r(t) is differentiable and that the limit, which is
labeled r´(t), will be called the derivative. It is also called the
velocity vector.
It doesn't take many examples to convince yourself that the (vector) derivative will exist exactly when all of the scalar functions a(t) and b(t) and c(t) are differentiable, and that the components of the resulting vector derivative will be the derivatives of these functions. That is, if a(t) and b(t) and c(t) are differentiable, then the vector function will also be differentiable, and r´(t)=<a´(t),b´(t),c´(t)>. Computing the derivative of such a vector function when the component functions are defined by familiar formulas involves nothing essentially new. I did a few examples.
Meaning of the derivative
The derivative balances out a large scalar, 1/h (when h is small),
with a very small secant vector, r(t+h)r(t). That the limit
exists is rather neat fact, that somehow the shrinking of the vector
stabilizes with the increasing of the scalar amount, and that the
direction tends to a fixed direction  this is not obvious, and one
should really not expect it!
The magnitude: speed
The magnitude of the velocity vector, r´(t), is called the
speed.
Velocity vector, tangent vector
I tried to argue, looking at a local picture of the path of a
particle, that the direction of the velocity vector is tangent to the
path of the particle. So the velocity vector is a vector
tangent to the path.
Tangent line to a helix
Look at the curve r(t)=<3cos(2t),3sin(2t),8t>. What is this?
The first two variables describe uniform circular motion. The radius
of the circle is 3, and that's the distance from the central axis of
this curve, which turns out to be a helix. The 2 changes the angular
velocity of the curve, and doubles it. The 8 affects the "pitch", the
angle of the helix, and also the distance between "loops". When 2t
changes by 2Pi, the curve passes around one loop. That means the
change in t is Pi, so the change in z is 8Pi.
Let's find the parametric equations of a line tangent to this helix
when t=Pi/2. The line must pass through
r(2)=<3cos(2[pi/2]),3sin(2[Pi/2]),8[Pi/2]>=<3,0,4Pi>. A
vector in the tangent direction can be gotten from the velocity
vector. So: r´(t)=<6sin(2t),6cos(2t),8> which, when
t=Pi/2, gives <0,6,8>. Therefore the parametric equations for
the line are:
x=3+0t
y=06t
z=4Pi+8t
To the right is a picture of both the helix and the tangent line just
specified.
Various formulas, especially product formulas
The derivative of u(t)·v(t) is
u´(t)·v(t)+u(t)·v´(t). The derivative of u(t)xv(t) is u´(t)xv(t)+u(t)xv´(t). And there's even another product
rule (multiplying a vector function by a scalar function and then
differentiating). Since the cross product is not commutative, the
order is important. We will use these formulas next time. They are
very important in describing the dynamics of motion of a point in
space.
This course is like ...
I get confused about the meaning of metaphor and
simile. One reference I found declares
Metaphor is often confused with simile, the difference being that the metaphor draws a parallel between concepts, while the simile points to poetic similarities.So I remarked in class that dot and cross product were sort of like rivets and screws (I worked on this, darn it!) and that the whole course was sort of like building a car. A car has thousand and thousands (tens of thousands) of parts. We're just starting. So stay calm. (A rivet has symmetry, fastens from both sides, and a screw has handedness, just like cross product).
I did problems 15 and 28 in section 12.4 since I wanted to show some direct examples of cross product computations.
Problem 13
What is the cross product of <(1/3),1,(1/3)> and
<1,1,2>? We must compute:
 i j k  det(1/3) 1 (1/3)=det 1 (1/3)idet(1/3) (1/3)j+det(1/3) 1 i=  1 1 2  1 2   1 2   1 1  = [(1·2(1/3)·(1)]i[((1/3)·2(1/3)·(1)]i+[((1/3)·(1)1·(1)]i=(7/3)ij+(2/3)kProblem 28
The direction should be backwards in y and the magnitude should be 9/2. The cross product therefore must be (9/2)j.
Mostly I think I know what a line is and what a plane is. Here is a picture with one of both of them displayed. But what happens if we have several of them, and we want to deal very precisely with them? It is useful, then, to have algebraic ways of describing and computing with lines and planes.
Specifying a line
The following information will usually be the most convenient way to
specify a line in this course:
Example
If p=(3,9,5) and v=<4,7,6>, then the vector pq will be
<x3,y9,z5>. Suppose t is a scalar. The vector equation pq=tv works
out to the following three scalar equations:
x=3+4t
y=9+7t
z=5+6t
So these are parametric equations for the straight line
determined by the data p and v. Notice where the 3 chunks of p's
information go (no t's!) and where the 3 chunks of v's information
(the direction) go (with the t's).
Other parametric representations
I then played with these equations for a while. For example, the
equations
x=3+12t
y=9+21t
z=5+18t
represent the same line. I just multiplied the direction v by 3. This doesn't
change the actual geometric line which is parameterized by the equations: every
point in "this" line is in the other line, and, similarly, the other way
around. And
x=34t
y=9+7t
z=56t
represents also the same line: I multiplied the direction by 1 but still
we will get the same total collection of points: the same straight line. And,
perhaps more complicated:
x=7+4t
y=2+7t
z=11+6t
This is also the same line. Notice, please, that the direction vector
is the same, but what I have changed is the "base point",
(7,2,11). But if you consider the original parameterization, and put
t=1 in those equations, we will get (7,2,11). So, in fact, the
descriptions sort of "overlay" one another: the v's are the same, and
the we've just happened to start the parameterizations at different
points on the same line. In the next class, we will consider in detail
the kinetic aspect of the parameterization, but right now I tell you
that the lines (the collection of all the points described by these
triples of equations) are the same.
There is a plethora of different parameterizations for the same
straight line.
Plethora officially means "A superabundance; an excess."
Symmetric equations Some manipulation gives another way to represent the line. x=3+4t becomes t=(x3)/4 y=9+7t becomes t=(y9)/(7) z=5+6t becomes t=(z5)/6 so that we have (x3)/4=(y9)/(7)=(z5)/6. Such a collection of equations is called the symmetric form of the line. As far as I know, we won't describe lines this way in this course.
Two points determine a line
Example 
Specifying a plane
In this course, the data specifying a plane will also turn out to be a
point and a vector. What I'm about to describe may not look
immediately useful, but it will be, really. So what we will need is
What condition(s) on x and y and z will guarantee that q=(x,y,z) is on the plane? Well, the vector pq should be perpendicular to n. We can check perpendicularity easily with dot product. An example may help.
Example
Suppose p=(4,5,7) and n is the vector <11,2,31>. The vector pq
is <x4,y5,z(7)> and pq is perpendicular to n exactly when pq·n=0. And that is
11(x4)+2(y5)+31(z(7))=0
This is a fine equation. You don't need to "clean it up". Of
course, it can be rewritten (if you must!) as
11x+2y+31z=44+10217=163.
Again, there are many equivalent equations for one geometric plane since there are lots of vectors normal to one plane, and lots of points on one plane. And we can do the reverse:
Reversing ...
6x+3y+7z=20 is the equation of a plane.
What is a vector normal to this plane? If you
followed the previous discussion, I hope you can see that the
components of such a vector are the coefficients of x and y and z in
order. So n=<6,3,7> is a vector normal to this plane.
What are all (nonzero) vectors normal to this plane? These vectors
are the nonzero scalar multiples of <6,3,7>.
What are the coordinates of a point on the plane 6x3y+7z=20?
Part of the problem in answer this question is that there are so
many possibly answers! For example, (0,0,20/7) is one answer, and
so is (1,7/3,1), and so is .... uhhh ... (1,000, 1,000,
8,980/7). There are many points on the plane.
Three points determine a plane Again, Euclid declared that 3 points in space should determine a plane. So can we find an algebraic description of a plane through the points p=(3,3,9) and q=(5,2,1) and r=(4,2,2)? We need to find a vector n normal to the plane. Well, the vectors pq and (say) qr are in the plane's direction. The crossproduct of these two vectors will be perpendicular to the plane. So: pq=<8,1,4> and qr=<9,0,1>, and  i j k  det8 1 4 =i(836)j(9)k=i+44j+9k  9 0 1 is a normal vector. So 1(x3)+44(y3)+9(z(9))=0 is an equation of this plane. (We would get equivalent equations if we used q or r instead of p, or if we computed with pqxpr instead of pqxqr.) 
Distance of a point to a plane
Suppose the point q is not on a plane, P. How can we find the
distance from q to P? I did this in a rather clumsy way in class,
mostly because I wanted to practice how to use line and plane
equations. Let me start with specific information:
Suppose
q=(2,1,3) and P is the plane specified by the equation
5x7y+6z=10.
First, is q on the plane, P? Well, we can check this by substituting
the coordinates of q into the equation presented for P:
5·27·1+6·3=107+18=21, and 21 is not 10, so q is
not on P.
My strategy for finding the distance was to write the parametric
equations for a line through q perpendicular to P, see where the line
intersected P, and then compute the distance from q to that point. So
let me try to carry this out.
Another way ... There are frequently different and equally valid solution strategies. Here is another way to do this problem: (2,1,3) is not on the plane 5x7y+6z=10. I can find a point on the plane just by "guess" (as I did earlier, and I will guess for something convenient!): r=(2,0,0). Then the vector rq=<0,1,3> points from the plane to q. Now if I project rq onto the normal vector I will get a vector whose length is the distance from the plane to the point. This "projection" is just what I did last time when I found the parallel part of a vector. So n=>5,7,6> is still a vector normal to the plane. So we just compute rq·n/n=(7+18)/sqrt{110}=11/sqrt{110}, which is the same answer. This is more direct and more efficient. The following material was not done in class but maybe showing you some examples will help you with homework.
Parallel lines Line A Line B x=5+t x=30+t y=3+2t y=9+2t z=1+3t z=1+3tare parallel. Well, certainly their directions are multiples of one another (the multiple is 1). But why are they "distinct"? The point (5,3,1) is on the first line. For it to be on the second line, uhhh ... 30+t=5, so t=25. The second equation is then y=9(2)25=59, and this is not 3. So (5,3,1) is not on the other line. These are parallel lines.
Parallel planes
Skew lines
Intersection of two planes
The planes are not parallel because the normal vectors (<3,1,0> and
<2,2,1>, respectively) are not scalar multiples of one another. A
vector in the direction of the line will be perpendicular to both of
these normal vectors. The cross product gives us such a vector:
<3,1,0>x<2,2,1> is i3j+4k. We also
need a point on both planes. Well, if z=0 then we must find x and y
which satisfy both 3xy=4 and 2x2y+0=1. I can do this: double the
first equation and subtract the second equation. The result is 4x=7,
so x=7/4 and then the first equation becomes 3(7/4)y=4 so y=5/4. And
(7/4,5/4,0) is a point on the line. Therefore a set of parametric
equations for the line is
Parametric equations for a plane 
The easy (?) product
Suppose v=ai+bj+ck=<a,b,c> (the text uses these angle brackets,
and I will try to use them but probably I will get excited and replace
them with ordinary parentheses once in a while) and w=di+ej+fk. Then
I'll define the dot product or the scalar product or the inner product
(all of these phrases are used but I will mostly use "dot product") of
v and w to be the number
ad+be+cf. Our notation for this will be
v·w. Some textbooks use (v,w) or
<v,w>.
This doesn't look too impressive, but it turns out to occur enormously
frequently in many applications (so many, as I mentioned in class,
that most computer languages have devoted special attention to how to
evaluate it rapidly, and, indeed, the chips of many CPU's have special
"microcode" to evaluate inner products efficiently). Why would such
things occur in "real life"?
The farmer's market
New Jersey is the
Garden State. In the summer and early fall, almost every Friday I
go to a local farmer's market and buy fresh New Jersey fruits and
vegetables. So the task facing the cashier when I come up, hands full
of stuff, is to figure out what I owe. The logic of the cost
computation might be tabulated as follows:
Type of purchase  Apples  Eggplant  Garlic  Onions  Peppers  Potatoes  Zucchini 

Price per pound  $1.50  $1  $2.50  $1.25  $1.50  $1  75¢ 
Weight of item in pounds  3  1.2  .3  4  2  5  1.2 
Now really look at the table. If you compute the total cost,
you'd better not confuse the order of the items. Each row has an
7tuple of numbers: this is an element of R^{7}. Let's call
the second row the cost vector, C, and the third row
will be the weight vector, W. What will the total cost
of this lovely purchase? Clearly it is
C·W dollars:
(1.5)(3)+(1)(1.2)+(2.5)(.3)+(1.25)(4)+(1.5)(2)+(1)(5)+(.75)(1.2).
I'm not particularly interested in the number, more in the logic.
A hobby or two?
Everyone should have a hobby, something silly to occupy their time
when they should be working. One of my hobbies, since I started
teaching calculus, is to go through a day and try to identify every
use of the derivative idea. There are lots and lots of them. Another
hobby, perhaps even more fruitful (pun intended!) is to identify every
darn use of inner product in a typical day. There are many uses. But I
am supposed to concentrate on three dimensions, so let's get back to
work.
Looking at lengths again
Start with v=ai+bj+ck so that
v=srqt{a^{2}+b^{2}+c^{2}}. If w=di+ej+fk,
then w=srqt{d^{2}+e^{2}+f^{2}}. If I draw
the vectors v and w with their tails on the origin, then a vector,
identified in the picture with "?", completes a triangle, going from
the head of w to the head of v. Well, that vector isn't too
mysterious. I know that w+?=v, so ?=vw. In terms of components, we
can write vw=(ad)i+(be)j+(cf)k. The length of vw is
sqrt{(ad)^{2}+(be)^{2}+cf)^{2}}. You can
check easily that the length of vw and the lengths of v and w are not
related in any simple way.
Really the squares of the lengths
With all the square roots, maybe we should be looking at the squares
of the lengths. Also the algebra will be easier!
Then
v^{2}=a^{2}+b^{2}+c^{2}
and
w^{2}=d^{2}+e^{2}+f^{2}
and
vw^{2}=(ad)^{2}+(be)^{2}+(cf)^{2}.
We can "expand" the last term to get
vw^{2}=a^{2}2ad+d^{2}+b^{2}2be+e^{2}+c^{2}2cd+e^{2}.
That is exciting.
The obstacle to making it work
Well, if you compare terms, the sum
v^{2}+w^{2} is the same as vw^{2}
except for 2ad2be2cf=2(ad+be+cf). So computations with lengths and
squares of lengths will have to include that "mess". Let me forget the
2, which isn't that essential. What an accident: the obstacle to
making things ridiculously simple is (essentially) the dot product.
Color change? I'll write material with this background color if I just did not have time to disucss it in class but if I think it is useful enough that students might profit from seeing it.
The dot product and its basic properties
Example with a little table _{·}  w_{1}  w_{2}  v_{1}  3  7  v_{2}  4  2Could we then compute something like (4v_{1}3v_{2})·(5w_{1}2w_{2}). Here I would expect people to distribute or take advantage of linearity on both the left and righthad sides. The process might go like this: (4v_{1})·(5w_{1}2w_{2})+(3v_{2})·(5w_{1}2w_{2}) (4v_{1})·(5w_{1})+(4v_{1})·(2w_{2})+(3v_{2})·(5w_{1})+(3v_{2})·(2w_{2}) (4·5)v_{1}·w_{1}+(4·2)v_{1}·w_{2}+(3·5)v_{2}·w_{1}+(3·2)v_{2}·w_{2} (4·5)(3)+(4·2)7+(3·5)(4)+(3·2)(2) And the answer seems to be .... 60566012=68. I think this is correct. I make more errors in such computations when no one is watching me. 
The law of cosines lurks ...
There's a formula which generalizes the Pythagorean Theorem's
formula. For the triangle shown above, here it is:
vw^{2}=v^{2}+w^{2}2v wcos(theta)
Here theta is the angle opposite the side vw. If theta is Pi/2, the
cosine term is 0 and we get back Pythagoras. Here is the geometric view of the areas of
the squares on the sides of a triangle.
The geometry of the dot product
Since we know that
vw^{2}=v^{2}+w^{2}2v wcos(theta)
and we know that
vw^{2}=a^{2}2ad+d^{2}+b^{2}2be+e^{2}+c^{2}2cd+e^{2}=v^{2}+w^{2}2v·w. We
see that (cancelling the 2's)
v·w=v wcos(theta) where theta is the angle between v and w.
Notice that we can now see if v and w are not 0, then v and w are
orthogonal (perpendicular, normal) exactly when v·w=0.
Determining an angle
Suppose v=<3,2,1> and w=<5,2,6>. What is the angle between v and w?
Well, v·w=15+46=13, v=sqrt(14), and w=sqrt(65). Therefore
cos(theta)=13/[sqrt(14)sqrt(65)]. This gives approximately
(calculatorassisted computation here!) 64.55 degrees or (better, I
guess) 1.126 radians.
Resolving a vector into perpendicular and parallel parts
Let me continue using the v=<3,2,1> and w=<5,2,6> from the previous
question. I'd like to show you how to write v as a sum of two
vectors, v_{} and v_{⊥} where v_{⊥}
is perpendicular to w and v_{} is parallel to w. This will
be useful several times in the course, and is also important in
physics.
How long is v_{}? If you look at the picture, the v_{} is part of a right triangle. It is the adjacent leg and v is the hypoteneuse. Theta is the angle between them. Therefore v_{}=vcos(theta)=v·w/w. We did some of these computations just above, so v_{}=13/sqrt(65). How can we get the correct direction? So we want a vector whose length is 13/sqrt(65) and whose direction is the direction of w. Well, we can create a unit vector (a vector of length 1) in the direction of w: that will be w/w which is <5,2,6>/sqrt(65). And now we adjust this unit vector by stretching it, to get [13/sqrt(65)]<5,2,6>/sqrt(65). This is about all I did in class, but let me finish things here. The vector is (13/65)<5,2,6>=<1,26/65,78/65>. Since v=v_{}+v_{⊥}, we know that v_{⊥}=vv_{}=<3,2,1><1,26/65,78/65>=<2,104/65,143/65>.
Checking the answer
Well, v_{⊥} should be perpendicular to w. So let's
compute v_{⊥}·w=<2,104/65,143/65>·<5,2,6>=10+(208/65)(859/65)=(650+208858)/65=(858858)/65=0,
so these vectors are indeed orthogonal.
Comment I admit (informative to you and irritating to me) that
I did all these computations by hand, and I had to do them three
times to get them to come out correctly.
Another product, introduced geometrically
I defined the dot product by a rather simple algebraic formula. The
cross product has a complicated algebraic formula, and maybe it is
easier to beguin by defining it geometrically.
The cross product (also called the vector or outer product) takes two
vectors, v and w, and produces vxw, another vector. Since this is a
vector, it is specified by its magnitude and direction.
Magnitude of vxw
Look at the the vectors v and w. There is a parallelogram which has
sides v and w. The area (a nonnegative quantity) of that
parallelogram is the magnitude of vxw.
I forgot to say in class that there is an easy formula for the area in
terms of the lengths of the vectors and the angle between them. The
area is v wsin(theta). You get this by mulitplying the base
by the altitude in the plane of the parellelogram.
Direction of vxw
Take your right hand and extend your
thumb. Curl up your fingers. Insert your hand (I guess by though, not
physically) so that your fingers go from v to w. Then
your thumb will be (sort of) pointed perpendicular to the plane
containing v and w. That perpendicular direction is the direction of
vxw.
The only reason that such a weird product would be considered is that it is useful. We will see some geometric uses of it in this course and most of you will see uses in mechanics (torque), eletromagnetism, etc.
The i,j,k multiplication table
So before anyone could yell too much I computed the cross product
"multiplication table" for i and j and k.
This wasn't too hard,
although getting the directions correct involved some thought
and some physical contortions. These are unit vectors and all mutually
perpendicular. If the vectors in the product are the same, then the
parallelogram involved collapses to a line segment and has no area. If
the vectors are different, then the parallelgram is a square with 1
unit sides, and has area 1. The important thing is the sign of the
answer and its direction. I tried to get that correct. If you were not
in class or you were confused, I urge you to try to get some of the
entries in this table yourself.
x  i  j  k  i  0  k  j  j  k  0  i  k  j  i  0
Weird stuff
Things to notice:
ixj=k and jxi=k. The cross product is not necessarily commutative.
(ixj)xj=kxj=1 and ix(jxj)=ix0=0. The cross product is not
necessarily associative.
This means that you can't necessarily rearrange or regroup cross
products. This is annoying. The cross product does not resemble many
other products you have seen.
Then the cross product in terms of components is ... I then did the following awful computation, which you should never do: If v=ai+bj+ck and w=di+ej+fk, then vxw= (ai+bj+ck)x(di+ej+fk)=ad(ixi)+ae(ixj)+af(ixk)+bd(jxi)+be(jxj)+bf(jxk)+cd(kxi)+ce(kxj)+cf(kxk). I filled in all the values from the multiplication table above and got the following mess: vxw=ad(0)+ae(k)+af(j)+bd(k)+be(0)+bf(i)+cd(j)+ce(i)+cf(0)=(bfce)i+(cdaf)j+(aebd)k 
This last formula can be expressed in a very neat way if you know about determinants. It turns out that
 i j k  vxw=det a b c   d e f here's a very brief discussion of determinants in the textbook. I will probably evaluate this determinant most of the time by "expanding along the first row" but you may use any valid method.
Why is what I've done correct? The definition I gave for cross product was geometric. Some of the properties of cross product are easy from the geometric definition. But I've just used linearity (or distributed the sum over the cross product), something like this: (v_{1}+v_{2})xw=(v_{1}xw)+(v_{2}xw). Why is this true? This is not clear (!) to me at all from the geometric definition. Here is a paper which shows why you can distribute using geometry. Figure 8 on page 9 of this paper is a picture which should convince (is supposed to convince) you that cross product does have the desired algebraic property. I hope you will find it convincing. I find it difficult to really understand. 
Who am I?
I'm a faculty member in the Rutgers Math Department. Much of my
professional work has involved differential equations involving
functions of more than one variable. I actually like the
content of this course, and I may get almost incoherent with
excitement and pleasure when describing some of the subject
matter. Certainly I will draw pictures that seem to mean something to
me, but they may be totally incomprehensible to almost everyone
else. I apologize in advance.
Diary
I will try to write a diary for this course, since some
students have told me it is helpful. I am teaching an extra course
this semester, so I don't know how much time I will have. I will use
lots of material from the diary I wrote the last time I taught
251. This is not "plagiarism". I wrote the material, I acknowledge
using it, and I've given myself permission. For your information, it
isn't usually the text that takes much time, but rather the darn
pictures. Lots of time is needed to create good pictures!
Who are you?
Most of the students in these sections are
engineering students. Many engineering students (especially in
Mechanical, Chemical, and Biochem) will go on to take Math 421 where
all the material of this course is used. Other students are from such
disciplines as chemistry, computer science, mathematics, meteorology,
and physics. Certain biological science majors take this course,
principally because they will take physical chemistry, and the
language of several variable calculus is essential in
pchem. Additionally, some students may be in this class for "fun" (if
that's possible) but may be interested in a business career. It turns
out that there's a huge developing field of mathematical finance and
the ideas of this course are, again, essential tools.
What will we study?
I tried to briefly discuss the background of the
course. The link indicated has more information.
Then we went on to "review" analytic geometry of one and two and even three dimensions. I put the word in quotes because most students who have taken courses in applied science and engineering have already seen the material we will discuss in 2 or 3 or even 4 other courses! The poor math majors may never have seen this before so they especially should read the text.
R^{1}
This is supposed to be the real line. There's an origin and a
unit length, and the conventional choice is to make the positive
direction go off to the right. Each point on the line corresponds to a
real number. The distance between points on the line which correspond
to the real numbers a and b is defined to be ab.
R^{2}
And here is the plane (I don't think I've ever heard anyone
call it the "real plane"). The simplest situation is what's shown
here. Two straight lines as axes, perpendicular to each. Every point
corresponds to a unique ordered pair of real numbers. There is a set
unit length on each coordinate axis, and almost always right is
positive and up is positive. Later in the course we will consider
situations where the coordinate axes are not necessarily
perpendicular. And even where what corresponds to the axes are not
straight. I hope that tilted axes could make sense if, for
example, you were interested in looking at crystals which did not have
rectangular symmetry.
R^{3}
In three dimensions, the geometric situation is specified by three
mutually perpendicular lines. Points will correspond to ordered
triples of real numbers. In almost every situation you are likely to
encounter, these axes will be righthanded. Righthanded means
... means ... means ... well, it means what I sketched to the right
(heh heh). The three dimensional world is complicated, and you will
see that the choice of righthandedness has some interesting
consequences.
Chirality or handedness
A discussion of Chirality is linked.
Here is a short
discussion of parity in chemistry and physics. And here is a Wikipedia
article on enantiomers, which are compounds that are
nonsuperimposable mirror images of one another. This article is a bit
more technical, but has many examples of pharmaceutical products which
are mirror images.
Perilous pictures
Here is my attempt to draw the point with coordinates (3,2,1).
The green "path" with 's sort of tells how to find the
point. I walk (?) from (0,0,0) to (1,0,0), then to (1,3,0), and
finally to (1,3,2). Each is
supposed to represent one unit of the path. I think that the
complications of perspective and reducing the "object" from three
dimensions to a twodimensional representation may make even such a
simple picture difficult to understand. If you weren't told that the
point was supposed to be (1,3,2), would you have guessed those were
the coordinates? Sigh.
I will try to draw good pictures, but I won't always
succeed. And even if I draw pictures which I believe are suitable, you
may well not agree. Pictures are complicated.
Some simple geometry and equations What points in R^{3} have y=2? Certainly (0,2,0), on the yaxis, is the only point on the yaxis with y=2. We can move up and down (changing zcoordinate) and sideways (changing xcoordinate). We can even change both. The collection of points we get is a plane, perpendicular to the yaxis through the point (0,2,0). An attempt at a picture is to the right. You don't need to agree that this is a good picture. In this course I will probably regard the following technical words as synonymous (the words mean the same) most of the time: perpendicular orthogonal normal  
What points in R^{3} have z<5? This is a an open halfspace. The plane z=5 is the boundary of this region and the region does not include the boundary. Later in the course I'll try to discuss the use of "open" and "boundary" in more detail. Here the boundary is a plane normal to the zaxis through the point (0,0,5). The plane, however, is not included in the region specified by z<5. What I've attempted to show in the picture is an open halfspace. 
Let's imagine a brick in R^{3} with its sides parallel to the xy and yz and xz planes. Maybe this could be called a rightangled parallelopiped. Suppose one corner (also called a vertex) has coordinates (a,b,c) and the diagonally opposite corner has coordinates (d,e,f).  
Suppose now we look at the corner displayed in the
accompanying picture, the corner illustrated with . What are the coordinates of that
point? If you can "see" what the picture is supposed to show, then the designated corner is just moved "out" parallel to the direction of the yaxis. The xcoordinate doesn't change at all, and the height from the xyplane (which is the zcoordinate) also doesn't change. Therefore the coordinates of are certainly (a,?,c). What about the second coordinate of the point? If you look at the picture with me, you can see that the entire face or side of the solid between and the point (d,e,f) has constant yvalue: it is perpendicular to the yaxis. Therefore the mystery middle coordinate must be the same as the middle coordinate of (d,e,f), and so =(a,e,c). 

Now let us play the same game with another vertex (or
corner). So the designation of
has been moved to the corner shown. What are the coordinates of this
point? This point is on the same face or side of the solid so it
certainly shares the second or ycoordinate with both (d,e,f) and
(a,e,c), so =(?_{1},e,?_{2}). Well, the
third coordinate, ?_{2}, must be c since the point
we're looking at is certainly on the bottom face of the solid. The
first coordinate might be the most puzzling, but both (d,e,f) and are on the "back" face (that is part
of the plane x=d) of the solid, so the xcoordinate we're looking for
is d. Therefore this is (d,e,c).
A good exercise for you is to figure out the coordinates of some of the other corners of the brick. 

Now I'd like to compute the length of some of the edges of the brick. For example, what is the length of the line segment connecting (a,b,c) and (a,e,c)? This is the distance between these two points. Notice that the only coordinate which is varying is the middle one. This side is a line segment which is parallel to the yaxis. We can measure length along it only by thinking about the onedimensional distance between b and e. That distance is be. The reason for the absolute value signs is that distance is supposed to be nonnegative.  
We can try to do the same thing with the edge connecting the points with coordinates (a,e,c) and (d,e,c). Here the only coordinate difference is in the first, x, coordinate. So distances again should be measured as if they were in one dimension, along the parallel xaxes. The distance we want is ad.  
Now we can be even more bold. (Well, bold for a math class.)
We can find the distance from (a,b,c) to (d,e,c). Well, look carefully
and see that there is a right triangle whose hypotenuse is the
distance wanted, and whose "legs" have distances we already know. Then
Pythagoras declares that the length of the hypotenuse is 

I'm not going to try to put that formula in the picture. I
will just label the distance Blah. Now I want the distance
along one of the main diagonals of the brick, from (a,b,c) to
(d,e,f). There is, as shown, another right triangle. The hypotenuse is
the distance I want, and the two legs have distances we know: one is
Blah and the other is ... the other is: well, the same ideas as
before (since the endpoints differ in only one coordinate) tell me
that the distance is cf. Therefore (again Pythagorus) the
distance between the farthest apart corners is 
The Euclidean distance
The (Euclidean) distance between (a,b,c) and (d,e,f) is sqrt((ad)^{2}+(be)^{2}+(cf)^{2}).
A bunch of comments
A sphere of radius 3
What algebraic condition on a point (x,y,z) is equivalent to the
geometric statement that the point lies on a sphere of radius 3
centered at (0,0,0)? This means that the distance from (x,y,z) to
(0,0,0) should be 3, or
sqrt((x0)^{2}+(y0)^{2}+(z0)^{2})=3 which of
course is the same as x^{2}+y^{2}+z^{2}=9.
Completing the square to "recognize" a sphere
We could of course "reverse" the process if we're given an equation
(or at least an equation of the correct form). For example, the equation
x^{2}3x+y^{2}+4y+z^{2}7=0
represents a sphere. What is its center and its radius? The key
algebraic maneuver here is completing the square, and
everything works very much as in two dimensions.
x^{2}3x >
x^{2}+2(3/2)x >
x^{2}+2(3/2)x+(3/2)^{2}(3/2)^{2} >
(x3/2)^{2}(3/2)^{2}
y^{2}+4y >
y^{2}+2(2y) >
y^{2}+2(2y)+2^{2}2^{2} >
(y+2)^{2}2^{2}
We don't have any term with z to the first power. So the equation
becomes:
(x3/2)^{2}(3/2)^{2}+(y+2)^{2}2^{2}+z^{2}7=0
which is the same as
(x3/2)^{2}+(y+2)^{2}+(z0)^{2}=(3/2)^{2}+2^{2}+7
and we can "read off" that the center is (3/2,2,0) and the radius is
sqrt((3/2)^{2}+2^{2}+7). It is easy to make mistakes
with minus signs when identifying the center. It is also easy to
forget to take the square root when identifying the radius. Oh well.
If we had different (but positive) numbers in front of the squares
(x^{2} and y^{2} and z^{2}) then we'd get an
eggshaped object, an ellipsoid. We'll see more about this
later.
Vectors 

Quantities which are not vectors have the strange adjective, scalars. So both Mass and Temperature are scalar quantities, and are measured by numbers (both positive and negative) and have no intrinsic direction. 
Equal vectors
A vector can also be specified by its head and tail.
Two vectors will be called equal (maybe the official word is
equivalent) if when the tail of one of them is moved to the tail of
the other, then their heads are in the same place.
Vectors are used to do algebra in more than one dimension. This is because algebra has really been successful, and the interaction between algebra and geometry has paid off well in both directions. Therefore we'll need to add and multiply vectors. Efforts to multiply run into serious difficulties, as we will see next time. Addition is neat and "everything" works well. As I mentioned in class, the generally accepted definition for vector addition models some situations which can be experienced and measured in "real life" with forces, velocities, etc. The word resultant is sometimes used in connection with this definition.
Definition of vector addition
First suppose there are two vectors, v and w, which are roaming, free and happy in R^{3}.  We take the vector w and drag it so that the tail of w is at the same point as the head of v.  The vector v+w is now defined using the geometric display we just arranged. v+w the vector whose tail is at the tail of v and whose head is at the head of w. 
Properties of vector addition
Zero and minus
The zero vector has its head equal to its tail. If v is a vector, then
v is the vector whose length is v's but whose direction is the
reverse: so the head of v is the tail of v and the tail of v is the
head of v. Huh. Say that fast.
I am especially proud of the image of the zero vector in the picture
to the right. I wanted to show the special beauty of the zero
vector. I tried several different angles. (This is a joke!)
Then v+(v)=0, and v+0=0 etc. for all vectors.
Scalar multiplication
In this course the scalars will be real numbers. Scalars are
things that multiply vectors. Many of the students in the course will
see later situations where the scalars are complex numbers. In the
case of the vectors arising in computer science and electrical
engineering, collections of scalars occur which are much less
familiar.
If c is a scalar and v is a vector, then the vector cv is defined by
the following:
If c>0 the direction of the vector cv is the same as the
direction of v, and the length of cv is the length of v multiplied by
c.
If c=0 then cv is the zero vector.
If c<0 the direction of the vector cv is the same as the
direction of v and the length of cv is the length of v
multiplied by c. (The absolute value makes sure that lengths stay
positive.)
Here are pictures of v and v and 3v and 2v.
Important vectors Some vectors are more important than others, especially if you have a coordinate system. The vectors i and j and k are vectors of length 1 (such vectors are called unit vectors) which are parallel to the coordinate axes x and y and z, respectively.  
Now take any vector v and move it so that the tail of v is at the
origin, (0,0,0). Then the head of v will be at some point, (a,b,c). If
you think about the geometry I hope you can convince yourself that
v=ai+bj+ck. We have written v as a sum of its components. The textbook also uses the notation v=<a,b,c> for this. 
The head and the tail produce the vector (algebraically)
Suppose the point P has coordinates
(x_{1},y_{1},z_{1}) and the point Q has
coordinates (x_{2},y_{2},z_{2}). Then the
vector from P to Q (so P is the tail and Q is the head) is just
(x_{2}x_{1})i+(y_{2}y_{1})j+(z_{2}z_{1})k.
I hope that you can draw a picture convincing yourself of that,
or you can look in the text.
Example The vector from P=(1,2,3) to Q=(5,6,2) is
(51)i+(62)j+(2[3])k=6i+4j+5k. It is easy to make mistakes with
signs in this computation.
How long is a vector?
I will use length and norm and magnitude to mean
the same thing about a vector: its length. If v=ai+bj+ck, then we can
think of the tail of v as sitting at (0,0,0) and the head of v, at
(a,b,c). The length formula we got then says that the length must be
sqrt(a^{2}+b^{2}+c^{2}). So if you know the
components, the magnitude can be computed. In the textbook we are
using the magnitude is written v. Please note that in some books
and some situations, this quantity is just written v.
NO! NO, NO, NO!!!
Suppose that v=<1,2,3> and w=<2,3,2>. Then v+w=<1,1,5>. And
v=sqrt(14) (approximately 3.74), w=sqrt(17) (approximately 4.12),
and v+w=sqrt(27) (approximately 5.19). In general, these numbers
need not be related in any simple fashion, aside from the Triangle
Inequality (see the textbook, please).
Next time I will explore how these lengths can be explained, as we discuss multiplication of vectors.
After the class ...
I think the classroom is badly suiting to teaching math. Most of the
blackboards are barely visible to most of the chairs. Next time I will
try to teach at the narrow end of the room. I did investigate if any
other class rooms are available. Exactly one other room on College
Avenue is vacant at each class time, and both of those two rooms are
the same geometry as the one we have. They are both classrooms in a
river dorm! The physical plant at Rutgers is overused.
Start reading chapter 12, please. Do the problems.
Maintained by greenfie@math.rutgers.edu and last modified 9/5/2008.